Proving $\sum_{n=0}^{\infty }\frac{\sin^4(4n+2)}{(2n+1)^2}=\frac{5\pi ^2}{16}-\frac{3\pi }{4}$ I dont have an idea to prove it because of exist $\sin(4n+2)^4$
$$\sum_{n=0}^{\infty }\frac{\sin^4(4n+2)}{(2n+1)^2}=\frac{5\pi ^2}{16}-\frac{3\pi }{4}$$
 A: One first calculates$$\sin\left(\theta\right)^{4}=\frac{1}{8}\cos\left(4\theta\right)-\frac{1}{2}\cos\left(2\theta\right)+\frac{3}{8}.$$
 Then$$\sum_{n=0}^{+\infty}\frac{\sin\left(2\left(2n+1\right)\right)^{4}}{\left(2n+1\right)^{2}}=\frac{1}{8}\sum_{n=0}^{+\infty}\frac{\cos\left(8m\right)}{m^{2}}-\frac{1}{2}\sum_{n=0}^{+\infty}\frac{\cos\left(4m\right)}{m^{2}}+\frac{9}{32}\zeta\left(2\right)$$
 where we changed the index $2n+1\mapsto m
 .$
Now the answer can be found in this link :
Series $\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}$.
A: Since:
$$\sin x = \frac{1}{2i}\left(e^{ix}-e^{-ix}\right)\tag{1}$$
we have:
$$\sin^4 x = \frac{1}{16}\left(e^{4ix}+e^{-4ix}-4e^{2ix}-4e^{2ix}+6\right)=\frac{1}{8}\left(\cos(4x)-4\cos(2x)+3\right).\tag{2} $$
It follows that:
$$\begin{eqnarray*} S &=& \sum_{n\geq 0}\frac{\sin^4(4n+2)}{(2n+1)^2}\\&=&\frac{1}{8}\sum_{n\geq 0}\frac{\cos(16n+8)}{(2n+1)^2}-\frac{1}{2}\sum_{n\geq 0}\frac{\cos(8n+4)}{(2n+1)^2}+\frac{3}{8}\sum_{n\geq 0}\frac{1}{(2n+1)^2}\tag{3}\end{eqnarray*}$$
where $\sum_{n\geq 0}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$. Now it happens that:
$$ f(x) = \sum_{n\geq 0}\frac{\cos((2n+1)x)}{(2n+1)^2}\tag{4}$$
is the Fourier series of the triangle wave with period $2\pi$ and amplitude $2\cdot \frac{\pi^2}{8}$, hence:
$$ f(4)=\pi-3\frac{\pi^2}{8}, \qquad f(2)=-\frac{\pi}{2}+\frac{\pi^2}{8} \tag{5}$$
and:
$$ S = \frac{1}{8}f(4)-\frac{1}{2}f(2)+\frac{3}{8}f(0)=\color{red}{\frac{5\pi^2}{16}-\frac{3\pi}{4}}\tag{6}$$
as wanted.

Update. Once proved that $\sin^4(2x)=\frac{1}{2}\left(1-\cos(2x)\right)-\frac{1}{8}\left(1-\cos(4x)\right),$ we have:
$$ S = \sum_{n\geq 0}\frac{\sin^4(2n)}{n^2}\cdot\frac{1-\cos(\pi n)}{2}=\sum_{n\geq 0}\frac{\sin^2(n)\sin^2\left(\frac{\pi}{2}n\right)}{n^2}-\frac{1}{4}\sum_{n\geq 0}\frac{\sin^2(2n)\sin^2\left(\frac{\pi}{2}n\right)}{n^2} $$
and the last sums can be computed through Parseval's theorem. 
From the Gudermannian function we know that:
$$\sum_{n\geq 0}\sin\left(\frac{\pi}{2}n\right)\frac{\sin(nx)}{n}=\frac{1}{2}\operatorname{arctanh}(\sin x)=\frac{1}{4}\log\left(\frac{1+\sin x}{1-\sin x}\right) $$
and we are left with:
$$\frac{2}{\pi}\int_{0}^{\pi}\left(\frac{1}{4}\log\left(\frac{1+\sin x}{1-\sin x}\right)\right)^2\,dx =\frac{2}{\pi}\int_{0}^{1}\log^2\left(\frac{1+t}{1-t}\right)\frac{dt}{1+t^2}=\frac{2}{\pi}\int_{0}^{+\infty}\frac{\log^2 z}{1+z^2}\,dz$$
where the last integral equals $\frac{\pi^3}{8}$ by the residue theorem.The other integral is even easier.
