\Sin12^\circ. \Sin24^\circ. \Sin48^\circ. \Sin84^\circ = ?

\Cos20^\circ. \Cos40^\circ. \Cos60^\circ. \Cos80^\circ

\Sin20^\circ. \Sin40^\circ. \Sin60^\circ. \Sin80^\circ

SSC CGL Model Paper 5

Question : 59 of 100

Marks: +1, -0

\Sin12°. \Sin24°. \Sin48°. \Sin84° = ?

Solution:

\Sin12°. \Sin24°. \Sin48°. \Sin84°

={ (\sin12°.\sin48°.\sin72°)(\sin24°,\sin36°.\sin84°)}/{\sin72° . \sin36°}

=1/4\sin36°×1/4\sin72°×1/{\sin72° . \sin36°}

=1/16

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