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Letting $\xi=e^{\frac{2\pi i}{2n}}$, a primitive root of unity $z^{2n}=1$. Then, $\sin\left(\frac{\pi k}n\right)=\frac{\xi^{k}-\xi^{-k}}{2i}$ and hence \begin{align} \sum_{k=0}^{n-1}\sin^m\left(\frac{\pi k}n\right) &=\frac1{(2i)^m}\sum_{k=0}^{n-1}\sum_{j=0}^m\binom{m}j(-1)^j\xi^{-jk}\xi^{(m-j)k} \\ &=\frac1{(2i)^m}\sum_{j=0}^m(-1)^j\binom{m}j\sum_{k=0}^{n-1}\xi^{(m-2j)k} \\ &=\frac1{(2i)^m}\sum_{j=0}^m(-1)^j\binom{m}j\frac{\xi^{(m-2j)n}-1}{\xi^{m-2j}-1} \\ &=\frac1{(2i)^m}\sum_{j=0}^m(-1)^j\binom{m}j\frac{\xi^{mn}-1}{\xi^{m-2j}-1}. \end{align} The outcome is parity-dependent. For example, if $m\rightarrow 2m$ is even then the average value is $$\frac1n\sum_{k=0}^{n-1}\sin^{2m}\left(\frac{\pi k}n\right) =\frac1{2^{2m}}\binom{2m}m.$$ By the way, $$\frac1{\pi}\int_0^{\pi}\sin^{2m}x\,dx=\frac1{2^{2m}}\binom{2m}m.$$ I believe, a similar computation can be worked out for $m$ odd.