De Moivre's Theorem

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De Moivre's Theorem

De Moivre's theorem can be expressed very succinctly as $$ e^{ni\theta} = \left(e^{i\theta}\right)^n $$

A more immediately useful form is $$ \cos (n\theta) + i\sin(n\theta) = \left(\cos (\theta) + i\sin (\theta)\right)^n $$

The proof of De Moivre's theorem in the second form can easily be established by induction, using the compound angle formulae

De Moivre's Theorem 1

Determine the values of the coefficients, $\alpha,\; \beta,\; \gamma,\;\delta$ such that the following is an identity $$ \cos(6\theta) \equiv \alpha \cos^6\theta + \beta \cos^4\theta + \gamma \cos^2\theta + \delta $$
solution - press button to display

Let $n=6$, then De Moivre's theorem becomes $$ \cos(6\theta) + i\sin(6\theta) = \left(\cos(\theta) + i\sin(\theta)\right)^6 $$

It follows that

$$\begin{align} \cos(6\theta) &= \Re\left(\cos(\theta) + i\sin(\theta)\right)^6\\ &= \cos^6\theta + ^6\!C_2(i)^2\cos^4\theta\sin^2\theta +^6\!C_4(i)^2\cos^2\theta\sin^4\theta + (i)^6\sin^6\theta \\ &= \cos^6\theta - 15\cos^4\theta\sin^2\theta + 15\cos^2\theta\sin^4\theta -\sin^6\theta \end{align}$$

By applying the Pythagorean identities, we can rewrite this as

$$\begin{align} \cos(6\theta) &= \cos^6\theta - 15\cos^4\theta(1-\cos^2\theta) + 15\cos^2\theta(1-\cos^2\theta)^2 -(1-\cos^2\theta)^3\\ &= 32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1\end{align} $$

 

De Moivre's Theorem 2

Determine the values $\alpha,\;\beta$ and $\gamma$ such that the following is an identity $$ \sin(6\theta) \equiv \sin(2\theta)\left(\alpha \sin^4\theta   +\beta\sin^2\theta + \gamma\right) $$
solution - press button to display

$$ \begin{align} \sin(6\theta) &= \Im\left(\cos\theta + i\sin\theta\right)^6 \\ &= ^6\!C_1\cos\theta\sin^5\theta - ^6\!C_3\cos^3\theta\sin^3\theta + ^6\!C_5\cos^5\theta\sin\theta \\ &= \cos\theta\sin\theta\left(6\sin^4\theta - 20\cos^2\theta\sin^2\theta + 6\cos^4\theta\right) \\ &= \sin(2\theta)\left(3\sin^4\theta - 10\cos^2\theta\sin^2\theta + 3\cos^4\theta\right) \\ &= \sin(2\theta)\left(16\sin^4\theta - 16\sin^2\theta + 3\right) \end{align} $$