If in Delta ABC,2b^2 = a^2 + c^2,then frac{si | vedclass

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(D) Given $2b^2 = a^2 + c^2$. Using the identity $\sin 3B = 3\sin B - 4\sin^3 B$,we have $\frac{\sin 3B}{\sin B} = 3 - 4\sin^2 B$. Using $\sin^2 B = 1

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$\left( \frac{c^2 - a^2}{2ca} \right)^2$

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(D) Given $2b^2 = a^2 + c^2$. Using the identity $\sin 3B = 3\sin B - 4\sin^3 B$,we have $\frac{\sin 3B}{\sin B} = 3 - 4\sin^2 B$. Using $\sin^2 B = 1 - \cos^2 B$,we get $3 - 4(1 - \cos^2 B) = 4\cos^2 B - 1$. From the Law of Cosines,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$. Since $b^2 = \frac{a^2 + c^2}{2}$,we substitute this into the expression for $\cos B$: $\cos B = \frac{a^2 + c^2 - \frac{a^2 + c^2}{2}}{2ac} = \frac{\frac{a^2 + c^2}{2}}{2ac} = \frac{a^2 + c^2}{4ac}$. Now,substitute $\cos B$ back into $4\cos^2 B - 1$: $4\left( \frac{a^2 + c^2}{4ac} \right)^2 - 1 = 4 \cdot \frac{(a^2 + c^2)^2}{16a^2c^2} - 1 = \frac{(a^2 + c^2)^2}{4a^2c^2} - 1$. $= \frac{(a^2 + c^2)^2 - 4a^2c^2}{4a^2c^2} = \frac{(c^2 - a^2)^2}{4a^2c^2} = \left( \frac{c^2 - a^2}{2ac} \right)^2$.

If in $\Delta ABC$,$2b^2 = a^2 + c^2$,then $\frac{\sin 3B}{\sin B} = $

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