\(\cos ^2 48^{\circ}-\sin ^2 12^{\circ}=\) __________ , if \(\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}\)

\(\text {Let } A=\cos ^2 48^{\circ}-\sin ^2 12^{\circ} \)
\( =\cos ^2\left(30^{\circ}+18^{\circ}\right)-\sin ^2\left(30^{\circ}-18^{\circ}\right) \)
\( =\left[\cos 30^{\circ} \cos 18^{\circ}-\sin 30^{\circ} \sin 18^{\circ}\right]^2 \)
\( =\left[\frac{\sqrt{3} \cos 30^{\circ} \cos 18^{\circ}-\sin 18^{\circ}}{2}\right]^2-\) \(\left[\frac{\left.\cos 18^{\circ}-\sqrt{3} \cos 30^{\circ}\right]^{\circ}}{2}\right]^2 \)
\( =\frac{3 \cos ^2 18^{\circ}+\sin ^2 18-2 \sqrt{3} \sin 18^{\circ} \cos 18^{\circ}}{4} \)
\(\quad-\frac{\cos ^2 18^{\circ}+3 \sin ^2 18-2 \sqrt{3} \sin 18^{\circ} \cos 18^{\circ}}{4} \)
\( =\frac{\cos ^2 18^{\circ}-\sin ^2 18^{\circ}}{2}\)
Note that \(\sin 18^{\circ}=\frac{\sqrt{5}-1}{4} \Rightarrow \cos ^2 18^{\circ}=\frac{5+\sqrt{5}}{8}\)
\(\text {and } \sin ^2 18^{\circ}=\frac{3-\sqrt{5}}{8} \)
\( \therefore \mathrm{A}=\frac{\frac{5+\sqrt{5}-3+\sqrt{5}}{8}}{2}=\frac{1+\sqrt{5}}{8}\)