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 $n.$
 A: The Chebyshev polynomials of the second kind give you
$$
\sin(n \theta) = U_{n-1}(\cos \theta) \sin \theta
.
$$
In this formula, put $\theta = \arctan x$ and use $\cos \theta = \dfrac{1}{\sqrt{1+x^2}}$ and $\sin \theta = \dfrac{x}{\sqrt{1+x^2}}$.
A: 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:
$$\sin(n\tan^{-1}(x))=\frac{i}{2}\left(\frac{\sum_{k=0}^{n} \binom{n}{k}(-ix)^k-\sum_{k=0}^{n} \binom{n}{k}(ix)^k}{(x^2+1)^\frac{n}{2}}\right)$$
$$\sin(n\tan^{-1}(x))=\frac{i}{2}\left(\frac{\sum_{k=0}^{n} \binom{n}{k}x^k((-i)^k-i^k)}{(x^2+1)^\frac{n}{2}}\right)$$
$$\sin(n\tan^{-1}(x))=\frac{1}{(x^2+1)^\frac{n}{2}}\sum_{k=0}^{n} \binom{n}{k}x^k\frac{i((-i)^k-i^k)}{2}$$
$$\sin(n\tan^{-1}(x))=\frac{1}{(x^2+1)^\frac{n}{2}} \left(\binom{n}{1}x^1-\binom{n}{3}x^3+\binom{n}{5}x^5+\dots\right)$$
$$\sin(n\tan^{-1}(x))=\frac{1}{(x^2+1)^\frac{n}{2}} \sum_{k=0}^{n}\binom{n}{k}\cos\left((n-1)\frac{\pi}{2}\right)x^k$$
A: Too long for a comment
Assume $x$ is real.
Let $S_n(x)=\sin(n\operatorname{arctan}(x))$.
Denote $\operatorname{arctan}$ as $g$.
Let $T_n(x)=e^{ing(x)}$.
Sloppily, $$T_n(x)=\left(e^{-ig(x)}\right)^{-n}$$
Let $f(x)=e^{-ig(x)}$.
Then, $$\frac{dT_n(x)}{df(x)}=-nT_{n+1}$$
Thus we obtain a recursive relation:
$$T_{n+1}=-\frac in\cdot e^{i\operatorname{arctan}(x)}(1+x^2)\cdot T_n’(x)$$
Also, $$S_{n}(x)=\Im T_n(x)$$
