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?

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
$$\sin(nu)=\sum_{k=1}^{(n+1)/2}(-1)^{k-1}\binom n{2k-1}\cos^{n-2k+1}u\sin^{2k-1}u\ .$$
