It is known that $\cos {{20}^{\circ }}$ is one of the roots of the cubic equation $8x^3-6x-1=0$. Is it possible not to use $i$ to give a closed form of $\cos {{20}^{\circ }}$? $$ \cos {{20}^{\circ }}={{2}^{\text{-}\frac{4}{3}}}\left( \sqrt[3]{1+\sqrt{3}i}+\sqrt[3]{1\text{-}\sqrt{3}i} \right) $$ By using de Moivre's formula，it can be told that $r=\sqrt[\alpha]{a+bi}+\sqrt[\alpha]{a-bi}$ for any $a,b,\alpha \in \mathbb{R} , r$ is a real number. But when apply it to the expression above, I just get $\cos {{20}^{\circ }}$ again. I cannot make any progress. Can we just use rational numbers and nth roots to express $\cos {{20}^{\circ }}$ , or some real numbers cannot be expressed without imaginary unit (if so, can it be proved?).
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6$\cos20^{\circ}$ is a closed form. But it can be proved that, to express it in radicals, nonreal numbers are needed. The usual proof goes via Galois Theory, and won't make much sense to you if you haven't studied that topic. Search term: casus irreducibilis. – Gerry Myerson Feb 02 '25 at 06:39
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2Even worse, there are three cube roots of a complex number, so this only give $\cos 20$ when you pick the usual branch of the cube root. – Thomas Andrews Feb 02 '25 at 06:55
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1See How to express $\cos(20^\circ)$ with radicals of rational numbers? and When $\cos{n^{\circ}}$ can be expressed in real radicals? and this 26 September 2005 sci.math post. – Dave L. Renfro Feb 02 '25 at 09:35