'Strange' trigonometric roots of $x^5-4x^4+2x^3+5x^2-2x-1$ - could someone explain?

Question

This quintic equation has $5$ real roots:

$$x^5-4x^4+2x^3+5x^2-2x-1=0 \tag{1}$$

The roots are, from left to right:

$$x_1=\frac{\cos \frac{19}{22} \pi}{\cos \frac{1}{22} \pi}$$

$$x_2=\frac{\cos \frac{9}{22} \pi}{\cos \frac{19}{22} \pi}$$

$$x_3=\frac{\cos \frac{7}{22} \pi}{\cos \frac{5}{22} \pi}$$

$$x_4=\frac{\cos \frac{1}{22} \pi}{\cos \frac{7}{22} \pi}$$

$$x_5=\frac{\cos \frac{5}{22} \pi}{\cos \frac{9}{22} \pi}$$

I found these roots numerically, using ISC. The equation was found on Wikipedia in a different form (for $x-4/5$), no solutions were provided.

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I can derive this equation for each root individually. But I don't see how do all five roots 'fit' together.

Why ${1,5,7,9,19}$? Why there is no $3/22, 13/22$ or $17/22$ in any of the arguments?

In any case, I would be grateful for the explanation for how these roots all fit together.

Answer 1

Using the equality $\cos(\pi \pm x)=-\cos(x)$ you get

$$x_1=-\frac{\cos \frac{3 }{22} \pi}{\cos \frac{1}{22} \pi} \\ x_2=-\frac{\cos \frac{9}{22} \pi}{\cos \frac{3}{22} \pi} \\ x_3=-\frac{\cos \frac{15}{22} \pi}{\cos \frac{5}{22} \pi} \\ x_4=-\frac{\cos \frac{21}{22} \pi}{\cos \frac{7}{22} \pi} \\ x_5=-\frac{\cos \frac{27}{22} \pi}{\cos \frac{9}{22} \pi} $$

Note that the identity $$\cos(3x)= \cos(x) [2 \cos(2x)-1]$$ gives you a nicer form for the roots. With this form, after the proper substitution you should be able to reduce your polynomial to the minimal polynomial of $\cos(\frac{\pi}{11})$, which will explain where the roots are coming.

Answer 2

We just have to find the minimal polynomial of $$ \alpha = \frac{\cos\frac{19\pi}{22}}{\cos\frac{\pi}{22}}=-\frac{\cos\frac{3\pi}{22}}{\cos\frac{\pi}{22}}=3-4\cos^2\frac{\pi}{22}=1-2\cos\frac{\pi}{11} \tag{1}$$ then find the conjugate roots. But it is well-known that the minimal polynomial of $\cos\frac{2\pi}{m}$ has degree $\frac{\varphi(m)}{2}$ (i.e. $5$ in our case) and the algebraic conjugates of $\cos\frac{2\pi}{m}$ are $\cos\frac{2\pi k}{m}$ with $\gcd(k,m)=1$, so the claim is straightforward.