Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 7 - 7.2 - Verifying Trigonometric Identities - 7.2 Exercises - Page 521: 59


The identity is verified. $tan^{-1}(sin\frac{x-1}{4})=\frac{x-1}{\sqrt {16-(x-1)^2}}$

Work Step by Step

Make $z=\frac{x-1}{4}$ $y=sin^{-1}z$, domain: $-1\leq z\leq1$, range: $-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}$: $z=sin~y$ $z^2=sin^2y$ $1-z^2=1-sin^2y$ $1-z^2=cos^2y$ $cos~y=+\sqrt {1-z^2}~~$ (Use only the $+$ because $-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}$) $tan^{-1}(sin\frac{x-1}{4})=tan(sin^{-1}z)=tan~y=\frac{sin~y}{cos~y}=\frac{z}{\sqrt {1-z^2}}=\frac{\frac{x-1}{4}}{\sqrt {1-(\frac{x-1}{4})^2}}=\frac{\frac{x-1}{4}}{\sqrt {1-\frac{(x-1)^2}{16}}}\frac{4}{4}=\frac{x-1}{\sqrt {16-(x-1)^2}}$
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