A product and a formula

Using the following identity ( equation 23 ),

$\begin{aligned} \frac {\sin x}x & = \prod_{n=1}^\infty \left(1-\frac {x^2}{n^2\pi^2}\right) \\ \frac 3\pi & = \prod_{n=1}^\infty \left(1-\frac {a^2}{n^2\pi^2}\right) \\ \implies \frac {\sin a}a & = \frac 3\pi \\ \implies a & = \frac \pi 6 \approx \boxed{0.524} \end{aligned}$

If $0 < t < \pi$ , then $\displaystyle \cot (t) = \frac{1}{t} + \sum_{n = 1}^{\infty} \frac{2t}{t^2 - n^2 \cdot \pi^2}$ . Then, if $0 < \alpha < x < \pi$ , $\displaystyle \log \frac{\sin (x)}{x} - \log \frac{\sin (\alpha)}{\alpha} = \int_{\alpha}^x (\frac{\cos (t)}{\sin (t)} - \frac{1}{t}) \, dt = \int_{\alpha}^x \sum_{n = 1}^{\infty} \frac{2t}{t^2 - n^2 \cdot \pi^2} =$ $= \displaystyle \sum_{n = 1}^{\infty} \log \left ( \frac{x^2 - n^2 \cdot \pi^2 }{\alpha^2 - n^2 \cdot \pi^2} \right ) \Rightarrow$ as $\alpha \rightarrow 0, \log \frac{\sin (x)}{x} = \sum_{n = 1}^{\infty} \log \left ( \frac{x^2 - n^2 \cdot \pi^2 }{ - n^2 \cdot \pi^2} \right ) \Rightarrow \frac{\sin (x)}{x} = \prod_{n = 1}^\infty \left ( 1 - \frac{x^2}{n^2 \pi^2} \right ) \Rightarrow$ $\frac{3}{\pi} = \frac{ \frac{1}{2}}{\frac{\pi}{6}} = \frac{\sin (\frac{\pi}{6})}{\frac{\pi}{6}} = \prod_{n = 1}^\infty \left ( 1 - \frac{a^2}{n^2 \pi^2} \right ) \Rightarrow a = \frac{\pi}{6}$