$\prod_{k=0}^{\infty} \cos(x \cdot 2^{-k}).$ 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.

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.
