Answer
$\frac{\sqrt {10}}{10}$
Work Step by Step
$\cos A$ is positive as $A$ is in the fourth quadrant.
$\cos A=\sqrt {1-\sin^{2}A}=\sqrt {1-(-\frac{3}{5})^{2}}=\frac{4}{5}$
$\sin \frac{A}{2}=\pm\sqrt {\frac{1-\cos A}{2}}=\pm\sqrt {\frac{1-\frac{4}{5}}{2}}$
$=\pm\frac{\sqrt {10}}{10}$
As $270^{\circ}\leq A\leq360^{\circ}$,
$\frac{270^{\circ}}{2}\leq \frac{A}{2}\leq\frac{360^{\circ}}{2}$ or
$135^{\circ}\leq\frac{A}{2}\leq180^{\circ}$
$\frac{A}{2}$ terminates in the second quadrant.
The sine function is positive in the second quadrant.
So, $\sin\frac{A}{2}=\frac{\sqrt {10}}{10}$
