If A+B+C=2S,then sin(S-A)+sin(S-B)-sin C= | vedclass

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(B) Given: $A+B+C=2S$,which implies $S-C = \frac{A+B}{2}$ and $C = 2S-(A+B)$.Consider the expression: $\sin(S-A)+\sin(S-B)-\sin C$.Using the sum-to-pr

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$4 \sin \frac{S-A}{2} \sin \frac{S-B}{2} \sin \frac{C}{2}$

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(B) Given: $A+B+C=2S$,which implies $S-C = \frac{A+B}{2}$ and $C = 2S-(A+B)$.Consider the expression: $\sin(S-A)+\sin(S-B)-\sin C$.Using the sum-to-product formula $\sin x + \sin y = 2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}$:$= 2 \sin \left(\frac{2S-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]$Since $C = 2S-A-B$,then $\frac{C}{2} = S - \frac{A+B}{2}$.$= 2 \sin \frac{C}{2} \left[ \cos \left(\frac{B-A}{2}\right) - \cos \left(S - \frac{A+B}{2}\right) \right]$Using $\cos x - \cos y = -2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}$:$= 2 \sin \frac{C}{2} \left[ -2 \sin \left(\frac{S-A}{2}\right) \sin \left(\frac{S-B}{2}\right) \right]$$= 4 \sin \frac{S-A}{2} \sin \frac{S-B}{2} \sin \frac{C}{2}$.

If $A+B+C=2S$,then $\sin(S-A)+\sin(S-B)-\sin C=$

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