103 Trigonometry Problems

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4 Solutions to Introductory Problems 1. [AMC12 1999] Let x be a real number such that sec x − tan x = 2. Evaluate sec x + tan x. Solution: Note that 1 = sec 2 x − tan 2 x = (sec x + tan x)(sec x − tan x). Hence sec x + tan x = 1 2 . 2. Let 0 ◦ 0. Thus t 3 1 > tan θ>0, we have t 1 < 1 t 3 >t 1 > t 2 .
84 103 Trigonometry Problems 3. Compute (a) sin 12 π , cos 12 π , and tan 12 π ; (b) cos 4 24 π − sin4 24 π ; (c) cos 36 ◦ − cos 72 ◦ ; and (d) sin 10 ◦ sin 50 ◦ sin 70 ◦ . Solution: (a) By the double-angle and addition and subtraction formulas,wehave cos π ( π 12 = cos 3 − π ) = cos π 4 3 cos π 4 + sin π 3 sin π 4 = 1 √ √ √ √ √ 2 3 2 2 + 6 2 · 2 + 2 · 2 = . 4 Similarly, we can show that sin 12 π √ √ = 6− 2 4 . It follows that tan 12 π = √ √ √ 6− √ 2 = 2 − √ 3. 6+ 2 (b) By the double-angle and addition and subtraction formulas, we obtain cos 4 π 24 − π ( sin4 24 = cos 2 π 24 + π )( sin2 cos 2 π 24 24 − π ) sin2 24 (c) Note that = 1 · cos π √ √ 2 + 6 12 = . 4 cos 36 ◦ − cos 72 ◦ = 2(cos 36◦ − cos 72 ◦ )(cos 36 ◦ + cos 72 ◦ ) 2(cos 36 ◦ + cos 72 ◦ ) = 2 cos2 36 ◦ − 2 cos 2 72 ◦ 2(cos 36 ◦ + cos 72 ◦ ) . By the double-angle formulas, the above equality becomes cos 36 ◦ − cos 72 ◦ = cos 72◦ + 1 − cos 144 ◦ − 1 2(cos 36 ◦ + cos 72 ◦ ) = cos 72◦ + cos 36 ◦ 2(cos 36 ◦ + cos 72 ◦ ) = 1 2 . This fact can also be seen in an isosceles triangleABC with AB = AC, BC = 1, and A = 36 ◦ . Point D lies on side AC with ̸ ABD = ̸ DBC. We leave it to the reader to show that BC = BD = AD = 1, AB = 2 cos 36, and CD = 2 cos 72, from which the desired result follows.
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About the Authors Titu Andreescu re
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Titu Andreescu University of Wiscon
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vi Contents Vectors 41 The Dot Prod
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viii Preface Throughout MOSP, full
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Abbreviations and Notation Abbrevia
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103 Trigonometry Problems
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Further Reading 1. Andreescu, T.; F
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Further Reading 213 23. Fomin, D.;
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