Viete's product is $$ \frac{2}{\pi}=\frac{\sqrt{2}}{2} * \frac{\sqrt{2+\sqrt{2}}}{2} * \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}*... $$
Show that
$$ \frac{3}{\pi} = \frac{\sqrt{2+\sqrt{3}}}{2} * \frac{\sqrt{2+\sqrt{2+\sqrt{3}}}}{2} * \frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}{2}*... $$
This uses Euler's product with $\theta = \frac{\pi}{2}$ and my thoughts were to find which $\theta=\frac{3}{\pi}$ but I'm not having any luck.