<s>[INST] Proving $\cos A \cdot \cos 2 A \cdot \cos 4 A \cdots \cos 2^{n-1} A = \frac{\sin 2^n A}{2^n \sin A}$ Just a bit of background on the question:
When proving:
$$\cos\frac{\pi}{15}\cdot \cos\frac{2\pi}{15} \cdot \cos\frac{3\pi}{15}\cdot \cos\frac{4\pi}{15} \cdot \cos\frac{5\pi}{15} \cdot \cos\frac{6\pi}{15}\cdot \cos\frac{7\pi}{15} = \frac{1}{128}$$
The following formula was used:

$$\cos A \cdot \cos 2A \cdot \cos 2^2A \cdot \cos 2^3A\cdot \cdots \cdot \cos2^{n-1}A = \frac{\sin 2^n A}{2^n \sin A}$$

I'm just interested to see how this is derived. The question assumes that this formula is memorised and known beforehand, but it looked interesting to me since I'd actually never come across it before.
If anyone's interested, I'll link full proof for the question here
Cheers in advance! 
 [/INST] Turn the double-angle formula for sine "inside out".  Put in
$$\begin{align}\cos(A)& =\frac{\sin(2A)}{2·\sin(A)} \\[6pt]
\cos(2A)&=\frac{\sin(4A)}{2·\sin(2A)}
\end{align}$$
etc., and use telescoping. 
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