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: 47

Answer

$\sin^{-1}2$ does not exist.

Work Step by Step

Solve for radians, then convert to degrees. $y=\sin^{-1}x$ Domain: $[-1, 1] $ Range: $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ Quadrants (unit circle): I and IV -------- $y$ is the number from $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ such that $\sin y=2.$ But, such a y does not exist, as sine can not be greater than 1. More directly, $2$ is not in the domain of $\sin^{-1}x$, so $\sin^{-1}2$ does not exist.
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