Trigonometry (11th Edition) Clone

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

Chapter 5 - Test - Page 250: 9d


$cos(\frac{\theta}{2}) = \sqrt{\frac{1}{5}}$

Work Step by Step

If $90^{\circ} \lt \theta \lt 180^{\circ}$, then the angle $\theta$ is in quadrant II. Then $45^{\circ} \lt \frac{\theta}{2} \lt 90^{\circ}$, so $\frac{\theta}{2}$ is in quadrant I. $cos(\frac{\theta}{2}) = \sqrt{\frac{1+cos~\theta}{2}}$ $cos(\frac{\theta}{2}) = \sqrt{\frac{1+(-\frac{3}{5})}{2}}$ $cos(\frac{\theta}{2}) = \sqrt{\frac{(\frac{2}{5})}{2}}$ $cos(\frac{\theta}{2}) = \sqrt{\frac{1}{5}}$
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.