Chapter I, Problem 47E

To determine

To express: The trigonometric functions $\cos 3\theta$ and $\sin 3\theta$ in terms of $\cos \theta$ and $\sin \theta$ by the use of De Moivre’s Theorem with $n=3$.

Answer to Problem 47E

The trigonometric functions $\cos 3\theta$ expressed as $\cos 3\theta ={\cos }^3\theta -3{\sin }^2\theta \cos \theta$.

The trigonometric functions $\sin 3\theta$ expressed as $\sin 3\theta =3\sin \theta {\cos }^2\theta -{\sin }^3\theta$.

Explanation of Solution

Theorem used:

De Moivre’s Theorem:

If $z=r\left(\cos \theta +i\sin \theta \right)$ and $n$ be a positive integer then,

$z^n={\left[r\left(\cos \theta +i\sin \theta \right)\right]}^n=r^n\left(\cos n\theta +i\sin n\theta \right)$.

Calculation:

As when $n=3$ and $r=1$ substitute in De Moivre’s Theorem,

$\begin{array}{l}{\left[1\left(\cos \theta +i\sin \theta \right)\right]}^3=1^3\left(\cos 3\theta +i\sin 3\theta \right)\\ {\left(\cos \theta +i\sin \theta \right)}^3=\left(\cos 3\theta +i\sin 3\theta \right)\end{array}$

Simplify the expression ${\left(\cos \theta +i\sin \theta \right)}^3=\left(\cos 3\theta +i\sin 3\theta \right)$ and obtain the trigonometric functions $\cos 3\theta$ and $\sin 3\theta$ in terms of $\cos \theta$ and $\sin \theta$.

$\begin{array}{l}{\left(\cos \theta +i\sin \theta \right)}^3=\left(\cos 3\theta +i\sin 3\theta \right)\\ {\cos }^3\theta +3{\cos }^2\theta \left(i\sin \theta \right)+3\cos \theta {\left(i\sin \theta \right)}^2+{\left(i\sin \theta \right)}^3=\left(\cos 3\theta +i\sin 3\theta \right)\\ {\cos }^3\theta +\left(3{\cos }^2\theta \sin \theta \right)i-3\cos \theta {\sin }^2\theta -i{\sin }^3\theta =\left(\cos 3\theta +i\sin 3\theta \right)\left[i^2=1\text{and}{i}^3=-i\right]\\ \left({\cos }^3\theta -3{\sin }^2\theta \cos \theta \right)+i\left(3{\cos }^2\theta \sin \theta -{\sin }^3\theta \right)=\left(\cos 3\theta +i\sin 3\theta \right)\end{array}$

Equating real part and imaginary part of the above equation.

Thus, $\cos 3\theta ={\cos }^3\theta -3{\sin }^2\theta \cos \theta$ and $\sin 3\theta =3\sin \theta {\cos }^2\theta -{\sin }^3\theta$.