We have,
$$\big(\cos(\tfrac{2\pi}{5})^{1/2}+(-\cos(\tfrac{4\pi}{5}))^{1/2}\big)^2 = \tfrac{1}{2}\left(\tfrac{-1+\sqrt{5}}{2}\right)^3\tag{1}$$
$$\big(\cos(\tfrac{2\pi}{7})^{1/3}+\cos(\tfrac{4\pi}{7})^{1/3}+\cos(\tfrac{6\pi}{7})^{1/3}\big)^3 = \frac{5-3\cdot 7^{1/3}}{2}\tag{2}$$
$$\big(\cos(\tfrac{2\pi}{11})^{1/5}+\cos(\tfrac{4\pi}{11})^{1/5}+\dots+\cos(\tfrac{10\pi}{11})^{1/5}\big)^5 = x?\tag{3}$$
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).