Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - Chapter Review - Review Exercises - Page 506: 47



Work Step by Step

Recall: $\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$ Hence, $\cos(\frac{5\pi}{12})\\ =\cos{75^\circ}\\ =\cos{(45^\circ+30^\circ)}\\ =\cos{45^\circ}\cos{30^\circ}-\sin{45^\circ}\sin{30^\circ}\\ =\frac{\sqrt3}{2}\frac{\sqrt2}{2}-\frac{1}{2}\frac{\sqrt2}{2}\\ =\frac{\sqrt6-\sqrt2}{4}$
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.