Challenging Problems in Trigonometry: The Mathematic Series 9781311853929, 1311853928

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Challenging Problems in Trigonometry Book One Copyright 2015 Demetrios P. Kanoussis All rights reserved. No part of this publication may be reproduced, distributed, or transmitted in any form or by any means, electronic or mechanical, without the prior written permission of the publisher, except in the case of brief quotations and certain other noncommercial uses permitted by copyright law.

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Published by Golden Ratio Publications at Smashwords, www.goldenratiopublications.com

First Edition, February 2015 ISBN: 9781311853929 Cover photo: James, desktopwallpapers4.me Cover design: Vicky Karistinou, GRP

Challenging Problems in Trigonometry – Book One Part One: Fundamental Formulas in Trigonometry Part Two: Problems Part Three: Solution to problems About the author

Part One: Fundamental Formulas in Trigonometry

The main trigonometric functions are the following,

1. sin2 x + cos2 x = 1

2. sin (x + y) = sin x ∙ cos y + sin y ∙ cos x

3. sin (x − y) = sin x∙ cos y − sin y ∙ cos x

4. cos (x + y) = cos x ∙ cos y − sin x ∙ sin y

5. cos (x − y) = cos x ∙ cos y + sin x ∙ sin y

10. sin 2x = 2 ∙ sin x ∙ cos x

11. cos 2x = cos2 x − sin2 x = 2∙cos2 x − 1 = 1 − 2∙sin2 x

22. sin 3x = 3 ∙ sin x − 4 ∙ sin3 x

23. cos 3x = 4 ∙ cos3 x − 3 ∙ cos x

Fig.1. A triangle ABC, with the circumcircle (C) with radius R, the in-circle (c) with radius r, and the three ex-circles (Ca), (Cb) and (Cc), having radii ra, rb and rc respectively.

In any triangle ABC, having angles A, B and C (A + B + C = π), and sides a, b and c (see Fig.1), we have:

27. a ∙ b ∙ c = 4 ∙ E ∙ R where E is the area of ABC

31. If ℓA, ℓB and ℓC are the internal bisectors of a triangle ABC, then

32. If tA, tB and tC are the external bisectors of a triangle ABC, then

33. If mA, mB and mC are the three medians of a triangle, then

34. If i = −1, then eix = cos x + i sin x, and e−ix = cos x − i sin x (Euler’s Formula)

36. If z = r ∙ {cos θ + i sin θ}, is a complex number z, in polar form, then

(De Moivre Formula)

37. Every complex number z = r ∙{cos θ + i sin θ}, r > 0, has exactly n different, nth roots, given by

38. If x is measured in radians, then

The hyperbolic functions, sinh x (hyperbolic sine of x), cosh x (hyperbolic cosine of x), tanh x (hyperbolic tangent of x), and coth x (hyperbolic cotangent of x), are defined as follows:

42. cosh2 x − sinh2x = 1

The Trigonometric Functions and the Hyperbolic Functions, are also defined for complex values of their arguments. If z = x + i y is a complex number, then

45. Contrary to the fact that the functions sin x and cos x are bounded, for Real x, ( ), the bounded, i.e.

and

46. sin(ix) = i sinh x, cos(ix) = cosh x

47. sinh(ix) = i sin x, cosh(ix) = cos x

and , as

, for complex z, are not .

Part Two: Problems

1. Solve the equation sin x + sin 2x + sin 3x = 0

2. Solve the equation sin 5x = 16 ∙ sin5 x

3. If cos x

0, show that (3∙cos x + sec x)2 − 12

0

4. Solve the equation tan x ∙ tan 3x + tan π/4 = 0

5. If tan x + tan y + tan z = tan x ∙ tan y ∙ tan z, then show that x + y + z = n ∙ π, where n is an integer number.

6. If

then show that tan2z = tan x ∙ tan y

7. If A, B, C and D are the angles of a convex qudrilateral, show that

8. Given that 4 sin 2x + 3 cos 2x = 3, find tan x/2

9. Show that

10. a) Show the identity cot x − cot 2x = csc 2x b) Making use of (a), show that

where κ is any positive integer.

11. Find sin(18o) and cos(36o)

12. Express sin(5x) in terms of the sin x, and cos(5x) in terms of the cos x.

13. Evaluate the sums, S1 = sin x + sin(x+a) + sin(x+2a) +…+ sin(x+(n−1)a), and S2 = cos x + cos(x+a) + cos(x+2a)+…+ cos(x+(n−1)a).

14. Show that cot 81o − cot 63o − cot 27o + cot 9o = 4

15. If tan−1x represents the principal branch of the inverse tangent function, show that

provided that x > 0, y > 0, and xy < 1.

16. Evaluate the sum

, and then evaluate the sum of the infinite series

17. a) Show that cot {cot−1(2κ−1) − cot−1(2κ+1)} = 2κ2 b) Making use of (a), evaluate the sum

(The cot−1 represents the principal branch of the inverse cotangent function).

