<s>[INST] How to deduce the following complex number problem I am stuck with the following problem that says:

Using the result $$x^n-1=(x^2-1)\prod_{k=1}^{(n-2)/2}[x^2-2x\cos \frac{2k\pi}{n}+1],$$ if $n$ be an even positive integer,  deduce that $$\sin \frac{\pi}{32}\sin \frac{2\pi}{32}\sin \frac{3\pi}{32}.........\sin \frac{15\pi}{32}=\frac{1}{2^{13}}$$

Can someone point me in the right direction with some explanation? Thanks in advance for your time.
 [/INST] This doesn't use the given information, but it does give some intuition of what you are trying to show.
Consider $x^8 - 1$
The roots of unity are equally spaced around the unit circle. 

$x^8-1 = (x-1)(x-r_1)(x-r_2)\cdots(x-r_7) = (x-1)(x^7 + x^6 + \cdots + 1)$
with $r_k = (\cos \frac {2k\pi}{8}, i\sin \frac{2k\pi}{8})$
$|(x^7 + x^6 + \cdots + 1)| = |x-r_1||x-r_2|\cdots|x-r_7|$
letting $x = 1$
$8 = |1-r_1||1-r_2|\cdots|1-r_7|$
$|1 - r_k| = 2\sin\frac {k\pi}{8}$  (highlighted in the figure for the case $k=3$)
$8 = 2^2\sin\frac {\pi}{8}\sin\frac {2\pi}{8}\sin\frac {5\pi}{8}\cdots\sin\frac {7\pi}{8}\\
\sin\frac {\pi}{8}\sin\frac {2\pi}{8}\cdots\sin\frac {7\pi}{8} = 2^{-4}$
Not a whole lot changes when we consider $x^{32} - 1$
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