How to evaluate $\sum\limits_{k=0}^{n-1} \sin^t(\pi k/2n)$? How to evaluate $\displaystyle\sum_{k=0}^{n-1} \sin^t\left(\frac{\pi k}{2n}\right)$?
$t,n$ are constants $\in \Bbb{Z}$.
My try:
$$\begin{align}
\zeta:=e^{i\pi/2n} \implies & \sum_{k=0}^{n-1}\sin^t\left(\frac{\pi k}{2n}\right) \\
= &\frac{1}{2^t}\sum_{k=0}^{n-1}\left(\zeta^k- \zeta^{-k}\right)^{t} \\
= &\frac{1}{2^t}\sum_{k=0}^{n-1}\left(\zeta^{-kt}\right)\left(\zeta^{2k}-1\right)^{t}\\
\end{align}$$
How to proceed from here?
If a closed-form solution is not possible, can we still work out cases such as $t = n, t = 2n, $ etc?
EDIT: Sangchul Lee provided an answer for even t. Still looking for a solution for odd t as well.

This integral remind me of Matsubara sum. Let me use contour integral to provide another perspective of the case $t = 2s$. 
Define $a_n = \sum_{k = 1}^{n-1} \sin ^{t} \left( \frac{\pi k }{n} \right)  $. What OP is looking for is
\begin{equation}
\sum_{k=1}^{n-1} \sin^{t} \left(  \frac{\pi k }{2n} \right)  = \frac{1}{2} \left( \sum_{k=1}^{2n-1} \sin^{t} \left(  \frac{\pi k }{2n} \right) -1  \right)  = \frac{1}{2} ( a_{2n } - 1 )
\end{equation}
One can extend the summation to all the roots of $\zeta^{2n} = 1$ if $t = 2s$
\begin{equation}
a_n = \frac{1}{2} \sum_{k=1}^{2n}  \sin^{2s} \left( \frac{\pi k }{n} \right )  = \frac{1}{2} \sum_{ k = 1}^{2n}   f( \zeta_k  )  \qquad \zeta_k = \exp( \frac{ik\pi }{n})
\end{equation} 
where 
\begin{equation}
f(z ) = \left( \frac{z - \frac{1}{z}}{2i} \right)^{2s} 
\end{equation}
The summation can be regarded as the sum of the residue of the follow contour integral
\begin{equation}
 \sum_{ k = 1}^{2n}   f( \zeta_k  ) =  \frac{2n}{2\pi i }\oint_{C}   \frac{z^{2n-1}}{ z^{2n} - 1} f( z)  dz
\end{equation}
where the integrand is specifically designed such that
\begin{equation}
\text{Res} ( \frac{ 2n z^{2n-1}}{z^{2n} - 1} f(z) , \zeta_k ) =   2n \frac{z^{2n-1}(z - \zeta_k)}{z^{2n} - \zeta^{2n}_k} f(\zeta_k )  \Big|_{z = \zeta_k } = f(\zeta_k ) 
\end{equation}
Here the closed contour should enclose all $\zeta_k$ but avoid the singularities of $f$, which can be taken as two concentric circles as shown in the following figure(red spots are those roots $\xi_k$). 

The inner contour gives (minus) the residue at $0$, and the outer contour gives (minus) the residue at $\infty$, hence
\begin{equation}
\begin{aligned}
a_n &= n \left[ -\text{Res}( \frac{z^{2n-1}}{ z^{2n} - 1} f( z) , 0 ) - \text{Res}( \frac{z^{2n-1}}{ z^{2n} - 1} f( z) , \infty ) \right]  \\
\end{aligned}
\end{equation}
Convert the residue of $\infty$ to $0$
\begin{equation}
\begin{aligned}
a_n =& n \left[ \text{Res}( \frac{z^{2n-1}}{ 1 - z^{2n} } f( z) , 0 ) + \text{Res}( \frac{\frac{1}{z}}{1 - z^{2n} } f( \frac{1}{z}) , 0 ) \right]  \\
=& n \left[ \text{Res}( \frac{1}{z}\frac{z^{2n} + 1 }{ 1 - z^{2n} } f(z)  , 0 ) \right] \\
=& n \left[ \text{Res}(  \frac{1}{z} ( 1 + 2\sum_{k=1}^{\infty} z^{2nj} ) f(z) , 0 ) \right] \\ 
\end{aligned}
\end{equation}
Finally, expand $f(z)$ and extract the terms with power $0, -2nj, -4nj$ etc,
\begin{equation}
\begin{aligned}
  a_n &= n \frac{(-1)^s}{2^{2s}} \left( (-1)^s {2s \choose s }   + 2  \sum_{0 < nj < s } (-1)^{s+ nj}  {2s \choose s+nj  } \right)\\
      &= \frac{n}{2^{2s}}   \sum_{ -s < nj < s } (-1)^{nj}  {2s \choose s + nj }   \\
\end{aligned}
\end{equation}
This is the same as what @Sangchul Lee obtained. 
