Given that, $\alpha<\beta<\gamma<\delta$ and 

$\sin \alpha=\sin \beta=\sin \gamma=\sin \delta=\mathrm{k}$

Also $\alpha, \beta, \gamma, \delta$ are smallest positive angles, satisfying above two conditions. 

$\therefore $ we can take $\beta=\pi-\alpha, \gamma=2 \pi+\alpha, \delta=3 \pi-\alpha$

The given expression

$=4 \sin \frac{\alpha}{2}+3 \sin \left(\frac{\pi}{2}-\frac{\alpha}{2}\right)$

$+2 \sin \left(\pi+\frac{\alpha}{2}\right)+\sin \left(\frac{3 \pi}{2}-\frac{\alpha}{2}\right)$

$=4 \sin \frac{\alpha}{2}+3 \cos \frac{\alpha}{2}-2 \sin \frac{\alpha}{2}-\cos \frac{\alpha}{2}$

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

$ = 2\sqrt {\left\{ {{{\left( {\sin \frac{\alpha }{2} + \cos \frac{a}{2}} \right)}^2}} \right\}} $

$=2 \sqrt{(1+\sin \alpha)}=2 \sqrt{(1+\mathrm{k})}$