If A+B+C=pi,then sin A-sin B+sin C= | vedclass

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(C) Given,$A+B+C=\pi$. We need to evaluate $\sin A - \sin B + \sin C$. Using the sum-to-product formula $\sin A + \sin C = 2 \sin \left(\frac{A+C}{2}\

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$4 \sin \left(\frac{A}{2}\right) \cos \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right)$

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(C) Given,$A+B+C=\pi$. We need to evaluate $\sin A - \sin B + \sin C$. Using the sum-to-product formula $\sin A + \sin C = 2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)$ and $\sin B = 2 \sin \frac{B}{2} \cos \frac{B}{2}$: Since $A+C = \pi - B$,we have $\frac{A+C}{2} = \frac{\pi}{2} - \frac{B}{2}$,so $\sin \left(\frac{A+C}{2}\right) = \cos \frac{B}{2}$. Thus,$\sin A + \sin C = 2 \cos \frac{B}{2} \cos \left(\frac{A-C}{2}\right)$. Now,$\sin A - \sin B + \sin C = 2 \cos \frac{B}{2} \cos \left(\frac{A-C}{2}\right) - 2 \sin \frac{B}{2} \cos \frac{B}{2}$. $= 2 \cos \frac{B}{2} \left[ \cos \left(\frac{A-C}{2}\right) - \sin \frac{B}{2} \right]$. Since $\frac{B}{2} = \frac{\pi}{2} - \frac{A+C}{2}$,$\sin \frac{B}{2} = \cos \left(\frac{A+C}{2}\right)$. $= 2 \cos \frac{B}{2} \left[ \cos \left(\frac{A-C}{2}\right) - \cos \left(\frac{A+C}{2}\right) \right]$. Using $\cos X - \cos Y = 2 \sin \left(\frac{X+Y}{2}\right) \sin \left(\frac{Y-X}{2}\right)$: $= 2 \cos \frac{B}{2} \left[ 2 \sin \frac{A}{2} \sin \frac{C}{2} \right] = 4 \sin \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$.

If $A+B+C=\pi$,then $\sin A-\sin B+\sin C=$

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