Answer
$sin~(arccos~u) = \sqrt{1-u^2}$
Work Step by Step
Let $~~\theta = arccos~u$
Then $~~cos~\theta = u$
$sin~\theta = \sqrt{1-cos^2~\theta}$
$sin~\theta = \sqrt{1-u^2}$
Therefore, $~~sin~(arccos~u) = \sqrt{1-u^2}$
Trigonometry (10th Edition)
by Lial, Margaret L.; Hornsby, John; Schneider, David I.; Daniels, Callie
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: 99
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