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Question: \[\sin 12{^\circ}\sin 48{^\circ}\sin 54{^\circ} =\]...

sin⁡12∘sin⁡48∘sin⁡54∘=\sin 12{^\circ}\sin 48{^\circ}\sin 54{^\circ} =

A

1/16

B

1/32

C

1/8

D

1/4

Answer

1/8

Explanation

Solution

sin⁡12osin⁡48osin⁡54o=12{cos⁡36o−cos⁡60o}cos⁡36o\sin 12^{o}\sin 48^{o}\sin 54^{o} = \frac{1}{2}\left\{ \cos 36^{o} - \cos 60^{o} \right\}\cos 36^{o}

=12[5+14−12][5+14]=12[5−14][5+14]= \frac{1}{2}\left\lbrack \frac{\sqrt{5} + 1}{4} - \frac{1}{2} \right\rbrack\left\lbrack \frac{\sqrt{5} + 1}{4} \right\rbrack = \frac{1}{2}\left\lbrack \frac{\sqrt{5} - 1}{4} \right\rbrack\left\lbrack \frac{\sqrt{5} + 1}{4} \right\rbrack

=5−132=432=18= \frac{5 - 1}{32} = \frac{4}{32} = \frac{1}{8}.