Answer
$-\dfrac{\sqrt{10}}{10}$
Work Step by Step
$\because A$ is in $QIII \hspace{30pt} \therefore \cos{A}$ is negative
$\cos{A} = -\sqrt{1-\sin^2{A}} = -\sqrt{1-(-\dfrac{3}{5})^2} = -\dfrac{4}{5}$
$180 < A < 270$
$90 < \dfrac{A}{2} < 135$
Using the half angle formula
$\cos{\dfrac{A}{2}}= -\sqrt{\dfrac{1+\cos{A}}{2}} = -\sqrt{\dfrac{1-\dfrac{4}{5}}{2}}$
$\cos{\dfrac{A}{2}} = -\dfrac{\sqrt{10}}{10}$
