Without using tables, find $\;$ $\sin \left(\dfrac{\alpha}{2}\right)$, if $\sin \alpha = \dfrac{-5}{13}$ and $\pi < \alpha < \dfrac{3 \pi}{2}$
Given: $\;$ $\pi < \alpha < \dfrac{3 \pi}{2}$ $\implies$ $\dfrac{\pi}{2} < \dfrac{\alpha}{2} < \dfrac{3 \pi}{4}$
$\implies$ $\alpha$ lies in the third quadrant $\implies$ Both $\sin \alpha$ and $\cos \alpha$ are negative.
And, $\;$ $\dfrac{\alpha}{2}$ lies in the second quadrant $\implies$ $\sin \left(\dfrac{\alpha}{2}\right)$ is positive.
Given: $\;$ $\sin \alpha = \dfrac{-5}{13}$
$\therefore \;$ $\cos \alpha = \sqrt{1- \sin^2 \alpha} = \sqrt{1 - \left(\dfrac{-5}{13}\right)^2} = \dfrac{\pm 12}{13}$
$\because \;$ $\alpha$ is in the third quadrant $\implies$ $\cos \alpha = \dfrac{-5}{13}$
Now, $\;$ $\cos \alpha = 1 - 2 \sin^2 \left(\dfrac{\alpha}{2}\right)$
$\therefore \;$ $\sin^2 \left(\dfrac{\alpha}{2}\right) = \dfrac{1 - \cos \alpha}{2} = \dfrac{1 - \left(\dfrac{-12}{13}\right)}{2} = \dfrac{25}{26}$
$\implies$ $\sin \left(\dfrac{\alpha}{2}\right) = \dfrac{\pm 5}{\sqrt{26}}$
$\because \;$ $\dfrac{\alpha}{2}$ is in the second quadrant $\implies$ $\sin \left(\dfrac{\alpha}{2}\right) = \dfrac{+5}{\sqrt{26}}$