# How do you solve and find the value of cos(tan^-1sqrt3)?

##### 1 Answer
Dec 27, 2016

$\cos \left({\tan}^{- 1} \left(\sqrt{3}\right)\right) = \textcolor{g r e e n}{\frac{1}{2}}$

#### Explanation:

The standard definition of the inverse tan function implies an angle either Quadrant I or II.

The angle $\theta = {\tan}^{- 1} \left(\sqrt{3}\right)$ can be represented by means of a right triangle in standard position with opposite side of length $\sqrt{3}$ and adjacent side of length $1$.
By the Pythagorean Theorem this implies a hypotenuse of $\sqrt{\left({\sqrt{3}}^{2} + {1}^{2}\right)} = \sqrt{4} = 2$

Using this triangle and the definition of $\cos$ as $\text{adjacent"/"hypotenuse}$
we have
$\textcolor{w h i t e}{\text{XXX}} \cos \left({\tan}^{- 1} \left(\sqrt{3}\right)\right) = \frac{1}{2}$