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.
 A: 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. 
A: Here is an answer when $t = 2s$ is even.

Let $n, s$ be non-negative integers. Then with $\zeta = \mathrm{e}^{i\pi/n}$ and $\omega = \zeta^2 = \mathrm{e}^{2i\pi/n}$, we have 
\begin{align*}
2^{2s} \sum_{k=0}^{n-1} \sin^{2s} \left(\frac{\pi k}{n} \right)
&= \sum_{k=0}^{n-1}  \left( \frac{\zeta^k + \zeta^{n-k}}{i} \right)^{2s} \\
&= (-1)^{-s} \sum_{k=0}^{n-1}\sum_{l=0}^{2s} \binom{2s}{l} \zeta^{(2s-l)k+l(n-k)} \\
&= (-1)^{-s} \sum_{k=0}^{n-1}\sum_{l=0}^{2s} \binom{2s}{l} (-1)^l \omega^{(s-l)k}.
\end{align*}
Interchanging the order of summation, we have
\begin{align*}
2^{2s} \sum_{k=0}^{n-1} \sin^{2s} \left(\frac{\pi k}{n} \right)
&= \sum_{l=0}^{2s} (-1)^{l-s}  \binom{2s}{l} \left( \sum_{k=0}^{n-1} (\omega^{s-l})^k \right) \\
&= \sum_{l=0}^{2s} (-1)^{l-s}  \binom{2s}{l} \left( n \cdot \mathbf{1}_{\{ l \equiv s \ (\mathrm{mod} \ n)\}} \right) \\
&= n \sum_{j} (-1)^{nj}  \binom{2s}{s+nj},
\end{align*}
where the last summation runs over all integers $j$ such that $-\frac{s}{n} \leq j \leq \frac{s}{n}$.

Examples. As special cases, plugging $s = n$ yields
$$ \sum_{k=0}^{n-1} \sin^{2n} \left(\frac{\pi k}{n} \right) = \frac{n}{2^{2n}} \left[ \binom{2n}{n} + (-1)^n 2 \right], $$
and similarly, replacing $n$ by $2n$ and plugging $s = n$ yields
$$ \sum_{k=0}^{n-1} \sin^{2n} \left(\frac{\pi k}{2n} \right) = \frac{n}{2^{2n}} \binom{2n}{n} - \frac{1}{2}. $$


Addendum. Here is an intuition on why we expect the sum to be simplified. If $t = 2s$ is even, then we can expand $\sin^{2s} x$ into a linear combination of $1, \cos 2x, \cos 4x, \cdots, \cos 2sx$. So in our case, we may write
$$ \sin^{2s} \left(\frac{\pi k}{n}\right) = \sum_{j=0}^{s} a_j \cos \left(\frac{2\pi j k}{n}\right). $$
Now, as you sum this over $k = 0, \cdots, n-1$, all the cosine terms will cancel out except when $j$ is a multiple of $n$. This means that
$$ \sum_{k=0}^{n-1} \sin^{2s} \left(\frac{\pi k}{n}\right) = n (a_0 + a_n + a_{2n} + \cdots). $$
So it is enough to identify $a_j$ for $j$'s multiple of $n$. This can be done by expanding $\sin^{2s} x$ using complex exponential and the binomial theorem. This is essentially what we did in the computation above.
On the other hand, this trick does not work for odd $t$. Indeed, when $t = 2s+1$ is odd, we can write
$$ \sin^{2s+1} \left(\frac{\pi k}{n}\right) = \sum_{j=0}^{s} a_j \sin \left(\frac{\pi (2j+1)k}{n}\right) $$
for some constants $a_0, \cdots, a_n$. (In fact, $a_j = (-1)^j 2^{-2s} \binom{2s+1}{s-j}$.) Now summing both sides for $k = 0, \cdots, n-1$,
\begin{align*}
\sum_{k=0}^{n-1}\sin^{2s+1} \left(\frac{\pi k}{n}\right)
&= \sum_{j=0}^{s} a_j \sum_{k=0}^{n-1} \sin \left(\frac{\pi (2j+1)k}{n}\right) \\
&= \frac{1}{2^{2s}} \sum_{j=0}^{s} (-1)^j \binom{2s+1}{s-j} \cot \left(\frac{\pi(2j+1)}{2n}\right).
\end{align*}
It is hard for me to believe that this will ever simplify except for some nice $s$.

