Trigonometry [RENTAL EDITION] (12th Edition) [12 ed.] 0135924189, 9780135924181

Table of contents :
Trigonometry 12th Edition [Margaret L. Lial]
MyLab Math
Half Title
Title Page
Copyright
Dedication
Contents
Preface
Chapter 0 Algebra Review
Chapter 1 Trigonometric Functions
Chapter 2 Acute Angles and Right Triangles
Chapter 3 Radian Measure and the Unit Circle
Chapter 4 Graphs of the Circular Functions
Chapter 5 Trigonometric Identities
Chapter 6 Inverse Circular Functions and Trigonometric Equations
Chapter 7 Applications of Trigonometry and Vectors
Chapter 8 Polar Equations, and Parametric Equations
Answers to Selected Exercises
Photo Credits
Index

Citation preview

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Interactive Exercises Mylab Math's interactive exercises mirror those in the textbook but are programmed to allow you unlimited practice, lead ing to mastery. Most exercises include learning aids, such as "Help Me Solve This," "View an Example," and ''Video," and they offer helpful feedback when you enter incorrect answers.

Instructional Videos Instructional videos cover all examples from the text and can conveniently be played on any mobile device. These videos are especial ly helpful if you miss a class or just need further explanation.

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Interactive eText The Pearson eText gives you access to your textbook anytime, anywhere. In addition to note taking, highlighting, and bookmarking, the Pearson eText App offers access even when offline.

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pearson.com/mylab/math

Trigonometry TWELFTH EDITION

This page intentionally left blank

Margaret L. Lial American River College

John Hornsby University of New Orleans

David I. Schneider University of Maryland

Callie J. Daniels St. Charles Community College

@ Pearson

Please contact https://support.pearson.com/getsupport/s/contactsupport with any queries on this content. Copyright© 202 1, 2017, 20 13 by Pearson Education, Inc. or its affi liates, 22 1 River Street, Hoboken, NJ 07030. All Rights Reserved. Manufactured in the United States of America. This publication is protected by copyright, and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise. For information regarding permissions, request forms, and the appropriate contacts within the Pearson Education Global Rights and Permissions department, please visit www.pearsoned.com/permissions/. Acknowledgments of th ird-party content appear on page C-1 , which constitutes an extension of this copyright page. PEARSON, ALWAYS LEARNING, and MYLAB are exclusive trademarks owned by Pearson Education, Inc. or its affiliates in the U.S. and/or other countries. Unless otherwise indicated herein, any third-party trademarks, logos, or icons that may appear in thi s work are the property of their respective owners, and any references to third-party trademarks, logos, ico ns, or other trade dress are for demonstrative or descriptive purposes only. Such references are not intended to imply any sponsorship, endorsement, authorization, or promotion of Pearson's products by the owners of such marks, or any relationship between the owner and Pearson Education, Inc., or its affi liates, authors, licensees, or distributors.

Library of Congress Cataloging-in-Publication Data Cataioging-in-Publication Data is avai lable on fi le at the Library of Congress.

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ISBN 13: 978-0- 13-5924 18- 1 ISBN 10: 0- 13-592418-9

This text is dedicated to you-the student. We hope that it helps you achieve your goals. Remember to show up, work hard, and stay positive. Everything else will take care of itself. The Lial Author Team

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Contents Preface

xiii

Resources for Success

(R

J

xviii

Algebra Review

1

Basic Concepts from Algebra

2

Sets of Numbers • Add itive Inverses • Absolute Value • Inequality and Order on a Number Line

Real Number Operations and Properties

12

Operations on Real Numbers • Distance between Points on a Number Line • Exponents • Order of Operations • Properties of Rea l Numbers

Exponents, Polynomials, and Factoring

24

Ru les for Exponents • Zero as an Exponent • Polynomials • Addition, Subtraction, and Mult iplication of Polynomials • Factoring Polynomials

Rational Expressions

34

Rational Expressions • LowestTerms of a Rational Expression • Multi plication and Division • Addition and Subtraction • Complex Fractions

Radical Expressions

44

Square Roots • Cube, Fourth, and Higher Roots • Product and Quotient Ru les for Radica ls • Operations with Rad icals • Rational izing Denominators • Simplified Radicals

Equations and Inequalities

56

Basic Terminology of Equations • Linear Equations • Quadratic Equations • Inequalities • Linear Inequalit ies and Interval Notation • Three-Part Inequalities

Rectangular Coordinates and Graphs

67

The Rectangular Coordinate System • The Distance Fo rmu la • The M idpoint Formula • Equations in Two Variables • Ci rcles

Functions

76

Relations and Functions • Domain and Range • Determining Whether Relations Are Functions • Function Notation • Increasing, Decreasing, and Constant Functions • Continuity

Graphing Techniques

88

Stretching and Shrinking • Reflecting • Symmetry • Even and Odd Functions • Translatio ns Test Prep

104 • Review Exercises

112 • Test

116

vii

viii

CONTENTS

1

J

Trigonometric Functions

Angles

119

120

BasicTerminology • Degree Measure • Standard Position • Cotermina l Ang les

Angle Relationships and SimilarTriangles

128

Geometric Properties • Triangles Chapter 1 Quiz (Sections 1.1-1.2)

Trigonometric Functions

139

140

The Pythagorean Theorem and the Distance Formula • Trigonometric Functions • Quadrantal Angles

Using the Definitions of the Trigonometric Functions

148

Reciprocal Identities • Signs and Ranges of Function Va lues • Pythagorean Identities • Quotient Identities Test Prep

157 • Review Exercises

160 • Test

163

Acute Angles and Right Triangles 165

2

Trigonometric Functions of Acute Angles

166

Right-Triangle-Based Definitions of the Trigonometric Functions • Cofunctions • How Function Va lues Change as Ang les Change • Trigonometric Function Values of Special Angles

Trigonometric Functions of Non-Acute Angles

174

Reference Ang les • Special Ang les as Reference Angles • Determination of Angle Measures with Special Reference Angles

Approximations of Trigonometric Function Values Calcu lator Approximations of Trigonometric Function Va lues • Calcu lator Approximations of Angle Measures • An Appl ication Chapter 2 Quiz (Sections 2.1-2.3)

189

Solutions and Applications of Right Triangles

190

Historical Background • Significant Digits • SolvingTriangles • Angles of Elevation or Depression

Further Applications of Right Triangles

200

Historical Background • Bearing • Further Appl ications Test Prep

209 • Review Exercises

211 • Test

215

182

ix

CONTENTS

Radian Measure and the Unit Circle 211

3

Radian Measure

218

Radian Measure • Conversions between Degrees and Rad ians • Trigonometric Function Values of Angles in Radians

Applications of Radian Measure

224

Arc Length on a Circle • Area of a Sector of a Circle

The Unit Circle and Circular Functions

235

Circular Functions • Va lues of the Circular Functions • Determining a Number w ith a Given Circular Function Value • Applications of Ci rcu lar Functions • Function Values as Lengths of Line Segments Chapter 3 Quiz (Sections 3.1-3.3)

245

Linear and Angular Speed

246

Linear Speed • Angu lar Speed Test Prep

( 4

J

252 • Review Exercises

254 • Test

257

Graphs of the Circular Functions

Graphs of the Sine and Cosine Functions

259

260

Periodic Functions • Graph of the Sine Function • Graph of the Cosine Function • Techn iques for Graphing, Amplitude, and Period • Connecting Graphs with Equations • A Trigonometric Model

Translations of the Graphs of the Sine and Cosine Functions 273 Horizontal Trans lations • Vertica l Translations • Combinations of Translations • A Trigonometric Model Chapter 4 Quiz (Sections 4.1-4.2)

284

Graphs of the Tangent and Cotangent Functions

284

Graph of theTangent Function • Graph of the Cotangent Function • Techniques for Graphing • Connecting Graphs with Equations

Graphs of the Secant and Cosecant Functions

293

Graph of the Secant Function • Graph of the Cosecant Function • Techniques for Graphing • Connect ing Graphs with Equations • Addition of Ordinates Summary Exercises on Graphing Circular Functions

Harmonic Motion

301

301

Simple Harmonic Motion • Damped Oscil latory Motion Test Prep

308 • Review Exercises

310 • Test 313

X

CONTENTS

5

J

Trigonometric Identities

Fundamental Identities

315

316

Fundamental Identities • Uses of the Fundamental Identities

Verifying Trigonometric Identities

322

Strategies • Verifying Identities by Working with One Side • Verifying Identities by Working with Both Sides

Sum and Difference Identities for Cosine

331

Difference Identity for Cosine • Sum Identity for Cosine • Cofunction Identities • Applications of the Sum and Difference Identities • Verifying an Identity

Sum and Difference Identities for Sine and Tangent

341

Sum and Difference Identities for Sine • Sum and Difference Identities forTangent • Applications of the Sum and Difference Identities • Verifying an Identity Chapter 5 Quiz (Sections 5.1-5.4)

350

Double-Angle Identities

350

Double-Angle Identities • An Application • Product-to-Sum and Sum-to-Product Identities

Half-Angle Identities

358

Half-Angle Identities • Applications of the Half-Angle Identities • Verifying an Identity Summary Exercises on Verifying Trigonometric Identities Test Prep

366 • Review Exercises

368 • Test

365

370

Inverse Circular Functions and Trigonometric Equations 371

6

Inverse Circular Functions 372 Inverse Functions • Inverse Sine Function • Inverse Cosine Function • Inverse Tangent Function • Other Inverse Circular Functions • Inverse Function Values

Trigonometric Equations I

388

Linear Methods • Zero-Factor Property Method • Quadratic Methods • Trigonometric Identity Substitutions • An Application

Trigonometric Equations II

395

Equations with Half-Angles • Equations with Multiple Angles • An Application Chapter 6 Quiz (Sections 6.1-6.3)

402

Equations Involving Inverse Trigonometric Functions 402 Solution for x in Terms of y Using Inverse Functions • Solution of Inverse Trigonometric Equations Test Prep

409 • Review Exercises

411 • Test

413

CONTENTS

xi

Applications of Trigonometry and Vectors 415

7

Oblique Triangles and the Law of Sines

416

Congruency and Oblique Triang les • Derivation of the Law of Sines • Solutions of SAA and ASA Triang les (Case 1) • Area of a Triangle

The Ambiguous Case of the Law of Sines

426

Description of the Ambiguous Case • Solutions of SSA Triangles (Case 2) • Analyzing Data for Possible Number of Triang les

The Law of Cosines

432

Derivation of the Law of Cosines • Solutions of SAS and SSS Triangles (Cases 3 and 4) • Heron's Formula for the Area of a Triangle • Derivation of Heron's Formula Chapter 7 Quiz (Sections 7.1- 7.3)

445

Geometrically Defined Vectors and Applications

446

Basic Terminology • The Equilibrant • Incline Applications • Navigation Appl ications

Algebraically Defined Vectors and the Dot Product

456

Algebraic Interpretation of Vectors • Operations w ith Vectors • The Dot Product and the Angle between Vectors Summary Exercises on Applications of Trigonometry and Vectors Test Prep

8

466 • Review Exercises

465

469 • Test 473

Complex Numbers, Polar Equations, and Parametric Equations 475 Complex Numbers

476

Basic Concepts of Complex Numbers • Comp lex Solutions of Quadratic Equations (Part 1) • Operations on Complex Numbers • Complex Solutions of Quadratic Equations (Part 2) • Powers of i

Trigonometric (Polar) Form of Complex Numbers

486

The Complex Plane and Vector Representation • Trigonometric (Polar) Form • Converting between Rectangular and Trigonometric (Polar) For ms • An Application of Complex Numbers to Fractals

The Product and Quotient Theorems

492

Products of Comp lex Numbers in Trigonometric For m • Quotients of Complex Numbers in Trigonometric Form

De Moivre's Theorem; Powers and Roots of Complex Numbers 498 Powers of Complex Numbers (De Moivre'sTheorem) • Roots of Complex Numbers Chapter 8 Quiz (Sections 8.1 - 8.4)

505

xii

CONTENTS

Polar Equations and Graphs

505

Polar Coordinate System • Graphs of Polar Equations • Conversion from Polar to Rectangu lar Equations • Classification of Po lar Equati ons

Parametric Equations, Graphs, and Applications

518

Basic Concepts • Parametric Graphs andThei r Rectangular Equivalents • T he Cycloid • Applications of Parametric Equations Test Prep

526 • Review Exercises

Answers to Selected Exercises Photo Credits C-1 Index 1-1

A-1

529 • Test

532

Preface

WELCOME TO THE 12TH EDITION In the twelfth edition of Trigonometry, we continue our ongoing commitment to

providing the best possible text to help instructors teach and students succeed. In this edition, we have remained true to the pedagogical style of the past while staying focused on the needs of today's students. Support for all classroom types (traditional, corequisite, flipped , hybrid, and online) may be found in this classic text and its supplements backed by the power of Pearson's MyLab Math. In this edition, we have drawn on the extensive teaching experience of the Lial team, with special consideration given to reviewer suggestions. General updates include enhanced readability as we continually strive to make math understandable for students, updates to our extensive list of applications and real-world mathematics problems, use of color in displays and side comments, and coordination of exercises and their related examples. The authors understand that teaching and learning mathematics today can be a challenging task. Some students are prepared for the challenge, while other students require more review and supplemental material. This text is written so that trigonometry students with varying abilities and backgrounds will all have an opportunity for a successful learning experience. The Lial team believes this to be our best edition of Trigonometry yet, and we sincerely hope that you enjoy using it as much as we have enjoyed writing it. Additional texts in this series are as follows. College Algebra, Thirteenth Edition College Algebra and Trigonometry, Seventh Edition Precalculus, Seventh Edition

HIGHLIGHTS OF NEW CONTENT • New Chapter R provides a thorough review of the basic algebraic concepts that trigonometry students need in order to succeed. Topics include real number operations and properties, rules for exponents, operations on polynomials and factoring, rational and radical expressions, solving equations and inequalities, circles, functions, domain, range, and graphing techniques. Instructors may choose to cover review topics from Chapter R at the beginning of their course or to insert these topics as-needed in a just-in-time fashion. Either way, students who are under-prepared for the demands of trigonometry, as well as those who need a quick review, will benefit from the material contained here. It is our hope that this chapter, which also includes a comprehensive end-of-chapter Test Prep, would serve as a valuable resource throughout all course types for students and instructors alike. • The exercise sets were a key focus of this revision, and Chapters 3 and 4 are among the chapters that have benefitted. Specifically, Section 3.2 Applications of Radian Measure has additional exercises devoted to finding arc length and area of a sector. Section 3.4 Linear and Angular Speed now contains new applications of linear and angular speed. Chapter 4 includes updated real data applications and additional Connecting Graphs with Equations exercises, as well as new applications of harmonic motion.

xiii

xiv

PREFACE

• Proofs of identities in Chapter S now feature a drop-down style for increased clarity and student understanding. Based on reviewer requests, Section 6.4 Equations Involving Inverse Trigonometric Functions includes new exercises in which solutions of inverse trigonometric equations are found. • In response to reviewer suggestions, Section 7.5 Algebraically Defined

Vectors and the Dot Product has new exercises on finding the angle between two vectors, determining magnitude and direction angle for a vector, and identifying orthogonal vectors. Additionally, Chapter 8 contains new exercises requiring students to graph polar and parametric equations and give parametric representations of plane curves.

FEATURES OF THIS TEXT SUPPORT FOR LEARNING CONCEPTS We provide a variety of features to support students' learning of the essential topics of trigonometry. Explanations that are written in understandable terms, figures and graphs that illustrate examples and concepts, graphing technology that supports and enhances algebraic manipulations, and real-life applications that enrich the topics with meaning all provide opportunities for students to deepen their understanding of mathematics. These features help students make mathematical connections and expand their own knowledge base. • Examples Numbered examples that illustrate the techniques for working

exercises are found in every section. We use traditional explanations, side comments, and pointers to describe the steps taken-and to warn students about common pitfalls. Some examples provide additional graphing calculator solutions, although these can be omitted if desired. • Now Try Exercises Following each numbered example, the student is

directed to try a corresponding odd-numbered exercise (or exercises). This feature allows for quick feedback to determine whether the student understands the principles illustrated in the example. • Real-Life Applications We have included hundreds of real-life applica-

tions, many with data updated from the previous edition. They come from fields such as sports, biology, astronomy, geology, music, and environmental studies. • Function Boxes Special function boxes offer a comprehensive, visual

introduction to each type of trigonometric function and also serve as an excellent resource for reference and review. Each function box includes a table of values, traditional and calculator-generated graphs, the domain, the range, and other special information about the function. These boxes are assignable in MyLab Math. • Figures and Photos Today's students are more visually oriented than ever

before, and we have updated the figures and photos in this edition to promote visual appeal. Guided Visualizations with accompanying exercises and explorations are available and assignable in My Lab Math. • Cautions and Notes Text that is marked llAuii(tl~I warns students of

common errors, and NOTE comments point out explanations that should receive particular attention. • Looking Ahead to Calculus These margin notes offer glimpses of how the

topics currently being studied are used in calculus.

PREFACE

XV

• Use of Graphing Technology We have integrated the use of graphing calculators where appropriate, although this technology is completely optional and can be omitted without loss of continuity. We continue to stress that graphing calculators support understanding but that students must first master the underlying mathematical concepts. Exercises that require the use of a graphing calculator are marked with the icon

rn.

SUPPORT FOR PRACTICING CONCEPTS This text offers a wide variety of exercises to help students master trigonometry. The extensive exercise sets provide ample opportunity for practice and increase in difficulty so that students at every level of understanding are challenged. The variety of exercise types promotes mastery of the concepts and reduces the need for rote memorization. • Concept Preview Each exercise set begins with a group of CONCEPT PREVIEW exercises designed to promote understanding of vocabulary and basic concepts in each section. These exercises are assignable in MyLab Math and provide support especially for hybrid, online, and flipped courses. • Exercise Sets In addition to traditional drill exercises, this text includes and multiplewriting exercises, optional graphing calculator exercises choice, matching, true/false, and completion exercises. Those marked Concept Check focus on conceptual thinking. Connecting Graphs with Equations exercises challenge students to write equations that correspond to given graphs. Video solutions for select problems are available in MyLab Math.

rn,

• Relating Concepts Exercises Appearing at the end of selected exercise sets, these groups of exercises are designed so that students who work them in numerical order will follow a line of reasoning that leads to an understanding of how various topics and concepts are related. All answers to these exercises appear in the student answer section, and these exercises are assignable in MyLab Math.

SUPPORT FOR REVIEW AND TEST PREP Ample opportunities for review are found both within the chapters and at the ends of chapters. Quizzes interspersed within chapters provide a quick assessment of students' understanding of the material presented up to that point in the chapter. Chapter Test Preps provide comprehensive study aids to help students prepare for tests. • Quizzes Students can periodically check their progress with in-chapter quizzes that appear in all chapters, beginning with Chapter 1. All answers, with corresponding section references, appear in the student answer section. These quizzes are assignable in MyLab Math. • Summary Exercises These sets of in-chapter exercises give students the all-important opportunity to work mixed review exercises, requiring them to synthesize concepts and select appropriate solution methods. • End-of-Chapter Test Prep Following the final numbered section in each chapter, the Test Prep provides a list of Key Terms, a list of New Symbols (if applicable), and a two-column Quick Review that includes a section-bysection summary of concepts and examples. This feature concludes with a comprehensive set of Review Exercises and a Chapter Test. The Test Prep, Review Exercises, and Chapter Test are assignable in My Lab Math.

xvi

PREFACE

Get the most out of 1> Pearson •

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Mylab Math is tightly integrated with author style, offering a range of author-created resources, to give students a consistent experience.

Preparedness Preparedness is one of the biggest challenges in many math courses. Pearson offers a variety of content and course options to support students with just-in-time remed iation and key-concept review as needed.

Integrated Review in Mylab Math Integrated Review can be used in corequisite courses or simply to help students who enter a course without a full understanding of prerequisite skills and concepts. Premade, editable Integrated Review assignments are available to assign in the Assignment Manager. Integrated Review landing pages (shown below) are visible by default at the start of most chapters, providing objective-level review. •Students begin each chapter by completing a Skills Check to pinpoint which topics, if any, they need to review:· ·· ··· · · • • · . • . • Chapter 5 lntegroted Review elnt Contet\ts • • •• ,

Chapters

• Personalized review homework provides extra support for students who need it on just the topics ...they didn't master in the preceding Skills Check. ...

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Skilb ll-t.w Homcworti: • • •. . , Cornplllwyour pwsof'lllllz:t means "is greater than."

6

-4 > -6 , and

12 > 5, 9 > -2,

-> 5 0

All are true.

In each case, the symbol ''points" toward the lesser number. The number line in Figure 8 shows the graphs of the numbers 4 and 9. On the graph, 4 is to the left of 9, so 4 < 9. The lesser of two numbers is always to the left of the other on a number line. ~----

0

3

2

4

4 b if a is to the right of b.

Determining Order on a Number Line

Use a number line to compare -6 and 1 and to compare -5 and -2. As shown on the number line in Figure 9, -6 is located to the left of 1. For this reason, -6 < 1. Also, 1 > -6. From Figure 9, -5 < -2, or - 2 > - 5. In each case, the symbol points to the lesser number.

SOLUTION

-6

-5

-4

-3

-2

- I

0

Figure 9

t/' Now Try Exercises 69 and 71 .

R.1 Basic Concepts from Algebra

9

The following table summarizes this discussion. Results about Positive and Negative Numbers Words

Symbols

Every negative number is less than 0. Every positive number is greater than 0.

If a is negative, then a If a is positive, then a

< 0. > 0.

0 is neither positive nor negative.

In addition to the symbols

~,

, the symbol s

:5

and

~

are often used.

Inequality Symbols Symbol

Meaning

*


is greater than

:S

is less than or equal to

~

is greater than or equal to

Example

3 #7 -4< - 1

is less than

3 > -2 6 ::; 6

-8

~- JO

With the symbol :5, if either the < part or the = part is true, then the inequality is true. This is also the case with the ~ symbol.

( R.1

4

Exercises CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.

1. The set { 0, I , 2, 3, . . . } describes the set of _ __ 2. The set containing no elements is the _ _ _ , symbolized _ __ 3. The opposite, or negative, of a number is its _ __ 4. The distance on a number line from a number to 0 is the _ _ _ of the number. 5. If the real number a is to the left of the real number b on a number line, then a b. 6. CONCEPT PREVIEW Match each number from Column I with the Jetter or letters of the sets of numbers from Column II to which the number belongs. There may be more than one correct choice, so give all correct choices. I (a) 0

(b) 34 9

(c) -4 (e)

II

Vi3

(d)

v36

(f) 2.16

A. Natural numbers

B. Whole numbers

C. Integers

D. Rational numbers

E. Irrational numbers

F. Real numbers

CONCEPT PREVIEW Work each problem.

7. Identify the set { l , ~,

b, :f?, .. .}as finite or infinite. Is 3 an element of the set?

8. Use set notation and write the elements belonging to the set {x Ix is a natural number Jess than 6 }.

10

CHAPTER R A lgebra Review

9. Evaluate - ( - 1). 10. Give (a) the additive inverse and (b) the absolute value of the number 10. List the elements in each set. See Example 1.

11. { x Ix is a whole number less than 6 }

12. { m Im is a whole number less than 9}

13. { z Iz is a natural number greater than 4}

14. {y Iy is a natural number greater than 8}

15. { z Iz is an integer less than or equal to 4}

16. {p Ip is an integer less than 3 }

17. {a Ia is an even integer greater than 8}

18. { k Ik is an odd integer less than I }

19. {p Ip is a number whose absolute value is 4} 20. { w Iw is a number whose absolute value is 7 } 21. { x Ixis an irrational number that is also rational } 22. { rl r is a number that is both positive and negative} Use set-builder notation to describe each set. See Example 2. (More than one descrip tion is possible.)

23. {2, 4, 6, 8}

24. { 11, 12, 13, 14 }

25. { 4, 8, 12, 16, .. . }

26. { . . . , - 6, - 3, 0, 3, 6, . . . }

Concept Check Let A = {I , 2, 3, 4, 5, 6} , B = { I, 3, 5 }, C ={ I, 6 }, andD = {4 }. Find each set.

no

(b) B n C

(c) B nA

(d) c n A

28. (a) A UB

(b) BUD

(c) B U C

(d) C UD

27. (a) A

12

5

. r:::

I

, /,;:: }

.

Let A = { -6, -4, -8, - v 3, 0, 4, 1, 27T, 3, v 12 . List all the elemen ts of A that belong to each set. See Example 3 .

29. Natural numbers

30. Whole numbers

31. Integers

32. Rational numbers

33. Irrational numbers

34. Real numbers

List all the elements ofeach set that are (a) natural numbers, (b) whole numbers, (c) integers, (d) rational numbers, (e) irrational numbers, (f) real numbers. See Example 3.

35.

x=

{

- 9, -

36. y = { -8, -

6 • r::: 21 75} v 6, - 0.7, 0, 7' v 7, 4.6, 8, 2 ' 13, 5

. r;

• r;

v 5, -

3 • r::: 13 40 } 0.6, 0, 4' v 3, 7T, 5, 2' 17, 2

Determine whether each statem ent is true or false. If it is f a lse, tell why. See Example 4.

37. Every integer is a whole number.

38. Every natural number is an integer.

39. Every irrational number is an integer.

40. Every integer is a rational number.

41. Every natural number is a whole number.

42. Some rational numbers are irrational.

43. Some rational numbers are whole numbers. 44. Some real numbers are integers. 45. The absolute value of any number is the same as the absolute value of its additive inverse. 46. The absolute value of any nonzero number is positive. 47. Concept Check For what value(s) of x is

lxl

=

4 true?

R.1 Basic Concepts from Algebra

11

48. Concept Check Match each expression in parts (a)-(d) with its value in choices A-D. Choices may be used once, more than once, or not at all. I

II

(a) -(-4)

(b) l-41

A. 4

B. - 4

(c) -1-4 1

(d) -1-(-4) 1

C. Both A and B

D. Neither A nor B

Give (a) the additive inverse and (b) the absolute value of each number. 49. 6

so.

-12

6 5

51.

52. 0.16

Evaluate each expression. See Example 5. 53. l-8 1

54. l- 19 1

55.

56.

58. -1 12 1

59. - 1-21

60. - l -61

1%1

l~I

57. - 15 1 62. - 112.4 I

61. - 14.51

Sea level refers to the surface of the ocean. The depth ofa body of water can be expressed as a negative number, representing average depth infeet below sea level. The altitude of a mountain can be expressed as a positive number, indicating its height infeet above sea level. The table gives selected depths and altitudes.

Body of Water

Average Depth in Feet (as a negative number)

Altitude in Feet (as a positive number)

Mountain

-14,040

Denali

20,310

South China Sea

-4802

Point Success

14,164

Gulf of California

-2375

Matlalcueyetl

14,636

Caribbean Sea

-8448

Rainier

14,410

Steele

16,624

Pacific Ocean

-12,800

Indian Ocean

Data from The World Almanac and Book of Facts.

63. List the bodies of water in order, deepest to shallowest. 64. List the mountains in order, shortest to tallest. 65. True or false: The absolute value of the depth of the Pacific Ocean is greater than the absolute value of the depth of the Indian Ocean. 66. True or false: The absolute value of the depth of the Gulf of California is greater than the absolute value of the depth of the Caribbean Sea. Use a number line to determine whether each statement is true or false. See Example 6. 67. -6 -3

70. -3 > -1

71. 3 > -2

72. 6 > -3

73. -3 > -3

74. -5 < - 5

Rewrite each statement with > so that it uses < instead. Rewrite each statement with < so that it uses >. See Example 6. 75. 6 > 2

76. 5

>1

78. -6 < 1

79. -5 > - 10

77. -9 - 12

Use an inequality symbol to write each statement. 81. 7 is greater than - 1.

82. -4 is less than 10.

83. 5 is greater than or equal to 5.

84. -6 is less than or equal to -6.

85. 5 + 0 is not equal to 0.

86. 5

+ 14 is greater than or equal to 19.

12

CHAPTER R Algebra Review

Simplify each inequality if needed. Then determine whether the statement is true or false.

87. 0::; -5 90. 10

~

93. 2. 5

88. - 11

91. -6 < 7

10 ~

4+6

96. -1-4 I ::; -4

R.2

~

0

89. 7::; 7

+3

92. -7 < 4

94. 8 + 7::; 3. 5 97. -8 >

+

1

- 1- 3 I ~ - 3

95.

98. - 10 > - 1-4 I

-l-61

Real Number Operations and Properties

•

Operations on Real Numbers

•

Distance between Points on a Number Line

•

Exponents

•

Order of Operations

•

Properties of Real Numbers

Operations on Real Numbers Recall that the answer to an addition problem is a sum. The answer to a subtraction problem is a difference. The answer to a multiplication problem is a product, and the answer to a division problem is a quotient. Other phrases also indicate these operations. Words or Phrases That Indicate Operations on Real Numbers Operation

Word or Phrase

Addition

Sum, added to, more than, increased by, plus

Subtraction

Difference of, subtracted from, less than, decreased by, minus

Multiplication

Product of, times, twice, triple, of, percent of

Division

Quotient of, divided by, ratio of

When performing operations on real numbers, we use sign rules. Sign Rules for Operations on Real Numbers Operation

Examples

To add two numbers with the same sign, add their absolute values. The sum has the same sign as the given numbers.

2+7=9

-2 + (-7) = -9

To add two numbers with different signs, find the absolute values of the numbers, and subtract the lesser absolute value from the greater. The sum has the same sign as the number with the greater absolute value.

- 12

+5 =

4+(-10)=-6 -3

To subtract a number b from another number a, change to addition and replace b by its additive inverse (opposite), -b.

+8=

5

5 - 7 = 5 + (-7) = -2 -8 - 4 = -8

a - b =a+ (-b)

-7

+ ( -4)

- 17 - ( - 4) = - 17

+4=

= - 12 - 13

Then use the sign rules for addition. To multiply or divide two numbers with the same sign, multiply or divide their absolute values. The answer will be positive.

5(7) = 35, 35 - =7 5 '

-35 =7

To multiply or divide two numbers with different signs, multiply or divide their absolute values. The answer will be negative.

-8(9) = -72,

6(-7) = - 42

- 45 -=-5

9

- 5( -7) = 35

-5

'

36 -=-4

-9

R.2 Real Number Operations and Properties

Consider the relationship between the quotient 15 -;- 3 and the product 15 · ~.

Reciprocals Number 2

-6, or - 1 7

15 15 -;- 3 = - = 5 3

-2 6

TT

Reciprocal 5

-5

13

I

-6 II

0.05

7 20

0

None

The reciprocal of a nonzero number a is~. Reciprocals have a product of I.

and

1 15 15. - = - = 5

3

3

Thus division can be written in terms of multiplication using the multiplicative inverse (reciprocal). For all real numbers a and b (b ~ 0) , a

+b=

~ = a .

!. .

b

b

To divide a by b, multiply a (the dividend) by the rec iprocal of b (the divisor).

t,

For this reason, the sign rules for division are the same as the sign rules for multiplication. Consider the quotient~. To find ~, we look for a number n that when multiplied by b will give a. For example, ~ = 2 because 2 · 5 = 10. What happens if we try to divide a nonzero number by 0? For example, to find we look for some number n that when multiplied by 0 will give 4. But n · 0 will always give a product of 0, never 4. Therefore, division by 0 is undefined. On the other hand,~ = 0 because 0 · 4 = 0. Thus, dividing 0 by any nonzero number will always yield 0. Trying to divide 0 by 0 is problematic for another reason. If§= n , then n · 0 = 0. This statement is true for all numbers n, which means that § is not unique. We say that§ is indeterminate -that is, there is no unique answer. This discussion can be summarized as follows.

5,

Division Involving 0

For any nonzero real number a,

a. 0 - IS undefined, 0 a

EXAMPLE 1

= 0,

and

0 . .

.

0 IS mdetermmate.

Adding and Subtracting Real Numbers

Find each sum or difference. (a) -6

+ (-3)

(b) -2.3

+ 5.6

(c) -12 - 4

SOLUTION

(a) -6

+ (-3)

=

- (l-6 1+ l-31)

=

- (6 + 3)

=

-9

(b) -2.3

Add the absolute values. The numbers have the same sign, both negative, so the sum will be negative.

+ 5.6

=

I5.6 I - I2.3 I

=

5.6 - 2.3

=

3.3

r----= r

t (c) -12 - 4 = -12

The numbers have different signs. Subtract the lesser absolute value from the greater. Change to additio n. The additive inverse of 4 is - 4.

+ ( - 4 ) = -16

14

CHAPTER R A lgeb ra Review

(d)

~-

(-%) 5

3

6

8

To subtrac t a - b, add the additive inverse (opposite) of b to a.

=- +20

9

24

24

The least common denominator is 24.

=- +-

54

20 3 3

9

6 . 4 = 24; 8 . 3 = 24

29

Add numerators.

24

Keep the same denominator.

t/' Now Try Exercises 11, 23, and 29. EXAMPLE 2

Multiplying and Dividing Real Numbers

Find each product or quotient where possible. (a) -3(-9)

(b) 6( -9)

-9

(e) -

(f )

0

(c) -0.05(0.3)

0 - 12

(g)

-12

(d)

-~(~)

(h)

4

2 3 5

--

9

SOLUTION

(a) - 3( - 9) = 27

(b) 6( - 9) = - 54

(c) - 0.05(0.3) = - 0.015

The numbers have the same The numbers have different signs, so the product is negative. sign, so the product is positive.

(d)

- ~(~) 6

Multiply numerators.

36

Multiply denominators.

1

. . l 6 W nte in owest terms; 36 =

6 (e)

- 9

0

is undefined.

I· 6 I 6--:--() = 6

- 12

0 - 12

(f) - = O

(g) - = - 3 4

This is true because

The numbers have different signs, so the quotient is negative.

0(- 12)

=

0.

2 (h)

3

This is a complex fraction. A complex fraction has a fraction in the numerator, the denominator, or both.

5 9 =

-~ (- ~)

~ = a · ~; Multiply by - ~, the reciprocal of the divisor - ~.

18

Multiply numerators.

15

Multiply denominators.

6 5

1s 6 . 3 6 . . I W nte in owest terms; 15 = ~ = 5

t/' Now Try Exercises 35, 41, 53, and 59.

R.2 Real Number Operations and Properties

15

Distance Between Points on a Number Line Using absolute value and subtraction of real numbers, we can find the distance between two points on a number line.

Distance Between Points on a Number Line

If P and Q are points on a number line with coordinates a and b, respectively, then the distance d(P, Q) between them is given by the following. Q

p

d(P,Q)

= lb - al

or d(P,Q)

= la - bi

'-------v------

a

d(P, Q)

Figure 10

b

That is, the distance between two points on a number line is the absolute value of the difference of their coordinates in either order. See Figure 10.

Finding the Distance between Two Points

EXAMPLE 3

Find the distance between -5 and 8. SOLUTION

Use the first formula in the preceding box, with a = -5 and b = 8.

d(P, Q)

= Ib - a I = J8 - (-5) I = J8 + 5 J = J13 J = 13

Using the second formula in the box, we obtain the same result. d(P, Q)

= Ia - b I = I (-5) - 8 J = J-13 J = 13 . / Now Try Exercise 65.

Exponents Any collection of numbers or variables joined by the basic operations of addition, subtraction, multiplication, or division (except by 0), or the operations of raising to powers or taking roots, formed according to the rules of algebra, is an algebraic expression.

-2x2

l5y 2y - 3'

+ 3x - - '

~ ; 3 v m -

64,

(3a

+ b) 4

Algebraic expressions

The expression 2 3 is an exponential expression, or exponential, where the 3 indicates that three factors of 2 appear in the corresponding product. The number 2 is the base, and the number 3 is the exponent. Exponent: 3

/ 23 = 2. 2. 2 = 8

/

Base: 2

Three factors of2

Exponential Notation

If n is any positive integer and a is any real number, then the nth power of a is written using exponential notation as follows.

a"= a· a· a· ... ·a n factors of a

Read a" as "a to the nth power'' or simply "a to the nth."

16

CHAPTER R Algebra Review

EXAMPLE 4

Evaluating Exponential Expressions

Evaluate each exponential expression, and identify the base and the exponent. (b) ( -6) 2

(a) 4 3

(c) -6 2

(d) 4. 32

(e) (4 · 3) 2

SOLUTION

(a) 4 3 = 4 · 4 · 4 = 64

The base is 4 and the exponent is 3.

'--v--" 3 factors of 4

The base is -6 and the exponent is 2.

(b) (-6 )2 = (-6)(-6) = 36

(c) - 62 = -(6 . 6) = -36

-===:

Notice that parts (b) and (c) are different.

The base is 6 and the exponent is 2. (d) 4 . 32 = 4 . 3 . 3 = 36 (e) (4·3) 2 = 122 = 144

The base is 3 and the exponent is 2.

~(4·3)2 7"4·3 2 )

The base is 4 · 3, or 12, and the exponent is 2. t/ Now Try Exercises 71, 73, and 77.

When an expression involves more than one opera-

Order of Operations

tion symbol, such as 5 · 2

+ 3, we use the following order of operations.

Order of Operations

If grouping symbols such as parentheses, square brackets, absolute value bars, or fraction bars are present, begin as follows.

Step 1 Work separately above and below each fraction bar. Step 2 Use the rules below within each set of parentheses or square brackets. Start with the innermost set and work outward. If no grouping symbols are present, follow these steps.

Step 1 Simplify all powers and roots. Work from left to right. Step 2 Do any multiplications or divisions in order. Work from left to right. Step 3 Do any negations, additions, or subtractions in order. Work from left to right.

EXAMPLE 5

Using Order of Operations

Evaluate each expression. (a) 6 -;- 3

+ 23 • 5

(b) (8

+ 6) -;- 7 . 3 - 6

4 + 32 (c) 6 - 5 · 3

SOLUTION

6 -;- 3

(a)

+ 23 • 5

Mu ltiply or divide in order

E valuate the expone ntial. Divide.

from left to right.

=

2 + 40

=42

Multipl y. Add.

R.2 Rea l Number Operations and Properties

(b) (8

+ 6)

-7-

7.3- 6

14

-7-

7·3- 6

=

17

Work inside the parentheses.

Be carefu l to divide before multiplying here.

= 2 ·3-6

Divide.

= 6 -6

Multiply.

=O

Subtract.

4 + 32 (c) 6 - 5 · 3

Work separately above and below the fraction bar.

4+ 9 6 - 15

Evaluate the exponential and multiply.

13

Add and subtract.

-9 13 9

II" Now Try Exercises 85, 91, and 93.

EXAMPLE 6

Using Order of Operations

Evaluate each expression for x = -2, y = 5, and z = -3. (a) -4x2

-

7y

2(x - 5) 2 + 4y (b) z+4

+ 4z

SOLUTION

-4x 2

(a)

-

7y

+ 4z

= -4( - 2)2 Use parentheses around substituted values to avoid errors.

(b)

-

7(5 )

+ 4( - 3)

Substitute: x = -2, y = 5, and z = - 3.

=

-4(4) - 7(5) + 4(-3)

Evaluate the exponential.

=

-16 - 35 - 12

Multiply.

=

-63

Subtract.

2(x - 5) 2 + 4y z+4

2( - 2 - 5) 2 + 4(5) - 3+4 2(-7) 2 + 4(5) -3 +4 2(49) + 4(5) -3 + 4 98

+ 20

Substitute: x

=-

2, y

= 5, and z = -

3.

Work inside the parentheses.

Evaluate the exponential. Multiply in the numerator. Add in the denominator.

= 118

'f

Add; = a

II" Now Try Exercises 97 and 103.

18

CHAPTER R A lgebra Review

Recall the following basic properties.

Properties of Real Numbers Properties of Real Numbers

Let a, b, and c, represent real numbers. Property

Description

Closure Properties a + b is a real number. ah is a real number.

The sum or product of two real numbers is a real number.

Commutative Properties a+b=b+a ab = ba

The sum or product of two real numbers is the same regardless of their order.

Associative Properties (a + b) + c =a + (b

The sum or product of three real numbers is the same no matter which two are added or multiplied first.

+ c)

= a(bc)

(ab)c

Identity Properties There exists a unique real number 0 such that

a

+

=a

0

and

+a

0

The sum of a real number and 0 is that real number, and the product of a real number and 1 is that real number.

= a.

There exists a unique real number 1 such that

a· 1 =a

1 ·a= a.

and

Inverse Properties There exists a unique real number -a such that

a

+ ( - a) =

0

and

- a

+a=

0.

The sum of any real number and its negative is 0, and the product of any nonzero real number and its reciprocal is 1.

If a ¥ 0, there exists a unique real I

number asuch that 1

1

a · - = 1 and - · a = 1. a a Distributive Properties a(b + c) = ab + ac

a(b - c)

= ab -

ac

Multiplication Property of Zero O·a=a·O=O

The product of a real number and the sum (or difference) of two real numbers equals the sum (or difference) of the products of the first number and each of the other numbers. The product of a real number and 0 is 0.

t«llim~I With the commutative properties, the order changes, but with

the associative properties, the grouping changes. Commutative Properties

Associative Properties

(x+4) + 9=(4+x) +9 7 . (5 . 2) = (5 . 2) . 7

(x +4) + 9 = x + (4 + 9) 7 . (5 . 2 ) = (7 . 5) . 2

R.2 Real Number Ope rations and Properties

19

Figure 11 helps to explain the distributive property. The area of the entire region shown can be found in two ways, as follows.

4

4( 5 + 3)

4(5) + 4(3) =20 + 12= 32

or

5 3 Geometric Model of the Distributive Property

4(8) = 32

=

The result is the same. This means that

4(5 + 3) = 4(5) + 4(3).

FIGURE 11

EXAMPLE 7

Using the Distributive Property

Rewrite each expression using the distributive property and simplify, if possible. (a) 3(x+y)

(b) -(m - 4n)

(c) 4x

13 (45 32 )

+ 8x

{d ) -

- m - - n - 27

SOLUTION

~

(b) - (m - 4n)

(a) 3(x+y) =

3x

+ 3y

=

Distributive property

- 1( m _ 4n)

(c) 4x

+ 8x =

+ ( - 1) ( - 4n )

= - 1 ( m) = -m

Be careful wit h t he negative signs.

+ 4n

~ 1 (4 23 ) 3 5

(d) -

(4 + 8)x

= 12x

- m - - n - 27

=

Use the distributive property in reverse. Add inside the parentheses.

±(~m) + ±( -~n) + ±(-27) 4

= -

15

1 m- - n-9 2

II' Now Try Exercises 117, 121, and 123. [IK!llilll~!

