Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 231: 67

Answer

$$\cos4x+\cos8x=2\cos6x\cos 2x$$

Work Step by Step

$$A=\cos4x+\cos8x$$ The sum-to-product identity that will be applied here is $$\cos X+\cos Y=2\cos(\frac{X+Y}{2})\cos(\frac{X-Y}{2})$$ Therefore, A would be $$A=2\cos[\frac{4x+8x}{2}]\cos[\frac{4x-8x}{2}]$$ $$A=2\cos(\frac{12x}{2})\cos(\frac{-4x}{2})$$ $$A=2\cos 6x\cos(-2x)$$ We know that $\cos(-X)=\cos X$, which means $\cos(-2x)=\cos 2x$. Therefore, $$A=2\cos6x\cos 2x$$
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