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In triangle $A B C$, the value of $\sin 2 A+\sin 2 B+\sin 2 C$ is equal to
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Verified Answer
The correct answer is:
$4 \sin A \cdot \sin B \cdot \sin C$
We know that $A+B+C=180^{\circ}$ (in $\triangle A B C$ )
Now, $\sin 2 A+\sin 2 B+\sin 2 C$
$\begin{aligned}
& =2 \sin (A+B) \cos (A-B)+2 \sin C \cos C \\
& =2 \sin (\pi-C) \cos (A-B)+2 \sin C \cos (\pi-A+B) \\
& =2 \sin C \cos (A-B)-2 \sin C \cos (A+B) \\
& =2 \sin C\{\cos (A-B)-\cos (A+B)\} \\
& =2 \sin C\{2 \sin A \sin B\}=4 \sin A \sin B \sin C .
\end{aligned}$
Now, $\sin 2 A+\sin 2 B+\sin 2 C$
$\begin{aligned}
& =2 \sin (A+B) \cos (A-B)+2 \sin C \cos C \\
& =2 \sin (\pi-C) \cos (A-B)+2 \sin C \cos (\pi-A+B) \\
& =2 \sin C \cos (A-B)-2 \sin C \cos (A+B) \\
& =2 \sin C\{\cos (A-B)-\cos (A+B)\} \\
& =2 \sin C\{2 \sin A \sin B\}=4 \sin A \sin B \sin C .
\end{aligned}$
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