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?

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)
$$sin(nx)=(-1)^{\frac{n-1}{2}} T_n(\sin (x))$$ 
$$sin(nx)=(-1)^{\frac{n}{2}-1} \cos (x) U_{n-1}(\sin (x))$$
