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:
- Write down de Moivre’s theorem replacing n with the relevant integer.
- Expand (\cos{\theta} + i \sin{\theta})^n, remembering that i^2=-1.
- Equate real or imaginary parts.
- 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:
- 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.
- Take both sides to the power of n and expand.
- 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
- 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}.
- Prove the following identities using de Moivre’s theorem:
- \sin2\theta=2\cos\theta\sin\theta
- \cos2\theta=1+2\cos^2\theta
- \sin3\theta=3\sin\theta-4\sin^3\theta
- \sin 5\theta=5\sin{\theta}-20\sin^3{\theta}+16\sin^5{\theta}
- \cos 4\theta = 1-8\cos^2{\theta}+8\cos^4{\theta}
- Prove the following identities using de Moivre’s theorem:
- \cos^2{\theta}= \frac{1}{2} (1+\cos{2\theta})
- \sin^3{\theta}=\frac{1}{4} (3\sin{\theta}-\sin{3\theta})
- \cos^4{\theta}=\frac{1}{8} (4 \cos{2\theta} + \cos{4\theta}+3)
- \sin^5{\theta}=\frac{1}{16}(10\sin\theta-5\sin{3\theta}+\sin{5\theta})