TITLE: Proving $\sum_{n=0}^{\infty }\frac{\sin^4(4n+2)}{(2n+1)^2}=\frac{5\pi ^2}{16}-\frac{3\pi }{4}$
QUESTION [6 upvotes]: I dont have an idea to prove it because of exist $\sin(4n+2)^4$
$$\sum_{n=0}^{\infty }\frac{\sin^4(4n+2)}{(2n+1)^2}=\frac{5\pi ^2}{16}-\frac{3\pi }{4}$$

REPLY [2 votes]: 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 $\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}$.