<s>[INST] Closed form for $\prod\limits_{l=1}^\infty \cos\frac{x}{3^l}$ Is there any closed form for the infinite product $\prod_{l=1}^\infty \cos\dfrac{x}{3^l}$? I think it is convergent for any $x\in\mathbb{R}$.
I think there might be one because there is a closed form for $\prod_{l=1}^\infty\cos\dfrac{x}{2^l}$ if I'm not wrong.
 [/INST] Let $\displaystyle P=\prod_{i=1}^n\cos\left(\frac x{2^i}\right)$.
$\displaystyle P=\cos\left(\frac x{2}\right)\cos\left(\frac x{4}\right)\cos\left(\frac x{8}\right)\cdots\cos\left(\frac x{2^n}\right)$
$\displaystyle P\sin\left(\frac x{2^n}\right)=\cos\left(\frac x{2}\right)\cos\left(\frac x{4}\right)\cos\left(\frac x{8}\right)\cdots\cos\left(\frac x{2^n}\right)\sin\left(\frac x{2^n}\right)$
$\displaystyle P\sin\left(\frac x{2^n}\right)=\frac12\,\cos\left(\frac x{2}\right)\cos\left(\frac x{4}\right)\cos\left(\frac x{8}\right)\cdots\cos\left(\frac x{2^{n-1}}\right)\sin\left(\frac x{2^{n-1}}\right)$
...
$\displaystyle P\sin\left(\frac x{2^n}\right)=\frac1{2^{n-1}}\,\cos\left(\frac x{2}\right)\sin\left(\frac x2\right)$
$\displaystyle P\sin\left(\frac x{2^n}\right)=\frac1{2^{n}}\,\sin(x)$
$\displaystyle P=\frac{\sin(x)}{2^{n}\sin\left(\frac x{2^n}\right)}$
$\displaystyle \lim_{n\to\infty}P=\lim_{n\to\infty}\frac{\sin(x)}x\frac{2^{-n}x}{\sin\left( 2^{-n}x \right)}=\frac{\sin(x)}x$
However, I am not able to solve the one with $3$ in the denominator.
</s>