Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.4 Double-Angle and Half-Angle Identities - 7.4 Exercises - Page 693: 44


$2\displaystyle \cos\frac{13x}{2}\cdot\cos\frac{3x}{2}$

Work Step by Step

Sum-to-Product: $\displaystyle \cos A+\cos B=2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2})$ ---------- $\displaystyle \frac{A+B}{2}=\frac{5x+8x}{2}=\frac{13x}{2}$ $\displaystyle \frac{A-B}{2}=\frac{5x-8x}{2}=-\frac{3x}{2}$ $\displaystyle \cos 5x+\cos 8x=2\cos\frac{13x}{2}\cdot\cos(-\frac{3x}{2})$ ...Even/odd identity: $\cos(-A)=\cos A$ $=2\displaystyle \cos\frac{13x}{2}\cdot\cos\frac{3x}{2}$
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