Minimal polynomial of $\cos(72°)$ over $\Bbb Q$ I'm asked to find the minimal polynomial over $\Bbb Q$ for $\mathrm{cos}(72°)$: given the complex number $z=\mathrm{cos}(72°)+i\mathrm{sin}(72°)$, one can say that $z^5=1$. Since $$(a+bi)^5=a^5+5a^4bi-10a^3b^2-10a^2b^3i+5ab^4+b^5i$$ I concluded that $\mathrm{cos}(72°)^5-10\mathrm{cos}(72°)^3\mathrm{sin}(72°)^2+5\mathrm{cos}(72°)\mathrm{sin}(72°)^4=1$, so if I replace $\mathrm{cos}(72°)$ with $x$ and $\mathrm{sin}(72°)$ with $\sqrt {1-x^2}$, I obtain the polynomial $p(x)=16x^5-20x^3+5x-1$. 
Now, it's clear that $1$ is a root, and dividing $p(x)$ by $x-1$ gives as result $p'(x)=16x^4+16x^3-4x^2-4x+1$. However I don't know how to show that $p'(x)$ is irreducible, since I can't use Eisenstein and the non existence of rational roots doesn't imply the irriducibility. Thank you in advance
 A: The rational root test quickly shows that the polynomial
$$q=16x^4+16x^3-4x^2-4x+1=y^4+2y^3-y^2-2y+1,$$
where $y=2x$, does not have any linear factors over $\Bbb{Q}$. So it is either irreducible, or it is of the form
$$q=(ay^2+by+c)(dy^2+ey+f),$$
for some $a,b,c,d,e,f\in\Bbb{Q}$, and in fact $a,b,c,d,e,f\in\Bbb{Z}$ by Gauss' lemma. Expanding the product leads to the system of equations
\begin{eqnarray*}
ad&=&1\\
ae+bd&=&2\\
af+be+cd&=&-1\\
bf+ce&=&-2\\
cf&=&1.
\end{eqnarray*}
The first and last show that without loss of generality $a=d=1$ and $c=f=\pm1$, yielding
\begin{eqnarray*}
e+b&=&2\\
be\pm2&=&-1\\
\pm(b+e)&=&-2,
\end{eqnarray*}
where the two $\pm$-signs agree. Comparing the first and last show that we must have the $-$-sign, yielding $b+e=2$ and $be=1$, so $b=e=1$ and hence
$$q=(y^2+y-1)^2=(4x^2+2x-1)^2.$$

From a slightly more abstract perspective; we see that the minimal polynomial of $z$ equals $z^4+z^3+z^2+z+1$ because
$$(z-1)(z^4+z^3+z^2+z+1)=z^5-1=0,$$
and $z\neq1$. Note that this polynomial is irreducible because
$$(z+1)^4+(z+1)^3+(z+1)^2+(z+1)+1=z^4+5z^3+10z^2+10z+5,$$
is Eisenstein w.r.t. the prime $5$. Now $2\cos(72^{\circ})=z+\bar{z}$, and the tower of fields
$$\Bbb{Q}\subset\Bbb{Q}(z+\bar{z})\subset\Bbb{Q}(z),$$
shows that $[\Bbb{Q}(z+\bar{z}):\Bbb{Q}]\leq2$ because $z+\bar{z}$ is fixed by complex conjugation. Hence the minimal polynomial of $z+\bar{z}$ has degree at most $2$.
A: Here Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$
we have proved:  $$\cos36^\circ-\cos72^\circ=\dfrac12$$
Now $$\cos36^\circ=-\cos(180^\circ-36^\circ)=-\cos(2\cdot36^\circ)=-(2\cos^272^\circ-1)$$
A: $$\cos72^{\circ}=\sin18^{\circ}=\frac{\sqrt{5}-1}{4}.$$
Thus, we obtain $$(4x+1)^2=5$$ or
$$4x^2+2x-1=0.$$
It's obvious that $4x^2+2x-1$ is irreducible, which says that we got the answer. 
