<s>[INST] $|\sin(\theta)|=\frac{2}{\pi}-\frac{4}{\pi} \cdot\sum_{m=1}^{\infty}\frac{\cos(2m\theta)}{4 m^2-1}\Rightarrow \sum_{m=1}^{\infty}\frac{1}{16m^2-1} =$? By the equation $|\sin(\theta)|=\frac{2}{\pi}-\frac{4}{\pi} \cdot \sum_{m=1}^{\infty}\frac{\cos(2m\theta)}{4m^2-1}$, how can I get the value of $\sum_{m=1}^{\infty}\frac{1}{16m^2-1}$?
What I tried:
If I substitute $\theta=\frac{\pi}{2}$, $|\sin(\frac{\pi}{2})|=\frac{2}{\pi}-\frac{4}{\pi} \cdot \sum_{m=1}^{\infty}\frac{\cos(m.\pi)}{4m^2-1}=1$. For odd $m$, $\cos(m.\pi)=-1$ and, for even $m$, $\cos(m.\pi)=1$.
If I substitute $\theta=0$, $|\sin(0)|=\frac{2}{\pi}-\frac{4}{\pi} \cdot \sum_{m=1}^{\infty}\frac{1}{4m^2-1}=0$.
$|\sin(\frac{\pi}{2})|+|\sin(0)|=\frac{4}{\pi}-\frac{4}{\pi} \cdot \sum_{m=1}^{\infty}\frac{1+\cos(m.\pi)}{4m^2-1}=1\Rightarrow\frac{4}{\pi}.(1-2.\sum_{n=1}^{\infty}\frac{1}{4.(2.n)^2-1})=1$
$\Rightarrow 1-2.\sum_{n=1}^{\infty}\frac{1}{16.n^2-1}=\frac{\pi}{4} \Rightarrow -2.\sum_{n=1}^{\infty}\frac{1}{16.n^2-1}=\frac{\pi-4}{4} \Rightarrow \sum_{n=1}^{\infty}\frac{1}{16.n^2-1}=\frac{4-\pi}{8}$
 [/INST] Hint. One may observe that
$$
\frac{\cos \left(2 m \times 0\right)+\cos \left(2 m\times\frac \pi 2\right)}{2} =
\begin{cases}
1,  & \text{if $m$ is even}, \\
0, & \text{if $m$ is odd}.
\end{cases}
$$ Then putting successively $\theta:=0$ and  $\theta:=\frac \pi 2$ in the given identity and averaging may help.
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