University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.6 - Inverse Functions and Logarithms - Exercises - Page 50: 67


(a) $\sin^{-1}\Big(\frac{-1}{2}\Big)=-\frac{\pi}{6}$ (b) $\sin^{-1}\Big(\frac{1}{\sqrt2}\Big)=\frac{\pi}{4}$ (c) $\sin^{-1}\Big(\frac{-\sqrt3}{2}\Big)=-\frac{\pi}{3}$

Work Step by Step

*Recall the definition of arcsine: $y=\sin^{-1}x$ is the number in $[-\pi/2,\pi/2]$ for which $\sin y=x$ (a) $$\sin^{-1}\Big(\frac{-1}{2}\Big)=a$$ According to the defintion: $$\sin a=-\frac{1}{2}\hspace{1cm}\text{and}\hspace{1cm}a\in[-\pi/2,\pi/2]$$ So $a=-\frac{\pi}{6}$. In other words, $$\sin^{-1}\Big(\frac{-1}{2}\Big)=-\frac{\pi}{6}$$ (b) $$\sin^{-1}\Big(\frac{1}{\sqrt2}\Big)=\sin^{-1}\Big(\frac{\sqrt2}{2}\Big)=a$$ According to the defintion: $$\sin a=\frac{\sqrt2}{2}\hspace{1cm}\text{and}\hspace{1cm}a\in[-\pi/2,\pi/2]$$ So $a=\frac{\pi}{4}$. In other words, $$\sin^{-1}\Big(\frac{1}{\sqrt2}\Big)=\frac{\pi}{4}$$ (c) $$\sin^{-1}\Big(\frac{-\sqrt3}{2}\Big)=a$$ According to the defintion: $$\sin a=\frac{-\sqrt3}{2}\hspace{1cm}\text{and}\hspace{1cm}a\in[-\pi/2,\pi/2]$$ So $a=-\frac{\pi}{3}$. In other words, $$\sin^{-1}\Big(\frac{-\sqrt3}{2}\Big)=-\frac{\pi}{3}$$
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