<s>[INST] 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.
 [/INST] How does step 2 happen:
multiplying the last factor by $\frac {\sin \frac{x}{2^{n-1}}}{\sin \frac{x}{2^{n-1}}}$
gives us
$\cos x  \cdot \cos \frac x2\cdots \cos \frac x{2^{n-2}}\cdot  \frac{\cos \frac{x}{2^{n-1}} \cdot \sin \frac{x}{2^{n-1}}}{\sin \frac{x}{2^{n-1}}}$
Double angle formula.
$\cos \frac{x}{2^{n-1}} \cdot \sin \frac{x}{2^{n-1}} = \frac 12 \sin \frac{x}{2^{n-2}}$
Applying this we get:
$\cos x  \cdot \cos \frac x2\cdots \cdot  \frac{\cos \frac x{2^{n-2}}\cdot\sin \frac{x}{2^{n-2}}}{2\sin \frac{x}{2^{n-1}}}$
And we can apply the double angle formula again.  And do it repeatedly until all of the $\cos$ factors have been devoured.
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