sin 20^{circ} cdot sin 40^{circ} cdot sin 60^ | vedclass

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(C) We use the identity $\sin A \cdot \sin(60^{\circ}-A) \cdot \sin(60^{\circ}+A) = \frac{1}{4} \sin 3A$. Setting $A = 20^{\circ}$,we have: $\sin 20^{

Vedclass · Accepted Answer

$\frac{3}{16}$

Vedclass · Answer

(C) We use the identity $\sin A \cdot \sin(60^{\circ}-A) \cdot \sin(60^{\circ}+A) = \frac{1}{4} \sin 3A$. Setting $A = 20^{\circ}$,we have: $\sin 20^{\circ} \cdot \sin(60^{\circ}-20^{\circ}) \cdot \sin(60^{\circ}+20^{\circ}) = \sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 80^{\circ} = \frac{1}{4} \sin(3 	imes 20^{\circ}) = \frac{1}{4} \sin 60^{\circ}$. Now,the expression becomes: $(\sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 80^{\circ}) \cdot \sin 60^{\circ} = (\frac{1}{4} \sin 60^{\circ}) \cdot \sin 60^{\circ} = \frac{1}{4} \sin^2 60^{\circ}$. Since $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,we have $\sin^2 60^{\circ} = \frac{3}{4}$. Thus,$\frac{1}{4} 	imes \frac{3}{4} = \frac{3}{16}$.

$\sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 60^{\circ} \cdot \sin 80^{\circ}$ is equal to

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