<s>[INST] 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?
 [/INST] Take a right-angled triangle with sides $1,x,\sqrt{1+x^2}$. The triangle exists by Pythagoras' Theorem.
The $\tan^{-1}(x)$ is its angle opposite to $x$. The cosine of that angle is $\frac{1}{\sqrt{1+x^2}}$.
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