Series representation of $\sin(nu)$ when $n$ is an odd integer? So, out of boredom and curiosity, today I came up with a series representation for $\sin(nu)$ when $n$ is an even integer:
$$\sin(nu) = \sum_{k=1}^\frac n2 \left(\left(-1\right)^{k-1}\binom{n}{-\left|2k-n\right|+n-1}\sin\left(u\right)^{2k-1}\cos\left(u\right)^{n-2k+1}\right)\;\mathtt {if}\;n\in 2\Bbb Z$$
I was working on a similar representation for when $n$ is an odd integer, but I'm having some difficulties. There doesn't seem to be much of a pattern. If it exists, could someone please point me in its direction? If it's impossible, could you provide me with the proof?
 A: Notice that, using Euler's Formula, for general $n$ a positive integer,
$$\begin{align}\sin(nx) &= \frac{e^{inx}-e^{-inx}}{2i}\\
&=\frac{(\cos{x} + i\sin{x})^n - (\cos{x}  - i\sin{x})^n}{2i}\\
&=\sum_{k=0}^{n}{\binom{n}{k}\frac{\cos^k{x}(i\sin{x})^{n-k} - \cos^k{x}(-i\sin{x})^{n-k}}{2i}}\\
&=\sum_{k=0}^{n}{\binom{n}{k}\cos^k{x}\sin^{n-k}{x}\frac{i^{n-k} - (-i)^{n-k}}{2i}}\\
&=\sum_{k=0}^{n}{\binom{n}{k}\cos^k{x}\sin^{n-k}{x}\sin{\frac{1}{2}(n-k)\pi}}\end{align}$$
I am ashamed to confess blatantly that this was taken (word for word) from here, the first link returned using the Google search query "Multiple Angle Formula".
A: Using complex methods and the binomial theorem,
$$\eqalign{\sin(nu)
  &={\rm Im}(\cos u+i\sin u)^n\cr
  &={\rm Im}\sum_{m=0}^n \binom nm (\cos u)^{n-m}(i\sin u)^m\ .\cr}$$
As only the terms for odd $m$ contribute to the imaginary part we can take $m=2k-1$ to give
$$\sin(nu)=\sum_{k=1}^{(n+1)/2}(-1)^{k-1}\binom n{2k-1}\cos^{n-2k+1}u\sin^{2k-1}u\ .$$
A: There are very nice and simple formulas using Chebyshev polynomials of the first kind $T_n$ (if $n$ is odd) and of the second kind $U_n$ (if $n$ is even)
$$sin(nx)=(-1)^{\frac{n-1}{2}} T_n(\sin (x))$$ 
$$sin(nx)=(-1)^{\frac{n}{2}-1} \cos (x) U_{n-1}(\sin (x))$$
