# Trigonometric way to find the solution of Basel problem

Let's generalise the whole thing first

Generally $\sum_{n=1}^∞ \dfrac{\sin(n\theta)}{n} = \dfrac{π-\theta}{2}$ Where $$0<\theta≤π$$

And consequently $\sum_{n=1}^∞ \dfrac{\cos(n\theta)}{n^2} = \sum_{n=1}^∞ \dfrac{1}{n^2} -\dfrac{π\theta}{2} +\dfrac{\theta^2}{4} =\zeta(2)-\dfrac{π\theta}{2} +\dfrac{\theta^2}{4}$ refer this solution

let's take this sum $\displaystyle\sum_{n=1}^∞ \dfrac{\sin^2(n\theta)}{n^2}$ It is $=\sum_{n=1}^∞\left( \dfrac{1}{2n^2} - \dfrac{\cos(2n\theta)}{2n^2} \right)$ $= \dfrac{\zeta(2)}{2} -\left (\dfrac{\zeta(2)}{2} +\dfrac{(2\theta)^2}{8} -\dfrac{π\theta}{2}\right )$ $= \dfrac{\theta(π-\theta)} {2}$ Now $\sum_{n=1}^∞ \dfrac{\sin^2(n\frac{π}{2})}{n^2} = \dfrac{ \frac{π}{2} (π-π/2)}{2} = \dfrac{π^2}{8}$ $\implies \dfrac{1}{1^2} +\dfrac{1}{3^2} +\dfrac{1}{5^2} +\cdots= \dfrac{π^2}{8}$ But this is all odd terms . $\dfrac{\zeta(2)}{4} +\sum_{n=0}^∞\dfrac{1}{(2n+1)^2} = \zeta(2)$ $\zeta(2)= \dfrac{4}{3} × \dfrac{π^2}{8} = \dfrac{π^2}{6}$

Note by Dwaipayan Shikari
1 week, 6 days ago

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