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If in a $\triangle A B C, 2 b^2=a^2+c^2$, then $\frac{\sin 3B}{\sin B}$ is equal to
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Verified Answer
The correct answer is:
$\left[\frac{c^2-a^2}{2 c a}\right]^2$
$\left[\frac{c^2-a^2}{2 c a}\right]^2$
$\frac{\sin 3 B}{\sin B}=\frac{3 \sin B-4 \sin ^3 B}{\sin B}$
$=3-4 \sin ^2 B$
$=3-4\left(1-\cos ^2 B\right)$
$=-1+\frac{4\left(a^2+c^2-b^2\right)^2}{4(a c)^2}$
$=-1+\frac{\left(\frac{a^2+c^2}{2}\right)^2}{(a c)^2}$
$=-1+\frac{\left(a^2+c^2\right)^2}{4(a c)^2}$
$=(\frac{c^2-a^2}{2 a c})^2$
$=3-4 \sin ^2 B$
$=3-4\left(1-\cos ^2 B\right)$
$=-1+\frac{4\left(a^2+c^2-b^2\right)^2}{4(a c)^2}$
$=-1+\frac{\left(\frac{a^2+c^2}{2}\right)^2}{(a c)^2}$
$=-1+\frac{\left(a^2+c^2\right)^2}{4(a c)^2}$
$=(\frac{c^2-a^2}{2 a c})^2$
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