Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 374: 28



Work Step by Step

Assume that $\csc\theta =\frac{20}{1}$, then by solving the triangle, we have: $$\sin\theta=\frac{1}{20}.$$ Now, since $\theta =\csc^{-1}20 $, we have $$\sin(\csc^{-1}20 )=\sin\theta=\frac{1}{20}.$$
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