Is there an easy way of proving $\prod_{k=1}^\infty \cos(x/2^k) = \sin(x)/x$? I just answered this question 
distribution of infinite sum of $\sum (2x_n -1)/2^n$
by using the formula in the title which I lifted off a random formula sheet on the internet. My question is, how do we derive this? I have never learnt how to sum infinite products like this. 
I believe there is also a formula for $\cosh$ (by Osborn's rule). A justification why this follows from the $\cos$ case would also be nice. 
 A: Hint: 
$$ \sin(2x) = 2\sin(x)\cos(x) $$ 
so 
$$
\cos \left(\frac{x}{2^k}\right) = 
\frac{\sin\left(\frac{x}{2^k}\right)}{2\sin\left(\frac{x}{2^{k+1}}\right)}
$$
A: Use formula $\displaystyle \cos x=\frac{e^{ix}+e^{-ix}}{2}$ and $a^2-b^2=(a-b)(a+b)$ $n$ times for $\displaystyle \frac{(e^{\frac{x}{2^n}}-e^{-\frac{x}{2^n}})}{\frac{x}{2^n}}\Pi_{k=1}^{n} \frac{e^{\frac{x}{2^k}}+e^{-\frac{x}{2^k}}}{2}=\frac{e^{ix}-e^{-ix}}{x}$
Because $\displaystyle \lim_{n \to\infty}\frac{(e^{\frac{x}{2^n}}-e^{-\frac{x}{2^n}})}{\frac{x}{2^n}}=2$ we have:
$\frac{e^{ix}-e^{-ix}}{x}\lim_{n \to \infty}\frac{(e^{\frac{x}{2^n}}-e^{-\frac{x}{2^n}})}{\frac{x}{2^n}}\Pi_{k=1}^{n} \frac{e^{\frac{x}{2^k}}+e^{-\frac{x}{2^k}}}{2}=\lim_{n \to \infty}\frac{(e^{\frac{x}{2^n}}-e^{-\frac{x}{2^n}})}{\frac{x}{2^n}} \Pi_{k=1}^{\infty}\cos \frac{x}{2^k}=2\Pi_{k=1}^{\infty}\cos \frac{x}{2^k}$
Finally $\frac{e^{ix}-e^{-ix}}{2x}=\frac{\sin x}{x}$
A: Let
$P_n=\prod\limits_{k=1}^n\cos(\frac{x}{2^{k}})$
Then we have:
$\sin(\frac{x}{2^{n}})P_n=\cos(\frac{x}{2}) \dots \cos(\frac{x}{2^{n}})\sin(\frac{x}{2^{n}})=\frac{1}{2}\cos(\frac{x}{2}) \dots \sin(\frac{x}{2^{n-1}})$
=$\dots$=$\frac{\sin(x)}{2^{n}}$
by applying double angle formula for $\sin(2x)$ repeatedly.
Then 
$P_n=\frac{\sin(x)}{2^{n}}\frac{1}{\sin(\frac{x}{2^{n}})}$
Now as $n \rightarrow \infty$ and $\sin(\frac{x}{2^{n}}) \rightarrow \frac{x}{2^{n}}$
and you obtain required result. You can produce rigourous arguments about limit part if you want, but this is essentially how you calculate infinite product.
