Question
In any triangle $$ABC,\sin \frac{A}{2}$$ is
A.
less than $$\frac{{b + c}}{a}$$
B.
less than or equal to $$\frac{{a}}{b + c}$$
C.
greater than $$\frac{{2a}}{a + b + c}$$
D.
None of these
Answer :
less than or equal to $$\frac{{a}}{b + c}$$
Solution :
$$\eqalign{
& \frac{a}{{b + c}} = \frac{{\sin A}}{{\sin B + \sin C}} \cr
& \frac{a}{{b + c}} = \frac{{2\sin \frac{A}{2}\cos \frac{A}{2}}}{{2\sin \frac{{B + C}}{2} \cdot \cos \frac{{B - C}}{2}}} = \frac{{\sin \frac{A}{2}}}{{\cos \frac{{B - C}}{2}}} \cr
& \therefore \,\,\sin \frac{A}{2} = \frac{a}{{b + c}}\cos \frac{{B - C}}{2} \leqslant \frac{a}{{b + c}}. \cr
& \frac{{2a}}{{a + b + c}} = \frac{{2\sin A}}{{\sin A + \sin B + \sin C}} = \frac{{2 \cdot 2\sin \frac{A}{2}\cos \frac{A}{2}}}{{4\cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2}}} \cr
& \frac{{2a}}{{a + b + c}} = \frac{{\sin \frac{A}{2}}}{{\cos \frac{B}{2}\cos \frac{C}{2}}} \geqslant \sin \frac{A}{2}. \cr} $$