<s>[INST] Finite summation $\sum _{k=0}^{n-1} \csc ^{2 m}\left(\frac{\pi (2 k+1)}{2 n}+y\right)$ How to prove
$$\sum _{k=0}^{n-1} \csc ^4\left(\frac{\pi  (2 k+1)}{2 n}+y\right)=(n \sec (n y))^4-\frac{2}{3} \left(n^2-1\right) (n \sec (n y))^2$$
Moreover, is there a closed-form for higher order summations? Any help will be appreciated.
 [/INST] Fix an $n$. By analytic continuation, it suffices to assume $y$ be a small positive number, so small such that $0<\frac{\pi(2k+1)}{2n}+y<\pi$ for $k=0,\cdots,n-1$.
Integrate $$f_m(z) = \tan (n(z-y)) \csc^{2m} z$$ around rectangle with vertices $\pm \infty i, \pi \pm \infty i$ (Indention has to be made at $0$ and $\pi$), integrals along two horizontal lines tend to $0$, integrals along two vertical lines cancel by periodicity $f_m(z)=f_m(z+1)$. 
$f_m(z)$ has poles at $$z=0 \qquad z = \frac{\pi(2k+1)}{2n}+y\quad \text{ for }k=0,\cdots,n-1$$
therefore
$$\text{Res}[f_m(z),z=0] -\frac{1}{n}\sum _{k=0}^{n-1} \csc^{2m}\left(\frac{\pi  (2 k+1)}{2 n}+y\right) = 0$$
Calculating residues proves your result, along with instances of higher powers:

$$\begin{aligned}&\sum _{k=0}^{n-1} \csc^{6}\left(\frac{\pi  (2 k+1)}{2
n}+y\right)=\\ &\frac{1}{15} n^2 \sec ^2(n y) \left(15 n^4 \sec ^4(n y)-15 n^4 \sec ^2(n y)+2 n^4+15 n^2 \sec ^2(n y)-10 n^2+8\right)\\ &\sum _{k=0}^{n-1} \csc^{8}\left(\frac{\pi  (2 k+1)}{2
n}+y\right)=\\ &\frac{1}{315} n^2 \sec ^2(n y) \left(315 n^6 \sec ^6(n
y)-420 n^6 \sec ^4(n y)+126 n^6 \sec ^2(n y)-4 n^6+420 n^4 \sec ^4(n
y)-420 n^4 \sec ^2(n y)+56 n^4+294 n^2 \sec ^2(n y)-196
n^2+144\right) \end{aligned}$$

Taking $y=0$, one can also easily deduce $\zeta(2n)$ from such finite trigonometric sums.
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