MCQ
$\left(cos^2 \ 48^0 - sin^2 \ 12^0\right) - \left(cos^2 \ 66^0 - sin^2 \ 6^0\right) = .....$
- ✓$\frac{1}{4}$
- B$\frac{1}{2}$
- C$\frac{\sqrt{5}}{4}$
- D$\frac{\sqrt{5} + 1}{8}$
$\left(cos^248^0 - sin^2 12^0\right) - \left(cos^2 66^0 - sin^26^0\right)$
$= cos \left(48 + 12\right)^0 . cos \left(48 - 12\right)^0 - cos \left(66 + 6\right)^0 . cos \left(66 - 6\right)^0$
$= cos 60^0 . cos36^0 - cos72^0 cos60^0$
$= \frac{1}{2} \times \frac{\sqrt{5} + 1}{4} - \frac{\sqrt{5} - 1}{4} . \frac{1}{2}$
$= \frac{\sqrt{5} + 1 - \sqrt{5} + 1}{8} = \frac{2}{8} = \frac{1}{4}$
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