a cos[(B-C)/2]=(b+c)sin A/2.

Question

Prove that : \(a\,cos\,(\frac{B-C}{2})=(b\,+\,c)\,sin\frac{A}{2}.\)

Answer 1

LHS = \(a\,cos\frac{B-C}{2}\)

\(=k\,sin\,A\,cos\frac{B-C}{2}\)

\(=k\,sin\,(B+C)\,cos\frac{B-C}{2}\)

[ ∴ A + B +C = 180° ∴ sin A = sin [180° − (B + C) ]

\(=\frac{cos\frac{C}{2}cos\frac{A-B}{2}}{sin\frac{C}{2}cos\frac{C}{2}}\)

\(=\frac{cos\frac{A-B}{2}}{sin\frac{C}{2}}\) = RHS

∴ LHS = RHS