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: 84


$\sin 2x=\frac{2\tan x}{1+\tan^2 x}$

Work Step by Step

Start with the right side: $\frac{2\tan x}{1+\tan^2 x}$ Use the identity $1+\tan^2 x=\sec^2 x$: $=\frac{2\tan x}{\sec^2 x}$ Rewrite everything in terms of sine and cosine: $=\frac{2*\frac{\sin x}{\cos x}}{\frac{1}{\cos^2 x}}$ Multiply top and bottom by $\cos^2 x$: $=\frac{2*\frac{\sin x}{\cos x}}{\frac{1}{\cos^2 x}}*\frac{\cos^2 x}{\cos^2 x}$ Simplify: $=\frac{2\sin x\cos x}{1}$ $=2\sin x\cos x$ $=\sin 2x$ Since this equals the left side, the identity has been proven.
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