Answer
$(cos~\theta+i~sin~\theta)^2 = cos~2\theta + i~sin~2\theta$
Work Step by Step
We can state the general form of De Moivre's theorem:
$[r~(cos~\theta+i~sin~\theta)]^n = r^n~(cos~n\theta + i~sin~n\theta)$
Then:
$(cos~\theta+i~sin~\theta)^2 = cos~2\theta + i~sin~2\theta$