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$2 \sin A \cos ^3 A-2 \sin ^3 A \cos A=$
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Verified Answer
The correct answer is:
$\frac{1}{2} \sin 4 A$
$\begin{aligned}
& 2 \sin A \cos ^3 A-2 \sin ^3 A \cos A \quad=2 \sin A \cos A\left(\cos ^2 A-\sin ^2 A\right) \\
& =2 \sin A \cos A \cos 2 A=\sin 2 A \cos 2 A=\frac{1}{2} \sin 4 A
\end{aligned}$
& 2 \sin A \cos ^3 A-2 \sin ^3 A \cos A \quad=2 \sin A \cos A\left(\cos ^2 A-\sin ^2 A\right) \\
& =2 \sin A \cos A \cos 2 A=\sin 2 A \cos 2 A=\frac{1}{2} \sin 4 A
\end{aligned}$
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