Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - 7.2 The Inverse Trigonometric Functions (Continued) - 7.2 Assess Your Understanding - Page 458: 71

Answer

$-\frac{3}{4}$

Work Step by Step

Step 1. Letting $f^{-1}(-\frac{3}{5})=sin^{-1}(-\frac{3}{5})=t$, we have $sin(t)=-\frac{3}{5}$ and $t$ in quadrant IV. Step 2. Form a right triangle with sides $y=3, r=5, x=\sqrt {r^2-y^2}=4$ with $|t|$ facing $y$. Step 3. We have $h(f^{-1}(-\frac{3}{5}))=tan(t)=-\frac{3}{4}$
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