Trigonometry (10th Edition)

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

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 260: 101

Answer

$cos~(arcsin~u) = \sqrt{1-u^2}$

Work Step by Step

Let $~~\theta = arcsin~u$ Then $~~sin~\theta = u$ $cos~\theta = \sqrt{1-sin^2~\theta}$ $cos~\theta = \sqrt{1-u^2}$ Therefore, $~~cos~(arcsin~u) = \sqrt{1-u^2}$
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.