Algebra and Trigonometry 10th Edition

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


$sin(arccos~x)=\sqrt {1-x^2}$

Work Step by Step

Let $u=arccos~x~~$ (Range: $0\lt u\lt\pi$) Then: $x=cos~u$ $sin^2u+cos^2u=1$ $sin^2u=1-cos^2u=1-x^2$ $sin~u=\sqrt {1-x^2}~~$ ($0\lt u\lt\pi$) $sin(arccos~x)=sin~u=\sqrt {1-x^2}$
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