# How do you find the exact value of \cos(tan^{-1}sqrt{3})?

Nov 12, 2014

Method 1

Let $\theta = {\tan}^{- 1} \sqrt{3}$.

By rewriting in terms of tangent,

$\Leftrightarrow \tan \theta = \sqrt{3} = \left\{\left(\text{Opposite")}/{("Adjacent}\right)\right\}$

So, we can let

$\left(\text{Opposite}\right) = \sqrt{3}$ and $\left(\text{Adjacent}\right) = 1$.

By Pythagorean Theorem,

$\left(\text{Hypotenuse}\right) = \sqrt{{\left(\sqrt{3}\right)}^{2} + {1}^{2}} = \sqrt{4} = 2$.

Hence,

$\cos \left({\tan}^{- 1} \sqrt{3}\right) = \cos \theta = \left\{\left(\text{Adjacent")}/{("Hypotenuse}\right)\right\} = \frac{1}{2}$

Method 2

$\cos \left({\tan}^{- 1} \sqrt{3}\right) = \cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$

I hope that this was helpful