Constructing the 11-gon by splitting an angle in five In "Angle Trisection, the Heptagon, and the Triskaidecagon", published in the American Mathematical Monthly in March 1988, Andrew Gleason discusses what regular polygons can be constructed with compass, straightedge and angle trisector. At the end of that article he notes that the angle p-sectors required for a regular n-gon are the odd primes p dividing $\varphi(n)$.
For the heptagon, which only requires an angle trisector, he gives the minimal polynomial of $2\cos(2\pi/7)$
$$x^3+x^2-2x-1$$
and transforms it into the Chebyshev polynomial expression
$$7\sqrt{28}(4\cos^3\theta-3\cos\theta)=7$$
leading to the final identity
$$\sqrt{28}\cos\left(\frac{\cos^{-1}(1/\sqrt{28})}{3}\right)=1+6\cos(2\pi/7).$$
I am interested in the hendecagon (11 sides), which requires an angle quinsector (that splits an angle into five equal parts).

Is there a similar transformation between the minimal polynomial for $2\cos(2\pi/11)$
  $$x^5+x^4-4x^3-3x^2+3x+1$$
  and the relevant Chebyshev polynomial
  $$\cos 5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta$$
  and how do I find it? If I had such a transformation, I could construct an exact hendecagon with the quinsector.

I have tried Tschirnhaus transforms on the former polynomial, without success.

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.