Be careful. The distributive property does not appl y when simplifying an expression such as the one that follows because there is no addition or subtraction involved.

(3x)(5)(y) =fa (3x)(5) · (3x)(y) See Example 8(d) on the next page.

Terms and Their Coefficients Numerical Coefficient -7

Term - 7y 34r 3

34

26x 5yz4

-26

-k = - l k

-1

-

Expressions such as 3x and 3y in Example 7(a) are examples of terms. A term is a number or the product of a number and one or more variables raised to powers. The numerical factor in a term is the numerical coefficient, or just the coefficient. Terms with exactly the same variables raised to exactly the same powers are like terms. Otherwise, they are unlike terms.

5p and -2lp

-6x 2 and 9x 2

Like terms

7y 3

Unlike terms

r = lr

l

3x

-g =

sx

3

3m and 16x t t

8

Different variables

x

Ix

3

I

3 = 3 = 3x

I

3

t

and - 3y

2

t

Different exponents on the same variable

Simplifying an expression such as 4x + 8x in E xample 7 (c) is called combining like terms. Only like terms may be combined.

20

CHAPTER R A lgebra Review

Using the Properties of Real Numbers

EXAMPLE 8

Simplify each expression.

+ 6y + y

(a) Sy - 8y

+ 4 - S (x + 1) - 8 4(3x - S) - 2(4x + 7)

(b) 3x

(d) 3x(S)(y)

(e)

(c) 8 - (3m + 2)

SOLUTION

+ 6y + y

(a) Sy - 8y =

Sy - 8y + 6y + ly

= (S =

- 8+6

+ 1 )y

Identity property Distributive property

4y

Add inside the parentheses.

~

(b) 3x+4 - S(x + l ) -~ =

3x + 4 - Sx - S - 8

Distributive property

=

3x - Sx + 4 - S - 8

Commutative property

=

-2x - 9

Combine like terms.

(c) 8 - (3m

+ 2)

=

8 - 1 ( 3m + 2)

Identity property

=

8 - 3m - 2

Distributive property

=

8 - 2 - 3m

Commutative property

=

6 - 3m

Combine like terms.

(d) 3x(S)(y) = [ 3x( S) ]y =

[3(x · S)]y

= [

3( Sx) ]y

= [ (3 =

· S )x ]y

(lSx)y

= 1S (xy) =

Order ofoperations Associative property Commutative property Associative property Multiply. Associative property

lSxy

Many of these steps are not usually written out. (e) 4(3x - S) - 2(4x

+ 7)

= 12x - 20 - 8x -

14

= 12x - 8x - 20 - 14 =

4x - 34

Distributive property Commutative property Combine like terms.

Like terms may be combined by adding or subtracting the coefficients of the terms and keeping the same variable factors.

v

Now Try Exercises 135, 143, and 147.

R.2 Real Number Operations and Properties

( R.2

21

;

Exercises CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.

I. The sum of two negative numbers is _ __

2. The product of two negative numbers is _ __ 3. The quotient formed by any nonzero number divided by 0 is _ _ _ , and the quotient formed by 0 divided by any nonzero number is _ __ 4. The _ _ _ property is used to change the order of two terms or factors, and the _ _ _ property is used to change the grouping of three terms or factors. 5. Like terms are terms with the same _ _ _ raised to the _ _ _ powers. 6. The numerical coefficient in the term -7yz2 is _ __ CONCEPT PREVIEW Evaluate each expression.

7.

8. 2. 5 - 10

1Q3

-7

2

9. 3a - 2b, for a = - 2 and b = -1

10. CONCEPT PREVIEW Rewrite the expression -7(x - 4y) using the distributive

property. Find each sum or difference. See Example 1.

11. - 6+( - 13)

12. -8+(- 16)

13. - 15 + 6

14. - 17 + 9

15. 13 + (-4)

16. 19+(-13)

7 3 17. --+3 4

5 4 18. --+6 9

19. -2.8 + 4.5

20. - 3.8 + 6.2

21. 4 - 9

22. 3 - 7

23. - 6 - 5

24. -8 - 17

25. 8 - (-13)

26. 12 - ( - 22)

27. - 12.31 - (-2.13)

28. - 15.88 - (-9.42 )

1- - (- ~) 3

~ - (- ~) 4

29 . 10

30. 14

31. l-8 - 61

32. l-7 - 15 I

33. - 2 - l-41

34. 16- l-131

Find each product or quotient where possible. See Example 2.

35. -8(-5)

36. -20(-4)

37. 5(-7)

38. 6( - 9)

39. 4(0)

40. 0( -8)

41.

- ~e~) 2 25

44. _

_3__(- 22)

42.

-*G~)

43.

-~( -

294)

45. - 0.06(0.4)

46. -0.08(0.7)

- 24 47. -4

- 45 48. -9

49. 100 -25

50. 150 -30

0 51. -8

0 52. - 14

11

4

22

CHAPTER R Algebra Review

5 S3. 0

13 S4. 0

SS. -

_!Q + 17

4

.

23

5

S8. -5-

5

12 13 S9.4 3

13

7 6 60. 2 3

31T

62. -1-

63.

.!3.) 5

12 13

5 S7. - 3

S6 - 22 -;- (- 33)

( -

21T

61. -

2 3 (Leave 1T in the answer.)

-7.2 o.s

- 4.5 64 - • 0.9

2 (Leave 1T in the answer.)

Find the given distances between points P, Q, R, and Son a number line, with coordinates -4, - 1, 8, and 12, respectively. See Example 3. 6S. d(P, Q)

67. d(Q, R)

66. d(P, R)

68. d(Q, S)

Concept Check Evaluate each exponential expression. 69. (a) 82

(b) -8 2

(c) ( -8)2

(d) -(-8) 2

70. (a) 4 3

(b) - 4 3

(c) (-4) 3

(d) -(-4) 3

Evaluate each expression. See Example 4. 71. -24

72. -3 5

73. (-2) 4

74. (-2) 6

7S. (-3) 5

76. (-2) 5

77. -2 . 34

78. - 4. 53

79. Concept Check Frank's grandson was asked to evaluate the following expression.

9

+ 15 -;- 3

Frank gave the answer as 8, but his grandson gave the answer as 14. The grandson explained that the answer is 14 because of the "Order of Process rule," which says that when evaluating expressions, we proceed from right to left rather than left to right. (Note: This is a true story.) (a) Whose answer was correct for this expression, Frank's or his grandson's? (b) Was the reasoning for the correct answer valid? Explain. 80. Concept Check Problems like this one occasionally appear on social media. Simplify this expression. 7+7+7+7X7-7

Evaluate each expression. See Example 5.

81. 12

+ 3. 4

84. 9. 4 - 8-;- 2

82. 15

+ 5. 2

83. 6 . 3 - 12 -;- 4

8S. 10 + 30 -;- 2 . 3

86. 12 + 24 -;- 3 . 2

87. 5 - 7 . 3 - (-2)3

88. - 4 - 3 . 5 - (-3) 3

89. 18 - 4 2 + 5 - (3 - 7)

90. 10 - 2 2 + 9 - ( 1 - 8)

91. - 4(9 -8)+(-7)(2)3

92. 6(-5) - (-3)(2) 4

93.

-8 + (-4)( -6)-;- 12 4 - (-3)

94.

15-;- 5 • 4-;- 6 - 8

-6 - (-5) - 8-;- 2

23

R.2 Rea l Number Operations and Properties

Evaluate each expression for p = -4, q = S, and r = - 10. See Example 6. 95. 3p - 2r

q+r 99. - q+p 102.

97. -p 2

96. Sr - 6p 100.

____]!f__

103.

3p - 2r

-

7q + r 2

p+r p+q

101.

-(p + 2) 2

-

98. - p 2

Sr 2p - 3r

---

-(q - 6)2

3r

104.

2- q

2q + r

-

-

2p

4-p

Identify the property illustrated in each statement. Assume all variables represent real numbers. 105. 6 . 12 + 6 . lS = 6( 12 + lS )

106. S(m+4)=Sm+32

1 107. (t - 6) . ( -- ) = 1, if t - 6 # 0 t- 6

2 +m 2 -m 108. - - · - - = l , ifm # 2or-2 2-m 2+m

+ 0 = 7.S - y

109. (7.S - y)

111. S(t + 3) = (t

115. S +

+ 3) · S

s(

113. (Sx)G) =

110. 1 · (3x - 7) = 3x - 7

x·

112. - 7 + (x + 3) = (x + 3) + (-7)

~)

114. (3S+99)+ 1=3S +(99+ 1)

V3 is a real number.

116. S1T is a real number.

Rewrite each expression using the distributive property and simplify, Example 7.

if possible. See

117. 2(m + p)

118. 3(a+b)

119. - 12(x-y)

120. - 1O(p - q)

121. -(2d - !)

122. -(3m - n)

123. Sk + 3k

124. 6a +Sa

125. 7r - 9r

126. 4n - 6n

127. a+ 7a

128. s + 9s

129. x + x

130. a+ a

131. 2(x - 3y + 2z)

132. S(3x + y - Sz )

133 -3(16 - y + -32 z - -40)

.s

9

27

1

134. -4(20m +Sy - 32z)

9

Simplify each expression. See Example 8. 135. - 12y + 4y + 3y + 2y 137. - 6p +

s-

4p + 6 + llp

139. 3(k + 2) - Sk + 6 + 3 141. 10 - (4y

+ S)

136. - Sr - 9r + Sr - Sr 138. -Sx - 12 + 3x - Sx + 9 140. S(r - 3) 142. 6 - (9y

+ 6r - 2r + 4

+ S)

143. 10x(3)(y)

144. Sx(6)(y)

2 145. -3( 12w)(7z)

146. -6( 1Sw)(Sz)

147. 3(m - 4) - 2(m + 1)

148. 6(a - S) - 4(a + 6)

149. 0.2S(S + 4p) - O.S(6 + 2p )

150. 0.4(10 - Sx) - O.S(S + lOx)

s

24

CHAPTER R Algebra Review

R.3

Exponents, Polynomials, and Factoring

•

Rules for Exponents

•

Zero as an Exponent

•

Polynomials

•

Addition, Subtraction, and Multiplication of Polynomials

•

Factoring Polynomials

Rules for Exponents The notation a"' (where m is a positive integer and a is a real number) means that a appears as a factor m times. In the same way, a" (where n is a positive integer) means that a appears as a factor n times.

Rules for Exponents

For all positive integers m and n and all real numbers a and b, the following rules hold true. Rule

Example

Description

Product Rule am. a" = am+11

22 . 23 = (2. 2)(2 . 2. 2)

When multiplying powers of like bases, keep the base and add the exponents.

= 22+3 =

25

Power Rule 1

(a"' )n =a"'"

To raise a power to a power, multiply the exponents.

( 4 5)3 = 45 • 45 • 4 5 = 45+5+5 = 45· 3 = 415

Power Rule 2 (ab}m = ambm

(7x) 3

=

(7x)(7x)(7x)

=

(7 · 7 · 7)(x · x · x)

To raise a product to a power, raise each factor to that power.

Power Rule 3

(~)4 (~)(~)(~)(~)

(~)m = ~: (b #=

=

O}

3.3.3.3 5.5.5.5 34

To raise a quotient to a power, raise the numerator and the denominator to that power.

54

EXAMPLE 1

Using the Product Rule

Simplify each expression. (a) y4 • y 1

(b)

(6z 5 )(9z3 )(2z)

SOLUTION (a) y 4 • y 7 = y 4 + 7 = y 11

(b)

Product rule: Keep the base and add the exponents.

(6z )(9z 3 )(2z) 5

= (6

· 9 · 2) · ( z5 z3z1)

= 108z 5 + 3+ 1 = 108z

9

Commutative and associative properties; z =

z1

Multiply. Apply the product rule. Add.

II' Now Try Exercises 13 and 17.

R.3

EXAMPLE 2

Exponents, Polynomials, and Factoring

25

Using the Power Rules

Simplify. Assume all variables represent nonzero real numbers.

-2m6)5 (d) ( t2z

25)3

(c) ( b4 SOLUTION

(a) ( 53) 2 = 53(2 ) = 56

Power rule I

(b) (34x2)3

(c)

= (3 4)3(x 2) 3 Power rule 2 = 34 ( 3lx 2 ( 3) =

(~:y (25)3 (b4)3

Power rule I

312x6

2 15

bl2

Power rule 3

Power rule I

-2m6) 5

(d) ( - t 2z

(-2m6)5 ( t2z)5

Power rule 3

( _ 2 )5(m6)5 (t2)5z5 -32m30 t1025

Power rule 2

Evaluate (- 2 ) 5. Then use Power rule I .

32m30 t/' Now Try Exercises 23, 25, 29, and 31 .

C«uOCll~I The expressions mn2 and (mn) 2 are not equivalent. The second

power rule can be used only with the second expression: (mn) 2 = m 2n2.

Zero as an Exponent

A zero exponent is defined as follows.

Zero Exponent

For any nonzero real number a, a 0

= 1.

That is, any nonzero number with a zero exponent equals 1.

To illustrate why a 0 is defined to equal 1, consider the product

a" · a0 , for

a ¥= 0.

We want the definition of a 0 to be consistent so that the product rule applies. Now apply this rule.

a" • a 0 = a 11 +0 = a" The product of a" and a0 must be a", and thus a0 is acting like the identity element 1. So, for consistency, we define a 0 to equal 1. (0° is undefined.)

26

CHAPTER R Algebra Review

EXAMPLE 3

Usin the Definition of a 0

Evaluate each expression. (b) ( -4 )0

(a) 4°

(d)

-(-4) 0

(e)

(c) -4°

(7r) 0

SOLUTION (a) 4° = 1 (c)

- 4°

(b) ( - 4 )0 = 1

Base is 4.

= - ( 4°) = - 1

(e) (7r)O= l,r#O

Base is - 4.

(d) - ( - 4) 0 = - ( 1) = - 1

Base is 4.

Baseis-4.

Base is7r.

v

Now Try Exercise 35.

Polynomials A polynomial is a term or a finite sum of terms, with only whole number exponents permitted on the variables. If the terms of a polynomial contain only the variable x, then the polynomial is a polynomial in x.

5x 3

-

8x 2 + 7x - 4,

9p 5

-

6

3,

Polynomials

The terms of a polynomial cannot have variables in a denominator.

9x 2

-

4x +

6

-

x

Not a polynomial

The degree of a term with one variable is the exponent on the variable. For example, the degree of 2x3 is 3, and the degree of 17x (that is, 17x 1) is 1. The degree of a polynomial is the greatest degree of any term in the polynomial. For example, the polynomial

4x 3

-

2x 2

-

3x + 7

has degree 3

because the greatest degree of any term is 3. A nonzero constant such as -6, equivalent to -6x 0 , has degree 0. (The polynomial 0 has no degree.) Polynomials are often written with their terms in descending order (or descending degree). The term of greatest degree is first, the one of next greatest degree is next, and so on. Recall that some polynomials are given special names. •

A polynomial containing exactly three terms is a trinomial.

•

A two-term polynomial is a binomial.

•

A single-term polynomial is a monomial. Addition, Subtraction, and Multiplication of Polynomials '

Adding and

subtracting polynomials is similar to simplifying expressions.

Adding and Subtracting Polynomials

To add polynomials, combine like terms. To subtract polynomials, change the sign of each term in the subtrahend (second polynomial) and add the result to the minuend (first polynomial)that is, add the opposite of each term of the second polynomial to the first polynomial.

Expo ne nts, Polynomials, a nd Factoring

R.3

27

Adding and Subtracting Polynomials

EXAMPLE 4

Add or subtract, as indicated. (a) (2y 4

-

(b) -4(x2

+ y) + (4y 4 + 7y 2 + 6y) + 3x - 6) - (2x 2 - 3x + 7) 3y 2

SOLUTION

(a) (2y 4 = =

+ y) + (4y 4 + 7y 2 + 6y~ ( 2 + 4 )y 4 + ( - 3 + 7)y 2 + ( 1 + 6 )y Add coefficients of like terms. Work inside the parentheses. 6y 4 + 4y 2 + 7y

-

(b) -4(x2 = =

3y 2

+ 3x - 6) - (2x 2 -

+ 7)

3x

-4 (x 2 + 3x - 6) - 1(2x2

-

3x + 7)

- 4(x 2 ) - 4(3x) - 4 (-6) - 1(2x 2 )

-a= - la

l (-3x) - 1(7)

-

+ 24 - 2x 2 + 3x - 7

= - 4x 2

-

12x

= -6x2

-

9x + 17

Distributive property Multiply. Combine like terms.

t/ Now Try Exercises 37 and 39.

Multiplying Polynomials

To multiply two polynomials, multiply each term of the first polynomial by each term of the second polynomial. Then combine like terms. To multiply two binomials, use the FOIL method. Also recall the special product rules. For x and y , the following hold true. (x + y ) 2 = x 2 + 2xy + y 2 }

(x _ y )2 = x2 _ 2xy

+ y )(x - y)

(x

+ y2

= x2 -

y2

Square of a binomial

Product of a sum and difference of two terms

Multiplying Polynomials

EXAMPLE 5

Find each product. (a) (4y - 1)(3y

+ 2)

(b) (3x

+ Sy)(3x -

Sy)

(c) (2t+3) 2

(e) (3x- 4 ) 3

(d) (Sx - 1) 2 SOLUTION

(a) (4y - 1)(3y + 2)

+ 4y(2)

=

4y(3y)

=

12y 2

=

12y 2 + Sy - 2

(b) ( 3x

+ Sy )( 3x -

= (3x) 2 =

+ 8y -

9x2

-

3y - 2

FOIL method - Multiply First terms, Outer terms, Inner terms, Last terms. Multiply. Combine like terms.

Sy ),A(ab)2 = a2b2, not ab2.)

-(Sy) 2 2Sy 2

- 1(3y) - 1(2)

(x+y)(x - y)=x2-y 2

28

CHAPTER R Algebra Review

(c) (2t + 3)2

= (2t) 2 + 2(2t)( 3) + 32

(x

+ y)2 =

x 2 + 2xy

+ y2

Remember the middle term.

= 4t 2 + 12t + 9 (d) (Sx - 1)2

=(5x) 2 -2(5x)( 1)+ 12

(x - y )2 = x 2 - 2xy + yz

= 25x 2

(5x )2 =

(e) (3x -

lOx

-

+ 1

52x 2 = 25x 2

4) 3

=(3x-4)(3x-4) 2 = (3x - 4 )(9x2 - 24x

= 3x(9x 2 =

+ 16)

Square 3x - 4.

24x + 16) - 4(9x 2

72x2

27x 3 -

= 27x 3

-

a3 = a · a2

36x2

+ 48x -

24x + 16)

-

+ 96x - 64

Multiply.

108x2 + 144x - 64

-

Distributive property

Combine like terms.

r/ Now Try Exercises 45, 49, 53, and 59. Factoring Polynomials

Recall that factoring involves writing a polynomial

as a product.

Guidelines for Factoring a Polynomial

Question 1 Is there a common factor other than 1? If so, factor it out. Question 2

How many terms are in the polynomial? Two terms: The polynomial is a binomial. Is it a difference of squares or a sum or difference of cubes? If so, factor using the appropriate rule. x 2 - y 2 = (x + y)(x - y) Difference of squares x3 x3

y3

+ y3

= (x = (x

- y )(x2 + xy + + y)(x2 - xy +

y 2) y 2)

Difference of cubes Sum of cubes

Three terms: The polynomial is a trinomial. Is it a perfect square trinomial? If so, factor as follows.

= (x 2xy + y2 = (x

x2 + 2xy + y2

+ y)2

x2 -

- y)2

.

.

Perfect square trrnom1als

If the trinomial is not a perfect square trinomial, use one of the

following methods. • To factor x 2 + bx •

+ c, find two integers whose product is c and whose sum is b, the coefficient of the middle term. To factor ax2 + bx + c, use the FOIL method in reverse, and try various combinations of the factors until the correct middle term is found.

Four terms: Try to factor by grouping. Question 3

Can any factors be factored further? If so, factor them.

R.3

29

Expone nts , Polynomials, and Factoring

Factoring Polynomials

EXAMPLE 6

Factor each polynomial completely. (a) 6x2y 3 - l 2x3y 2 (b) 4y 2 - lly + 6 (d) 100t2 - 81 (e) 4x2 + 20xy + 25y 2

(c) 3x 2 - 27x

+ 42

(f) 1000x3 - 27

+ 4y - 2

(g) 6xy - 3x SOLUTION

(a) 6x2y3 - l2x3y2 =

6x 2y 2(y) - 6x 2y 2( 2x)

6x 2y 2 is the greatest common factor.

=

6x 2y 2(y - 2x)

Distributive property

(b) To factor this polynomial, we must find values for integers a, b, c, and din

such a way that 4y 2 - lly = ( ay

+6 + b) ( cy + d).

FOIL method

Using the FOIL method, we see that ac = 4 and bd = 6. The positive factors of 4 are 4 and 1 or 2 and 2. Because the middle term has a negative coefficient, we consider only negative factors of 6. The possibilities are -2 and -3 or - 1 and -6. Now we try various arrangements of these factors until we find one that gives the correct coefficient of y. (2y - 1)(2y - 6) = 4y 2 - l4y + 6

(2y - 2)(2y - 3) = 4y 2 - lOy + 6

(y-2)(4y-3) = 4y 2 - l l y + 6

Incorrect

Incorrect

Correct

Therefore, 4y2 - ll y CHECK

+ 6 factors as (y - 2)(4y - 3).

(y - 2)(4y - 3) = 4y 2 - 3y - 8y = 4y 2 - l l y

(c) 3x 2

-

+ 14)

3(x 2

=

3(x - 7 )(x - 2) -

9x

-

Original polynomial

Factor out the common factor. Factor the trinomial.

81

=

( 10t)2

=

( lOt

-

92

+ 9 )( 10t - 9)

(e) In 4x2 + 20xy 2

+ 6 ./

FOIL method

27x + 42

=

(d) l00t 2

+6

Difference of squares x2

-

y 2 = (x

+ y)(x - y)

+ 25y 2 , the terms 4x 2 and 25y 2, which can be written as 2

(2x) and (5y ) , are both perfect squares, so this trinomial might factor as a perfect square trinomial. Try to factor 4x2

+ 20xy + 25y 2 as (2x + 5y )2.

30

CHAPTER R Algebra Review

Take twice the product of the two terms in the squared binomial

CHECK

(2x + 5y )2 . 2 · 2x · Sy = 20xy ~ Middle term of 4x 2 + 20.xy + 25y 2 Twice

_j j

t_ Last term

First term

Because 20.xy is the middle term of the trinomial, 4x 2 (f) 1000x3

+ 20.xy + 25y 2

27

-

= ( 1Ox ) 3 = ( lOx

33

Difference of cubes

- 3 )[ ( lOx ) 2 + lOx (3 ) + 3 2 J

(10x - 3)(100x2 + 30x + 9)

=

(2x + 5y) 2 .

factors as

x 3 - y 3 = (x - y) (x 2 + xy + y 2) ( 10x) 2 = 102x 2 = 100x2

(g) Because there are four terms, try factoring by grouping.

6xy - 3x + 4y - 2 =

(6.xy - 3x ) + (4y - 2)

Group the terms.

=

3x( 2y - 1) + 2( 2y - 1)

Factor each group.

=

(2y - 1) (3x + 2)

Factor out 2y- I.

v' Now Try Exercises 65, 71 , 77, 79, 87, and 95.

t1MuiCll~I After factoring a polynomial, it is wise to do the following.

1. Check that the product of all the factors does indeed yield the original polynomial.

2. Check that the original polynomial has been factored completely.

( R.3

4

Exercises CONCEPT PREVIEW Fill in the blank to correctly complete each sentence.

1. The polynomial 2x 5

x

-

+ 4 is a trinomial of degree _ __

2. A polynomial containing exactly one term is a(n) _ __ 3. A polynomial containing exactly two terms is a(n) _ __ 4. When multiplying powers of like bases, as in y 2 • y 3, keep the base and _ __ the exponents.

5. To raise a power to a power, as in ( a 2 ) 3,

___

the exponents.

CONCEPT PREVIEW Work each problem.

6. Match each polynomial in Column I with its factored form in Column II. I

II

(b) x 2

+ lOxy + 25y - 10.xy + 25y 2

(c) x 2

-

(a) x

2

2

(d) 2Sy

2Sy 2 2

-

x

2

A. (x

B. (x

+ 5y )(x - Sy ) + Sy) 2

C. (x - 5y )2 D. (Sy+ x)( Sy - x )

R.3 Exponents, Polynomials, and Factoring

31

7. Match each polynomial in Column I with its factored form in Column II. II

I

(a)

8x 3 -

(b) 8x 3

A. (3 - 2x)(9 + 6x + 4x 2 ) B. (2x - 3) (4x 2 + 6x + 9 ) C. (2x + 3)(4x2 - 6x + 9)

27

+ 27

(c) 27 - 8x 3

8. Which of the following is the correct factorization of 6x 2 + x - 12?

A. ( 3x + 4) (2x + 3)

B. (3x - 4)(2x - 3 )

C.

D. (3x - 4)(2x + 3)

(3x + 4)(2x - 3)

9. Which of the following is the correct complete factorization of x 4

A. (x 2 - l )(x 2 + 1) C. (x 2

-

-

l?

B. (x 2 + l )(x + l )(x - 1) D. (x- 1)2(x+ l )2

1) 2

10. Which of the following is the correct factorization of x 3 + 8? A. (x+ 2)3 B. (x+ 2 )(x2 + 2x+ 4 )

C. (x+ 2 )(x2 - 2x+ 4 )

D. (x + 2)(x2

4x + 4)

-

Simplify each expression. See Example 1.

12. (3y 4 ) (-6y 3 ) 13. n6

•

n4 • n

14. a 8 • a 5 • a

15. 93

•

95

16. 42 • 48

17. (-3m4 )(6m2)(-4m 5 )

18. (-8t3 )(2t6 )(-5t 4 )

19. ( 5x 2y)(-3x3y4)

20. (-4xy 3 )(7x 2y)

21. G mn

}sm 2 2

n

22. (35m4 n) ( -*mn 2)

)

Simplify each expression. Assume all variables represent non zero real numbers. See Examples 2 and 3.

23. (22)5

24. (64) 3

25. (-6x 2) 3

26. (- 2x 5)5

27. -(4m3n0)2

28. -(2x 0 y 4 ) 3

(::)3

30. ( :4Y

31. ( - 4m2)4 tp2

32. ( -:2n4

33. - ( x3:5 )°

34. -

29.

y

( p2q3 ) o - 3 r

Match each expression in Column I with its equivalent in Column II. See Example 3. I

35. (a) 6°

II

A. 0

I

36. (a) 3p 0

II

A. 0

(b) -6°

B. l

(b ) -3p0

B. 1

(c) ( - 6) 0 (d) -(-6)0

c.

(c) (3p ) 0

c.

- 1 D. 6 E. -6

(d) (-3p ) 0

- 1 D. 3 E. - 3

32

CHAPTER R A lg ebra Review

Add or subtract, as indicated. See Example 4.

37. (Sx 2

-

4x + 7) + (-4x2 + 3x- S)

38. (3m 3

-

3m 2 + 4) + ( -2m3

39. 2( 12y 2 40. 3(8p 2 41. (6m4

Sy+ 6) - 4(3y 2

-

Sp) - S(3p 2

-

-

m2 + 6)

-

4y + 2)

2p + 4)

3m2 + m) - (2m3 + Sm 2 + 4m) + (m2

-

42. -(8x 3 + x - 3) + (2x 3 + x 2 )

-

m)

-

(4x2 + 3x - 1)

Find each product. See Example 5.

43. 4x 2 (3x 3 + 2x2 45. (4r - 1)(7r

Sx + 1)

-

+ 2)

+ 1)(2x - 7)

47. (3x

49. (2m + 3)(2m - 3) 51. (4x 2

Sy)(4x 2 + Sy)

-

+ 2n) 2

53. (4m

44. 2b 3 (b 2 -4b+3) 46. (Sm - 6)(3m + 4)

48. (Sz + 3)(2z - 3) 50. (8s - 3t)(8s + 3t)

+ n)(2m3 -

52. (2m 3

n)

54. (a - 6b) 2

55. (Sr - 3t 2 ) 2

56. (2z4 -3y) 2

57. (x + l )(x + l )(x - l )(x - 1)

58. ( t + 4) ( t + 4) ( t - 4) (t - 4)

59. (y + 2) 3

60. (z - 3) 3

+ S) 3

61. (2x

62. (4x - 1) 3

63. (q - 2) 4

64. (r

+ 3)4

Factor each polynomial completely. See Example 6.

65. 40ab - 16a

67. 8x 3y 4 69. x 2

66. 2Sxy - lSy

+ 12x2y 3 + 36xy 4

70. x 2

2x - lS

-

71. 2x 2

-

9x - 18

73. 6a 2

-

lla

68. 10m5n + 4m2n3 + 18m3n2

+x

- 12

72. 3x 2 + 2x - 8

+4

74. 8h2

-

2h - 21

75. 3m2 + 14m + 8

76. 9y 2

-

18y + 8

77. 4x 2 - 28x + 40

78. 2x 2

-

18x + 36

79. 36t 2

-

2S

80. 49r 2

-

9

81. 2Ss4

-

9t2

82. 36z2

-

8ly 4

84. 2St 2

+ 90t + 81

83. 16t 2 + 24t + 9 85. 4m 2p - 12mnp

+ 9n2p

86. 16p2r - 40pqr + 2Sq2r

+ 27

87. x 3 + 1

88. x 3

89. 8t 3 + 12S

90. 27s 3 + 64

91. t 6

92. w 6

-

12S

-

27

R.3 Exponents, Polynomials, and Factoring

93. t 4

-

95. Sxt

94. r 4

1

+

1Sxr + 2yt + 6yr

96. 3am

97. 6ar + 12br - Sas - lObs 99. 4x

2

+

12xy

+ 9y

2

81

+

18mb

+ 2an +

12nb

98. 7mt + 35ms - 2nt - lOns

100. 8l t 2

1

-

-

33

+ 36ty + 4y 2 - 9

(Modeling) Solve each problem.

101. Geometric Modeling Consider the figure, which is a square divided into two squares and two rectang les.

x

y y

x

(a) The length of each side of the largest square is x + y. Use the formula for the area of a square to write the area of the largest square as a power. (b) Use the formulas for the area of a square and the area of a rectangle to write the area of the largest square as a trinomial that represents the sum of the areas of the four figures that make it up. (c) Explain why the expressions in parts (a) and (b) must be equivalent. (d) What special product formula from this section does this exercise reinforce

geometrically?

102. Geometric Modeling Use the figure to geometrically support the distributive property. Write a short paragraph explaining this process.

Relating Concepts For individual or collaborative investigation (Exercises 103-106)

The special products can be used to perform selected multiplications. On the left, we use (x + y )(x - y) = x 2 - y 2 . On the right, (x - y) 2 = x 2 - 2xy + y 2. 51

x 49 = (50 + 1) (50 - 1) = 50 2

-

12

47 2

= (50 -

3) 2

= 502 - 2(50)(3)

= 2500 - 1

= 2500 - 300

= 2499

= 2209

+ 32

+9

Use the sp ecial products to evaluate each expression.

103. 99 x 101

104. 63

x 57

105. 1022

106. 7 12

34

CHAPTER R Algebra Review

( R.4

q

Rational Expressions

•

Rational Expressions

•

LowestTerms of a Rational Expression

•

Multiplication and Division

•

Addition and Subtraction

•

Complex Fractions

Rational Expressions The quotient of two polynomials P and Q, with Q ¥- 0, is a rational expression.

x+6 x + 2'

+ 7p - 4 5p 2 + 20p

(x+6)(x+4) (x+2)(x+4)'

2p2

Rational expressions

The domain of a rational expression is the set of real numbers for which the expression is defined. Because the denominator of a fraction cannot be 0, the domain consists of all real numbers except those that make the denominator 0. We find these numbers by setting the denominator equal to 0 and solving the resulting equation. For example, in the rational expression

x+6 x+ 2 ' the solution to the equation x + 2 = 0 is excluded from the domain. The solution is -2, so the domain is the set of all real numbers x not equal to -2. { x Ix ¥- -2}

Set-builder notation

If the denominator of a rational expression contains a product, we determine the domain with the zero-factor property, which states that ab = 0 if and only if a = 0 orb = 0 or both equal 0. EXAMPLE 1

Finding the Domain

Find the domain of the rational expression.

(x+6)(x+4) (x+2)(x+4) SOLUTION

(x+2)(x+ 4 ) =0 x+2=0 x = -2

or x

Set the denominator equal to 0.

+4= 0

or

Zero-fac tor property

x = -4

Solve each equation.

The domain is the set of real numbers not equal to -2 or -4, written

{x lx ¥- -2, -4 }. tl" Now Try Exercises 11 and 13. Lowest Terms of a Rational Expression A rational expression is written in lowest terms when the greatest common factor of its numerator and its denominator is 1. We use the following fundamental principle of fractions to write a rational expression in lowest terms by dividing out common factors.

Fundamental Principle of Fractions

ac be

a

=b

(b

::/=

O,c

::/=

0)

R.4 Rational Expressions

35

Writing Rational Expressions in LowestTerms

Write each rational expression in lowest terms. 2x 2 + 7x - 4 (a) 5x2 + 20x

6 - 3x (b)

x2 -

4

SOLUTION

2x2 + 7x - 4 (a) 5x 2 + 20x (2x - l ) (x Sx (x

+ 4)

+ 4)

2x - 1

Factor.

Divide out the common factor.

Sx

To determine the domain, we find values of x that make the original denominator 5x2 + 20x equal to 0, and exclude them.

+ 20x = Sx(x + 4) = or x + 4 = 5x 2

Sx = 0 x= 0

0

Set the denominator equal to 0.

0

Factor.

0

Zero-factor property

x = -4

or

Solve each equation.

The domain is { x Ix ¥- 0, -4}. From now on, we will assume such restrictions when writing rational expressions in lowest terms. (b) 6 - 3x x2 - 4

3(2 - x) Factor.

(x+2)(x-2)

3(2 - x) (- 1) (x+2) (x-2) (- 1) 3(2 - x) (-1) (x+2) (2-x)

2 - x and x - 2 are opposites. Multiply numerator and denominator by - 1.

(x - 2)(- 1)= - x+ 2 =2 - x

~Be ca reful w it h signs.)

-3

Divide out the common factor.

x+2 LOOKING AHEAD TO CALCULUS

A standard problem in calculus is

Working in an alternative way would lead to the equivalent result _ x

3 _ 2.

investigating what value an expression 1

such as :

t/' Now Try Exercises 23 and 27.

~ / approaches as x

approaches I. We cannot do this by simply substituting I for x in the expression since the result is the indeterminate form§. When we factor the numerator and write the expression in lowest terms, it becomes x

+ I.

Then, by substituting I for x, we obtai n I

+ I = 2, which is the limit of ·:'~ /

as x approaches I.

CIA!Qim~I The fundamental principle requires a pair of common factors, one in the numerator and one in the denominator. Only after a rational expression has been factored can any common factors be divided out. For example,

2x + 4 6

2 (x + 2) 2.3

x +2 3

Factor first, and then divide.

36

CHAPTER R Algebra Review

Multiplication and Division We multiply and divide rational expressions using the same properties used for multiplying and dividing fractions.

Multiplying and Dividing Rational Expressions

For fractions~ and~ (b # 0, d # 0), the following hold true.

a c ac a c a d - ·- = and - + - = - · bd bd b d be

(c -:/= 0)

That is, to find the product of two fractions, multiply their numerators to find the numerator of the product. Then multiply their denominators to find the denominator of the product. To divide two fractions, multiply the dividend (the first fraction) by the reciprocal of the divisor (the second fraction).

EXAMPLE 3

Multiplying or Dividing Rational Expressions

Multiply or divide, as indicated. 2y 2 27 (a) -

(c)

·-

9

3p

8y

2

3m 2 - 2m - 8 3m + 2 (b) 3m 2 + 14m + 8. -3m_+_ 4

5

+ l lp - 4

24p 3 - 8p 2

9p

24p 4

x 3 - y3 2x + 2y + xz + yz x2 - y2 2x2 + 2y2 + zx2 + zy2

+ 36 -

(d) - - . - - - - - - -

36p 3

SOLUTION

2y2

(a)

27 -9 · -8y5 2y 2 . 27 9. 8y 5

Multiply fractions.

2 . 9 . 3 . y2

9 . 2 . 4 . y 2 • y3 3 4y3

Factor.

Lowest terms

As shown here, it is generally easier to divide out any common factors in the numerator and denominator before performing the actual multiplication. 3m2 - 2m - 8 3m + 2 (b) 3m2 + 14m + 8 . -3,- n-+4

(m - 2)(3m + 4)

3m + 2

(m + 4)(3m + 2)

3m + 4

(m - 2) (3m + 4 ) (3m + 2) (m + 4) (3m + 2) (3m + 4 ) m-2 m+4

Factor.

Multiply fractions.

Lowest terms

R.4 Rational Expressions

(c)

3p2 + l lp - 4 24p 3 - 8p 2

-i-

37

9p + 36 24p 4 - 36p 3

(p + 4)(3p - 1) 8p 2(3p - 1) (p + 4) (3p - 1) 8p 2(3p - 1)

9(p + 4) 12p3 (2p - 3)

Factor.

12p3 (2p - 3)

Multiply by the reciprocal of the divisor.

9(p + 4 )

12p 3 (2p - 3) 9. 8p 2

Divide out common factors. Multiply fractions.

3 . 4 . p 2 • p(2p - 3) 3 . 3. 4 . 2. p 2

Factor.

p(2p - 3) Lowest terms

6

x3 _ y3 x2 _ y2

(d) - - - ·

2x + 2y + xz + yz 2x2 + 2y2 + zx2 + zy2

(x - y )(x 2 + xy + y 2 ) (x+y)(x-y) (x - y) (x 2 + xy + y 2 ) (x+y) (x-y)

2(x+y)+z(x+y) 2(x2 + y2) + z(x2 + y2)

Factor. Group terms and factor.

(x + y) (2 +z) (x2 + y2) (2 + z)

Factor by grouping. Divide out common factors. Multiply fractions.

x2 + xy + y2 x2 + y2

tl" Now Try Exercises 33, 43, and 47.

Addition and Subtraction We add and subtract rational expressions in the same way that we add and subtract fractions.

Adding and Subtracting Rational Expressions

i

For fractions and

J (b ~ 0, d ~ 0 ), the following hold true. and

a b

c ad - be =--d bd

To add (or subtract) two fractions in practice, find their least common denominator (LCD) and change each fraction to one with the LCD as denominator. The sum (or difference) of their numerators is the numerator of their sum (or difference), and the LCD is the denominator of their sum (or difference). Finding the Least Common Denominator (LCD) Step 1 Write each denominator as a product of prime factors. Step 2 Form a product of all the different prime factors. Each factor should have as exponent the greatest exponent that appears on that factor.

38

CHAPTER R A lgeb ra Review

EXAMPLE 4

Adding or Subtracting Rational Expressions

Add or subtract, as indicated. y 8 (b) - - + - -

5 1 + 9x2 6x

(a) -

y-2

2-y

3 (c) x 2

+x

- 2

x2

x - 12

-

SOLUTION

5

1

(a) 9x2 + 6x

Step 1 Write each denominator as a product of prime factors. 9x 2 = 32

•

x2

6x = 2 1 • 3 1

• XI

Step 2 For the LCD, form the product of all the prime factors, with each factor having the greatest exponent that appears on it. Greatest exponent on 3 is 2.

t

t

Greatest exponent on x is 2.

LCD = 2 1 • 32 • x 2 =

18x 2

Write the given expressions with this denominator, and then add.

-

5

9x2

1

+-

=

6x

·2 1 · 3x -5 2 - + --9x · 2

10

= --

18x 2

6x · 3x

3x

+ - -2 18x

10 + 3x 18x 2

LCD = 18x 2

Multiply.

Add the numerators.

There is no common factor in the numerator and the denominator, so the answer is in lowest terms. A lways check to see that the answer is in lowest terms. y 8 (b) - - + - -

y- 2

We arbitrarily choose y - 2 as the LCD.

2- y

y

8( - 1)

y-2

(2-y) (- 1)

= -- + ----y

-8

y-2

y-2

= -- + -y- 8 y- 2

Multiply the second expression by - 1 in both the numerator and the denominator.

Simplify.

Add the numerators.

We could use 2 - y as the common denominator instead of y - 2. In this case, we multiply the first expression by - 1 in both the numerator and denominator.

R.4 Rational Expressions

y-2

8 2-y

=

y( -1 )

y

39

--+-+ _ 8_

(y - 2) ( - 1 )

-y 2-y

= --

Multiply the first expression by - I in both the numerator and the denominator.

2 - y

8 2-y

+--

8- y

Simplify.

This equivalent expression results.

2- y 3 (c) x 2

+x- 2

x2

-

x - 12

3 (x-l)(x+2)

Factor. The LCD is

(x+3)(x-4)

(x - i )(x + 2) (x + 3)(x - 4).

3(x + 3)(x - 4) (x - l )(x + 2) (x + 3)(x - 4)

l (x - l )(x + 2) (x + 3)(x - 4) (x - l) (x + 2)

3(x 2 - x - 12) - (x 2 + x - 2) (x - l)(x + 2)(x + 3)(x - 4)

Multiply in the numerators, and then subtract them.

A

3x 2 - 3x - 36 - x 2 - x + 2 (x - l )(x + 2)(x + 3)(x - 4) 2x2 - 4x - 34 (x - l )(x + 2)(x + 3)(x - 4)

Be careful with sig ns.)

Distributive property

Combine like terms in the numerator.

tl' Now Try Exercises 53, 61, and 67.

tC\ijim~I When subtracting fractions where the secondfraction has

more than one term in the numerator, as in Example 4(c), be sure to distribute the negative sign to each term.

Complex Fractions The quotient of two rational expressions is a complex fraction. There are two methods for simplifying a complex fraction.

EXAMPLE 5

Simplifying Complex Fractions (Method 1)

Simplify each complex fraction.

2 3 (a) 1

5 9 1

-+-

-+4

12

3 x

-+ 6 (b) x

1

-+4

8

Method 1 for simplifying complex fractions uses the definition of division. First, we add terms in the numerator and denominator.

