Evaluate $\prod_{k=1}^{\infty}\left(2\cos\left(\frac{\pi}{3^{k+1}}\right)-1\right)$ from polynomial

Question

I found the problem online (without solution) that to evaluate $$\prod_{k=1}^{\infty}\left(2\cos\left(\frac{\pi}{3^{k+1}}\right)-1\right)$$ given that we know how to solve polynomial $$4x^3-3x+1 = 0$$

Right now I know that $$0<\prod_{k=1}^{\infty}\left(2\cos\left(\frac{\pi}{3^{k+1}}\right)-1\right)\leq 1$$ due to the fact that $\lim_{k\rightarrow\infty}\cos\left(\frac{\pi}{3^{k+1}}\right) = 1$ so we should have something like $0<(0.8...)...(\text{almost 1})...<1$ (my guess is that our answer is around $0.8$ to $0.9$) and I also know that the solution for the polynomial is $x = -1, \frac{1}{2}$.

One thing I found out interesting is that the polynomial can be express as $(x+1)(2x-1)^2$ in which is quite the same with our product if we let $x = \cos\left(\frac{\pi}{3^{k+1}}\right)$.

So how can I solve this problem using the polynomial? Could anyone provides me a hint or some guideline to solve this problem?

Answer 1

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 $$\sqrt{\frac{c_0+1}{c_\infty+1}}=\sqrt{\frac{\cos\frac\pi3+1}{2}}=\cos\frac\pi6=\frac{\sqrt{3}}2.$$

Answer 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 $.

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