Too many cosines!

$f(x) = \displaystyle \prod_{n=2}^{x} \cos\left(\dfrac{\pi}{2^{n}}\right)$
$f(x) = \displaystyle \dfrac{\sin\left(\frac{\pi}{2}\right)}{2^{x}\sin\left(\dfrac{\pi}{2^{x+1}}\right)} = \dfrac{2^{-x}}{\sin\left(\dfrac{\pi}{2^{x+1}}\right)}$
$\displaystyle \lim_{ x \to \infty}f(x) = \dfrac{2}{\pi} \approx 0.636$

This is a famous formula represented as,
$\dfrac{2}{\pi} = \dfrac{1}{\sqrt{2}}\sqrt{\dfrac{\sqrt{2} +1}{2\sqrt{2}}} \ldots$ known as Walli's formula if I am not wrong.

$\prod_{n=2}^{x} \cos\left(\dfrac{\pi}{2^{n}}\right)=\frac{1}{2^{x-1} \sin\frac{\pi}{2^x}}$ We can show this by induction. The base case of $x=2$ is straightforward. For the iterative step, we can use the double angle identity for sin(x). $\prod_{n=2}^{x+1} \cos\left(\dfrac{\pi}{2^{n}}\right)= \cos\frac{\pi}{2^{x+1}} \prod_{n=2}^{x} \cos\left(\dfrac{\pi}{2^{n}}\right)= \cos\left(\dfrac{\pi}{2^{x+1}}\right) \frac{1}{2^{x-1} \sin\frac{\pi}{2^x}}=\frac{1}{2^{x} \sin\frac{\pi}{2^{x+1}}}$ This completes the induction. Taking the limit as x goes to infinity of the product that we just proved gives us $\prod_{n=2}^{\infty} \cos\left(\dfrac{\pi}{2^{n}}\right)=\dfrac {2}{\pi}$