Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Review Exercises - Page 291: 34


$cos~(arctan~\frac{u}{\sqrt{1-u^2}}) = \sqrt{1-u^2}$

Work Step by Step

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