Have a infinite problem

so lets write a recursive relation: $x_1=\sqrt{\dfrac{1}{2}} =\cos\left(\dfrac{\pi}{4}\right), x_n=\sqrt{\dfrac{1+x_{n-1}}{2}}$ i claim $x_n=\cos\left(\dfrac{\pi}{2^{n+1}}\right)$ . we can prove this by induction. we see the base case is true. we use the half angle identities to show $x_n=\sqrt{\dfrac{1+x_{n-1}}{2}}=\sqrt{\dfrac{1+\cos\left(\dfrac{\pi}{2^{n}}\right)}{2}}=\cos\left(\dfrac{\pi}{2^{n+1}}\right)$ that proves that, now lets take the product. $\lim_{n\to \infty}\prod_{k=2}^n \cos\left(\dfrac{\pi}{2^{k}}\right)=\lim_{n\to \infty}\dfrac{2^{n-1} \sin\left(\dfrac{\pi}{2^{n}}\right)\prod_{k=2}^n \cos\left(\dfrac{\pi}{2^{k}}\right)}{2^{n-1} \sin\left(\dfrac{\pi}{2^{n}}\right)}$ by using double angle identity,we have $1=\sin\left(\dfrac{\pi}{2}\right)=2\cos\left(\dfrac{\pi}{4}\right)\sin\left(\dfrac{\pi}{4}\right)=4\cos\left(\dfrac{\pi}{4}\right)\cos\left(\dfrac{\pi}{8}\right)\sin\left(\dfrac{\pi}{8}\right)=2^{n-1}\cos\left(\dfrac{\pi}{4}\right)\cos\left(\dfrac{\pi}{8}\right)....\cos\left(\dfrac{\pi}{2^{n}}\right)\sin\left(\dfrac{\pi}{2^{n}}\right)$ so the limit becomes $\lim_{n\to \infty}\dfrac{2^{n-1} \sin\left(\dfrac{\pi}{2^{n}}\right)\prod_{k=2}^n \cos\left(\dfrac{\pi}{2^{k}}\right)}{2^{n-1} \sin\left(\dfrac{\pi}{2^{n}}\right)}=\lim_{n\to \infty}\dfrac{1}{2^{n-1} \sin\left(\dfrac{\pi}{2^{n}}\right)}$ using small-angle approximation we have the limit $\lim_{n\to \infty}\dfrac{1}{2^{n-1} \sin\left(\dfrac{\pi}{2^{n}}\right)}=\lim_{n\to \infty}\dfrac{1}{2^{n-1} \left(\dfrac{\pi}{2^{n}}\right)}=\boxed{\dfrac{2}{\pi}}$