Answer
$-\frac{4\sqrt 2}{9}$
Work Step by Step
To apply the formula for $\sin 2A$, we must first find $\sin A$.
$\sin A=\pm \sqrt {1-\cos^{2}A}=\pm\sqrt {1-(-\frac{1}{3})^{2}}$
$=\pm\sqrt {\frac{8}{9}}=\pm\frac{2\sqrt 2}{3}$
Because $A$ terminates in quadrant II, $\sin A$ is positive.
$\sin A=\frac{2\sqrt {2}}{3}$
Now, $\sin 2A=2\sin A\cos A=2(\frac{2\sqrt 2}{3})(-\frac{1}{3})=-\frac{4\sqrt 2}{9}$
