# How do you evaluate  cos(arctan(2) +arctan(3))?

Apr 16, 2016

$\cos \left(\arctan \left(2\right) + \arctan \left(3\right)\right) = - \frac{\sqrt{2}}{2}$

#### Explanation:

Consider the following two right angled triangles:

Then:

$\left\{\begin{matrix}\tan \left(\alpha\right) = 2 \\ \tan \left(\beta\right) = 3\end{matrix}\right.$

Notice that:

$\left\{\begin{matrix}\sin \left(\alpha\right) = \frac{2}{\sqrt{5}} \\ \cos \left(\alpha\right) = \frac{1}{\sqrt{5}} \\ \sin \left(\beta\right) = \frac{3}{\sqrt{10}} \\ \cos \left(\beta\right) = \frac{1}{\sqrt{10}}\end{matrix}\right.$

The sum of angles formula for $\cos$ tells us:

$\cos \left(\alpha + \beta\right) = \cos \left(\alpha\right) \cos \left(\beta\right) - \sin \left(\alpha\right) \sin \left(\beta\right)$

$= \frac{1}{\sqrt{5}} \frac{1}{\sqrt{10}} - \frac{2}{\sqrt{5}} \frac{3}{\sqrt{10}}$

$= \frac{1 - 6}{\sqrt{50}} = - \frac{5}{5 \sqrt{2}} = - \frac{1}{\sqrt{2}} = - \frac{\sqrt{2}}{2}$

Jul 16, 2016

$\cos \left({\tan}^{-} 1 2 + {\tan}^{-} 1 3\right)$

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

$= \cos \left({\tan}^{-} 1 \left(- 1\right)\right)$

$= \cos \left({\tan}^{-} 1 \left(- \tan \left(\frac{\pi}{4}\right)\right)\right)$

$= \cos \left({\tan}^{-} 1 \left(\tan \left(\pi - \frac{\pi}{4}\right)\right)\right)$

$= \cos \left(\pi - \frac{\pi}{4}\right)$

$= - \cos \left(\frac{\pi}{4}\right)$

$= - \frac{1}{\sqrt{2}}$