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: 77

Answer

$-\sqrt {15}$

Work Step by Step

Step 1. Letting $g^{-1}(-\frac{1}{4})=cos^{-1}(-\frac{1}{4})=t$, we have $cos(t)=-\frac{1}{4}$ and $t$ in quadrant II. Step 2. Form a right triangle of sides $x=1, r=4, y=\sqrt {r^2-x^2}=\sqrt {15}$ with $\pi-|t|$ facing $y$. Step 3. We have $h(g^{-1}(-\frac{1}{4}))=tan(t)=-\sqrt {15}$
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