Prove DeMoivre's theorem using complex numbers

Prove DeMoivre’s theorem,
$(\cos \theta + i \sin \theta)^n = \cos (n \theta) + i \sin (n \theta),$

for all $\theta \in \mathbb{R}$ and all $n \in \mathbb{Z}_{>0}$ .
Prove the triple angle formulas for sine and cosine,
$\sin (3 \theta) = 3 \cos^2 \theta \sin \theta - \sin^3 \theta, \qquad \cos (3 \theta) = \cos^3 \theta - 3 \cos \theta \sin^2 \theta,$

by letting $n = 3$ in part (a).

Proof. Since $\cos \theta + i \sin \theta = e^{i \theta}$ we have
$(\cos \theta + i \sin \theta)^n = (e^{i \theta})^n = e^{ni \theta} = \cos (n \theta) + i \sin (n \theta). \qquad \blacksquare$
Letting $n = 3$ , we first apply DeMoivre’s theorem to get
$(\cos \theta + i \sin \theta)^3 = \cos (3 \theta) + i \sin (3 \theta).$

On the other hand, we can expand the product,

$(\cos \theta + i \sin \theta)^3 = \cos^3 \theta + 3 i \cos^2 \theta \sin \theta - 3 \cos \theta \sin^2 \theta - i \sin^3 \theta$

Equating real and imaginary parts from the two expressions we obtain the requested identities:

$\cos (3 \theta) = \cos^3 \theta - 3 \cos \theta \sin^2 \theta \quad \text{and} \quad \sin (3 \theta) = 3 \cos^2 \theta \sin \theta - \sin^3 \theta.$