TITLE: How does one show that $\prod_{n=4}^{\infty}\left(1-2\sin\left({\pi\over {2^n}}\right)^2\right)={2\sqrt{2}\over \pi}?$
QUESTION [5 upvotes]: Consider the infinite product $(1)$

$$\prod_{n=4}^{\infty}\left(1-2\sin\left({\pi\over {2^n}}\right)^2\right)=P\tag1$$
  How does one show that $P={2\sqrt{2}\over \pi}?$

An attempt:
$$1-2\sin\left({\pi\over {2^n}}\right)^2=\sin\left({\pi\over {2^n}}\right)^2+\cos\left({\pi\over {2^n}}\right)^2-2\sin\left({\pi\over {2^n}}\right)^2=\cos\left({\pi\over {2^{n-1}}}\right)$$
$$\prod_{n=4}^{\infty}\left(\cos\left({\pi\over {2^{n-1}}}\right)\right)=P\tag2$$
Not sure where to go from $(2)$

REPLY [3 votes]: Hint. From $(2)$ one may just use$$
\cos{\left(\frac {x}{2^n}\right)}=\frac12 \cdot \frac{\sin{\left(\frac {x}{2^{n-1}}\right)}}{\sin{\left(\frac {x}{2^n}\right)}}
$$ then factors telescope.