I am learning Galois Theory by myself and studying its glorious applications. In the section of Straightedge and compass I got stuck at the following: Let be an angle whose cosine is equal to $1/9$. How do I show that the cosine of $\theta /3$ satisfies a polynomial equation of degree 3, and that it is irreducible over $\mathbb{Q}$. Why the angle cannot be trissected by ruler and compass?
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Using the product to sum formula, we have
$$ \cos^2(\theta/3)=\frac{1}{2}[1+\cos(2\theta/3)]. $$
Therefore, we have $$ \cos^3(\theta/3)=\frac{1}{2}[1+\cos(2\theta/3)]\cos(\theta/3)=\frac{1}{2}\cos(\theta/3)+\frac{1}{2}\cos(2\theta/3)\cos(\theta/3). $$
Using the product to sum formula again, we have $$ \cos^3(\theta/3)=\frac{1}{2}\cos(\theta/3)+\frac{1}{4}[\cos(\theta/3)+\cos(\theta)]=\frac{3}{4}\cos(\theta/3)+\frac{1}{4}\cdot\frac{1}{9}. $$
Therefore, $\cos(\theta/3)$ satisfies the polynomial $x^3-3x/4-1/36=0$.
Triplication formula: $$ \cos(3\alpha)=4\cos^3\alpha-3\cos\alpha $$ (this can be derived directly from the addition formulas). If $\alpha=\theta/3$ and we set $r=\cos(\theta/3)$, then we get $$ \frac{1}{9}=4r^3-3r $$ that becomes $$ 36r^3-27r-1=0 $$ For simplicity, let $s=6r$, so we get $$ \frac{1}{6}s^3-\frac{9}{2}s-1=0 $$ or $$ s^3-27s-6=0 $$ The polynomial $X^3-27X-6$ satisfies Eisenstein's criterion for the prime $3$, hence it is irreducible over the rational numbers.
It follows that $s$ has degree $3$ over the rationals and the same is true for each of its rational multiples, including $r=\cos(\theta/3)$.
A consequence of Wantzel's theorem is that a necessary condition for an angle to be constructible with ruler and compass is that its cosine is an algebraic number having degree a power of $2$.