See the attached picture. If a point P(x, y) is on the terminal side of an angle theta.
[tex]\cos(\theta) = \frac{x}{r} [/tex]
In this case, x = -2 and r = 3. Use the Pythagorean Theorem to find the value of y.
[tex]x^2 + y^2 =r^2 \\ 4+y^2=9 \\ y^2 = 5 \\ y=\sqrt{5} [/tex]
Note that you pick the positive square root for y because the point is in Quadrant II.
Now, [tex]\sin(\theta) = \frac{y}{r} = \frac{\sqrt{5}}{3} [/tex]
