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

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Question

(C) We use the identity $\sin 	heta \sin(60^\circ - 	heta) \sin(60^\circ + 	heta) = \frac{1}{4} \sin 3	heta$.Given expression: $E = \sin 20^\circ

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$3$

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(C) We use the identity $\sin 	heta \sin(60^\circ - 	heta) \sin(60^\circ + 	heta) = \frac{1}{4} \sin 3	heta$.Given expression: $E = \sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ$.Substitute $\sin 60^\circ = \frac{\sqrt{3}}{2}$:$E = \frac{\sqrt{3}}{2} (\sin 20^\circ \sin 40^\circ \sin 80^\circ)$.Using the identity with $	heta = 20^\circ$:$\sin 20^\circ \sin(60^\circ - 20^\circ) \sin(60^\circ + 20^\circ) = \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{4} \sin(3 	imes 20^\circ) = \frac{1}{4} \sin 60^\circ$.Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$,we have:$\sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{4} 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8}$.Now substitute this back into the expression for $E$:$E = \frac{\sqrt{3}}{2} 	imes \frac{\sqrt{3}}{8} = \frac{3}{16}$.

$\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = $ (in $/16$)

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