Simplifying a radical-trigonometric expression for the hendecagon angle

Question

This question is related to my very first question on this site, on constructing the hendecagon. The Gleason paper I referred to states the following identities, which lead to constructions of a regular heptagon and $13$-gon with compass/straightedge/angle trisector: $$1+6\cos\frac{2\pi}7=\sqrt{28}\cos\left(\frac13\cos^{-1}\frac1{\sqrt{28}}\right)\tag A$$ $$1+12\cos\frac{2\pi}{13}=\sqrt{13}+\sqrt{104-8\sqrt{13}}\cos\left(\frac13\tan^{-1}\frac{\sqrt3(\sqrt{13}+1)}{7-\sqrt{13}}\right)\tag B$$ Using GAP's RadiRoot package I have derived $$1+10\cos\frac{2\pi}{11}=11(z\omega + (-2+2z-z^4)\omega^2 + (12+6z+10z^2+3z^4)\omega^3 + (26+16z+41z^2+6z^3)\omega^4)\tag1$$ where $$z=e^{2i\pi/5}\qquad\omega=\frac1{\sqrt{11}}e^{i\theta}\qquad\theta=-\frac15\cos^{-1}-\frac{979+275\sqrt5}{4\cdot11^{5/2}}$$ So a hendecagon can be consructed by compass, straightedge and one angle quinsection (cf. here): construct $z$ with compass and straightedge, construct $\omega$ using the quinsector and perform basic arithmetic operations until you get $\cos\frac{2\pi}{11}$. However, the construction implied by $(1)$ would be very long and tedious.

Can $(1)$ be simplified to something like $(\rm A)$ and $(\rm B)$, where $1+10\cos\frac{2\pi}{11}$ equals a low-degree (preferably linear) polynomial in $\cos\theta$ with real coefficients? Would the result lead to an easily stated explicit Gleason-type construction?

Answer 1

Let $\phi$ be the golden ratio. The two quintics in $x$ and $y$ as described here,

$$x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1=0\tag1$$ $$y^5+5\left(\frac{-7+\sqrt5}{\;10\phi^2}\right)y^3+5\left(\frac{-7+\sqrt5}{\;10\phi^2}\right)^2y+\left(\frac{-830+139\sqrt5}{\;11(\phi\sqrt5)^6}\right)=0\tag2$$

have roots,

$$x = 2\cos\big(\tfrac{2\pi}{11}\big),\quad y = \frac2{\phi}\sqrt{\frac{7-\sqrt5}{10}}\,\cos\,(z)$$

where,

$$z =\frac15\arccos\left(\frac{-15+32\sqrt5}{110}\sqrt{\frac{10}{7-\sqrt5}}\right)$$

and $(x,y)$ have the quadratic relation,

$$y = \frac{x^2+(\phi^2/\sqrt5)x-(\phi^2/\sqrt5)}{x+\phi^2}$$

Thus, using the quadratic formula, one can then solve for $x$ and answer the OP's question whether "... $\cos\frac{2\pi}{11}$ equals a low-degree polynomial in $\cos\theta$ with real coefficients".