$$
\begin{aligned}
& \text { Let } A=\cos ^2 48^{\circ}-\sin ^2 12^{\circ} \\
& =\cos ^2\left(30^{\circ}+18^{\circ}\right)-\sin ^2\left(30^{\circ}-18^{\circ}\right) \\
& =\left[\cos 30^{\circ} \cos 18^{\circ}-\sin 30^{\circ} \sin 18^{\circ}\right]^2 \\
& =\left[\frac{\sqrt{3} \cos 30^{\circ} \cos 18^{\circ}-\sin 18^{\circ}}{2}\right]^2-\left[\frac{\left.\cos 18^{\circ}-\sqrt{3} \cos 30^{\circ}\right]^{\circ}}{2}\right]^2 \\
& =\frac{3 \cos ^2 18^{\circ}+\sin ^2 18-2 \sqrt{3} \sin 18^{\circ} \cos 18^{\circ}}{4} \\
& \quad-\frac{\cos ^2 18^{\circ}+3 \sin ^2 18-2 \sqrt{3} \sin 18^{\circ} \cos 18^{\circ}}{4} \\
& =\frac{\cos ^2 18^{\circ}-\sin ^2 18^{\circ}}{2}
\end{aligned}
$$

Note that $\sin 18^{\circ}=\frac{\sqrt{5}-1}{4} \Rightarrow \cos ^2 18^{\circ}=\frac{5+\sqrt{5}}{8}$
$$
\begin{aligned}
& \text { and } \sin ^2 18^{\circ}=\frac{3-\sqrt{5}}{8} \\
\therefore \quad & \mathrm{A}=\frac{\frac{5+\sqrt{5}-3+\sqrt{5}}{8}}{2}=\frac{1+\sqrt{5}}{8}
\end{aligned}
$$