Evaluating the expression $\cos(\frac{1}{2}\tan^{-1}(-\frac{4}{3}))$ I can prove that $\cos(\tan^{-1}(x)) = +\dfrac{1}{\sqrt{1+x^2}}$, set $y=\tan^{-1}(x)$ and therefore $y=\cos^{-1}(\frac{1}{\sqrt{1+x^2}})$
Dividing both sides by $2$: 
$\dfrac{y}{2}=\dfrac{1}{2}\cos^{-1}(\frac{1}{\sqrt{1+x^2}})$ 
$\cos(\dfrac{1}{2}\tan^{-1}(x))=\cos(\dfrac{1}{2}\cos^{-1}(\frac{1}{\sqrt{1+x^2}}))$ 
$\cos(\dfrac{1}{2}\tan^{-1}(-\frac{4}{3}))=\cos(\dfrac{1}{2}\cos^{-1}(\frac{3}{5}))$ 
This is my approach to the problem, are there any better ways to simplify the initial expression and perhaps find an approximation even without using a calculator?


Let $\theta = \arctan\left(-\dfrac 43 \right)$
Then, from the picture, $\cos \theta = \dfrac 35$ and $\theta$ is in the fourth quadrant.
So $\cos(\frac 12 \theta) 
= \sqrt{\dfrac{1 + \cos \theta}{2}}
= \sqrt{\dfrac{1 + \frac 35}{2}}
= \dfrac{2}{\sqrt 5}$