SOLUTION

40

CHAPTER R A lgeb ra Review

2 3

5 9

1

1

3 x

-+ 6

- +(a)

(b) x

- +4

1

- +4

12

3 6 x -+.-

2 3 5 3 3 9 1 3 1 - · - +4 3 12

- · - +-

6

8

1

x

l 8

2 2

x 4

For Method I, work to obtain single fractions in the numerator and in the denominator.

x

-· - +-

5

6x x

3 x

-+9 9

-+-

-+-

3

1

12

12

- +-

11

2x

1

8

8

3

+ 6x x

9 4

2x

12

+1 8

Next, use the definition of division, and multiply by the reciprocal of the divisor.

11

12

9

4

=- · -

Write in lowest terms.

2x+l 8

3+6x x

4

11

9 ~12

+ 6x

3

8

x

2x

+

11 . 3 . 4

3(1+2x) · 8

3 .3 .4

x (1 + 2x)

1

24 x

11

3

vl' Now Try Exercises 73 and 77. EXAMPLE 6

Simplifying Complex Fractions (Method 2)

Simplify each complex fraction.

6-~ (a)

a a+ 1

1

--+-

k

(b)

5 k

1

a

1

- + --

l +-

a

a+ 1

SOLUTION

(a) Method 2 for simplifying complex fractions uses the identity property for multiplication. We multiply both numerator and denominator by the LCD of all the fractions, here k.

5

6- -

k 5 l +k

kf D k{J__~})

Distribute k to a// terms within the parentheses.

6k- k(i) 6k- 5 k+ k(i) k+5

R.4 Rational Expressions

a

41

1

--+(b)

a+ 1

a

1

1

a

a+ 1

-+-_!_)a(a + 1) ( _a_+ a+ 1 a ( _!_ a

For Method 2, multiply both numerator and denominator by the LCD of all the fractions, a (a + 1).

1 - ) a(a +I ) a+ 1

+ -

1 a

a a+ l 1

- - (a)(a + 1) + - (a)(a + 1) Distributive property

1

- (a)(a + 1) + - - (a)(a + 1) a a+ 1 a2 + (a+ 1) (a+ l )+a

Multiply.

a2 +a+ 1 2a

+

Combine like terms.

1

t/' Now Try Exercises 79 and 91.

( R.4

;

Exercises CONCEPT PREVIEW Fill in the blank to correctly complete each sentence.

1. The quotient of two polynomials in which the denominator is not equal to 0 is a 2. The domain of a rational expression consists of all real numbers except those that make the equal to 0. 3. In the rational expression ~ ~ ~, the domain cannot include the number _ __ 4. A rational expression is in lowest terms when the greatest common factor of its numerator and its denominator is _ __ CONCEPT PREVIEW Perform the indicated operation, and write each answer in

lowest terms. 2x

5.

10

S . x2 4

9

8. - - - - x-y x-y

6.

y3 ~ ~ .

9.

-+-

8

4

2x 5

x 4

7.

3

7

x

x

- +7

8

10. - - x2 y

Find the domain of each rational expression. See Example 1. 11.

--

x+3 x-6

12.

2x-4 x+7

13.

14.

9x + 12 (2x + 3)(x - 5)

15.

12 x + 5x + 6

16.

17.

x2 - 1 -x+ l

18.

2

x 2 - 25 --x- 5

19.

3x + 7 (4x + 2)(x - l ) x2

-

3 5x - 6

x3 - 1 -x- l

42

CHAPTER R Alg ebra Review

20. Concept Check Use specific values for x and y to show that, in general, ~ . 1ent to x +I not eqmva

+ ~ is

Y.

Write each rational expression in lowest terms. S ee Example 2 .

21. 24.

8x2 + 16x

22.

4x 2

-8(4 - y)

36y 2 + 72y 9y2

3(3 - t) 23. - - - - (t+S)(t-3)

20r + 10 26. - - 30r + lS

8k + 16 2S.-9k + 18

(y + 2 )(y - 4)

r2 - r - 6 r 2 + r - 12

27.

m 2 - 4m + 4 m2 + m - 6

28.

30.

6y 2 + 1ly + 4 3y 2 + 7y + 4

x 3 + 64 31.--

8m2 + 6m - 9 29. - - - - 16m2 - 9 y 3 - 27 32. - -3y-

x+4

Multiply or d ivide, as indicated. See E xample 3. 1Sp 3 12p ·• 9p2 10p3

8r 3 Sr 2 34. ·6r 9r 3

2k + 8 3k + 12 3S. - - - 7 - - -

x2 + x 2S 37.--·--

y3 + y2 49 38 - - · - . 7 y4 + y3

33 -

Sm+ 2S

36.

6m

10

+ 30

s

12

6

xy

4a + 12 a2 - 9 39. - - - -7- - - - 2 2a - 10 a - a - 20

41.

43

40.

p 2 - p - 12

p 2 - 9p + 20 . ----p2 - 2p - 1S p 2 - 8p + 16

42.

4r - 12 12r - 16

x 2 + 2x - l S x 2 + 1Ix+ 30

x 2 + 2x - 24 x 2 - 8x + IS

y2 + y - 2 -'- y 2 + 3y + 2 44. y2 + 3y - 4 . y2 + 4y + 3

m 2 + 3m + 2 m 2 + Sm + 6 2 · m + Sm + 4 · m 2 + lOm + 24

•

6r - 18

_2_ _ __

9r + 6 r - 24

x 3 + y3 x2 _ y2 4S - - - . - - - - • x 3 _ y3 x2 + 2xy + y2

47

+y

2

x2 _ y2 . x2 _ xy + y2 _,_ x3 + y3 46. (x-y)2 x2-2xy+y2. (x-y)4

xz - xw + 2yz - 2yw 4z + 4w + xz + wx · ------z2 - w2 16 - x2

ac + ad + be + bd 48. - - - - - a2 - b1

a3 - b3 2a 2 + 2ab

+ 2b 2

Add or subtract, as indicated. S ee Example 4.

3

s

2k

3k

8 so. -+ -3

49.

- +-

S2.

-+-+-

Sp

8

s

9

3p

4p

2p

s

S3.

4p

1

b

a

a

-+-2

SS. - - - 12x2y 6xy

7 2 - S6 - . l 8a3b2 9ab

?x + 8 x +4 S8. - - - - 3x + 2 3x + 2

S9. - - + - -

11

l

l

x+z

x-z

2x

2x - y

y - 2x

4

I

x + l

x2 - x + l

-- +

4

m

S4.

-+-2 z z

6m

+-+-

3

x

17y+3 - lOy- 18 S7. - - - - - - - 9y + 7 9y + 7 m+l m- 1 60. - - + - m-1 m+ l

m-4

63. - - - - -

6S.

2

Sm

-

4 2 62. - - - - p-q q-p

3 1 61. - - - - a-2 2-a x+ y

1

Sl.

Sm

64. - - - - 3m - 4 4 - 3m 12

- 3 x + l

66.

s

2

x + 2

x 2 - 2x + 4

-- +

60

- 3 x + 8

43

R.4 Rational Expressions

p 68. __ 2 _ _ __ 2p - 9p - 5

3x x 67" -x 2_ + _ x___ l_2 - -x 2- -- 1-6

2p 6p 2

-

p - 2

Simplify each complex fraction. See Examples 5 and 6.

69.

70.

m n

2 3 1 4

3r 4

a+ 1

~+~ 83. y x x y

2 a2

1

-

3

77.

y 2 2+ y

80.

1

1- x

5

2 x

3x

+-

2

2 -~

1

x

+l

9 5

l

1+ 79.

I

5 74. 2

- +3

76. - - - - -

8 r

6

- +-

2(-D ]-(-D2

6+ -

r x r

5 8 73. 7 3

75.

82.

71.

- +-

n

78.

6 5

4

9

72.

y

5

4 3 2

4

z2 -

25

y

x

x

84. y

y x

z+ 5

81.

85.

~+ ~ y x

I

I

a

b

- +-

l

- 1

86.

a

1

1

x

y

88.

+3

2

-

+x

91.

2 2 - -1 -x

y

89. - - -] 1 - -1+b

I

y

x

b

2+ 90.

+

x

87. - - - -

1

1 + -1- b

y+ 3 y

y y- I

4

x+4

y- 1

x

92.

1

-- + -

x x- 2

3 x- 2 1

-- + -

y

x

(Modeling) Distance from the Origin of the Nile River Th e Nile River in Africa is about 4000 mi long. The Nile begins as an outlet of Lake Victoria at an altitude of 7000 ft above sea level and empties into the Mediterranean Sea at sea level (0 ft). The distance from its origin in thousands of miles is related to its height above sea level in thousands off eet (x) by the following formula.

7-x

Distance = - - - - 0.639x + 1.75

For example, when the river is at an altitude of 600 ft, x = 0.6 (thousand), and the distance from the origin is Distance = 0.

7 - 0.6 (0. ) + 1. 75 6

639

= 3,

which represents 3000 mi.

(Data from The World Almanac and Book of Facts.) 93. What is the distance from the origin of the Nile when the river has an altitude of 7000 ft? 94. What is the distance, to the nearest mile, from the origin of the Nile when the river has an altitude of 1200 ft?

44

CHAPTER R Algebra Review

(Modeling) Cost-Benefit Model for a Pollutant Jn situations involving environmental pollution, a cost-benefit model expresses cost in terms of the percentage of pollutant removed from the environment. Suppose a cost-benefit model is expressed as 6.7x y= 100 -x'

where y is the cost in thousands of dollars of removing x percent of a certain pollutant. Find the value of y for each given value of x. 95. x = 75 (75%)

LR.5

J

96. x=95 (95% )

4

Radical Expressions

•

Square Roots

•

Cube, Fourth, and Higher Roots

•

Product and Quotient Rules for Radicals

•

Operations with Radicals

•

Rationalizing Denominators

•

Si mplified Radicals

Square Roots Recall that 6 2 = 36. We say that "6 squared equals 36." The opposite (or inverse) of squaring a number is taking its square root.

\/36 =

because 62 = 36.

6,

The positive square root, or principal square root, of a number is written with the symbol

V . For example, the positive square root of 121 is 11. v m = 11

11 2 = 121

The symbol - V is used for the negative square root of a number. For example, the negative square root of 121 is - 11. - v m = - 11

( -1 1)2 = 121

V

The radical symbol always represents the positive square root (except that = 0 ). The number inside the radical symbol is the radicand, and the entire expression - radical symbol and radicand- is a radical.

Vo

Radical symbol

Radicand

Va / ~

Radical

An algebraic expression containing a radical is a radical expression. We summarize our discussion of square roots as follows.

Square Roots of a Let a be a positive real number.

Va is the positive, or principal, square root of a. -Va is the negative square root of a. For nonnegative a, the following hold true.

Va · Va = (Va )2 = Also,

Vo= 0.

a

and

-Va · ( -Va) = ( -Va )2 =

a

45

R.5 Radical Expressions

EXAMPLE 1

Finding Square Roots

Find each square root. (a)

v'l44

(b)

-Vsl

(c)

~

SOLUTION

(a)

v'l44 =

12, because 122 = 144.

(b) The symbol Vsl = 9, (c)

{4 2 \j 9 = 3

-Vsl represents the negative square root of 81. Because Vsl = - 9. 2

) ( 0.9 )

(d) -

_±7

(16 = \j 49

VQ.81 =

(e)

:

~:~,· 0.9J

0.9

tl' Now Try Exercises 13, 15, 19, and 23.

Not every number has a real number square root. For example, there is no real number that can be squared to obtain - 36. (The square of a real number can never be negative.) Because of this, \1'=36 is not a real number. If a is a negative real number, then

Va is not a real number.

As noted previously, when the square root of a positive real number is squared, the result is that positive real number. (Also, (Vo )2 =

0.)

Squaring Radical Expressions

Find the square of each radical expression. (a)

VJ3

(b)

-\/29

(c)

\/P2+J

(d)

v3 3

SOLUTION

(a) The square of VJ3 is ( VJ3) = 13. 2

(b) ( -

\/29 )2 =

29

Definition of square root

The square of a negative number is positive.

(d)

(v3)2= (\/3)2 2 3

3

3 9

1 3

tl' Now Try Exercises 27, 29, 31, and 33. Cube, Fourth, and Higher Roots Finding the square root of a number is the inverse (opposite) of squaring a number. In a similar way, there are inverses to finding the cube of a number and to finding the fourth (or greater) power of a

number. These inverses are, respectively, the cube root,

%, and the fourth

root, "f[;;,. Similar symbols are used for other roots.

\'Ya The nth root of a, written%, is a number whose nth power equals a . That is,

..v"r a = b

means

b" = a.

46

CHAPTER R Algebra Review

In

efa, the number n is the index, or order, of the radical. Radical symbol

Index

~~/

Radicand

~

Radical

We could write ~ instead of Va, but the simpler symbol using tomary because the square root is the most commonly used root.

Va is cus-

Finding Cube, Fourth, and Higher Roots

EXAMPLE 3

Find each root. (a)

\o/64

(b)

(d)

-€2

(e)

'1125

#

(c)

fu

(f)

~ 0.0016

SOLUTION

'1125 =

(a)

\o/64 =

(c)

fu = 2, because 24 = 16. -'\1'32 = - 2 This symbol represents the negative fifth root of 32.

(d)

4, because 4 3 = 64.

(b)

(e)

\j/8 27 = ~3' because (~) 3

(t)

~ = 0.2, because (0.2) 4 = 0.0016.

3

3 =

5, because 53 = 125.

~27 t/' Now Try Exercises 35, 41, 45, and 49.

Product and Quotient Rules for Radicals

V4 · V9 = 2 · 3 = 6

and

Consider the following.

~ = \/36 = 6

V4 · V9 = ~

This shows that

The result here is a particular case of the product rule for radicals. The quotient rule for radicals is similar.

Rules for Radicals

If a and b are nonnegative real numbers, then the following rules hold true.

Product Rule The product of two square roots is the square root of the product.

Quotient Rule

fa=

Va

'/"h Vb

(b

* 0)

The square root of a quotient is the quotient of the square roots.

R.5 Radical Expressions

47

Using the Rules for Radicals

EXAMPLE 4

Use the product and quotient rules for radicals to rewrite each expression. (a)

\/2 . \/3

(b)

V7 . V7

(c)

~

21Vi5 (d)

' r,;

9v3

SOLUTION

(a)

\/2 . \/3

(b)

V7 . V7

=~

=v49

=v6

=7

(c)

\j{i434

(d) 27Vi5

9\/3 27

Vi5

=9· V3

Multiplication of fractions

=~· ii

Quotient rule

=3Vs

Divide.

tl" Now Try Exercises 51, 57, 59, and 67.

A square root radical is simplified when no perfect square factor other than 1 remains under the radical symbol. EXAMPLE 5

Simplifying Radicals

Simplify each radical. (a)

\/20

(b)

vn.

(c)

-\/300

(d) Vi5

SOLUTION

(a)

= = =

(b)

20 has a perfect square factor of 4.

\/20

\/4-:-s

Factor; 4 is a perfect square.

\!4 · Vs Product rule in the form ~ = Va · Vb 2Vs v4=2

vn.

Look for the greatest perfect square factor of 72.

=

~

Factor; 36 is a perfect square.

=

\/36 · \/2

Product rule

=

6Vl

V36 =

6

48

CHAPTER R A lgeb ra Review

We could also simplify for pairs of like factors.

VTI by factoring 72 into prime factors and looking

= Y2·2· 2·3·3

(c)

Factor into primes.

=

2 . 3 . V2

~ = 2; yl3:3 = 3

=

6\/2

The result is the same.

-V30o = -

~

Factor; 100 is a perfect square.

VWo · V3 Product rule -10V3 VlOo = JO

= =

(d) Because 15 has no perfect square factors (except 1),

Vi5 cannot be simplified.

II" Now Try Exercises 69, 75, and 77. Operations with Radicals We add or subtract radicals using the distributive property, ac +be = (a+ b)c.

sV3 + 6V3 = (8 + 6) V3 = 14V3

2vll - 7vll

Distributive property

= (2 - 7) vll = -5 vll

Add.

Distributive property Subtract.

Only like r ad icals-those that are multiples of the same root of the same number-can be combined in this way. EXAMPLE 6

Adding and Subtracting Radical Expressions

Add or subtract, as indicated.

YlO - T\f W Vi8- V27

(a) 5

(b)

(c)

(d )

3\/2 + Vs 2Vi2 + 3V75

SOLUTION (a)

(b)

5Yl0 - 7Yl0 = ( 5 - 7) YlO = -2Yl0

These are like radicals. Distributi ve property Subtract.

3\/2 + Vs = 3\/2 + ~ = 3\/2 + \/4 · \/2 =3 V2 +2 V2 = =

(3 + 2) \/2 5\/2

Factor; 4 is a perfect square. Product rule

v4=2 Distributive property Add.

R.5 Radi cal Ex pressions

49

(c)

Stop here. These are unlike radicals. Th ey cannot be combined.

(d)

\/9-:-:3

=

yl9":"2 -

=

\/9 · \/2 - \/9 · \13

Product rule

=

3 \/2 -3 \13

v9=3

2\/12 + 3\/75 = 2 \/4-:-:3 + 3 ~ = 2 ( v4 · \13) + 3( \/25 · \13) = 2(2\/3) + 3(sV3) = 4 \13 + 1sv3 Think: (4 + 15)VJ = 19\/3

Factor; 9 is a perfect square.

Factor; 4 and 25 are perfect squares. Product rule

\/4 =

2; V25 = s

Multiply. Add like radicals.

t/' Now Try Exercises 85, 91, 93, and 95.

Multiplying radicals is similar to multiplying polynomials.

EXAMPLE 7

Multiplying Radicals

Multiply. (a)

V5(2 + v6)

(c) (

V10 + v3)( VIO- v3)

(b) (

Vs + 3) (v6 + 1)

(d) (

V7 -

3)

2

SOLUTION

(a)

V5(2 + v6) = Vs · 2 + Vs · v'6 = 2Vs + v3Q

(b) (

Distributive property: a(b

+ c)

=

ab + ac

Commutative property ; product rule

Vs + 3)( v6 + 1) First

Outer

Inner

Last

~~~~

= Vs · v6 + Vs · 1 + 3 · v6 + 3 · 1 FOIL method This result ca nnot be = v30 + Vs + 3\/6 + 3 simplified f urther. (c) (

V10 + v3)( VIO- v3) =vw · VIO-vw · v3+v3 · VIO-v3 · v3 =

10 - 3

= 7

Product rule;

-V30 + V30 =

FOIL method

0

Subtract.

When we multiply conjugates, such as ( VlO result is a rational number.

+ \13)( VlO - \13), the

50

CHAPTER R Algebra Review

(d) (

v7 - 3 ) 2 = (v7- 3)(v7- 3) = v7 · v7- 3v7- 3v7 + 3 · 3 = 7- 6v7 + 9 Be ca reful. These terms cannot be = 16 - 6v7

a 2 = a· a FOIL method Multiply. Combine like terms. Add.

combined.

v

Now Try Exercises 103, 107, 109, and 113.

NOTE In Example 7(d), we could have used the formula for the square of a binomial to obtain the same result.

(v7 - 3 )2 =(v7) - 2(v7)(3) +3 = 7 - 6v7 + 9 = 16 - 6v7 2

2

(x - y)2 = x2 - 2xy +y2 Apply the exponents. Multiply. Add.

Although calculators make it fairly easy to

Rationalizing Denominators

divide by a radical in an expression such as ~, it is sometimes easier to work with radical expressions if the denominators do not contain radicals. Rationalizing a Denominator

The process of changing a denominator from one with a radical to one without a radical is called rationalizing the denominator.

The value of the radical expression is not changed. Only the form is changed, because the expression has been multiplied by a form of 1.

EXAMPLE 8

Rationalizing Denominators

Rationalize each denominator. (a) -

12 v8

9

(b) -

V6

(c)

3 1+

(d)

V2

V2-V3 Vs+V3

SOLUTION

9

(a)

V6 =

3V6. - is

-

2

equ iva lent

to~V6.

9 V6 \/6 · \/6

9\/6 6

3\/6 2

Multiply by

~

=

I.

In the denominator,

v6 · v6 = \/36 =

. .m lowest terms; 69 = 32--:--:l · 3 3 Wnte = 2

6.

51

R.5 Radical Expressions

(b) -

12

The denominator could be rationalized by multiplying by but simplifying the denominator first is more direct.

v8 12

Vs = V4 . V2 = 2V2

2V2 =

12 V2 2V2 . V2

Multiply by

~=

l.

Multiply.

12\!2

2V2. V2 = 2\/4 = 2. 2= 4

4

=3V2

. .m 1owest terms; 412 = 3 W nte

3

(c)

1+

Again, we are multiplying by a form of 1.

V2 1- V2 1 + V2 1-V2 3

The denominator is now a rati onal number. Either form can be given as t he answer.

(d)

Vs,

=

V2,

(1+ V2)(1- V2) = 12- (V2)2 = 1- 2

- 1 =

Multiply the numerator and denominator by 1 the conjugate of the denominator.

= - 1

-3(1-v'2) -3 + 3\!2

a

=-t =

-a

Distributive property

V2- v3 v5+v3 V2-v3 v5 - v3 v5+v3 v5 - v3 V10-v6-Vi5 + 3 5- 3

V10-v6-Vi5 + 3 2

Multiply the numerator and v3. denominator by

Vs -

Multi ply.

Subtract in the denominator.

II" Now Try Exercises 117, 123, 127, and 133.

Simplified Radicals An expression with radicals is simplified when the following conditions are satisfied.

Conditions for a Simplified Square Root Radical

1. No perfect square factor other than I remains under the radical symbol. 2. The radicand has no fractions. 3. No denominator contains a radical.

4. All indicated operations have been performed (if possible).

52

CHAPTER R A lgeb ra Review

Some radical expressions to be simplified involve complex fractions. EXAMPLE 9

Simplifying Radical Expressions

Simplify.

Vt3 (a)

V2

4

(b)

v3

2

V2

1- -

4

2

SOLUTION

Vt3 (a)

4

v3 4

Vt3

=- - 4-

4

·

v3

Vt3 v3 Vt3 v3 =-v3 · v3

Multiply by the reciprocal of the denominator.

Divide out the common factor.

M ultiply by

~=

I.

v39 3

V2 2

(b) - - -

V2

1- -

2

2t j) V2 2-V2 V2 2+ V2 2-V2 2+ V2 2V2+2 4-2

2(V2+ 1) 2 =

V2+ 1

Multiply both numerator and denominator by the LCD of all the fractions, 2.

M ultiply.

M ultiply the numerator and denominator by the conjugate of the denominator.

(2 - \/2)(2 + \/2) = 22- (\/2)2 = 4- 2 Factor in the numerator. Subtract in the denominator. Divide out the common factor.

2+ \/2,

R.5 Radi cal Ex pressions

~

=~ --+ -~

by ~=

Multiply

Product rule:

=

~2 +4v'3

53

1.

Va · Vb =

~

Multiply.

. ru ie:·Vrav;, b = Vb QuotJent tl"' Now Try Exercises 135, 141, and 143.

;

LR.5 I Exercises

CONCEPT PREVIEW Match each expression in Column I with the equivalent choice in Column II. Answers may be used once, more than once, or not at all.

I

II

1.

-Vl6

2.

3.

Vii

4.

5.

fu

6.

v'=t6 Vii -'1'64

A. 3

B. -8

c.

D. 9

2

E. -4

F. Not a real number

CONCEPT PREVIEW Perform the operations mentally, and write the answers without

doing intermediate steps. 7. 9.

v2s + V64 V6 . V6

8.

Vs . v2

10. (

v28 - Vl4)( v28 + Vl4)

Find each square root. See Example 1. 11.

Vi

12.

v4

13.

V10o

14.

v40o

15.

-\/36

16.

-\/64

17.

-v'2s6

18.

-Vi96

-1*

22. -

V-W

26.

19.

23.

'{25

20. ~

21.

VoM

24.

VQ.i6

25.

(4

(49 \j 36

V=-64

54

CHAPTER R A lgebra Review

Find the square of each radical expression. See Example 2.

27.

v59

Vi9

28.

V2

Vs 32 . - 7

31. - 3-

29.

-Vi9

33. v'3x2

30.

+4

-v59

34. v'9y 2

+3

Find each root. See Example 3.

36.

V343

37.

39.

V2i6 eft2%

40.

ef625

41.

43.

%4

44. ~

3S.

47.

-/£

48.

-Vsii -fu

38. 42.

4S.~

-~

46. ~

vo.Oo1

49.

-ViOOO -ef256

so. vo.i2s

Use the product and quotient rules for radicals to rewrite each expression. See Example 4.

Sl. v'3 . Vs

S2. v'3 . V7

V2 . \/i5

S3.

V2 . v'il V2 . v'8

SS.

y'3 · Vii

S6.

S7. v'3 . v'3

S8.

v'6. v'6

S9.

~

61.

!lo

62.

~

Vs 6S. v2o 20

30v'i0 --v12 5 2

68. sov2o 2v'i0

S4.

60.

5

V75

63. v'3

64.

V7

66.

50

v'63 63

67.

H. J* 16

2

Simplify each radical. See Example 5.

69. 73. 77.

v'24 V75 -v'l60

71.

74.

v'44 v48

78.

-Vl2s

79.

70.

81. 3V2.7

82. 9Vs

7S.

v'45 v'i45 -V7o0

83. 5Vs0

72.

Vii

76.

v'LJo

80.

-v'60o

84. 6v'4o

Add or subtract, as indicated. See Example 6.

8s. 2v'3 + sv'3

86. 6Vs + 8Vs

88. 6V7-V7

89.

v'6 + v'6

90.

92.

Vl4 + v'l7 v'45 + 4v2o 6 v'18 - 5 v'32

93. 5 v'3 +

91.

v'6 + V7

94. 3 V2 +

v'50

97. 2Vs0 - s vn 100.

-4V75 + 3vlz

9S. 98. 10i.

±V288 + ~vn.

87. s v'3 - v'3

96.

v'11 + v'11

vl2 v'24 + 6 v'54

99. -s\/32 + 2v'98 102.

~Vii + ~ v48

R.5 Radi cal Expressions

55

Multiply. See Example 7. 104.

105.

v6(3 +Vi) V3(Vi2- 4)

106.

Vi0(5 + V3) vs(vm - 6)

107.

(Vi+ 1)(V3 + 1)

108.

(V3 + 3)(\/5 + 2)

103.

109. ( 111.

V2 - V3)( V2 + V3)

(1V3 + V5)(3V3 - 2\/5)

113. (

Vs + 2 )

115. ( V2l -

2

V5)

2

110. (

V7 + Vl4)( V7 - Vl4)

112. (

V7 - V11)(2V7 + 3Vll)

114. (

V11 - 1)

116. (

v6 - V2)

2

2

Rationalize each denominator. See Example 8. 118.

3 V2

119.

121.

4 v6

122. ~

6

117.

Vs

120.~

Vi5

123.

126.

18 V27 27 15 v45 45

124.

127.

6

129.

132.

Vs5 + V33 Vs- 1 Vs 1+ 5

130.

133.

24 \/ls 18 3 4+ Vs 12 v66 + V33 V2-V3 v66 - Vs5

5 Vs ViO 12

125.

vn 72

128.

4 5+ v6

131.

134.

V3 + 1 1-V3 V5+v6 V33- V22

Simplify. See Example 9.

135.

V2 3 V7 3

138.

v6 4 Vs 12

141.

V3 2 1- -V3 2

144.

)1-,v:

136.

Vs 2 vl3 13 2

137.

10

1

-

139.

142.

145.

2 1- -Vs 2 Vs 3 1- -Vs 3

µ

V7 5 V3 1 2 1+-V2 2 -

140.

143.

146.

µ

gl_

56

CHAPTER R Algebra Review

Relating Concepts For Individual or Group Work (Exercises 147-150)

In calculus, it is sometimes desirable to rationalize a numerator. To do this, we multiply the numerator and the denominator by the conjugate of the numerator. For example,

6-Vl 4

6-Vl 6 + V2 4 6 + V2

36 - 2

34

17

4(6 +Vi)

4(6 +Vi)

2(6+V2)"

Rationalize each numerator.

147.

6-\/3 8

149. 2 V10 + 30

148. 2Vs - 3 2

V7

150.

v's + 3Vl 26

q

( R.6 I Equations and Inequalities •

Basic Terminology of Equations

•

Linear Equations

•

Quadratic Equations

•

Inequalities

•

Linear Inequalities and Interval Notation

•

Three·Part Inequalities

Basic Terminology of Equations

An equation is a statement that two

expressions are equal.

x

+ 2 = 9,

l lx

= Sx + 6x, x 2

-

2x - 1

=0

Equations

To solve an equation means to find all numbers that make the equation a true statement. These numbers are the solutions, or roots, of the equation. A number that is a solution of an equation is said to satisfy the equation, and the solutions of an equation make up its solution set. Equations with the same solution set are equivalent equations. For example,

x

= 4, x + 1 = 5, and 6x + 3 = 27 are equivalent equations

because they have the same solution set, { 4 }. However, the equations

x2 = 9

and x = 3

are not equivalent

because the first has solution set { - 3, 3} while the solution set of the second is { 3}. One way to solve an equation is to rewrite it as a series of simpler equivalent equations using the addition and multiplication properties of equality. Let a, b, and c represent real numbers. If a If a

= b, then a + c = b + c. = band c * 0, then ac = be.

These properties can be extended: The same number may be subtracted from each side of an equation, and each side may be divided by the same nonzero number, without changing the solution set. Linear Equations We use the addition and multiplication properties of equality to solve linear equations.

Linear Equation in One Variable

A linear equation in one variable is an equation that can be written in the form

ax

+b

= 0,

where a and b are real numbers and a ""' 0.

R.6 Equations and Inequ alities

57

A linear equation is a first-degree equation because the greatest degree of the variable is 1. 3x +

\/2 =

Vx + 2 =

EXAMPLE 1

3 0, 4x = 12,

5,

1 x

0.5(x + 3) = 2x - 6

-8, x 2

- =

Linear equations

+ 3x + 0.2 = 0

Nonlinear equations

Solving a Linear Equation

Solve 4x - 2x - 5 = 3

+ 6x.

The goal is to isolate x on one side of the equation.

SOLUTION

+ 6x 3 + 6x 3 + 6x - 6x

4x - 2x - 5 = 3 2x - 5 = 2x - 5 - 6x =

Combine like terms. Subtract 6x from each side.

- 4x - 5 = 3

Combine like terms.

-4x - 5 + 5 = 3 + 5

Add 5 to each side.

- 4x = 8

Combine like terms.

-4x

8

-4

-4

Divide each side by - 4.

x = - 2 CHECK

Substitute -2 for x in the original equation.

A check of t he solution is recommended.

4x - 2x - 5

3 + 6x

=

4( - 2) - 2( - 2) - 5 Jo 3 -8

+ 6( - 2)

+ 4 - 5 Jo 3 - 12 -9 = -9 ./

The solution set is { - 2 }.

EXAMPLE 2

Let x= - 2. M ultiply. True

t/' Now Try Exercise 15.

Solving a Linear Equation

Solve 3(2x - 4) = 7 - (x SOLUTION

Original equation

+ 5).

~ ~ 3(2x-4)=7 - (x+5~

6x - 12 = 7 - x - 5

Distributive property

6x - 12 = 2 - x

Combine like terms.

6x - 12

+x

= 2 -

x

+x

Add x to each side.

7x - 12 = 2

Combi ne like terms.

7x - 12 + 12 = 2 + 12

Add 12 to each side.

7x = 14 7x

14

7

7

x=2

Combine like terms. Divide each side by 7.

58

CHAPTER R A lgebra Review

CHECK

3(2x - 4)

3(2·2-

+ 5) + 5)

(x

= 7 -

4) J, 7 - (

2

3 ( 4 - 4) J, 7 - ( 7)

Original equation Let x = 2. Work inside the parentheses.

0 = 0 ,/

True

II" Now Try Exercise 19.

The solution set is { 2} .

EXAMPLE 3

Solving a Linear Equation with Fractions

2x+4 1 1 7 Solve - - - + - x = - x - - . 3 2 4 3 SOLUTION

Distribute to all te rms within the parentheses.

2x+4 1 1 7 - - - + - x= - x- 3 2 4 3

1~~x)~ 1~D

12(2x; 4) + 12(±x)

=

Multiply by 12, the LCD of the fractions.

12(±x)- 12(~)

4(2x + 4)

+ 6x = 3x 8x + 16 + 6x = 3x l 4x + 16 = 3x -

28

Multiply.

28

Distributive property

28

Combine like terms.

llx = -44

Subtract 3x . Subtract 16.

x= - 4

Divide each side by 11.

2x + 4 1 1 7 --- + - x = - x - 3 2 4 3

CHECK

Distributive property

2( - 4 ) + 4 1 - - - + - ( -4 ) 3

Original equation

=-41 (-4) ?

2

7 3

-

-34 + ( -2) =-1 - 37

Let x = - 4.

?

10 10 = - ,/ 3 3

- -

Simplify on each side.

True

II" Now Try Exercise 25.

The solution set is { -4} .

An equation satisfied by every number that is a meaningful replacement for the variable is an identity.

3(x + 1)

=

3x + 3

Identity

An equation that is satisfied by some numbers but not others is a conditional

equation. 2x = 4

Conditional equation

The equations in Examples 1-3 are conditional equations. An equation that has no solution is a contradiction.

x= x

+

1

Contradiction

R.6 Equations and Inequal ities

59

Identifying Types of Equations

EXAMPLE 4

Determine whether each equation is an identity, a conditional equation, or a contradiction. Give the solution set. (a) -2(x

+ 4) + 3x =

x- 8

(b) 5x - 4 = 11

(c) 3(3x - 1) = 9x

+7

SOLUTION

+ 4) + 3x = -2x - 8 + 3x =

(a) -2(x

x- 8 x- 8

Distributive property

x- 8= x- 8

Combine like terms.

0= 0

Subtract x. Add 8.

When a true statement such as 0 = 0 results, the equation is an identity, and the solution set is {all real numbers }. (b) 5x - 4 = 11

5x = 15

Add 4 to each side.

x = 3

Divide each side by 5.

This is a conditional equation that is true when x = 3. Its solution set is { 3}. (c) 3(3x-1)=9x+7

9x - 3 = 9x

+7

-3 = 7

Distributive property Subtract 9x.

When a false statement such as - 3 = 7 results, the equation is a contradiction, and the solution set is the empty set, symbolized 0 . II" Now Try Exercises 33, 35, and 39.

Quadratic Equations '

A quadratic equation is defined as follows.

Quadratic Equation

A quadratic equation is an equation that can be written in the form

ax2

+ bx + c

= 0,

where a, b, and care real numbers and a '/:- 0. The given form is called standard form.

A quadratic equation is a second-degree equation-that is, an equation with a squared variable term and no terms of greater degree. x 2 = 25,

4x 2 + 4x - 5 = 0,

3x 2 = 4x - 8

Quadratic equations

When the expression ax2 +bx+ c in a quadratic equation is easily factorable over the real numbers, the equation can be solved using factoring and the following zero-factor property. If a and b are real numbers with ab = 0, then a = 0 or b both equal 0.

= 0 or

60

CHAPTER R A lgebra Review

EXAMPLE 5

Using the Zero-Factor Property

+ 7x =

Solve 6x 2

3.

6x 2 + 7x = 3 ~ Don't factor out x here.)

SOLUTION

+ 7x - 3 = ( 3x - 1 ) ( 2x + 3) = 6x 2

0

Standard form

0

Factor.

3x - 1 =0

or

3x = 1

or

2x= -3

x=3

or

3 x= - 2

2x+3=0

6x 2 + 7 x

CHECK

= 3

Zero-factor property Solve each equation.

Original equation Letx = - 32 .

6

7

9

3

?

-+- =3 3=3 ~

T~

3=3 ~

T~

Both values check because true statements result. The solution set is { -

~, ~ }.

II" Now Try Exercise 43.

A quadratic equation written in the form x 2 = k, where k is a constant, can be solved using the square root property. If x 2

= k,

That is, the solution set of x 2

{ Vk, -'\/k}, EXAMPLE 6

then

x

= Vk

or

x

= - Vk.

= k is

which may be abbreviated

{

±'\/k}.*

Using the Square Root Property

Solve each quadratic equation. (b) (x-4) 2 = 12

(a) x 2 = 17 SOLUTION

x 2 = 17

(a)

x= ±

W

Square root property

The solution set is { ± \/U }. (b)

(x-4) 2 = 12 x - 4 = ± \/i2

Generalized square root property

± \/i2 Add 4. 4 ± 2\/3 v'i2 =

x= 4 x= * The symbol

~=

± is read "positive or negative" or "plus or minus."

V4. v3 = 2v'3

R.6 Equations and Inequalities

(x - 4) 2 = 12

CHECK

( 4 + 2\/:3 - 4 )2 Jo 12

Let x= 4 + 2 \/3.

Original equation

(4 - 2\/:3 -4)2Jo 12

=

Let x = 4 - 2\/3.

( -2\/:3)2 Jo 12

(2\/:3)2Jo 12

12

61

12 ./

True

12 = 12 ./

v

The solution set is { 4 ± 2\/:3}.

True

Now Try Exercises 53 and 57.

Any quadratic equation can be solved using the quadratic formula, which says that the solutions of ax 2 + bx + c = 0, where a ¥= 0, are given by

x=

-b

Solve

SOLUTION

-

4ae

This formula is derived in algebra courses.

2a

Using the Quadratic Fonnula

EXAMPLE 7

x2 -

± Vb2

4x = -2. x2

-

Write in standard form. Here a = 1, b = - 4, and c = 2.

4x + 2 = 0 -b ±

Vb2 -

4ac

x=-------

Quadratic formula

2a

The fraction bar extends under - b.

x= x=

-( -4) ± v( -4)2 2( 1) 4

\/16-=8

±

2

4± 2\/2 x=---2

x=

-

4( 1)(2)

Substitute a = 1, b = - 4, and c = 2.

Simplify.

v'i6=8 = Vs =

yl4":2 =

v4 . v'2 = 2 v2

2(2±\/2)

( Factor first, then divide.

Factor out 2 in the numerator.

2

Y

x=2±\/2

Divide out the common factor.

v

The solution set is { 2 ± \/2 }.

Now Try Exercise 65.

Inequalities An inequality says that one expression is greater than, greater than or equal to, less than, or less than or equal to another. As with equations, a value of the variable for which the inequality is true is a solution of the inequality, and the set of all solutions is the solution set of the inequality. Two inequalities with the same solution set are equivalent. Inequalities are solved with the properties of inequality, which are similar to the properties of equality. Let a, b, and c represent real numbers.

+ e < b + e. If a 0, then ae ,

:S,

or

2:

> be.

results in similar properties. (Restrictions on c remain the same.)

62

CHAPTER R Algebra Review

Multiplication may be replaced by division in Properties 2 and 3. Always remember to reverse the direction of the inequality symbol when multiplying or dividing by a negative number. Linear Inequalities and Interval Notation The definition of a linear inequality is similar to the definition of a linear equation.

Linear Inequality in One Variable A linear inequality in one variable is an inequality that can be written in the form

ax + b

> O,*

where a and b are real numbers and a ¥- 0. * The symbol > can be replaced w ith < , s, or

x- 6

~

0,

2 .

3x + 5 < 17, x ;::=: -2

Linear inequalities

Solving linear inequalities is similar to solving linear equations.

EXAMPLE 8

Solve 3x

Solving a Linear Inequality

+ 5 < 17.

SOLUTION

3x + 5

< 17 3x + 5 - 5 < 17 - 5 3x < 12 3x

Subtract 5. Combine like terms.

12

3 ;)

~'-'-1,....,....,--;-+-'--;-"'-+-~~ x

§ ~

(c)

(d)

Y

y

rs( T

- x •

~ ~

x

•

SOLUTION

(a) The domain is the set of x-values, { - 1, 0, 1, 4, 5 }. The range is the set of y-values, { - 3, -1 , 1, 2 }. (b) The x-values of the points on the graph include all numbers between -4 and

4, inclusive. They-values include all numbers between -6 and 6, inclusive. The domain is [ - 4, 4). The range is [ -6, 6).

Use interval notation.

(c) The arrowheads indicate that the line extends indefinitely left and right, as well as up and down. Therefore, both the domain and the range include all real numbers, which is written

(-00, 00).

Interval notation for the set of all real numbers

(d) The arrowheads indicate that the graph extends indefinitely left and right, as

well as upward. The domain is ( - oo, oo). Because there is a least y-value, - 3, the range includes all numbers greater than or equal to - 3, written [ - 3, oo).

v

Now Try Exercise 29.

Determining Whether Relations Are Functions Because each value of x leads to only one value of y in a function, any vertical line must intersect the graph in at most one point. This is the vertical line test for a function.

Vertical Line Test

If every vertical line intersects the graph of a relation in no more than one

point, then the relation is a function.

R.8 Functions

79

The graph in Figure 29(a) represents a function because each vertical line intersects the graph in no more than one point. The graph in Figure 29(b) is not the graph of a function because there exists a vertical line that intersects the graph in more than one point. y

y

This is the graph of a function. Each x-value corresponds to only one y-value.

This is not the graph of a function. The same x-value corresponds to two different y-values.

(a)

(b) Figure 29

Using the Vertical Line Test

EXAMPLE 4

Use the vertical line test to determine whether each relation graphed in Example 3 is a function. We repeat each graph from Example 3, this time with vertical lines drawn through the graphs.

SOLUTION

(a)

(b)

y

y

(5, 2) ~ : ~:~ ~ +0 -+-+-'~-+-...._ x

[ (0'(11)

(4,

~3)

y

(c)

•

x

y

(d)

•

•

•

x

•

•

•

x

. .

•

The graphs of the relations in parts (a), (c), and (d) pass the vertical line test because every vertical ljne intersects each graph no more than once. Thus, these graphs represent functions.

•

The graph of the relation in part (b) fai ls the vertical line test because the same x-value corresponds to two different y-values. Therefore, it is not the graph of a function. tl"' Now Try Exercises 25 and 27.

The vertical line test is a simple method for identifying a function defined by a graph. Deciding whether a relation defined by an equation is a function, as well as determining the domain and range, is more difficult. The next example gives some hints that may help.

80

CHAPTER R Algebra Review

EXAMPLE 5

Identifying Functions, Domains, and Ranges

Determine whether each relation defines y as a function of x, and give the domain and range. 5 (d) y = - (a) y = x + 4 (b) y = Vh-=-1 x - 1 SOLUTION

(a) In the defining equation (or rule), y = x + 4, y is always found by adding 4 to x. Thus, each value of x corresponds to just one value of y, and the relation defines a function. The variable x can represent any real number, so the domain is {x lxisareal number },

or

(-00,00).

Because y is always 4 more than x, y also may be any real number, and so the range is (-oo, oo) .

\/h-=-1,

(b) For any choice of x in the domain of y = there is exactly one corresponding value for y (the radical is a nonnegative number), so this

equation defines a function. The quantity under the radical sign cannot be negative-that is, 2x - 1 must be greater than or equal to 0.

2x - 1 ;:::: 0

y

2x ;:::: 1

1

x ;:::: 2

Solve the inequality. Add I. Divide by 2.

The domain of the function is [ ~, oo ) . Because the radical must represent a 0

I

2

2

4

6

nonnegative number, as x takes values greater than or equal to~, the range is {y Iy ;:::: 0}, or [ 0, oo) . See Figure 30.

Figure 30

(c) The ordered pairs ( 16, 4) and ( 16, -4) both satisfy the equation y 2 = x . There exists at least one value of x - for example, 16-that corresponds to

two values of y, 4 and -4, so this equation does not define a function. Because x is equal to the square of y, the values of x must always be nonnegative. The domain of the relation is [ 0, oo) . Any real number can be squared, so the range of the relation is ( - oo, oo) . See Figure 31.

y

4

2

(d) Given any value of x in the domain of ~0-t==t~..f===l~*"-x

5 y= - 1,

-2

x-

-4

we find y by subtracting 1 from x, and then dividing the result into 5. This process produces exactly one value of y for each value in the domain, so this equation defines a function.

Figure 31

The domain of y = y

5 Range -4

x .:_ 1

includes all real numbers except those that

make the denominator 0. We find these numbers by setting the denominator equal to 0 and solving for x.

I ~= x=l

:

x= 1

~omain I 2 I I I I

Figure 32

x- 1 =0

6

Add I .

Thus, the domain includes all real numbers except 1, written as the interval ( -oo, 1) U ( 1, oo ) . Values of y can be positive or negative, but never 0, because a fraction cannot equal 0 unless its numerator is 0. Therefore, the range is the interval ( - oo, 0) U (0, oo ) , as shown in Figure 32.

ii" Now Try Exercises 33, 35, and 39.

R.8

LOOKING AHEAD TO CALCULUS

One of the most important concepts in

= f(x) ,

y

function, is defined using func tion

!~f(x) = L

81

Function Notation When a function f is defined with a rule or an equation using x and y for the independent and dependent variables, we say, "y is a function of x" to emphasize that y depends on x . We use the notation

calculus, that of the limit of a notation:

Functions

called function notation, to express this and read f (x) as "f of x," or "fat x." The letter f is the name given to this function. For example, if y = 3x - 5, we can name the function f and write

(read "the limit of f(x) as x approaches

a is equal to L") means that the values of f(x) become as close as we wish to L when we choose values of x

f (x) = 3x - 5.

Note that f (x) is just another name for the dependent variable y. For example, if y = f(x) = 3x - 5 and x = 2, then we find y, or f(2), by replacing x with 2.

sufficiently close to a.

f( 2)=3 · 2 -5

Letx=2.

f(2)

Multiply, and then subtract.

= 1

The statement "In the function f , if x = 2, then y = l " represents the ordered pair ( 2, 1) and is abbreviated with function notation as follows.

f( 2)

=

1

The symbol f(2) is read "f of 2" or "fat 2." Function notation can be illustrated as follows. Name of the function Defining expression \

~

y = f (x ) = 3x - 5

~ ~

Value of the function

Using Function Notation

EXAMPLE 6

Let f(x) = -x (a)

Name of the independent variable

2

+ Sx - 3 and g(x)

f(2)

= 2x

(b) f(q)

+ 3. Find each of the following. (c) g(a + 1)

SOLUTION

+ Sx - 3 f( 2) = - 2 + 5 · 2 - 3 f( 2) = - 4 + 10 - 3

Replacexwith 2.

f( 2)

Add and subtract.

f( x ) = - x 2

(a)

2

Be careful here. -

22

=

-( 2)2

= - 4

= 3

Apply the exponent and multiply.

Thus, f(2) = 3, and the ordered pair (2, 3) belongs to f. (b) f (x ) = - x 2

+ Sx

- 3

f( q ) = - q2 + Sq - 3 (c)

Replacexwithq.

g(x )=2x +3 g( a g( a g( a

+ 1) + 1) + 1)

+ 1) + 3 2a + 2 + 3 2a + 5

= 2( a

Replacexwith a + I.

=

Distributive property

=

Add.

II' Now Try Exercises 45, 53, and 59.

Functions can be evaluated in a variety of ways, as shown in Example 7.

82

CHAPTER R Algebra Review

Using Function Notation

EXAMPLE 7

For each function, find f ( 3) . (a) (c)

f(x)

=

3x - 7 f

.

Domain

(b)

f

(d)

Range

=

{(-3, 5), (0, 3), (3 , 1), (6, - 1)} y

®=® SOLUTION

(a)

f( x ) = 3x - 7 f( 3) = 3(3 ) - 7

Replacexwith 3.

f( 3) = 2

Simplify.

f(3) Figure 33

y

=

2 indicates that the ordered pair (3, 2) belongs to f.

{(-3, 5), (0, 3), (3, 1), (6, - 1)}, we want f(3), the y-value of the ordered pair where x = 3. As indicated by the ordered pair ( 3, 1), when x = 3, y = 1, so f( 3) = 1.

(b) For f =

(c) In the mapping, repeated in Figure 33, the domain element 3 is paired with 5

in the range, so f( 3) = 5. (d) To evaluate f(3) using the graph, find 3 on the x-axis. See Figure 34. Then move up until the graph off is reached. Moving horizontally to the y-axis gives 4 for the corresponding y-value. Thus, f( 3) = 4. II' Now Try Exercises 61, 63, and 65. Figure 34

Increasing, Decreasing, and Constant Functions Informally speaking, a function increases over an open interval of its domain if its graph rises from left to right on the interval. It decreases over an open interval of its domain if its graph falls from left to right on the interval. It is constant over an open interval of its domain if its graph is horizontal on the interval. For example, consider Figure 35.

•

The function increases over the open interval ( -2, 1) because they-values continue to get larger for x-values in that interval.

•

The function is constant over the open interval ( 1, 4) because the y-values are always 5 for all x-values there.

•

The function decreases over the open interval ( 4, 6) because in that interval they-values continuously get smaller.

The intervals refer to the x-values where the y-values increase, decrease, or are constant. The formal definitions of these concepts follow.

y

-2 I

0

I 2 I

4

6

I

I

1. - . -1. _ . .1. . _ . _1

y is I increasing. I

y is I y is decreasing. constant. I

Figure 35

R.8 Functions

83

Increasing, Decreasing, and Constant Functions

Suppose that a function f is defined over an open interval I and that x 1 and x 2 are in/. (a) f increases over I if, whenever x 1 < x 2 , f(x 1) < f(x 2 ). (b) f decreases over I if, whenever x 1 < x 2, f (x 1) > f(x 2 ) . (c) f is constant over I if, for every x 1 and x 2 , f(x 1) = f(x 2 ) .

Figure 36

illustrates these ideas.

y

y

y- f(x) f(xz)

f(x1)

I

f(x,)-

I

f(xz)

y

h:_ -

