Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.5 Inverse Circular Functions - 7.5 Exercises - Page 708: 36


$\displaystyle \csc^{-1}(-\frac{1}{2})$ does not exist

Work Step by Step

$y=\csc^{-1}x$ Domain: $(-\infty, -1]\cup[1, \infty)$ Range: $[-\displaystyle \frac{\pi}{2},0)\cup(0,\frac{\pi}{2}]$ --------------- $-\displaystyle \frac{1}{2}\not\in(-\infty, -1]\cup[1, \infty)$. So, $-\displaystyle \frac{1}{2}$ is not in the domain of $\csc^{-1}x$, meaning that it is not in the range of $\csc x.$ (There is no y from $[-\displaystyle \frac{\pi}{2},0)\cup(0,\frac{\pi}{2}]$ such that $\displaystyle \csc y=\frac{1}{\sin y}=-\frac{1}{2},$ as this would mean that $\sin y=-2 < -1) $ $\displaystyle \csc^{-1}(-\frac{1}{2})$ does not exist
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