Answer
$\frac{\sqrt 3-2\sqrt 2}{6}$
Work Step by Step
Step 1. Using the given figures, we have $x=\sqrt {4-1}=\sqrt 3$ and $y=\sqrt {1-1/9}=-\frac{2\sqrt 2}{3}$
Step 2. We have $sin(\alpha)=\frac{1}{2}, cos(\alpha)=\frac{\sqrt 3}{2}, tan\alpha=\frac{\sqrt 3}{3}, sin(\beta)=-\frac{2\sqrt 2}{3}, cos(\beta)=\frac{1}{3}, tan\beta=-2\sqrt 2$
Step 3. $g(\alpha-\beta)=cos(\alpha-\beta)=cos((\alpha)cos(\beta)+sin((\alpha)sin(\beta)=(\frac{\sqrt 3}{2})(\frac{1}{3})+(\frac{1}{2})(-\frac{2\sqrt 2}{3})=\frac{\sqrt 3-2\sqrt 2}{6}$