Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Review Exercises - Page 249: 51


$\frac{sin^2~x}{2-2~cos~x} = cos^2~\frac{x}{2}$

Work Step by Step

$\frac{sin^2~x}{2-2~cos~x} = \frac{1-cos^2~x}{2(1-cos~x)}$ $\frac{sin^2~x}{2-2~cos~x} = \frac{(1-cos~x)(1+cos~x)}{2(1-cos~x)}$ $\frac{sin^2~x}{2-2~cos~x} = \frac{1+cos~x}{2}$ $\frac{sin^2~x}{2-2~cos~x} = \sqrt{\frac{1+cos~x}{2}}~\sqrt{\frac{1+cos~x}{2}}$ $\frac{sin^2~x}{2-2~cos~x} = (cos~\frac{x}{2})~(cos~\frac{x}{2})$ $\frac{sin^2~x}{2-2~cos~x} = cos^2~\frac{x}{2}$
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.