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?

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
