I came across this question in the following form:
Compute the following infinite product
$$\left[\sin (x)\cos \left(\frac{x}{2}\right)\right]^{1/2}\cdot \left[\sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{4}\right)\right]^{1/4}\cdot \left[\sin \left(\frac{x}{4}\right) \cos \left(\frac{x}{8}\right)\right]^{1/8} \cdots$$
And I have been unable to solve it or to find similar problems online, which might guide me to the solution.
However, I was able to convert it into a nested radical form, but after that I'm out of ideas. Could really use a bit of help. Thanks!
Here's the nested radical form I mentioned:
$$\sqrt{\sin (x) \cos \left(\frac{x}{2}\right) \sqrt{\sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{4}\right) \sqrt {\sin \left(\frac{x}{4}\right) \cos \left(\frac{x}{8}\right) \ldots}}}$$
Use this formula $2\sin x \cos x = \sin 2x$ recursively $$ \lim _{N \to \infty} \prod _{n=0} ^{N} \frac{S^{2^{-n-1}}_{x 2^{-n}} C^{2^{-n-1}}_{x 2^{-n-1}} 2^{2^{-n-1}} S^{2^{-n-1}}_{x 2^{-n-1}} }{ 2^{2^{-n-1}} S^{2^{-n-1}}_{x 2^{-n-1}} } = \lim _{N \to \infty} \prod _{n=0} ^{N} \frac{S^{2^{-n}}_{x 2^{-n}} } {2^{2^{-n-1}} S^{2^{-n-1}}_{x 2^{-n-1}} } $$ I believe you can start from here.