Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Review Exercises - Page 708: 39


$\frac{\sqrt 3}{2}$

Work Step by Step

Using the double-angle formula $cos2\theta=cos^2\theta-sin^2\theta$, we have $cos^215^\circ -sin^215^\circ=cos2(15^\circ)=cos30^\circ=\frac{\sqrt 3}{2}$
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.