Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - Chapter 7 Test Prep - Review Exercises - Page 737: 69


$-\cot \frac{x}{2}=\frac{\sin 2x+\sin x}{\cos 2x-\cos x}$

Work Step by Step

Simplify the left side: $-\cot \frac{x}{2}$ Write cotangent as 1 divided by tangent: $=-\frac{1}{\tan \frac{x}{2}}$ Simplify: $=-\frac{1}{\frac{1-\cos x}{\sin x}}$ $=-\frac{\sin x}{1-\cos x}$ $=\frac{\sin x}{\cos x-1}$ Simplify the right side: $\frac{\sin 2x+\sin x}{\cos 2x-\cos x}$ Expand using the double-angle identities: $=\frac{2\sin x\cos x+\sin x}{2\cos^2-1-\cos x}$ $=\frac{2\sin x\cos x+\sin x}{2\cos^2-\cos x-1}$ Factor: $=\frac{\sin x(2\cos x+1)}{(2\cos x+1)(\cos x-1)}$ $=\frac{\sin x}{\cos x-1}$ Since the left side and the right side both simplify to $\frac{\sin x}{\cos x-1}$, they are equal to each other, and the identity has been proven.
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