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: 37


$cos~\frac{\theta}{2} = \frac{1}{2}$

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 the angle $\frac{\theta}{2}$ is in quadrant I. We can find the value of $cos~\frac{\theta}{2}$: $cos~\frac{\theta}{2} = \sqrt{\frac{1+cos~\theta}{2}}$ $cos~\frac{\theta}{2} = \sqrt{\frac{1+(-\frac{1}{2})}{2}}$ $cos~\frac{\theta}{2} = \sqrt{\frac{(\frac{1}{2})}{2}}$ $cos~\frac{\theta}{2} = \sqrt{\frac{1}{4}}$ $cos~\frac{\theta}{2} = \frac{1}{2}$
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