If $\alpha, \beta \in [0,\pi]$ then the minimum value of $\sin(\frac{\alpha +\beta}{2})$ is... Problem :  If $\alpha, \beta \in [0,\pi]$ then the minimum value of $\sin(\frac{\alpha +\beta}{2})$ is 
a)  $\frac{\sin\alpha +\sin\beta}{2}$ 
b)  $|\sin\alpha -\sin\beta|$  
c) $\frac{\cos\alpha +\cos\beta}{2}$ 
d) $|\cos\alpha -\cos\beta|$ 
Solution : 
Using $$\sin(A+B)= \sin A \cos B +\sin B \cos A  $$ we can write $$\sin(\frac{\alpha+ \beta}{2}) =\sin\alpha/2 \cos\beta /2 +\sin\beta/2 \cos\alpha/2 \qquad(i)$$
Now using Arithemtic Mean $\geq $ Geometric Mean  for $(i)$ we get : 
$$\Rightarrow \sin(\frac{\alpha +\beta}{2}) \geq 2\sqrt{\frac{1}{4}\sin\alpha \sin\beta}\quad \Rightarrow \quad  \sin(\frac{\alpha +\beta}{2}) \geq \sqrt{\sin\alpha \sin\beta}$$
Now how to proceed further in this I am not getting the clue please guide thanks..

Using Werner Formula,
$$2\sin\frac{A+B}2\cos\frac{A-B}2=\sin A+\sin B$$
But,
$$2\sin\frac{A+B}2\cos\frac{A-B}2\le2\sin\frac{A+B}2$$ as for $\displaystyle A,B\in[0,\pi]; 0\le \cos\frac{A-B}2\le1$(why?)
$$\implies2\sin\frac{A+B}2\ge\sin A+\sin B$$
Please  try Werner formula, with $\displaystyle2\sin\frac{A+B}2\sin\frac{A-B}2$
