# How do you evaluate cos(2Arctan(3/4))?

$\cos \left(2 \arctan \left(\frac{3}{4}\right)\right) = \frac{7}{25.}$
Let $\arctan \left(\frac{3}{4}\right) = \theta$, sothat, by defn. of $\arctan$ function, $\tan \theta = \frac{3}{4}$, and, $\theta \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right) = \left(- \frac{\pi}{2} , 0\right) \cup \left\{0\right\} \cup \left(0 , \frac{\pi}{2}\right) .$
Note that, $\tan \theta = \frac{3}{4} > 0 , \theta \notin \left(- \frac{\pi}{2} , 0\right]$, but, $\theta \in \left(0 , \frac{\pi}{2}\right) .$
Now, reqd. value $= \cos \left(2 \arctan \left(\frac{3}{4}\right)\right) = \cos 2 \theta = \frac{1 - {\tan}^{2} \theta}{1 + {\tan}^{2} \theta} = \frac{1 - {\left(\frac{3}{4}\right)}^{2}}{1 + {\left(\frac{3}{4}\right)}^{2}} = \frac{16 - 9}{16 + 9} = \frac{7}{25.}$