In a triangle ABC,sin 2A + sin 2B + sin 2C = | vedclass

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(A) In $	riangle ABC$,$A+B+C = \pi \Rightarrow A+B = \pi - C$. $\sin 2A + \sin 2B + \sin 2C = 2 \sin(A+B) \cos(A-B) + 2 \sin C \cos C$. Since $\sin(A

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$4 \sin A \sin B \sin C$

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(A) In $	riangle ABC$,$A+B+C = \pi \Rightarrow A+B = \pi - C$. $\sin 2A + \sin 2B + \sin 2C = 2 \sin(A+B) \cos(A-B) + 2 \sin C \cos C$. Since $\sin(A+B) = \sin(\pi - C) = \sin C$,we have: $= 2 \sin C \cos(A-B) + 2 \sin C \cos C$. $= 2 \sin C [\cos(A-B) + \cos C]$. Since $\cos C = \cos(\pi - (A+B)) = -\cos(A+B)$,we have: $= 2 \sin C [\cos(A-B) - \cos(A+B)]$. Using the identity $\cos(A-B) - \cos(A+B) = 2 \sin A \sin B$: $= 2 \sin C [2 \sin A \sin B] = 4 \sin A \sin B \sin C$.

In a $\triangle ABC$,$\sin 2A + \sin 2B + \sin 2C =$

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