sin 12^circ sin 48^circ sin 54^circ = | vedclass

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Question

(c) $\sin \,{12^o}\,\sin \,{48^o}\,\sin \,{54^o} = \frac{1}{2}\,\left\{ {\cos {{36}^o} - \cos {{60}^o}} \right\}\,\cos \,{36^o}$

$ = \frac{1}{2}\,\

Vedclass · Accepted Answer

$1/8$

Vedclass · Answer

(c) $\sin \,{12^o}\,\sin \,{48^o}\,\sin \,{54^o} = \frac{1}{2}\,\left\{ {\cos {{36}^o} - \cos {{60}^o}} \right\}\,\cos \,{36^o}$

$ = \frac{1}{2}\,\left[ {\frac{{\sqrt 5 + 1}}{4} - \frac{1}{2}} \right]\,\left[ {\frac{{\sqrt 5 + 1}}{4}} \right] $

$= \frac{1}{2}\,\left[ {\frac{{\sqrt 5 - 1}}{4}} \right]\,\left[ {\frac{{\sqrt 5 + 1}}{4}} \right]$

$ = \frac{{5 - 1}}{{32}} = \frac{4}{{32}} = \frac{1}{8}$.

$\sin 12^\circ \sin 48^\circ \sin 54^\circ = $

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