Question

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Let } A=\cos ^2 48^{\circ}-\sin ^2 12^{\circ} \ & =\cos ^2\left(30^{\circ}+18^{\circ}ight)-\sin ^2\left(30^{\circ}-18^{\circ}ight) \ & =\left[\cos 30^{\circ} \cos 18^{\circ}-\sin 30^{\circ} \sin 18^{\circ}ight]^2 \ & =\left[\frac{\sqrt{3} \cos 30^{\circ} \cos 18^{\circ}-\sin 18^{\circ}}{2}ight]^2-\left[\frac{\left.\cos 18^{\circ}-\sqrt{3} \cos 30^{\circ}ight]^{\circ}}{2}ight]^2 \ & =\frac{3 \cos ^2 18^{\circ}+\sin ^2 18-2 \sqrt{3} \sin 18^{\circ} \cos 18^{\circ}}{4} \ & \quad-\frac{\cos ^2 18^{\circ}+3 \sin ^2 18-2 \sqrt{3} \sin 18^{\circ} \cos 18^{\circ}}{4} \ & =\frac{\cos ^2 18^{\circ}-\sin ^2 18^{\circ}}{2} \end{aligned}$ Note that $\sin 18^{\circ}=\frac{\sqrt{5}-1}{4} \Rightarrow \cos ^2 18^{\circ}=\frac{5+\sqrt{5}}{8}$ $\begin{aligned} & 	ext { and } \sin ^2 18^{\circ}=\frac{3-\sqrt{5}}{8} \ 	herefore \quad & \mathrm{A}=\frac{\frac{5+\sqrt{5}-3+\sqrt{5}}{8}}{2}=\frac{1+\sqrt{5}}{8} \end{aligned}$

Solution