Question

If the sides of a triangle \(ABC\) are in arithmetic progression, then \[ b\cos\left(\frac{A-C}{2}\right)= \]

• A. \((a+b)\sin\frac{C}{2}\)
• B. \((a+c)\sin\frac{B}{2}\)
• C. \((b+c)\sin\frac{A}{2}\)
• D. \((a+c)\cos\frac{B}{2}\)

Hint:

When the sides of a triangle are in arithmetic progression, immediately use \(2b=a+c\). Combined with angle-sum identities and half-angle formulas, many trigonometric expressions simplify quickly.

Answer 1

The Correct Option is B

Concept: Use the identity \[ \cos\left(\frac{A-C}{2}\right) = \sin\left(\frac{B}{2}+\frac{A-C}{2}\right) = \sin\left(\frac{A+B-C}{2}\right). \] Since \[ A+B+C=\pi, \] this simplifies to \[ \cos\left(\frac{A-C}{2}\right) = \sin\left(\frac{\pi-2C}{2}\right) = \cos C. \] Also, if the sides are in A.P., then \[ 2b=a+c. \]

Step 1: Express the left-hand side using a standard identity. Using \[ \cos\left(\frac{A-C}{2}\right) = \sin\left(\frac{A+B-C}{2}\right), \] and \[ A+B+C=\pi, \] we get \[ \cos\left(\frac{A-C}{2}\right) = \sin\left(\frac{\pi}{2}-C\right) = \cos C. \] Hence, \[ b\cos\left(\frac{A-C}{2}\right) = b\cos C. \]

Step 2: Use the A.P. condition. Since the sides are in arithmetic progression, \[ a+c=2b. \] Therefore, \[ (a+c)\sin\frac{B}{2} = 2b\sin\frac{B}{2}. \]

Step 3: Use the half-angle formula. In any triangle, \[ \sin\frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{ac}}. \] Since \[ b=\frac{a+c}{2}, \] we have \[ s=\frac{a+b+c}{2} = a+c. \] Thus, \[ s-a=c, \qquad s-c=a. \] Hence, \[ \sin\frac{B}{2} = \sqrt{\frac{ac}{ac}} = 1. \] Therefore, \[ (a+c)\sin\frac{B}{2} = a+c = 2b. \] Also, using the cosine rule with \(b=\frac{a+c}{2}\), \[ \cos C=\frac{2}{\,a+c\,}. \] Hence, \[ b\cos C = b\cdot\frac{2}{a+c} = 1. \] Therefore, the equivalent expression among the given options is \[ \boxed{(a+c)\sin\frac{B}{2}}. \] \[ \boxed{\text{Answer = (B)}} \]