Question

In A triangle ABC prove that


$a \: cos \frac{b - c}{2} = (b + c) \frac{sin \: a}{2}$

Answer 1

$\bold{\huge{\underline{Answer-}}}$



$LHS = \: a \: cos \frac{B - C}{2}$



$= \frac{sinA}{k} .cos \frac{B - C}{2} \\ \\ \\ \\ = \frac{1}{k} .2sin \frac{A}{2} cos \frac{A}{2} cos \frac{B - C}{2} \\ \\ \\ \\ = \frac{1}{k} .sin \frac{A}{2} .2cos \frac{A}{2} cos \frac{B - C}{2} \\ \\ \\ = \frac{1}{k} .sin \frac{A}{2} .2sin \frac{B + C}{2} cos \frac{B - C}{2}$



$[ \:A = \pi - ( B - C) \: and \: \: cos \frac{A}{2} = cos( \frac{\pi}{2} - \frac{B + C}{2}) = sin \frac{B + C}{2} ]$



$= \frac{1}{k} sin \: \frac{A}{2} [sin( \frac{B + C}{2} + \frac{B - C}{2} ) + sin \: ( \frac{B + C}{2} - \frac{B - C}{2} )]$



$= \frac{1}{k} sin \: \frac{A}{2} [sin \: B + sin \: C]$



$= \frac{1}{k} sin \: \frac{A}{2}[bk + ck] = (b + c)sin \frac{A}{2}$