16 sin(20^{circ}) sin(40^{circ}) sin(80^{circ | vedclass

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Question

(B) We use the identity $\sin(	heta) \sin(60^{\circ}-	heta) \sin(60^{\circ}+	heta) = \frac{1}{4} \sin(3	heta)$.Given expression: $16 \sin(20^{\cir

Vedclass · Accepted Answer

$2 \sqrt{3}$

Vedclass · Answer

(B) We use the identity $\sin(	heta) \sin(60^{\circ}-	heta) \sin(60^{\circ}+	heta) = \frac{1}{4} \sin(3	heta)$.Given expression: $16 \sin(20^{\circ}) \sin(40^{\circ}) \sin(80^{\circ})$.This can be written as $16 \sin(20^{\circ}) \sin(60^{\circ}-20^{\circ}) \sin(60^{\circ}+20^{\circ})$.Using the identity with $	heta = 20^{\circ}$:$= 16 	imes \left( \frac{1}{4} \sin(3 	imes 20^{\circ}) ight)$.$= 4 \sin(60^{\circ})$.$= 4 	imes \frac{\sqrt{3}}{2} = 2 \sqrt{3}$.

$16 \sin(20^{\circ}) \sin(40^{\circ}) \sin(80^{\circ})$ is equal to

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