Question

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

Answer 1

We know that

${\cos }^{2}A-{\sin }^{2}B=\cos (A+B)\cos (A-B)$

Now,

${\cos }^{2}48-{\sin }^{2}12=\cos (48+12)\cos (48-12)$

$=\cos 60\cos 36$

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

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