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

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(B) Using the sine rule,we have $b = 2R \sin B$ and $c = 2R \sin C$. Substituting these into the expression: $b^2 \sin 2C + c^2 \sin 2B = (2R \sin B)^

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$4\Delta$

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(B) Using the sine rule,we have $b = 2R \sin B$ and $c = 2R \sin C$. Substituting these into the expression: $b^2 \sin 2C + c^2 \sin 2B = (2R \sin B)^2 (2 \sin C \cos C) + (2R \sin C)^2 (2 \sin B \cos B)$ $= 8R^2 \sin^2 B \sin C \cos C + 8R^2 \sin^2 C \sin B \cos B$ $= 8R^2 \sin B \sin C (\sin B \cos C + \cos B \sin C)$ $= 8R^2 \sin B \sin C \sin (B + C)$ Since $A + B + C = \pi$,we have $\sin (B + C) = \sin (\pi - A) = \sin A$. $= 8R^2 \sin A \sin B \sin C$ $= 2(2R \sin B)(2R \sin C) \sin A$ $= 2bc \sin A$ Since the area of the triangle $\Delta = \frac{1}{2} bc \sin A$,we have $bc \sin A = 2\Delta$. Therefore,$2bc \sin A = 2(2\Delta) = 4\Delta$.

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

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