Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Test - Page 244: 4

Answer

$\frac{\sqrt{6}-\sqrt{2}}{4}$

Work Step by Step

$cos~\frac{5\pi}{12}$ $=cos~(\frac{3\pi}{12}+\frac{2\pi}{12})$ $=cos~(\frac{\pi}{4}+\frac{\pi}{6})$ $=cos~\frac{\pi}{4}~cos~\frac{\pi}{6} - sin~\frac{\pi}{4}~sin~\frac{\pi}{6}$ $=(\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) - (\frac{\sqrt{2}}{2})(\frac{1}{2})$ $=\frac{\sqrt{6}-\sqrt{2}}{4}$
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.