Find the value of infinite product $(2\cos(\frac{\pi}{9})-1)(2\cos(\frac{\pi}{27})-1)\cdot\cdot\cdot(2\cos(\frac{\pi}{3^{n+1}})-1)\cdot\cdot\cdot$ [duplicate]

Well, here's how to use the polynomial and the factorization you found (thanks for doing the dirty work! ;). You can write the triplication formula also in this way: $$\cos3\theta=4\cos^3\theta-3\cos\theta.$$ With $$c_k=\cos\frac\pi{3^{k+1}},$$ you'll have $$4\,c^3_k-3\,c_k=c_{k-1},$$ and adding $1$ on both sides, $$(c_k+1)(2\,c_k-1)^2=c_{k-1}+1,$$ i.e. $$2\,c_k-1=\sqrt{\frac{c_{k-1}+1}{c_k+1}}.$$ So we get another telescoping product, and its value is $$\sqrt{\frac{c_0+1}{c_\infty+1}}=\sqrt{\frac{\cos\frac\pi3+1}{2}}=\cos\frac\pi6=\frac{\sqrt{3}}2.$$

Hint: $\cos 3 \theta = \cos \theta ( 2 \cos 2 \theta - 1)$.

Thus, we can use a telescoping series to evaluate the product (fill in the missing gap) to calculate a final value of $ \cos \frac{\pi}{6} = \frac{ \sqrt{3}}{2} \approx 0.866 $.

I'm not quite sure how it relates to solving the cubic (or even if that is relevant).