<s>[INST] What is the value of this product: $\prod_{n=1}^\infty \;\frac{3}{1+2 \cos(\frac{\pi}{3^n})} \;=\;? $ emphasized text$$\left[\, \prod_{n=1}^\infty \;\frac{3}{1+2 \cos\left(\frac{\pi}{3^n}\right)}\, \right]\, =\;? $$
Where $\;[\, .]\;$ denotes the integral part function.  
$\mathbf {My Attempt}$
I tried to confine the n-th term to get 
$$\frac12 \le \cos\left(\frac{\pi}{3^n}\right) \lt 1$$
$$ 1 \lt \;\frac{3}{1+2 \cos\left(\frac{\pi}{3^n}\right)}\; \le \frac32 $$
But this didn't help as the product now is bounded below but unbounded above $\rightarrow \infty$
Using Wolfram Alpha, the product approaches 1.5708. So, the floor is 1.  
I tried also to bound product from above using 
$$\left(\frac{\pi}{3^n}\right) \lt \left(\frac{\pi}{2^n}\right)$$
$$\cos\left(\frac{\pi}{3^n}\right) \gt \cos\left(\frac{\pi}{2^n}\right) \quad(\cos x \text { is decreasing on } ]0, \frac{\pi}{3}]
)$$
And use the Telescoping product
$$\cos\left(\frac{\pi}{2^n}\right)=\frac{\sin\left(\frac{\pi}{2^{n-1}}\right)}{2\sin\left(\frac{\pi}{2^n}\right)}$$
But this doesn't help much.
Any hint?
 [/INST] $$1+2\cos2x=1+2(1-2\sin^2x)=\dfrac{\sin3x}{\sin x}=\dfrac{f(n+1)}{f(n)}$$
where $f(m)=\sin(3^mx)$
Here $3^mx=?$
Related:$\cos x(2\cos2x-1)=\cos3x$
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