General angle quintisection is impossible with straightedge and compass ($96c^5-120c^3+30c-5=0$)

Consider the constructible angle whose cosine is $5/6$ and sine is $\sqrt{11}/6$. If one fifth of this angle were constructible, its cosine $c$ would satisfy the displayed polynomial, by the identity $\cos 5\theta =16{\cos }^5\theta -20{\cos }^3\theta +5\cos \theta$. The identity follows by taking the real part of $(\cos \theta +i\sin \theta {)}^5$ and substituting ${\sin }^2\theta =1-{\cos }^2\theta$.

The polynomial is irreducible by Eisenstein's criterion at five, so $c$ has degree five over the rational numbers. The initial angle field $\mathbb{Q}(\sqrt{11})$ already has degree two, and adjoining any finite collection of constructible numbers gives total degree over the rational numbers a power of two. Such a field cannot contain the degree-five element $c$, by the tower law for field extensions. Hence this constructible angle cannot be divided into five equal angles by straightedge and compass construction. Some particular angles can be quintisected; the obstruction disproves a construction valid for all angles.