Evaluate $\sum_{r=0}^n \binom{n}{r}\sin rx \cos (n-r)x$ 
Evaluate
$$ \sum_{r=0}^n \left[\binom{n}{r}\cdot\sin rx \cdot \cos (n-r)x\right] $$

I tried to use binomial identities, but since there are trigonometric terms, I don't have the idea how to approach it.
Can anyone please help?
 A: Hint:
Since $${n\choose r}={n\choose {n-r}}\qquad \text{and}\qquad \sin(a+b)=\sin a\cos b+\sin b\cos a$$ we have $${n\choose r}\sin rx\cdot\cos (n-r)x+{n\choose {n-r}}\sin(n-r)x\cdot \cos rx={n\choose r}\sin nx$$
A: Let, 
$\text{S} =\displaystyle \sum_{r=0}^n \left[\dbinom{n}{r}\cdot\sin (rx) \cdot \cos (n-r)x\right]$    
$=\dfrac{1}{2}\displaystyle \sum_{r=0}^n \left[\dbinom{n}{r}\cdot2\sin (rx)  \cos (n-r)x\right]$
$= \dfrac{1}{2}\displaystyle \sum_{r=0}^n \dbinom{n}{r}\cdot (\sin(nx)+\sin(2r-n)x)$   
$= 2^{n-1}\sin(nx)+\displaystyle\dfrac{1}{2}\sum_{r=0}^n \dbinom{n}{r}\cdot\sin(2r-n)x$     
Now, 
$\text{J}=\displaystyle\sum_{r=0}^n \dbinom{n}{r}\cdot\sin(2r-n)x $    
$=\displaystyle\sum_{r=0}^n \dbinom{n}{n-r}\cdot\sin[2(n-r)-n]x $     $\left(\because \displaystyle\sum_{r=0}^n f(r) = \sum_{r=0}^n f(n-r)\right)$  
$=-\displaystyle\sum_{r=0}^n \dbinom{n}{r}\cdot\sin(2r-n)x $    
$\implies \text{J}=-\text{J}$   
$\implies \text{J}=0$
$\therefore S=\boxed{2^{n-1}\sin(nx)}$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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Note that
  $\ds{\pars{a - \ol{a}}\pars{b + \ol{b}} = ab + a\ol{b} - \ol{a}b - \ol{a}\ol{b} =2\ic\,\Im\pars{ab + a\ol{b}}}$

\begin{align}
&\color{#f00}{%
\sum_{r = 0}^{n}{n \choose r}\sin\pars{rx}\cos\pars{\bracks{n - r}x}} =
\sum_{r = 0}^{n}{n \choose r}{\expo{\ic rx} - \expo{-\ic rx} \over 2\ic}\,
{\expo{\ic\pars{n - r}x} + \expo{\ic\pars{r - n}x} \over 2}
\\[3mm] = &\
-\,{1 \over 4}\,\ic\pars{2\ic}\Im\sum_{r = 0}^{n}{n \choose r}\pars{%
\expo{\ic n x} + \expo{-\ic nx}\expo{2\ic rx}}
\\[3mm] = &\
\half\,\Im\bracks{\expo{\ic nx}\sum_{r = 0}^{n}{n \choose r} +
\expo{-\ic nx}\sum_{r = 0}^{n}{n \choose r}\pars{\expo{2\ic x}}^{r}} =
\half\,\Im\bracks{\expo{\ic nx}2^{n} +
\expo{-\ic nx}\pars{1 + \expo{2\ic x}}^{n}}
\\[3mm] = &\
\half\,\Im\bracks{\expo{\ic nx}2^{n} +
2^{n}\cos^{n}\pars{x}} = \color{#f00}{2^{n - 1}\sin\pars{nx}}
\end{align}
