Answer
$\frac{2\sqrt 5}{5}$
Work Step by Step
As $A$ is in QII, $\cos A$ is negative.
$\cos A=-\sqrt {1-\sin^{2}A}$
$=-\sqrt {1-(\frac{4}{5})^{2}}=-\sqrt {\frac{9}{25}}$
$=-\frac{3}{5}$
$\sin \frac{A}{2}=\pm\sqrt {\frac{1-\cos A}{2}}$
$=\pm \sqrt {\frac{1-(-\frac{3}{5})}{2}}=\pm \sqrt {\frac{4}{5}}=\pm\frac{2\sqrt 5}{5}$
We need to know where $\frac{A}{2}$ terminates to determine the sign of $\sin \frac{A}{2}$.
Given that $A$ is in QII:
Or $90^{\circ}\lt A\lt180^{\circ}$
Then $\frac{90^{\circ}}{2}\lt\frac{A}{2}\lt\frac{180^{\circ}}{2}$
That is, $45^{\circ}\lt\frac{A}{2}\lt90^{\circ}$
So $\frac{A}{2}$ terminates in the first quadrant. In the first quadrant, sine is positive. Therefore,
$\sin \frac{A}{2}=\frac{2\sqrt 5}{ 5}$
