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If $A+B+C=2 S$, then $\sin (S-A)+\sin (S-B)-\sin C=$
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Verified Answer
The correct answer is:
$4 \sin \frac{S-A}{2} \sin \frac{S-B}{2} \sin \frac{C}{2}$
Consider, $\sin (S-A)+\sin (S-B)-\sin C$
$$
\begin{aligned}
& =2 \sin \left(\frac{2 S-(A+B)}{2}\right) \cos \left(\frac{B-A}{2}\right)-\sin C \\
& =2 \sin \left(\frac{C}{2}\right) \cos \left(\frac{B-A}{2}\right)-2 \sin \frac{C}{2} \cos \frac{C}{2} \\
& =2 \sin \frac{C}{2}\left[\cos \left(\frac{B-A}{2}\right)-\cos \frac{C}{2}\right] \\
& =2 \sin \frac{C}{2}\left[-2 \sin \left(\frac{B-A+C}{2}\right) \sin \left(\frac{B-A-C}{2}\right)\right] \\
& =4 \sin \frac{C}{2} \sin \left(\frac{B+C-A}{4}\right) \sin \left(\frac{A+C-B}{4}\right) \\
& =4 \sin \frac{C}{2} \sin \left(\frac{2 S-2 A}{4}\right) \sin \left(\frac{2 S-2 B}{4}\right) \\
& =4 \sin \frac{C}{2} \sin \left(\frac{S-A}{2}\right) \sin \left(\frac{S-B}{2}\right)
\end{aligned}
$$
$$
\begin{aligned}
& =2 \sin \left(\frac{2 S-(A+B)}{2}\right) \cos \left(\frac{B-A}{2}\right)-\sin C \\
& =2 \sin \left(\frac{C}{2}\right) \cos \left(\frac{B-A}{2}\right)-2 \sin \frac{C}{2} \cos \frac{C}{2} \\
& =2 \sin \frac{C}{2}\left[\cos \left(\frac{B-A}{2}\right)-\cos \frac{C}{2}\right] \\
& =2 \sin \frac{C}{2}\left[-2 \sin \left(\frac{B-A+C}{2}\right) \sin \left(\frac{B-A-C}{2}\right)\right] \\
& =4 \sin \frac{C}{2} \sin \left(\frac{B+C-A}{4}\right) \sin \left(\frac{A+C-B}{4}\right) \\
& =4 \sin \frac{C}{2} \sin \left(\frac{2 S-2 A}{4}\right) \sin \left(\frac{2 S-2 B}{4}\right) \\
& =4 \sin \frac{C}{2} \sin \left(\frac{S-A}{2}\right) \sin \left(\frac{S-B}{2}\right)
\end{aligned}
$$
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