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: 68

Answer

$$\sin9x-\sin3x=2\cos 6x\sin3x$$

Work Step by Step

$$A=\sin9x-\sin3x$$ The sum-to-product identity that will be applied here is $$\sin X-\sin Y=2\cos(\frac{X+Y}{2})\sin(\frac{X-Y}{2})$$ Therefore, A would be $$A=2\cos[\frac{9x+3x}{2}]\sin[\frac{9x-3x}{2}]$$ $$A=2\cos(\frac{12x}{2})\sin(\frac{6x}{2})$$ $$A=2\cos 6x\sin3x$$
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