Question

In a triangle ABC, the expression \[ (b+c) \sin \frac{A}{2} \] is equal to:

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

Hint:

Use half-angle formulas in a triangle and simplify using expressions involving other sides and angles.

Answer 1

The Correct Option is B

Step 1: Recall half-angle formula in a triangle.
\[ \sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}, \quad s = \frac{a+b+c}{2} \]

Step 2: Express \((b+c) \sin \frac{A}{2}\).
\[ (b+c) \sin \frac{A}{2} = (b+c) \sqrt{\frac{(s-b)(s-c)}{bc}} \]

Step 3: Simplify using half-angle identities.
\[ (b+c) \sin \frac{A}{2} = \sqrt{\frac{(b+c)^2 (s-b)(s-c)}{bc}} \]

Step 4: Express in terms of \(a, B, C\).
\[ (b+c)^2 (s-b)(s-c) = a^2 \cos^2 \frac{B-C}{2} \cdot bc \quad (\text{from known triangle formulas}) \]

Step 5: Take square root.
\[ (b+c) \sin \frac{A}{2} = a \cos \frac{B-C}{2} \]

Step 6: Final conclusion.
\[ \boxed{(b+c) \sin \frac{A}{2} = a \cos \frac{B-C}{2}} \]