The original question was if $\sum_{k=0}^{5} {}^5C_k\sin(kx)\cos(5-k)x=N\sin(5x)$ then what is the value of N?
I plugged $\frac{\pi}{2}$ in place of x and N came out to be 16. But is there any algebraic way to prove this. I tried expanding $(\sin(x)+i\cos(x))^5$ but eventually got stuck.
On
Since $$ 2\sin(kx)\cos((n-k)x)=\sin(nx)+\sin((n-2k)x) $$ we have $$ 2\sum_{k=0}^n\binom{n}k\sin(kx)\cos((n-k)x)=\sin(nx)\sum_{k=0}^n\binom{n}k+\sum_{k=0}^n\binom{n}k\sin((n-2k)x) $$ Now $\sum_{k=0}^n\binom{n}k=2^n$ and \begin{align*} &\sum_{k=0}^n\binom{n}k\sin((n-2k)x)\\ &=\frac12 \sum_{k=0}^n\left[\binom{n}k\sin((n-2k)x)+\binom{n}{n-k}\sin((n-2(n-k))x)\right]\\ &=\sum_{k=0}^n\frac12\binom{n}k\left[\sin((n-2k)x)+\sin(-(n-2k)x)\right]=0 \end{align*} Thus $$ \sum_{k=0}^n\binom{n}k\sin(kx)\cos((n-k)x)=2^{n-1}\sin(nx) $$
Let's consider more generally, $$S=\sum_{k=0}^N\binom Nk\sin kx\cos(N-k)x.$$ Then \begin{align} S&=\frac1{4i}\sum_{n=0}^N\binom Nk(e^{ikx}-e^{-ikx})(e^{i(N-k)x}+e^{-i(N-k)x})\\ &=\frac1{4i}\sum_{n=0}^N\binom Nk(e^{iNx}+e^{i(2k-N)x}-e^{i(N-2k)x}-e^{-iNx})\\ &=\frac1{4i}(2^Ne^{iNx}+e^{-iNx}(1+e^{2ix})^N-e^{iNx}(1+e^{-2ix})^N-2^Ne^{-iNx})\\ &=\frac1{4i}(2^Ne^{iNx}+(e^{-ix}+e^{ix})^N-(e^{ix}+e^{-ix})^N-2^Ne^{-iNx})\\ &=\frac{2^{N-1}}{2i}(e^{iNx}-e^{-iNx})\\ &=2^{N-1}\sin Nx. \end{align}