summation of binomial series

Question

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.

Answer 1

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 $$\boxed{S = 2^{n-1}\sin(nx)} $$