<s>[INST] how can we show $\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + …$? Let $f(x) = \frac4\pi \cdot (\sin x + \frac13 \sin (3x) + \frac15 \sin (5x) + \dots)$. If for $x=\frac\pi2$, we have
$$f(x) = \frac{4}{\pi} ( 1 - \frac13 +\frac15 - \frac17 + \dots) = 1$$
then obviously : 
$$ 1 - \frac13 +\frac15 - \frac17 + \dots=\frac{\pi}{4}$$
Now how can we prove that:
$$\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + \dots$$
 [/INST] From the Basel Problem, we have
$$\frac{\pi^2}{6} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2}\dots$$
$$\frac{\pi^2}{24} = \frac{\pi^2}{6\cdot 2^2} = \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2}\dots$$
so that
$$\begin{align}\frac{\pi^2}{8} &= \frac{\pi^2}{6} - \frac{\pi^2}{24}\\&=\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \dots \end{align}$$
</s>