De Moivre's formula

Prerequisite knowledge:
Expressing $e^x, \sin(x), \cos(x)$ as infinite series
Basic trigonometry
Complex number
Binomial theorem

$$\begin{align}e^{i\theta}&=1+i\theta+\dfrac{(i\theta)^2}{2!}+\dfrac{(i\theta)^3}{3!}+\dfrac{(i\theta)^4}{4!}+\dfrac{(i\theta)^5}{5!}+\cdots\\
&=1+i\theta-\dfrac{\theta}{2!}-\dfrac{i\theta}{3!}+\dfrac{\theta^4}{4!}+\dfrac{i\theta^5}{5!}+\cdots\\
&=(1-\dfrac{\theta^2}{2!}+\dfrac{\theta^4}{4!}-\cdots)+i(\theta-\dfrac{\theta^3}{3!}+\dfrac{\theta^5}{5!}+\cdots)\\
&=\cos\theta+i\sin\theta \end{align}$$
Similarly, $e^{in\theta}=\cos n\theta+i\sin n\theta=(\cos\theta+i\sin\theta)^n$.

Using Binomial theorem, we have $$\begin{align}&(\cos\theta+i\sin\theta)^n\\&=\cos^n\theta+n\cos^{n-1}\theta\:i\sin\theta+\frac{n(n-1)}{2}\cos^{n-2}\theta\:i^2\sin^2\theta+\cdots\\&=\cos^n\theta-\frac{n(n-1)}{2}\cos^{n-2}\theta\sin^2\theta+\frac{n(n-1)(n-2)(n-3)}{4!}\cos^{n-4}\theta\sin^4\theta+\cdots\\&+i(n\cos^{n-1}\theta\sin\theta-\frac{n(n-1)(n-2)}{3!}\cos^{n-3}\theta\sin^3\theta+\cdots)\\&=\cos n\theta+i\sin n\theta.\end{align}$$ Therefore, $$\begin{cases}\cos n\theta=\cos^n \theta-\frac{n(n-1)}{2}\cos^{n-2}\theta\sin^2\theta+\cdots \\ \sin n\theta=n\cos^{n-1}\theta\sin\theta-\frac{n(n-1)(n-2)}{3!}\cos^{n-3}\theta\sin^3\theta+\cdots.\end{cases}$$

$z=e^{i\theta}=\cos\theta+i\sin\theta,\qquad \frac{1}{z}=e^{-i\theta}=\cos\theta-i\sin\theta\\ z+\frac{1}{z}=2\cos\theta,\qquad (z+\frac{1}{z})^n=2^n\cos^n\theta\\ z-\frac{1}{z}=2i\sin\theta,\qquad (z-\frac{1}{z})^n=i^n\:2^n\sin^n\theta\\ \star z^n+\frac{1}{z^n}=2\cos n\theta\\ \star z^n-\frac{1}{z^n}=2i\sin n\theta$

Application of De Moivre's formula:
1. Proving trigonometric identities
Example 1:


Example 2:


Example 3:
Express $\cos^5 \theta$ in terms of $\cos\theta$.
Recall $$2^n\cos^n\theta=(z+\frac{1}{z})^n.$$
Substitute $n=5$, $$2^5\cos^5\theta=(z+\frac{1}{z})^5.$$
Using Binomial theorem, $$\begin{align}(z+\frac{1}{z})^5&=z^5+5z^3+10z+\frac{10}{z}+\frac{5}{z^3}+\frac{1}{z^5}\\&=z^5+\frac{1}{z^5}+5(z^3+\frac{1}{z^3})+10(z+\frac{1}{z}).\end{align}$$ Therefore, $\cos^5\theta=\dfrac{1}{2^4}(\cos 5\theta+5\cos 3\theta+10 \cos\theta)$.

2. Finding complex roots
[later]