~~~--+~~~~ x

0

X1

Xz

Whenever x 1 < x 2, and/(x 1) /(x 2 ), f is decreasing.

For every x 1and x 2, if/(x 1) = /(x2), then f is constant.

(b)

(c)

Figure 36

NOTE To decide whether a function is increasing, decreasing, or constant over an interval, ask, "What does y do as x goes from left to right?" Our definition of increasing, decreasing, and constant function behavior applies to open intervals of the domain, not to individual points.

EXAMPLES

Detennining Increasing, Decreasing, and Constant Intervals

Figure 37 shows the graph of a function. Determine the largest open intervals of the domain over which the function is increasing, decreasing, or constant. y

( I , 8)

Figure 37

"What is happening to they-values as the x-values (domain values) are getting larger?" Moving from left to right on the graph, we see the following:

SOLUTION

• •

Over the open interval (-co, -2), the y-values are decreasing .

•

Over the open interval ( 1, co), the y-values are constant (and equal to 8) .

Over the open interval ( -2, 1), the y-values are increasing .

Therefore, the function is decreasing on (-co, -2), increasing on (-2, 1), and constant on ( 1, co). tl"'

Now Try Exercise 75.

84

CHAPTER R Algebra Review

Continuity The graph of a straight line may be drawn by hand over any interval of its domain without picking the pencil up from the paper. In mathematics we say that a function with this property is continuous over any interval. The formal definition of continuity requires concepts from calculus, but we can give an informal definition at the precalculus level.

Continuity (Informal Definition)

y

0

2

The function is discontinuous at x = 2.

Figure 38

A function is a continuous function over an interval of its domain if its hand-drawn graph over that interval can be sketched without lifting the pencil from the paper.

If a function is not continuous at a point, then it has a discontinuity there. shows the graph of a function with a discontinuity at the point where x = 2. Figure 38

Determining Intervals of Continuity

EXAMPLE 9

Determine the intervals of the domain over which each function in continuous. y

Figure 39

is

y

(a)

(b)

Figure 39

SOLUTION

The function in

Figure 39(a)

is continuous over its entire domain , = 3. Thus,

( -oo, oo) . The function in Figure 39(b) has a point of discontinuity at x it is continuous over the intervals (-oo, 3) and (3, oo).

II" Now Try Exercises 77 and 81 .

( R.8

q

Exercises CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.

1. The domain of the relation { (3, 5), ( 4, 9), ( 10, 13)} is _ __

2. The range of the relation { (3, 5), ( 4, 9), ( 10, 13)} is _ __ 3. The equation y = 4x - 6 defines a function with independent variable dependent variable _ __

4. The function y

=

4x - 6 includes the ordered pair (6, _ ) .

5. For the function f(x) = - 4x + 2, f(-2) = 6. For the function g(x) =

Vx, g(9) =

_ _ __

and

R.8

Functions

85

y

Vx has domain _ __ 8. The function g(x) = Vx has range _ __ 7. The function g(x) =

(3, I )

9. The largest open interval over which the function graphed here increases is _ __ 10. The largest open interval over which the function graphed here decreases is _ __

Determine whether each relation defines a function . See Example 1. 11. { (5, 1), (3, 2), (4, 9), (7, 8)}

13.

{(2,4), (0,2), (2, 6)}

14. {(9, -2) , (-3, 5), (9, !)}

15. {(-3, l), (4, l), (-2, 7)}

17.my 3 7 10

12. { (8, 0) , (5, 7), (9, 3), (3, 8)}

16. {(- 12, 5),(-10,3),(8,3)}

18.

-4 -4 -4

x

-4 0 4

y

V2 V2 V2

Determine whether each relation defines a function, and give the domain and range. See Examples 1-4. 19. { ( 1, 1), (1, - 1), (0,0),

(2, 4), (2, - 4) }

20. { (2, 5), (3, 7), (3, 9), (5, 11 )} 21.

22.

23.f:y 0

0

- 1

1 2

-2

24.my 0 1 2

y

25.

0 - 1 -2

y

26.

x

27.

)'

28.

)'

x

86

CHAPTER R Algebra Review

29.

30.

y

y

Determine whether each relation defines y as a function ofx. Give the domain and range. See Example 5.

31. y

=

x2

34. x = y 4 37. y =

Vx

32. y = x 3

33. x = y 6

35. y = 2x - 5

36. y = -6x + 4

38. y =

-Vx

39. y=~ -7

2

40. y=~

41. y= - x-3

42. y= - x-5

43. Concept Check Choose the correct answer: For function f , the notation f(3) means A. the variable f times 3, or 3f. B. the value of the dependent variable when the independent variable is 3. C. the value of the independent variable when the dependent variable is 3. D. f equals 3. 44. Concept Check Give an example of a function from everyday life. (Hint: Fill in the depends on , so is a function of .) blanks:

Let f(x) = -3x + 4 and g(x) necessary. See Example 6.

46. f(-3)

45. f(O) 49.

= -x2 + 4x + 1. Find each of the following.

1(D

50.

1(-D

Simplify

47. g(-2)

48. g( 10)

51. gG)

52. g(-±)

53. f(p)

54. g(k)

55. f(-x)

56. g(-x)

57. f(x + 2)

58. f(a + 4 )

59. f(2m - 3)

60. f(3t - 2)

if

For each function, find (a) f(2) and (b) f(- 1). See Example 7.

61. f = { ( - 1, 3 ), (4, 7), ( 0, 6), (2, 2)}

62. f = { (2, 5 ), (3, 9) , ( - 1, 11 ), ( 5, 3)}

63.

f

64.

f

65.

y

66.

y

R.8 Functions

87

Use the graph of y = f(x) to find each function value: (a)f(-2),

(b)f(O),

(c)f(l) , and

(d)f(4).

See Example 7(d). 67.

y

68.

y

x

y

69.

y

70.

x

x

Determine the largest open intervals of the domain over which each function is (a) increasing, (b) decreasing, and (c) constant. See Example 8. 71.

72. y

y

(-1, 3) 2

0

x

x

-3

2

2 -2

73.

74. y

y

(3, 1)

75.

76. y

y

(-2, 0)

(2, 0) X

~+-t--+--+..,-1--+-+-+-t--- X

(-1, -3)

(1 ,-3)

(0, -4)

88

CHAPTER R Algebra Review

Determine the intervals of the domain over which each function is continuous. See Example9. 77.

78. y

79. y

y

(0, 3) -+-+-+-t-co,,t-t-t--t-+-- x

80.

81. y

( R.9

82. y

y

q

Graphing Techniques

•

Stretching and Shrinking

•

Reflecting

•

Symmetry

•

Even and Odd Functions

•

Translations

Graphing techniques presented in this section show how to graph functions that are defined by altering the equation of a basic function. NOTE Rec all that Ia I is the absolute value of a number a. if a is positive or 0 if a is negative

Thus I2 I = I2 I and I-2 I = I2 1. We use absolute value functions to illustrate the graphing techniques in Example 1. Stretching and Shrinking

y

*

x

-

Graph of the absolute value function

We begin by considering how the graphs of

y = af(x) and y = f(ax) compare to the graph of y = f(x), where a> 0.

EXAMPLE 1

Stretching or Shrinking Graphs

Graph each function. (a) g(x) = 2lx l

1 (b) h(x) = 2 lx l

(c) k(x) = l2x l

SOLUTION

(a) Comparing the tables of values for f(x) = Ix I and g(x) = 21 x I in Figure 40 on the next page, we see that for corresponding x-values, they-values of g are each twice those of f. The graph of f(x) = Ix I is vertically stretched. The graph of g(x ), shown in blue in Figure 40, is narrower than that of f(x) , shown in red for comparison.

89

R.9 Graphing Techniques

x

lxl

f(x) =

-2

g(x ) =

2

4

- I

I

0

0

I 2

I 2

2 0 2 4

2lxl

y

g(x )

=2lxl

0

-3

Figure 40

t

(b) The graph of h(x) = Ix I is also the same general shape as that of f (x), but for corresponding x-values, they-values of h are each half those of f. The coefficient is between 0 and 1 and causes a vertical shrink. The graph of h(x) is wider than the graph of f (x), as we see by comparing the tables of values. See Figure 41.

t

x

f(x)

= lxl

h(x )

= ! lxl

-2

2

-I

I

2

0

0

0

y

4 I

f(x) =

lxl

I

2 2

2

l -4

-2

x

0

2

4

Figure 41

(c) Use the property of absolute value that states I ab I = Ia I · I b I to rewrite l2x l.

k( x)

=

12x I = 121 · Ix I = 21x I

y

Therefore, the graph of k(x) = I 2x I is the same as the graph of g(x) = 2 lx I in part (a). This is a horizontal shrink of the graph off(x) = Ix 1- See Figure 40.

t/' Now Try Exercises 15 and 17.

Vertical Stretching or Shrinking of the Graph of a Function Vertical stretching a> 1 y

Suppose that a > 0. If a point (x, y ) lies on the graph of y = point (x, ay) lies on the graph of y = af(x ).

f (x ), then the

(a) If a> 1, then the graph of y = af(x) is a vertical stretching of the graph of y = f(x). (b) If 0 < a < 1, then the graph of y = af (x) is a vertical shrinking of the graph of y = f(x).

Vertical shrinking 0 0): y = f(x)

+c

y = f(x) - c

y

= f(x + c)

y = f(x - c)

ofy =

Graphing Techniques

97

Vertical and hori zontal translations are summarized in the table, where f is a fu nction and c is a positive number.

f(x) by

c Units: up down left

EXAMPLE 8

Using More Than One Transformation

Graph each function. (b) h(x) = l2x - 4 1

(a) f(x) =- Ix + 3 1+ 1

rig ht

1 (c) g(x) = - - x 2 2

+4

SOLUTION

(a) To graph f (x) = - Ix + 3 I + 1, the lowest point on the graph of y = Ix I is translated to the left 3 units and up I unit. The graph opens down because of the negative sign in front of the absolute value expression, making the lowest point now the highest point on the graph, as shown in Figure 57. The graph is symmetric with respect to the line x = -3. y

J O

Figure 26

Now we use the definitions of the trigonometric functions.

-4

4

5

5

-3 5

sine=-= - -

cose

5 5 csce = - = - -4 4

sece =

= -

3 5

= --

5

5

-3

3

-= --

-4 4 = -3 3 -3 3 cote= - = -4 4 tane = -

, / Now Try Exercise 17. y

We can find the six trigonometric functions using any point other than the origin on the terminal side of an angle. To see why any point can be used , refer to Figure 21, which shows an angle and two distinct points on its terminal side. Point P has coordinates (x, y), and point P ' (read "P-prime") has coordinates (x', y'). Let r be the length of the hypotenuse of triangle OPQ, and let r' be the length of the hypotenuse of triangle OP' Q' . Because corresponding sides of similar triangles are proportional, we have

e

0

Q

Q'

Figure 27

y

y'

r

r

,.

Corresponding sides are proportional.

e 7

Thus sin = is the same no matter which point is used to find it. A similar result holds for the other five trigonometric functions. We can also find the trigonometric function values of an angle if we know the equation of the line coinciding with the terminal ray. Recall from algebra that the graph of the equation

Ax

+ By = 0

Linear equation in two variables x and y

is a line that passes through the origin ( 0, 0 ). If we restrict x to have on!y non positive or only nonnegative values, we obtain as the graph a ray with endpoint at the origin. For example, the graph of x + 2y = 0 , x ;:::: 0, shown in Figure 28, is a ray that can serve as the terminal side of an angle in standard position. By choosing a point on the ray, we can find the trigonometric function values of the angle.

e

y

x + 2y =

o, x ~ o

t

Figu re 28

143

1.3 Trigonometric Funct ions

EXAMPLE 3

Finding Function Values of an Angle

Find the six trigonometric function values of an angle the terminal side of e is defined by x + 2y = 0, x 2: 0.

e in standard position, if

The angle is shown in Figure 29. We can use any point except (0, 0) on the terminal side of e to find the trigonometric function values. We choose x = 2 and find the corresponding y-value.

y

SOLUTION

+ 2y = 0, 2 + 2y = 0

x

x 2: 0 Let x = 2.

2y = -2 x+ 2y = 0, x

t

~

0

y

Figure 29

=

Subtract 2.

-1

Divide by 2.

The point (2, - I) lies on the terminal side, and so r = v'22 Now we use the definitions of the trigonometric functions. . y -1 -1 Vs Vs sme = - = - - = - - . - - = - - r Vs Vs Vs 5 Multiply by

cos

e=

x

2

2

Vs

2Vs

r

Vs

Vs

Vs

5

+ ( - I )2

=

Vs.

~ , a form of I,

to rationalize the denominators.

- = -- = -- · -- = --

y -1 1 tan e = - = - = - x 2 2

csc e = !:._ = Vs = - Vs y - I

sece =

r x

Vs

- = --

2

x

2

cot e = - = - = -2 y -1 II" Now Try Exercise 51 .

Recall from algebra that when the equation of a line is written in the form y

= mx

+ b,

Slope-intercept form

the coefficient m of x gives the slope of the line. In Example 3, the equation + 2y = 0 can be solved for y to obtain y = The slope of this line is Notice also that tan e =

x

-k.

In general, it is true that m NOTE

-k.

-kx.

= tan 6.

The trigonometric function values we found in Examples 1-3 are

exact. If we were to use a calculator to approximate these values, the decimal results would not be acceptable if exact values were required.

y

x=O (0, 1)

x = -1

y=l r=l

y=O

_ r___ x (1, 0) = 1 _ _o.,...___ _______

x=O y

=1 r =1

x

(-1, 0)

=-1

y=O (0, -1)

r=l Figure 30

Ouadrantal Angles If the terminal side of an angle in standard position lies along the y-axis, any point on this terminal side has x-coordinate 0. Similarly, an angle with terminal side on the x-axis has y-coordinate 0 for any point on the terminal side. Because the values of x and y appear in the denominators of some trigonometric functions, and because a fraction is undefined if its denominator is 0, some trigonometric function values of quadrantal angles (i.e., those with terminal side on an axis) are undefined. When determining trigonometric function values of quadrantal angles, Figure 30 can help find the ratios. Because any point on the terminal side can be used, it is convenient to choose the point one unit from the origin, with r = 1. (Later we will extend this idea to the unit circle.)

144

CHAPTER 1 Trigon ometric Funct ions

To find the function values of a quadrantal angle, determine the position of the terminal side, choose the one of these four points that lies on this terminal side, and then use the definitions involving x, y, and r. Finding Function Values of Ouadrantal Angles

EXAMPLE 4

Find the values of the six trigonometric functions for each angle. (a) an angle of 90° (b) an angle

e in standard position with terminal side passing through (-3, 0)

SOLUTION HOR11AL FLOAT AUTO REAL 0£GP.££ HP

0

sin(90)

(a) Figure 31 shows that the terminal side passes through (0, 1). Sox= 0, y = 1, and r = 1. Thus, we have the following.

..................................................~ . cos C90)

sin 90° =

t:aii'c 90»····· ········· ·· ······· ··· ·· ······· ·0··· ..........................................!;r:r:Rr:.

l1 =

1

cos 90° =

1 csc 90° = - = 1 1

sec 90° =

A calculator in degree mode returns the correct values for sin 90° and cos 90°. The screen shows an ERROR message for tan 90°, because 90° is not in the domain of the tangent function.

l0 = 0 1

0

tan 90° =

1

0

Undefined

0 cot 90° = - = 0 1

Undefined

y

y

(0, 1)

0

90° x

0

(-3,0)

Figure 31

x

0

Figure 32

(b) Figure 32 shows the angle . Here, x = -3, y = 0 , and r = 3, so the trigonometric functions have the following values. .

0 3

Sin

8=

csc

e = 03

- =

e = --3 = 3

0

cos

Undefined

3 sece=-= -1

- 1

-3

0 -3

tane = - = 0

cote=

-3

0

Undefined

Verify that these values can also be found using the point ( - 1, 0). t/' Now Try Exercises 23, 67, 69, and 71 .

The conditions under which the trigonometric function values of quadrantal angles are undefined are summarized here. Conditions for Undefined Function Values y

(0, 1)

x = -1

Identify the terminal side of a quadrantal angle. x =O y =l r=l

y =O (1, 0)

r =1

~--,__~~o+-~~----- x

(- 1, 0)

x =O y =-1 r =1

x

=1

•

If the terminal side of the quadrantal angle lies along the y-axis, then the tangent and secant functions are undefined.

•

If the terminal side of the quadrantal angle lies along the x-axis, then the cotangent and cosecant functions are undefined.

y =O r=l (0, - 1)

Figure 30 (repeat ed)

The funct ion values of some commonly used quadrantal angles, 0°, 90°, 180°, 270°, and 360°, are summarized in the table on the next page. They can be determined when needed by using Figure 30 and the method of Example 4(a).

1.3 Trigonometric Functions

145

For other quadrantal angles such as -90°, -270°, and 450°, first determine the coterminal angle that lies between 0° and 360°, and then refer to the table entries for that particular angle. For example, the function values of a -90° angle would correspond to those of a 270° angle. Function Values of Quadrantal Angles (}

oo 90° 180° 270° 360°

sin(}

cos(}

0

tan(}

cot(}

1

0

Undefined

Undefined

0

Undefined

0

Undefined

- 1

Undefined

1

0

0

- 1

- 1

0

Undefined

0

0

1

0

Undefined

csc (}

sec(}

Undefined

1

1

Undefined - 1

Undefined

1

The values given in this table can be found with a calculator that has trigonometric function keys. Make sure the calculator is set to degree mode.

TI-84 Plus

Figure 33

tWQim~I One of the most common errors involving calculators in trigonometry occurs when the calculator is set for radian measure, rather than degree measure. Be sure to set your calculator to degree mode. See Figure 33.

;

Exercises CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.

1. The Pythagorean theorem for right triangles states that the sum of the squares of the lengths of the legs is equal to the square of the _ _ __ 2. In the definitions of the sine, cosine, secant, and cosecant functions, r is interpreted geometrically as the distance from a given point (x, y) on the terminal side of an angle (J in standard position to the _ _ __ 3. For any nonquadrantal angle 6, sin (J and csc (J will have the _ _ _ _ _ _ _ sign. (same I opposite) 4. If cot (J is undefined , then tan (J = - -· 5. If the terminal side of an angle (J lies in quadrant III, then the values of tan (J and cot (J are , and all other trigonometric function values are (positive I negative) (positive/ negative) 6. If a quadrantal angle (J is coterminal with 0° or 180°, then the trigonometric functions and are undefined. CONCEPT PREVIEW The terminal side of an angle (J

in standard position passes through the point ( -3, -3 ). Use the figure to find the following values. Rationalize denominators when applicable. 7. r 9. cos (J

8. sin (J 10. tan

(J

y

146

CHAPTER 1 Trigonometric Functions

Sketch an angle (} in standard position such that (} has the least positive measure, and the given point is on the terminal side of 0. Then find the values of the six trigonometric functions for each angle. Rationalize denominators when applicable. See Examples 1, 2, and 4. 11. (S, -12)

12. (- 12, -S)

13. (3,4)

14. (-4, -3)

15. ( -8, lS)

16. ( lS,-8)

17. (-7, -24)

18. (-24, -7)

19. (0, 2)

20. (O,S)

21. (-4, 0)

22. (-S,O)

23. (0, -4)

24. (0, -3)

25. ( 1,

21.

(v2. Vi)

28.

(-v2, -v2)

\/3)

26. (-1 , v3)

29. (-2v3,-2)

30. (-2v3,2)

Concept Check Suppose that the point (x, y) is in the indicated quadrant. Determine whether the given ratio is positive or negative. Recall that r = a sketch may help.)

Vx

2

+ y 2 . (Hint: Drawing

x 31. II, r

y 32. III, r

y 33. IV, x

x 34. IV, y

y 35. II, r

x 36. III, r

x 37. IV, r

y 38. IV, r

x 39. II, y

y 40. II, x

y 41. III, x

42. III,~ y

43. III,!_ x

44. III,!_ y

45. I,~ y

y 46. I, x

y 47. I, r

x 48. I, r

r 49. I , x

r 50. I, y

An equation of the terminal side of an angle (} in standard position is given with a restriction on x. Sketch the least positive such angle 0, and find the values of the six trigonometric functions of O. See Example 3.

51. 2x + y

54. -Sx - 3y = 0, x 57. x

52. 3x + Sy = 0, x

= 0, x ~ 0

+y =

0, x

~

:S

0

0

55. -4x + 7y = 0, x

:S

58. x - y

0

60. v3x + y = 0, x

~

:S

0

53. -6x - y = 0, x 0

= 0, x ~ 0

61. x = 0, y

~

0

56. 6x - Sy = 0, x

:S

~

0 0

= 0, x :S 0

59. -v3x + y 62. y = 0, x

:S

0

Find the indicated function value. If it is undefined, say so. See Example 4.

63. cos 90°

64. sin 90°

65. tan 180°

66. cot 90°

67. sec 180°

68. csc 270°

69. sin(-270°)

70. cos( -90°)

71. cot S40°

72. tan 4S0°

73. csc( -4S0°)

74. sec(-S40°)

75. sin 1800°

76. cos 1800°

77. csc 1800°

78. cot 1800°

79. sec 1800°

80. tan 1800°

81. cos( -900°)

82. sin(-900°)

83. tan( -900°)

84. How can the answer to Exercise 83 be given once the answers to Exercises 81 and 82 have been determined? Use trigonometric function values of quadrantal angles to evaluate each expression.

85. cos 90° + 3 sin 270°

86. tan 0° - 6 sin 90°

87. 3 sec 180° - S tan 360°

88. 4 csc 270° + 3 cos 180° 2

89. tan 360° + 4 sin 180° + S (cos 180°) 2

91. (sin 180°) +( cos 180°)

2

90. S (sin 90°) 2 + 2 (cos 270°)2 2

92. (sin 360°) +(cos 360°)

2

-

tan 360°

1.3 Trigonometric Functions

93. (sec 180°) 2

-

3 (sin 360°) 2 + cos 180°

147

94. 2 sec 0° + 4 (cot 90°) 2 + cos 360°

95. -2(sin 0°) 4 + 3 (tan 0°)2

96. -3 (sin 90°) 4 + 4(cos 180°) 3

97. [sin(-90°)]2 + [cos ( -90°) ]2

98. [cos(-180°)]2 + [sin (- 180°) ]2

If n is an integer, n · 180° represents an integer multiple of 180°, (2n + 1) · 90° represents an odd integer multiple of 90°, and so on. Determine whether each expression is equal to 0, 1, or - 1, or is undefined.

99. cos [ (2n + 1) · 90°]

102. sin[270° + n · 360°]

100. sin[ n • 180°]

101. tan[n · 180°]

103. tan[ (2n + 1) · 90°]

104. cot[ n · 180°]

105. cot[ (2n + 1) · 90°]

106. cos [ n · 360° J

107. sec[ (2n + 1) · 90°]

108. csc [n · 180°]

Concept Check In later chapters we will study trigonometric functions of angles other than quadrantal angles, such as 15°, 30°, 60°, 75°, and so on. To prepare for some important concepts, provide conjectures in each exercise. Use a calculator set to degree mode.

109. The angles 15° and 75° are complementary. Determine sin 15° and cos 75°. Make a conjecture about the sines and cosines of complementary angles, and test this hypothesis with other pairs of complementary angles. 110. The angles 25° and 65° are complementary. Determine tan 25° and cot 65°. Make a conjecture about the tangents and cotangents of complementary angles, and test this hypothesis with other pairs of complementary angles. 111. Determine sin 10° and sin ( - 10°). Make a conjecture about the sine of an angle and the sine of its negative, and test this hypothesis with other angles. 112. Determine cos 20° and cos( -20°). Make a conjecture about the cosine of an angle and the cosine of its negative, and test this hypothesis with other angles. ~

Set a TI graphing calculator to parametric and degree modes. Use the window values shown in the first screen, and enter the equations shown in the second screen. The corresponding graph in the third screen is a circle of radius 1. Trace to move a short distance around the circle. Jn the third screen, the point on the circle corresponds to the angle T = 25°. Because r = 1, cos 25° is X = 0.90630779 and sin 25° is Y = 0.42261826. HlkMflL FLOltT RUJO ll:EHL OtlfRtE M"

WI NDOW Tmin=0 Tmax=360 TsteP=l Xmin=-1.8 Xmax=l.8 Xscl=l Ymin=-1.2 Ymax=l. 2 Ysc l gl

a

HIJIMfll. FLOIT AUIO Pl:EFIL OtO:CE "" ~1otl

Plot2

a

HCIRNRL FLDl"l l flUTO ll:ERL Ot(iREE NP'

a

Plo U

l \Xu llc os (Tl Y t T B~i n(l)

1\ X2 T=

Y2T = 1\X> r=

Y3 T=

l \ X4T =

Y4T= 1 \ XST=

Use this information to answer each question.

113. Use the right- and left-arrow keys to move to the point corresponding to 20° (T = 20). Approximate cos 20° and sin 20° to the nearest thousandth.

= 0.766? For what angle T, 0° :ST :S 90°, is sin T = 0.574?

114. For what angle T, 0° :S T :s 90°, is cos T 115.

116. For what angle T, 0° :s T :s 90°, does cos T equal sin T? 117. As T increases from 0° to 90°, does the cosine increase or decrease? What about the sine? 118. As T increases from 90° to 180°, does the cosine increase or decrease? What about the sine?

148

CHAPTER 1 Trigonometric Functions

1.4 •

Using the Definitions of the Trigonometric Functions Identities are equations that are true for all values of the variables for which all expressions are defined.

Reciprocal Identities

• Signs and Ranges of Function Values •

Pythagorean Identities

2(x

+ 3)

+6

= 2x

(x

+ y )2 = x 2 + 2xy + y 2

Identities

• Quotient Identities

Recall the definition of a reciprocal.

Reciprocal Identities

1 The reciprocal of a nonzero number a is - . a Examples: The reciprocal of 2 is

&, and the reciprocal of !J is 1f.There is no

reciprocal for 0 because &is undefined. The definitions of the trigonometric functions in the previous section were written so that functions in the same column were reciprocals of each other. Because sin (} = ~ and csc (} = ~ , 1 sin (} = - - and csc (}

1

csc (} = - -, sin(}

provided sin (} ¥- 0.

Also, cos(} and sec(} are reciprocals, as are tan (} and cot e. The reciprocal identities hold for any angle (} that does not lead to a 0 denominator.

Reciprocal Identities

a

HIRMflL FlOflT AUTO ll:Efll OtlliRCE MF'

For all angles (} for which both functions are defined, the following identities hold true.

1

si•On " " ' ' '! ' ' " ' " " " " " " " " " " " " ' ' ...... ....

~

~.

. smfJ

1 =csc (J

csc (J

= --:-0 Sill

-1

cos (J

1 = --(J sec

tan (J

= cot (J

1

sec fJ

1 =cos 6

cotfJ

1 =tan 6

'i~~~i·:180')') :i' ' '' "'''"'''' '' ' ''' ''' ' '' ' '

1

(a) HIRMflL FLDflT l'IUTO REHL OCCiREE M,

0

1

COs 0, tan e < 0

ethat satisfies the given

(b) cos e < 0, sec e < 0

SOLUTION

e

> 0 in quadrants I and II and tan both conditions are met only in quadrant II.

(a) Because sin y

e < 0 in quadrants II and IV,

(b) The cosine and secant functions are both negative in quadrants II and III, so (0, r)

in this case

e could be in either of these two quadrants.

v" Now Try Exercises 43 and 49.

y

(r, 0) x

(a)

Figure 35(a) shows an angle e as it increases in measure from near 0° toward 90°. In each case, the val ue of r is the same. As the measure of the angle increases, y increases but never exceeds r, soy :s r. Dividing both sides by the

positive number r gives~ ::; 1. In a similar way, angles in quadrant IV as in Figure 35(b ) suggest that

y

y

-1 < -

(r, 0)

-

x

y

r'

y

so

-1 ::;

and

- 1

S

Similarly,

- 1

s cos (J s 1.

r

::; 1

sin (J S 1.

*

= sin {j for any angle 0.

T he tangent of an angle is defined as~· It is possible that x

(0,-r)

(b)

Figure 35

< y, x

= y, or

x > y. Thus , ~ can take any value, so tan (J can be any real nu m ber, as can cot 8.

1.4 Using the Definitions of the Trigonometric Funct ions

151

The functions sec fJ and csc fJ are reciprocals of the functions cos fJ and sin fJ, respectively, making sec (J :s; - 1

or

sec (J ;;::: 1

and

csc (J :s; - 1

or

csc (J ;;::: 1.

In summary, the ranges of the trigonometric functions are as follows.

Ranges of Trigonometric Functions

Trigonometric Function of (}

Range (Set-Builder Notation)

Range (Interval Notation)

sin(}, cos(}

{yilYl~I}

[ - 1, 1J

tan 6, cot (}

{ y Iy is a real number}

(-oo, oo)

sec(}, csc (}

{yilYl2:1}

(-oo, -1] U [ I ,oo)

EXAMPLE 4

Determining Whether a Value Is in the Range of a Trigonometric Function

Determine whether each statement is possible or impossible. (a) sin fJ = 2.5

(b) tan fJ = 110.47

(c) sec fJ = 0 .6

SOLUTION

(a) For any value of fJ, we know that - 1 s; sin fJ s; 1. Because 2.5 impossible to find a value of fJ that satisfies sin fJ = 2.5.

> 1, it is

(b) The tangent function can take on any real number value. Thus, tan fJ = 110.47 is possible. (c) Because I sec fJ I 2: 1 for all fJ for which the secant is defined, the statement sec fJ = 0.6 is impossible. ./' Now Try Exercises 53, 57, and 59.

The six trigonometric functions are defined in terms of x , y, and r , where the Pythagorean theorem shows that r 2 = x 2 + y 2 and r > 0. With these relationships, knowing the value of on ly one function and the quadrant in which the angle lies makes it possible to find the values of the other trigonometric functions.

EXAMPLE 5

Finding All Function Values Given One Value and the Quadrant

Suppose that angle fJ is in quadrant II and sin fJ = ~. Find the values of the five remaining trigonometric functions. SOLUTION Choose any point on the terminal side of angle fJ . For simplicity, since sin fJ = ~' choose the point with r = 3.

sin fJ =

32

y

2

r

3

Given value Substitute ~ for sin 6.

152

CHAPTER 1 Trigonometric Functions

Because ~ = ~ and r

3, it follows that y = 2. We must find the value of x .

=

+ y 2 = r2 x2 + 22 = 32

x2

Pythagorean theorem Substitute.

+4= 9

x2

Apply exponents.

x2 = 5 ( Remember both roots.\.

"'x =

y

x

=-vs

, r;

v 5 or

x= -

Square root property: If x 2 = k, then x =

vk or x = - Vk.

- Vs so that the

2)

- Vs

Vs

x cos f) = - = - - = - - r 3 3

r=3 ()

0

v5

Because fJ is in quadrant II, x must be negative. Choose x = point ( -Vs, is on the terminal side of fJ. See Figure 36.

y =2

-vs

Subtract 4. , r;

3 3 Vs 3Vs -Vs Vs Vs 5 2 2 Vs 2Vs - - = - -- . -- = - -- Vs Vs Vs 5 - Vs Vs - - = - --

r secfJ = - = - - = - - - · - - = - - x

2

-2

tan

y

f) = - =

x

Figure 36

x cot fJ = - =

2

y

These have rati onalized denominators.

2

r 3 csc f) = - = y 2

v

Now Try Exercise 75.

We now derive three new identities.

Pythagorean Identities

x2

+ y2

= r2

x2

y2

r2

Pythagorean theorem

-r2 + -r2 = -r2

Divide by r 2 . Power rule for exponents; a y;;"' = (a)"' b

(cos 11) 2 and cos 2 II are equivalent forms.

(cos fJ) 2 + (sinfJ )2 = l

+

sin2 ()

cos

2

()

cosO=~.sinO=~

=1

Apply exponents; commutative property

Starting again with x 2 + y 2 = r 2 and dividing through by x 2 gives the following. x2

y2

r2

-x 2 + -x2 = -x2 1+

( ~x)2 =

Divide by x 2 .

( -xr )2

Power rule for exponents

+ (tan 8) 2 = (sec 8) 2 tan2 () + 1 = sec2 ()

l

tan 0 = ~. sec 0 = ~ Apply exponents; commutative property

Similarly, dividing through by y 2 leads to another identity. 1

+ cot2 ()

= csc2 ()

These three identities are the Pythagorean identities because the original equation that led to them, x 2 + y 2 = r 2, comes from the Pythagorean theorem.

1.4 Using the Definitions of the Trigonometric Fu nctions

153

Pythagorean Identities

For all angles () for which the function values are defined, the following identities hold true. sin2 (J

+

=1

cos2 (J

+

tan2 (J

1

= sec2 (J

1

+

cot2 (J

= csc2 (J

We give only one form of each identity. However, algebraic transformations produce equivalent forms. For example, by subtracting sin2 () from both sides of sin2 () + cos2 () = 1, we obtain an equivalent identity. cos2 (J

=1

-

sin2 (J

Alternative form

It is important to be able to transform these identities quickly and also to recognize their equivalent forms.

Consider the quotient of the functions sin ()and cos e,

Quotient Identities

for cos() # 0. LOOKING AHEAD TO CALCULUS

. Sill ()

The reciprocal, Pythagorean, and

y ~

y

-- = - = cos () ! r

quotient identities are used in calculus

7

X -

r

y

y

r

= - · - = - = tan () r x x

to find derivatives and integrals of trigonometric functions. A standard technique of integration called

Similarly,~~::= cot(), for sin()# 0. Thus, we have the quotient identities.

trigonometric substitution relies on the Pythagorean identities.

Quotient Identities

For all angles () for which the denominators are not 0, the following identities hold true. sin (J

cos (J

- - =tan (J cos 0

-:-Sill 0

Using Identities to Find Function Values

EXAMPLE 6

Find sin() and tan e, given that cos() = SOLUTION

~ and sin() > 0.

Start with the Pythagorean identity that includes cos().

sin2 sin2 ()

= cot (J

()

+ cos2 () = 1

Pythagorean identity

~ )2 = 1

Replace cos (J with -

+(-

sin2 ()

+ l_ = 16

1

13 sin2 () = 16

.

sm() = Choose th e correct sign here.

Vt3 +- 4Vt3

sin()= - 4

'{3.

v'3

Square - 4 . 3

Subtract T6 .

Take square roots. Choose the positive square root because sin (J is positive.

154

CHAPTER 1 Trigonometric Functions

To find tan 8, use the values of cos 8 and sin 8 and the quotient identity for tan 8.

Vt3 v3

v3 . v3 =

= -

v39

- - 3-

Rationalize the denominator.

II" Now Try Exercise 79. CIMuiCll~I In exercises like Examples 5 and 6, be careful to choose the correct sign when square roots are taken. Refer as needed to the diagrams preceding Example 2 that summarize the signs of the functions.

Using Identities to Find Function Values

EXAMPLE 7

Find sin 8 and cos 8, given that tan 8 =~and 8 is in quadrant III. Because 8 is in quadrant III, sin 8 and cos 8 will both be negative. sin O and tan 8 = 43, then sm · 8 = - 4 and · · · It 1s temptmg to say t hat smce tan 8 = cos 0 cos 8 = -3 . This is incorrect, however-both sin 8 and cos 8 must be in the interval [ - 1, 1] . We use the Pythagorean identity tan2 8 + 1 = sec 2 8 to find sec 8, and then

SOLUTION

the reciprocal identity cos 8 = se~ 0 to find cos 8. tan2 8

+ 1 = sec 2 8

Pythagorean identity

(~)2 + 1 = sec 8

4

2

tan 8 = 3

16

- + 1 =sec2 8

4

Square 3·

9

25

- = sec2 8 9

Be ca reful to choose t he correct sign here.

