We are entertaining polynomials with roots, all unique, on the curve $\Upsilon_s = \{{( 1-\cos[\theta])^{-s} \exp(i \theta)} \ | \ \theta \in \mathbb{R} \}$, where $s>0.$ $\Upsilon_s$ looks like a sewing needle, with two asymptotes to the positive real axis connected by a bend about the origin; if you think of it in polar co-ordinates, it has a pole at $\theta = 0$.
Let the monic polynomials $P_n$ of degree $n$ have roots on $\Upsilon_1$ at $ \theta = 2 \pi \frac{j}{n+1}, \ j= 1, 2, ..., n. $
When $n$ is odd, say $n = 2m-1$, we pair up the conjugate roots to get $P_n(z) =P_{2m-1}(z) = (z+ \frac{1}{2}) \ \Pi_{t=1}^{m-1}(z^2+\frac{2\cos{\frac{\pi t}{m}}}{1+\cos{\frac{\pi t}{m}}}z +\ (1+\cos{\frac{\pi t}{m}})^{-2} ).$ As an example, $P_7(z)=z^7-\frac{7}{2}z^6+7z^5 +\frac{17}{2}z^4+14z^3+14z^2+8z+2$, which we can find, albeit with great tedium, from $\cos{\frac{\pi}{8}}=\frac{1}{2}\sqrt{2+\sqrt{2}}.$
I would like a more efficient way to compute this, or just to demonstrate that $P_n$ is always in $\mathbb{Q}[z]$. Any hints?