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If $\mathrm{A}+\mathrm{B}+\mathrm{C}=180^{\circ}$, then what is $\sin 2 \mathrm{~A}-\sin 2 \mathrm{~B}-\sin 2 \mathrm{C}$ equal
to?
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to?
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Verified Answer
The correct answer is:
$-4 \sin A \cos B \cos C$
$\begin{aligned} & \sin 2 \mathrm{~A}-\sin 2 \mathrm{~B}-\sin 2 \mathrm{C} \\ &=2 \cos (\mathrm{A}+\mathrm{B}) \cdot \sin (\mathrm{A}-\mathrm{B})+\sin (2 \mathrm{~A}+2 \mathrm{~B}) \\ &=2 \cos (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B})+2 \sin (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}+\mathrm{B}) \\ &=2 \cos (\mathrm{A}+\mathrm{B})[\sin (\mathrm{A}-\mathrm{B})+\sin (\mathrm{A}+\mathrm{B})] \\ &=-2 \cos \mathrm{C}[2 \sin \mathrm{A} \cdot \cos \mathrm{B}] \\ &=-4 \sin \mathrm{A} \cos \mathrm{B} \cos \mathrm{C} \end{aligned}$
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