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. \(\sin^5{\theta}=\frac{1}{16}(10\sin\theta-5\sin{3\theta}+\sin{5\theta}) \)