summation of binomial series Can someone pls provide the approach/solution to this problem
$$\sum_{r=1}^n \begin{pmatrix}n\\r\end{pmatrix}\sin(rx) \cos((n-r)x)$$
I have tried to simplify the sine and cosine terms and then arrived at this part but am unable to solve any further...
$$\sum_{r=1}^n \begin{pmatrix}n\\r\end{pmatrix}\frac {\sin(2rx-nx)}{2}+2^{n-1} \sin(nx)-\frac{\sin(nx)}{2}$$
Also it would be very helpful if you can advice on how to approach such questions.

For answering the question first observe the fact that $\displaystyle \binom{n}{k} = \binom{n}{n-k}$
let $$\begin{align} S &= \sum_{r =\color{red}0}^{n}\binom{n}{r}\sin(rx)\cos((n-r)x)\\
S&=\sum_{r =\color{red}0}^{n}\binom{n}{n-r}\sin((n-r)x)\cos(rx) = \sum_{r=1}^{n}\binom{n}{n-r}\sin((n-r)x)\cos(rx) + \sin(0x)\cos(nx) \\
2S &= \sum_{r =\color{red}0}^{n}\binom{n}{r}\sin(rx + (n-r)x)\\
2S &= \sin(nx)\sum_{r = 0}^{n}\binom{n}{r}  \end{align}$$
Or $$\boxed{S = 2^{n-1}\sin(nx)} $$
