Find the value of $\cos ^{2} 10^{\circ}+\cos ^{2} 50^{\circ}-\sin 40^{\circ} \sin 80^{\circ}$.
The basic method for this problem is to reduce the degree and transform sums and differences into products, constructing solutions based on structural characteristics.

Solution 1: Note that $\sin 40^{\circ}=\cos 50^{\circ}, \sin 80^{\circ}=\cos 10^{\circ}$, and the trigonometric expression is symmetric with respect to $\cos 10^{\circ}$ and $\cos 50^{\circ}$. Therefore, we can construct a symmetric substitution to find the value.
Let $\cos 10^{\circ}=a+b, \cos 50^{\circ}=a-b$, then $a=\frac{1}{2}\left(\cos 10^{\circ}+\cos 50^{\circ}\right)=\frac{\sqrt{3}}{2} \cos 20^{\circ}$,
$$
\begin{array}{l}
b=\frac{1}{2}\left(\cos 10^{\circ}-\cos 50^{\circ}\right)=\frac{1}{2} \sin 20^{\circ} . \text { So the original expression }=\cos ^{2} 10^{\circ}+\cos ^{2} 50^{\circ}- \\
\cos 50^{\circ} \cos 10^{\circ}=(a+b)^{2}+(a-b)^{2}-(a-b)(a+b)=a^{2}+3 b^{2}=\left(\frac{\sqrt{3}}{2} \cos 20^{\circ}\right)^{2} \\
+3\left(\frac{1}{2} \sin 20^{\circ}\right)^{2}=\frac{3}{4} .
\end{array}
$$

Solution 2: Using $\cos ^{2} 10^{\circ}+\sin ^{2} 10^{\circ}=1, \cos ^{2} 50^{\circ}+\sin ^{2} 50^{\circ}=1$, construct a dual model to solve.

Let $A=\cos ^{2} 10^{\circ}+\cos ^{2} 50^{\circ}-\sin 40^{\circ} \sin 80^{\circ}, B=\sin ^{2} 10^{\circ}+\sin ^{2} 50^{\circ}-$ $\cos 40^{\circ} \cos 80^{\circ}$, then $A+B=2-\cos 40^{\circ}, A-B=\cos 20^{\circ}+\cos 100^{\circ}+\cos 120^{\circ}=$ $\cos 40^{\circ}-\frac{1}{2}$, thus we find $A=\frac{3}{4}$.