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: 90


$\cot 4\theta=\frac{1-\tan^2 2\theta}{2\tan 2\theta}$

Work Step by Step

Start with the left side: $\cot 4\theta$ Rewrite cotangent as the inverse function of tangent: $=\frac{1}{\tan 4\theta}$ Rewrite $\tan 4\theta$ as $\tan (2*2\theta)$ and use the double-angle identity: $=\frac{1}{\tan (2*2\theta)}$ $=\frac{1}{\frac{2\tan 2\theta}{1-\tan^2 2\theta}}$ Multiply top and bottom by $1-\tan^2 2\theta$: $=\frac{1}{\frac{2\tan 2\theta}{1-\tan^2 2\theta}}*\frac{1-\tan^2 2\theta}{1-\tan^2 2\theta}$ $=\frac{1-\tan^2 2\theta}{2\tan 2\theta}$
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