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: 62


$tan[sin^{-1}(-\frac{\sqrt 2}{2})]=-1$

Work Step by Step

Let $u=sin^{-1}(-\frac{\sqrt 2}{2})$. Then: $sin~u=-\frac{\sqrt 2}{2}$ The range $sin^{-1}~x$ is $-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}$. So, since $sin^{-1}(-\frac{\sqrt 2}{2})\lt0$, then $-\frac{\pi}{2}\lt u\lt0$ (Fourth Quadrant) $sin^2u+cos^2u=1$ $cos^2u=1-sin^2u$ $cos^2u=1-(-\frac{\sqrt 2}{2})^2=1-\frac{2}{4}=\frac{1}{2}$ $cos~u=\frac{1}{\sqrt 2}=\frac{1}{\sqrt 2}\frac{\sqrt 2}{\sqrt 2}=\frac{\sqrt 2}{2}$ (Fourth Quadrant) $tan[sin^{-1}(-\frac{\sqrt 2}{2})]=tan~u=\frac{sin~u}{cos~u}=\frac{-\frac{\sqrt 2}{2}}{\frac{\sqrt 2}{2}}=-1$
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