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Mathematics
cos 48° ⋅ cos 12°=
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Q. $\cos 48^{\circ} \cdot \cos 12^{\circ}=$
AP EAMCET
AP EAMCET 2020
A
$\frac{(3-\sqrt{5})}{8}$
B
$\frac{(3+\sqrt{5})}{4}$
C
$\frac{(3+\sqrt{5})}{2}$
D
$\frac{(3+\sqrt{5})}{8}$
Solution:
$\cos 48^{\circ} \cdot \cos 12^{\circ}=\frac{1}{2}\left(2 \cos 48^{\circ} \cdot \cos 12^{\circ}\right)$
$=\frac{1}{2}\left[\cos 60^{\circ}+\cos 36^{\circ}\right]=\frac{1}{2}\left[\frac{1}{2}+\frac{\sqrt{5}+1}{4}\right]$
$=\left(\frac{3+\sqrt{5}}{8}\right)$