Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.5 Inverse Circular Functions - 7.5 Exercises - Page 709: 75

Answer

$$\frac{{\sqrt 7 }}{{\text{3}}}$$

Work Step by Step

$$\eqalign{ & \tan \left( {\arccos \frac{3}{4}} \right) \cr & {\text{Let }}\theta = \arccos \frac{3}{4}{\text{, thus}} \cr & \cos \theta = \frac{3}{4} \cr & {\text{Recall that }}\cos \theta = \frac{{{\text{adjacent side}}}}{{{\text{hypotenuse}}}} \cr & {\text{adjacent side}} = 3 \cr & {\text{hypotenuse}} = 4 \cr & {\text{opposite side }} = \sqrt {{4^2} - {3^2}} = \sqrt 7 \cr & {\text{Then}} \cr & \tan \theta = \tan \left( {\arccos \frac{3}{4}} \right) = \frac{{{\text{opposite side}}}}{{{\text{adjacent side}}}} \cr & \tan \left( {\arccos \frac{3}{4}} \right) = \frac{{\sqrt 7 }}{{\text{3}}} \cr} $$
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