Add.

5 - - =sec8 3

Choose the negative square root because sec 8 is negative when 8 is in quadrant III.

3 - - = cos 8 5

Secant and cosine are reciprocals.

Now we use this value of cos 8 to find sin 8. sin2 8 = 1 - cos2 8 sin2 8 = 1 - ( sin2 8 = 1 16 sin2 8 = ( Again, be carefu l. ~

25

4

sin 8 = - -

5

_2_ 25

~)

Pythagorean identity (alternative form) 2

3

cos8 = - 5

Square - ~ .

Subtract.

Choose the negative square root.

II" Now Try Exercise 77.

155

1.4 Using the Definitions of the Trigonometric Functions

NOTE Example 7 can also be worked by sketching e in standard position in quadrant III, finding r to be 5, and then using the definitions of sin e and cos e in terms of x, y, and r. See Figure 37. When using this method, be sure to choose the correct signs for x and y as determined by the quadrant in which the terminal side of e lies. This is analogous to choosing the correct signs after applying the Pythagorean identities.

1.4

y

=

x -3 y = -4

r=S

Figure 37

Exercises CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.

1

1. Given cos () = se~ 0 , two equivalent forms of this identity are sec () = - - and cos() . = 1. 2. Given tan()

=co~ 0 , two equivalent forms of this identity are cot() = - -1

and

tan()· - - - = 1. 3. For an angle () measuring 105°, the trigonometric functions and _ __ are positive, and the remaining trigonometric functions are negative. 4. If sin () > 0 and tan () > 0, then () is in quadrant _ __ CONCEPT PREVIEW Determine whether each statement is possible or impossible.

5. sin () =

1

2, csc () = 2

6. tan() = 2, cot()= -2

8. cos()= 1.5

9. cot() = - 1.5

7. sin() > 0, csc () < 0 10. sin2 () + cos2 () = 2

Use the appropriate reciprocal identity to find each function value. Rationalize denominators when applicable. See Example 1. 11. sec (), given that cos () = ~

12. sec (), given that cos() = ~

13. csc 0, given that sin () = - ~

14. csc 0, given that sin() =

15. cot 0, given that tan () = 5

16. cot 0, given that tan () = 18

17. cos 0, given that sec () = - ~

18. cos(), given that sec () =

. () , given . 19. sm that csc () = 2Vs

. () , given . that csc () = -v24 20. sm 3

21. tan 0, given that cot() = -2.5

22. tan 0, given that cot() = -0.01

23. sin 0, given that csc () = 1.25

24. cos 0, given that sec () = 8

-f:J -!,f-

25. Concept Check What is wrong with the following item that appears on a trigonometry test?

"Find sec 0, given that cos () =

%."

26. Concept Check What is wrong with the statement tan 90° =

co/90o?

156

CHAPTER 1 Trigonometric Functions

Determine the signs of the trigonometric functions of an angle in standard position with the given measure. See Example 2.

27. 74°

28. 84°

29. 218°

30. 195°

31. 178°

32. 125°

33. -80°

34. - 15°

35. 855°

36. 1005°

37. - 345°

38. - 640°

Identify the quadrant (or possible quadrants) of an angle () that satisfies the given conditions. See Example 3.

39. sin() > 0, csc () > 0

40. cos () > 0, sec () > 0

41. cos() > 0, sin () > 0

42. sin () > 0, tan () > 0

43. tan () < 0, cos () < 0

44. cos () < 0, sin () < 0

45. sec () > 0, csc () > 0

46. csc () > 0, cot () > 0

47. sec() < 0, csc () < 0

48. cot () < 0, sec () < 0

49. sin () < 0, csc () < 0

50. tan () < 0, cot () < 0

51. Why are the answers to Exercises 41 and 45 the same? 52. Why is there no angle () that satisfies tan () > 0, cot () < O? Determine whether each statement is possible or impossible. See Example 4.

53. sin() = 2

54. sin() = 3

55. cos() = - 0.96

56. cos () = - 0.56

57. tan () = 0.93

58. cot() = 0.93

59. sec () = - 0.3

60. sec () = - 0.9

61. csc () = 100

62. csc () = - 100

63. cot()= - 4

64. cot()= - 6

Use identities to solve each of the following. Rationalize denominators when applicable. See Examples 5-7.

65. Find cos (), given that sin () = ~and () is in quadrant II. 66. Find sin 0, given that cos () = ~and() is in quadrant IV.

-1 and() is in quadrant IV. 68. Find sec 0, given that tan () = v;1 and() is in quadrant III. 69. Find tan 0, given that sin() = 1and() is in quadrant II. 67. Find csc 0, given that cot() =

70. Find cot(), given that csc () = - 2 and() is in quadrant III. 71. Find cot 0, given that csc () = - 1.45 and () is in quadrant III. 72. Find tan 0, given that sin() = 0.6 and() is in quadrant II. Give all six trigonometric function values for each angle 0. Rationalize denominators when applicable. See Examples 5-7.

73. tan () =

-lj , and () is in quadrant II

74. cos () = - ~, and() is in quadrant III

'{5, and () is in quadrant I 77. cot() = '{3, and() is in quadrant I 79. sin()= '{2, and cos() < 0

78. csc () = 2, and () is in quadrant II

81. sec () = - 4, and sin () > 0

82. csc () = - 3, and cos()> 0

83. sin() = 1

84. cos() = 1

75. sin () =

76. tan () =

\/3, and ()is in quadrant III Vs

80. cos () = 8 , and tan () < 0

157

CHAPTER 1 Test Prep

Work each problem. 85. Derive the identity 1

+ cot2 0 =

csc2 Oby dividing x 2

+ y 2 = r 2 by y 2 .

'd . cos 0 = cot u. {) . the quotient . 86. D enve 1 entity sin 0 87. Concept Check True or false: For all angles 0, sin 0 false, give an example showing why.

+ cos 0 =

1. If the statement is

= ~~~ :, if cot 0 = ~ with 0 in quadrant I, = I and sin 0 = 2. If the statement is false, explain why.

88. Concept Check True or false: Since cot 0 then cos 0

Concept Check Suppose that 90° < 0 < 180°. Find the sign of each function value.

w

89. sin W

90. csc

93. cot( 0 + 180°)

94. tan(O + 180°)

0 91. tan 2

0 92. cot2

95. cos(-0)

96. sec( -0)

Concept Check Suppose that -90° < 0 < 90°. Find the sign of each function value. 0 97. cos 2

98. sec

101. sec( -0)

0

2

99. sec(O + 180°)

102. cos( -0)

103. cos(O - 180°)

100. cos(O

+ 180°)

104. sec(O - 180°)

Concept Check Find a solution for each equation. 105. tan (30 - 4°) = cot(S:- 80) 107. sin( 40

+ 2°) csc(30 + S0) =

1

109. Concept Check The screen below was obtained with the calcu lator in degree mode. Use it to j ustify that an angle of 14,879° is a quadrant II angle. HIRMAL fLOl'IT RUTO IURL OCGRCE M,

0

cos C14879l

106. cos( 60

+ s o) =

sec( 401+ 1So)

108. sec(W

+ 6°) cos(SO + 3°)

= 1

110. Concept Check The screen below was obtained with the calculator in degree mode. In which quadrant does a 1294° angle lie?

HIJIMAL FLOl'IT AUTO REAL OE,REE 11,.

tanC1294 l

0

.............................:,.!\!M~l!?!i71!7 .

51ii'(i294'i···································

····· ··········· ·······'·~?.!\!'!~?.?.!!?.~ .

.............................:,.~~?~?7?\'!~~.

s inl14879l

• 6745085168

Chapter 1 Test Prep Key Terms 1.1

line line segment (or segment) ray endpoint (of a ray) angle side (of an angle) vertex (of an angle) initial side

terminal side positive angle negative angle acute angle right angle obtuse angle straight angle complementary angles (complements)

1.2

supplementary angles (supplements) standard position (of an angle) quadrantal angle coterminal angles vertical angles parallel lines transversal

1.3

similar triangles congruent triangles sine (sin) cosine (cos) tangent (tan) cotangent (cot) secant (sec) cosecant (csc)

158

CHAPTER 1 Trigonometric Functions

New Symbols 1

8

right angle symbol (for a right triangle) Greek Jetter theta

degree minute second

Quick Review Concepts

Examples

Angles

TYpes of Angles Two positive angles with a sum of 90° are complementary angles. Two positive angles with a sum of 180° are supplementary angles. 1 degree = 60 minutes (1° = 60') 1 minute = 60 seconds {1' = 60")

70° and 90° - 70° = 20° are complementary. 70° and 180° - 70° = 110° are supplementary. 15° 30' 45" 30°

45° + -60 3600

= 15°+ -

= 15.5125°

Coterminal angles have measures that differ by a multiple of 360°. Their terminal sides coincide when in standard position.

30 ' ·

1° 60 •

,,

45 .

30°

= 60 and

1°

-

45°

3600" - 3600 '

Decimal degrees

The ac ute angle (} in the figure is in standard position. If(} measures 46°, find the measure of a positive and a negative coterminal angle.

y

46° + 360° = 406° 46° - 360° = -3 14°

Angle Relationships and Similar Triangles

Vertical angles have equal measures.

Find the measures of angles I , 2, 3, and 4.

When a transversal intersects two parallel lines, the following angles formed have equal measure: •

alternate interior angles,

•

alternate exterior angles, and

•

corresponding angles.

Interior angles on the same side of a transversal are supplementary.

m and n are parallel lines.

12x - 24 + 4x + 12 = 180 16x - 12 = 180

x = 12

Interior angles on the same side of a transversal are supplementary.

Angle 2 has measure 12 · 12 - 24 = 120°. Angle 3 has measure 4 · 12 + 12

= 60°.

Angle 1 is a vertical angle to angle 2, so its measure is 120°. Angle 4 corresponds to angle 2, so its measure is 120°. Angle Sum of a Triangle The sum of the measures of the angles of any tri angle is 180°.

The measures of two angles of a triangle are 42° 20' and 35° 10'. Find the measure of the third angle, x. 42° 20 ' + 35° 10' +x= 180° 77° 30' + x = 180° x = 102° 30'

The sum of the angles is 180°.

159

CHAPTER 1 Test Prep

Concepts

Examples

Similar triangles have corresponding angles with the same measures and corresponding sides proportional.

Find the unknown side length.

~,

Congruent triangles are the same size and the same shape.

50 ~

= 50

+ 15

Corresponding sides of similar triangles are proportional.

50

20

50x = 1300

Property of proportions

x= 26

1.3

15

Divide by 50.

Trigonometric Functions

Trigonometric Functions Let (x, y) be a point other than the origin on the terminal side of an angle () in standard position. The distance from the point to the origin is r =

v'x2

If the point ( -2, 3) is on the terminal side of an angle () in standard position , find the values of the six trigonometric functions of e.

Here x = -2 and y = 3, so

+ y2.

r=

The six trigonometric functions of() are defined as follows. y

v'(-2) 2 + 32 = v'4+9 = vl3.

3vl3

cos()=- - - -

vl3

sec()= - - -

~

x cosO = r

tan 8 = - (x -:F 0)

sin()=

cscO=:_

r sec 8 = x

cot 8 = ~

csc() = - 3

sinO =

r

y

(y

* 0)

x y

(x -:F 0)

(y

* 0)

2vl3

13

13

3

tan()= - 2

vl3

2

cot ()= - -

2

3

See the summary table of trigonometric function values for quadrantal angles in this section.

1.4

Using the Definitions of the Trigonometric Functions

If cot() = - 23, find tan ().

Reciprocal Identities 1 sinO = --,, csc"

csc 8 =

1

-:--a SID v

1 cosO = --,, sec.,

1 tanO = --,, cot.,

1 sec 8 = --,, cos v

1 cotO = --,, tan v

Find sin() and tan(), given that cos() =

Pythagorean Identities sin2 8

+ cos2 8 1

tan2 8

= 1

+

1 1 3 tan()=--=--= - cot() -~ 2

2

+1

= sec2 8

sin2

()

+ cos2 () =

2

cot 8 = csc 8 sin

2 ()

+( .

sm2

1

Pythagorean identity

~ Y= 1 ()

+ -3

25

'{3 and sin() < 0.

Replace cos (J with

'{3.

v'3

= 1

Square 5 ·

22 sin2 () = 25

v'22

sin()= - - 5

3

Subtract 25 .

Choose the negative root.

160

CHAPTER 1 Trigonometric Functions

Concepts

Examples

Quotient Identities

To find tan (), use the values of sin () and cos() from the

sin (J - - = tan(J cos 0

cos (J ---:---- = cot (J SID 0

preceding page and the quotient identity tan () = ~~~ ~.

'v22 c:: 'v22 c:: sin() - -v22 5 5tan()= - - = - - - =- - - · - - = - - cos () v3 5

V3

5

Signs of the Trigonometric Functions x < 0, y > O, r > 0

y

= -

V3

Vii V3 V66 V3 . V3 = - - 3Simplify the complex fraction, and rationalize the denominator.

x > O,y > 0, r > 0

II

Sine and cosecant positive

All functions positive

---------+--------- x xO,yO

IV Cosine and secant positive

Identify the quadrant(s) of any angle () that satisfies sin () < 0, tan () > 0. Because sin () < 0 in quadrants III and IV , and tan () > 0 in quadrants I and III, both conditions are met only in quadrant III.

Review Exercises 1. Give the measures of the complement and the supplement of an angle measuring 35°.

Find the angle of least positive measure that is coterminal with each angle.

2. -5 1°

3. - 174°

4. 792°

Solve each problem. 5. Rotating Propeller The propeller of a speedboat rotates 650 times per min. Through how many degrees does a point on the edge of the propeller rotate in 2.4 sec? 6. Rotating Pulley A pulley is rotating 320 times per min. Through how many degrees does a point on the edge of the pulley move in ~ sec?

Convert decimal degrees to degrees, minutes, seconds, and convert degrees, minutes, seconds to decimal degrees. If applicable, round to the nearest second or the nearest thousandth of a degree.

7. 119° 08' 03"

8. 47° 25' 11"

9. 275. 1005°

Find the measure of each marked angle.

11.

12.

10. -61.5034°

CHAPTER 1 Review Exercises

161

14.

13.

m and n are parallel.

Solve each problem. 15. Length of a Road A camera is located on a satellite with its lens positioned at C in the figure. Length PC represents the distance from the lens to the film PQ, and BA represents a straight road on the ground. Use the measurements given in the figure to find the length of the road. (Data from Kastner, B., Space Mathematics, NASA.)

Not to scale

16. Express (} in terms of a and {3.

Find all unknown angle measures in each pair of similar triangles. 17.

p

~N

z

18.

M

86°

Q ~s

l~R

x

Find the unknown side lengths in each pair of similar triangles.

19.

q

20.

~

V , 50

16

In each figure, there are two similar triangles. Find the unknown measurement. Give any approximation to the nearest tenth. 21.

22.

11

7

23. Length of a Shadow If a tree 20 ft tall casts a shadow 8 ft long, how long would the shadow of a 30-ft tree be at the same time and place?

162

CHAPTER 1 Trigonometric Functions

Find the six trigonometric function values for each angle. Rationalize denominators when applicable.

24.

25.

y

26.

y

y

8

++--+.....-+--+-+'-+-+-i--. x (-2, 0) 0

Find the values of the six trigonometric functions for an angle in standard position having each given point on its terminal side. Rationalize denominators when applicable.

27. (3, -4)

28. (9, -2)

29. (-8, 15 )

30. (1, - 5)

31. ( 6\/3, -6)

32. ( -2Vz, 2V2)

An equation of the terminal side of an angle 0 in standard position is given with a restriction on x. Sketch the least positive such angle 0, and find the values of the six trigonometric functions of 0.

33. 5x - 3y = 0, x :=::: 0

34. y = - 5x, x

:5

35. 12x + 5y = 0, x :=::: 0

0

Complete the table with the appropriate function values of the given quadrantal angles.

If the value is undefined, say so. (J

36.

180°

37.

-90°

sin (J

cos (J

tan (J

cot (J

sec (J

csc (J

38. Concept Check If the terminal side of a quadrantal angle lies along the y-axis , which of its trigonometric functions are undefined?

Give all six trigonometric function values for each angle 0. Rationalize denominators when applicable.

39. cos 0 = - ~, and 0 is in quadrant III 41. sec 0 = -

Vs, and 0 is in quadrant II

43. sec 0 = ~, and 0 is in quadrant IV

40. sin 0 =

1,

and cos 0 < 0

42. tan 0 = 2, and 0 is in quadrant III 44. sin 0 = - ~, and 0 is in quadrant III

45. Determine whether each statement is possible or impossible. 2

(a)sec0=-3

(b)tanO=l.4

(c)cos0=5

46. Concept Check If, for some particular angle 0, sin 0 < 0 and cos 0 > 0, in what quadrant must 0 lie? What is the sign of tan O?

Solve each problem. 47. Swimmer in Distress A lifeguard located 20 yd from the water spots a swimmer in distress. The swimmer is 30 yd from shore and 100 yd east of the lifeguard. Suppose the lifeguard runs and then swims to the swimmer in a direct line, as shown in the figure. How far east from his original position will the lifeguard enter the water? (Hint: Find the value of x in the sketch.)

CHAPTER 1

Test

163

48. Angle through Which the Celestial North Pole Moves At present, the north star Polaris is located very near the celestial north pole. However, because Earth is inclined 23.5°, the moon 's gravitational pull on Earth is uneven. As a result, Earth slowly precesses (moves in) like a spinning top, and the direction of the celestial north pole traces out a circ ular path once every 26,000 yr. See the figure. For example, in approximately A.D. 14,000 the star Vega-not the star Polaris-will be located at the celestial north pole. As viewed from the center C of this circ ular path, calculate the angle (to the nearest second) through which the celestial north pole moves each year. (Data from Zeilik, M., S. Gregory, and E. Smith, Introductory Astronomy and Astrophysics, Second Edition, Saunders College Publishers.) 49. Depth ofa Crater on the Moon The depths of unknown craters on the moon can be approximated by comparing the lengths of their shadows to the shadows of nearby craters with known depths. The crater Aristillus is 11,000 ft deep, and its shadow was measured as 1.5 mm on a photograph. Its companion crater, Autolycus, had a shadow of 1.3 mm on the same photograph. Use similar triangles to determine the depth of the crater Autolycus to the nearest hundred feet. (Data from Webb, T., Celestial Objects for Common Telescopes, Dover Publications.) 50. Height of a Lunar Peak The lunar mountain peak Huygens has a height of 2 1,000 ft. The shadow of Huygens on a photograph was 2.8 mm, while the nearby mountain Bradley had a shadow of 1.8 mm on the same photograph. Calculate the height of Bradley. (Data from Webb, T., Celestial Objects for Common Telescopes, Dover Publications.)

Chapter 1

Test 1. Give the measures of the complement and the supplement of an angle measuring 67 °.

Find the measure of each marked angle. 2.

3.

4. (-3x+ 5)

6.

0

(32 - 2x)0

7.

(8x)

0

( 12x + 40)

m and n are parallel.

Perform each conversion. 8. 74° 18' 36" to decimal degrees 9. 45.2025° to degrees, minutes, seconds

0

164

CHAPTER 1 Trigonometric Funct ions

Solve each problem. 10. Find the angle of least positive measure that is coterminal with each angle. (b) -80°

(a) 390°

(c) 810°

11. Rotating Tire A tire rotates 450 times per min. Through how many degrees does a point on the edge of the tire move in 1 sec?

12. Length of a Shadow If a vertical pole 30 ft tall casts a shadow 8 ft long, how long would the shadow of a 40-ft pole be at the same time and place?

13. Find the unknown side lengths in this pair of similar triangles.

y

Sketch an angle () in standard position such that () has the least positive measure, and the given point is on the terminal side of O. Then find the values of the six trigonometric functions for the angle. If any of these are undefined, say so. 15. (0, -2)

14. (2, -7)

Work each problem. 16. Draw a sketch of an angle in standard position having the line with the equation 3x - 4y = 0, x :S 0, as its terminal side. Indicate the angle of least positive measure (), and find the values of the six trigonometric functions of e. 17. Complete the table with the appropriate function values of the given quadrantal angles. If the value is undefined, say so. ()

sin 8

cos ()

tan 8

cot 8

sec 8

csc ()

90° -360° 630° 18. If the terminal side of a quadrantal angle lies along the negative x-axis, which two of its trigonometric fu nction values are undefined?

19. Identify the possible quadrant(s) in which () must lie under the given conditions. (a) cos() > 0, tan()> 0

(b) sin()< 0, csc () < 0

(c) cot() > 0, cos() < 0

20. Determine whether each statement is possible or impossible.

(a) sin () = 1.5

(b) sec () = 4

21. Find the value of sec () if cos () =

(c) tan() = 10,000

-12.

22. Find the five remaining trigonometric function values of() if sin () = ~ and () is in quadrant II.

2

Acute Angles and Right Triangles

..~·

T •

Trigonometry is used in safe roadway design to provide sufficient visibility around curves as well as a smoothflowing, comfortable ride.

166

CHAPTER 2 Acute Angles and Right Triangles

2.1

Trigonometric Functions of Acute Angles

•

Right-Triangle-Based Definitions of the Trigonometric Functions

•

Cofunctions

•

How Function Values Change as Angles Change

•

Trigonometric Function Values of Special Angles

y

x

Right-Triangle-Based Definitions of the Trigonometric Functions Angles in standard position can be used to define the trigonometric functions. There is also another way to approach them: as ratios of the lengths of the sides of right triangles. Figure 1 shows an acute angle A in standard position. The definitions of the trigonometric function values of angle A require x, y, and r. As drawn in Figure 1, x and y are the lengths of the two legs of the right triangle ABC, and r is the length of the hypotenuse. The side of lengthy is the side opposite angle A , and the side of length x is the side adjacent to angle A. We use the lengths of these sides to replace x and y in the definitions of the trigonometric functions, and the length of the hypotenuse to replace r, to obtain the following right-triangle-based definitions. In the definitions , we use the standard abbreviations for the sine, cosine, tangent, cosecant, secant, and cotangent functions.

Right-Triangle-Based Definitions of Trigonometric Functions

x

Let A represent any acute angle in standard position.

Figure 1

y

side opposite A

sin A

=r =

cos A

=r =

x

side adjacent to A hypotenuse

tanA

= x = side adjacent to A

y

side opposite A

hypotenuse

r

hypotenuse

cscA

= y = side opposite A

sec A

= x = side adjacent to A

cot A

=y =

r

hypotenuse

x

side adjacent to A side opposite A

NOTE We will sometimes shorten wording like "side opposite A" to just "side opposite" when the meaning is obvious.

B

A~c7 24

Figure 2

.

Slll

EXAMPLE 1

Finding Trigonometric Function Values of an Acute Angle

Find the sine, cosine, and tangent values for angles A and B in the right triangle in Figure 2. The length of the side opposite angle A is 7, the length of the side adjacent to angle A is 24, and the length of the hypotenuse is 25.

SOLUTION

side opposite A = ----hypotenuse

side adjacent cosA=----hypotenuse

7 25

24 25

side opposite tanA=----side adjacent

7 24

The length of the side opposite angle B is 24, and the length of the side adjacent to angle Bis 7. sin B =

24 25

cos B =

7 25

tanB =

24

7

Use the right-triangle-based definitions of the trigonometric functions.

t/'

Now Try Exercise 7.

2.1 Trigonometric Functions of Acute Angles

167

NOTE The cosecant, secant, and cotangent ratios are reciprocals of the sine, cosine, and tangent values, respectively, so in Example 1 we have cscA csc B =

;~ A b

25 24

secB =

25 24

25

7

cotA and

24

=?

cot B =

7 . 24

Figure 3

.

Whenever we use A, B, and C to name angles in a right triangle, C will be the right angle.

Figure 3

sec A=

shows a right triangle with acute angles A and B and a right angle at C. The length of the side opposite angle A is a, and the length of the side opposite angle Bis b. The length of the hypotenuse is c. By the Also, cos B = ~c . Thus, we have the following. preceding definitions, sin A = ~. c Cofunctions

B

25

=?

smA Similarly, tan A

a

= b = cot B

= -ac = cosB and

secA =

c

b=

cscB.

In any right triangle, the sum of the two acute angles is 90°, so they are complementary . In Figure 3, A and Bare thus complementary, and we have established that sin A = cos B. This can also be written as follows. sin A = cos (90° - A)

B = 90° - A

This is an example of a more general relationship between cofunction pairs. sine, cosine tangent, cotangent secant, cosecant

I

Cofooohoo ' ' ' "

Cofunction Identities

For any acute angle A, the following hold true. sin A = cos(90° - A) sec A = csc(90° - A) tan A = cot(90° - A) cosA = sin(90° - A) cscA = sec(90° - A) cotA = tan(90° - A) The cofunction identities state the following.

Co/unction values of complementary angles are equal. EXAMPLE 2

Writing Functions in Terms of Cofunctions

Write each function in terms of its cofunction. (a) cos 52°

(b) tan 71°

(c) sec 24°

SOLUTION

(a)

Cofunctions

cos 52° = sin(90° - 52°) =sin 38°

cos A = sin (90° - A)

Complementary angles

(b) tan 71° = cot(90° - 71°) =cot 19°

(c) sec 24° = csc 66° t/' Now Try Exercises 25 and 27.

168

CHAPTER 2

Acute Angles and Right Triangles

EXAMPLE 3

Solving Equations Using Cofunction Identities

Find one solution for each equation. Assume all angles involved are acute angles. (a) cos( 0

+ 4°)

=

sin(30

+ 2°)

tan(W - 18°) = cot(O

(b)

+ 18°)

SOLUTION

(a) Sine and cosine are cofunctions, so cos( 0

+ 4°) = sin(30 + 2°) is true if

the sum of the angles is 90°.

( 0 + 4°) + (30 + 2°) 40

=

90°

+ 6° = 90° 40

=

84°

0 = 21 °

Complementary angles Combine like terms. Subtract 6° from each side. Divide by 4.

(b) Tangent and cotangent are cofunctions.

( 20 - 18°)

+ (0 + 18°)

=

90°

30 = 90° 0 = 30°

Complementary angles Combine like terms. Divide by 3.

tl"' Now Try Exercises 31 and 33.

How Function Values Change as Angles Change Figure 4 shows three right triangles. From left to right, the length of each hypotenuse is the same, but angle A increases in measure. As angle A increases in measure from 0° to 90°, the length of the side opposite angle A also increases.

A~~J

A ~y x

x

y

x

sin A = ~ r

As A increases, y increases. Because r is fixed, sin A increases.

Figure 4

In the ratio .

side opposite hypotenuse

smA= --~--

y r'

as angle A increases, the numerator of this fraction also increases, while the denominator is fixed. Therefore, sin A increases as A increases from 0° to 90°. As angle A increases from 0° to 90°, the length of the side adjacent to A decreases. Because r is fixed, the ratio ::'r: decreases. This ratio gives cos A , showing that the values of cosine decrease as the angle measure changes from 0° to 90°. Finally, increasing A from 0° to 90° causes y to increase and x to decrease, making the values of ~ = tan A increase. A similar discussion shows that as A increases from 0° to 90°, the values of sec A increase, while the values of cot A and csc A decrease.

2.1 Trigonometric Functions of Acute Ang les

169

Comparing Function Values of Acute Angles

EXAMPLE 4

Determine whether each statement is true or false. (b) sec 56° ~ sec 49°

(a) sin21°> sinl8° SOLUTION

(a) In the interval from 0° to 90°, as the angle increases, so does the sine of the

angle. This makes sin 21° > sin 18° a true statement. (b) For fixed r, increasing an angle from 0° to 90° causes x to decrease. There-

fore, sec (J =

!:. x

increases. The statement sec 56° ~ sec 49° is false. tl"

2 Equilateral triangle (a)

Trigonometric Function Values of Special Angles Certain special angles, such as 30°, 45°, and 60°, occur so often in trigonometry and in more advanced mathematics that they deserve special study. We start with an equilateral triangle, a triangle with all sides of equal length. Each angle of such a triangle measures 60°. Although the results we will obtain are independent of the length, for convenience we choose the length of each side to be 2 units. See Figure 5(a). Bisecting one angle of this equilateral triangle leads to two right triangles, each of which has angles of 30°, 60°, and 90°, as shown in Figure 5(b). An angle bisector of an equilateral triangle also bisects the opposite side. Thus the shorter leg has length 1. Let x represent the length of the longer leg.

22 = 12 + x2

Pythagorean theorem

4 = 1+x

Apply the exponents.

2

3 = x2

(b)

Figure 5

Subtract 1 from each side. Square root property; Choose the positive root.

V3=x 30°- 60° right triangle

Now Try Exercises 41 and 47.

summarizes our results using a 30° - 60° right triangle. As shown in the figure, the side opposite the 30° angle has length 1. For the 30° angle,

Figure 6

hypotenuse = 2,

side opposite = l ,

side adjacent = \/3.

Now we use the definitions of the trigonometric functions. side opposite sin 30° = - - - - hypotenuse side adj acent cos 30° = - - - - hypotenuse

Figure 6

tan

300

=

1 2 \/3 2

side opposite l l \/3 \/3 side adj acent = \/3 = \/3 . :\13 = - 3-

2 csc 30° = - = 2 1

Rationalize the denominators.

V3

0 2 2 2\/3 sec 30 = - - = - - · - - = - \/3 \/3\/3 3 cot 30° = \/3 = \/3 1

170

CHAPTER 2 Acute Angles and Right Triangles

Finding Trigonometric Function Values for 60°

EXAMPLE 5

Find the six trigonometric function values for a 60° angle. SOLUTION

.

sm60° =

Refer to

Figure 6

to find the following ratios.

2V3

cos 60° =

1

2

tan 60° =

I

Figure 6 (repeated)

2 2V3 V3

1

V3 l cot 60° = - - = - V3 3

2 sec 60° = - = 2

csc 60° = - - = - 3

V3 = V3

1

II" Now Try Exercises 49, 51, and 53.

NOTE The results in Example 5 can also be found using the fact that cofunction values of the complementary angles 60° and 30° are equal.

We find the values of the trigonometric functions for 45° by starting with a 45°- 45° right triangle, as shown in Fi gur e 7. This triangle is isosceles. For simplicity, we choose the lengths of the equal sides to be 1 unit. (As before, the results are independent of the length of the equal sides.) If r represents the length of the hypotenuse, then we can find its value using the Pythagorean theorem.

I

+

12 I 45 '- 45' right triangle

12 = r 2

Pythagorean theorem

2 = r2

Apply the exponents and add.

\/2 =

Figure 7

r

Choose the positive root.

Now we use the measures indicated on the 45°-45° right triangle in

\/2

1

sin 45° = - - = - \/2 2 csc 45° =

Figu re 7.

1 \/2 cos 45° = - - = - \/2 2

tan 45° = - = 1 1

\/2 = \/2

1 cot 45° = - = 1 1

\/2 = \/2

sec 45° =

1

1

1

Function values for 30°, 45°, and 60° are summarized in the table that follows . Function Values of Special Angles (J

sin (J

cos (J

30°

-

1

V3 -

45°

2

V2 -

V2 -

V3

-

2

60°

2

-

2

tan (J

cot (J

sec (J

csc (J

V3

--

2'\/3 3

2

1

1

V2

V2

V3

-V3

2

2'\/3 -

V3

-

3

2

1

2

3

3

NOTE To reproduce this table quickly, it is important to learn the values of sin 30°, sin 45°, and sin 60°. Then the rest of the table can be completed using the reciprocal, cofunction, and quotient identities.

171

2.1 Trigonometric Functions of Acute Angles

A calculator can find trigonometric function values at the touch of a key. Why then do we spend so much time finding values for special angles? We do this because a calculator gives only approximate values in most cases instead of exact values. A scientific calculator gives the following approximation for tan 30°. tan 30° = 0.57735027

"" means "is approximately equal to."

Earlier, however, we found the exact value.

V3

tan 30° = - - Exact value 3 ~ Figure 8 shows mode display options for the TI-84 Plus. Figure 9 displays the output when evaluating the tangent, sine, and cosine of 30°. (The calculator should be in degree mode to enter angle measure in degrees.)

O

HORHAL FLOAT AUTO REAL DEGREE HP CLASSIC SCI EHG

,.

tanC30) .5773502692

8123~56789

~ RA DIAH

I

sin.c3if>·······································

.

• PARAHETRI C POLAR SEQ DDT-THICK THIH OOT-THIH SIHUL

.5 c:c;5·c30»················ ······················· .............................. '·~RR!i!~~~!i!~~ .

o.+b \.. r•""C8\.)

HORIZOHTAL GRAPH-TABLE fRACTIOHTYPE•!i'l'l!l u n,d ftHSMERS : rll!ll!J DEC fRAC-APPROX GO TO 2HO fORHAT GRAPH:l:I!J YES STAT OiftGHOSTICS: l)]Il OH STAT MIZAROS: ' arr SET CLOCK •

Figure 8

•

Figure 9

;

Exercises

CONCEPT PREVIEW Match each trigonometric function in Column I with its value in

Column II. Choices may be used once, more than once, or not at all. I

II

1. sin 30°

2. cos 45°

A.

V3

B. 1

3. tan 45°

4. sec 60°

D.

V3 2

E.

5. csc 60°

G.

6. cot 30°

2

H.

1

c. 2

2V3 3

V2 2

F.

I.

Find exact values or expressions for sin A, cos A, and tan A. See Example I. 7.

21 ~ 20

8.

45 A A

9.

10. n

m

A

A V

V3 3

V2

172

CHAPTER 2 Acute Angles and Right Triangles

Suppose ABC is a right triangle with sides of lengths a, b, and c and right angle at C. B

A

~· b

C

Use the Pythagorean theorem to find the unknown side length. Then find exact values of the six trigonometric functions for angle B. Rationalize denominators when applicable. See Example 1. 11. a = 5, b = 12

12. a= 3, b = 4

13. a= 6, c = 7

14. b = 7, c = 12

lS. a = 3, c = 10

16. b = 8, c = 11

17. a= 1, c = 2

18. a =

Vl, c = 2

19. b = 2, c = 5

20. Concept Check Give the six cofunction identities. Write each function in terms of its cofunction. Assume all angles involved are acute angles. See Example 2. 21. cos 30°

22. sin 45°

23. csc 60°

24. cot 73°

2S. sec 39°

26. tan 25.4°

27. sin 38.7°

28. cos( 8 + 20°)

29. sec( 8 + 15°)

30. Concept Check With a calculator, evaluate sin(90° - 8) and cos 8 for various values of 8. (Check values greater than 90° and less than 0°.) Comment on the results. Find one solution for each equation. Assume all angles involved are acute angles. See Example 3. 31. tan a= cot( a+ 10°)

32. cos 8 = sin (28 - 30°)

33. sin(28 + 10°) = cos(38 - 20°)

34. sec(f3 + 10°) = csc(2{3 + 20°)

3S. tan(3B + 4°) =cot( SB - 10°)

36. cot(58

+ 2°)

38. cos(28

+ 50°)

+ 5°)

37. sin( 8 - 20°) = cos(28 39. sec(3{3

+

10°) = csc(f3

+ 8°)

40. csc(f3

+ 40°)

= tan(28

+ 4°)

= sin(28 - 20°)

= sec(f3 - 20°)

Determine whether each statement is true or false. See Example 4. ~

41. sin 50° > sin 40°

42. tan 28°

tan 40°

43. sin 46° < cos 46° (Hint: cos 46° = sin 44°)

44. cos 28° < sin 28° (Hint: sin 28° =cos 62°)

4S. tan41 ° sec 30°

48. csc 20° < csc 30°

Give the exact value of each expression. See Example 5. 49. tan 30°

so.

cot 30°

Sl. sin 30°

S2. cos 30°

S3. sec 30°

S4. csc 30°

SS. csc 45°

S6. sec 45°

S7. cos 45°

S8. cot 45°

S9. tan 45°

60. sin 45°

61. sin 60°

62. cos 60°

63. tan 60°

64. csc 60°

173

2. 1 Trigonometric Funct ions of Acute Ang les

Concept Check Work each problem.

65. What value of A between 0° and 90° will produce the output shown on the graphing calculator screen?

HIRMAL FLDllT FRAC REftL

0£~REE

M,

a

J3

T

siii"IA'i· · ················ '·~!i!i!~~~!i!~!!. ................................!!§§!i!7.~~!i!~!!.

66. A student was asked to give the exact value of sin 45°. Using a calculator, he gave the answer 0.707 1067812. Explain why the teacher did not give him credit. 67. Find the equation of the line that passes through the origin and makes a 30° angle with the x-axis. 68. Find the equation of the Une that passes through the origin and makes a 60° angle with the x-axis. 69. What angle does the Une y =

v'3x make with the positive x-axis?

70. What angle does the Une y =

~ x make with the positive x-axis?

71. Consider an equilateral triangle with each side having length 2k.

(a) What is the measure of each angle? (b) Label one angle A. Drop a perpendicular from A to the side opposite A. Two 30° angles are formed at A, and two right triangles are formed. What is the length of the sides opposite the 30° angles?

2k

(c) What is the length of the perpendicular in part (b)? (d) From the results of parts (a)-(c), complete the following statement:

In a 30°-60° right triangle, the hypotenuse is always _ _ times as long as the shorter leg, and the longer leg has a length that is _ _ times as long as that of the shorter leg. Also, the shorter leg is opposite the _ _ angle, and the longer leg is opposite the _ _ angle. 72. Consider a square with each side of length k.

k

(a) Draw a diagonal of the square. What is the measure of each

angle formed by a side of the square and this diagonal? (b) What is the length of the diagonal?

(c) From the results of parts (a) and (b), complete the following statement:

k

In a 45°-45° right triangle, the hypotenuse has a length that is _ _ times as Jong as either leg.

Find the exact value of the variables in each figure.

73.

74.

x a

d

76.

75.

7

15

p

m

174

CHAPTER 2 Acute Angles and Right Triangles

Find a formula for the area of each figure in terms of s.

77.

78.

~ 79. With a graphing calculator, find the coordinates of the point of intersection of the

graphs of y = x and y = ~- These coordinates are the cosine and sine of what angle between 0° and 90°? 80. Concept Check Suppose we know the length of one side and one acute angle of a 30°-60° right triangle. Is it possible to determine the measures of all the sides and angles of the triangle?

Relating Concepts For individual or collaborative investigation (Exercises 87-84) y

The figure shows a 45° central angle in a circle with radius 4 units. To find the coordinates of point P on the circle, work Exercises 81-84 in order. 81. Sketch a line segment from P perpendicular to the x-axis. 82. Use the trigonometric ratios for a 45° angle to label the sides of the right triangle sketched in Exercise 81. y

83. Which sides of the right triangle give the coordinates of point P? What are the coordinates of P? 84. The figure at the right shows a 60° central angle in a circle of radius 2 units. Follow the same procedure as in Exercises 81- 83 to find the coordinates of Pin the figure.

2.2

Trigonometric Functions of Non-Acute Angles

•

Reference Angles

•

Special Angles as Reference Angles

•

Determination of Angle Measures with Special Reference Angles

Reference Angles Associated with every nonquadrantal angle in standard position is an acute angle called its reference angle. A reference angle for an angle 8, written 8', is the acute angle made by the terminal side of angle 8 and the x-axis.

NOTE

Reference angles are always positive and are between 0° and 90°.

Figure 10 shows several angles 8 (each less than one complete counterclockwise revolution) in quadrants II, III, and IV, respectively, with the reference angle 8' also shown. In quadrant I, 8 and 8' are the same. If an angle 8 is negative or has measure greater than 360°, its reference angle is found by first finding its coterrninal angle that is between 0° and 360°, and then using the diagrams in Figure 10.

2.2 Trigonometric Functions of Non-Acute Angles

y

IJ in quadrant II

y

y

IJ in quadrant lll

IJ in quadrant IV

175

Figure 10

tMJQICtl~I

A common error is to find the reference angle by using the termjnal side of fJ and the y-axis. The reference angle is always found with reference to the x-axis.

EXAMPLE 1

Finding Reference Angles

Find the reference angle for each angle. (a) 2 18°

(b) 1387°

SOLUTION

(a) As shown in Figure 11 (a), the positive acute angle made by the terminal side of this angle and the x-axis is 218° - 180° = 38°. For fJ = 218°, the reference angle fJ' = 38°. y

21s0

-

y

1so0 = 38°

360° - 307° = 53°

(a)

(b)

Figure 11

(b) First find a coterminal angle between 0° and 360°. Divide 1387° by 360° to

obtain a quotient of about 3.9. Begin by subtracting 360° three times (because of the whole number 3 in 3.9). 1387° - 3 . 360° =

1387° - 1080°

Multiply.

=

307°

Subtract.

The reference angle for 307° (and thus for 1387°) is 360° - 307° = 53°. See Figure 11 (b).

.I" Now Try Exercises 5 and 9.

176

CHAPTER 2 Acute Ang les and Right Triangles

The preceding example suggests the following diagram. Reference Angle (} ' for (}, where 0° Q II

Y

e' = 1so

Q III

Y

e' =e- 1so

0 -

< (}
0°, and the safe stopping distance S for a given speed limit. In addition, Lis dependent on the vertical alignment of the headlights. Headlights are usually pointed upward at a slight angle a above the horiwntal of the car. Using these quantities, for a 55 mph speed limit, L can be modeled by the formula ( 02 - 0 I )s2

L =

200(h +S tan a) '

where S < L. (Data from Mannering, F., and W Kilareski, Principles of Highway Engineering and Traffic Analysis, Second Edition, John Wiley and Sons.)

91. Compute length L, to the nearest foot, if h s = 336 ft. 92. Repeat Exercise 91 with a = 1.5°.

= 1.9 ft, a= 0.9°, 0 1 = -3°, 02 = 4°, and

CHAPTER 2

Chapter 2

Quiz

Quiz

189

(Sections 2.1-2.3)

Solve each problem. 1. Find exact values of the six trigonometric functions for angle A in the right triangle.

~24 32

A

2. Complete the table with exact trigonometric function values. (J

sin (J

cos (J

tan (J

sec (J

cot (J

csc (J

30° 45° 60° 3. Find the exact value of each variable in the figure.

x

y

4. Area of a Solar Cell A solar cell converts the energy of sunlight directly into electrical energy. The amount of energy a cell produces depends o n its area. Suppose a solar cell is hexagonal, as shown in the fi rst fi gure o n the right. Express its area .stl in terms of sin 8 and any side x. (Hint: Consider one of the six equilateral triangles from the hexagon. See the second fi gure on the right.) (Data from Kastner, B., Space Mathematics, NASA.)

