Question

In $\triangle ABC$, with usual notations, if $a, b, c$ are in A. P. Then $a \cos^2 (\frac{C}{2}) + c \cos^2 (\frac{A}{2}) =$

• A. $\frac{3a}{2}$
• B. $\frac{3c}{2}$
• C. $\frac{3b}{2}$
• D. $\frac{3abc}{2}$

Answer 1

Answer: (C)

$\frac{3b}{2}$

Answer 2

1. Use the Half-Angle Formula:

   In any triangle, one useful formula is:

   $\cos^2\left(\frac{A}{2}\right) = \frac{s(s-a)}{bc}, \quad \cos^2\left(\frac{C}{2}\right) = \frac{s(s-c)}{ab}$,

   where $s = \frac{a+b+c}{2}$ is the semiperimeter.

2. Substitute into the Expression:

   $a\cos^2\left(\frac{C}{2}\right) + c\cos^2\left(\frac{A}{2}\right) = a\frac{s(s-c)}{ab} + c\frac{s(s-a)}{bc}$

   This simplifies to:

   $\frac{s(s-c)}{b} + \frac{s(s-a)}{b} = \frac{s[(s-c)+(s-a)]}{b}$.

   Notice that

   $(s-c)+(s-a) = 2s - (a+c)$.

   Since $s = \frac{a+b+c}{2}$, we have:

   $2s = a+b+c \quad \text{so} \quad 2s - (a+c) = b$.

   Therefore,

   $a\cos^2\left(\frac{C}{2}\right) + c\cos^2\left(\frac{A}{2}\right) = \frac{s\cdot b}{b} = s$.

3. Use the A.P. Condition:

   Since $a$, $b$, $c$ are in A.P., we have:

   $b = \frac{a+c}{2} \quad \Rightarrow \quad a+b+c = (a+c)+b = 2b+b = 3b$.

   Hence,

   $s = \frac{a+b+c}{2} = \frac{3b}{2}$.

Therefore, $a\cos^2\left(\frac{C}{2}\right)+c\cos^2\left(\frac{A}{2}\right)= \frac{3b}{2}$.