Addendum 2 (Just for fun). Although no longer simpler than the original sum, one can also prove that for complex $t$ with $\Re(t) > 0$ and for positive integer $n$ the following formula holds:
$$\sum_{k=0}^{n-1} \sin^{t} \left(\frac{\pi k}{n} \right) = \frac{n}{2^{t}} \sum_{j=-\infty}^{\infty} (-1)^{nj}  \binom{t}{\frac{t}{2}+nj}. $$
Here, $\binom{n}{k} = \frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}$ is the extended binomial coefficient.
A: I found this long ago:

$$\begin{aligned}
  {S_t}(n) &= \sum\limits_{k = 0}^n {{{\sin }^p}} (ak)\\
  {\text{For odd p}}&:\\
  {S_t}(n) &= \frac{1}{{{2^p}}}\sum\limits_{k = 0}^{\frac{{p - 1}}{2}} {{{( - 1)}^k}} C(p,\frac{{p - 1}}{2} - k){\text{Long}}\\
  {\text{Long}} &= \left[ {\cot \left( {\frac{a}{2}(2k + 1)} \right) - \csc \left( {\frac{a}{2}(2k + 1)} \right)\cos \left( {\frac{a}{2}(4kn + 4k + 4n + 1)} \right)} \right]\\
  {\text{For even p}}&:\\
  {S_t}(n) &= \frac{1}{{{2^p}}}\left[ {\left( {\frac{1}{2} + n} \right)C(p,\frac{p}{2}) + \sum\limits_{k = 1}^{\frac{p}{2}} {{{( - 1)}^k}} C(p,\frac{p}{2} - k)\csc (ak)\sin (ak(2n + 1))} \right]\\ 
\end{aligned}$$

$C(n,k) = \left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right)$, just as a short to input $\LaTeX$.
I try to find my proof but in vain, and I am seeking for a new elegent proof.

So, for this question:
Let $p \to t,n \to n - 1,a \to \pi /2n$
It's:
$$\begin{aligned}
  {S_p}(n) &= \sum\limits_{k = 0}^{n - 1} {{{\sin }^t}} \left( {\frac{\pi }{{2n}}k} \right)\\
  {\text{For odd n}}&:\\
  {S_p}(n) &= {2^{ - t}}\sum\limits_{k = 0}^{\frac{{t - 1}}{2}} {{{( - 1)}^k}C(t,\frac{1}{2}( - 2k + t - 1)){\text{Long}}} \\
  {\text{Long}} &= \csc \left( {\frac{{2\pi k + \pi }}{{4n}}} \right)\left( {\sin \left( {\frac{{\pi (k(4n - 2) - 1)}}{{4n}}} \right) + \cos \left( {\frac{{2\pi k + \pi }}{{4n}}} \right)} \right)\\
  {\text{For even n}}&:\\
  {S_p}(n) &= {2^{ - t}}\left( {\left( {n - \frac{1}{2}} \right)C(t,\frac{t}{2}) + \sum\limits_{k = 1}^{\frac{t}{2}} {{{( - 1)}^k}} C(t,\frac{t}{2} - k){\text{Long}}} \right)\\
  {\text{Long}} &= \sin \left( {\frac{{\pi (k(2(n - 1) + 1))}}{{2n}}} \right)\csc \left( {\frac{{\pi k}}{{2n}}} \right)\\ 
\end{aligned}$$
It we want to find some nice closed-form , the $\text{Long}$ part should be simple.
