Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Section 6.7 Product-to-Sum and Sum-to-Product Formulas - 6.7 Assess Your Understanding - Page 524: 10

Answer

$\dfrac{1}{2} [\cos (2 \theta) -\cos (8 \theta) ] $

Work Step by Step

Recall the Product to Sum Identities: $a) \sin x \cos y =\dfrac{1}{2} [\sin (x-y) +\sin (x+y)] \\ b) \sin x \sin y =\dfrac{1}{2} [\cos (x-y) -\cos (x+y)] \\ c) \cos x \cos y =\dfrac{1}{2} [\cos (x-y) +\cos (x+y)]$ We also know that $\cos$ and $\sec$ are even trigonometric functions. This implies that $f (-x)=f(x) \implies \cos(-x)=\cos(x)$ By identity $(b)$, we have: $\sin 3 \theta \ \sin 5 \theta=\dfrac{1}{2} [\cos (3 \theta- 5 \theta) -\cos (3 \theta+5 \theta) ] \\=\dfrac{1}{2} [\cos (-2 \theta) -\cos (8 \theta) ] \\=\dfrac{1}{2} [\cos (2 \theta) -\cos (8 \theta) ] $
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