Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

by Sullivan III, Michael

Published by Pearson

ISBN 10: 0-32193-104-1

ISBN 13: 978-0-32193-104-7

Chapter 7 - Applications of Trigonometric Functions - Section 7.2 The Law of Sines - 7.2 Assess Your Understanding - Page 555: 68

Answer

$-\frac{\sqrt {15}}{7}$

Work Step by Step

Let $cos^{-1}(-\frac{7}{8})=t$ (t in quadrant II), we have $cos(t)=-\frac{7}{8}$ and $sin(t)=\frac{\sqrt {64-49}}{8}=\frac{\sqrt {15}}{8}$, thus $tan(t)=-\frac{\sqrt {15}}{7}$

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