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: 70


(a) $\arcsin(-1)=-\frac{\pi}{2}$ (b) $\arcsin\Big(-\frac{1}{\sqrt2}\Big)=-\frac{\pi}{4}$

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) $$\arcsin(-1)=a$$ According to the defintion: $$\sin a=-1\hspace{1cm}\text{and}\hspace{1cm}a\in[-\pi/2,\pi/2]$$ So $a=-\frac{\pi}{2}$. In other words, $$\arcsin(-1)=-\frac{\pi}{2}$$ (b) $$\arcsin\Big(-\frac{1}{\sqrt2}\Big)=\arcsin\Big(-\frac{\sqrt2}{2}\Big)=a$$ 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, $$\arcsin\Big(-\frac{1}{\sqrt2}\Big)=-\frac{\pi}{4}$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.