Express $\cos2\theta$ in terms of $\cos$ and $\sin$ (De Moivre's Theorem) Use De Moivre's to express $\cos2\theta$ in terms of powers of $\sin$ and $\cos$.
What I have is:
$\cos2\theta + i\sin2\theta\\
= (\cos\theta + i \sin\theta)^2\\
= \cos^2\theta + 2 \cos\theta ~i \sin\theta + (i \sin)^2\theta\\
= \cos^2\theta + i(2\cos\theta \sin\theta) - \sin^2\theta\\
= \cos^2\theta - \sin^2\theta + i(2\cos\theta \sin\theta)
$
so $\cos2\theta = \cos^2\theta - \sin^2\theta$
Is this correct?

Yes, 
$$\cos^2\theta - \sin^2\theta = (\cos\theta)(\cos\theta) - (\sin\theta)(\sin\theta)$$
$$=\cos(\theta + \theta) = \cos(2\theta)$$
