The Trouble Of Squared Sines

Relevant wiki: Fourier Series

Consider a function, $g_x (z) : \left[ 0, \pi \right] \rightarrow \mathbb{R}$ , where $\displaystyle 0<x< \frac{\pi}{2}$ is a parameter which is defined as follows
$g_x (z) = \begin{cases} \frac{\pi}{4} & 0 < z < 2x \\ 0 & 2x < z \leq \pi\\ \end{cases}$ The function can be expressed as a Fourier sine series as $\displaystyle \sum_{n=1}^{\infty} \frac{\sin^2 (nx)}{n} \sin(nz)$ .
Thus over the domain, we can write,
$\displaystyle g_x (z) = \sum_{n=1}^{\infty} \frac{\sin^2 (nx)}{n} \sin(nz)$
$\displaystyle \Rightarrow \int_0^{2y} g_x (z) dz = \sum_{n=1}^{\infty} \frac{\sin^2 (nx)}{n} \int_0^{2y} \sin(nz) dz$
The RHS evaluates to $\displaystyle 2 \sum_{n=1}^{\infty} \frac{\sin^2 (nx) \sin^2 (ny)}{n^2}$ which is actually $2$ times the required sum for $x = \dfrac{\pi}{12}$ and $y = \dfrac{\pi}{16}$ .

The evaluation of the LHS actually depends upon the relative values of $x$ and $y$ . If $x > y$ , then, $\displaystyle \int_0^{2y} g_x (z) dz = \int_0^{2y} \frac{\pi}{4} dz = 2 \times \frac{\pi}{4}y$
Otherwise, if $x < y$ , then, $\displaystyle \int_0^{2y} g_x (z) dz = \int_0^{2x} \frac{\pi}{4} dz = 2 \times \frac{\pi}{4}x$

Thus, LHS is actually $\displaystyle 2 \times \frac{\pi}{4} \text{min}(x,y)$ .

Therefore, $\displaystyle \sum_{n=1}^{\infty} \frac{\sin^2 (nx) \sin^2 (ny)}{n^2} = \frac{\pi}{4} \text{min}(x,y)$ .

$\displaystyle \sum_{n=1}^{\infty} \frac{\sin^2 \left( \frac{n \pi}{12} \right) \sin^2 \left( \frac{n \pi}{16} \right)}{n^2} = \frac{\pi}{4} \text{min}\left( \frac{n \pi}{12} , \frac{\pi}{16} \right) = \boxed{\frac{\pi ^2}{64} }$ .
Compare and substitute to get the final answer as $128$ .