Find the limit of $\lim_{j\rightarrow \infty}\sin^2(2^{j-1}\pi f)\prod_{i=0}^{j-2} \cos^2(2^i\pi f)$

Question

Can anyone give a hint on how to see if the following has a limit? $f$ stands for frequency.

$$\lim_{j\rightarrow \infty}\sin^2(2^{j-1}\pi f)\prod_{i=0}^{j-2} \cos^2(2^i\pi f)$$

I've tried a few different things, but nothing seems to be working. Any help?

Answer 1

Hint:

Since $\sin (2x)=2\sin x\cos x$ we have $$\cos x =\frac{\sin (2x)}{\sin x}\qquad\text{for }\sin x \neq 0$$ then, if $\;2^if\notin \mathbb{Z}$ for every $i\in\mathbb{Z}$, $$\prod_{i=0}^{j-2} \cos^2(2^i\pi f)=\prod_{i=0}^{j-2} \frac{\sin^2(2^{i+1}\pi f)}{\sin^2 (2^i\pi f)}=\frac{\sin^2(2^{j-1}\pi f)}{\sin^2(\pi f)}$$