18. Solve the equation, cot−1 x + cot−1(10−x) = cot−1(x−1)

19. Show that the quantity Q = x∙sin2a + y∙sin2a + z∙cos2a, where

,

lies between

(The x, y, z are real numbers)

20. If A, B and C are the angles of a triangle ABC, and sin 4A + sin 4B + sin 4C = 0, then show that the ABC is an orthogonal triangle.

21. If A, B and C are the angles of a triangle ABC, and sin2 A + sin2 B + sin2 C = 2, then show that the ABC is an orthogonal triangle.

22. Eliminate the angle φ between the equations, x = 3∙sin φ − sin 3φ, y = cos 3φ + 3∙cos φ.

23. Given that cos x = cos y ∙ cos z + sin y ∙ sin z ∙ cos a cos y = cos x ∙ cos z + sin x ∙ sin z ∙ cos b cos z = cos x ∙ cos y + sin x ∙ sin y ∙ cos c where the angles x, y, z, a, b, c are between 0 and π, show that

24. If show that

, .

25. Eliminate x and y from the equations, (a a∙sin2 x + b∙cos2 x = 1 a∙cos2 y + b∙sin2 y = 1 a∙tan x = b∙tan y

b)

Part Three: Solution to Problems

1. Trigonometric equation is an equation relating the trigonometric numbers (i.e. sin, cos, tan, …) of an unknown arc or angle x. Solving the trigonometric equation, means to find all angles x, which identically satisfy the given equation. For example, sin x + cos x = 1/3, or tan x + tan 2x = cos 2x etc. are trigonometric equations. Due to the periodicity of the trigonometric functions, as a rule every trigonometric equation has an infinite number of solutions, contrary to the polynomial equations, for example, where the number of solutions (roots), is limited. The usual method to solve a complicated trigonometric equation, is to reduce it down to some fundamental form of trigonometric equation, i.e. sin x = a, −1

a

1 , or

cos x = b, −1

b

1 , or

tan x = c, −

0, y > 0 and xy < 1

.

Having thus defined the A, B and C, we want to show that A + B = C. Let us firstly show that

or making use of eqs. (*), (**), (***),

which is true, therefore, we have proved that tan(A + B) = tan C

A + B = C + κ∙π, (****)

where κ is some integer. From the eqs. (*), (**) and (***), we have, respectively,

and adding together, we obtain,

or, making use of eq. (****),

The only integer, satisfying this inequality, is κ=0, therefore ep. (****) becomes, A + B = C, and this completes the proof.

16. We will start with the formula (7),

and apply it for tan x = κ + 1, (i.e. x = tan−1(κ+1)), and tan y = κ, (i.e. y = tan−1 (κ)). In this case, we have,

For κ = 1, 2, 3,…, n, we obtain successively,

Adding together these formulas, and taking into consideration the cancellations involved, one easily obtains,

(since tan π/4 = 1). Passing to the limit, as n

17. a) Let cot−1(2κ−1) = Α and cot−1(2κ+1) = Β

, we obtain,

cot A = 2κ − 1,

cot B = 2κ + 1.

Making use of formula (9), we obtain

b) From part (a), one obtains easily that cot−1(2κ−1) − cot−1(2κ+1) = cot−1(2κ2). (*) In order to evaluate the sum, Sκ = cot−1(2∙12) + cot−1(2∙22) + … + cot−1(2∙κ2), we will make use of eq. (*). According to this equation,

cot−1(2∙12) = cot−1(1) − cot−1(3), cot−1(2∙22) = cot−1(3) − cot−1(5), cot−1(2∙32) = cot−1(5) − cot−1(7),

cot−1(2∙κ2) = cot−1(2κ−1) − cot−1(2κ+1). Adding together, we can easily see the cancellations involved, (telescopic series), leading to the final result, Sκ = cot−1(1) − cot−1(2κ+1) = π/4 − cot−1(2κ+1).

18. Starting with the given equation, taking the cotangent of both sides, and making use of formula (9), one obtains,

which constitute the roots of the given equation. Notice: We have used the obvious fact that, cot (cot−1 x) = x, cot (cot−1 (10−x)) = 10 − x, and cot (cot−1 (x−1)) = x − 1.

19. The functions sin2 a, cos2 a and sin 2a, can all be expressed, rationally, in terms of the tan a. Indeed,

(formula (12)). The given expression Q, can now be expressed in terms of the tan a, as follows,

or, if we make the substitution s = tan a,

or equivalently, s2 (x−Q) + 2ys + z − Q = 0.

(*)

This, quadratic for s equation, must have real roots, (for s), i.e. its discriminant, Δ must be positive or zero, i.e. Δ = 4y2 − 4∙(x−Q)(z−Q)

0,

which in turn means that −Q2 + (x+z)Q − xz + y2

0.

(**)

The quantity Q, therefore, should lie between the two roots of this quadratic for Q, polynomial. The two roots Q1, and Q2, are

and similarly,

Therefore,

, and this completes the proof.

