$\prod_{k=0}^{\infty} \cos(x \cdot 2^{-k}).$ Task is to find $$\prod_{k=0}^{\infty} \cos(x \cdot 2^{-k}).$$ 
I tried to make it with double-angle formula:
$\prod_{k=0}^{\infty} \cos(x \cdot 2^{-k}) = \frac{\prod_{k=0}^\infty \sin(x2^{1-k})}{2^\infty \cdot \prod_{k=0}^\infty \sin(x \cdot 2^{-k})} $
I'm sad to admit, that I'm stuck with that one.
So I would appreciate any help, such as pointing the right direction.
 A: For the partial products we have
$$\begin{align}
\prod_{k = 0}^N \cos \frac{x}{2^k} &= \frac{\sin \frac{x}{2^N}\cos\frac{x}{2^N}}{\sin \frac{x}{2^N}} \prod_{k=0}^{N-1} \cos \frac{x}{2^k}\\
&= \frac{\sin \frac{x}{2^{N-1}}}{2\sin \frac{x}{2^N}} \prod_{k=0}^{N-1} \cos \frac{x}{2^k}\\
&= \frac{\sin \frac{x}{2^{N-1}}\cos \frac{x}{2^{N-1}}}{2\sin \frac{x}{2^N}} \prod_{k=0}^{N-2} \cos \frac{x}{2^k}\\
&= \frac{\sin \frac{x}{2^{N-2}}}{2^2\sin \frac{x}{2^N}} \prod_{k=0}^{N-2} \cos \frac{x}{2^k}\\
&\qquad\qquad \vdots\\
&= \frac{\sin \frac{x}{2^{N-N}}}{2^N\sin \frac{x}{2^N}}\cos \frac{x}{2^0}\\
&= \frac{\sin x\cos x}{2^N\sin (x\cdot 2^{-N})}.
\end{align}$$
From that, the limit is easily found.
A: Using the identity
$$
\cos(x2^{-k})=\frac{\sin(x2^{1-k})}{2\sin(x2^{-k})}
$$
we get
$$
\begin{align}
\prod_{k=0}^n\cos(x2^{-k})
&=\prod_{k=0}^n\frac{\sin(x2^{1-k})}{2\sin(x2^{-k})}\\
&=\frac1{2^{n+1}}\frac{\prod\limits_{k=-1}^{n-1}\sin(x2^{-k})}{\prod\limits_{k=0}^n\sin(x2^{-k})}\\
&=\frac1{2^{n+1}}\frac{\sin(2x)}{\sin(x2^{-n})}\\[12pt]
\end{align}
$$
Now use the limit
$$
\lim_{x\to0}\frac{\sin(x)}{x}=1
$$
