Explanation

(b)
Given, $\alpha < \beta < \gamma < \delta$
Also, $\sin{\alpha = \sin{\beta = \sin{\gamma = \sin{\delta = k}}}}$
$\therefore\ \ \ \beta = \pi - \alpha,\ \ \gamma = 2\pi + \alpha,\ \ \ \ \delta = 3\pi - \alpha$
$Now,\ 4\sin{\frac{\alpha}{2} + 3{sin\ }\frac{\beta}{2} + 2\sin\frac{\gamma}{2} + \sin\frac{\delta}{2}}$
$4\sin{\frac{\alpha}{2} + 3{sin\ }\left( \frac{\pi - \alpha}{2} \right) + 2\sin\left( \frac{2\pi + \alpha}{2} \right) + \sin\left( \frac{(3\pi - \alpha)}{2} \right)}$
$= 4\sin{\frac{\alpha}{2} + 3\cos{\frac{\alpha}{2} - 2\sin{\frac{\alpha}{2} - \cos\frac{\alpha}{2}}}}$
$= 2\sin{\frac{\alpha}{2} + 2\cos\frac{\alpha}{2}}$
$= 2\sqrt{\left( \sin{\frac{\alpha}{2} + \cos\frac{\alpha}{2}} \right)^{2}}$
$= 2\sqrt{\sin^{2}{\frac{\alpha}{2} + \cos^{2}{\frac{\alpha}{2} + 2\sin{\frac{\alpha}{2}\cos\frac{\alpha}{2}}}}}$
$= 2\sqrt{1 + \sin\alpha} = 2\sqrt{1 + k}$