Equal sins

It is given that $\alpha < \beta < \gamma < \delta$ and $\sin \alpha = \sin \beta = \sin \gamma = \sin \delta = k$ . $\implies \beta = \pi - \alpha$ , $\gamma = 2\pi + \alpha$ and $\delta = 3\pi - \alpha$ . Therefore, we have:

$\begin{aligned} S & = 4 \sin \frac \alpha 2 + 3 \sin \frac \beta 2 + 2 \sin \frac \gamma 2 + \sin \frac \delta 2 \\ & = 4 \sin \frac \alpha 2 + 3 \sin \left( \frac \pi 2 - \frac \alpha 2 \right) + 2 \sin \left(\pi + \frac \alpha 2 \right) + \sin \left( \frac {3\pi} 2 - \frac \alpha 2 \right) \\ & = 4 \sin \frac \alpha 2 + 3 \cos \frac \alpha 2 - 2 \sin \frac \alpha 2 - \cos \frac \alpha 2 \\ & = 2 \sin \frac \alpha 2 + 2 \cos \frac \alpha 2 \end{aligned}$

From Weierstrass or tangent half-angle substitution, we have $\sin \alpha = \frac {2t}{1+t^2} = k$ , where $t = \tan \frac \alpha 2$ . Also that $\sin \frac \alpha 2 = \frac t{\sqrt{1+t^2}}$ and $\cos \frac \alpha 2 = \frac 1{\sqrt{1+t^2}}$ . Then, we have:

$\begin{aligned} S & = 2 \sin \frac \alpha 2 + 2 \cos \frac \alpha 2 \\ & = 2 \left( \frac t{\sqrt{1+t^2}} + \frac 1{\sqrt{1+t^2}} \right) \\ & = 2 \left( \frac {1+t}{\sqrt{1+t^2}} \right) = 2 \sqrt{\frac {(1+t)^2}{1+t^2}} \\ & = 2 \sqrt{\frac {1+2t + t^2}{1+t^2}} = 2 \sqrt{1 + \color{#3D99F6}{\frac {2t}{1+t^2}}} \\ & = 2\sqrt{1+\color{#3D99F6}{\sin \alpha}} = \boxed{2\sqrt{1+\color{#3D99F6}{k}}} \end{aligned}$