Find the exact value of sin2θ given that cosθ= -12/13 and θ is in Quadrant II.
$\displaystyle cos(\theta) = -\frac{12}{13}$
tells us that $\displaystyle sin(\theta) = \pm \sqrt{1 - cos^2(\theta)} = \pm \frac{5}{13}$
Since we know that $\displaystyle \theta$ is in QII where sine is positive, we know to use the + sign.
Now
$\displaystyle sin(2\theta) = 2~sin(\theta)~cos(\theta)$
and you can take it from there.
-Dan