Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.7 - Inverse Trigonometric Functions - Exercise Set - Page 626: 58


$- \sqrt {15}$

Work Step by Step

Suppose $\theta =\cos^{-1} (-\dfrac{1}{4})$ This gives: $\cos \theta=-\dfrac{1}{3}$ Since, $ r=\sqrt {x^2+y^2}$ This implies that $ y=\sqrt {r^2-x^2}=\sqrt {4^2-(1)^2}= \sqrt {15}$ Therefore, we have $\tan \theta =\tan [\cos^{-1} (-\dfrac{1}{4})]=- \sqrt {15}$
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