Closed form for $\prod\limits_{l=1}^\infty \cos\frac{x}{3^l}$

If we start from $$ \cos(x) = \prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2 \pi^2}\right) \tag{1}$$ we have:

$$ \log\cos(x) = -\sum_{n\geq 0}\sum_{m\geq 1}\frac{4^m x^{2m}}{m(2n+1)^{2m} \pi^{2m}}=-\sum_{m\geq 1}\frac{(4^m-1)\zeta(2m)\,x^{2m}}{m\pi^{2m}}\tag{2} $$ hence: $$ \sum_{l\geq 1}\log\cos\left(\frac{x}{3^l}\right)=-\sum_{m\geq 1}\frac{(4^m-1)\zeta(2m)}{(9^m-1)m\pi^{2m}}x^{2m}\tag{3} $$ and $$ \prod_{l\geq 1}\cos\left(\frac{x}{3^l}\right) = \color{red}{\exp\left(-\sum_{m\geq 1}\frac{(4^m-1)\zeta(2m)}{(9^m-1)m\pi^{2m}}x^{2m}\right)}\tag{4}$$

does not simplify much further.