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: 68


$csc[arccos(x-1)]=\frac{1}{\sqrt {-x^2+2x}}$

Work Step by Step

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