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

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 $(2)$

Answer 1

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.