In triangle ABC,the value of sin 2A + sin 2B | vedclass

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(A) In $\Delta ABC$,we have $A + B + C = 180^\circ$.Consider the expression $\sin 2A + \sin 2B + \sin 2C$.Using the sum-to-product formula $\sin X + \

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

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(A) In $\Delta ABC$,we have $A + B + C = 180^\circ$.Consider the expression $\sin 2A + \sin 2B + \sin 2C$.Using the sum-to-product formula $\sin X + \sin Y = 2\sin(\frac{X+Y}{2})\cos(\frac{X-Y}{2})$,we get:$\sin 2A + \sin 2B = 2\sin(A+B)\cos(A-B)$.Since $A+B = 180^\circ - C$,then $\sin(A+B) = \sin(180^\circ - C) = \sin C$.So,$\sin 2A + \sin 2B = 2\sin C \cos(A-B)$.Now,$\sin 2C = 2\sin C \cos C$.Since $C = 180^\circ - (A+B)$,then $\cos C = \cos(180^\circ - (A+B)) = -\cos(A+B)$.So,$\sin 2C = -2\sin C \cos(A+B)$.Substituting these back into the expression:$\sin 2A + \sin 2B + \sin 2C = 2\sin C \cos(A-B) - 2\sin C \cos(A+B)$$= 2\sin C [\cos(A-B) - \cos(A+B)]$.Using the identity $\cos(A-B) - \cos(A+B) = 2\sin A \sin B$,we get:$= 2\sin C [2\sin A \sin B] = 4\sin A \sin B \sin C$.

In triangle $ABC$,the value of $\sin 2A + \sin 2B + \sin 2C$ is equal to

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