Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 260: 79

Answer

$tan~(arccos~\frac{3}{4}) = \frac{\sqrt{7}}{3}$

Work Step by Step

$\theta = arccos(\frac{3}{4})$ $cos~\theta = \frac{3}{4} = \frac{adjacent}{hypotenuse}$ Note that $\theta$ is in quadrant I. We can find the value of the opposite side: $opposite = \sqrt{4^2-3^2} = \sqrt{7}$ We can find the value of $tan~\theta$: $tan~\theta = \frac{opposite}{adjacent}$ $tan~\theta = \frac{\sqrt{7}}{3}$ Therefore, $tan~(arccos~\frac{3}{4}) = \frac{\sqrt{7}}{3}$
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