Find the value of $\cos \frac{2 \pi}{5}+\cos \frac{4 \pi}{5}$.
Analysis from the perspective of basic methods and construction methods.

Solution 1 (Reverse use of sum-to-product formulas)
$$
\cos \frac{2 \pi}{5}+\cos \frac{4 \pi}{5}=2 \cos \frac{\pi}{5} \cos \frac{3 \pi}{5}=-2 \cos \frac{\pi}{5} \cos \frac{2 \pi}{5},
$$

Multiplying the numerator and denominator by $4 \sin \frac{\pi}{5}$, and using the double-angle formula twice, we get the value as $-\frac{1}{2}$.
Solution 2 (Solving by constructing a conjugate expression)
Let $x=\cos \frac{2 \pi}{5}+\cos \frac{4 \pi}{5}, y=\cos \frac{2 \pi}{5}-\cos \frac{4 \pi}{5}, x y=\cos ^{2} \frac{2 \pi}{5}-\cos ^{2} \frac{4 \pi}{5}=$ $\frac{1}{2}\left(1+\cos \frac{4 \pi}{5}\right)-\frac{1}{2}\left(1+\cos \frac{8 \pi}{5}\right)=\frac{1}{2}\left(\cos \frac{4 \pi}{5}-\cos \frac{2 \pi}{5}\right)=-\frac{1}{2} y$, canceling $y(y>$
$0)$ gives $x=-\frac{1}{2}$.
Solution 3 (Constructing an equation by self-substitution)
$x=\cos \frac{2 \pi}{5}+\cos \frac{4 \pi}{5}=\cos \frac{2 \pi}{5}-\cos \frac{\pi}{5}<0$, squaring gives $x^{2}=\cos ^{2} \frac{2 \pi}{5}+\cos ^{2} \frac{4 \pi}{5}$
$+2 \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5}=\frac{1}{2}\left(1+\cos \frac{4 \pi}{5}\right)+\frac{1}{2}\left(1+\cos \frac{8 \pi}{5}\right)+2 \frac{\sin \frac{4 \pi}{5} \sin \frac{8 \pi}{5}}{2 \sin \frac{2 \pi}{5} \cdot 2 \sin \frac{4 \pi}{5}}=\frac{1}{2}$ $+\frac{1}{2}\left(\cos \frac{4 \pi}{5}+\cos \frac{8 \pi}{5}\right)=\frac{1}{2}+\frac{1}{2}\left(\cos \frac{4 \pi}{5}+\cos \frac{2 \pi}{5}\right)=\frac{1}{2}+\frac{x}{2}$, thus $x^{2}=\frac{1}{2}+$ $\frac{1}{2} x$, solving gives $x=-\frac{1}{2}$.
Solution 4 (Constructing a similar equation)

Let $\cos x+\cos 2 x=\cos \frac{2 \pi}{5}+\cos \frac{4 \pi}{5}$, then $\cos x=\cos \frac{2 \pi}{5}, \cos \frac{4 \pi}{5}$ simultaneously satisfy this similar equation. Using the double-angle formula, we get the quadratic equation $2 \cos ^{2} x+\cos x-\left(1+\cos \frac{2 \pi}{5}+\cos \frac{4 \pi}{5}\right)=0$, which indicates that $\cos \frac{2 \pi}{5}, \cos \frac{4 \pi}{5}$ are the roots of the equation $2 y^{2}+y-\left(1+\cos \frac{2 \pi}{5}+\cos \frac{4 \pi}{5}\right)=0$, and are all the roots. By the relationship between roots and coefficients, we get $\cos \frac{2 \pi}{5}+\cos \frac{4 \pi}{5}=-\frac{1}{2}$.