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: 63

Answer

$$\cos 4x-\cos 2x=-2\sin3x\sin x$$

Work Step by Step

$$A=\cos 4x-\cos 2x$$ The sum-to-product identity that will be applied here is $$\cos X-\cos Y=-2\sin(\frac{X+Y}{2})\sin(\frac{X-Y}{2})$$ Therefore, A would be $$A=-2\sin(\frac{4x+2x}{2})\sin(\frac{4x-2x}{2})$$ $$A=-2\sin(\frac{6x}{2})\sin(\frac{2x}{2})$$$$A=-2\sin3x\sin x$$
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