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If $\sin \mathrm{A} \sin \left(60^{\circ}-\mathrm{A}\right) \sin \left(60^{\circ}+\mathrm{A}\right)=\mathrm{k} \sin 3 \mathrm{~A}$, then what isk
equal to?
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equal to?
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Verified Answer
The correct answer is:
$1 / 4$
$\sin A \cdot \sin \left(60^{\circ}-\mathrm{A}\right) \sin \left(60^{\circ}+\mathrm{A}\right)=\mathrm{k} \sin 3 \mathrm{~A}$
$\begin{aligned} \Rightarrow & \sin \mathrm{A} \cdot \frac{\sin 3 \mathrm{~A}}{4 \sin \mathrm{A}}=\mathrm{k} \cdot \sin 3 \mathrm{~A} \\ & \quad\left[\because \sin \left(60^{\circ}+\mathrm{A}\right) \cdot \sin \left(60^{\circ}-\mathrm{A}\right)=\frac{\sin 3 \mathrm{~A}}{4 \sin \mathrm{A}}\right] \\ \Rightarrow & \frac{\sin 3 \mathrm{~A}}{4}=\mathrm{k} \cdot \sin 3 \mathrm{~A} \\ \therefore \mathrm{k} &=\frac{1}{4} \end{aligned}$
$$
$\begin{aligned} \Rightarrow & \sin \mathrm{A} \cdot \frac{\sin 3 \mathrm{~A}}{4 \sin \mathrm{A}}=\mathrm{k} \cdot \sin 3 \mathrm{~A} \\ & \quad\left[\because \sin \left(60^{\circ}+\mathrm{A}\right) \cdot \sin \left(60^{\circ}-\mathrm{A}\right)=\frac{\sin 3 \mathrm{~A}}{4 \sin \mathrm{A}}\right] \\ \Rightarrow & \frac{\sin 3 \mathrm{~A}}{4}=\mathrm{k} \cdot \sin 3 \mathrm{~A} \\ \therefore \mathrm{k} &=\frac{1}{4} \end{aligned}$
$$
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