In triangle ABC,if a cos^2 frac{C}{2} + c cos | vedclass

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(A) Given the equation: $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$ Using the half-angle formulas $\cos^2 \frac{C}{2} = \frac{s(s-c)}

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$2b = a + c$

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(A) Given the equation: $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$ Using the half-angle formulas $\cos^2 \frac{C}{2} = \frac{s(s-c)}{ab}$ and $\cos^2 \frac{A}{2} = \frac{s(s-a)}{bc}$: $a \left( \frac{s(s-c)}{ab} \right) + c \left( \frac{s(s-a)}{bc} \right) = \frac{3b}{2}$ $\frac{s(s-c)}{b} + \frac{s(s-a)}{b} = \frac{3b}{2}$ $\frac{s}{b} (s - c + s - a) = \frac{3b}{2}$ Since $2s = a + b + c$,we have $2s - a - c = b$: $\frac{s}{b} (b) = \frac{3b}{2} \Rightarrow s = \frac{3b}{2}$ $\frac{a + b + c}{2} = \frac{3b}{2} \Rightarrow a + b + c = 3b$ $a + c = 2b$

In $\triangle ABC$,if $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$,then

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