Continued Cosine Product. Is there a way to evaluate, 
$$
\large \cos x \cdot \cos \frac{x}{2} \cdot \cos \frac{x}{4} ... \cdot \cos \frac{x}{2^{n-1}} \tag*{(1)}
$$
I asked this to one of my teachers and what he told is something like this, 
Multiply and divide the last term of $(1)$  with  $\boxed{\sin \frac{x}{2^{n-1}}}$
So,
$$
\large \frac{\cos \frac{x}{2^{n-1}} \cdot \sin \frac{x}{2^{n-1}}}{\sin \frac{x}{2^{n-1}}} \\ \tag*{(2)}
$$
$$
\large \implies \frac{\sin (2x)}{2^n \cdot \sin \frac{x}{2^{n-1}}} \\
$$
$$
\large \implies \frac{\sin (2x)}{2^n \cdot \frac{\sin \frac{x}{2^{n-1}}}{\frac{x}{2^{n-1}}} \cdot \frac{x}{2^{n-1}}} \tag*{(3)}
$$
Now, as $n \to \infty$ , we have $x \to 0$,
Using this, $\lim$ we have, 
$$
\boxed{ \lim_{x \to 0} \frac{\sin x}{x} = 1}
$$
Using this in $(3)$, we have, 
$$
\large \boxed{\frac{\sin (2x)}{2x}} \tag*{(4)}
$$
All the steps sort of make sense. My doubts are, 


*

*How do I do this for other trigonometric ratios?

*How does the step 2 happen? 


I need help looking into it more intuitionally. 
Please provide necessary reading suggestions. 
Regards.

By the sine duplication formula $\sin(2z)=2\sin(z)\cos(z)$ we have $\cos(z)=\frac{1}{2}\cdot\frac{\sin(2z)}{\sin(z)}$.
In particular
$$ \prod_{k=0}^{n-1}\cos\left(\frac{x}{2^k}\right)=\frac{1}{2^n}\prod_{k=0}^{n-1}\frac{\sin\frac{x}{2^{k-1}}}{\sin\frac{x}{2^k}}=\frac{\sin(2x)}{2^n\sin\frac{x}{2^{n-1}}} \tag{1} $$
is a simple telescopic product, that can be written in the form
$$ \frac{\sin(2x)}{2x}\cdot\frac{x}{2^{n-1}\sin\frac{x}{2^{n-1}}}\tag{2}$$
and since $\lim_{w\to 0}\frac{\sin w}{w}=\lim_{w\to 0}\frac{w}{\sin w}=1$ it follows that
$$ \prod_{k\geq 0}\cos\left(\frac{x}{2^k}\right)=\frac{\sin(2x)}{2x}.\tag{3}$$
In general, as soon as we have a differentiable function $f(x)$ such that $f(2x)=2 g(x)\,f(x)$ and $f(0)=0, f'(0)\neq 0$ we have the identity
$$ \prod_{k\geq 0}g\left(\frac{x}{2^k}\right) = \frac{f(2x)}{2x}\cdot \frac{1}{f'(0)}\tag{4} $$
by de l'Hospital theorem.
