Constructing the 11-gon by splitting an angle in five

The question essentially asks about transforming solvable equations from one form to another.

I. Cubic

Using just a linear transformation, the general cubic $P(x)=0$ can be transformed to the form,

$$y^3+3ay+b = 0\tag1$$

with solution,

$$y_k = 2\sqrt{-a}\;\cos\left(-\tfrac{2\pi\,k}{3}+\tfrac{1}{3}\,\arccos\big(\tfrac{-b}{2\sqrt{-a^3}}\big)\right)\tag2$$

for $k=0,1,2$. Undoing the transformation establishes a relation between the roots $x,y$.

II. Quintic

Similarly, an appropriate Tschirnhausen transformation can transform a solvable quintic $P(x)=0$ to the Demoivre form (essentially the Chebyshev polynomial mentioned by the OP),

$$y^5+5ay^3+5a^2y+b = 0\tag3$$

with analogous solution,

$$y_k = 2\sqrt{-a}\;\cos\left(-\tfrac{2\pi\,k}{5}+\tfrac{1}{5}\,\arccos\big(\tfrac{-b}{2\sqrt{-a^5}}\big)\right)\tag4$$

for all five roots $y_k$. A cubic Tschirnhausen gives us three degrees of freedom to transform a solvable quintic to Demoivre form.

III. Transformations

For $p=7$:

$$x=2\cos\big(\tfrac{2\pi}{7}\big)\tag5$$

$$\color{blue}{y=3x+1} = 2\sqrt{7}\cos\left(\tfrac{1}{3}\,\cos^{-1}\big(\tfrac{1}{2\sqrt{7}}\big)\right)=4.7409\dots$$

then $x,y$ solves,

$$x^3+x^2-2x-1=0$$ $$y^3-21y-7=0$$

For $p=11$:

Let $\phi=\tfrac{1+\sqrt{5}}{2}$ be the golden ratio.

$$x=2\cos\big(\tfrac{2\pi}{11}\big)\tag6$$

$$\color{blue}{y=x^3-\phi\,x^2-\tfrac{7+\sqrt{5}}{2}x+\tfrac{5+4\sqrt{5}}{5}} = 2\,\phi\sqrt{\tfrac{11}{5}}\cos\left(-\tfrac{6\pi}{5}+\tfrac{1}{5}\,\cos^{-1}\big(\tfrac{-89-25\sqrt{5}}{44\sqrt{11}}\big)\right)=-4.7985\dots$$

then $x,y$ solves,

$$x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1=0$$ $$y^5-5ay^3+5a^2y+b=0$$

where $a=\tfrac{11}{5}\phi^2,\;\;b=\tfrac{11(125+89\sqrt{5})}{250}\phi^5$.

Thus as you can see, the transformation (in blue) which relates the quintic roots $x,y$ is more complicated than the cubic version, but is nonetheless doable in radicals.

If the object is to construct a regular hendecagon, it can be done more simply without going through an angle quinsection. Benjamin and Snyder proved the existence of a construction using a marked ruler and compasses in 2014 (BENJAMIN, ELLIOT; SNYDER, C. Mathematical Proceedings of the Cambridge Philosophical Society156.3 (May 2014): 409-424.; http://dx.doi.org/10.1017/S0305004113000753 ).

The basic premises of the construction are as follows:

1) It's based on the properties of "conchoid-circle" constructions where we position the marked straightedge to paas through a fixed point $P$, with one mark on a line $l$ and the other on a circle $K$.

2) With this type of construction, we define a "signed distance" $z$. This is the distance between $P$ and the mark on $K$, with a negative sign if that mark lies vetween $P$ and the other mark which is on $l$, a positive sign otherwise.

3) Then $z$ satisfies a sextic equation whose coefficients satisfy certain relationships called the "verging theorem".

4) The quintic equation given here for the hendecagonal cosines is converted to a sextic equation that satisfies the verging theorem by (4.1) defining $z=ux$ for some scale factor $u$, and (4.2) introducing an additional root $\eta$ with the appropriate value relative to $u$. Then, all the geometric parameters needed to determine $l$ and $K$ may be expressed and constructed in terms of this scale factor $u$.

5) Now to the heart of the matter. It looks like we have to solve a seventh-degree equation for that parameter $u$. But, "a miracle occurs" (the authors' own words); the equation for $u$ is reducuble and all we are left with is cubic factor equation (with integer coefficients), which can be solved by an auxiliary marked ruler construction.

6) So $z=ux$ has a construction with a marked ruler and compasses because it solves a sextic equation that satisfies the verging theorem, and $u$ has such a construction as well because it is obtained from a cubic equation in $Z[u]$; and so $x=2 cos(2\pi m/11)$ has one too.

7) Now for the parameters. For $u$, choose the one real root of $u^3+2u^2+2u+2=0$. For the construction of $z=ux$: $P=(0,0)$, $l$ is the line $x=-u-1$ where the length unit is the distance between the marks (conventional in this type of construction), $K$ is centeted at $(u(u-1)/2,-(u^2+3u+1)/2)$ and passes through $(-u-2,0)$. One orientation of the ruler is along the $x$ axis, that is the "extra" root of the sextic; the other roots on $K$ with proper distance signs, see (2), give the roots for $z$. Note that the authors do not give the formulas that way, I did some algebra of my own to get everything in terms of $u$.