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: 48


$2\cos 6x\cdot\sin 3x$

Work Step by Step

Sum-to-Product: $\displaystyle \sin A-\sin B=2\cos(\frac{A+B}{2})\sin(\frac{A-B}{2})$ ---------- $\displaystyle \frac{A+B}{2}=\frac{9x+3x}{2}=6x$ $\displaystyle \frac{A-B}{2}=\frac{9x-3x}{2}=3x$ $\sin 9x-\sin 3x=2\cos 6x\cdot\sin 3x$
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