For A, B and C,if A+B+C=0,then sin(2A) + sin( | vedclass

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(C) Given $A+B+C=0$,we have $A+B = -C$. Taking the sine of both sides,$\sin(A+B) = \sin(-C) = -\sin(C)$. Now,consider the expression $\sin(2A) + \sin(

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$-4 \sin(A) \sin(B) \sin(C)$

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(C) Given $A+B+C=0$,we have $A+B = -C$. Taking the sine of both sides,$\sin(A+B) = \sin(-C) = -\sin(C)$. Now,consider the expression $\sin(2A) + \sin(2B) + \sin(2C)$: $\sin(2A) + \sin(2B) + \sin(2C) = 2 \sin(A+B) \cos(A-B) + 2 \sin(C) \cos(C)$ Since $A+B = -C$,then $\cos(A+B) = \cos(-C) = \cos(C)$. Substituting $\sin(A+B) = -\sin(C)$ and $\cos(C) = \cos(A+B)$: $= 2(-\sin(C)) \cos(A-B) + 2 \sin(C) \cos(A+B)$ $= -2 \sin(C) [\cos(A-B) - \cos(A+B)]$ Using the identity $\cos(x-y) - \cos(x+y) = 2 \sin(x) \sin(y)$: $= -2 \sin(C) [2 \sin(A) \sin(B)]$ $= -4 \sin(A) \sin(B) \sin(C)$.

For $A, B$ and $C$,if $A+B+C=0$,then $\sin(2A) + \sin(2B) + \sin(2C)$ is equal to

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