Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.4 Double-Angle and Half-Angle Identities - 7.4 Exercises: 55

Answer

$\displaystyle \frac{\sqrt{10}}{4}$

Work Step by Step

$0 < x < \displaystyle \frac{\pi}{2}\qquad/\div 2$ $0 < \displaystyle \frac{x}{2} < \frac{\pi}{4}$, so, $\displaystyle \frac{x}{2}$ is in quadrant I, where the cosine is positive. Half-Angle Identity: $\displaystyle \cos\frac{x}{2}=\pm\sqrt{\frac{1+\cos x}{2}}=+\sqrt{\frac{1+\frac{1}{4}}{2}}$ $=\displaystyle \sqrt{\frac{\frac{5}{4}}{2} }=\frac{\sqrt{5}}{\sqrt{8}}\cdot\frac{\sqrt{8}}{\sqrt{8}}=\frac{\sqrt{40}}{8}$ $=\displaystyle \frac{2\sqrt{10}}{8}$ $=\displaystyle \frac{\sqrt{10}}{4}$
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.