Answer
$sin^{-1}(1) = \boxed{\frac{\pi}{2}}$
Work Step by Step
Domain of $sin^{-1}(x)$ is $[-1, 1]$ and the range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Since $sin(\frac{\pi}{2}) = 1$, $sin^{-1}(1) = \frac{\pi}{2}$.
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
Published by
Pearson
ISBN 10:
0321947347
ISBN 13:
978-0-32194-734-5
Chapter 1 - Functions - 1.4 Trigonometric Functions and Their Inverse - 1.4 Exercises - Page 48: 47
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