Answer
$sin(arccos~x)=\sqrt {1-x^2}$
Work Step by Step
Let $u=arccos~x~~$ (Range: $0\lt u\lt\pi$)
Then: $x=cos~u$
$sin^2u+cos^2u=1$
$sin^2u=1-cos^2u=1-x^2$
$sin~u=\sqrt {1-x^2}~~$ ($0\lt u\lt\pi$)
$sin(arccos~x)=sin~u=\sqrt {1-x^2}$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 6 - 6.6 - Inverse Trigonometric Functions - 6.6 Exercises - Page 485: 67
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