I was working through some roots of unity questions when this question stumped me.
Find all the roots of $z^{2n}=1$
Hence, factorise $z^{2n-1}+z^{2n-2}+...+z^3+z^2+z+1$ into quadratic factors with real coefficients.
Hence find the exact value of $$\cos(\frac \pi n)\cos(\frac {2\pi} n)\cos(\frac{3\pi}n)...\cos\left(\frac{\frac{(n-1)} 2 \pi }n\right)$$
I think i've worked out the first part and the the beginning of the second i.e. $$z^{2n}=1$$ $$z=\text{cis}\left(\frac{k\pi} n\right)\qquad k=0,1,2...,2n-1\\\therefore z^{2n}-1=0\\=(z+1)(z^{2n-1}+z^{2n-2}+z^{2n-3}+...+z^2+z+1)\\=(z+1)(z-\text{cis}(2\pi))(z-cis(-2\pi))...(z-cis\left(\frac{2n-1} {2n} \right))$$
I'm completely lost on where to go from here. I think the next step has to do with multiplying the conjugates together but where do I get them from?
Any help is greatly appreciated!
Since $z^{2n}-1$ has roots $\exp\frac{ik\pi}{n}$ for $0\le k\le2n-1$,$$z^{2n}-1=\prod_{k=0}^{2n-1}\left(z-\exp\frac{ik\pi}{n}\right)=(z-1)(z+1)\prod_{j=1}^{n-1}\left(z-\exp\frac{ik\pi}{n}\right)\left(z-\exp\frac{-ik\pi}{n}\right),$$where we have used$$\exp\frac{i(2n-k)\pi}{n}=\exp\frac{-ik\pi}{n}.$$Thus$$\sum_{l=0}^{2n-1}z^l=\frac{z^{2n}-1}{z-1}=(z+1)\prod_{j=1}^{n-1}\left(z^2-2z\cos\frac{j\pi}{n}+1\right)$$(note there's one spare linear factor, as we're factorizing a polynomial of odd parity). Substituting $z=i$, the left-hand side is the sum of $2n$ consecutive powers of $i$ with $n$ odd, so simplifies to $1+i$, which is just the right-hand side's linear factor. So$$1=\prod_j\left(-2i\cos\frac{j\pi}{n}\right)\implies\prod_j\cos\frac{j\pi}{n}=(-2i)^{1-n}.$$This product of cosines is the product we want, squared, times $(-1)^{(n-1)/2}=(-i)^{1-n}$, since $\cos(\pi-x)=-\cos x$. So the final result is $2^{(1-n)/2}$.