$\renewcommand\Re{\mathop{\mathfrak{Re}}}$$\renewcommand\Im{\mathop{\mathfrak{Im}}}$By considering $z^5=1$, it is possible to calculate the exact value of $\cos\frac{2\pi}{5}$ using the fact that $$1+\omega + \omega^2 +\omega^3+\omega^4=0,$$ where $1,\omega , \omega^2, \dots $ are the roots of the equation, so $$\Re(\omega) + \Re(\omega^2) + \Re(\omega^3)+ \Re(\omega^4)=-1.$$
This can then be simplified using the symmetry of the roots of unity, and a quadratic can be made and then solved so show that $$\cos\frac{2\pi}{5}=\frac{-1+\sqrt{5}}{4}.$$
I was wondering if the same sort of approach could be used to find the exact value of $\sin\frac{2\pi}{5}$, however don't think i can really simplify this using symetrry as $\Im(\omega) \ne \Im(\omega^2)$. Any insights to how it could be done would be great.
Best Answer
The problem with reading off the imaginary parts is $\sin(2\pi-x)=-\sin x$ makes the fact that the sines sum to $0$ trivial. But we can use what you've already established. Since $\cos\frac{2\pi}{5}=\frac{\sqrt{5}-1}{4}$,$$\sin\frac{2\pi}{5}=\sqrt{1-\cos^2\frac{2\pi}{5}}=\sqrt{\frac{5+\sqrt{5}}{8}}$$so$$\sin\frac{4\pi}{5}=2\sin\frac{2\pi}{5}\cos\frac{2\pi}{5}=\sqrt{\frac{5-\sqrt{5}}{8}}.$$Then use $\sin\frac{6\pi}{5}=-\sin\frac{4\pi}{5},\,\sin\frac{8\pi}{5}=-\sin\frac{2\pi}{5}$.