Question

Find the value of ${\sin }^2{48}^{\circ }-{\cos }^2{12}^{\circ }$

Answer 1

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