What is the minimal polynomial equation with integral coefficients that the area of the regular 11-gon with side lengths 1 satisfies? Let $q_n(x)$ be the minimal polynomial for the area of the regular $n$-gon. What is $q_{11}(x)$?
The following are the simpler cases, and $x$ is given by $16 (area^2) $:
For the corresponding triangle the polynomial is $p_3(x) = x-3$, for the square it is $p_4(x) = x - 16$, for the pentagon it is $p_5(x) = x^2-50x+125$, for the hexagon it is $p_6(x) =x-108$.
What is e.g.  $p_{11} (x)$? What is $p_n(x)$ in general? The question follows from a discussion on the Robbins' formula: http://youtu.be/OeZ6LsZHKcA
Robbins' formulas are meant to generalize Heron's and Brahmagrupta's formula to general cyclic n-gons (convex and concave alike). So far the formulas for n up to 9 have been discovered. A viewer of that video suggested we can do regular n-gons first. So for the pentagon case, if  you substitute all five $distance^2$ with 1, then you will  have the following factorized form: $(x^2-50x+125)(x-3)^5 = 0$. This is the same as $p_5(x)(p_3(x))^5 = 0$.  This 5 is a result from combinatoric arguments on  the degenerate cases: http://youtu.be/alFkaEZ4cZQ
So how do you generalize?
 A: Edited and corrected
If I set $$\kappa_n(x)=(x-i\sqrt{1-x^2})^{\frac{\phi(n)}{2}}\Phi_n(x+i\sqrt{1-x^2})$$ and then calculate $$f_n(x)=\kappa_{4n}(\frac{x}{2n})\kappa_{4n}(-\frac{x}{2n})$$ then I get at least for odd primes $n$ $$f_n(x)=\frac{(-1)^{\frac{\phi(n)}{2}}}{n^{\phi(n)}}\prod_{k=1}^{\frac{\phi(n)}{2}}(x^2-(2n\sin(\frac{2\pi}{n}))^2)$$ So the 16*area^2 should be a root of $f(\sqrt{x})$. I have not yet done some CAS program checks though.
The correction is necessary as then $\frac{\phi(4n)}{2}=\phi(n)$ and so $f_n(x)$ is a polynomial then because of the degree correpondence of the cyclotomic polynomials. I will test this for e.g. n=5 asap.
Edit 1 The product formula for $f_n(x)$ above is for $n=p$ odd prime. For odd $n$
I set $E_n=\{k|1\le k<n,\ \text{k being coprime to $n$}\ (k,,)=1\}$ and then ( not yet fully proved ) $$f_n(x)=\frac{(-1)^{\frac{\phi(n)}{2}}}{n^{\phi(n)}}\prod_{k\in E_n}(x^2-(2n\sin(\frac{2\pi}{n}))^2)$$
Edit 2 I hope to have proved correctly that $\kappa_{4n}(x)$ is an even polynomial. So it results in $$f_n(x)=\kappa_{4n}^2(\frac{x}{2n})$$ being a square of the irreducible polynomial.
Edit 3 Please excuse that I work contrary to the problem posted on the case with the n-gon lying on the unit circle. Just now I dont know whether there are connections to the case of side length 1. I hope to be able to give further results on my work if available.
A: Meanwhile I also did work on the case of the side of the polygonal being equal to 1. My results agree with the ones given by user Blue. I use the polynomial given as 
$$p_n(x)=\Phi_n(1)\prod_{k=1,(k,n)=1}^{n}(x-\cot(\frac{k\pi}{n}))$$ 
Then one gets the expression/"interesting factors" of user Blue when one calculates
$$n^{\phi(n)}p_n(\frac xn)$$
where $\phi(n)$ is the Euler totient function and $\Phi_n(x)$ the n-th cyclotomic polynomial.
