# 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..

• If $\alpha, \beta \in [0,\pi]$, then … $\frac {\alpha + \beta}{2}$ is an angle between 0, and $\pi$. Then the minimum of the sine of it is 0. Therefore , I think the question should be re-phrased. – Mick Jul 12 '14 at 18:42

$$2\sin\frac{A+B}2\cos\frac{A-B}2=\sin A+\sin B$$
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 $\displaystyle2\sin\frac{A+B}2\sin\frac{A-B}2$