The value of frac{4 sin 9^{circ} sin 21^{circ | vedclass

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Question

(C) Let the expression be $E = \frac{4 \sin 9^{\circ} \sin 21^{\circ} \sin 39^{\circ} \sin 51^{\circ} \sin 69^{\circ} \sin 81^{\circ}}{\sin 54^{\circ}

Vedclass · Accepted Answer

$\frac{1}{8}$

Vedclass · Answer

(C) Let the expression be $E = \frac{4 \sin 9^{\circ} \sin 21^{\circ} \sin 39^{\circ} \sin 51^{\circ} \sin 69^{\circ} \sin 81^{\circ}}{\sin 54^{\circ}}$.Using the identity $\sin 	heta \sin(60^{\circ}-	heta) \sin(60^{\circ}+	heta) = \frac{1}{4} \sin 3	heta$,we group the terms:$E = \frac{4}{\sin 54^{\circ}} [(\sin 9^{\circ} \sin 51^{\circ} \sin 69^{\circ}) \cdot (\sin 21^{\circ} \sin 39^{\circ} \sin 81^{\circ})]$.Note that $\sin 51^{\circ} = \sin(60^{\circ}-9^{\circ})$ and $\sin 69^{\circ} = \sin(60^{\circ}+9^{\circ})$,so $\sin 9^{\circ} \sin 51^{\circ} \sin 69^{\circ} = \frac{1}{4} \sin(3 	imes 9^{\circ}) = \frac{1}{4} \sin 27^{\circ}$.Similarly,$\sin 21^{\circ} \sin 39^{\circ} \sin 81^{\circ} = \sin 21^{\circ} \sin(60^{\circ}-21^{\circ}) \sin(60^{\circ}+21^{\circ}) = \frac{1}{4} \sin(3 	imes 21^{\circ}) = \frac{1}{4} \sin 63^{\circ}$.Substituting these back into $E$:$E = \frac{4}{\sin 54^{\circ}} \cdot \frac{1}{4} \sin 27^{\circ} \cdot \frac{1}{4} \sin 63^{\circ} = \frac{\sin 27^{\circ} \sin 63^{\circ}}{4 \sin 54^{\circ}}$.Since $\sin 63^{\circ} = \cos 27^{\circ}$,we have $E = \frac{\sin 27^{\circ} \cos 27^{\circ}}{4 \sin 54^{\circ}} = \frac{2 \sin 27^{\circ} \cos 27^{\circ}}{8 \sin 54^{\circ}} = \frac{\sin 54^{\circ}}{8 \sin 54^{\circ}} = \frac{1}{8}$.

The value of $\frac{4 \sin 9^{\circ} \sin 21^{\circ} \sin 39^{\circ} \sin 51^{\circ} \sin 69^{\circ} \sin 81^{\circ}}{\sin 54^{\circ}}$ is equal to:

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