sin 12^circ sin 24^circ sin 48^circ sin 84^ci | vedclass

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Question

(A) We need to evaluate $S = \sin 12^\circ \sin 24^\circ \sin 48^\circ \sin 84^\circ$. Using the identity $\sin 	heta \sin(60^\circ - 	heta) \sin(60

Vedclass · Accepted Answer

$\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ $

Vedclass · Answer

(A) We need to evaluate $S = \sin 12^\circ \sin 24^\circ \sin 48^\circ \sin 84^\circ$. Using the identity $\sin 	heta \sin(60^\circ - 	heta) \sin(60^\circ + 	heta) = \frac{1}{4} \sin 3	heta$: $S = \sin 12^\circ \sin 48^\circ \sin 84^\circ \sin 24^\circ = \sin 12^\circ \sin(60^\circ - 12^\circ) \sin(60^\circ + 12^\circ) \sin 24^\circ$ $S = \frac{1}{4} \sin(3 	imes 12^\circ) \sin 24^\circ = \frac{1}{4} \sin 36^\circ \sin 24^\circ$. Now,consider option $A$: $P = \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$. Using the identity $\cos 	heta \cos(60^\circ - 	heta) \cos(60^\circ + 	heta) = \frac{1}{4} \cos 3	heta$: $P = \cos 20^\circ \cos(60^\circ - 20^\circ) \cos(60^\circ + 20^\circ) \cos 60^\circ = \frac{1}{4} \cos(3 	imes 20^\circ) 	imes \frac{1}{2} = \frac{1}{8} \cos 60^\circ = \frac{1}{8} 	imes \frac{1}{2} = \frac{1}{16}$. Calculating $S = \frac{1}{4} \sin 36^\circ \sin 24^\circ = \frac{1}{8} [2 \sin 36^\circ \sin 24^\circ] = \frac{1}{8} [\cos(36^\circ - 24^\circ) - \cos(36^\circ + 24^\circ)] = \frac{1}{8} [\cos 12^\circ - \cos 60^\circ]$. Since $S = \frac{1}{16}$ and $P = \frac{1}{16}$,the correct option is $A$.

$\sin 12^\circ \sin 24^\circ \sin 48^\circ \sin 84^\circ = $

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