# How do you find the exact value of cos[arctan(-8/15)]?

Feb 5, 2016

Ok guy i want you to imagine a triangle rectangle with one side $a$ and another side $b$, so the hypotenuse $c$ is equal to $\sqrt{{a}^{2} + {b}^{2}}$

You have an angle $\theta$ is opposite side is $a$ and his adjacent side is $b$

So you know trigonometry and

$\tan \left(\theta\right) = \frac{a}{b}$

$- \tan \left(\theta\right) = - \frac{a}{b}$

and you know $\tan \left(\theta\right)$ is odd so

$\tan \left(- \theta\right) = - \frac{a}{b}$

So you can say

$- \theta = \arctan \left(- \frac{a}{b}\right)$

again, you know the trigonometry and know

$\cos \left(\theta\right) = \frac{b}{c}$

but you know $- \theta = \arctan \left(- \frac{a}{b}\right)$ and the cosinus is an even function so you don't care if you put $- \theta$ or $\theta$ so

$\cos \left(\arctan \left(- \frac{a}{b}\right)\right) = \frac{b}{c}$