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: 43

Answer

$-2\sin 3x\cdot\sin x$

Work Step by Step

Sum-to-Product: $\displaystyle \cos A-\cos B=-2\sin(\frac{A+B}{2})\sin(\frac{A-B}{2})$ ---------- $\displaystyle \frac{A+B}{2}=\frac{4x+2x}{2}=3x$ $\displaystyle \frac{A-B}{2}=\frac{4x-2x}{2}=x$ $\cos 4x-\cos 2x=-2\sin 3x\cdot\sin x$
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