Answer
$\frac{10-2\sqrt 5}{15}$
Work Step by Step
Step 1. Use the Addition Formula, $cos(\alpha+\beta)=cos\alpha cos\beta - sin\alpha sin\beta$
Step 2. For the left triangle shown in the figure of the problem, we have $sin\alpha=\frac{1}{\sqrt {2^2+1^2}}=\frac{\sqrt 5}{5}$ and $cos\alpha=\frac{2}{\sqrt {2^2+1^2}}=\frac{2\sqrt 5}{5}$
Step 3. Similarly, for the right triangle in the figure, we have $sin\beta=\frac{2}{3}$ and $cos\beta=\frac{\sqrt {3^2-2^2}}{3}=\frac{\sqrt 5}{3}$
Step 4. Use the results of Steps 2 and 4, we get from Step 1 that $cos(\alpha+\beta)=\frac{2\sqrt 5}{5}\times\frac{\sqrt 5}{3}-\frac{\sqrt 5}{5}\times \frac{2}{3}=\frac{10-2\sqrt 5}{15}$
