Algebraic equation (find $b$ and $c$) The goal is, given the field extension $\,\mathbb{Q}\subset\mathbb{C}$, to find the minimal polynomial for the element
$$\eta=\cos\left(\frac{2\pi}{5}\right)$$



I define the element
$$\xi=\cos\left(\frac{2\pi}{5}\right)+i\sin\left(\frac{2\pi}{5}\right)$$
and due to $\,\left|\xi\right|=1$, I rewrite $\,\eta\,$ as
$$\eta=\frac{\xi+\overline{\xi}}{2}=\frac{1}{2}\left(\xi+\frac{1}{\xi}\right)$$
The next step is to find $\,b,c\in\mathbb{Q}\,$ such that $\,f\left(\eta\right)=0$, where $\,f(x)=x^2+bx+c\in\mathbb{Q}\left[X\right].$ So, I am trying to solve (unsuccessfully) the following equation
$$\frac{1}{4}\left(\xi+\frac{1}{\xi}\right)^2+\frac{b}{2}\left(\xi+\frac{1}{\xi}\right)+c=0$$ 

I know the solution is $\,b=\frac{1}{2},\,c=-\frac{1}{4}\,$ but when it comes to obtaining those values myself, I fail again and again. HELP!
Note $\xi = e^{2i\pi/5}$. Since $\xi$ is a complex fifth root of unity, we know $\xi^4 + \xi^3 + \xi^2 + \xi + 1 = 0$. Now divide through by $\xi^2$ to get $$\xi^2 + \xi + 1 + \frac{1}{\xi} + \frac{1}{\xi^2} = 0$$ Hence $$\left(\xi+ \frac{1}{\xi}\right)^2 - 2 + \left(\xi + \frac{1}{\xi}\right) + 1 = 0$$ and so $$4\eta^2 + 2\eta - 1 = 0$$ and since $\eta\not\in \mathbb{Q}$, the minimal polynomial of $\eta$ is $4x^2 +2x - 1$, or making it monic $x^2 + \frac{1}{2} x - \frac{1}{4}$, as per your solution.
Steps with CAS (Maple, Mathematica, Maxima, ...):


$x=\cos\frac{2\pi}{5}$
$\arccos(x)=\frac{2\pi}{5}$
$5\arccos(x)=2\pi$
$\cos(5\arccos(x))=1$
$5 x\, {{\left( 1-{{x}^{2}}\right) }^{2}}+{{x}^{5}}-10 {{x}^{3}}\, \left( 1-{{x}^{2}}\right) =1$
$\left( x-1\right) \, {{\left( 4 {{x}^{2}}+2 x-1\right) }^{2}}=0$


Then $x=\cos\frac{2\pi}{5}$ is solution of $4x^2+2x-1=0$. 

This method can apply for other elements. For example, $x=\cos\frac{2\pi}{15}$ is solution of 

$$16x^4-8x^3-16x^2+8x+1=0$$