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

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 $(2)$

• Multiply $$P_N=\prod_{n=4}^N\cos\left({\pi\over {2^{n-1}}}\right)$$ by $$\sin\left({\pi\over {2^{N-1}}}\right)$$ and use repeatedly the identity $$\sin x\cos x=\frac12\sin (2x)$$ to get $$2^{N-3}\sin\left({\pi\over {2^{N-1}}}\right)P_N=\sin\left({\pi\over {4}}\right)$$ and conclude using the asymptotics when $x\to0$, $$\frac{\sin x}x\to1$$
– Did
Commented Feb 15, 2017 at 12:52
• Infinite product formulae. $$\frac{\sin{x}}{x}=\prod_{n=1}^{\infty}\cos\frac{x}{2^n}$$ Commented Feb 15, 2017 at 12:57

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.