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.
Note the fact that $$ \cos \frac{x}{2^k} = \frac12 \cdot \frac{\sin (2^{1-k} x)}{\sin(2^{-k}x)}, $$ and we have $$ \prod_{k = 1}^n \cos \frac{x}{2^k} = \frac{1}{2^n} \cdot \frac{\sin x}{\sin(2^{-n}x)} = \frac{2^{-n}x}{\sin(2^{-n}x)} \cdot \frac{\sin x}{x}. $$ For all $x$, as $n \to \infty$, we have $$ \lim_{n \to \infty} \prod_{k = 1}^n \cos \frac{x}{2^k}= \frac{\sin x}{x} \cdot\lim_{n \to \infty} \frac{2^{-n}x}{\sin(2^{-n}x)} = \frac{\sin x}{x}. $$
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 $\frac{\frac{1}{8}}{1- \frac{1}{4}} = \frac{1}{6}$.
