Question

In $△ABC$, if $A,B$ and $C$ represent the angles of a triangle, then the maximum value of $\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2}$ is

• A. $\frac{1}{2}$
• B. $1$
• C. $\frac{3}{2}$
• D. $3$

Answer 1

The correct option is C $\frac{3}{2}$

Let $\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2}=k$
$\Rightarrow 2\sin \frac{A+B}{4}\cos \frac{A-B}{4}+\cos \frac{A+B}{2}=k\Rightarrow 2{\sin }^2\frac{A+B}{4}-2\cos \frac{A-B}{4}\sin \frac{A+B}{4}+k-1=0$
Since, $\sin \frac{A+B}{4}$ is real
$\therefore 4{\cos }^2\frac{A-B}{4}-8(k-1)\ge 0\Rightarrow 2(k-1)\le {\cos }^2\frac{A-B}{4}\le 1\Rightarrow k\le \frac{3}{2}$