Trigonometry (10th Edition)

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

Chapter 6 - Quiz (Sections 6.1-6.3) - Page 276: 4a

Answer

$cos~(tan^{-1}~(\frac{4}{5})) = \frac{5\sqrt{41}}{41}$

Work Step by Step

$\theta = tan^{-1}(\frac{4}{5})$ $tan~\theta = \frac{4}{5} = \frac{opposite}{adjacent}$ Note that $\theta$ is in quadrant I. We can find the value of the hypotenuse: $hypotenuse = \sqrt{4^2+5^2} = \sqrt{41}$ We can find the value of $cos~\theta$: $cos~\theta = \frac{adjacent}{hypotenuse}$ $cos~\theta = \frac{5}{\sqrt{41}}$ $cos~\theta = \frac{5}{\sqrt{41}}~\frac{\sqrt{41}}{\sqrt{41}}$ $cos~\theta = \frac{5\sqrt{41}}{41}$ Therefore, $cos~(tan^{-1}~(\frac{4}{5})) = \frac{5\sqrt{41}}{41}$
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