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: 61


$csc(cos^{-1}~\frac{\sqrt 3}{2})=\frac{1}{sin~u}=2$

Work Step by Step

Let $u=cos^{-1}~\frac{\sqrt 3}{2}$. Then: $cos~u=\frac{\sqrt 3}{2}$ The range $cos^{-1}~x$ is $0\lt x\lt \pi$. So, since $0\lt cos^{-1}~\frac{\sqrt 3}{2}\lt\frac{\pi}{2}$, then $0\lt u\lt\frac{\pi}{2}$ (First Quadrant) $sin^2u+cos^2u=1$ $sin^2u=1-cos^2u$ $sin^2u=1-(\frac{\sqrt 3}{2})^2=1-\frac{3}{4}=\frac{1}{4}$ $sin~u=\frac{1}{2}$ (First Quadrant) $csc(cos^{-1}~\frac{\sqrt 3}{2})=csc~u=\frac{1}{sin~u}=2$
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