# How do you evaluate cos[arctan(-5/12)]?

Mar 3, 2016

$\cos \left[\arctan \left(- \frac{5}{12}\right)\right] = \pm \frac{12}{13}$

#### Explanation:

$\arctan \left(- \frac{5}{12}\right)$ means an angle $\theta$ whose $\tan \theta = - \frac{5}{12}$. As such we have to find $\cos \theta$.

$\cos \theta = \frac{1}{\sec} \theta = \sqrt{\frac{1}{\sec} ^ 2 \theta} = \sqrt{\frac{1}{1 + {\tan}^{2} \theta}}$

Hence

$\cos \theta = \sqrt{\frac{1}{1 + {\left(- \frac{5}{12}\right)}^{2}}} = \sqrt{\frac{1}{1 + \frac{25}{144}}}$

= $\sqrt{\frac{1}{\frac{169}{144}}} = \sqrt{\frac{144}{169}} = \pm \frac{12}{13}$

Hence $\cos \left[\arctan \left(- \frac{5}{12}\right)\right] = \pm \frac{12}{13}$