1. Problem: Given $\cos \theta = -\frac{15}{17}$ and $\theta$ is in Quadrant II, find $\sin(2\theta)$. 2. Formula: Use the double-angle identity for sine: $$\sin(2\theta) = 2 \sin \theta \cos \theta$$ 3. Since $\cos \theta = -\frac{15}{17}$ and $\theta$ is in QII, $\sin \theta$ is positive. Use Pythagorean identity: $$\sin^2 \theta + \cos^2 \theta = 1$$ $$\sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(-\frac{15}{17}\right)^2 = 1 - \frac{225}{289} = \frac{64}{289}$$ $$\sin \theta = \sqrt{\frac{64}{289}} = \frac{8}{17}$$ 4. Calculate $\sin(2\theta)$: $$\sin(2\theta) = 2 \times \frac{8}{17} \times \left(-\frac{15}{17}\right) = 2 \times \frac{8 \times (-15)}{289} = \frac{-240}{289}$$ 5. Final answer: $$\sin(2\theta) = -\frac{240}{289}$$