Let $c=\cos \frac{\pi}{5}$.
Define the polynomials on $\mathbb C$ as $f(z)=z^2-2cz+1, g(z)=z^4-z^3+z^2-z+1.$
Property
If $a\in \mathbb C$ satisfies $f(a)=0,$ then $g(a)=0.$
This is true, the reason is as follows.
Let $a\in \mathbb C$ satisfies $f(a)=0$. Since $f(a)=a^2-2ca+1=0$, I get $$a=c\pm \sqrt{c^2-1}=c\pm i \sqrt{1-c^2}=\cos \dfrac{\pi}{5}\pm i\sin\dfrac{\pi}{5}=e^{\pm \frac{\pi}{5}i}.$$
Noting $g(z)=\dfrac{z^5+1}{z+1}$ for $z\neq -1,$ I get $g(a)=\dfrac{a^5+1}{a+1}=\dfrac{\left(e^{\pm \frac{\pi}{5}i}\right)^5+1}{a+1}=0.$
But if I do as follows, the contradiction arises.
Note $g(z)=f(z)q(z)+r(z)$, where $q(z)=z^2+(2c-1)z+2c(2c-1), r(z)=-2c(1+2c-4c^2)z+1+2c-4c^2=(-2cz+1)(1+2c-4c^2)$.
Let $f(a)=0.$
Then $g(a)=f(a)q(a)+r(a)=0+r(a)=r(a).$
So if $g(a)=0$ holds, $r(a)=(-2ca+1)(1+2c-4c^2)=0$ holds.
Since $1+2c-4c^2\neq 0,$ I get $-2ca+1=0$, so $a=\dfrac{1}{2c}$.
But this contradicts $f(a)=0.$
$f(\frac{1}{2c})=\left(\dfrac{1}{2c}\right)^2-2c \cdot \dfrac{1}{2c}+1=\dfrac{1}{4c^2}\neq 0$.
So, I think I mistook somewhere. If you find the point where I mistook, I'd like you to tell me that.
If $c=\cos \frac{\pi}{5}$, then $4c^2-2c-1=0$ holds. Since $\phi=\frac{1+\sqrt 5}{2}$ and we know that $\phi=2\cos (\pi/5)$ we have that $$4(\phi/2)^2-2(\phi/2)-1=\phi^2-\phi-1=0$$
EDIT: Another proof using field theory
Let $\zeta=e^{2\pi i/10}$ a 10-th root of unity. And note that $2c=\zeta+\zeta^{-1}$. Since the minimal polynomial of $\zeta$ over rationals is $\Phi_{10}(t)=t^4-t^3+t^2-t+1$ and $[\mathbb{Q}(c):\mathbb{Q}]=\frac{\phi(10)}{2}=2$, we can find the minimal polynomial of $c$ over the rationals. $$t^{-2}\Phi_{10}(t)=t^2-t+1-t^{-1}+t^{-2}=(t^2+t^{-2})-(t+t^{-1})+1=(t+t^{-1})^2-(t+t^{-1})-1$$ Plugging the value $t=\zeta$ we get that $$(2c)^2-2c-1=4c^2-2c-1=0$$ So the minimal polynomial of $c$ over rationals is $4x^2-2x-1$