20. Starting with sin 4A + sin 4B + sin 4C = 0 2sin (2A+2B) ∙ cos (2A−2B) + 2sin (2C) ∙ cos (2C) = 0

(*)

Since A, B and C are the angles of a triangle, A+B+C=π

Α+ Β = π − C

2A + 2B = 2π − 2C, i.e.

sin (2A + 2B) = sin (2π − 2C) = sin (−2C) = −sin (2C), and eq. (*), becomes, 2∙sin (2C)∙{ cos(2C) − cos(2A − 2B)} = 0, or equivalently, by virtue of formula (21), 4∙sin (2C) ∙ sin(A−B−C) ∙ sin(A+C−B) = 0. (**) Since B + C = π − A

A − (B+C) = A − (π−A) = 2A−π,

And sin(A−B−C) = sin(2A−π) = −sin(2A), and similarly, sin(A+C−B) = sin(π−2Β) = sin(2Β). In the light of the last two equations, eq. (**), becomes, sin (2A) ∙ sin (2B) ∙ sin (2C) = 0, which means that one of the factors is zero, for example, let sin(2A)=0, i.e. 2A=π or A=π/2, and this completes the proof.

21. From the given equation, sin2 A + sin2 B + sin2 C = 2

sin2 C = 1 − sin2 A +1 − sin2 B

sin2 C = cos2 A + cos2 B, and using the formula (11),

1 + cos 2C = −(cos 2A+cos 2B) 2 ∙ cos2 C = −2 ∙ cos(A+B) ∙ cos(A−B) Since A + B = π − C

(*)

cos(A+B) = −cos C,

So eq. (*) becomes, cos2 C = cos C ∙ cos(A−B)

cos C ∙ { cos C−cos(A−B) } = 0,

or equivalently,

(**)

Since B+C = π−A

, therefore,

and similarly,

Equation (**), becomes finally, cos A ∙ cos B ∙ cos C = 0, From which one of the factors must necessarily be zero, for example, let cos A=0, i.e. A=π/2, and the proof is completed.

22. Making use of the formulas (22) and (23), we have, x = 3sin φ − sin 3φ = 4 sin3 φ, and y = cos 3φ + 3cos φ = 4cos3 φ.

(*) (**)

From eq. (*) we obtain x2/3 = 42/3 ∙ sin2 φ, While from eq. (**), we obtain y2/3 = 42/3 ∙ cos2 φ. Adding together, the last two equations, results in x2/3 + y2/3 = 42/3, (since sin2 φ + cos2 φ = 1).

23. From the first of the given equations, we have

and squaring both sides,

and if we substitute sin2 y and sin2 z by 1−cos2 y and 1−cos2 z, respectively, and then divide both sides by sin2 x, we obtain,

(*) We notice that the right side of eq. (*), remains unchanged, if we perform a cyclic permutation on { x, y, z }, which means that

or since all angles are between 0 and π, (and therefore all sines are positive numbers),

24. Let

where, we have defined 1/m, to be the common value of the equal fractions. Therefore, a = m ∙ cos x b = m ∙ cos(x+φ) c = m ∙ cos(x+2φ) d = m ∙ cos(x+3φ) Then,

25. If we express the sines and the cosines, in terms of the tangents, we have,

a ∙ tan2 x + b = 1 + tan2 x

(a−1) ∙ tan2 x = 1 − b,

(*)

and quite similarly, from the second equation, we obtain, (b − 1) ∙ tan2 y = 1 − a

(**)

From the third equation, a∙tan x = b∙tan y

and taking into consideration eqs. (*) and (**), we obtain,

which is easily simplified to the following, (a − b)∙(a + b −2∙a∙b) = 0, or since a−b

0,

a + b − 2∙a∙b = 0, which is the result of elimination of x and y, among the given equations. ### End of current e-book

About the author Demetrios P. Kanoussis, Ph.D. [email protected] https://www.facebook.com/public/Demetrios-Kanoussis

Dr. Kanoussis is a professional Electrical Engineer with Ph.D in Engineering and MSc in Mathematics from the Tennessee Technological University in USA, and BS in Electrical Engineering from N.T.U.A in Greece. He has long teaching experience on the fields of Applied Mathematics and Electrical Engineering. As a professional engineer, he has been actively involved in the design and implementation of various projects, mainly in the area of Integrated Control Systems. His scientific research in Mathematics and Engineering is published in many, high impact, international journals. Additionally to his teaching and research, he is the author of textbooks in Electric Circuit Theory and Applied Mathematics. He’s one of the founders of the Golden Ratio Publications effort.

More titles by Demetrios P. Kanoussis Algebraic Equations, e-book, February 2015, Golden Ratio publications Direct Current Circuit Analysis (D.C. Circuit Analysis), e-book, March 2015, Golden Ratio publications (soon to be released) Introduction to Electric Circuit Analysis, paperback, August 2013, Vogiatzi publications (Greek edition) Topics in Applied Mathematics, paperback, November 2011, Giourdas publications (Greek edition)