The De Moivre’s Theorem

Question

Express cos (5θ) and sin (5θ) in terms of cos (θ) and sin (θ).

Answer

Using the De Moivre’s Theorem:

\texrm{(1)}\hspace{10px}\displaystyle{(\cos \theta &+ \mathbf{i} \sin \theta)^{n} &= \cos n\theta + \mathbf{i} \sin n\theta}

where \mathbf{i} is an imaginary number and has a value of \sqrt{&−1}, or \mathbf{i}^2 &= −1, we can write

\texrm{(2)}\hspace{10px}\displaystyle{\cos 5\theta + \mathbf{i} \sin 5\theta = (\cos \theta + \mathbf{i} \sin \theta)^{5}}

Expanding the right hand side (RHS) of the equation (2) gives

\texrm{(3)} \begin{alignat*}{2} {(\cos \theta + \mathbf{i} \sin \theta)^{5}} = & { \cos^{5} \theta + 5(\cos^{4} \theta)(\mathbf{i} \sin \theta)} \\ { } & + {10(\cos^{3} \theta)(\mathbf{i}^{2} \sin^{2} \theta) + 10(\cos^{2} \theta)(\mathbf{i}^{3} \sin^{3} \theta)} \\ { } & + {5(\cos \theta)(\mathbf{i}^{4} \sin^{4} \theta) + \mathbf{i}^{5} \sin^{5} \theta} \end{align*}

Noting that \mathbf{i}^{2} = −1, therefore, \mathbf{i}^{3} = − \mathbf{i}, \mathbf{i}^{4} = 1, and \mathbf{i}^{5} = \mathbf{i}, Eq. (3) can be written as

\texrm{(4)} \begin{alignat*}{2} {(\cos \theta + \mathbf{i} \sin \theta)^{5}} =& { \cos^{5} \theta + 5(\cos^{4} \theta)(\mathbf{i} \sin \theta)}\\ { } & - {10(\cos^{3} \theta)(\sin^{2} \theta) - 10(\cos^{2} \theta)(\mathbf{i} \sin^{3} \theta)}\\ { } & + {5(\cos \theta)(\sin^{4} \theta) + \mathbf{i} \sin^{5} \theta} \end{align*}

Substitute the left hand side (LHS) of Eq. 4 with \cos 5\theta + \mathbf{i} \sin 5\theta (De Moivre’s Theorem) and grouping the real part (real) and imaginary part (im) on the right hand side (RHS), we have

\texrm{(5)} \begin{alignat*}{2} {\cos 5\theta + \mathbf{i} \sin 5\theta} = & {\left(\cos^{5} \theta - 10\cos^{3} \theta \sin^{2} \theta+ 5\cos \theta \sin^{4} \theta\right)}\\ { } & + {\mathbf{i} \left(5\cos^{4} \theta \sin \theta - 10\cos^{2} \theta \sin^{3} \theta + \sin^{5} \theta\right)} \end{align*}

For Eq. (5) to be valid the real and im parts on the LHS and RHS of the equation must be respectively equal, that is

\texrm{(6a)} \begin{alignat*}{2} {\cos 5\theta} = & {\cos^{5} \theta - 10\cos^{3} \theta \sin^{2} \theta + 5\cos \theta \sin^{4} \theta } \end{align*}

and

\texrm{(6b)} \begin{alignat*}{2} {\sin 5\theta} = & {5\cos^{4} \theta \sin \theta - 10\cos^{2} \theta \sin^{3} \theta + \sin^{5} \theta} \end{align*}

Making use of the trigonometric identities \sin^{2} \theta = 1 – \cos^{2} \theta, and \sin^{4} \theta = (1 – \cos^{2} \theta)^{2} = 1 – 2 \cos^{2} \theta + \cos^{4} \theta, Eq. (6a) becomes

\texrm{(7)} \begin{alignat*}{2} {\cos 5\theta} = & {\cos^{5} \theta - 10\cos^{3} \theta (1 - \cos^{2} \theta) + 5\cos \theta (1 - 2\cos^{2} \theta + \cos^{4} \theta)} \end{align*}

which can be further simplified to give the expression for \cos 5\theta in terms of \cos \theta as:

\texrm{(8)} \begin{alignat*}{2} {\cos 5\theta} = & {16 \cos^{5} \theta - 20 \cos^{3} \theta + 5 \cos \theta} \end{align*}

Similarly, making use of the trigonometric identity \cos^{2} \theta = 1 – \sin^{2} \theta, expression for \sin 5\theta in terms of \sin \theta is

\texrm{(9)} \begin{alignat*}{2} {\sin 5\theta} = & {16 \sin^{5} \theta - 20 \sin^{3} \theta + 5 \sin \theta } \end{align*}