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.

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?
