How can I compute $\tan(.5\arctan(x))$? The plot for this function appears to be in the form of $\alpha*\arctan(\beta*x)$ but I've no clue how to go about simplifying the expression.
 A: You have to compute $y=\tan\left( \frac{\arctan{(x)}}{2} \right)$. Let's call $\theta=\arctan(x)$, so you have to compute $y=\tan\left(\frac{\theta}{2}\right)$.
You should write $\tan\left(\frac{\theta}{2}\right)$ in terms of $\tan\left(\theta\right)$, say you use:
$$\tan\left(\frac{\theta}{2}\right)=\frac{1-\cos\left(\theta\right)}{\sin\left(\theta\right)} = \frac{1}{\sin\left(\theta\right)}-\frac{1}{\tan\left(\theta\right)}$$
but from $$\cos^2\left(\theta\right)+\sin^2\left(\theta\right)=1$$ we get that $$\frac{1}{\tan^2\left(\theta\right)}+1 = \frac{1}{\sin^2\left(\theta\right)}$$
which is the same of
$$
\frac{1}{\sin\left(\theta\right)}=\sqrt{1+\frac{1}{\tan^2\left(\theta\right)}}\quad 0\lt\theta\lt\pi \Rightarrow x \gt 0 \\
$$
$$
\frac{1}{\sin\left(\theta\right)}=-\sqrt{1+\frac{1}{\tan^2\left(\theta\right)}}\quad -\pi\lt\theta\lt 0 \Rightarrow  x \lt 0\\
$$
so finally we can write
$$\tan\left(\frac{\theta}{2}\right)= \frac{1}{\sin\left(\theta\right)}-\frac{1}{\tan\left(\theta\right)}=\pm\sqrt{1+\frac{1}{\tan^2\left(\theta\right)}}-\frac{1}{\tan\left(\theta\right)}$$
This is very helpfull because  $\theta=\arctan(x)$ so $\tan(\arctan(x))=x$ and
$$y=\tan\left( \frac{\arctan{(x)}}{2} \right)=+\sqrt{1+\frac{1}{x^2}}-\frac{1}{x}\quad x\gt 0$$
$$y=\tan\left( \frac{\arctan{(x)}}{2} \right)=-\sqrt{1+\frac{1}{x^2}}-\frac{1}{x}\quad x\lt 0$$
A: $$y=\tan(0.5 \arctan(x))$$
$$\arctan(y)= \frac12 \arctan(x)$$
$$x=\tan(2 \arctan(y))$$
using formula of $\tan(2\theta)$
$$x= \frac{2 \tan(\arctan(y))}{1-\tan^2(\arctan(y))}=\frac{2y}{1-y^2}$$
Then you can find $y$. 

For the form of $y=\tan(\beta \arctan(x))$, in general, there is no simplification.
Update
Thanks to Jeevan Devaranjan for his comment mentioning that 
$$\tan(\beta \arctan x)=(\frac1i)(\frac{(1+ix)^\beta-(1-ix)^\beta}{(1+ix)^\beta+(1-ix)^\beta})$$
A: Recall that $\tan(a - b) = \frac{\tan(a) - \tan(b)}{1 + \tan a \tan b}$. Also let $c =   \tan({\frac{\arctan x}{2}})$.
\begin{align}
c &= \tan(\arctan x - \frac{\arctan x}{2})\\
&= \frac{\tan(\arctan x) - \tan(\frac{\arctan x}{2})}{1 + \tan(\arctan x)\tan(\frac{\arctan}{2})}\\
&= \frac{x - c}{1 + xc}\\
c^2x + c &= x- c\\
c^2x + 2c - x & = 0
\end{align}
So by the quadratic formula(note that we take the positive root since due to the domain of the inverse tangent function)
\begin{equation}
c = \frac{-2 + \sqrt{4  - 4(x)(-x)}}{2x} = -\frac{1}{x} + \sqrt{1 + \frac{1}{x^2}}
\end{equation}
Therefore
\begin{equation}
\tan(\frac{\arctan x}{2}) = -\frac{1}{x} + \sqrt{1 + \frac{1}{x^2}}
\end{equation}
A: As shown in this answer, $\tan(x/2)=\frac{\sin(x)}{1+\cos(x)}$. Then, since $\sin\left(\tan^{-1}(x)\right)=\frac{x}{\sqrt{1+x^2}}$ and $\cos\left(\tan^{-1}(x)\right)=\frac1{\sqrt{1+x^2}}$, we get
$$
\tan\left(\tfrac12\tan^{-1}(x)\right)=\frac{x}{\sqrt{1+x^2}+1}
$$
This works for all $x\in\mathbb{R}$.
