Evaluating the infinite product $ \prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) $ How does one evaluate 
$ \displaystyle\prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) $?
Seems fairly straightforward, as I just plugged in some initial values $n = 1, 2, 3,\dotsc$
$n = 1$
$ \sin(y)= 2\sin(\frac{y}{2})\cos(\frac{y}{2})$
$ \cos(\frac{y}{2}) = \sin(y)/2\sin(\frac{y}{2}) $
$n = 2$
$\sin(\frac{y}{2})= 2\sin(\frac{y}{4})\cos(\frac{y}{4}) $
$\cos(\frac{y}{4}) = \sin(\frac{y}{2})/2\sin(\frac{y}{4}) $
Hence,
$ \prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) = \sin(y)/2\sin(\frac{y}{2}) \times \sin(\frac{y}{2})/2\sin(\frac{y}{4}) ... \times \sin(\frac{y}{2^{n-1}})/2\sin(\frac{y}{2^n})  $ 
Then some denominators/numerators cancel, reducing the expression into
$ \prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) = \sin(y)/2^n\sin(\frac{y}{2^n})$
Is this expression no longer reducible? 
Because apparently, somehow, the denominator seems to reduce to $y$
How does that work? 

Let's take a look at the finite product first:
$$\prod_{i=0}^k \cos\frac{x}{2^i} = \frac1{2^k}\frac{\sin 2 x}{\sin \frac{x}{2^k}}$$
which is the standard trick of multiplying the product by $\sin \frac{x}{2^k}$.
Now take limit as $k \rightarrow +\infty$ and you get:
$$\prod_{i=0}^\infty \cos\frac{x}{2^i} = \frac{\sin{2 x}}{x}$$
