sin 20^circ sin 40^circ sin 60^circ sin 80^ci | vedclass

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Question

(C) We know that $\sin 60^\circ = \frac{\sqrt{3}}{2}$.Expression $= \frac{\sqrt{3}}{2} \sin 20^\circ \sin 40^\circ \sin 80^\circ$.Using the identity $

Vedclass · Accepted Answer

$3/16$

Vedclass · Answer

(C) We know that $\sin 60^\circ = \frac{\sqrt{3}}{2}$.Expression $= \frac{\sqrt{3}}{2} \sin 20^\circ \sin 40^\circ \sin 80^\circ$.Using the identity $\sin 	heta \sin(60^\circ - 	heta) \sin(60^\circ + 	heta) = \frac{1}{4} \sin 3	heta$:Here,$	heta = 20^\circ$,so $\sin 20^\circ \sin 40^\circ \sin 80^\circ = \sin 20^\circ \sin(60^\circ - 20^\circ) \sin(60^\circ + 20^\circ) = \frac{1}{4} \sin(3 	imes 20^\circ) = \frac{1}{4} \sin 60^\circ = \frac{1}{4} 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8}$.Therefore,the total expression $= \frac{\sqrt{3}}{2} 	imes \frac{\sqrt{3}}{8} = \frac{3}{16}$.

$\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = $

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