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.

$$
\alpha = \frac{\omega+\omega^4}{2}\\
\alpha^2 = \frac{\omega^2 + \omega^3 + 2}{4}\\
2 \alpha + 4 \alpha^2 = \omega+\omega^4 + \omega^2 + \omega^3 + 2\\
4 \alpha^2 + 2 \alpha -1 = 0\\
$$
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 $\omega$.
