Question

${\cos }^2{48}^{\circ }-{\sin }^2{12}^{\circ }$ is equal to

Answer 1

Let $J=\cos {}^{2}48-\sin {}^{2}12$

$=\cos (48+12)\cos (48-12)$

$=\cos 60\cos 36$

$=\frac{1}{2}\times \left(\frac{\sqrt{5}+1}{4}\right)$

$J=\frac{\sqrt{5}+1}{8}$