# Trigonometric identities with de Moivre’s theorem

## Multiple angle formulas

We can use de Moivre’s theorem to express $$\sin{n\theta}$$ or $$\cos{n\theta}$$ as powers of $$\sin{\theta}$$ or $$\cos{\theta}$$. The steps to do this are:

1. Write down de Moivre’s theorem replacing $$n$$ with the relevant integer.
2. Expand $$(\cos{\theta} + i \sin{\theta})^n$$, remembering that $$i^2=-1$$.
3. Equate real or imaginary parts.
4. Use the identity $$\sin^2{\theta} + \cos^2{\theta}= 1$$ to simplify.

Worked exercise: Given that $$(x+y)^5 = x^5 + 5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5,$$ show that $$\cos{5\theta} = 5 \cos{\theta}-20\cos^3{\theta}+16\cos^5{\theta}.$$
Solution: By de Moivre’s theorem, we have \begin{align*} \cos 5\theta + i \sin 5\theta &=(\cos\theta + i \sin\theta)^5\\ &= \cos^5 \theta +5i \cos^4 \theta \sin \theta-10 \cos^3 \theta \sin^2 \theta\\ &\quad -10 i\cos^2 \theta \sin^3 \theta + 5 \cos\theta \sin^4 \theta + i \sin^5 \theta.\end{align*} Taking the real part of both sides and using the Pythagorean identity, we have \begin{align*} \cos 5\theta &= \cos^5{\theta}-10 \cos^3 \theta \sin^2 \theta+5 \cos\theta \sin^4 \theta \\ &= \cos^5 \theta-10 \cos^3 \theta (1-\cos^2 \theta) +5 \cos\theta (1-\cos^2 \theta)^2 \\ &= 5 \cos{\theta}-20\cos^3{\theta}+16\cos^5{\theta}. \end{align*}

## Power reduction formulas

We can also express $$\sin^n \theta$$ and $$\cos^n \theta$$ in terms of sines and cosines or multiple angles. To do this we need to know an identity.

Theorem 1. If $$z=\cos\theta+i\sin\theta$$ and $$m$$ is an integer, $$z^m+\frac{1}{z^m}=2\cos{m\theta}$$ and $$z^m-\frac{1}{z^m}=2i\sin{m\theta}.$$

Proof. By de Moivre’s theorem, we have $$z^m = \cos m\theta+i\sin m\theta$$ and \begin{align*} z^{-m} &= \cos(-m\theta)+i\sin (-m\theta) \\ &= \cos m\theta-i \sin m\theta. \end{align*} Taking the sum and the differences gives our result.

Using these identities, we can determine $$\cos^n \theta$$ and $$\sin^n \theta$$ by:

1. Let $$m=1$$ in Theorem 1 to see that $$z+\frac{1}{z}=2\cos{\theta}$$ or $$z-\frac{1}{z}=2i \sin{\theta}$$ depending on whether you are looking for $$\cos^n \theta$$ or $$\sin^n \theta$$.
2. Take both sides to the power of $$n$$ and expand.
3. Repeatedly apply Theorem 1 to simplify.

This method is better explained through an example

Example. We will prove that $$\sin^4 \theta = \frac{1}{8} (3-4\cos 2\theta + \cos 4\theta).$$ Start by noting that $$z-\frac{1}{z}=2i \sin{\theta}$$, and take both sides to the power of $$4$$ to see that \begin{align*} (2i \sin{\theta})^4&=\left(z-\frac{1}{z}\right)^4\\ 16\sin^4 \theta &=z^4-4z^2 +6-\frac{4}{z^2} +\frac{1}{z^4} \\ &=\left(z^4+\frac{1}{z^4}\right)-4\left(z^2+\frac{4}{z^2}\right) +6 .\end{align*} We now apply Theorem 1 with $$m=2$$ and $$m=4$$ to see that \begin{align*} 16\sin^4 \theta &= 2\cos 4 \theta-8\cos{2\theta}+6 \\ \sin^4 \theta&= \frac{1}{8} (3-4\cos 2\theta + \cos 4\theta).\end{align*}

## Problems

1. Prove that if $$z =\cos\theta+i\sin\theta$$, then $$z^m+\frac{1}{z^m}=2\cos{m\theta}$$ and $$z^m-\frac{1}{z^m}=2\sin{m\theta}$$.
2. Prove the following identities using de Moivre’s theorem:
1. $$\sin2\theta=2\cos\theta\sin\theta$$
2. $$\cos2\theta=1+2\cos^2\theta$$
3. $$\sin3\theta=3\sin\theta-4\sin^3\theta$$
4. $$\sin 5\theta=5\sin{\theta}-20\sin^3{\theta}+16\sin^5{\theta}$$
5. $$\cos 4\theta = 1-8\cos^2{\theta}+8\cos^4{\theta}$$
3. Prove the following identities using de Moivre’s theorem:
1. $$\cos^2{\theta}= \frac{1}{2} (1+\cos{2\theta})$$
2. $$\sin^3{\theta}=\frac{1}{4} (3\sin{\theta}-\sin{3\theta})$$
3. $$\cos^4{\theta}=\frac{1}{8} (4 \cos{2\theta} + \cos{4\theta}+3)$$
4. $$\sin^5{\theta}=\frac{1}{16}(10\sin\theta-5\sin{3\theta}+\sin{5\theta})$$