<s>[INST] Prove that $\frac{\sin x}{x}=(\cos\frac{x}{2}) (\cos\frac{x}{4}) (\cos \frac{x}{8})...$ How do I prove this identity:
$$\frac{\sin x}{x}=\left(\cos\frac{x}{2}\right) \left(\cos\frac{x}{4}\right) \left(\cos \frac{x}{8}\right)...$$
My idea is to let 
$$y=\frac{\sin x}{x}$$
and
$$xy=\sin x$$
Then use the double angle identity $\sin 2x=2\sin x \cos x$ and its half angle counterparts repeatedly.  I see some kind of pattern, but I can't seem to make out the pattern and complete the proof.
 [/INST] Partial answer:
The Taylor series of $\sin x$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!} \cdots$, so:
$$\frac{\sin x}{x} = 1 - \frac{x^2}{3!}+\frac{x^4}{5!} \cdots.$$
Meanwhile, $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}$, and therefore:
$$\cos(\frac{x}{2^1}) = 1 - \frac{x^2}{2^2 \cdot 2!} + \frac{x^4}{2^4 \cdot 4!} - \frac{x^6}{2^6 \cdot 6!} \cdots$$
$$\cos(\frac{x}{2^2}) = 1 - \frac{x^2}{2^4 \cdot 2!} + \frac{x^4}{2^8 \cdot 4!} - \frac{x^6}{2^{12} \cdot 6!} \cdots$$
$$\cos(\frac{x}{2^3}) = 1 - \frac{x^2}{2^6 \cdot 2!} + \frac{x^4}{2^{12}\cdot 4!} - \frac{x^6}{2^{18} \cdot 6!} \cdots$$
and so on.
The constant term of the infinite product is $1$.
There is only one way to make an $x^2$ term: by multiplying the constant and the $x^2$ term. The coefficient of the $x^2$ term is:
$$-\frac{1}{2^2 \cdot 2!} - \frac{1}{2^4 \cdot 2!} - \frac{1}{2^6 \cdot 2!},$$
which is a geometric series with $a = \frac{1}{8}$, and $r = \frac{1}{4}$, which the formula gives as $\frac{\frac{1}{8}}{1- \frac{1}{4}} = \frac{1}{6}$.
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