MCQ
${\cos ^2}48^\circ - {\sin ^2}12^\circ = $
- A$\frac{{\sqrt 5 - 1}}{4}$
- ✓$\frac{{\sqrt 5 + 1}}{8}$
- C$\frac{{\sqrt 3 - 1}}{4}$
- D$\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$
$\therefore \,\,{\cos ^2}{48^o} - {\sin ^2}{12^o} = \cos \,\,{60^o}\,.\,\cos \,\,{36^o}$
$ = \frac{1}{2}\,\left( {\frac{{\sqrt 5 + 1}}{4}} \right) = \frac{{\sqrt 5 + 1}}{8}.$
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