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.
 A: 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 $x^{32} - 1$
A: Fixing $n=32$, we can rewrite the equation like this:
$$\frac{x^{32}-1}{x^2-1} = \prod_{k=1}^{15}\left(x^2+1-2x\cos\frac{k\pi}{16}\right)$$
for all $x\neq\pm1$.
We will take the limit as $x\to 1$ of both sides of the equality.
First, we have (using L'Hôpital's rule)
$$\lim\limits_{x\to 1} \frac{x^{32}-1}{x^2-1} = \lim\limits_{x\to 1}\frac{32x^{31}}{2x} = 16$$
Then, plugging in $x=1$, we have
$$x^2+1-2x\cos\frac{k\pi}{16} = 2-2\cos\frac{k\pi}{16} = 4\cdot\frac{1-\cos(k\pi/16)}{2} = 4\sin^2\frac{k\pi}{32}$$
The product becomes
$$\prod_{k=1}^{15}4\sin^2\frac{k\pi}{32}= 4^{15}\left(\prod_{k=1}^{15}\sin\frac{k\pi}{32}\right)^2$$
Note that the limit as $x\to 1$ of the product is equal to its value at $x=1$ because the expression is continuous (it is the product of continuous functions).
Now, equating the limits we get
$$16 = 4^{15}\left(\prod_{k=1}^{15}\sin\frac{k\pi}{32}\right)^2$$
As all the factors in the product are positive, we can take the square root of both sides and we get
$$\sin \frac{\pi}{32}\sin \frac{2\pi}{32}\sin \frac{3\pi}{32}\cdots\sin \frac{15\pi}{32}= \prod_{k=1}^{15}\sin\frac{k\pi}{32} =\frac{1}{2^{13}}$$
A: It is very straightforward to use complex numbers for this issue.
Proof of the first formula (though not asked) : Let $n=2m$. 
The roots of $x^{2m}-1=0$ are the 2m-th roots of unity ; they can be written in the following way : 
$$\omega_k=e^{ik \pi/m}, \ \ \ \ \ k=0,1,... (2m-1) $$
Thus, we can decompose: 
$$x^{2m}-1=(x-\omega_0)(x-\omega_1) \cdots (x-\omega_{2m-1})\tag{1}$$
Let us group these factors 2 by 2 in this way :
$$(x-\omega_k)(x-\omega_{2m-k})=x^2-(e^{ik\pi/m }+e^{i(2m-k)\pi/m })x+ e^{i \pi k/m}.e^{i(2m-k)\pi/m}$$
which gives, knowing that $e^{2i\pi}=1,$
$$x^2-\underbrace{(e^{ik\pi/m }+e^{-ik\pi/m})}_{2 \cos(k\pi/m)}x+ 1 \tag{2}$$
using Euler formula.
Then it suffices to use (2) in (1)  giving identity
$$x^{2m}-1=(x^2-1)\prod_{k=1}^{m-1}(x^2-2x\cos \frac{k\pi}{m}+1)\tag{3}$$
i.e., your identity, because the product of factors $(x-1)(x-(-1))=(w-\omega_0)(x-\omega_m)$ has been set apart.
2nd formula :
Factorizing $x$ in each 2nd degree factor on the RHS of (3) gives :
$$x^m(x^{m}-\tfrac{1}{x^m})=x^m(x-\tfrac{1}{x})\prod_{k=1}^{m-1}(x-2\cos \frac{k\pi}{m}+\tfrac{1}{x})\tag{4}$$
Simplifying both sides by $x^{m}$, (4) becomes :
$$\frac{x^{m}-\tfrac{1}{x^m}}{x-\tfrac{1}{x}}=\prod_{k=1}^{m-1}(x-2\cos \frac{k\pi}{m}+\tfrac{1}{x})\tag{5}$$
identity $\frac{a^m-b^m}{a-b}=a^{m-1}+a^{m-2}b+...+b^{m-1}$ ($m$ terms) allows us to write the LHS of (5) under the form 
$$x^{m-1} + x^{m-3} + ... \tfrac{1}{x^{m-3}}+\tfrac{1}{x^{m-1}} \ \ (m \  \text{factors})$$
Let us now take $m=16$. Let $x$ tends to $1$, 


*

*the LHS of (5) tends to $16=4^2$, whereas 

*the RHS of (5) tends to the product of $m$ factors. Using classical formula $1-\cos2a=2 \sin^2a$, this RHS is equal to : 
$$4^{15}\Pi_{k=1}^{15}(\sin^2(k\pi/32))$$
Equating RHS and LHS, and taking the square root of both sides, we get the desired formula.
