# If 0<(alpha),(beta),(gamma)<(pi)/2 prove that sin(alpha)+sin(beta)+sin(gamma)>sin(alpha+beta+gamma)?

##### 1 Answer
Aug 5, 2018

$\sin \left(\alpha\right) + \sin \left(\beta\right) + \sin \left(\gamma\right) - \sin \left(\alpha + \beta + \gamma\right)$

$= 2 \sin \left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha - \beta}{2}\right) - \left(\sin \left(\alpha + \beta + \gamma\right) - \sin \left(\gamma\right)\right)$

$= 2 \sin \left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha - \beta}{2}\right) - \left(2 \cos \left(\frac{\alpha}{2} + \frac{\beta}{2} + \gamma\right) \sin \left(\frac{\alpha + \beta}{2}\right)\right)$

$= 2 \sin \left(\frac{\alpha + \beta}{2}\right) \left[\cos \left(\frac{\alpha - \beta}{2}\right) - \cos \left(\frac{\alpha}{2} + \frac{\beta}{2} + \gamma\right)\right]$

$= 2 \sin \left(\frac{\alpha + \beta}{2}\right) \left[2 \sin \left(\frac{\alpha + \gamma}{2}\right) \sin \left(\frac{\beta + \gamma}{2}\right)\right] > 0$

As it is given

$0 < \alpha , \beta \mathmr{and} \gamma < \frac{\pi}{2}$ we have,

$\left(\frac{\alpha + \beta}{2}\right) , \left(\frac{\alpha + \gamma}{2}\right) \mathmr{and} \left(\frac{\beta + \gamma}{2}\right)$ arr acute angles and sine of all must be $> 0$

Hence

$\sin \left(\alpha\right) + \sin \left(\beta\right) + \sin \left(\gamma\right) > \sin \left(\alpha + \beta + \gamma\right)$