How to evaluate $\lim\limits_{n\to \infty}\prod\limits_{r=2}^{n}\cos\left(\frac{\pi}{2^{r}}\right)$

Question

How do I evaluate this limit ? $$\lim_{n\to \infty}\cos\left(\frac{\pi}{2^{2}}\right)\cos\left(\frac{\pi}{2^{3}}\right)\cdots\cos\left(\frac{\pi}{2^{n}}\right)$$

I assumed it is using this formaula $\displaystyle \cos(A)=\sqrt{\frac{1+\cos(2A)}{2}}$ But I am stuck

Answer 1

Hint: $\cos x = \dfrac{\sin (2x)}{2\sin x}$

Answer 2

Thanks ah-huh-moment for the hint

Using his formula $\cos x = \dfrac{\sin (2x)}{2\sin x}$, we can expand the product of cosines as

$$\frac{\sin(2\pi/2^{2})}{2\sin(\pi/2^{2})}.\frac{\sin(2\pi/2^{3})}{2\sin(\pi/2^{3})}....\frac{\sin(2\pi/2^{n})}{2\sin(\pi/2^{n})}$$

$$\require{cancel}\underbrace{\frac{\sin(2\pi/2^{2})}{\cancel{2\sin(\pi/2^{2})}}.\frac{\cancel{\sin(2\pi/2^{3})}}{\cancel{2\sin(\pi/2^{3})}}....\frac{\cancel{\sin(2\pi/2^{n})}}{2\sin(\pi/2^{n})}}_\text{n-1 terms}$$

After cancellation $$\frac{\sin(2\pi/2^{2})}{2^{n-1}\sin(\pi/2^{n})}=\frac{\sin(\pi/2)}{2^{n-1}\sin(\pi/2^{n})}$$

Now by rearranging $$\frac{\sin(\pi/2)}{2^{n}\sin(\pi/2^{n})} = \frac{2}{\pi}.\frac{(\pi/2^{n})}{\sin(\pi/2^{n})}$$

Now $$\lim_{n\to\infty, \frac{\pi}{2^{n}}\to0} \frac{2}{\pi}.\frac{(\pi/2^{n})}{\sin(\pi/2^{n})} = \frac{2}{\pi} \because \lim_{x\to0} \frac{\sin(x)}{x} =1 $$