<s>[INST] A Viéte-like infinite product for $\sin (\pi/7)$ In the article "The unruly $\sin (\pi/7)$ of Samuel Moreno, the following infinite product is given:

Also, the same article shows that $\sin (\pi/7)$ is equal to the infinite nested radical
$ \sin (\pi/7) = \frac{1}{2}\sqrt{2 - \sqrt{2 - \sqrt{2 - \sqrt{2 + \sqrt{2 - \sqrt{2 - \sqrt{2 + \sqrt{2 - \sqrt{2 - \sqrt{2+...}}}}}}}}}}$ 
The question is: are these two expressions fundamentally different for $\sin(\pi/7)$, or is it possible to rewritte the infinite nested radical into that infinite product by simple algebraic manipulations?
 [/INST] With the help of Half angle cosine formula it is possible to express many sine and cosine angles as finite or infinite nested square roots of 2 here.
In this perspective $2\sin\frac{\pi}{7}$ can be solved as follows
$2\sin\frac{\pi}{7}$ = $\sqrt{2-2\cos\frac{2\pi}{7}}$ = $\sqrt{2-\sqrt{2+2\cos\frac{4\pi}{7}}}$ = $\sqrt{2-\sqrt{2-2\cos\frac{3\pi}{7}}}$ = $\sqrt{2-\sqrt{2-\sqrt{2+2\cos\frac{6\pi}{7}}}}$ = $\sqrt{2-\sqrt{2-\sqrt{2-2\cos\frac{\pi}{7}}}}$ = $\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+2\cos\frac{2\pi}{7}}}}}$
One can observe repetition of $2\cos\frac{2\pi}{7}$ subsequently and $\therefore$ $2\sin\frac{\pi}{7}$ can be represented simply as

$$\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+...2\cos\frac{2\pi}{7}}}}}}}}$$ with repeating pattern $--+$ infinitely and first $-$ as single nonrepeating sign

But getting infinite product has to be worked out
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