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

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(C) Given $A+B+C=2S$,so $S-C = S-(2S-A-B) = A+B-S$. Also $2S-C = A+B$.The expression is $E = \sin(S-A) \cos(S-B) - \sin(S-C) \cos S$.Multiply and divi

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$\cos A \sin B$

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(C) Given $A+B+C=2S$,so $S-C = S-(2S-A-B) = A+B-S$. Also $2S-C = A+B$.The expression is $E = \sin(S-A) \cos(S-B) - \sin(S-C) \cos S$.Multiply and divide by $2$: $E = \frac{1}{2} [2 \sin(S-A) \cos(S-B) - 2 \sin(S-C) \cos S]$.Using $2 \sin X \cos Y = \sin(X+Y) + \sin(X-Y)$,we get:$E = \frac{1}{2} [(\sin(2S-A-B) + \sin(B-A)) - (\sin(2S-C) + \sin(S-C-S))]$.Since $2S-A-B = C$ and $2S-C = A+B$,we have:$E = \frac{1}{2} [\sin C + \sin(B-A) - \sin(A+B) + \sin C]$.This expression simplifies to $\sin B \cos A$.

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

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