Binomial Summations 5

Let $S = \sum_{k=0}^n \left[\binom nk \sin(kx)\cos\left((n-k)x\right)\right]\ \ \cdots(1)$

Now, since $^n\!C_r =\ ^n\! C_{n-r}$

$S$ can be reversed to reach $S = \sum_{k=0}^n \left[\binom nk \sin((n-k)x)\cos\left(kx\right)\right]\ \ \cdots(2)$

Adding $(1)$ and $(2)$ and using $\sin(A)\cos\left(B\right)+\cos\left(A\right)\sin(B)=sin\ (A+B)$ $\begin{aligned}2S &= \sum_{k=0}^n\binom nk \sin nx\\&=2^n\sin nx\end{aligned}$

$\Huge\displaystyle \therefore \boxed{2S = 2^{n}\sin\ nx}$