If sin^{2}(10^{circ}) sin(20^{circ}) sin(40^{ | vedclass

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(C) Let $S = \sin^{2}(10^{\circ}) \sin(20^{\circ}) \sin(40^{\circ}) \sin(50^{\circ}) \sin(70^{\circ})$.Using the identity $\sin(	heta) \sin(60^{\circ

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(C) Let $S = \sin^{2}(10^{\circ}) \sin(20^{\circ}) \sin(40^{\circ}) \sin(50^{\circ}) \sin(70^{\circ})$.Using the identity $\sin(	heta) \sin(60^{\circ}-	heta) \sin(60^{\circ}+	heta) = \frac{1}{4} \sin(3	heta)$,we have $\sin(10^{\circ}) \sin(50^{\circ}) \sin(70^{\circ}) = \frac{1}{4} \sin(30^{\circ}) = \frac{1}{8}$.Substituting this into the expression:$S = \sin(10^{\circ}) \cdot \sin(20^{\circ}) \sin(40^{\circ}) \cdot [\sin(10^{\circ}) \sin(50^{\circ}) \sin(70^{\circ})]$$S = \sin(10^{\circ}) \cdot \sin(20^{\circ}) \sin(40^{\circ}) \cdot \frac{1}{8}$.Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$:$S = \frac{1}{8} \sin(10^{\circ}) \cdot \frac{1}{2} [\cos(20^{\circ}) - \cos(60^{\circ})] = \frac{1}{16} \sin(10^{\circ}) [\cos(20^{\circ}) - \frac{1}{2}]$$S = \frac{1}{16} \sin(10^{\circ}) \cos(20^{\circ}) - \frac{1}{32} \sin(10^{\circ})$.Using $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:$S = \frac{1}{32} [\sin(30^{\circ}) + \sin(-10^{\circ})] - \frac{1}{32} \sin(10^{\circ})$$S = \frac{1}{32} [\frac{1}{2} - \sin(10^{\circ})] - \frac{1}{32} \sin(10^{\circ}) = \frac{1}{64} - \frac{1}{16} \sin(10^{\circ})$.Comparing with $\alpha - \frac{1}{16} \sin(10^{\circ})$,we get $\alpha = \frac{1}{64}$.Thus,$16 + \alpha^{-1} = 16 + 64 = 80$.

If $\sin^{2}(10^{\circ}) \sin(20^{\circ}) \sin(40^{\circ}) \sin(50^{\circ}) \sin(70^{\circ}) = \alpha - \frac{1}{16} \sin(10^{\circ})$,then $16 + \alpha^{-1}$ is equal to

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