Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 680: 23


See the explanation below.

Work Step by Step

To verify the given identity, $\sin 2\theta =\frac{2\tan \theta }{1+{{\tan }^{2}}\theta }$ Recall the Trigonometric Identities and apply below, $\begin{align} & \sin 2\theta =2\sin \theta \cos \theta \\ & \tan \theta =\frac{\sin \theta }{\cos \theta } \\ & \cos \theta =\frac{1}{\text{sec}\theta } \\ & 1+{{\tan }^{2}}\theta =\text{se}{{\text{c}}^{2}}\theta \\ \end{align}$ Consider the left side of the given expression, $\sin 2\theta =2\sin \theta \cos \theta $ Multiply and divide by $\cos \theta $. $\begin{align} & \sin 2\theta =2\sin \theta \cos \theta \times \frac{\cos \theta }{\cos \theta } \\ & =2\frac{\sin \theta }{\cos \theta }\times {{\cos }^{2}}\theta \\ & =2\tan \theta {{\cos }^{2}}\theta \\ & =2\tan \theta \frac{1}{{{\sec }^{2}}\theta } \end{align}$ Substitute $1+{{\tan }^{2}}\theta $ for $\text{se}{{\text{c}}^{2}}\theta $. $\sin 2\theta =\frac{2\tan \theta }{1+{{\tan }^{2}}\theta }$ Hence, it is proved that the given identity $\sin 2\theta =\frac{2\tan \theta }{1+{{\tan }^{2}}\theta }$ holds true.
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