Transform the following expression into a product:
$\sin \alpha + \sin 60^\circ + \sin \left(\alpha + 60^\circ\right)$
Given: $\;$ $\sin \alpha + \sin 60^\circ + \sin \left(\alpha + 60^\circ\right)$
$= \sin \alpha + \dfrac{\sqrt{3}}{2} + \sin \alpha \cos 60^\circ + \cos \alpha \sin 60^\circ$
$= \sin \alpha + \dfrac{\sqrt{3}}{2} + \dfrac{1}{2} \times \sin \alpha + \dfrac{\sqrt{3}}{2} \times \cos \alpha$
$= \dfrac{3}{2} \times \sin \alpha + \dfrac{\sqrt{3}}{2} \left(1 + \cos \alpha\right)$
$= \dfrac{3}{2} \times 2 \sin \left(\dfrac{\alpha}{2}\right) \cos \left(\dfrac{\alpha}{2}\right) + \dfrac{\sqrt{3}}{2} \times 2 \cos^2 \left(\dfrac{\alpha}{2}\right)$
$= \sqrt{3} \cos \left(\dfrac{\alpha}{2}\right) \left[\sqrt{3} \sin \left(\dfrac{\alpha}{2}\right) + \cos \left(\dfrac{\alpha}{2}\right)\right]$
$= 2 \sqrt{3} \cos \left(\dfrac{\alpha}{2}\right) \left[\dfrac{\sqrt{3}}{2} \sin \left(\dfrac{\alpha}{2}\right) + \dfrac{1}{2} \cos \left(\dfrac{\alpha}{2}\right)\right]$
$= 2 \sqrt{3} \cos \left(\dfrac{\alpha}{2}\right) \left[\sin \left(\dfrac{\pi}{6}\right) \cos \left(\dfrac{\alpha}{2}\right) + \cos \left(\dfrac{\pi}{6}\right) \sin \left(\dfrac{\alpha}{2}\right)\right]$
$= 2 \sqrt{3} \cos \left(\dfrac{\alpha}{2}\right) \sin \left(\dfrac{\pi}{6} + \dfrac{\alpha}{2}\right)$