In a triangle ABC, a^2 sin 2C + c^2 sin 2A is | vedclass

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

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$4\Delta$

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(B) Using the sine rule, $a = 2R \sin A$ and $c = 2R \sin C$. Substituting these into the expression: $a^2 \sin 2C + c^2 \sin 2A = (2R \sin A)^2 (2 \sin C \cos C) + (2R \sin C)^2 (2 \sin A \cos A)$ $= 8R^2 \sin^2 A \sin C \cos C + 8R^2 \sin^2 C \sin A \cos A$ $= 8R^2 \sin A \sin C (\sin A \cos C + \cos A \sin C)$ $= 8R^2 \sin A \sin C \sin(A + C)$ Since $A + B + C = 180^{\circ}$, $\sin(A + C) = \sin B$. $= 8R^2 \sin A \sin B \sin C$ Using the area formula $\Delta = \frac{abc}{4R}$, we know $abc = 4R\Delta$. Also, $\sin A = \frac{a}{2R}$, $\sin B = \frac{b}{2R}$, $\sin C = \frac{c}{2R}$. So, $8R^2 \cdot \frac{a}{2R} \cdot \frac{b}{2R} \cdot \frac{c}{2R} = \frac{abc}{R} = \frac{4R\Delta}{R} = 4\Delta$.

In a $\triangle ABC$, $a^2 \sin 2C + c^2 \sin 2A$ is equal to (in $\Delta$)

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