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.

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}
