$S i n 12^{0} . S i n 24^{0} . S i n 48^{0} . S i n 84^{0} =$

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$S i n 20^{0} . S i n 40^{0} . S i n 60^{0} . S i n 80^{0}$

$C o s 20^{0} . C o s 40^{0} . C o s 60^{0} . C o s 80^{0}$

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1/2

answer is A.

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Detailed Solution

$\begin{array}{rcl} \sin 12^{0} \sin 24^{0} \sin 48^{0} \sin 84^{0} \\ = & \frac{1}{4} (2 \sin 48^{0} \sin 12^{0}) (2 \sin 84^{0} \sin 24^{0}) \\ = & \frac{1}{4} (\cos 36^{0} - \cos 60^{0}) (\cos 60^{0} - \cos 108^{0}) \\ = & \frac{1}{4} (\frac{\sqrt{5} + 1}{4} - \frac{1}{2}) (\frac{1}{2} + \frac{\sqrt{5} - 1}{4}) \\ = & \frac{1}{4} (\frac{\sqrt{5} - 1}{4}) (\frac{\sqrt{5} + 1}{4}) \\ = & \frac{1}{4} (\frac{5 - 1}{16}) = \frac{1}{16} \end{array}$

$\begin{array}{rcl} \cos 20^{0} \cos 40^{0} \cos 60^{0} \cos 80^{0} \\ = & \frac{1}{2} \cos 20^{0} \cos (60^{0} - 20^{0}) \cos (60^{0} + 20^{0}) \\ = & \frac{1}{2} \frac{1}{4} \cos 3 (20^{0}) \\ = & \frac{1}{8} \cdot \cos 60^{0} \\ = & \frac{1}{8} \cdot \frac{1}{2} = \frac{1}{16} \end{array}$