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?
 A: 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.
A: 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 $\frac{1}{\sqrt{1+x^2}}$.
A: 
Let $\theta = \arctan\left(-\dfrac 43 \right)$
Then, from the picture, $\cos \theta = \dfrac 35$ and $\theta$ is in the fourth quadrant.
So $\cos(\frac 12 \theta) 
= \sqrt{\dfrac{1 + \cos \theta}{2}}
= \sqrt{\dfrac{1 + \frac 35}{2}}
= \dfrac{2}{\sqrt 5}$
A: Consider the following :
$$\begin{align}
&\implies\frac{\arctan\left(\frac{-4}{3}\right)}{2}=y \\
&\implies2y=\arctan(\frac{-4}{3})\\
&\implies\tan(2y)=-4/3\\
&\implies\frac{2\tan y}{1-\tan^2y}=-4/3\\ 
&\implies2\tan^2y-3\tan y-2=0\\
\end{align}$$
Solving this equation we get-
$$\tan y=-1/2$$and $$\tan y=2$$
Using this we can find the value of $\cos y$ which will give the value of the expression. It comes out to be:
$$\cos y=\frac{1}{\sqrt{5}}$$
Hope my answer is helpful. Sorry for any errors and please let me know if there are some. 
