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