If \(\cos(\theta) = -\frac{1}{3}\), \(\theta\) in Quadrant III, find the exact value of \(\sin(2\theta)\).

Step 1 ：Step 1: Use the Pythagorean Identity \(\sin^2(\theta) + \cos^2(\theta) = 1\) to find \(\sin(\theta)\)

Step 2 ：Since \(\cos(\theta) = -\frac{1}{3}\) and \(\theta\) is in Quadrant III where sine is negative, solving for \(\sin(\theta)\), we get \(\sin(\theta) = -\sqrt{1 - \left(-\frac{1}{3}\right)^2} = -\sqrt{1 - \frac{1}{9}} = -\sqrt{\frac{8}{9}} = -\frac{2\sqrt{2}}{3}\)

Step 3 ：Step 2: Use the double angle identity \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\) to find \(\sin(2\theta)\)

Step 4 ：Substitute \(\sin(\theta) = -\frac{2\sqrt{2}}{3}\) and \(\cos(\theta) = -\frac{1}{3}\) into the identity, we get \(\sin(2\theta) = 2\left(-\frac{2\sqrt{2}}{3}\right)\left(-\frac{1}{3}\right) = \frac{4\sqrt{2}}{9}\)