Find exact values of the six trigonometric functions for each angle. Rationalize denominators when applicable.

s.

135°

6. - 150°

Find all values of 8,

7. 1020°

if 8 is in the interval [0°, 360°) and has the given function value.

V3

8. sin8 = - 2

9. sec 8 =

-v2

Use a calculator to approximate the value of each expression. Give answers to six decimal places. 10. sin 42° 18'

11. sec (-212° 12')

Find a value of 8 in the interval [ 0°, 90°) that satisfies each statement. Write each answer in decimal degrees to six decimal places. 12. tan 8 = 2.6743210

13. csc 8 = 2.3861147

Determine whether each statement is true or false. 14. sin( 60° + 30°) = sin 60° + sin 30°

15. tan (90° - 35°) = cot 35°

190

CHAPTER 2 Acute Angles and Right Triangles

2.4

Solutions and Applications of Right Triangles

•

Historical Background

•

Significant Digits

•

Solving Triangles

•

Angles of Elevation or Depression

Historical Background The beginnings of trigonometry can be traced back to antiquity. Figure 19 shows the Babylonian tablet Plimpton 322, which provides a table of secant values. The Greek mathematicians Hipparchus and Claudius Ptolemy developed a table of chords, wh ich gives values of sines of angles between 0° and 90° in increments of 15 minutes. Until the advent of scientific calculators in the late 20th century, tables were used to find function values that we now obtain with the stroke of a key. Applications of spherical trigonometry accompanied the study of astronomy for these ancient civilizations. Until the mid-20th century , spherical trigonometry was studied in undergraduate courses. See Figure 20. An introduction to applications of the plane trigonometry studied in this text involves applying the ratios to sides of objects that take the shape of right triangles. PLANE TRJGONOMETRY

----

ALLV' A'-D •"CO:f

Plimpton 322 Figure 19

Significant Digits ' A number that represents the result of counting, or a number that results from theoretical work and is not the result of measurement, is an exact number. There are 50 states in the United States. In this statement, 50 is an exact number. Most values obtained for trigonometric applications are measured values that are not exact. Suppose we quickly measure a room as 15 ft by 18 ft. See Figure 21. To calculate the length of a diagonal of the room, we can use the Pythagorean theorem.

;;;;

d ;; ;;; ;;

;

18 ft

.

Figure 21

Figure 20

15 ft

d 2 = 15 2

d2= d=

+ 182

549

Vs49

d = 23.430749

Pythagorean theorem Apply the exponents and add. Square root property; Choose the positive root.

Should this answer be given as the length of the diagonal of the room? Of course not. The number 23.430749 contains six decimal places, while the original data of 15 ft and 18 ft are accurate only to the nearest foot. In practice, the results of a calculation can be no more accurate than the least accurate number in the calculation. Thus, we should indicate that the diagonal of the 15-by18-ft room is approximately 23 ft.

2.4 Solutions and Applications of RightTriangles

191

If a wall measured to the nearest foot is 18 ft long, this actually means that the wall has length between 17 .5 ft and 18.5 ft. If the wall is measured more accurately as 18.3 ft long, then its length is really between 18.25 ft and 18.35 ft. The results of physical measurement are only approximately accurate and depend on the precision of the measuring instrument as well as the aptness of the observer. The digits obtained by actual measurement are significant digits. The measurement 18 ft is said to have two significant digits; 18.3 ft has three significant digits. In the following numbers, the significant digits are identified in color.

408

21.5

18.00

6.700

0.0025

0.09810

7300

Notice the following. •

18.00 has four significant digits. The zeros in this number represent measured digits accurate to the nearest hundredth.

•

The number 0.0025 has only two significant digits, 2 and 5, because the zeros here are used only to locate the decimal point.

•

The number 7300 causes some confusion because it is impossible to determine whether the zeros are measured values. The number 7300 may have two, three, or four significant digits. When presented with this situation, we assume that the zeros are not significant, unless the context of the problem indicates otherwise.

To determine the number of significant digits for answers in applications of angle measure, use the following table. Significant Digits for Angles

Angle Measure to Nearest

Examples

Write Answer to This Number of Significant Digits

Degree

62°, 36°

two

Ten minutes, or nearest tenth of a degree

52° 30', 60.4°

three

Minute, or nearest hundredth of a degree

81 ° 48', 71.25°

four

Ten seconds, or nearest thousandth of a degree

10° 52' 20", 21.264°

five

To perform calculations with measured numbers , start by identifying the number with the least number of significant digits. Round the final answer to the same number of significant digits as this number. Remember that the answer is no more accurate than the least accurate number in the calculation. Solving Triangles To solve a triangle means to find the measures of all the angles and sides of the triangle. As shown in Figure 22, we use a to represent the length of the side opposite angle A , b for the length of the side opposite angle B, and so on. In a right triangle, the letter c is reserved for the hypotenuse. B

When we are solving triangles, a labeled sketch is an important aid.

Figure 22

192

CHAPTER 2 Acute Angles and Right Triangles

Solving a Right Triangle Given an Angle and a Side

EXAMPLE 1

Solve right triangle ABC, if A = 34° 30' and c = 12.7 in. B

a

A

b

SOLUTION To solve the triangle, find the measures of the remaining sides and angles. See Figure 23. To find the value of a , use a trigonometric function involving the known values of angle A and side c. Because the sine of angle A is given by the quotient of the side opposite A and the hypotenuse, use sin A.

a sin A = c

c

Figure 23

.

sm A =

a sin 34° 30' = - 12.7

side opposite hypotenuse

A = 34° 30 ', c = 12.7

a= 12.7 sin 34° 30'

Multiply by 12.7 and rewrite.

a= 12.7 sin 34.5°

Convert to decimal degrees.

a = 12.7(0.56640624)

Use a calculator.

a = 7 .19 in.

Three significant digits

Assuming that 34° 30' is given to the nearest ten minutes, we rounded the answer to three significant digits. To find the value of b, we could substitute the value of a just calculated and the given value of c in the Pythagorean theorem. It is better, however, to use the information given in the problem rather than a result just calculated. If an error is made in finding a, then b also would be incorrect. And, rounding more than once may cause the result to be less accurate. To find b, use cos A. Cos A cos 34° 30'

= ~

side adjacent COS A =

C

b

hypotenuse

A = 34° 30 ', c = 12.7

= --

12.7

b = 12.7 cos 34° 30'

Multiply by 12.7 and rewrite.

b = 10.5 in.

Three significant digits

Once b is found, the Pythagorean theorem can be used to verify the results. All that remains to solve triangle ABC is to find the measure of angle B.

LOOKING AHEAD TO CALCULUS The derivatives of the parametric equations x = f (t) and y = g(t) often

A

+B

=

90°

A and B are comple mentary angles.

34° 30'

+B

=

90°

A = 34° 30'

B = 89° 60' - 34° 30'

Rewrite 90°. Subtract 34° 30 ' .

B = 55° 30'

Subtract degrees and minutes separately.

II' Now Try Exercise 25.

represent the rate of change of physical quantities, such as velocities. When x and y are related by an equation, the derivati ves are r elated rates because a change in one causes a related change in the other. Detennining these rates in calculus often requires solving a right triangle.

NOTE In Example 1, we could have found the measure of angle B first and then used the trigonometric function values of B to find the lengths of the unknown sides. A right triangle can usually be solved in several ways, each producing the correct answer.

To maintain accuracy, always use given information as much as possible, and avoid rounding in intermediate steps.

2.4 Solutions and Applications of RightTriangles

193

Solving a Right Triangle Given Two Sides

EXAMPLE 2

Solve right triangle ABC, if a = 29.43 cm and c = 53.58 cm.

• • 29A3

~ ts:_~ C

We draw a sketch showing the given information, as in One way to begin is to find angle A using the sine function. SOLUTION

B

b Figure 24

A

.

a

A= c

Slll

S

in A

= side opposite hypotenuse

29.43 sin A= 53.58

a = 29.43, c = 53.58

sin A = 0.5492721165

Use a calculator.

A = sin-

=

A

1

( 0.5492721165)

33.32°

Figure 24.

Use the inverse sine function . Four significant digits 33.32° = 33° + 0.32( 60')

A = 33° 19'

The measure of Bis approximately 90° -33° 19' =56° 41 ' .

90° =89° 60 '

We now find b using the Pythagorean theorem.

+ b2 = c 2 29.43 2 + b 2 = 53.58 2 a2

2

b = 53.58

2

Pythagorean theorem

a= 29.43, c = 53.58 -

29.43

b = V2004.6915 b = 44.77 cm

2

Subtract 29.432 . Simplify on the right; square root property

Choose the positive square root.

t/'

Now Try Exercise 35.

Angles of Elevation or Depression In applications of right triangles, the angle of elevation from point X to point Y (above X) is the acute angle formed by ray XY and a horizontal ray with endpoint at X. See Figure 25(a). The angle of depression from point X to point Y (below X) is the acute angle formed by ray XY and a horizontal ray with endpoint X. See Figure 25(bJ.

x

Horizontal (b)

(a) Figure 25

t411iim~I

Be careful when interpreting the angle of depression. Both the angle of elevation and the angle of depression are measured between the line of sight and a horizontal line.

194

CHAPTER 2 Acute A ngles and Right Triangles

To solve applied trigonometry problems, follow the same procedure as solving a triangle. Drawing a sketch and labeling it correctly in Step 1 is crucial.

Solving an Applied Trigonometry Problem

Step 1 Draw a sketch, and label it with the given information. Label the quantity to be found with a variable. George Polya (1887-1985) Polya, a native of Budapest, Hungary, wrote more than 250 papers and a number of books. He proposed a general outline for solving applied problems in his classic book How to Solve It.

Step 2 Use the sketch to write an equation relating the given quantities to the variable. Step 3 Solve the equation, and check that the answer makes sense.

EXAMPLE 3

Finding a Length Given the Angle of Elevation

At a point A, 123 ft from the base of a flagpole, the angle of elevation to the top of the flagpole is 26° 40'. Find the height of the flagpole.

A

~ 123 ft

SOLUTION

Step 1 See Figure 26. The length of the side adjacent to A is known, and the length of the side opposite A must be found. We will call it a. Step 2 The tangent ratio involves the given values. Write an equation.

Figure 26

side opposite tanA= - - - - side adjacent

a tan 26° 40' = - 123 Step3

Tangent ratio

A = 26° 40 '; side adjacent = 123

a= 123 tan 26° 40 '

Multiply by 123 and rewrite.

a = 123(0.50221888)

Use a calculator.

a = 61.8 ft

Three significant digits

The height of the flagpole is 61.8 ft.

EXAMPLE 4

t/' Now Try Exercise 53.

Finding an Angle of Depression

From the top of a 210-ft cliff, David observes a lighthouse that is 430 ft offshore. Find the angle of depression from the top of the cliff to the base of the lighthouse. As shown in Figure 27, the angle of depression is measured from a horizontal line down to the base of the lighthouse. The angle of depression and angle B, in the right triangle shown, are alternate interior angles whose measures are equal. We use the tangent ratio to solve for angle B.

SOLUTION

tan B =

210 430

Figure 27

B = tan- 1( B = 26°

Tangent ratio

210 ) 430

Use the inverse tangent function. Two significant digits

t/' Now Try Exercise 55.

2.4 Solutions and Applications of RightTriangles

195

;

Exercises

CONCEPT PREVIEW Match each equation in Column I with the app ropriate right tri-

angle in Column II. In each case, the goal is to find the value of x.

II

I

1. x = 5 cot 38°

A.

2. x = 5 cos 38° 3. x = 5 tan 38° 4. x = 5 csc 38°

c.

B.

,~ 5~ ,~ x

5

E.

D.

~ ~

5. x = 5 sin 38° 6. x = 5 sec 38°

x

F.

5~

5

Concept Check Refer to the discussion of accuracy and significant digits in this section to answer the f ollowing.

7. Lake Ponchartrain Causeway The world's longest bridge over a body of water (continuous) is the causeway that joins the north and south shores of Lake Ponchartrain, a saltwater lake that lies north of New Orleans, Louisiana. It consists of two parallel spans. The longer of the spans measures 23.87 mi. State the range represented by thi s number. (Data from www.worldheritage.org) 8. Mt. Everest When Mt. Everest was first surveyed, the surveyors obtained a height of 29,000 ft to the nearest foot. State the range represented by this number. (The surveyors thought no one would believe a measurement of 29,000 ft, so they reported it as 29,002.) (Data from Dunham, W., The Mathematical Universe, John Wiley and Sons.) 9. Vehicular Tunnel The E. Johnson Memorial Tunnel in Colorado, which measures 8960 ft, is one of the longest land vehicular tunnels in the United States. What is the range of this number? (Data from The World Almanac and Book of Facts.) 10. WNBA Scorer Women's National Basketball Association player Breanna Stewart of the Seattle Storm was the WNBA's top scorer for the 201 8 regular season, with 742 points. Is it appropriate to consider this number between 741.5 and 742.5? Why or why not? (Data from www.wnba.com) 11. If h is the actual height of a building and the height is measured as 58.6 ft, then lh - 58.6 1 $ - - · 12. If w is the actual weight of a car and the weight is measured as 1542 lb , then lw- 1542 1 ,,; - - · Solve each right triangle. When two sides are given, give angles in degrees and minutes. See Examples 1 and 2.

13.

a

c

y

14.

B

b

A

:i 1

x

47.8° y

z

196

CHAPTER 2 Acute Angles and Right Triangles

15.

16.

M

B

~359km A

17.

18.

A

C

b

68.5 142°

I

B

·~

LJ56851 0, follow these steps.

To graph y

Step 1 Find the period,

2 ; .

Start at 0 on the x-axis, and lay off a distance of

2 ; .

Step 2 Divide the interval into four equal parts. (See the Note preceding Example 2.)

Step 3 Evaluate the function for each of the five x-values resulting from Step 2. The points will be maximum points, minimum points, and x-intercepts.

Step 4 Plot the points found in Step 3, and join them with a sinusoidal curve having amplitude Ia I. Step 5 Draw the graph over additional periods as needed.

EXAMPLE 4

Graphing y = a sin bx

Graph y = -2 sin 3x over one period using the preceding guidelines. SOLUTION

Step 1 For this function, b

=

over the interval [ 0,

3, so the period is

2 ; . The

function will be graphed

2 ;].

Step 2 Divide the interval [ 0,

2 ; ]

into four equal parts to obtain the x-values

d 21T 0 ,6, 3•2, an 3 · 1T

1T

1T

Step 3 M ake a table of values determined by the x-values from Step 2. y 2-rr

x

0

"'6

"'3

"'2

3x

0

2"'

'TT

2 'TT

sin3x

0

I

0

T -]

-2 sin 3x

0

-2

0

2

0

3

3-rr

2

0 - I

y = - 2 sin 3r

-2

Figure 9

Step4 Plot the points (0, 0), (~ ,

-2), (I,o), (I,2), and (2; ,o) , and join

them with a sinusoidal curve having amplitude 2. See Figure 9.

Step 5 The graph can be extended by repeating the cycle. Notice that when a is negative, the graph of y across the x-axis of the graph of y = Ia I sin bx.

= a sin bx is a reflection r/ Now Try Exercise 29.

4.1 Graphs of the Sine and Cosine Funct ions

267

Graphing y = a cos bx (Where b Is a Multiple of 11)

EXAMPLE 5

Graph y = - 3 cos 1TX over one period. SOLUTION 2 ;

Step 1 Here b = 1T and the period is the interval [ 0, 2 J.

=

2, so we will graph the function over

Step 2 Dividing [ 0, 2 J into four equal parts yields the x-values 0,

k, 1, ~ , and 2.

Step 3 Make a table using these x-values. x

y

J =-3

COS 1TX

2

7TX

0

COS 7TX -3COS 1TX

0

2

-I

I

0

1

2 7r

3

2

2

37r

2

1T

1

0

-1

0

1

-3

0

3

0

-3

T

21T

Step 4 Plot the points (0, -3) , (k, 0 ), (1, 3) , (~, 0 ), and (2, -3), and join them with a sinusoidal curve having amplitude

I- 3 I =

3. See Figure 10.

-2

Step 5 The graph can be extended by repeating the cycle. -3

Figure 10

Notice that when b is an integer multiple of TT , the first coordinates of the x-intercepts of the graph are rational numbers. tl" Now Try Exercise 37. Connecting Graphs with Equations EXAMPLE 6

Determining an Equation for a Graph

Determine an equation of the form y = a cos bx or > 0, for the given graph.

Y

y = a sin bx, where b

SOLUTION This graph is that of a cosine function that is reflected across its horizontal axis, the x-axis. The amplitude is half the distance between the maximum and minimum values. 1

- [2 - (-2)] 2

1

- (4) 2

=

=

2

The amplitude

lal is 2.

Because the graph completes a cycle on the interval [ 0, 41T J, the period is 41T. We use this fact to solve for b. 21T

41T = -

b

41Tb = 27T b=

21

27' Pen.od = b

Multiply each side by b. Divide each side by 41T.

An equation for the graph is 1 y = -2 cos - x.

J

x-axis reflection

2

""" Horizontal stretch

tl" Now Try Exercise 41.

268

CHAPTER 4 Graphs of t h e Circular Functions

A Trigonometric Model Sine and cosine functions may be used to model many real-l ife phenomena that repeat their values in a cyclical, or periodic, manner. Average temperature in a certain geographic location is one such example.

Interpreting a Sine Function Model

EXAMPLE 7

The average temperature (in °F) at Mould Bay, Canada, can be approximated by the function

f(x)

=

34 sin [ : (x - 4.3)

l

where x is the month and x = 1 corresponds to January, x = 2 to February, and so on. (a) To observe the graph over a two-year interval, graph f in the window [O, 25 ] by [ -45, 45 ]. (b) According to this model, what is the average temperature during the month of May? (c) What would be an approximation for the average annual temperature at Mould Bay? SOLUTION (a)

The graph of f(x) = 34 sin [ ~ (x - 4.3) ] is shown in Figure 11. Its amplitude is 34, and the period is 27T

-

'!!_

6

HORHAL FLOAT AUTO REAL RAOIAH HP 45 V1=3~Sin 0 and down Ic I units if c < 0. The horizontal translation (phase shift) is to the right d units if d > 0 and to the left Id units if d < 0. J

EXAMPLE 5

Graphing y = c + a sin [b(x -

Graph y = -1

+ 2 sin( 4x + 'TT) over two periods.

SOLUTION We use Method 1. We first write the expression on the right side of the equation in the form c + a sin [ b(x - d) J.

y = -1

+ 2 sin( 4x + 'TT),

y = -1

or

+ 2 sin [ 4 ( x + : ) ]

Rewrite by factoring out 4.

Step 1 Find an interval whose length is one period. 0

:S

0

$

4( x

+: ) 1T

x+ -

:S

'TT $-

4

'TT --$

Three-part inequality

Divide each part by 4.

2 'TT $-

x

4

2'Tr

Subtract ~ from each part.

4

Step 2 Divide the interval [ - ~ , ~ ] into four equal parts to obtain these x-values. 'TT

4'

'TT

8,

'TT

O,

8,

'TT

4

Key x-values

278

CHAPTER 4 Graphs of the Circu lar Functions

Step 3 Make a table of values. y

1T

x x+~

4(x

+ ~)

sin[4{x

+

2sin[4{x y

- 1

=-1+2 sin(4x+ 1T)

+

~)]

+

~) ]

2 sin ( 4x

+

rr)

1T

- 4

- 3

0

0

1T

1T

8

4

1T

1T

8

4

3rr

1T

2

8 3rr

0

1T

2

1T

T

21T

0

1

0

- I

0

0

2

0

-2

0

- 1

I

- 1

-3

- 1

Figure 18

Steps 4 and 5 Plot the points found in the table and join them with a sinusoidal curve. Figure 18 shows the graph, extended to the right and left to include two full periods. II" Now Try Exercise 59.

A Trigonometric Model For natural phenomena that occur in periodic patterns (such as seasonal temperatures, phases of the moon , or heights of tides), a sinusoidal function will provide a good approximation of a set of data points.

ModelingTemperature with a Sine Function

EXAMPLE 6

The maximum average monthly temperature in New Orleans, Louisiana, is 83°F, and the minimum is 53°F. The table shows the average monthly temperatures. The scatter diagram for a two-year interval in Figure 19 strongly suggests that the temperatures can be modeled with a sine curve.

Month

OF

Jan

Apr

53 57 63 69

May

77

Nov

June

82

Dec

Feb Mar

Month

OF

July

83 83 80 71 63 56

Aug Sept Oct

Data from National Cli matic Data Center.

HDRHAL FLOAT AUTO REAL DEGREE HP

a

85 D

D

D

D

0 ..... .. . . . . . ... .... ... ... 25 50

Figure 19

(a) Using only the maximum and minimum temperatures, determine a function of the form

f(x)

=

a sin [ b(x - d) ] + c,

where a, b, c, and dare constants,

that models the average monthly temperature in New Orleans. Let x represent the month, with January corresponding to x = 1. (b) On the same coordinate axes, graph f for a two-year period together with the actual data values found in the table. (c) Use the sine regression feature of a graphing calculator to determine a second model for these data.

4.2 Translations of the Graphs of the Sine and Cosine Funct ions

279

SOLUTION

(a) We use the maximum and minimum average monthly temperatures to find

the ampJjtude a.

a=

83 - 53 = 15 2

Amplitude a

For c, we find the average of the maximum and minimum temperatures.

c = 83

+ 53

= 68

Vertical translation c

2

Because temperatures repeat every 12 months, b can be found as follows. HD'-t1flL FlQflT JllUTO llUL P.fllOiflH NP

D 21T

12= - b ,

0 . . . . . . . . . . . . . . . . . . . . . 25 50

Figure 20

HD'-t1fll rIX2 flUTI lt[fll ltflOiflH HP

so

1T

b= 6

Solve for b.

The coldest month is January, when x = 1, and the hottest month is July, when x = 7. A good choice ford is 4 because April, when x = 4, is located at the midpoint between January and July. Also, notice that the average monthly temperature in April is 69°F, which is very close to the value of the vertical translation, c. The average monthly temperature in New Orleans is modeled closely by the following eq uation.

D

f(x ) =asin [b(x-d)J +c

~

'='=a•sin(b>e+c1+d a=lS .14 b=.52 c= -2.11

f(x) = 15 sin [ ~ (x - 4)] + 68

d=69 . 67

Substitute for a, b, c, and d.

(b) Figure 20 shows two iterations of the data points from the table, along with

the graph of y = 15 sin [ ~(x - 4) ]

(a) HDltt1fll nx2 flUTI lt[fll ltflOiflH HP

0

SS Y&sl5.l'tllt(t.5Vl:-Ul>•H.•7

+ 68. The graph of y = 15 sin ~x + 68

is shown for comparison. (c) We used the given data for a two-year period and the sine regression capability of a graphing calculator to produce the model described in Figure 21(a).

0 ....... .. . . . . . . . . . . . . 25 50

(b)

f(x) = 15.14 sin(0.52x - 2.11) + 69.67 Its graph, along with the data points, is shown in

Figure 21(b).

t/"

Figure 21

Now Try Exercise 63.

;

Exercises CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.

1. The graph of y = sin (x

+ ~) is obtained by shifting the graph of y = sin x

__

unit(s) to the _ _ _ __ (right/left) 2. The graph of y = cos(x - ~) is obtained by shifting the graph of y =cos x _ _ unit(s) to the _ _ _ __ (right/ left)

280

CHAPTER 4 Graphs of the Circular Functio ns

3. The graph of y = 4 sin x is obtained by stretching the graph of y = sin x vertically by a factor of _ _ . 4. The graph of y = - 3 sin x is obtained by stretching the graph of y = sin x by a factor of _ _ and reflecting across the _ _ -axis.

5. The graph of y = 6 + 3 sin x is obtained by shifting the graph of y = 3 sin x _ _ unit(s) _ _ _ __ (up/ down) 6. The graph of y = - 5 + 2 cos x is obtained by shifting the graph of y = 2 cos x _ _ unit(s) _ _ _ __ (up/down) 7. The graph of y = 3 + 5 cos ( x + ~) is obtained by shifting the graph of y = cos x horizontally _ _ unit(s) to the

, stretching it vertically by a factor of (right/left) _ _ ,and then shifting it vertically _ _ unit(s) _ _ _ __ (up/down) 8. The graph of y = -2

+ 3 cos ( x - ~) is obtained by shifting the graph of y

= cos x

horizontally _ _ unit(s) to the _ _ _ _ _ , stretching it vertically by a factor of (right/ left) _ _ ,and then shifting it vertically _ _ unit(s) _ _ _ __ (up/down) Concept Check Match each function with its graph in choices A- I. (One choice will not be used.)

12. y = cos ( x

+ ~)

13. y = 1 + sin x

15. y = 1 + cos x

A.

)'

14. y = -1

+ sin x

16. y = - 1 + cos x

B.

c.

)'

y

2 --1~----l----'-- X

-0 4----if--"-..«.~ 2l--~ x

-1

'IT

"

0

- I

-2

7r

w

17. The graphs of y =sin x this is so.

2w

2w

+ I and y = sin(x + l ) are NOT the same. Explain why

4.2 Translations of the Graphs of the Sine and Cosine Functions

281

18. Concept Check Which one of the two graphs, y = sin x + 1 or y = sin(x + 1), is the same as the graph of y = 1 + sin x?

Concept Check Match each function in Column I with the appropriate description in Column II. I

II

= 2, period = I, phase shift = ~ amplitude = 3, period = 7r, phase shift = 2 2 amplitude = 4, period = ; , phase shift = ~

19. y = 3 sin(2x - 4)

A. amplitude

20. y = 2 sin(3x - 4 )

B.

21. y = - 4 sin(3x - 2)

C.

22. y = -2 sin(4x - 3)

D. amplitude

= 2, period = 2; , phase shift = 1

Concept Check Fill in each blank with the word right or the word left.

23. If the graph of y = cos x is translated horizontally to the ____ I units, it will coincide with the graph of y = sin x. 24. If the graph of y = sin x is translated horizontally to the ____ coincide with the graph of y = cos x.

I units, it will

Connecting Graphs with Equations Each function graphed is of the form y = c + cos x, y = c + sin x, y = cos(x - d), or y = sin(x - d), where dis the least possible positive value. Determine an equation of the graph.

25.

26.

y

27.

y

y

2

2 2 x

0 -I

- I

-3

-2

29.

y 2

-I

x

0

-2

28.

x

0 21T

-2

30.

y

y

2

x

x

x

0 -I

-2

-2

-2

For each function, give the amplitude, period, vertical translation, and phase shift, as applicable. See Examples 1-5.

31. y

= 2 sin(x +

33. y

7r) = - 41 cos 2x + 2

35. y

= 3 cos [

37. y

=2 -

7r)

(1

f ( DJ x-

sin ( 3x -

~)

f)

32. y

= 3 sin ( x +

34. y

. = -21 sm

36. y

= -cos [ 7r ( x -

(1

1

2x + 7r

)

DJ

38. y= - 1 + - cos(2x-37r ) 2

282

CHAPTER 4 Graphs of the Circular Functions

Graph each function over a two-period interval. See Examples 1and2.

41. y = sin ( x +

42. y = cos ( x +

~)

~)

43. y = 2 cos ( x -

~)

. (x - 2 37T) 44. y = 3 sm

Graph each function over a one-period interval. See Example 3.

45. y =

~ sin [ 2 ( x + ~)]

47. y = -4 sin(2x - 7T )

(1

1 cos z-x-4 7T) 49. y=z-

46. y = -

~cos [ 4 ( x + %) ]

48. y = 3 cos(4x + 7T ) 50. Y =

-±

sin (

~x + i)

Graph each function over a two-period interval. See Example 4.

51. y = - 3 + 2 sin x

52. y = 2 - 3 cos x

2 . 3 54. y = 1 - -sm -x 3 4

55. y = 1 - 2 cos

57. y = -2

I .

+ 2 sm 3x

53. y = -1 - 2 cos 5x

zx 1

. 1

56. y = - 3 + 3 sm 2x

58. y = 1 +

2 3

I x 2

- cos -

Graph each function over a one-period interval. See Example 5.

59.y=-3+2sin(x+%)

60. y=4-3cos(x- 7T )

(Modeling) Solve each problem. See Example 6.

63. Average Monthly Temperature The average monthly temperature (in °F) in Seattle, Washington, is shown in the table. (a) Plot the average monthly temperature over a two-year period, letting x = 1 correspond to January of the first year. Do the data seem to indicate a translated sine graph? (b) The highest average monthly temperature is 66°F, and the lowest average monthly temperature is 41 °F. Their average is 53.5°F. Graph the data together with the line y = 53.5. What does this line represent with regard to temperature in Seattle?

Month

OF Month

OF

Jan

42

July

66

Feb

43

Aug

66

Mar

47

Sept

61

Apr

50

Oct

53

May

56

Nov

45

June

61

Dec

41

Data from National Climatic Data Center.

(c) Approximate the amplitude, period, and phase shift of the translated sine wave. (d) Determine a function of the form f(x) = a sin[ b(x - d) J + c, where a , b, c, and d are constants, that models the data. (e) Graph f together with the data on the same coordinate axes. How well does f model the given data? ~ (f) Use the sine regression capability of a graphing calculator to find the equation of

a sine curve that fits these data (over a two-year interval).

283

4.2 Translatio ns of the Grap hs of the Sine and Cosine Funct io ns

64. Average Monthly Temperature The average monthly temperature (in °F) in Phoenix, Arizona, is shown in the table.

Month

OF

Month

OF

Jan

56

July

95

(a) Predict the average annual temperature.

Feb

60

Aug

94

(b) Plot the average monthly temperature over a two-year period, letting x = 1 correspond to January of the first year.

Mar

65

Sept

88

(c) Dete rmine a functio n of the form f(x) = a cos [ b(x - d) J + c, where a, b, c, and dare constants, that models the data.

Apr

73

Oct

77

May

82

Nov

64

June

91

Dec

55

Data from National Climatic Data Center.

(d) Graph f together with the data on the same coordinate axes. How well does f model the given data?

E@ (e) Use the sine regression capability of a graphing calculator to find the equation of a sine curve that fits these data (over a two-year interval).

rn (Modeling) Monthly Temperatures

A set of temperature data (in °F) is given in the tables for a particular location. (Data from www.weatherbase.com)

(a) Plot the data over a two-year interval. (b) Use sine regression to determine a model for the two-year interval. Let x = 1 represent January of the first year. (c) Graph the equationfrompart (b) together with the given data on the same coordinate axes. 65. Average Monthly Temperature, Buenos Aires, Argentina Jan

Feb

Mar

Apr

May

Jun

Jui

Aug

Sept

Oct

Nov

Dec

77.2

74.7

70.5

63.9

57.7

52.2

51.6

54.9

57.6

63 .9

69. 1

73.8

66. Average High Temper ature, Buenos Aires, Argentina Jan

Feb

Mar

Apr

May

Jun

Jui

Aug

Sept

Oct

Nov

Dec

86.7

83.7

79.5

72.9

66.2

60.1

58.8

63. 1

66.0

72.5

77.5

82.6

E@ (Modeling) Fractional Part of the Moon Illuminated The data in the tables give the fractional part of the moon that is illuminated during the month indicated. (Data from http://aa.usno.navy.mil) (a) Plot the data for the month. (b) Use sine regression to determine a model for the data. (c) Graph the equation from part (b) together with the given data on the same coordinate axes. 67. January 2019 2

Day

3

4

5

6

7

8

9

10

11

12

13

14

15

16

Fraction 0.22 0.14 0.08 0.03 0.0 I 0.00 0.0 I 0.04 0.09 0. 15 0.22 0.30 0.40 0.49 0.60 0. 70 Day

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

Fraction 0.79 0.87 0.94 0.98 1.00 0.98 0.94 0.87 0.78 0.68 0.57 0.47 0.36 0.27 0.19

68. November 2019 Day

I

2

3

4

5

6

7

8

9

10

11

12

13

14

Fraction 0.20 0.29 0.38 0.48 0.58 0.67 0.75 0.83 0.89 0.94 0.98 1.00 0.99 0.97 Day

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

Fraction 0.93 0.86 0.78 0.68 0.58 0.46 0.35 0.25 0. 15 0.08 0.03 0.00 0.00 0.03 0.08 0. 14

284

CHAPTER 4 Graphs of the Ci rcu lar Funct ions

Chapter 4

Quiz

(Sections 4 .1and4.2)

1. Give the amplitude, period, vertical translation, and phase shift of the function

y = 3 - 4 sin ( 2x + ~} Graph each.function over a two-period interval. Give the period and amplitude. 1

2. y = -4 sin x 5. y = -2 cos(x

3. y =- - cos 2x

4. y = 3 sin

2

+ ~)

6. y = 2 + sin (2x -

1TX

1

7. y = - 1 + 2 sin x

1T )

Connecting Graphs with Equations Each function graphed is of the form y = a cos bx or y =a sin bx, where b > 0. Determine an equation of the graph.

8.

9.

y

10.

y

(Modeling) Average Monthly Temperature The average temperature (in °F) at a certain location can be approximated by the fu nction f (x) = 12 sin[

~(x -

-· •

y

•

--

•

-

..,;?' -

3.9)] + 72,

where x = 1 represents January, x = 2 represents February, and so on. 11. What is the average temperature in April? 12. What is the lowest average monthly temper-

ature? What is the highest?

4.3

Graphs of the Tangent and Cotangent Functions

•

Graph of the Tangent Function

•

Graph of the Cotangent Function

•

Techniques for Graphing

•

Connecting Graphs with Equations

Graph of the Tangent Function Consider the table of selected points accompanying the graph of the tangent function in Figure 22 on the next page.

These points include special values between -

I

I and I. The tangent function

is undefined for odd multiples of and, thus, has vertical asymptotes for such values. A vertical asymptote is a vertical line that the graph approaches but does not intersect. As the x-values get closer and closer to the line, the function values increase or decrease without bound. Furthermore, because tan (-x) = - tan x,

Odd function

the graph of the tangent function is symmetric with respect to the origin.

4.3 Graphs of the Tangent and Cotangent Functions

x

y

71"

y

= tanx

- \/3 =

71"

- 3

285

y = tanx

- 1.7

-I

- 4 71"

V3 = -0.6 -3

0

0

71"

'{3 =

- 6

6 71"

1

71"

V3 =

4 3

0.6

1.7

y

Figure 22

The tangent function has period 'TT. Because tan x = ~~~ ~ , tangent values are 0 when sine values are 0, and are undefined when cosine values are 0. As x-values tangent values range from - oo to oo and increase throughincrease from - to out the interval. Those same values are repeated as x increases from to 3; , from 3 1T 51T 3 1T 3 1T • • • T to T , and so on. The graph of y = tan x from - 2 to T is shown m Figure 23.

I I,

y = tan x

Period:

I

1T

The graph continues in this pattern.

Figure 23

= tan x x =F- ( 2n + 1) I, where n is any integer}

Tangent Function Domain: { x x

J

y

71"

undefined

-4

71"

-]

0

0

-2

/(x)

Range: ( -

oo, oo)

y

x

71"

4 71"

2

undefined

f(x) = tanx, -

1T

1T

2 0, follow these steps. Step 1 Graph the corresponding reciprocal function as a guide, using a dashed curve. To Graph

Use as a Guide

y =a csc bx

y =a sin bx

y =a sec bx

y =a cos bx

Step 2 Sketch the vertical asymptotes. They will have equations of the form x = k, where k corresponds to an x-intercept of the graph of the guide function. Step 3 Sketch the graph of the desired function by drawing the typical Ushaped branches between the adjacent asymptotes. The branches will be above the graph of the guide function when the guide function values are positive and below the graph of the guide function when the guide function values are negative. The graph will resemble those in Figures 37 and 40 in the function boxes given earlier in this section.

Like graphs of the sine and cosine functions, graphs of the secant and cosecant functions may be translated vertically and horizontally. The period of both basic functions is 27T. EXAMPLE 1

Graphing y = a sec bx I

Graph y = 2 sec 2x. SOLUTION

Step 1 This function involves the secant, so the corresponding reciprocal function will involve the cosine. The guide function to graph is 1

y = 2 cos 2x. Using the guidelines given earlier, we find that this guide function has amplitude 2 and that one period of the graph lies along the interval that satisfies the following inequality. 1

0

~

-x 2

~

27T

0

~

x

~

47T

Multiply each part by 2.

Dividing the interval [ 0 , 47T ] into four equal parts gives these key points. (0,2),

(7r,0),

(27T, -2) ,

(37T,0) ,

(47T,2)

Keypoints

297

4.4 Graphs of t he Secant and Cosecant Funct ions

These key points are plotted and joined with a dashed red curve to indicate that this graph is only a guide. An additional period is graphed as shown in Figure 41(a). y

I I I I 2 I ,, /

- 47T

-31T' -21T

/ -1T

I '-" 1 I I

I I I I

I

1

n

HORHAL FLOAT AUTO REAL RADIAN HP

0

y

)',1

I y = 2 cos !x : is used as~ guide.

'

-2

I ,. I /"

I I

x

1T ' 21T ,_/31T I I I I

'.-

1

47T

- 47T

I I I I

,'"'~ii\

I I I I/

,n _ , n·

-31T' -21T

/ -1T

I I I

1

I I I

'

0

x

47T

I I

(a)

I I I /,,

I I

(b ) Figure 41

V1:2lcOSC define their sum to be

YJ = Y1 + Y2 ·

Evaluate y 1 and y 2 at the given value of x and show that their sum is equal to y 3 evaluated at x. Use the method of addition of ordinates. See Example 4. 45. y 1

= sin x, y2 = sin 2x; x- -6

47. y 1

= tan x, y2 = sec x ; x- -4

46. y 1 = cos x, y2

7T

7T

48. y 1

= cos 2x;

=cotx, y =cscx; 2

x

=I

Summary Exercises on Graphing Circular Functions These summary exercises provide practice with the various graphing techniques presented in this chapter. Graph each function over a one-period interval. 1. y

= 2 sin 'TTX I

3. y = -2 + 2 cos

'TT

4x 6. y

= 3 tan ( %x + 'TT)

Graph eachfanction over a two-period interval. x 7. y = -5 sin 3 9. y = 3 - 4 sin

[ 4.5

8. y

= 10 cos ( ~ +

10. y = 2 - sec [ 'TT(x - 3) ]

(%x +'TT)

;

Harmonic Motion

•

Simple Harmonic Motion

•

Damped Oscillatory Motion

%)

Simple Harmonic Motion In part A of Figure 46, a spring with a weight attached to its free end is in equilibrium (or rest) position. If the weight is pulled down a units and released (part B of the figure), the spring' s elasticity causes the weight to rise a units (a > 0) above the equilibrium position, as seen in part C, and then to oscillate about the equilibrium position. y

!_ A.

B.

a

-----

0

----C.

-a

Figure 46

If friction is neglected, this oscillatory motion is described mathematically by a sinusoid. Other applications of this type of motion include sound, electric current, and electromagnetic waves.

302

CHAPTER 4 Graphs of the Circular Functions

y

To develop a general equation for such motion, consider Figure 47. Suppose the point P(x, y) moves around the circle counterclockwise at a uniform angular speed w. Assume that at time t = 0, P is at (a, 0). The angle swept out by ray OP at time t is given by = wt. The coordinates of point P at time tare

(0,a)

e

x = a cos e = a cos wt

and

y=

a sin e = a sin wt.

As P moves around the circle from the point (a, 0), the point Q(O, y) oscillates back and forth along the y-axis between the points ( 0, a) and ( 0, -a). Similarly, the point R(x, 0) oscillates back and forth between (a, 0) and (-a, 0) . This oscillatory motion is simple harmonic motion. 2 The amplitude of the oscillatory motion is Ia I, and the period is ;: . The moving points P and Q or P and R complete one oscillation, or cycle, per period. The number of cycles per unit of time, called the frequency, is the reciprocal of the period, 2 where w > 0.

Figure 47

%,

Simple Harmonic Motion

The position of a point oscillating about an equilibrium position at time t is modeled by either

s(t) = a cos wt or s(t) = a sin wt, where a and w are constants, with w > 0. The amplitude of the motion is Ia I, the period is : , and the frequency is 2 oscillations per time unit.

%

EXAMPLE 1 y

I

~A.

Suppose that an object is attached to a coiled spring such as the one in Figure 46 (repeated in the margin). It is pulled down a distance of 5 in. from its equilibrium position and then released. The time for one complete oscillation is 4 sec.

m

~ ~

Modeling the Motion of a Spring

(a) Give an equation that models the position of the object at time t. (b) Determine the position at t

=

1.5 sec.

a

B.

-----

0

-----

-a

c.

Figure 46 (repeated)

(c) Find the frequency. SOLUTION

(a) When the object is released at t = 0, the distance of the object from the equilibrium position is 5 in. below equilibrium. Ifs( t) is to model the motion, then s(O) must equal -5. We use s(t) =a cos wt,

with a = -5.

We choose the cosine function here because cos w(O ) = cos 0 = 1, and -5 · 1 = -5. (Had we chosen the sine function, a phase shift would have been required.) Use the fact that the period is 4 to solve for w. 2'1T

-

w

= 4 'TT

w = -

2

Th e pen.od.1s 21T -;;; .

Solve for w.

Thus, the motion is modeled by s(t) = -5cos1t.

4.5 Harmonic Motion

303

(b) Substitute the given value oft in the equation found in part (a).

s( t ) s( l.5 )

= -

=

1T

5 cos 2 t

Equation from part (a)

-5 cos[; ( 1.5) ]

s(l.5) = 3.54 in.

Lett= 1.5.

Use a calculator.

Because 3.54 > 0, the object is above the equilibrium position. (c) The frequency is the reciprocal of the period, or~ oscillation per sec. t/' Now Try Exercise 7. EXAMPLE 2

Analyzing Harmonic Motion

Suppose that an object oscillates according to the model

s( t)

=

8 sin 3t,

where t is in seconds and s(t) is in feet. Analyze the motion. SOLUTION

The motion is harmonic because the model is

s(t) =a sin wt. Because a = 8, the object oscillates 8 ft in either direction from its starting 2 point. The period ; = 2.1 is the time, in seconds, it takes for one complete oscillation. The frequency is the reciprocal of the period, so the object completes 2~ = 0.48 oscillation per sec. t/' Now Try Exercise 17. Damped Oscillatory Motion In the example of the stretched spring, we disregard the effect of friction. Friction causes the amplitude of the motion to diminish gradually until the weight comes to rest. In this situation, we say that the motion has been damped by the force of friction. Most oscillatory motions are damped. For instance, shock absorbers are put on an automobile in order to damp oscillatory motion. Instead of oscillating up and down for a long while after hitting a bump or pothole, the oscillations of the car are quickly damped out for a smoother ride. The decrease in amplitude of a damped oscillatory motion usually follows the pattern of exponential decay. EXAMPLE 3

Analyzing Damped Oscillatory Motion

A typical example of damped oscillatory motion is provided by the function s(x) = e- x cos 21Tx.

