Precalculus (10th Edition)

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

Chapter 8 - Applications of Trigonometric Functions - Chapter Review - Review Exercises - Page 557: 5



Work Step by Step

The identity says: $\sin^2(\theta)+\cos^2(\theta)=1$. We also know that $\cos(\theta)=\sin(90^{\circ}-\theta)$. Hence, $\cos^2(40^{\circ})+\cos^2(50^{\circ}) \\=\cos^2(40^{\circ})+\sin^2(90^{\circ}-50^{\circ})\\ =\cos^2(40^{\circ})+\sin^2(40^{\circ}) \\=1$
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.