Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 264: 13



Work Step by Step

$$y=\sin^{-1}0$$ First, we see that the domain of inverse sine function is $[-1,1]$. $0$ lies in this range, so $\sin^{-1}0$ does exist. Also, it should be noted that the range of inverse sine function is $[-\frac{\pi}{2},\frac{\pi}{2}]$. In other words, $y\in[-\frac{\pi}{2},\frac{\pi}{2}]$. We can rewrite $y=\sin^{-1}0$ as $\sin y=0$ We see that $\sin 0=0$. Since $0\in[-\frac{\pi}{2},\frac{\pi}{2}]$, we can conclude $y=0$
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