(The number e = 2.718 is the base of the natural logarithm function.) We use x rather than t to match the variable for graphing calculators. ~ (a)

Provide a calculator graph of y 3 = e-x cos 21Tx, along with the graphs of ::5 x ::5 3.

Y1 = e- x and Y2 = -e-x for 0

(b) Describe the relationships among the three graphs drawn in part (a).

(c) For what values of x does the graph of y 3 touch the graph of y 1? (d) For what values of x does the graph of y 3 intersect the x-axis?

304

CHAPTER 4 Graphs of the Circular Functions

SOLUTION

(a) Figure 48 shows a TI-84 Plus graph of y 1, y 2 , and y 3 in the window [ 0, 3 J by [ -1, 1] . (b) The graph of y 3 is bounded above by the graph of y 1 and below by the graph of Y2- (The graphs of y 1 and y 2 are referred to as envelopes for the graph of YJ.)

(c) When 27Tx = 0, 27T, 47T, and 67T, cos 27Tx = 1. Thus, the value of e- x cos 27Tx is the same as the value of e-x when 27Tx = 0, 27T, 47T, and 67T-that is, when x = 0, 1, 2, and 3.

Figure 48

rr 3rr Srr 7rr 9rr d 2,cos 11 rr 27Tx= 0 .Th us, th egrap h o f y 3 (d) When 27Tx= 2,T,T•T•T,an . . h 1 3579 intersects the x-axis w en x = 4, 4, 4, 4, 4, and 4i! .

II' Now Try Exercise 39.

q

Exercises

CONCEPT PREVIEW An object in simple harmonic motion has position function s(t), in inches, from an equilibrium point, as follows, where t is time in seconds.

s(t) = 5 cos 2t 1. What is the amplitude of this motion?

2. What is the period of this motion?

4. What is s(O)?

3. What is the frequency?

6. What is the range of the graph of this function? (Modeling) Solve each problem. See Examples 1 and 2. 7. Spring Motion An object is attached to a coiled spring, as in Figure 46. It is pulled down a distance of 4 units from its equilibrium position and then released. The time for one complete oscillation is 3 sec. (a) Give an equation that models the position of the object at time t. (b) Determine the position at t = 1.25 sec to the nearest hundredth.

(c) Find the frequency. 8. Spring Motion Repeat Exercise 7, but assume that the object is pulled down a distance of 6 units and that the time for one complete oscillation is 4 sec. 9. Voltage of an Electrical Circuit The voltage E in an electrical circuit is modeled by E = 5 cos 1201Tt,

where t is time measured in seconds. (a) Find the amplitude and the period.

(b) Find the frequency.

(c) Find £ , to the nearest thousandth, when t

= 0, 0.03, 0.06, 0.09, 0.12.

(d) Graph E for 0 ~ t ~

fcJ .

10. Voltage of an Electrical Circuit The voltage E in an electrical circuit is modeled by

E = 3.8 cos 407rt, where t is time measured in seconds. (a) Find the amplitude and the period.

(b) Find the frequency.

(c) Find£, to the nearest thousandth, when t = 0.02, 0.04, 0.08, 0.12, 0.14. (d) Graph E for 0 ~ t ~

I w·

4.5 Harmonic Motion

305

11. Particle Movement Write the equation and then determine the amplitude, period, and frequency of the simple harmonic motion of a particle moving uniformly around a circle of radius 2 units, with the given angular speed. (a) 2 radians per sec

(b) 4 radians per sec

12. Spring Motion The height attained by a weight attached to a spring set in motion is

s(t) = -4 cos 8?Tt inches after t seconds. (a) Find the maximum height that the weight rises above the equilibrium position of

s(t) = 0. (b) When does the weight first reach its maximum height if t

2:

O?

(c) What are the frequency and the period?

13. Pendulum Motion What are the period P and frequency T of oscillation of a

t

pendulum of length ft? (Hint: P = 27TVf; , where Lis the length of the pendulum in feet and the period P is in seconds.) 14. Pendulum Motion In Exercise 13, how long should the pendulum be to have a period of 1 sec? 15. Spring Motion The formula for the up and down motion of a weight on a spring is given by

s(t)

=

asin~t.

If the spring constant k is 4, what mass m must be used to produce a period of 1 sec?

16. Spring Motion (See Exercise 15.) A spring with spring constant k = 2 and a I-unit mass m attached to it is stretched and then allowed to come to rest. (a) If the spring is stretched

t

ft and released, what are the amplitude, period, and frequency of the resulting oscillatory motion?

(b) What is the equation of the motion? 17. Spring Motion The position of a weight attached to a spring is

s(t) = - 5 cos 4?Tt inches after t seconds. (a) Find the maximum height that the weight rises above the equilibrium position of s(t) = 0.

(b) What are the frequency and period? (c) When does the weight first reach its maximum height? (d) Calculate and interpret s( 1.3) to the nearest tenth.

18. Spring Motion The position of a weight attached to a spring is

s(t) = -4 cos 10t inches after t seconds. (a) Find the maximum height that the weight rises above the equilibrium position of s(t) = 0.

(b) What are the frequency and period? (c) When does the weight first reach its maximum height?

(d) Calculate and interpret s( 1.466).

19. Spring Motion A weight attached to a spring is pulled down 3 in. below the equilibrium position. (a) Assuming that the frequency is ~ cycles per sec, determine a model that gives

the position of the weight at time t seconds. (b) What is the period?

306

CHAPTER 4 Graphs of the Circu lar Functions

20. Spring Motion A weight attached to a spring is pulled down 2 in. below the equilibrium position. (a) Assuming that the period is ~ sec, determine a model that gives the position of the weight at time t seconds. (b) What is the frequency?

21. Sound Wave When middle C is played on a piano, an oscilloscope produces a graph modeled by the function

y =a sin wt having frequency 261.6 oscillations per sec. (a) What is the value of w? (b) What is the period of the function? 22. Sound Wave When the A key above middle C is played on a piano, an oscilloscope produces a graph modeled by the function y = a sin 8807Tt,

where t is measured in seconds. Find the frequency of the sound wave. 23. Circadian Rhythm Body temperature is an example of a circadian rhythma 24-hour internally controlled biological cycle. Many circadian rhythms can be modeled by trigonometric functions. Find a function T of the form T(t) =a cos wt

that models the increase and decrease in body temperature from average temperature for a person whose temperature ranges from 97.7°F to 99.5°F. Let t= 0 correspond to 4 A.M., the time at which a person's body temperature is lowest. 24. Circadian Rhythm Suppose that body temperature for a person whose temperature ranges from 97.9°F to 99.3°F is in circadian rhythm, where t = 0 corresponds to 4 P.M., the time at which the person's body temperature is highest. Find a function Tof the form T(t) =a cos wt

that models the increase and decrease in body temperature from average temperature. 25. Oscillating Cork When a rock is thrown into a pond, waves travel away from the rock across the surface of the pond. Suppose that a nearby cork is bobbing up and down in simple harmonic motion. If the height of the middle of the cork above water level after t seconds is given by

s(t)

=

0.8sin l87Tt,

where s(t) is measured in inches, how many times per second will the cork bob up and down? 26. Tides Suppose that during a 12-hour period, water height at the end of a pier begins at an average tide level of 13 ft, rises to a high tide level of 18 ft after 3 hr, falls to 8 ft at low tide 6 hr later, and returns to average tide level after another 3 hr. Find a function of the form h(t) =a sin wt

that models the increase and decrease in water height from average tide level after t hours.

4.5 Harmonic Mot ion

307

(Modeling) Springs A weight on a spring has initial position s(O) and period P. (a) To model displacement of the weight, find a function s given by s(t) =a cos wt. (b) Evaluate s( I ). Is the weight moving upward, downward, or neither when t

= I?

Support the results g raphically or numerically.

= 5 in.;

= 1.5 sec

27. s( O) = 2 in.; P = 0.5 sec

28. s( O)

29. s( O) = -3 in. ; P = 0.8 sec

30. s( O) = - 4 in. ; P = 1.2 sec

P

~ (Modeling) Music A note on a piano has g iven freq uency F. Suppose the maximum

displacement at the center of the piano wire is g iven by s(O). Find constants a and w so that the equation s(t) =a cos wt models this displacement. Graph s in the viewing window [ 0, 0.05 ] by [ - 0.3, 0.3] .

= 0.11

31. F = 27.5; s(O) = 0.21

32. F= llO; s(O)

33. F = 55; s( O) = 0.14

34. F = 220; s(O) = 0.06

(Modeling) Damped Spring Motion A sp ring with a weight attached is pulled down and released. Because offriction and other resistive f orces, the amplitude is decreasing over time, and t seconds after the spring is released, its position in inches is given by the function s (t)

= - ll e- 0·2' cos 0.57Tt.

y

35. How far was the weight pulled down from the equilibrium position before it was released? 36. How far, to the nearest hundredth of an inch, is the weight from the equilibrium position after 6 sec? ~ 37. Graph the function on the interval [ 0, 12] by [ - 12, 12 ], and determine the values

for which the graph intersects the horizontal axis. 38. How many complete oscillations will the graph make during 12 sec?

~ (Modeling) Damped Oscillatory Motion Work each exercise. See Example 3.

39. Consider the damped oscillatory function s(x)

= 5e- 0·3'

COS

7TX.

(a) Graph the function Y3 = 5e-0·3' cos 7TX in the window [ 0, 3] by [ - 5, 5 ]. (b) The graph of which function is the upper envelope of the graph of y 3 ? (c) For what values of x does the graph of y 3 touch the graph of the function found in part (b)?

308

CHAPTER 4 Graphs of the Circular Functions

40. Consider the damped oscillatory function s(x) = lOe- x sin 27TX. (a) Graph the fu nction y 3 = lOe-x sin 27Tx in the window ( 0, 3 ] by ( - 10, 10

J.

(b) The graph of which function is the lower envelope of the graph of y 3?

(c) For what values of x does the graph of y 3 touch the graph of the function found in part (b)?

Chapter 4 Test Prep Key Terms 4.1

periodic function period sine wave (sinusoid) amplitude

4.2 phase shift

4.4 addition of ordinates

argument vertical asymptote

4.3

4.5

frequency damped oscillatory motion envelope

abscissa ordinate simple harmonic motion

Quick Review Examples

Concepts Graphs of the Sine and Cosine Functions

Translations of the Graphs of the Sine and Cosine Functions

Graph y = I + sin 3x.

Sine and Cosine Functions y

y

y

y = l+ sin 3x

2

---,,. 01--~ 7T__......~ 27T~+7T......_4+ 7T-- x

Domain: ( - oo, oo)

Domain: ( - oo, oo)

Range: ( - 1,l ]

Range:( - 1, l ]

Amplitude: l

Amplitude: I

Period: 27T

Period: 27T

The graph of

3 3

- I

amplitude: 1 . d 27T per10 : 3 vertical translation: up 1 unit Graph y = - 2 cos ( x

y = c +a sin [b(x - d)] or y = c +a cos[b(x - d)] ,

domain: ( - oo, oo) range: [ 0, 2 ]

+ I) .

y

y

with b > 0, has the following characteristics.

3

=- 2 cos (x + ~ )

1. amplitude Ia I . d 27T 2. peno b 3. vertical translation up c units if c > 0 or down Ic I units if c < O 4. phase shift to the right d units if d > 0 or to the left Id I units if d < o

-2' domain: ( - oo, oo )

amplitude: 2

range: [ -2, 2 ]

period: 27T phase shift: left

I units

CHAPTER 4 Test Prep

Concepts

309

Examples

Graphs of the Tangent and Cotangent Functions

Graph y = 2 tan x over a one-period interval.

Tangent and Cotangent Functions y

y

y

I

I ly = cotx I I

1y=tanx

I I

2

I I I

Domain: {x lx ¥- (2n

+ lH,

Domain:

{xlx ¥- n7T,

y = 2tan x

where n is any integer}

where n is any integer }

Range: ( -oo, oo)

Range: ( -oo, oo)

Period: 7T

Period: 7T

period: 7T domain:{x ix¥-(2n+

1 )~:,

where n is any integer } range: ( -oo, oo)

Graphs of the Secant and Cosecant Functions

Graph y = sec(x +~)over a one-period interval.

Secant and Cosecant Functions y

y

y

I

y =sec x

0-Iv

-+--+--,,+--+--ii---

~

I

x

Domain: {x lx ¥- (2n

ii

~ i"-

h

4

~/ +- h 1• Let(} be the angle of elevation to the top of the object. See the figure. (a) Show that d = (h 2

-

h 1) cot 6.

(b) Let h 2 = 55 and h 1 = 5. Graph d for the interval 0 < (} 48. (Modeling) Tides The figure shows a function f that models the tides in feet at Clearwater Beach, Flo rida, x hours after midnight. (Data from Pentcheff, D., WWW Tide and Current Predictor.)

I.

y

4

(14.7, 2.6)

-::;-

(2.4, 2.6)

"'

~

I

(27, 2.6)

c

(a) Find the time between high tides.

:.=- 2

(b) What is the difference in water levels between high tide and low tide?

l=

"'"'

"O

(8.7, 1.4)

I

(c) The tides can be modeled by the function

(2 1, 1.4) x

0

f(x) = 0.6cos[0.5ll (x - 2.4) ]

:S

4

+ 2.

Estimate the tides, to the nearest hundredth, when x = 10.

8 12 16 20 24 28

Time (in hours)

312

CHAPTER 4 Graphs of the Circular Functions

49. (Modeling) Maximum Temperatures The maximum afternoon temperature (in °F) in a given city can be modeled by the function t = 60 - 30 cos

X 'TT

6,

where t represents the maximum afternoon temperature in month x, with x = 0 representing January, x = 1 representing February, and so on. Find the maximum afternoon temperature, to the nearest degree, for each month. (a) January

(b) April

(c) May

(d) June

(e) August

e@ 50. (Modeling) Average Monthly Temperature The

OF

Month

Jan

24

July

74

Feb

28

Aug

72

Month

average monthly temperature (in °F) in Chicago, Illinois, is shown in the table.

(f) October

OF

(a) Plot the average monthly temperature over a two-year period. Let x = 1 correspond to January of the first year.

Mar

38

Sept

65

Apr

49

Oct

53

(b) To model the data, determine a function of the form f(x) =a sin ( b(x - d) J + c, where a, b, c, and dare constants.

May

59

Nov

40

June

69

Dec

28

(c) Graph f together with the data on the same coordinate axes. How well does f model the data?

Data from National Cli matic Data Center.

(d) Use the sine regression capability of a graphing calculator to find the equation of

a sine curve that fits these data (over a two-year interval). 51. (Modeling) Pollution Trends Circular functions can be used to model fluctuation of pollution levels as shown in the figure. The pollution level in a certain area might be given by

y = 7(1 - cos 2'1Tx)(x

HIRMRL FIM2 AUTO REiil RllOIAH HP'

a

+ 10) + 100e02',

where x is time in years, with x = 0 representin g ,,, y,.,. January 1 of the base year. July 1 of the same year would be represented by x = 0.5, October 1 of the following year would be represented by x = 1.75, and so on. (e is the base of the natural logarithm; e = 2. 718282) Find the pollution level on each date. (a) January 1, base year (See the figure.)

(b) July 1, base year

(c) January 1, following year

(d) July 1, following year

52. (Modeling) Lynx and Hare Populations The figure shows the populations of lynx and hares in Canada for the years 1847- 1903. The hares are food for the lynx. An increase in hare population causes an increase in lynx population some time later. The increasing lynx population then causes a decline in hare population. The two graphs have the same period. Canadian Lynx and Hare Populations - - Hare - - - Lynx 150,000 ~

"E

..0

::s

100,000

z

50,000

1850

1860

1870

1880

1890

1900

Year

(a) Estimate the length of one period. (b) Estimate the maximum and minimum hare populations.

CHAPTER 4

Test

313

An object in simple harmonic motion has position function s(t), in inches, from an equilibrium point, where t is time in seconds. Find the amplitude, period, and frequency. 53. s(t) = 4 sin 7rt

54. s(t) = 3 cos 2t

55. In Exercise 53, what does the frequency represent? Find the position of the object relative to the equilibrium point at 1.5 sec, 2 sec, and 3.25 sec. 56. In Exercise 54, what does the period represent? What does the amplitude represent?

Chapter 4

Test 1. Identify each of the following basic circular function graphs. (b)

(a) y

(c) y

y

J l I

I

_-2+-"--1h --7T-+--~>-10,..._+-h -+"--+-;--12 "- x

(d)

(f)

(e) y

y

y

- 27T

I I I I

2. Connecting Graphs with Equations Determine the simplest form of an equation for each graph. Choose b > 0, and include no phase shifts. (a)

(b)

y

y

2

7T

27T

37T

47T

-I

-2

3. Answer each question. (a) What is the domain of the cosine function? (b) What is the range of the sine function?

(c) What is the least positive value for which the tangent function is undefined? (d) What is the range of the secant function?

314

CHAPTER 4 Graphs of the Circular Functions

4. Consider the function y = 3 - 6 sin ( 2x +

I).

(a) What is its period? (b) What is the amplitude of its graph?

(c) What is its range? (d) What is they-intercept of its graph?

(e) What is its phase shift?

Graph each function over a two-period interval. Identify asymptotes when applicable. 5. y=sin(2x+7T) 7. y = 2

6. y = -cos 2x

+ cosx

8. y = - l + 2 sin ( x + 7T) 10. y = -2 - cot ( x - % )

9. y= tan(x-%) 11. y = -csc 2x

12. y = 3 CSC 7TX

(Modeling) Solve each problem. 13. Average Monthly Temperature The average monthly temperature (in °F) in San Antonio, Texas, can be modeled by the function

f(x) = 16.5 sin [ ~(x - 4) ] + 68.5, where x is the month, with x = l corresponding to January, x = 2 corresponding to February, and so on. (Data from National Climatic Data Center.) ~ (a) Graph f in the window [ 0, 25 J by [ 40, 90 J.

(b) Determine the amplitude, period, phase shift, and vertical translation of f.

(c) What is the average monthly temperature for the month of December? (d) Determine the minimum and maximum average monthly temperatures and the months when they occur.

(e) What would be an approximation for the average annual temperature in San Antonio? How is this related to the vertical translation of the sine function in the formula for f? 14. Spring Motion The position of a weight attached to a spring is

s(t) = -4 cos 87Tt inches after t seconds. (a) Find the maximum height that the weight rises above the equilibrium position of s( t ) = 0. (b) When does the weight first reach its maximum height if t 2: 0?

(c) What are the frequency and period? 15. Explain why the domains of the tangent and secant functions are the same, and then give a similar explanation for the cotangent and cosecant functions.

5

Trigonometric Identities

Electricity that passes through wires to homes and businesses alternates its direction on those wires and is modeled by sine and cosine functions.

316

CHAPTER 5 Trigonometric Identities

l 5.1

q Fundamental Identities

•

Fundamental Identities

•

Uses of the Fundamental Identities

y

Fundamental Identities Recall that a function is even if f( - x) = f(x) for all x in the do main of f , a nd a fun ction is odd if f( - x) = - f(x) for all x in the domain of f. We have used graphs to classify the trigonometric functions as even or odd. We can also use Figure 1 to do this. As suggested by the circle in Figure 1, an angle f) having the point (x, y) on its terminal side has a corresponding angle - f) with the point (x, -y) on its terminal side. From th e definition of sine, we see th at sin( - fJ ) and sin f) are negatives of each other. That is,

-y sin(-fJ)= r so

= -? = - sin 0 Figure 1

y sinfJ= -, r

and

sin(- 0)

sin(-0)

= - sin 0

Sine is an odd function.

This is an example of an identity, an equation that is satisfied by every value in the domain of its variable. Some examples from algebra follow. x 2 -y 2 =(x+y)(x-y)

+ y) = x 2 + 2xy + y 2 = x (x

+ xy (x + y )2

x2

Identities

Figure 1 shows an angle f) in quadrant II, but the same result holds for any quadrant. The figure also suggests the following identity for cosine.

x cos( - fJ) = r cos( - 0)

and

f)

in

x cos fJ = r

= cos 0

Cosine is an even function.

We use the identities for sin( - fJ ) and cos( - fJ ) to find tan( - fJ ) in terms of tan e. tan( - fJ) =

sin( - fJ ) cos( - fJ) tan(-0)

=

-sin fJ cos fJ

= -tanO

sin f) cos f) Tangent is an odd function.

The reciprocal identities are used to determine that cosecant and cotangent are odd functions and secant is an even function. These even-odd identities, together with the reciprocal, quotient, and Pythagorean identities, make up the fundamental identities.

In trigonometric identities, fJ can represent an angle in degrees or radians, or a real number.

NOTE

5.1 Fu ndame ntal Identit ies

317

Fundamental Identities

Reciprocal Identities 1

1

cot(}= - tan (J

sec(}= - cos (J

1

csc (J = --;---(} Sill

Quotient Identities sin (J tan(}= - cos 8

cos (J cot (J = -:--Sill 8

Pythagorean Identities sin2 (J

+ cos2 (J = 1

tan2 (J

+ 1 = sec2 (J

1

+ cot2 (J = csc2 (J

Even-Odd Identities sin( - 8) = - sin (J

cos( - 8) = cos (J

tan( - 8) = - tan (J

csc( - 8) = - csc (J

sec( - 8) = sec (J

cot( - 8) = - cot (J

NOTE We will also use alternative forms of the fundamental identities. For example, two other forms of sin2 (J + cos 2 (J = 1 are sin2 (J = 1 - cos2 (J and cos2 (J = 1 - sin 2 (}.

Uses of the Fundamental Identities We can use these identities to find the values of other trigonometric functions from the value of a given trigonometric function.

Finding Trigonometric Function Values Given One Value and the Quadrant

EXAMPLE 1

If tan f) = - ~ and

f)

is in quadrant II, find each function value.

(a) sec f)

(c) cot( -fJ)

(b ) sin f)

SOLUTION

(a) We use an identity that relates the tangent and secant functions.

tan2

f)

+ 1 = sec 2 f)

(-~)2 + 1

2

5

= sec f)

tan 0 = - 3

- + 1 = sec 2 f)

Square -3.

25

9

34

-

9

Choose the correct sign.

Pythagorean identity

-,J¥.

= sec 2 f)

=

sec f)

V34

sec f) = - - -

3

5

Add-' I

=

2

9

Take the negative square root because 0 is in quadrant II. S.imp 11·fy t he rad.1ca I : - 'V/34 9 = and rewrite.

v'34 V9 = 9

- -v'34 3- .

318

CHAPTER 5

Trigonometric Identities

sin 11 tan8 = - cos 11

(b)

Quotient identity that relates the tangent and sine functions

cos 11 tan 11 = sin 11

Multiply each side by cos 0.

1 ( -sec -11 )tan 11=sin11 3 ( -

Reciprocal identity 5

tan (} = - 3, and from part (a),

~) ( - ~) = sin 11

I secO=

I 3 3 V:i4 3v'34 v'34 = - v'34 = - v'34 ' v'34 = -34 . - -3-

s\/34

sinl1=34

(c)

cot(-11) =

1 (

tan - 11

)

1 cot(-11) = - - tan 11 1

cot( -11) = - - -

-( -~)

Multiply and rewrite.

Reciprocal identity that relates the tangent and cotangent functions Even-odd identity tan (}= - 3~ I

3 cot(-11) = 5

-( - ~) = I

v

7

s 3 3 3 = I. 5 = 5 Now Try Exercises 11, 19, and 31 .

tlAuiim~i

When taking the square root, be sure to choose the sign based on the quadrant of (J and the function being evaluated.

Writing One Trigonometric Function in Terms of Another

EXAMPLE 2

Write cos x in terms of tan x . SOLUTION

By identities, sec x is related to both cos x and tan x .

+ tan2 x

= sec 2 x

1 + tan 2 x

sec 2 x

1

y 1 = sin2 x

h~~

+ cos2 x

Rem ember both the positive and negative roots.

AUTO REAL RAOIAH HP

- - -2- = cos2 x 1 + tan x

Pythagorean identity Take reciprocals.

The reciprocal of sec2 x is cos2 x.

Take the square root of each side.

a

±1

cos x = -===== Yl + tan 2 x

± Yl + tan2 x cos x = - - - - -2 - 1 + tan x

111T 4 -+--+--+-+-+-t-- 0. To see what happens when b < 0, work Exercises 85-90 in order. 85. Use an even-odd identity to write y = sin( -2.x) as a function of 2.x. 86. How is the answer to Exercise 85 related to y = sin 2.x? 87. Use an even-odd identity to write y = cos(-4x) as a function of 4x. 88. How is the answer to Exercise 87 related to y = cos 4x? 89. Use the results from Exercises 85-88 to rewrite the following with a positive value of b. (a) y = sin(-4x)

(b) y = cos(-2.x)

(c) y = -5 sin( -3x)

90. Write a short response to this statement, which is often used by one of the authors of this text in trigonometry classes: Students who tend to ignore negative signs should enjoy graphing functions involving the cosine and the secant.

5.2

Verifying Trigonometric Identities

•

Strategies

•

Verifying Identities by Working with One Side

•

Verifying Identities by Working with Both Si des

Strategies One of the skills required for more advanced work in mathematics, especially in calculus, is the ability to use identities to write expressions in alternative forms. We develop this skill by using the fundamental identities to verify that a trigonometric equation is an identity (for those values of the variable for which it is defined).

C¥u0 Ctl~i The procedure for verifying identities is not the same as that for solving equations. Techniques used in solving equations, such as adding the same term to each side, and multiplying each side by the same term, should not be used when working with identities.

5.2 Verifying Trigonomet ric Ident ities

323

Hints for Verifying Identities LOOKING AHEAD TO CALCULUS

Trigonometric identities are used in calculus to simplify trigonometric expressions, determine derivatives of trigonometric functions, and change the form of some integrals.

1. Learn the fundamental identities. Whenever one side of a fundamental identity is given, the other side should come to mind. Also, be aware of equivalent forms of the fundamental identities. For example, sin2 (} = 1 - cos2 (} is an alternative form of sin 2 (} + cos2 (} = 1.

2. Try to rewrite the more complicated side of the equation so that it is identical to the simpler side.

3. It is sometimes helpful to express all trigonometric functions in the equation in terms of sine and cosine and then simplify the result. 4. Usually, any factoring or indicated algebraic operations should be performed. These algebraic identities are often used in verifying trigonometric identities. x 2 + 2xy

+ y2 = 2xy + y 2 =

(x

+ y )2

x2 -

(x - y ) 2

+ xy + y2) (x + y)(x2 - xy + y2) (x + y)(x - y)

x 3 - y3 = (x- y)(x2 x3

+ y3

=

x2 - y2 = For example, the expression sin2 x

+ 2 sin x + 1

can be factored as

(sin x

+ 1 ) 2.

The sum or difference of two trigonometric expressions can be found in the same way as any other rational expression. For example,

1

1

sin(}

cos(}

--+-1 . cos (}

1 · sin (} cos (} sin (}

----+----

Write with the LCD.

cos (} + sin (} sin(} cos(}

'!. + ~=a+ b

sin (} cos (}

c

c

c

5. When selecting substitutions, keep in mind the side that is not changing, because it represents the goal. For example, to verify that the equation tan2 x

+

1 1 = -cos2 x

is an identity, think of an identity that relates tan x to cos x. In this case, because sec x = co~ x and sec2 x = tan 2 x

+

1, the secant function is the

best link between the two sides. 6. If an expression contains 1 + sin x, multiplying both numerator and denominator by 1 - sin x would give 1 - sin2 x, which could be replaced with cos2 x. Similar procedures apply for 1 - sin x , 1 + cos x, and 1 - cosx.

Verifying Identities by Working with One Side ' A void the temptation to use algebraic properties of equations to verify identities.

One strategy is to work with one side and rewrite it to match the other side.

324

CHAPTER 5 Trigon ometric Identities

EXAMPLE 1

Verifying an Identity (Working with One Side)

Verify that the following equation is an identity. cot 8

+ 1 = csc 8( cos 8 + sin 8)

We use the fundamental identities to rewrite one side of the eq uation so that it is identical to the other side. The right side is more complicated, so we work with it, as suggested in Hint 2, and use Hint 3 to change all functions to expressions involving sine or cosine.

SOLUTION

Right side of given equation

For() = x, YI=

Y2 , ..

=

cotx +I csc x(cos x

+ sin x)

RAOIAH MP ·~ UTO REAL 4

csc 8 (cos 8

a

+ sin 8)

1

= - .- (cos 8 sm 8

.

+ sm 8)

cos 8

sin 8

sin 8

sin 8

I CSC () =sin 8

= -- + - -

Distributive property: a (b + c) = ab + ac

=cot 8 + 1

cosO sin 0

=

()· sin O = I cot ' sin 0

Left side of given equation

-4

The graphs coincide, which supports the conclusion in Example 1.

The right side is identical to the left side, so the given equation is an identity. v' Now Try Exercise 45.

EXAMPLE 2

Verifying an Ident it y (Working with One Side)

Verify that the following equation is an identity. tan 2 x( 1 + cot2 x) =

1 . 1 - sm2 x

We work with the more complicated left side, as suggested in Hint 2, and use the fundamental identities to obtain the right side.

SOLUTION

Left side of given equation

tan2 x(l

+ cot2 x) = tan2 x + tan2 x cot2 x

YI =

tan2 x(l I

+ cot2 x)

TO REAL RAOIAH MP

1

Distributive property

= tan2 x

+ tan2 x · tan2 x

1 cot2 x = - -

= tan2 x = sec2 x

+1

tan2 x · -tan-2 x = I

a cos2 x 1 - sin2 x

tan' x

1

Pythagorean identity 1 sec2 x = cos - -, x

Pythagorean identity

'--,,-------'

Right side of given equation -4

The left side is identical to the right side, so the given equation is an identity . The screen supports the conclusion in Example 2.

v' Now Try Exercise 49.

5.2 Verifying Trigonometric Identities

325

Verifying an Identity (Working with One Side)

EXAMPLE 3

Verify that the following equation is an identity. tan t - cot t - - - - - = sec 2 t - csc2 t sin t cos t SOLUTION

We transform the more complicated left side to match the right side.

tan t - cot t sin t cos t

Left side of given equation

tan t sin t cos t = tan t •

cot t sin t cos t

sin t cost

sin t = -- · cost sin t cos t cos 2 t

- cot t •

a - b

a

b

c

c

c

I

a

sin t cost

cost - -· sin t sin t cos t

b = a· b t

an

t =~ ·

cost'

cot t =COS/ smt

(This is Hint 3.) Multiply.

sin2 t

1 - -,

= sec2 t - csc2 t

cos- 1

1 = sec 2 I''sm -. -2 t = csc2 t

tl" Now Try Exercise 53. EXAMPLE 4

Verifying an Identity (Working with One Side)

Verify that the following equation is an identity. cos x 1 - sin x SOLUTION

1 +sin x cos x

We work on the right side, using Hint 6 in the list given earlier.

1 +sin x cos x

Right side of given equation

( I + sin x) ( I - sin x) cos x( 1 - sin x)

. l y by l m . th e 1orm " M u1tip (This is Hint 6.)

I - sin x 1 _ sin x ·

I - sin2 x cos x( 1 - sin x)

(x + y)(x - y) = x2

cos2 x cos x( I - sin x)

I - sin2 x = cos2 x

cos x . cos x cos x( I - sin x)

a 2 =a · a

cos x 1 - sin x

Divide out the common factor to write in lowest terms.

-

y2

tl" Now Try Exercise 59.

left

right

~"' // common third expression

Verifying Identities by Working with Both Sides If both sides of an identity appear to be equally complex, the identity can be verified by working independently on the left side and on the right side, until each side is changed into some common third result. Each step, on each side, must be reversible. With all steps reversible, the procedure is as shown in the margin.

326

CHAPTER 5

Trigonometric Identities

EXAMPLE 5

Verifying an Identity (Working with Both Sides)

Verify that the following equation is an identity. 1 + 2 sin a + sin2 a

seca+tana seca-tana

cos2 a

SOLUTION Both sides appear equally complex, so we verify the identity by changing each side into a common third expression. We work first on the left.

sec a+ tan a sec a - tan a

Left side of given equation

(sec a+ tan a) cos a (sec a - tan a) cos a

Multiply by 1 in the form ~~~ :.

sec a cos a + tan a cos a sec a cos a - tan a cos a

Distributive property

I +tan a cos a 1 - tan a cos a

sec a cos a= I

sin a 1 + - - • COSll' cos ll' tan a =

sin a 1 - - - • COSll' cos ll' 1 + sin a 1 - sin a

sin a cos a

Simplify.

On the right side of the original equation, we begin by factoring. 1 + 2 sin a + sin2 a cos2 a

Right side of given equation

( 1 + sin a ) 2

Factor the numerator; x 2 + 2xy + y 2 = (x + y )2

cos2 a ( 1 + sin a ) 2 1 - sin2 a

cos2 a = I - sin 2 a

(1 +sin a) 2 ( 1 + sin a ) ( 1 - sin a ) 1 +sin a 1 - sin a

Thus,

Factor the denominator; x 2 - y 2 = (x + y)(x - y) Write in lowest terms.

Left side of given equation

Common third expression

Right side of given equation

seca+tana seca- tana

1 + sin a 1 - sin a

1 + 2 sin a + sin2 a

cos2 a

and we have verified that the given equation is an identity. v' Now Try Exercise 75.

ilMIJICll~I Use the method of Example S only if the steps are reversible.

5.2 Verifying Trigonometric Identit ies

327

There are usually several ways to verify a given identity. Another way to begin verifying the identity in Example 5 is to work on the left as follows. seca+tana seca-tana

Left side of given equation in Example 5

1 cos a

sin a cos a

cos a

sin a cos a

-- + -Fundamental identities

1 +sin a cos a

Add and subtract fractions.

1 - sin a cos a 1 + sin a cos a 1 +sin a cos a

1 - sin a cos a

Simplify the complex fraction. Use the definition of division.

cos a 1 - sin a

Multiply by the reciprocal.

1 +sin a 1 - sin a

Multiply and write in lowest terms.

Compare this with the result shown in Example 5 for the right side to see that the two sides indeed agree.

Applying a Pythagorean Identity to Electronics

Tuners in radios select a radio station by adjusting the frequency. A tuner may contain an inductor L and a capacitor C, as illustrated in Figu re 3. The energy stored in the inductor at time t is given by

L(t)

An Inductor and a Capacitor

Figure 3

=

k sin2 2nFt

and the energy stored in the capacitor is given by

C(t) = k cos 2 2nFt, where Fis the frequency of the radio station and k is a constant. The total energy E in the circuit is given by

E(t) = L(t) + C(t). Show that Eis a constant function. (Data from Weidner, R., and R. Sells, Elementary Classical Physics, Vol. 2, Allyn & Bacon.) SOLUTION

E( t) = L( t) =

+ C( t) 2

k sin 21TFt + k cos 21TFt

= k [sin2 21TFt =

Given equation

2

k( 1)

= k

+

cos 2 21TFt ]

Substitute. Factor outk. sin2 (}

+ cos2 (} =

l (Here(}= 21TFt.)

Identity property

Because k is a constant, E( t) is a constant function. II' Now Try Exercise 105.

328

( s.2

CHAPTER 5 Trigonometric Identities

q

Exercises

To the student: Exercises 1- 44 are designed for practice in using the fundamental identities and applying algebraic techniques to trigonometric expressions. These skills are essential in verifying the identities that follow. CONCEPT PREVIEW Match each expression in Column I with its correct factorization

in Column II. I

II

+ y)(x2 - .xy + y 2 )

1. x2 - y2

A. (x

2. x3 - y3

B. (x + y)(x - y)

3. x3 + y3

C. (x+ y) 2

4. x2 + 2xy + y2

D. (x - y)(x2

+ xy + y 2 )

CONCEPT PREVIEW Fill in the blank(s) to correctly complete each fundamental identity.

5. sin 2 8 + cos2 () = _ __

6. tan2 8 + l = _ _ _

7. sin( -8) = _ __

8. sec( -8) = _ __ 1 cos() 10. cot 8 = - - = - -

1 sin 8 9. tan 8= - - = - -

Petform each indicated operation and simplify the result so that there are no quotients. l cot 8

11. cot()+ - -

secx CSCX 12. - - + - cscx secx

13. tan x( cot x + csc x)

14. cos {3( sec {3 + csc {3 )

l l +-15. - csc2 8 sec 2 8

cos x sin x 16. - - + - secx CSCX

17. (sin a - cos a) 2

18. (tan x +cot x) 2

19. ( l + sin t ) 2 + cos2 t

20. ( I + tan 8) 2 - 2 tan()

21. - - - 1 +cos x

22. - - - sin a - 1

1 - cos x

sin a+ 1

Factor each trigonometric expression. 23. sin2 8 - l

24. sec2 8 - 1

25. (sin x + I ) 2 - (sin x - I ) 2

26. (tan x + cotx) 2 - (tan x - cot x) 2

27. 2 sin2 x + 3 sin x + l

28. 4 tan2 {3 + tan {3 - 3

29. cos4 x + 2 cos2 x + l

30. cot4 x + 3 cot2 x + 2

31. sin 3 x - cos3 x

32. sin3 a + cos3 a

Each expression simplifies to a constant, a single function, or a power of a function . Use fundamental identities to simplify each expression. 33. tan 8 cos 8

34. cot a sin a

36. cot t tan t

37.

39. sec 2 x - 1

40. csc2 t - 1

sin2 x 41. - -2- + sin x csc x cos x

l 42. - - + cot a tan a tan2 a

l 43. 1 - - 2csc x

44. 1 - - 2sec x

sin {3 tan {3 cos {3

35. sec r cos r 38.

csc 8 sec 8 cot 8

1

5.2 Verifying Trig o nometric Identit ies

329

Verify that each equation is an identity. See Examples 1-5.

cot lJ 4S. - (} = coslJ csc 47.

I - sin2 {3 cos {3

tan a 46. - - =sin a sec a 48.

=cos {3

tan2 a + 1 =sec a sec a

49. cos2 lJ(tan2 lJ + 1) = 1

SO. sin2 {3( 1 + cot2 {3) = 1

Sl. cot lJ + tan lJ = sec lJ csc lJ

S2. sin2 a + tan2 a + cos2 a = sec 2 a

cos a sin a S3. - - + - - = sec 2 a - tan2 a sec a csc a

sin2 lJ S4. - - = sec lJ - cos lJ cos(}

SS. sin4 lJ - cos4 lJ = 2 sin2 lJ - 1

S6. sec4 x - sec2 x = tan4 x + tan2 x

S7. S9. 61.

I - cosx = (cot x - cscx) 2 1 + COSX cos(} sec lJ - 1

cos(}+ 1 tan2 lJ I - sin lJ

+

60. = 2 sec2 lJ

1 +sin lJ

S8. (sec a - tan a) 2 =

1 + tan a

62.

1 - sin a 1 +sin a

(sec lJ - tan lJ) 2 + 1 sec lJ csc lJ - tan lJ csc lJ sec a - tan a

= 2 tan lJ

=seca+ tana

csc lJ +cot lJ =cot lJ csc lJ tan lJ +sin lJ

63.

cot a+ 1 cot a - I

6S.

cos(} = I sin lJ cot lJ

66. sin2 lJ( I + cot2 lJ) - I = 0

67.

sec4 lJ - tan4 lJ = sec2 lJ - tan2 lJ sec2 lJ + tan2 (}

68.

sin4 a - cos4 a = 1 sin2 a - cos2 a

69.

tan2 t - 1 2 sec t

70.

cot2 t - 1 = 1 - 2 sin2 t I + cot2 t

I - tan a

tan t - cot t tan t + cot t

71. sin2 a sec 2 a + sin2 a csc 2 a = sec 2 a

64.

72. tan 2 a sin 2 a = tan 2 a + cos 2 a - 1

73.

sin x tan x + = cot x + sec x csc x 1 - cos x 1 + cos x

74.

sin lJ _sin lJ cos lJ = csc lJ( l + cosZlJ ) 1 - cos (} 1 + cos (}

7S.

I + cosx 1 - cosx

I - cos x = 4 cot x csc x 1 +cos x

76.

I + sin lJ 1 - sin lJ

1 - sin lJ = 4 tan lJ sec lJ 1 +sin lJ

77.

1 - sin lJ = sec2 lJ - 2 sec lJ tan lJ + tan2 lJ 1 +sin lJ

78. sin lJ + cos lJ = - ]

sin lJ cos lJ + ---1 - cot lJ 1 - tan lJ - ]

79. - - - - - + = 2 tan a tan a-seca tana+seca 80. ( 1 + sin x + cosx) 2 = 2( 1 + sin x)( I + cosx) 81. ( 1 - cos2 a) ( 1 + cos2 a) = 2 sin 2 a - sin4 a 82. (sec a + csc a) (cos a - sin a) = cot a - tan a 83.

1 - cosx = csc2 x - 2 csc x cot x + cot2 x 1 +cos x

330

CHAPTER 5 Trigonometric Identities

84.

l - cos 0 l +cos 0

= 2 csc 2 0 - 2 csc 0 cot 0 - I

85. (2 sinx + cosx) 2 + (2 cosx - sin x) 2 = 5 86. sin2 x( I + cot x) + cos2 x( I - tan x) + cot2 x = csc 2 x 87. sec x - cos x + csc x - sin x - sin x tan x = cos x cot x 88. sin3 0 + cos3 0 = (cos 0 + sin 0) ( l - cos 0 sin 0)

ru Graph each expression and use the graph to make a conjecture, predicting what might be an identity. Then verify your conjecture algebraically.

90. ( csc 0 + cot 0) (sec 0 - l )

89. (sec 0 + tan 0)( l - sin 0)

91.

cos 0 + l sin 0 +tan 0

92. tan 0 sin 0 + cos 0

ru Graph the expressions on each side of the equals symbol to determine whether the equation might be an identity. (Note: Use a domain whose length is at least 27T.) If the equation looks like an identity, then verify it algebraically. See Example 1. 93.

