Question

Show that:
${\cos }^2{48}^{\circ }-{\sin }^2{12}^{\circ }=\frac{\sqrt{5}+1}{8}$

Answer 1

${\cos }^{2}48-{\sin }^{2}12=\cos (48+12)\cos (48-12)=\cos \left({60}^0\right)\cos \left({36}^0\right)=\frac{1}{2}\times \frac{\sqrt{5}+1}{4}=\frac{\sqrt{5}+1}{8}$