One one hand, $\cos\frac{2\pi}{5}$ and $\sin\frac{2\pi}{5}$ are values of trigonometric functions; on the other hand, $\cos\frac{2\pi}{5} + i\sin\frac{2\pi}{5}$ is a root of algebraic equation $z^5-1=0$. Is it possible to use this algebraic relation to approximate the values of $\cos\frac{2\pi}{5}$ and $\sin\frac{2\pi}{5}$?
Progress: Roots of equation $z^5-1=0,z\in\mathbb{C}$ are $$\sqrt[5]{z_0}=1$$ $$\sqrt[5]{z_1}=-\sqrt[5]{-1}$$ $$\sqrt[5]{z_2}=(-1)^{\frac{2}{5}}$$ $$\sqrt[5]{z_3}=(-1)^{\frac{3}{5}}$$ $$\sqrt[5]{z_4}=(-1)^{\frac{4}{5}}$$
How to approximate $\cos\frac{2\pi}{5}$ and $\sin\frac{2\pi}{5}$ using these solutions?
First, we may factor $z^5-1$ as $$z^5-1=(z-1)(z^4+z^3+z^2+z+1)\tag{1}$$ The polynomial of $4$th degree in ($1$) is \begin{align} z^4+z^3+\frac{9}{4}z^2+z+1-\frac{5}{4}z^2&=\left(z^2+\frac{1}{2}z+1\right)^2-\frac{5}{4}z^2\\ &=\left(z^2+\frac{1+\sqrt{5}}{2}z+1\right)\left(z^2+\frac{1-\sqrt{5}}{2}z+1\right) \end{align} Then, ($1$) becomes $$z^5-1=(z-1)\left(z^2+\frac{1+\sqrt{5}}{2}z+1\right)\left(z^2+\frac{1-\sqrt{5}}{2}z+1\right)$$ It is known that quadratic polynomials with real coefficients and negative discriminant have complex conjugate roots, thus, by recognizing the signs of $\cos \left(\frac{2\pi}{5}\right)$ and $\cos \left(\frac{4\pi}{5}\right)$ it follows that
$$(z-e^{i2\pi/5})(z-e^{-i2\pi/5})=z^2-2\cos \left(\frac{2\pi}{5}\right)z+1=z^2+\frac{1-\sqrt{5}}{2}z+1\tag{2}$$
From ($2$) follows $$\color{blue}{\boxed{\cos \left(\frac{2\pi}{5}\right)=\frac{\sqrt{5}-1}{4}}\quad\text{and}\quad\boxed{\sin \left(\frac{2\pi}{5}\right)=\frac{\sqrt{10+2\sqrt{5}}}{4}}}$$