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: 89


$\frac{2\cos 2\theta}{\sin 2\theta}=\cot\theta-\tan\theta$

Work Step by Step

Start with the right side: $\cot\theta-\tan\theta$ Rewrite it in terms of sine and cosine: $=\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta}$ Get a common denominator: $=\frac{\cos\theta}{\sin\theta}*\frac{\cos\theta}{\cos\theta}-\frac{\sin\theta}{\cos\theta}*\frac{\sin\theta}{\sin\theta}$ $=\frac{\cos^2\theta}{\sin\theta\cos\theta}-\frac{\sin^2\theta}{\sin\theta\cos\theta}$ $=\frac{\cos^2\theta-\sin^2\theta}{\sin\theta\cos\theta}$ Use the identity $\cos^2\theta-\sin^2\theta=\cos 2\theta$: $=\frac{\cos 2\theta}{\sin\theta\cos\theta}$ Multiply top and bottom by $2$: $=\frac{2\cos 2\theta}{2\sin\theta\cos\theta}$ Use the identity $2\sin\theta\cos\theta=\sin 2\theta$: $=\frac{2\cos 2\theta}{\sin 2\theta}$ Since this equals the left side, the identity has been proven.
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