Why is $ 2 \cos(\frac{2π}{5}) + \left(2 \cos(\frac{2π}{5})\right)^2 = 1$? I was able to prove $(a)$, but I have trouble with $(b).$ Can somebody give me a hint?

Let $ \alpha := e^{
\frac{2πi}{5}}  \in \mathbb{C}$, $K:=\mathbb{Q}(\alpha) $ and $f(X) := X^4 + X^3 + X^2 + X + 1$
(a) Prove that $f(\alpha) = 0$.
(b) Let $\overline{\alpha}$ be the complex conjugate of $\alpha$ and let $ \beta := \alpha + \overline{\alpha}$. Prove that $\beta + \beta^2 = 1$ and find the minimal polynomial of $\beta$ over $\mathbb{Q}$.
(c) Show that $\mathbb{Q}(\sqrt{5}) = \mathbb{Q}(\beta) ⊂ K$.

I was able to prove (a) because of
$$ 0 =\frac{\alpha^5−1}{\alpha−1}=\alpha^4+\alpha^3+\alpha^2+\alpha+1,$$
but now I am struggling with (b). 
Well obviously: $\alpha = \text{cos} (\frac{2π}{5}) + i ~ \text{sin}(\frac{2π}{5}),$ so $\beta = 2 ~ \text{cos} (\frac{2π}{5})$
and therefore I have to show:
$$ 2 \text{cos} (\frac{2π}{5}) + \left(2 ~\text{cos} (\frac{2π}{5})\right)^2 = 1$$
But why is that? Or did I make a mistake?
My minimal polynomial over $\mathbb{Q}$ is then $ X^2 + X -1$.
 A: Note that $1,\alpha,\alpha^2,\alpha^3,\alpha^4$ are roots of $x^5-1=0$
Now we have $\alpha\bar{\alpha}=1\implies\alpha\bar{\alpha}=\alpha^5\implies\bar{\alpha}=\alpha^4$
Therefore $\beta=\alpha+\alpha^4$ and $\beta+\beta^2=\alpha+\alpha^4+\alpha^2+\alpha^8+2\alpha^5=2+\alpha+\alpha^2+\alpha^3+\alpha^4=2-1=\boxed{1}$
Hence proved
A: This has been known since at least the time of Euclid.
Draw isosceles $\triangle ABC$ with $C$ at the apex and the congruent base angles measuring twice the apex angle. Thus from the known angular sum in a triangle, $\angle CAB$ and $\angle ABC$ each measure $2\pi/5$ and $\angle BCA$ measures $\pi/5$.

Draw the angle bisector $\overline{CD}$ from the apex which also bisected the base and is perpendicular thereto.Then SOA CAH TOA applied to right $\triangle ACD$ will measure $\cos(2\pi/5)$ where the hypotenuse AC is taken as one unit. Thus $AB$ measures $2\cos(2\pi/5)$.
Next switch to the tight hand figure on the right where base angle bisector $\overline{AE}$ is drawn. This renders $\angle EAB$ congruent with $\angle BCA$ abd therefore, with the two triangles also sharing a common vertex angle at $B$, $\triangle BEA$ is similar to $\triangle ABC$.From the proportionality of corresponding sides we render
$\dfrac{BE}{AB}=\dfrac{AB}{CA}$
From the above, $CA=1$ and $AB=2\cos{2\pi/5}$, so
$BE = (2\cos(2\pi/5))^2$
Next with the angle bisectoon at $A$, $\angle ACE$ is congruent with $\angle EAC$ forcing $\triangle CEA$ to be isosceles with base $\overline{CE}$. Thus $EC=EA$, and since $\triangle BEA$ is also isosceles (similar to $\triangle ABC$), we further extend this equality:
$EC=EA=AB=2\cos(2\pi/5)$
So, from betweenness we ultimately obtain:
$BE+EC=BC$
$\color{blue}{(2\cos(2\pi/5))^2+2\cos(2\pi/5)=1.}$
