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?

Inverse trig functions should always be thought of as angles.  Let $\theta = \tan^{-1}\left(\frac{-4}{3}\right)$.  Then your expression is $\cos \frac{\theta}{2},$ which might suggest the half-angle identity $\cos \frac{\theta}{2} = \pm\sqrt{\frac{1+\cos \theta}{2}}.$    Since you've already evaluated $\cos \theta,$ you're done.
