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

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(B) Given $A+B+C = 2S$. Then $2S-A = B+C$,$2S-B = A+C$,and $2S-C = A+B$. The expression becomes $\sin(B+C) + \sin(A+C) + \sin(A+B) - \sin(2S)$. Using

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

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(B) Given $A+B+C = 2S$. Then $2S-A = B+C$,$2S-B = A+C$,and $2S-C = A+B$. The expression becomes $\sin(B+C) + \sin(A+C) + \sin(A+B) - \sin(2S)$. Using $\sin(B+C) = \sin(2S-A)$,the expression is $\sin(2S-A) + \sin(2S-B) + \sin(2S-C) - \sin(2S)$. Using the identity $\sin X + \sin Y + \sin Z - \sin(X+Y+Z) = 4 \sin \frac{X+Y}{2} \sin \frac{Y+Z}{2} \sin \frac{Z+X}{2}$,Let $X = 2S-A$,$Y = 2S-B$,$Z = 2S-C$. Then $X+Y+Z = 6S - (A+B+C) = 6S - 2S = 4S$. This does not match the standard identity directly. However,for $A+B+C = 2S$,the identity $\sin A + \sin B + \sin C - \sin(A+B+C) = 4 \sin \frac{A+B-C}{2} \sin \frac{B+C-A}{2} \sin \frac{C+A-B}{2}$ is not applicable. Actually,for $A+B+C = 2S$,$\sin(2S-A) + \sin(2S-B) + \sin(2S-C) - \sin(2S) = 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.

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

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