2 + 5 cosx sin x

sec2 x 94. l + cot2 x = - - - sec2 x - I

= 2 csc x + 5 cot x

l 96. - - - - + = sec 2 x 1 + sin x l - sin x

95. tan x - cot x = 2 sin2 x tan x + cotx

Show that the equation is not an identity by substituting a number fort. 98. ~=cost

97. sin(csc t) = l 99. csc t = V l + cot2 t

100. cost=

Vl - sin2 t

(Modeling) Work each problem.

101. Intensity of a Lamp According to Lambert's law, the intensity of light from a si ngle source on a flat surface at point P is given by

I = k cos2 0,

()

where k is a constant. (Data from Winter, C. , Solar Power Plants, Springer-Yerlag.) (a) Write I in terms of the sine function. (b) Why does the maximum value of I occur when 0 = O?

102. Oscillating Spring The distance or displacement y of a weight attached

p y

to an oscillating spring from its natural position is modeled by

y = 4 cos 27Tt, where t is time in seconds. Potential energy is the energy of position and is given by

where k is a constant. The weight has the greatest potential energy when the spring is stretched the most. (Data from Weidner, R., and R. Sells, Elementary Classical Physics, Vol. 2, Allyn & Bacon.) (a) Write an expression for P that involves the cosine function. (b) Use a fundamental identity to write Pin terms of sin 27Tt.

5.3 Sum and Difference Identities for Cosine

E:E1 (Modeling) Radio Tuners

331

See Example 6. Let the energy stored in the inductor be

given by L( t) = 3 cos2 6,000,000t and let the energy stored in the capacitor be given by

C( t) = 3 sin 2 6,000,000t, where t is time in seconds. The total energy E in the circuit is given by E(t) = L(t) + C(t). 103. Graph L, C, and E in the window (0, 10- 6 ] by [ - 1, 4 ], with Xscl = 10- 7 and Yscl = 1. Interpret the graph . 104. Make a table of values for L, C, and E starting at t = 0 , incrementing by 10- 1 . Interpret the results. 105. Use a fundamental identity to derive a simplified expression for E(t).

5.3 • •

Sum and Difference Identities for Cosine

Difference Identity for Cosine Sum Identity for Cosine

•

Cofunction Identities

•

Applications of the Sum and Difference Identities

•

Verifying an Identity

Several examples presented earlier have

Difference Identity for Cosine

shown that cos(A - B)

does not equal cos A - cos B.

For example, if A = ~ and B = 0, then

cos(A - B) while

=

cos ( ; - 0) = cos ; = 0, 7T

cos A - cos B = cos - - cos 0 = 0 - 1 = - 1. 2

To derive a formula for cos( A - B), we start by locating angles A and Bin standard position on a unit circle, with B U· ~)

¥ 90° o :":3

o

120 ( - 2, 2 ~ 4 t3so 6 1so0 (- I • 0) 1so0

60

(~. ~) 2

'1T

-

4504 !! 6

2 (../3 I) T' 2

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A-25

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Photo Credits

COVER:

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CHAPTER R: 1 Mdd/Shutterstock 5 Courtesy of Terry McGinnis Fargas/123RF 113 David Anderson/Shutterstock

43 Xavier

CHAPTER 1: 119 Owen Smith/123RF 131 Bill Gozansky/Alamy Stock Photo 137 Courtesy of Callie Daniels 138 Markus Gann/Shutterstock 139 Ingus Kruklitis/123RF CHAPTER 2: 165 Haveseen/123RF 186 Em Campos/Shutterstock 187 Jane Gould/ Alamy Stock Photo 188 VladKol/Shutterstock 189 Manfredxy/Shutterstock 190 (top left) Art Collection 3/Alamy Stock Photo; (top right) Artokoloro Quint Lox Limited/Alamy Stock Photo; (bottom) Didesign021 /Shutterstock 194 Associated Press 195 Courtesy of John Hornsby 199 Anna Omelchenko/Shutterstock 200 World History Archive/Alamy Stock Photo 201 Maximchuk/Shutterstock 207 Courtesy of Callie Daniels 213 (top) Serhii Kamshylin/123RF; (bottom) Great Stock/Shutterstock CHAPTER 3: 217 Vadim Sadovski/Shutterstock 224 Mikhail Bravin/123RF 227 Sam Wirzba/Design Pies lnc/Alamy Stock Photo 233 Lisic/Shutterstock 246 Sirtravelalot/ Shutterstock 251 Dmitry Morgan/Shutterstock 256 John Henshall/Alamy Stock Photo CHAPTER 4: 259 Shane Myers/123RF 278 ShengYing Lin/Shutterstock 284 Jon Nicholls Photography/Shutterstock 305 Alexander Sakhatovsky/Shutterstock 306 Mike Booth/Alamy Stock Photo 312 1558157/S hutterstock CHAPTER 5: 315 Studiovin/Shutterstock 327 Serggod/Shutterstock 340 Ysbrand Cosijn/Shutterstock 348 Hteam/Shutterstock 358 Lithian/123RF 363 US Navy Photo/ Alamy Stock Photo 364 Remik44992/123RF CHAPTER 6: 371 Georgii Dolgykh/123RF 383 EMPICS Sport/PA lmages/Alamy Stock Photo 394 EB Adventure Photography/S hutterstock 398 MindStudio/Pearson Education CHAPTER 7: 415 Noppon kobpimai/123RF 424 Richard Cavalleri/Shutterstock 432 Stephanie Kennedy/Shutterstock 436 Artokoloro Quint Lox Limited/Alamy Stock Photo 441 Inge Johnsson/Alamy Stock Photo 453 WitR/Shutterstock 454 Zdorov Kirill Vladimirovich/Shutterstock CHAPTER 8: 475 Georg Henrik Lehnerer/ 123RF 477 M ichaeljung/Shutterstock 489 Courtesy of Ross Anderson 498 Pearson Education, Inc. 503 Courtesy of Ross Anderson 504 Courtesy of Ross Anderson 516 NatUlrich/Shutterstock

C-1

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Index

A Abscissa, 298 Absolute value of a complex number, 487 definition of, 7 symbol for, 7, 88 Absolute value function , graph of, 88 Acute angles definition of, 120 trigonometric functions of, 14 1- 142, 166, 168 Acute triangle, 130, 417 Addition of complex numbers, 479 identity property for, 18 order of operations for, 16 of ordinates method, 298 of polynomials, 26-27 property of equality, 56 of radicals, 48- 49 of rational expressions, 37-39 of real numbers, 13-14 Additive inverse definition of, 12 of a real number, 6- 7, 12 Adjacent side to an angle, 166 Aerial photography, 425 Airspeed, 450 Algebraic expressions, 15 Alternate exterior angles, 129 Alternate interior angles, 129 Alternating current, 336, 401 Ambiguous case of the law of sines, 426 Amplitude of cosine function, 263 definition of, 263 of sine function, 263 Angle(s) acute, 120, 166 adjacent side to an, 166 alternate exterior, 129 alternate interior, 129 complementary, 12 1 corresponding, 129 coterminal, 123 critical, 412 definition of, 120 of depression, 193 direction angle for vectors, 456 of elevation, 193, 203 of inclination, 449 initial side of, 120 measure of, 120, 179, 183 negative, 120 obtuse, 120 opposite side to an, 166 phase, 257 positive, 120

quadrantal, 123 reference, 174-17 6 right, 120 side adjacent to, 166 side opposite, 166 significant digits for, 190-191 special, 169-170, 176-177, 179 standard position of, 123 straight, 120 subtend an, 230 supplementary, 121 terminal side of, 120 types of, 120 between vectors, 461 vertex of, 120 vertical, 128 Angle-Side-Angle (ASA) axiom, 4 16 Angle sum of a triangle, 130 Angular speed applications of, 247-248 definition of, 246 formula for, 246 Applied trigonometry problems, steps to solve, 194 Approximately equal to definition of, 171 symbol for, 17 1 arccos x, 376 arccot x, 378 arccsc x, 378 Archimedes, spiral of, 51 1-512 Arc length, 224 arcsec x, 378 arcsin x, 374 arctan x, 377 Area of a sector, 226-227 Area of a triangle deriving formula for, 420 Heron's formula for, 436 Argument of a complex number, 487 definition of, 273 of a fu nction, 273, 3 19 Associative properties, 18 Asymptote, vertical, 284 Axis horizontal, 486 imaginary, 486 polar, 505 real, 486 reflecting a graph across an, 90-91 symmetry with respect to an, 9 1-93 vertical, 486 x-, 67 y-, 67

B Base of an exponent, 15 Bearing, 200-202, 450

Beats, 400 Binomials, definition of, 26 Braces, 2 Brackets, 16 Braking distance, 187 British nautical mile, 413

c Calculators. See Graphing calculators Cardioid, 509 Cartesian coordinate system, 67 Cartesian equations, 508 Center of a circle, 72 Center-radius form of the equation of a circle, 72 Chords, table of, 190 Circadian rhythm, 306 Circle(s) arc length of, 224 center of, 72 circumference of, 2 18 defin ition of, 72 equation of, 72 graph of, 72-73 polar form of, 508-509, 513 radius of, 72 sector of, 226 unit, 72, 235 Circular functions applications of, 240 definition of, 235 domains of, 237 evaluating, 237 finding numbers with a given circular function value, 239 interpreting in calculus 72 Circumference of a circle, '2 18 cis (}, 487 Cissoid graphing, 515 with a loop, 515 Clinometer, 199, 213 Closed interval, 62 Closure properties, 18 Cloud ceiling, 199 Coefficient, definition of, 19 Cofunction identities 167 333-334 , , Cofunctions of trigonometric functions, 167 Columbus, Christopher, 200 Common factors in a rational expression, 34, 35 Commutative properties, 18 Complementary angles, 121 Complex conjugates, 481 Complex fractions defi nition of, 14, 39 simplifying, 39-41

1-1

1-2

INDEX

Complex nu mber(s) absolute value of, 487 argument of, 487 conjugate of, 48 1 definition of, 476 De Moivre's theorem for, 498 equality of, 476 graph of, 486 imaginary part of, 476 modulus of, 487 nonreal, 476 nth root of, 499 nth root theorem for, 500 operations on, 478 - 481 polar form of, 487 powers of, 498 product theorem for, 493 pure imaginary, 476 quotient theorem for, 494 real part of, 476 rectangular form of, 486 roots of, 499-502 standard fo rm of, 476, 486 trigonometric form of, 487 Complex plane imaginary axis of, 486 real axis of, 486 Components of a vector horizontal, 456 vertical, 456 Conditional equation, 58-59 Conditional trigonometric equations with half-angles, 395 linear methods for solving, 388 with multiple angles, 396- 397 satisfying, 388 solving with the quadratic formula, 390 steps to solve algebraically, 391 using trigonometric identities to solve, 390-39 1 zero-factor property method for solving, 389 Congruence axioms, 4 16 Congruent triangles, 130, 4 16 Conj ugate(s) of a complex number, 48 1 multiplication of, 49 Constant function, 82-83 Continuity of a function, 84 Continuous function, 84 Contradiction, 58 Coordinate(s) on a line, 4 photographic, 425 of a point, 4 Coordinate (xy) plane, 67 Coordinate system Cartesian, 67 polar, 260, 505 rectangular, 67, 505 Corresponding angles, 129 Cosecant fu nction characteristics of, 295 defi nition of, 14 1 domain of, 295 graph of, 295 inverse of, 378 period of, 295

range of, 15 1, 295 steps to graph, 296 Cosine fu nction amplitude of, 263 characteristics of, 262 definition of, 14 1 difference identity for, 33 1-332 domain of, 262 double-angle identity for, 350-35 1 graph of, 262 half-angle identity fo r, 359 hori zontal translation of, 275 inverse of, 376-377 period of, 262, 265 range of, 15 1, 262 steps to graph, 266, 277 sum identi ty for, 332 - 333 translating graphs of, 275, 276, 277 vertical translation of, 276 Cosines, law of applications of, 434-435 applying to solve a triangle, 432, 434-435 derivation and statement of, 432-433 Heron's formula and, 437- 439 Cost-benefit model, 44 Cotangent function characteristics of, 286 definition of, 14 1 domain of, 286 graph of, 286 horizontal translation of, 289 inverse of, 378 period of, 286 range of, 15 1, 286 steps to graph, 287 vertical translation of, 289 Coterminal angles, 123 Counting numbers, 2, 5 Critical angle, 4 I 2 Cube root(s) application of, 46 explanation of, 45 symbol for, 45 Curvature, degree of, 232, 364 Cycloid calculator graphing of, 520 definition of, 520

D Damped oscillatory motion, 303 Decimal degrees, 122 Decreasing function, 82- 83 Degree measure converting to radian measure, 2 18- 2 19 definition of, 120 Degree mode, calculators in, 144, 182 Degree of a polynomial, 26 Degree of a term, 26 Degree of curvature, 232, 364 Degree/radian relationship converting measures, 2 18-2 19 table of, 220 De Moivre, Abraham, 498 De Moivre's theorem, 498 Denominator(s) least common, 37 rationalizing, 50- 51, 143, 149, 152

Dependent variable, 76-77 Depression, angle of, 193 Derivatives of parametric equations, 192 of trigonometric functions, 323 Descending order, 26 Diastolic pressure, 270 Difference, definition of, 12 Difference identity application of, 334-337, 343-344 for cosine, 331-332 for sine, 341-342 for tangent, 342 Difference tones, 40 l Digits, significant, 190-191 Direction angle of a vector, 456 Discontinuity of the graph of a function, 84 Distance between points, 15 Distance formula, 68-69, 140 Distributive property, 18-1 9 Division of complex numbers, 48 I order of operations for, 16 of rational expressions, 36-37 of real nu mbers, 12, 14 by zero, 13 Domain(s) of circular functions, 237 of a function, 77, 80 of inverse circular fu nctions, 379 of a rational expression, 34 of a relation, 77, 80 Dot product of vectors definition of, 459 geometric interpretation of, 460 properties of, 459 Double-angle identities definition of, 350- 351 simplifying expressions using, 353 verifying, 353

E Eccentricity e, 516 Elements of a set definition of, 2 symbol for, 2 Elevation, angle of, 193 , 203 Ellipse, 519 Empty set definition of, 2 notation for, 2, 59 Endpoint of a ray, 120 Envelopes, 304 Equation(s) Cartesian, 508 of a circle, 72 with complex solutions, 477- 478 conditional, 58-59 conditional trigonometric. See Conditional trigonometric equations connecting graphs with, 267, 290, 298 definition of, 8, 56 equivalent, 56 fi nding ordered-pair solutions fo r, 70 first-degree, 57 graphs of, 7 1-72

INDEX

linear, 56-58 parametric, 192, 273, 518 polar, 508-513 quadratic, 59 rectangular, 508, 512 roots of, 56 second-degree, 59 solution of, 56 solution set of, 56 steps to graph, 71 trigonometric. See Trigonometric equations types of, 58-59 Equilateral triangle, 130 Equilibrant vector, 448 Equivalent equations, 56 Even function, 94-95, 262, 316 Even-odd identities, 316 Exact number, 190 Exponent(s) base of, I 5 definition of, 15 integer, 24 notation for, 15 rules for, 24-25 zero, 25 Exponential expressions definition of, 15 evaluating, 16 Expressions algebraic, 15 exponential, I 5-16 rational. See Rational expressions roots of, 16

F Factor(s) of a number, 15 of a rational expression, 34, 35 Factoring polynomials, 28-30 Finite set, 2 First-degree equation, 57 Four-leaved rose, 5 I 0 Fourth root, 45-46 Fractals, 489 Fraction(s) complex, 14, 39-41 finding least common denominator, 37 fundamental principle of, 34 unit, 248 Fraction bar, 16 Frequency definition of, 302 fundamental, 398 Function(s) absolute value, 88 argument of, 273, 319 circular. See Circular functions constant, 82-83 continuity of, 84 continuous, 84 cosecant. See Cosecant function cosine. See Cosine function cotangent. See Cotangent function decreasing, 82-83 definition of, 76 discontinuity of, 84 domain of, 77

even,94-95,262,316 horizontal translations of, 96 increasing, 82-83 integrals of, 355 inverse, 372-373 inverse cosecant, 378 inverse cosine, 376-377 inverse cotangent, 378 inverse secant, 378 inverse sine, 373-375 inverse tangent, 377 -378 limit of, 7, 81 notation, 81-82 odd, 94-95,261,316 one-to-one, 372 periodic, 260 range of, 77 secant. See Secant function sine. See Sine function sinusoidal, 278 tangent. See Tangent function trigonometric. See Trigonometric functions vertical line test for, 78-79 vertical translations of, 95-96 Fundamental frequency , 398 Fundamental identities, 316-317 Fundamental principle of fractions, 34 f(x) notation, 81

G Grade, 224 Grade resistance, 183 Graph(s) of circles, 72-73 of complex numbers, 486 compressed, 263 connecting with equations, 267, 290, 298 ofcosecantfunction, 295 of cosine function, 262 of cotangent function, 286 definition of, 4 of equations, 71- 73 of even and odd functions, 94 horizontal translation of, 96 of inverse cosecant function , 378 of inverse cosine function, 376-377 of inverse cotangent function, 378 of inverse functions, 373 of inverse secant function, 378 of inverse sine function, 374- 375 of inverse tangent function , 377- 378 of numbers, 4- 5 parametric, 518 of polar coordinates, 506 of polar equations, 508-5 13 reflecting, 90-91 of secant function , 294 shrinking, 88-90 of sine function, 26 1 stretched, 263 stretching, 88-90 summary of polar, 5 13 of tangent function, 285 translations of, 95-98 vertical translation of, 95 - 96

1-3

Graphing calculators analyzing the path of a projectile, 522 converting from trigonometric form to rectangular form, 487 degree mode in, 144, 182 finding magnitude and direction angle of vectors, 456 finding trigonometric function values with, 182 graphing a cycloid, 520 graphing a plane curve defined parametrically, 518 graphing of polar equations, 509- 5 13 sine regression feature, 278-279 solving problems involving angles of elevation, 203 solving trigonometric equations by linear methods, 388 ZTrig viewing window, 263 (is) Greater than symbol, 8- 9 Groundspeed, 450 Grouping symbols, I 6

H Half-angle identities application of, 360-361 defin ition of, 358 general statement about, 359 simplifying expressions using, 361 trigonometric equations with, 395 verifying, 361 Harmonic motion, 302 Heron of Alexandria, 436 Heron' s formula application of, 437 derivation of, 437-439 statement of, 436 Heron triangle, 444 Hipparchus, 190 Horizontal axis, 486 Horizontal component of a vector, 456 Horizontal line test for one-to-one functions, 372 Horizontal shrinking of a graph, 90 Horizontal stretching of a graph, 90 Horizontal translations of cosine function, 275 of cotangent function , 289 definition of, 96 of a graph, 96 graphing, 273-275 of sine function, 274 Hyperbola, 363 Hyperbolic spiral, 526 Hypotenuse, 140, 166 Hypotrochoids, 520

definition of, 476 powers of, 482 simplifying powers of, 482 i, j unit vectors, 459 Identities, 58 - 59, 316- 317. See also Trigonometric identities

1-4

INDEX

Identity property for addition, 18 for multiplication, 18, 40-41 Imaginary axis, 486 Imaginary numbers, 476 Imaginary part of a complex number, 476 Imaginary unit, 476 Impedance, 485, 497 Inclination, angle of, 449 Increasing function, 82-83 Independent variable, 76-77 Index of a radical, 46 Inequality(ies) definition of, 61 explanation of, 8-9 linear, 62-63 properties of, 61-62 summary of symbols, 9 symbols for, 8-9 three-part, 64 Infinite set, 2 Initial point of a vector, 446 Initial side of an angle, 120 Inner product of vectors, 459 Integer exponents, 24 Integers, definition of 4 5 Integrals of functions: 3S5 Intercept(s), 7 1 Intersection of sets 3 Interval(s) ' closed, 62 definition of, 62 notation, 62 open, 62 summary of types of, 63 Inverse additi ve,6-7, 12, 18 multiplicative, 13, 18 Inverse cosecant function definition of, 378 graph of, 378 Inverse cosine function definition of, 376 graph of, 376-377 Inverse cotangent function definition of, 378 graph of, 378 Inverse functions definition of, 372 graphs of, 373-378 notation for, 372 summary of concepts, 373 Inverse properties, 12-13, 18 Inverse secant function definition of, 378 graph of, 378 Inverse sine function definition of, 374 graph of, 374- 375 Inverse tangent function definition of, 377 graph of, 377-378 Inverse trigonometric equations, 403-405 Inverse trigonometric functions domains and ranges of, 379 equations containing, 403-405 graphs of, 374- 378

notation for, 372 summary of, 379 Irrational numbers 4 5 Isosceles triangle, '130

J Julia set, 489

L Lambert's law, 330 Latitude, 225 Law of cosines applications of, 434-435 applying to solve a triangle, 432 derivation and statement of 432-433 , Heron's formula and, 437-439 Law of sines ambiguous case of, 426 applications of, 418-419 derivation and statement of, 4 17 Law of tangents, 473 Least common denominator in fractions, 37 steps to find, 37 Legs of a right triangle, 140 Lemniscates, 511, 513 (is) Less than symbol, 8-9 Like radicals, 48 Like terms, 19 Lima9ons, 512, 5 13 Limit(s) definition of, 7 of a function, 7, 81 notation, 7, 8 1 . of rational expressions, 35 Lme(s). See also Line segment(s) definition of, 120 number, 4-5 parallel, I 28 polar form of, 508 Linear equation in one variable 56-58 Linear inequalities in one variable 62-63 , Linear methods for solving . trigonometric equations, 388 Linear speed applications of, 248 definition of, 246 formula for, 246 Line segment(s) definition of, 120 finding lengths of, 241 function values as lengths of, 240-241 Lissajous figure, 524 Locating points in a plane, 67 Longitude, 233 Lowest terms of a rational expression, 34

M Mach number, 363 Magnitude of a vector, 446 Mandelbrot set, 503 Maor, Eli, 200 Measure of an angle, 120, 179, 183

Members of a set, 2 Midpoint formula, 69 Minute, 122 Models, cost-benefit, 44 Modulus of a complex number, 487 Mollweide's formula, 473 Monomial, 26 MUiler, Johann, 200 Multiple-angle identity, deriving, 354 Multiple angles, trigonometric equations with, 396-397 Multiplication of complex numbers, 480 of conjugates, 49 identity property for, 18, 40-41 order of operations for, 16 of polynomials, 27 -28 of radicals, 49 -50 of rational expressions, 36-37 of real numbers, 12, 14 by zero, I 8 Multiplication property of equality 56 Multiplication property of zero 18, Multiplicative inverse of a real 'number 13, 18 ,

N Natural numbers, 2, 5 Nautical mile, 232 Negation, order of operations for 16 Negative angle, 120 ' Negative of a number, 6-7 Negative of a real number, 18, 45 Negative square roots, 44 Newton, 448 Newton' s formula, 473 Nonreal complex number, 476 Notation exponential, I 5 function, 81-82 interval, 62 inverse function, 372 limit, 7, 81 for ordered pair, 67 set-builder, 3, 63 for sets, 2, 59, 63 subscript, 67 (is) Not equal to explanation of, 8-9 symbol for, 8-9 nth root, notation for, 45 nth root of a complex number, 499 nth root theorem for complex numbers, 500 Null set definition of, 2 notation for, 2 Number(s) absolute value of, 7 - 8 additive inverse of, 6- 7 complex, 476 counting, 2, 5 exact, 190 factors of, 15 graphs of, 4-5 integers, 4, 5 irrational, 4, 5 natural, 2, 5

INDEX

negative of, 6- 7 opposite of, 6-7 pure imaginary, 476 rational, 4 - 5 real. See Real numbers reciprocal of, 13, 18 sets of, 5- 6 signed, 7 square of, 44 square roots of, 44 whole, 2, 5 Number line(s) coordinate of a point on, 4 definition of, 4 dista nce between points on, 15 graph of a point, 4 using, 4-5 Numerical coefficient, 19

0 Oblique triangle(s) congruency and, 4 16-417 data required for solving, 416 definition of, 4 16 finding area of, 420- 421 solving procedures for, 41 6-419 Obtuse angle, 120 Obtuse triangle, 130, 417 Odd fu nction, 94-95, 261, 316 Ohm 's law, 485 One-to-one function definition of, 372 horizontal line test for, 372 Open interval, 62 Operations order of, 16-17 on polynomials, 26-28 on radicals, 48-50 on real numbers, 12-14 Opposite of a number, 6- 7 Opposite of a vector, 447 Opposite side to an angle, 166 Orbital period of a satellite, 5 16 Ordered pair(s) definition, 67 finding in solutions of equations, 70 notation for, 67 Ordered triple(s), 68 Order of a radical, 46 Order of operations, 16- 17 Ordinates, addition of, 298 Origin definition of, 67 symmetry with respect to the, 93-94 Orthogonal vectors, 461

p Parabola, 72 Parallel lines definition of, 128 transversal intersecti ng, 128 Parallelogram rule for vectors, 447 Parallelograms, properties of, 447 Parametric equations definition of, 273 , 518 derivatives of, 192 of a plane curve, 518

Parentheses, 16 Path of a projectile, 522 Perfect triangle, 444 Period of cosecant fu nction, 295 of cosine fu nction, 262, 265 of cotangent function, 286 definition of, 260, 392 of secant fu nction, 294 of sine function, 26 1, 265 of tangent function, 285 Periodic function definition of, 260 period of, 260 Phase angle, 257 Phase shift, 273 Photographic coordinates, 425 Pi (7T), 5, 218 Plane c urve, 518 Plane trigonometry, 190 Plimpton 322 (tablet), 190 Point(s) coordinate on a li ne, 4 distance between, 15 locating in a plane, 67 Polar axis, 505 Polar coordinates of a point definition of, 260, 505 graph of, 506 Polar coordinate system definition of, 505 polar axis of, 505 pole of, 505 Polar equations calculator graphing of, 508-51 3 classifying, 512-513 converting to rectangular equations, 512 definition of, 508 graphs of, 508 - 5 13 Polar form of a complex number, 487 of lines and circles, 508-509 Polar grid, 509 Pole of a polar coordinate system, 505 Polya, George, 194 Polynomial(s) addition of, 26-27 definition of, 26 degree of, 26 factoring, 28 - 30 mul tiplication of, 27-28 subtraction of, 26-27 term of, 19 in x, 26 Position vector, 456 Positive angle, 120 Positive square roots, 44 Power rules for exponents, 24 Powers of complex numbers, 498 of i, 482 Principal square root, 44 Prism diopter, 2 13 Product, definition of, 12 Product rule for expone nts, 24 fo r radicals, 46-47

1-5

Product theorem for complex numbers, 493 Product-to-sum identities, 354-355 Properties of equality, 56 of inequality, 6 1- 62 of parallelograms, 447 of triangles, 130 Ptolemy, 190, 425 Pure imaginary numbers, 476 Pythagorean identities, 152-153, 3 17 Pythagorean theorem, 68, 140

a Quadrantal angles, 123 Quadrants, 67 Quadratic equation definition of, 59 standard form of, 59 Quadratic form ula, 61 Quotient, 12 Quotient identities, 153, 317 Quotient rule for radicals, 46-47 Quotient theorem for complex numbers, 494

R Radian, 2 18 Radian/degree relationship converting measures, 218-2 19 table of, 220 Radian measure applications of, 225 - 226 converting to degree measure, 2 18- 2 19 definition of, 21 8 Radical(s) adding, 48 - 49 explanation of, 44 index of, 46 like, 48 multiplication of, 49-50 operations on, 48-50 order of, 46 product rule fo r, 46- 47 quotient rule fo r, 46 - 47 simplifying, 47-48, 5 1-53 subtraction of, 48- 49 unlike, 49 Radical expressions conditions for simpl ified, 51 explanation of, 44 operations on, 48 - 50 squari ng of, 45 Radicand, 44 Radius of a circle, 72 Range(s) of a function, 77, 80 of inverse circular functions, 379 of a relation, 77, 80 of trigonometric functions, 151 Rational expression(s) addition of, 37-39 common factors of, 34, 35 definition of, 34 division of, 36-37 domain of, 34

1-6

INDEX

Rational expression(s) (Continued) least common denominator of, 37 limits in, 35 lowest terms of, 34 multiplication of, 36-37 subtraction of, 37-39 Rationalizing denominators, 50-5 1 Rational numbers, definition of, 4 - 5 Ray definition of, 120 endpoint of, 120 r cis (}, 487 Reactance, 485 Real axis, 486 Real numbers addition of, 13-14 additive inverse of, 7, 12, 18 definition of, 5 division of, 12, 14 multiplication of, 12, 14 multiplicative inverse of, 13, 18 negative of, 18, 45 operations on, 12- 14 properties of, 18- 20 reciprocals of, 13, 18 sign rules for operations on, 12 subtraction of, 13-14 summary of properties of, 18 Real part of a complex nu mber, 476 Reciprocal definition of, 148 of a real number, 13, 18 Reciprocal identities, 148- 149, 3 17 Rectangular coordinates, converting to polar coordinates, 506 Rectangular coordinate system, 67 Rectangular equations converting polar equations to, 5 12 definition of, 508 Rectangular form of a complex number converting to trigonometric form, 488 definition of, 486 Reduction form ula, 340 Reference angles definition of, 174 special angles as, 176-177 table of, 176 Reference arc, 235 Reflecting a graph, 90- 91 Refraction of light, 187 Regiomontanus, 200 Regression, sine, 278-279 Related rates, 192 Relation definition of, 76 domain of a, 77 range of a, 77 Resistance grade, 183 impedance and, 485 Resultant vectors, 447, 486 Right angle, 120 Right triangle(s) application of, 200 definition of, 130 hypotenuse of, 140, 166 legs of, 140

in Pythagorean theorem, 68 solving, 191 Right-triangle-based definitions of trigonometric functions, 166 Root(s) of an equation, 56 of an expression, 16 of a complex number, 499-502 cube, 45 - 46 fourth , 45-46 nth, 45 - 46 principal, 44 square, 44-45 Rose curve, 510, 5 13 Rules for exponents, 24- 25

s Sag curve, 188 Scalars, 446 Scalene triangle, 130 Secant function characteristics of, 294 definition of, 14 1 domain of, 294 graph of, 294 inverse of, 378 period of, 294 range of, 15 1, 294 steps to graph, 296 Second, 122 Second-degree equation, 59 Sector of a circle area of, 226-227 definition of, 226 Segment of a line. See Line segment(s) Semiperimeter, 436 Set(s) definition of, 2 elements of, 2 empty, 2 fin ite, 2 infinite, 2 intersection of, 3 members of, 2 notation for, 2, 59, 63 null,2 union of, 3 Set braces, 2 Set-builder notation, 3, 63 Shrinking a graph, 88-90 Side adjacent to an angle, 166 Side-Angle-Angle (SAA) axiom, 4 16 Side-Angle-Side (SAS) axiom, 4 16 Side opposite an angle, 166 Side-Side-Side (SSS) axiom, 416 Signed numbers, 7 Significant digits, 190- 191 Signs of trigonometric function values, 149 Similar triangles conditions for, 131 definition of, 130 Simple harmonic motion, 302 Simplifying complex fractions, 39-41 radicals, 47- 48, 51-53

Sine function amplitude of, 263 characteristics of, 261 definition of, 141 difference identity for, 341-342 domain of, 261 double-angle identity for, 350-35 1 graph of, 261 half-angle identity for, 359 horizontal translation of, 274 inverse of, 374-375 period of, 26 1, 265 range of, 15 1, 261 steps to graph, 266, 277 sum identity for, 341- 342 translating graphs of, 274, 277 Sine regression, 278- 279 Sines, law of ambiguous case of, 426 applications of, 418 - 419 derivation of, 417 Sine wave, 261 Sinusoid, 261 Sinusoidal function, 278 Snell's law, 187, 412-413 Solar constant, 272 Solution, of an equation, 56 Solution set, 56 Solving triangles, 191 , 418-419, 427-429, 434-436 Sound waves, 272, 340 Special angles finding angle measures with, 179 reference angles as, 176- 177 trigonometric function values of, 169-170 Speed angular, 246 - 248 linear, 246 - 248 Spherical trigonometry, 190 Spiral, hyperbolic, 526 Spiral of Archimedes, 511-512 Square of a number, 44 Square root(s) explanation of, 44 finding, 45 negative, 44 of a negative number, 45 of a positive number, 44 positive, 44 principal, 44 symbol for, 44 Square root property, 60 Squaring of radical expressions, 45 Standard form of a complex number, 476, 486 of a quadratic equation, 59 Standard position of an angle, 123 Statute mile, 232 Straight angle, 120 Stretching a graph, 88-90 Subscript notation, 67 Subtend an angle, 230 Subtense bar method, 202 Subtraction of complex numbers, 479 order of operations for, 16

INDEX

of polynomials, 26-27 of radicals, 48- 49 of rational expressions, 37-39 of real numbers, 13-14 of vectors, 447 Sum, defi nition of, 12 Sum identity application of, 334-337, 343-346 for cosine, 332-333 for sine, 34 1-342 for tangent, 342 Sum-to-product identities, 356 Superelevation, 186 Supplementary angles, 121 Swokowski, Earl, 355 Symbol(s) for cube root, 45 for elements of a set, 2 grouping, 16 for inequalities, 8-9 for square roots, 44 Symmetry, 9 1-94 Systolic pressure, 270

T Tangent function characteristics of, 285 definition of, 141 difference identity for, 342 domain of, 285 double-angle identity for, 351 graph of, 285 half-angle identity for, 359 horizontal translation of, 289 inverse of, 377-378 period of, 285 range of, 151 , 285 steps to graph, 287 sum identity for, 342 vertical translation of, 289 Tangents, law of, 473 Term(s) coefficient of, 19 definition of, 19 like, 19 of a polynomial, 19 unlike, 19 Terminal point of a vector, 446 Terminal side of an angle, 120 Three-part inequality, 64 Transit, 197 Translation(s) combinations of, 277-278 definition of, 95 of graphs, 95-98 horizontal, 96, 273-275, 289 summary of, 98 vertical, 95-96, 276, 289 Transversal definition of, 128 interior angles on same side of, 129 Triangle(s) acute, 130, 417 angle sum of, 130 applications of, 132 area of, 420- 421 , 436

congruent, 130, 416 equilateral, 130 Heron, 444 isosceles, 130 oblique, 4 16-421 obtuse, 130, 417 perfect, 444 properties of, 130 right. See Right triangle(s) scalene, 130 similar, 130, 131 solving, 191, 41 8-419, 427-429, 434- 436 types of, 130 Triangulation method, 443 Trigonometric equations conditional, 388 with half-angles, 395 inverse, 402 linear methods for solving, 388 with multiple angles, 396-397 quadratic methods for solving, 389-390 solving by squaring, 391 solving using the quadratic formu la, 390 solving by zero-factor property, 389 Trigonometric form of a complex number, 487 Trigonometric function s circular, 235 cofunctions of, 167, 333-334 combinations of translations of, 277 defi nitions of, 141 derivatives of, in calculus, 323 domains of, 237 interpreting in calculus, 72 inverses of, 372-379 ranges of, 151 right-triangle-based defi nitions of, 166 translations of, 273-278 Trigonometric function values of acute angles, 166, 168 for angles in radians, 221 finding with a calculator, 182 of nonquadrantal angles, 177 of quadrantal angles, 143-144 signs and ranges of, 149 of special angles, 169- 170 undefined, 144 Trigonometric identities cofunction , 167, 333-334 difference, 331-332, 334, 341-342 double-angle, 350-35 1, 353 even-odd, 3 16-3 17 fundamental, 316-3 17 half-angle, 358-361, 395 product-to-sum, 354-355 Pythagorean, 152-153, 317 quotient, 153, 317 reciprocal, 148, 317 solving conditional trigonometric equations using, 390-391 sum, 332-333, 334, 341-342

1-7

sum-to-product, 356 verifying, 323-327, 337, 346 Trigonometric models, 268, 278-279 Trigonometric substitution, 153 Trinomials, definition of, 26 Triples, ordered, 68 Trochoid, 520

u Umbra, 137 Union of sets, 3 Unit, imaginary, 476 Unit circle, 72, 235 Unit fractions, 248 Unit vector, 458 Unlike radicals, 49 Unlike terms, 19 Upper harmonics, 398, 401

v Variable(s) defin ition of, 2 dependent, 76-77 independent, 76-77 Vector(s) algebraic interpretation of, 456-457 angle between, 461 applications of, 449-451 components of, 456 direction angle for, 456 dot product of, 459 -461 equilibrant of, 448 horizontal component of, 456 i, j units, 459 initial point of, 446 inner product of, 459 magnitude of, 446 naming practices, 446 operations with, 458 opposite of, 447 orthogonal, 461 parallelogram rule for, 447 position, 456 quantities, 446 resultant of, 447, 486 symbol for, 446 terminal point of, 446 unit, 458 vertical component of, 456 zero, 447 Verifying trigonometric identities, 323 - 327, 337, 346 Vertex of an angle, 120 Vertical angles, 128 Vertical asymptote, 284 Vertical axis, 486 Vertical component of a vector, 456 Vertical line test, 78-79 Vertical shrinking of a graph, 89 Vertical stretching of a graph, 89 Vertical translations of cosine function, 276 of cotangent function, 289 defin ition of, 96 of a graph, 95-96

1-8

INDEX

Vertical translations (Continued) graphing, 276 of tangent function, 289

w Wave sine, 261 sound, 272,340 Whole numbers, 2, 5

x

z

x-axis, 67 x-intercept, 7 1 xy-plane, 67

Zero(s) division by, 13 multiplication by, 18 multiplication property of, 18 Zero exponent, 25 Zero-factor property, 34, 59-60, 389 Zero vector, 447 ZTrig viewing window, 263

v y-axis, 67 y-intercept, 7 1

3.1 Conversion of Angular Measure Degree/Radian Relat ionship:

3.2 Applications of Radian Measure

180° =

7r

Arc Length: s = r I!,

radians

Conversion Formulas: From Degrees Radians

To

Multiply by

Radians

-

3.4

2

Angular Speed w

7r

Linear Speed v s

Ii w=-

180

v= -

t

f

180°

Degrees

Ii in radians

s'l = .!_r2 1!, Ii in radians

Area of Sector:

--

(w in radians per

7r

unit time ,

r Ii v= -

r

e in

v = rw

radians)

5.5 Product-to-Sum and Sum-to-Product Identities

5.1 Fundamental Identities I

1

1

cot I! = - sec I! = - cscl! = - tan I! cos I! sin I! sin I! cos Ii tan Ii = - - cot Ii = - cos Ii sin I! sin2 I! + cos2 I! = I tan2 Ii + I = sec 2 I! I + cot2 Ii = csc2 I! sin( - Ii) = -sin I!

cos( - Ii) = cos I!

tan ( - Ii) = - tan I!

csc( - 1!) = - csc I!

sec( - 1!) = sec Ii

cot( - 1!) = - cot I!

cos A cos B =

I

1

sin A cos B = 2 (sin(A + B ) + sin(A - B) J

5.3, 5.4 Sum and Difference Identities

cos(A - B ) = cos A cos B + sin A sin B sin(A + B ) = sin A cos B + cos A sin B

I

2

[sin(A + B ) - sin(A - B)]

sin A + sin B = 2 sin ( -A+2

B) cos (A- -2- B)

sin A - sin B = 2 cos ( -A+2

B) sin (A- -2- B)

cos A + cos B = 2 cos ( -A+2

B) cos ( -A -2- B)

sin(A - B ) = sin A cos B - cos A sin B tan(A + B ) = tan A + tan B I - tan A tan B

cos A - cos B = - 2 sin ( -A+2

tanA - tan B tan(A - B ) = I + tan A tan B

5.3 Cofunction Identities

B) sin (A- -2- B)

5.5, 5.6 Double-Angle and Half-Angle Identities cos 2A = cos2A - sin2 A cos 2A = I - 2 sin2 A cos2A = 2cos2 A - I sin 2A = 2 sin A cos A

cos(90° - I!) = sin I! sin(90° - Ii) = cos I! tan (90° - Ii) =cot Ii

cos~2 =±~ v~

2 tan A tan 2A = I - tan2 A

cot(90° - I!) = tan I!

sin~= ±~ 1 - cosA

sec(90° - Ii) = csc Ii

2

csc(90° - Ii) = sec I!

tan ~ = + 2 -

2

7.1 Law of Sines

7.3 Law of Cosines

In any triangle ABC, with sides a, b, and c,

In any triangle ABC, with sides a, b, and c,

a

b

a

c

-- = sin A sin C'

and

b sin B

a 2 = b 2 + c 2 - 2bc cosA ,

c

sin c ·

and

c2

1 - cos A 1 + cos A

A I - cos A tan- = - - 2 sin A

A sin A tan - = 2 1 +cos A

-- = -sin A sin B'

B)J

sin A sin B = 2 ( cos( A - B ) - cos( A + B ) ]

cos A sinB = cos(A + B ) = cos A cos B - sin A sin B

I

"2 [cos( A + B ) + cos(A -

=

a2

+

b2 -

b 2 = a 2 + c 2 - 2ac cos B, 2ab cos C.

Area of a Triangle

Heron 's Area Formula

The area s'l of a triangle is given by half the product of the lengths of two sides and the sine of the angle between the two sides.

If a triangle has sides of lengths a, b, and c, with semiperimeter

s'l =

I

.

2 be sm A,

I

.

s'l = 2ab sm C,

I

.

s'l = 2ac sm B

s =&(a+ b + c), then the area s'l of the triangle is

s'l = Ys(s - a)(s - b )(s - c).

4.1-4.4 Trigonometric (Circular) Functions The graph of y = c +a sin [ b(x - d)] or y = c +a cos [ b(x - d ) ], where b > 0, has amplitude Ia I. period ¥-, a vertical translation up c units if c > 0 or down Ic I units if c < 0, and a phase shift to the right d units if d > 0 or to the left IdI un its if d < 0. The graph of y = a tan bx or y = a cot bx has period ~, where b > 0. y

y

y

y

y

y

I I i y=cotx

6.1 Inverse Trigonometric (Circular) Functions

y

y

y :!!.. 2

- - - - - -7r 2

------

- - - - _!!_

_ _ _OL_ _ _

2

y = cos-1 x or y arccos x

-I

y

=sin- 1 x

=

or y = arcsin x

"'

y

2

=arctan x

2

y

'.!!: 2

~

-2

\

0

y

y

y = csc 1 x or y = arccscx

y = cor- 1 x or y = arccotx

y = sec-1 x or y = arcsecx

x

~-

7r ------

7r

------

2

_:!!: 2

-2 - I

0

x 2

-----