How can I simplify $\tan(S\arctan(R))$ for real S and R? My attempt
$\begin{align}S\arctan R & = \frac{1}{i}\log\left(\frac{1}{\sqrt{1+R^2}}+\frac{R}{\sqrt{1+R^2}}i\right)^S\\ &= \frac{1}{i}\log\left(\frac{1}{1+R^2}\right)^\frac{S}{2}\sum_{n\ge 0}{S\choose n}(Ri)^n \end{align}$
But I have no idea how to simplify this expression.

Note that the question is essentially equivalent to asking for a simplification of $\tan(\alpha x)$ for a real number $\alpha$, since if we had an expression for $\tan(\alpha x)$ in terms of $\sin(x),\cos(x)$ and $\tan(x)$, then we can simply use the identities
$$\sin(\arctan(x))=\frac{x}{\sqrt{x^2+1}},\quad\cos(\arctan(x))=\frac{1}{\sqrt{x^2+1}},\quad\tan(\arctan(x))=x$$
to convert that into a formula for $\tan(\alpha\arctan(x))$. This question becomes one that has been asked many times before, and essentially the answer is that no, there is no nice simplification except for the cases when $\alpha$ is a positive integer. For more details, see this question. Note that the linked question addresses $\cos(\alpha x)$, but the idea is essentially the same for $\tan(\alpha x)$ as well.
