## Trigonometry (11th Edition) Clone

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