Vtyjkyuk

We have,

$$big(cos(tfrac2pi5)^1/2+(-cos(tfrac4pi5))^1/2big)^2 = tfrac12left(tfrac-1+sqrt52right)^3tag1$$

$$big(cos(tfrac2pi7)^1/3+cos(tfrac4pi7)^1/3+cos(tfrac6pi7)^1/3big)^3 = frac5-3cdot 7^1/32tag2$$

$$big(cos(tfrac2pi11)^1/5+cos(tfrac4pi11)^1/5+dots+cos(tfrac10pi11)^1/5big)^5 = x?tag3$$

Question: What degree is the minimal polynomial of $x$? Since the previous two are deg 2 and 3, I had assumed (3) would be deg 5, but Mathematica does not recognize it as a quintic with small coefficients, nor a 25th deg (even after using 500-decimal digit precision, though I am not sure of the latter result).

As some people said, the 2nd formula is easy to derive. In Maple, there are commands to get the minimal polynomial of LHS of the 2nd. The following method works for the 2nd formula, but not for the 3rd.

Prove that $sqrt[3]cosfrac2pi7 + sqrt[3]cosfrac4pi7 + sqrt[3]cosfrac6pi7 = sqrt[3]frac5-3sqrt[3]72$.

$textitProof$: $cosfrac2pi7,cosfrac4pi7,cosfrac6pi7$ are the three distinct real roots of $8x^3+4x^2-4x-1=0$.
Let $u_1=sqrt[3]cosfrac2pi7, u_2=sqrt[3]cosfrac4pi7, u_3=sqrt[3]cosfrac6pi7$ which are all real.
All roots of $8u^9+4u^6-4u^3-1=0$ are $u_1, t u_1, t^2 u_1, u_2, t u_2, t^2 u_2, u_3, t u_3, t^2 u_3$
where $t = e^2pi i/3$.

Let $(u-u_1)(u-u_2)(u-u_3)=u^3+a_2u^2+a_1u - 1/2$. We have
beginalign
(u-tu_1)(u-tu_2)(u-tu_3) &=u^3+ta_2u^2+t^2a_1u - 1/2, \
(u-t^2u_1)(u-t^2u_2)(u-t^2u_3) &=u^3+t^2a_2u^2+ta_1u - 1/2.
endalign
It follows from
beginalign
&8u^9+4u^6-4u^3-1 \
= & 8(u^3+a_2u^2+a_1u - 1/2)(u^3+ta_2u^2+t^2a_1u - 1/2)(u^3+t^2a_2u^2+ta_1u - 1/2)
endalign that
beginalign
a_2^3-3a_1a_2-2 &=0, \
4a_1^3+6a_1a_2+5 &= 0.
endalign
From these two equations, we get $a_1 = fraca_2^3-23a_2$
and $4a_2^9+30a_2^6+75a_2^3-32=0$.
Let $f(v)=4v^9+30v^6+75v^3-32$. We have $f'(v)=9v^2(2v^3+5)^2ge 0$. Thus, $f(v)=0$ has a unique real root.
It is easy to check that $v=-sqrt[3]frac5-3sqrt[3]72$ is a root of
$f(v)=0$. Thus, $a_2 = -sqrt[3]frac5-3sqrt[3]72$.
Thus, $u_1+u_2+u_3 = -a_2 = sqrt[3]frac5-3sqrt[3]72$. This completes the proof.

In general, suppose $x_1,x_2,x_3$ are the three distinct real roots of $x^3+bx^2+cx+d^3=0$,
where $b,c,d$ are rational numbers. We have
beginequation
sqrt[3]x_1+sqrt[3]x_2+sqrt[3]x_3 = sqrt[3]
-frac32Big(sqrt[3]Q+4sqrtDelta+sqrt[3]Q-4sqrtDeltaBig) -b-6d ,
endequation
where $Q = 24bd^2+36d^3+4bc+24cd, Delta = -4b^3d^3-27d^6+18bcd^3+b^2c^2-4c^3$.
Here, $Delta$ is the discriminant of $x^3+bx^2+cx+d^3=0$.

For the 3rd, similarly,
$x_k = cosfrac2kpi11, k=1,cdots, 5$ are the five distinct real roots of $32x^5+16x^4-32x^3-12x^2+6x+1=0$.
Let $u_1 = sqrt[5]cosfrac2pi11, u_2= sqrt[5]cosfrac4pi11, u_3 = sqrt[5]cosfrac6pi11, u_4 = sqrt[5]cosfrac8pi11, u_5= sqrt[5]cosfrac10pi11$.
All roots of $32u^25+16u^20-32u^15-12u^10+6u^5+1=0$ are
$t^j u_k, k=1,2,3,4, 5; j=0,1,2,3,4$ where $t= e^2pi i/5$.

Let $prod_k=1^5 (u-u_k) = u^5 + a_4u^4+a_3u^3+a_2u^2+a_1u+frac12$.
Similarly, we have four equations of
beginalign
0 & = a_4^5-5a_3a_4^3+5a_2a_4^2+5a_3^2a_4-5a_1a_4-5a_2a_3+2, \
0 & = 8a_1^5-20a_1^3a_2+10a_1^2a_3+10a_1a_2^2-5a_1a_4-5a_2a_3+1,\
0 & = -10a_1a_2a_4^3+10a_1a_3^2a_4^2+10a_2^2a_3a_4^2-10a_2a_3^3a_4+2a_3^5+10a_1^2a_4^2-10a_1a_2a_3a_4\
&quad-10a_1a_3^3-10a_2^3a_4+10a_2^2a_3^2-5a_3a_4^3+10a_1^2a_3+10a_1a_2^2+10a_2a_4^2+10a_3^2a_4\
&quad -15a_1a_4-15a_2a_3+7,\
0 & = -40a_1^3a_3a_4+40a_1^2a_2^2a_4+40a_1^2a_2a_3^2-40a_1a_2^3a_3+8a_2^5-40a_1^3a_2+20a_1^2a_4^2\
&quad-20a_1a_2a_3a_4-20a_1a_3^3-20a_2^3a_4+20a_2^2a_3^2+40a_1^2a_3+40a_1a_2^2+10a_2a_4^2\
&quad +10a_3^2a_4-30a_1a_4-30a_2a_3+13.
endalign

However, it seems no hope for determination of $a_4$.