Proving that the sum $\sum_{n=1}^{\infty }\frac{\sin^m(n\pi/2)}{n^2}=\frac{\pi^2}{8}$ Proving that the sum


  $$\sum_{n=1}^{\infty }\frac{\sin^m(n\pi/2)}{n^2}=\frac{\pi^2}{8}$$
  When $m$=integer even number


I know that the $\frac{\pi^2}{8}$ comes from $\sum_{n=0}^{\infty }\frac{1}{(2n+1)^2}$ and I know how to prove it but I don't know how to prove the above sum  especially with the power $m$
Let $m=2n$, where $n\in\mathbb{N}^{+}$. IOW, $m$ is a positive even number (that statement is false for $m=0$).

$$\begin{align}
s_{2n}
&amp;=\sum_{k=1}^{\infty}\frac{\sin^{2n}{\left(\frac{k\pi}{2}\right)}}{k^2}\\
&amp;=\sum_{k=1}^{\infty}\frac{\sin^{2n}{\left(\frac{(2k-1)\pi}{2}\right)}}{(2k-1)^2}+\sum_{k=1}^{\infty}\frac{\sin^{2n}{\left(\frac{(2k)\pi}{2}\right)}}{(2k)^2};~~\text{(split into even/odd terms)}\\
&amp;=\sum_{k=1}^{\infty}\frac{\sin^{2n}{\left(\frac{(2k-1)\pi}{2}\right)}}{(2k-1)^2}\color{red}{+\sum_{k=1}^{\infty}\frac{\sin^{2n}{\left(k\pi\right)}}{4k^2}};~~\text{(even terms vanish)}\\
&amp;=\sum_{k=0}^{\infty}\frac{\sin^{2n}{\left(\frac{(2k+1)\pi}{2}\right)}}{(2k+1)^2}\\
&amp;=\sum_{k=0}^{\infty}\frac{\left[\sin{\left(k\pi+\frac{\pi}{2}\right)}\right]^{2n}}{(2k+1)^2}\\
&amp;=\sum_{k=0}^{\infty}\frac{\left[\cos{\left(k\pi\right)}\right]^{2n}}{(2k+1)^2};~~~[\sin{(z+\pi/2)}=\cos{z}]\\
&amp;=\sum_{k=0}^{\infty}\frac{\left[(-1)^{k}\right]^{2n}}{(2k+1)^2}\\
&amp;=\sum_{k=0}^{\infty}\frac{1^{n}}{(2k+1)^2}\\
&amp;=\frac{\pi^2}{8}.~~\blacksquare\\
\end{align}$$
Hint: Determine the expression $\sin^m{(n\pi/2)}$. From this point of view you can do some machinary work for solving the series.

After some calculations you get the series
$$\sum _{k=1}^{\infty } \frac{1-(-1)^k}{2k^2}$$

Finally you have to compute the series.
Maybe some Fourier series are useful or you are able to split the series in smart partial series.