Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.6 - Inverse Trigonometric Functions - Exercises 7.6 - Page 420: 3

Answer

$a.\displaystyle \quad -\frac{\pi}{6}$ $b.\displaystyle \quad\frac{\pi}{4}$ $c.\displaystyle \quad -\frac{\pi}{3}$

Work Step by Step

$y=\sin^{-1}x$ is the number in $[-\pi/2, \pi/2]$ for which $\sin y=x.$ Use reference angles in the 1st quadrant, keeping in mind that sine is an odd function. $\left\{\begin{array}{llll} \sin\frac{\pi}{6}=\frac{1}{2} & \Rightarrow & \sin(-\frac{\pi}{6})=-\dfrac{1}{2} & \sin^{-1}(-\frac{1}{2})=-\frac{\pi}{6}\\ & & & \\ \sin\frac{\pi}{4}=\frac{1}{\sqrt{2}} & \Rightarrow & & \sin^{-1}(\frac{1}{\sqrt{2}})=\dfrac{\pi}{4}\\ & & & \\ \sin\frac{\pi}{3}=\frac{ \sqrt{3}}{2} & \Rightarrow & \sin(-\frac{\pi}{3})=-\frac{\sqrt{3}}{2} & \Rightarrow\sin^{-1}(-\frac{\sqrt{3}}{2})=-\dfrac{\pi}{3} \end{array}\right.$
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