How to express $\cos(20^\circ)$ with radicals of rational numbers?

Question

In showing that the trisection of an angle with ruler and compass is not possible in general one shows that $\cos(20^\circ)$ cannot be constructed (thus the angle $60^\circ$ cannot be trisected) by determining its minimal polynomial, which is $8x^3-6x-1$.

Solving $8x^3-6x-1=0$ yields a solution $x_1=\sqrt[3]{\frac{1}{16}+\frac{\sqrt{3}}{16}i}+\sqrt[3]{\frac{1}{16}-\frac{\sqrt{3}}{16}i}$. Expressing $\sqrt[3]{\frac{1}{16}+\frac{\sqrt{3}}{16}i}$ and $\sqrt[3]{\frac{1}{16}-\frac{\sqrt{3}}{16}i}$ in polar form yields $x_1=\cos(20^\circ)$.

Is it possible to express $\cos(20^\circ)$ with radicals without complex numbers?

Answer 1

$$ \cos (20^\circ) = \cos (\pi/9) = -\frac12 (-1)^{8/9} \left(1+(-1)^{2/9}\right) $$

Answer 2

The problem arises in the fact that the polynomial you got has 3 real roots and is irreducible over the rationals. It is an example of casus irreducibilis, which essentially shows that certain algebraic numbers whose minimal polynomial is of degree 3 or higher are real themselves but require complex numbers in order to be expressed in radicals.

This is pretty much as simple as you can get:

$$ \cos\left(20^{\circ}\right)=\frac{\sqrt[3]{1-i\sqrt{3}}+\sqrt[3]{1+i\sqrt{3}}}{2\sqrt[3]{2}} $$

...In which the imaginary parts simply cancel out leaving a real number. Again, you cannot remove the complex numbers from this expression.