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: 64


See the explanation below.

Work Step by Step

Let us consider the right side of the given expression: $\frac{{{\sin }^{2}}\alpha +1-{{\cos }^{2}}\alpha }{\sin \alpha \left( 1+\cos \alpha \right)}$ By using the trigonometric identity $\tan \frac{x}{2}=\frac{\sin x}{1+\cos x}$ and ${{\sin }^{2}}x=1-{{\cos }^{2}}x$, The above expression can be further simplified as: $\begin{align} & \frac{{{\sin }^{2}}\alpha +1-{{\cos }^{2}}\alpha }{\sin \alpha \left( 1+\cos \alpha \right)}=\frac{{{\sin }^{2}}\alpha +{{\sin }^{2}}\alpha }{\sin \alpha \left( 1+\cos \alpha \right)} \\ & =\frac{2{{\sin }^{2}}\alpha }{\sin \alpha \left( 1+\cos \alpha \right)} \\ & =2\left( \frac{\sin \alpha }{1+\cos \alpha } \right) \\ & =2\tan \frac{\alpha }{2} \end{align}$ Hence, the left side of the given expression is equal to the right side, which is $2\tan \frac{\alpha }{2}=\frac{{{\sin }^{2}}\alpha +1-{{\cos }^{2}}\alpha }{\sin \alpha \left( 1+\cos \alpha \right)}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.