Answer
$\frac{\sqrt 3}{3}$
Work Step by Step
$\cos \frac{A}{2}=\pm \sqrt {\frac{1+\cos A}{2}}$
$=\pm\sqrt {\frac{1+(-\frac{1}{3})}{2}}=\pm\sqrt {\frac{1}{3}}=\pm\frac{1}{\sqrt 3}=\pm\frac{\sqrt 3}{3}$
We need to know where $\frac{A}{2}$ terminates to determine the sign of $\cos\frac{A}{2}$.
$90^{\circ}\lt A\lt180^{\circ}$
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, cosine is positive. Therefore,
$\cos\frac{A}{2}=\frac{\sqrt 3}{3}$
