Further Complex Numbers and Trigonometry

Exponential Form of Trigonometric Functions

From Euler’s formula, substituting in $-\theta$,

$$\begin{array}{rl}{e}^{i\theta }& =\cos (\theta )+i\sin (\theta )\\ e^{-i\theta }& =\cos (-\theta )+i\sin (-\theta )\\ & =\cos (\theta )-i\sin (\theta )\end{array}$$

By combining these equations, we can form alternative equations for sine and cosine:

$$\begin{array}{rl}{e}^{i\theta }+e^{-i\theta }& =2\cos (\theta )\\ \cos (\theta )& =\frac{e^{i\theta }+e^{-i\theta }}{2}\end{array}\begin{array}{rl}{e}^{i\theta }-e^{-i\theta }& =2i\sin (\theta )\\ \sin (\theta )& =\frac{e^{i\theta }-e^{-i\theta }}{2i}\end{array}$$

This can be generalised; if we let $z=e^{i\theta }$ then $z^n=e^{in\theta }$ which leads to

$$\begin{array}{rl}{e}^{in\theta }+e^{-in\theta }& =2\cos (n\theta )\\ 2\cos (n\theta )& =z^n+\frac{1}{z^n}\end{array}\begin{array}{rl}{e}^{in\theta }-e^{-in\theta }& =2i\sin (n\theta )\\ 2i\sin (n\theta )& =z^n-\frac{1}{z^n}\end{array}$$

Multiple angle formulae

A complex number can be raised to a power both using the binomial expansion and de Moivre’s Theorem. These two results can then be equated and the real and imaginary components used to find multiple angle formulae for sine and cosine.

Example finding formulae for $\cos (3\theta )$ and $\sin (3\theta )$

Let $z=\text{cis}\theta =\cos \theta +i\sin \theta$. Using the binomial expansion:

$$\begin{array}{rl}{z}^3& =(\cos \theta +i\sin \theta {)}^3\\ & ={\cos }^3\theta +3i{\cos }^2\theta \sin \theta -3\cos \theta {\sin }^2\theta -i{\sin }^3\theta \end{array}$$

Using de Moivre’s Theorem:

$$z^3=\text{cis}(3\theta )=\cos (3\theta )+i\sin (3\theta )$$

Equating real and imaginary parts:

$$\begin{array}{rl}\cos (3\theta )& ={\cos }^3\theta -3\cos \theta {\sin }^2\theta \\ & ={\cos }^3\theta +3{\cos }^3\theta -3\cos \theta \\ & =4{\cos }^3\theta -3\cos \theta \end{array}\begin{array}{rl}\sin (3\theta )& =3{\cos }^2\theta \sin \theta -{\sin }^3\theta \\ & =3\sin \theta -3{\sin }^3\theta -{\sin }^3\theta \\ & =3\sin \theta -4{\sin }^3\theta \end{array}$$

Powers of trigonometric functions

The general exponential form of trigonometric functions can be expanded and collected in order to find powers of trigonometric functions in terms of multiple angle functions.

Example finding formulae for ${\cos }^5\theta$

Let $z=\text{cis}\theta$. Using the binomial expansion:

$$\begin{array}{rl}{\left(z+\frac{1}{z}\right)}^5& =z^5+5z^4\left(\frac{1}{z}\right)+10z^3{\left(\frac{1}{z}\right)}^2+10z^2{\left(\frac{1}{z}\right)}^3+5z{\left(\frac{1}{z}\right)}^4+{\left(\frac{1}{z}\right)}^5\\ & =z^5+5z^3+10z+10\left(\frac{1}{z}\right)+5\left(\frac{1}{z^3}\right)+\left(\frac{1}{z^5}\right)\\ & =\left(z^5+\frac{1}{z^5}\right)+5\left(z^3+\frac{1}{z^3}\right)+10\left(z+\frac{1}{z}\right)\\ & \\ (2\cos (\theta ){)}^5& =2\cos (5\theta )+5(2\cos (3\theta ))+10(2\cos (\theta ))\\ 32{\cos }^5(\theta )& =2\cos (5\theta )+10\cos (3\theta )+20\cos (\theta )\\ co{s}^5(\theta )& =\frac{1}{16}\cos (5\theta )+\frac{5}{16}\cos (3\theta )+\frac{5}{8}\cos (\theta )\end{array}$$

A similar method can be used by expanding ${\left(z-\frac{1}{z}\right)}^n$ to find ${\sin }^n(\theta )$.

Trigonometric Series

Some sums of trigonometric series can be calculated using the sum of the geometric series formed by the exponential form of the trigonometric functions (if one is formed).

Example sum of a trigonometric series

Show that $\sin (x)+\sin (2x)+\sin (3x)+\cdots +\sin (10x)=\frac{\sin (x)+\sin (10x)-\sin (11x)}{4{\sin }^2(\frac{x}{2})}$

Consider the geometric series

$$a=e^{ix}r=e^{ix}$$

We can sum the first 10 terms to get

$$e^{ix}+e^{2ix}+e^{3ix}+\cdots +e^{10ix}=\frac{e^{ix}(1-e^{10ix})}{1-e^{ix}}$$

As $\sin (nx)=\text{Im}(e^{inx})$,

$$\begin{array}{rl}\sin (x)+\sin (2x)+\sin (3x)+\cdots +\sin (10x)& =\text{Im}(e^{ix}+e^{2ix}+e^{3ix}+\cdots +e^{10ix})\\ & =\text{Im}\left(\frac{e^{ix}(1-e^{10ix})}{1-e^{ix}}\right)\\ & =\text{Im}\left(\frac{e^{ix}(1-e^{10ix})}{1-e^{ix}}\times \frac{1-e^{-ix}}{1-e^{-ix}}\right)\\ & =\text{Im}\left(\frac{e^{ix}-1-e^{11ix}+e^{10ix}}{1-e^{ix}-e^{-ix}+1}\right)\\ & =\text{Im}\left(\frac{e^{ix}-1-e^{11ix}+e^{10ix}}{2-2\cos x}\right)\\ & =\frac{\sin x-\sin 11x+\sin 10x}{2-2\cos x}\\ & =\frac{\sin x-\sin 11x+\sin 10x}{4{\sin }^2\left(\frac{x}{2}\right)}\end{array}$$