Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 1 - Section 1.5 - Inverse Functions and Logarithms - 1.5 Exercises - Page 66: 63

Answer

$(a) cos^{-1}(-1) = \pi $ $(b) sin^{-1}(0.5) = \frac{\pi}{6}$

Work Step by Step

$(a) cos^{-1}(-1) = \pi $ because $cos(\pi) = -1$ and $\pi$ lies between $0$ and $\pi$ $(b) sin^{-1}(0.5) = \frac{\pi}{6}$ because $sin(\frac{\pi}{6}) = 0.5$ and $\frac{\pi}{6}$ lies between $\frac{-\pi}{2}$ and $\frac{\pi}{2}$
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