Evaluate $\sum_{r=0}^n \binom{n}{r}\sin rx \cos (n-r)x$

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 $${n\choose r}\sin rx\cdot\cos (n-r)x+{n\choose {n-r}}\sin(n-r)x\cdot \cos rx={n\choose r}\sin nx$$

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 $

$\implies \text{J}=-\text{J}$

$\implies \text{J}=0$

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