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 $sin(\alpha)=\frac{1}{2}, cos(\alpha)=\frac{\sqrt 3}{2}, sin(\beta)=-\frac{2\sqrt 2}{3}, cos(\beta)=\frac{1}{3}$
Step 3. $g(\alpha+\beta)=cos(\alpha+\beta)=cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)=(\frac{\sqrt 3}{2})(\frac{1}{3})-(\frac{1}{2})(-\frac{2\sqrt 2}{3})=\frac{\sqrt 3+2\sqrt 2}{6}$
