Closed form for $\sin(n\arctan(x))$ Is there a closed form for the function $\sin(n\arctan x)$, perhaps where $n$ is restricted to being an integer, or if not, perhaps some special integers (such as triangular numbers or some other figurate numbers)?
From playing around with a few values, it seems that 
$$\sin\arctan(x)=\frac{x}{\sqrt{1+x^2}},~\sin(2\arctan x)=\frac{2x}{1+x^2},~\sin(3\arctan x)=\frac{3x-x^3}{(1+x^2)^{3/2}},$$
I can see that the denominator is $(1+x^2)^{\tfrac12n}$ but can't quite see the form of the numerator.
Motivation: This is motivated by an inconvenient but necessary change of coordinates from polar to Cartesian when a function involves not $\sin\theta$ but $\sin n\theta$ for some integer $n.$

We have that
$$\tan^{-1}(x)=\frac{i}{2}\log\left(\frac{1-ix}{1+ix}\right)$$
And
$$\sin(x)=\frac{i}{2}e^{-ix}-\frac{i}{2}e^{ix}$$
So
$$\sin(n\tan^{-1}(x))=\frac{i}{2}\exp\left(\frac{i}{2}n\log\left(\frac{1-ix}{1+ix}\right)\right)-\frac{i}{2}\exp\left(\frac{i}{2}n\log\left(\frac{1+x}{1-ix}\right)\right)$$
$$\sin(n\tan^{-1}(x))=\frac{i}{2}\left(\frac{(1-ix)^\frac{n}{2}}{(1+ix)^\frac{n}{2}}-\frac{(1+ix)^\frac{n}{2}}{(1-ix)^\frac{n}{2}}\right)$$
$$\sin(n\tan^{-1}(x))=\frac{i}{2}\left(\frac{(1-ix)^n-(1+ix)^n}{(x^2+1)^\frac{n}{2}}\right)$$
Now we can use the Binomial theorem:
$$\sin(n\tan^{-1}(x))=\frac{i}{2}\left(\frac{\sum_{k=0}^{n} \binom{n}{k}(-ix)^k-\sum_{k=0}^{n} \binom{n}{k}(ix)^k}{(x^2+1)^\frac{n}{2}}\right)$$
$$\sin(n\tan^{-1}(x))=\frac{i}{2}\left(\frac{\sum_{k=0}^{n} \binom{n}{k}x^k((-i)^k-i^k)}{(x^2+1)^\frac{n}{2}}\right)$$
$$\sin(n\tan^{-1}(x))=\frac{1}{(x^2+1)^\frac{n}{2}}\sum_{k=0}^{n} \binom{n}{k}x^k\frac{i((-i)^k-i^k)}{2}$$
$$\sin(n\tan^{-1}(x))=\frac{1}{(x^2+1)^\frac{n}{2}} \left(\binom{n}{1}x^1-\binom{n}{3}x^3+\binom{n}{5}x^5+\dots\right)$$
$$\sin(n\tan^{-1}(x))=\frac{1}{(x^2+1)^\frac{n}{2}} \sum_{k=0}^{n}\binom{n}{k}\cos\left((n-1)\frac{\pi}{2}\right)x^k$$
