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.
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 $x^{32} - 1$
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
$$\sin \frac{\pi}{32}\sin \frac{2\pi}{32}\sin \frac{3\pi}{32}\cdots\sin \frac{15\pi}{32}= \prod_{k=1}^{15}\sin\frac{k\pi}{32} =\frac{1}{2^{13}}$$
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.
