Find the minimal polynomial for $\cos(\frac{2\pi}{5})$ and $\sin(\frac{2\pi}{5})$ Let $\omega$ be the primitive 5th root of $1$, then $\cos(\frac{2\pi}{5}) = \frac{w+w^{-1}}{2}$ and $\sin(\frac{2\pi}{5}) = \frac{w-w^{-1}}{2i}$. How to find the minimal polynomial of  $\frac{w+w^{-1}}{2}$ then? (without using the Chebyshev polynomials)
Thanks.
 A: Below I'll show how to find 
(1) the minimal polynomial for $\omega+\omega^{-1}$,
(2) the minimal polynomial for $\cos\left(\dfrac{2\pi}{5}\right)$, and
(3) the minimal polynomial for $\sin\left(\dfrac{2\pi}{5}\right)$.
(1) First note that $\omega$ is a root of $x^5-1$, but not of $x-1$. Hence $\omega$ is a root of $\dfrac{x^5-1}{x-1}=x^4+x^3+x^2+x+1$. So we have that
$$\omega^4+\omega^3+\omega^2+\omega+1=0.$$
Since $\omega\ne0$, we can divide through by $\omega^2$ to obtain
$$\omega^2+\omega+1+\omega^{-1}+\omega^{-2}=0.$$
Which we can rewrite as
$$\left(\omega^2+2+\omega^{-2}\right)+\left(\omega+\omega^{-1}\right)+1-2=0.$$
So that 
$$\left(\omega+\omega^{-1}\right)^2+\left(\omega+\omega^{-1}\right)-1=0.$$
Hence $\omega+\omega^{-1}$ is a root of $x^2+x-1$. Since the roots of this polynomial are not rational, this polynomial is irreducible. Hence $x^2+x-1$ is the minimal polynomial of $\omega+\omega^{-1}$.
(2) $\cos\left(\dfrac{2\pi}{5}\right)=\dfrac{\omega+\omega^{-1}}{2}$, and $4\left(\dfrac{\omega+\omega^{-1}}{2}\right)^2+2\left(\dfrac{\omega+\omega^{-1}}{2}\right)-1=0$.
So the minimal polynomial for $\cos\left(\dfrac{2\pi}{5}\right)$ is $4x^2+2x-1$.
(3) The roots of $4x^2+2x-1$ are $\dfrac{-1\pm\sqrt{5}}{4}$, and $\cos\left(\dfrac{2\pi}{5}\right)>0$, so $\cos\left(\dfrac{2\pi}{5}\right)=\dfrac{\sqrt{5}-1}{4}$.
It follows that $\sin\left(\dfrac{2\pi}{5}\right)=\sqrt{\dfrac{\sqrt{5}+5}{8}}$.
For simplicity, let's let $\alpha=\sin\left(\dfrac{2\pi}{5}\right)=\sqrt{\dfrac{\sqrt{5}+5}{8}}$. So we have that
$$\alpha^2=\dfrac{\sqrt{5}+5}{8}.$$
$$8\alpha^2-5=\sqrt{5}.$$
$$64\alpha^4-80\alpha^2+25=5.$$
It follows that $\alpha$ is a root $16x^4-20x^2+5$, which is irreducible because of Eisenstein's criterion. So the minimal polynomial of $\sin\left(\dfrac{2\pi}{5}\right)$ is $16x^4-20x^2+5$.
A: Hint: square it, then make use of the fact that the 5 fifth roots of unity sum to 0.
A: $$
\alpha = \frac{\omega+\omega^4}{2}\\
\alpha^2 = \frac{\omega^2 + \omega^3 + 2}{4}\\
2 \alpha + 4 \alpha^2 = \omega+\omega^4 + \omega^2 + \omega^3 + 2\\
4 \alpha^2 + 2 \alpha -1 = 0\\
$$
Do $\alpha^n$ enough times that you see all the powers of $\omega^k$. Then you can see how to combine them that gives you the minimal polynomial for $\omega$.
A: Let $w=e^{2\pi i/5}$.  Then $w^5=1$ and $w\ne1$ so $w^4+w^3+w^2+w+1=0$, $w^4=w^{-1},$ and $w^3=w^{-2}$.
Also, $\cos(2\pi i/5)=\dfrac{w+w^{-1}}2$, 
so $\cos^2(2\pi i/5)=\dfrac{w^2+w^{-2}+2}4=\dfrac{w^2+w^3+2}4=\dfrac{-1-w-w^{-1}+2}4=\dfrac{1-2\cos(2\pi i/5)}{4}.$
Can you take it from here?
