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Question: \[\sin 12{^\circ}\sin 24{^\circ}\sin 48{^\circ}\sin 84{^\circ} =\]...

sin⁡12∘sin⁡24∘sin⁡48∘sin⁡84∘=\sin 12{^\circ}\sin 24{^\circ}\sin 48{^\circ}\sin 84{^\circ} =

A

cos⁡20∘cos⁡40∘cos⁡60∘cos⁡80∘\cos 20{^\circ}\cos 40{^\circ}\cos 60{^\circ}\cos 80{^\circ}

B

sin⁡20∘sin⁡40∘sin⁡60∘sin⁡80∘\sin 20{^\circ}\sin 40{^\circ}\sin 60{^\circ}\sin 80{^\circ}

C

315\frac{3}{15}

D

None of these

Answer

cos⁡20∘cos⁡40∘cos⁡60∘cos⁡80∘\cos 20{^\circ}\cos 40{^\circ}\cos 60{^\circ}\cos 80{^\circ}

Explanation

Solution

sin⁡12osin⁡24osin⁡48osin⁡84o\sin{}12^{o}\sin{}24^{o}\sin{}48^{o}\sin{}84^{o}

=14(2sin⁡12osin⁡48o)(2sin⁡24osin⁡84o)= \frac{1}{4}(2\sin{}12^{o}\sin{}48^{o})(2\sin{}24^{o}\sin{}84^{o})

=12(cos⁡36o−cos⁡60o)(cos⁡60o−cos⁡108o)= \frac{1}{2}(\cos{}36^{o} - \cos{}60^{o})(\cos{}60^{o} - \cos{}108^{o})

=14(cos⁡36o−12)(12+sin⁡18o)= \frac{1}{4}\left( \cos{}36^{o} - \frac{1}{2} \right)\left( \frac{1}{2} + \sin{}18^{o} \right)

=14{14(5+1)−12}{12+14(5−1)}=116= \frac{1}{4}\left\{ \frac{1}{4}(\sqrt{5} + 1) - \frac{1}{2} \right\}\left\{ \frac{1}{2} + \frac{1}{4}(\sqrt{5} - 1) \right\} = \frac{1}{16}

and cos⁡20ocos⁡40ocos⁡60cos⁡80o\cos{}20^{o}\cos{}40^{o}\cos{}60\cos{}80^{o}

=12[cos⁡(60o−20o)cos⁡20ocos⁡(60o+20o)]= \frac{1}{2}\lbrack\cos(60^{o} - 20^{o})\cos{}20^{o}\cos(60^{o} + 20^{o})\rbrack

=12[14cos⁡3(20o)]=18cos⁡60o=12×18=116= \frac{1}{2}\left\lbrack \frac{1}{4}\cos{}3(20^{o}) \right\rbrack = \frac{1}{8}\cos{}60^{o} = \frac{1}{2} \times \frac{1}{8} = \frac{1}{16}.