Does $$\prod_{m=1}^{2^{n-1}} \cos\left(2^{m-1} x\right) \stackrel{?}{=} \frac{\sin\left(2^nx\right)}{2^n\sin x}?$$ I want a more rigorous way to derive this than my approach:
Using that $\sin(2x)= 2\sin x\cos x$, we can easily show that $$\sin\left(2^m x\right) = 2^m\sin x\cos x \cos(2x)\cdots \cos\left(2^{m-1}x\right)$$ and the initial equality follows. What about something like $$\cos^4\left(2^n x\right)-\prod_{m=1}^{2^{n-1}} \cos\left(2^{m-1} x\right)?$$ is there a nice way to represent this? Thank you!