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Question: \[\sin 36{^\circ}\sin 72{^\circ}\sin 108{^\circ}\sin 144{^\circ} =\]...

sin⁡36∘sin⁡72∘sin⁡108∘sin⁡144∘=\sin 36{^\circ}\sin 72{^\circ}\sin 108{^\circ}\sin 144{^\circ} =

A

1/4

B

1/16

C

3/4

D

5/16

Answer

5/16

Explanation

Solution

sin⁡36osin⁡72osin⁡108osin⁡144o\sin{}36^{o}\sin{}72^{o}\sin{}108^{o}\sin{}144^{o}

=sin⁡236osin⁡272o=14{(2sin⁡236o)(2sin⁡272o)}= \sin^{2}36^{o}\sin^{2}{}72^{o} = \frac{1}{4}\left\{ (2\sin^{2}36^{o})(2\sin^{2}{}72^{o}) \right\}

=14{(1−cos⁡72o)(1−cos⁡144o)}= \frac{1}{4}\left\{ (1 - \cos{}72^{o})(1 - \cos{}144^{o}) \right\}

=14{(1−sin⁡18o)(1+cos⁡36o)}= \frac{1}{4}\left\{ (1 - \sin{}18^{o})(1 + \cos{}36^{o}) \right\}

=14[(1−5−14)(1+5+14)]=2016×14=516= \frac{1}{4}\left\lbrack \left( 1 - \frac{\sqrt{5} - 1}{4} \right)\left( 1 + \frac{\sqrt{5} + 1}{4} \right) \right\rbrack = \frac{20}{16} \times \frac{1}{4} = \frac{5}{16}.