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 694: 94


$\frac{\sin 2x}{2\sin x}=\cos^2 \frac{x}{2}-\sin^2 \frac{x}{2}$

Work Step by Step

Let's simplify the left side: $\frac{\sin 2x}{2\sin x}$ Use the identity $\sin 2x=2\sin x\cos x$: $=\frac{2\sin x\cos x}{2\sin x}$ $=\cos x$ Simplify the right side: $=\cos^2 \frac{x}{2}-\sin^2 \frac{x}{2}$ Use the identity $\cos 2A=\cos^2 A-\sin^2 A$, where $A=\frac{x}{2}$: $=\cos (2*\frac{x}{2})$ $=\cos x$ Since the left side and the right side are both equal to $\cos x$, they are equal to each other, and the identity has been proven.
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