Without using tables, calculate $\;$ $\sin 2 \alpha$, $\;$ if $\;$ $\sin \left(\dfrac{\alpha}{2}\right) + \cos \left(\dfrac{\alpha}{2}\right) = \dfrac{-1}{2}$ $\;$ and $\alpha$ belongs to the fourth quadrant.
$\because \;$ $\alpha$ is in the fourth quadrant $\implies$ $\sin \alpha$ is negative and $\cos \alpha$ is positive
Given: $\;$ $\sin \left(\dfrac{\alpha}{2}\right) + \cos \left(\dfrac{\alpha}{2}\right) = \dfrac{-1}{2}$
i.e. $\;$ $\left[\sin \left(\dfrac{\alpha}{2}\right) + \cos \left(\dfrac{\alpha}{2}\right)\right]^2 = \left(\dfrac{-1}{2}\right)^2$
i.e. $\;$ $\sin^2 \left(\dfrac{\alpha}{2}\right) + \cos^2 \left(\dfrac{\alpha}{2}\right) + 2 \sin \left(\dfrac{\alpha}{2}\right) \cos \left(\dfrac{\alpha}{2}\right) = \dfrac{1}{4}$
i.e. $\;$ $1 + \sin \left(2 \times \dfrac{\alpha}{2}\right) = \dfrac{1}{4}$
i.e. $\;$ $\sin \alpha = \dfrac{1}{4} - 1 = \dfrac{-3}{4}$
$\therefore \;$ $\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \left(\dfrac{-3}{4}\right)^2} = \dfrac{\pm\sqrt{7}}{4}$
$\because \;$ $\alpha$ is in the fourth quadrant, $\cos \alpha = \dfrac{\sqrt{7}}{4}$
Now, $\;$ $\sin 2 \alpha = 2 \sin \alpha \cos \alpha = 2 \times \left(\dfrac{-3}{4}\right) \times \dfrac{\sqrt{7}}{4} = \dfrac{-3 \sqrt{7}}{8}$