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

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(C) Using the Sine Rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = K$. Thus,$a = K \sin A$ and $b = K \sin B$. Substituting the

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$2ab \sin C$

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(C) Using the Sine Rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = K$. Thus,$a = K \sin A$ and $b = K \sin B$. Substituting these into the expression: $a^2 \sin 2B + b^2 \sin 2A = (K \sin A)^2 (2 \sin B \cos B) + (K \sin B)^2 (2 \sin A \cos A)$ $= 2K^2 \sin A \sin B (\sin A \cos B + \cos A \sin B)$ $= 2(K \sin A)(K \sin B) \sin(A + B)$ $= 2ab \sin(A + B)$ Since $A + B + C = \pi$,we have $\sin(A + B) = \sin(\pi - C) = \sin C$. Therefore,the expression equals $2ab \sin C$.

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

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