Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 8 - Cumulative Test for Chapters 6-8 - Page 621: 24


$-2 \sin 8x \sin x$

Work Step by Step

We know that $\cos a-\cos b=-2 \sin (\dfrac{a+b}{2}) \sin (\dfrac{a-b}{2})$ $\cos 9 x-\cos 7 x=-2 \sin (\dfrac{9x+7x}{2}) \sin (\dfrac{9x-7x}{2})$ This implies that $\cos 9 x-\cos 7 x=-2 \sin (\dfrac{16x}{2}) \sin (\dfrac{2x}{2})$ Therefore, $\cos 9 x-\cos 7 x=-2 \sin 8x \sin x$
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