How would you show that $\prod_{k=1}^\infty \cos( 2 \pi/3^k)$ is nonzero? Wolfram approximates it as about $-0.37$, and I have a guess that $$ \Big \vert \prod_{k=1}^\infty \cos( 2 \pi/3^k) \Big \vert\geq c \prod_{k=1}^\infty 3^{-1/k^2}, $$ although I cannot show it.
This product arises as the modulus of the characteristic function of the Cantor distribution.
On
It is well-known that an infinite product of positive terms $$ \prod_{n=1}^\infty (1-a_n) \ \text{where}\ 0 \le a_n < 1$$ converges (to a nonzero limit) if and only if $\sum_{n=1}^\infty a_n < \infty$. In this case $$a_k = 1 - \cos(2\pi/3^k) \sim \frac{(2\pi/ 3^{k})^2}{2}$$
For $x\approx 0$, we have $\cos x\approx 1-\frac 12x^2$ and $\ln\cos x\approx-\frac 12x^2 $. This allows you to compare $\sum\ln\cos(2\pi/3^k) $ with a nicely convergent series