<s>[INST] 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? 
 [/INST] Hint: Use the Taylor series and let $n\rightarrow \infty:$
$$2^n\sin\left(\frac{y}{2^n}\right)= 2^n\left(\frac{y}{2^n}-\left(\frac{y}{2^n}\right)^3/3! + \dots\right) = y - \frac{y^3}{3!2^{2n}} + \dots$$
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