Compute $\sum_{j=1}^k\cos^n(j\pi/k)\sin(nj\pi/k)$ Compute the series
$\sum_{j=1}^k\cos^n(j\pi/k)\sin(nj\pi/k)$
Hint: the answer is in fact 0
 A: Recall that $\cos(x)=(e^{ix}+e^{-ix})/2$ and $\sin(x)=(e^{ix}-e^{-ix})/2i$, so your sum can be rewritten as
$$\sum_{j=1}^k\biggl[\frac{e^{j\pi i/k}+e^{-j\pi i/k}}2\biggr]^n\,\biggl[\frac{e^{nj\pi i/k}-e^{-nj\pi i/k}}{2i}\biggr]\,.$$
Applying binomial theorem on the first factor (of each summand) your sum becomes
$$\begin{align*}
&\,\frac1{2^{n+1}i}\sum_{j=1}^k\sum_{r=0}^n\binom nr\bigl(e^{j\pi i/k}\bigr)^r\bigl(e^{-j\pi i/k}\bigr)^{n-r}\,\bigl[e^{nj\pi i/k}-e^{-nj\pi i/k}\bigr]\\[2mm]
&\,\frac1{2^{n+1}i}\sum_{r=0}^n\binom nr\sum_{j=1}^k\bigl[e^{2rj\pi i/k}-e^{2j(r-n)\pi i/k}\,\bigr]\,.
\end{align*}$$
Now $\sum_{j=1}^ke^{2rj\pi i/k}$ is the sum of a geometric progression, with sum equals to 
$$e^{2r\pi i/k}\frac{e^{2rk\pi i/k}-1}{e^{2r\pi i/k}-1}=0\,,$$
because $e^{2r\pi i}=1$; similarly,
$$\sum_{j=1}^ke^{2j(r-n)\pi i/k}=e^{2(r-n)\pi i/k}\frac{e^{2(r-n)k\pi i/k}-1}{e^{2(r-n)\pi i/k}-1}=0\,,$$
and so all the inner summands above are equal to $0$.
A: Hint: by symmetry $\sum (\zeta+\zeta^{-1})^n\zeta^n=\sum (\zeta+\zeta^{-1})^n\zeta^{-n}$ as $\zeta$ is summed over the $k$th roots of $1$.
