Answer
$\frac{\sqrt 3-2\sqrt 2}{6}$
Work Step by Step
1. $g(\alpha-\beta)=cos(\alpha-\beta)=cos\alpha cos\beta +sin\alpha sin\beta $
2. From the given figures, we have $sin\alpha=\frac{1}{2}, cos\alpha=\frac{\sqrt 3}{2}, sin\beta=-\frac{2\sqrt 2}{3}, cos\beta=\frac{1}{3}$
3. Thus $g(\alpha-\beta)=(\frac{\sqrt 3}{2})(\frac{1}{3})+(\frac{1}{2})(-\frac{2\sqrt 2}{3})=\frac{\sqrt 3-2\sqrt 2}{6}$
