$|\sin(\theta)|=\frac{2}{\pi}-\frac{4}{\pi} \cdot\sum_{m=1}^{\infty}\frac{\cos(2m\theta)}{4 m^2-1}\Rightarrow \sum_{m=1}^{\infty}\frac{1}{16m^2-1} =$?

Question

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 $\theta=0$, $|\sin(0)|=\frac{2}{\pi}-\frac{4}{\pi} \cdot \sum_{m=1}^{\infty}\frac{1}{4m^2-1}=0$.

$|\sin(\frac{\pi}{2})|+|\sin(0)|=\frac{4}{\pi}-\frac{4}{\pi} \cdot \sum_{m=1}^{\infty}\frac{1+\cos(m.\pi)}{4m^2-1}=1\Rightarrow\frac{4}{\pi}.(1-2.\sum_{n=1}^{\infty}\frac{1}{4.(2.n)^2-1})=1$

$\Rightarrow 1-2.\sum_{n=1}^{\infty}\frac{1}{16.n^2-1}=\frac{\pi}{4} \Rightarrow -2.\sum_{n=1}^{\infty}\frac{1}{16.n^2-1}=\frac{\pi-4}{4} \Rightarrow \sum_{n=1}^{\infty}\frac{1}{16.n^2-1}=\frac{4-\pi}{8}$

Answer 1

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.