Question
In a triangle $$ABC, a, b, c$$ are the lengths of its sides and $$A, B, C$$ are the angles of triangle $$ABC.$$ The correct relation is given by
A.
$$\left( {b - c} \right)\sin \left( {\frac{{B - C}}{2}} \right) = a\cos \frac{A}{2}$$
B.
$$\left( {b - c} \right)\cos \left( {\frac{A}{2}} \right) = a\sin \frac{{B - C}}{2}$$
C.
$$\left( {b + c} \right)\sin \left( {\frac{{B + C}}{2}} \right) = a\cos \frac{A}{2}$$
D.
$$\left( {b - c} \right)\cos \left( {\frac{A}{2}} \right) = 2a\sin \frac{{B + C}}{2}$$
Answer :
$$\left( {b - c} \right)\cos \left( {\frac{A}{2}} \right) = a\sin \frac{{B - C}}{2}$$
Solution :
Let us consider $$\frac{{b - c}}{a},$$ which is involved in each of the these options.
$$\eqalign{
& \frac{{b - c}}{a} = \frac{{\sin B - \sin C}}{{\sin A}} \cr
& = \frac{{2\cos \left( {\frac{{B + C}}{2}} \right)\sin \left( {\frac{{B - C}}{2}} \right)}}{{2\sin \frac{A}{2}\cos \frac{A}{2}}} \cr
& = \frac{{\sin \frac{A}{2}\sin \left( {\frac{{B - C}}{2}} \right)}}{{\sin \frac{A}{2}\cos \frac{A}{2}}} \cr
& = \frac{{\sin \left( {\frac{{B - C}}{2}} \right)}}{{\cos \frac{A}{2}}} \cr
& \therefore \,\,\left( {b - c} \right)\cos \frac{A}{2} = a\sin \left( {\frac{{B - C}}{2}} \right) \cr} $$