Euler's Formula

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Euler's Formula

Euler's formula is a fundamental identity in complex numbers, linking the exponential functions and trig functions: $$ e^{i\theta} = \cos(\theta) + i\sin(\theta) $$

A complete and robust proof tends to involve aspects of complex analysis, beyond the scope of these notes, but a sketch proof can be constructed from the power series for these functions, and assuming that the extend naturally complex numbers:

$$ \begin{align} e^{i\theta} &= 1 + (i\theta) + \frac{1}{2!}(i\theta)^2 + \frac{1}{3!}(i\theta)^3 + \dots \\ &= \sum_{p=0}^\infty \frac{1}{(2p)!}(i\theta)^{2p} + \sum_{p=0}^\infty \frac{1}{(2p+1)!}(i\theta)^{2p+1} \\ &= \sum_{p=0}^\infty \frac{1}{(2p)!}(-1)^p(\theta)^{2p} + i\sum_{p=0}^\infty \frac{1}{(2p+1)!}(-1)^p(\theta)^{2p+1} \\ &= \cos(\theta) + i\sin(\theta) \end{align} $$

As an immediate consequence of Euler's formula, we have $$\begin{align}\cos(\theta) &= \frac{e^{i\theta} + e^{-i\theta}}{2} \\ \sin(\theta) &= \frac{e^{i\theta} - e^{-i\theta}}{2i}\end{align} $$

Euler's formula 1

Determine the values, $\alpha,\;\beta,\;\gamma,\;\delta$ 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

From Euler's formula, we know that $\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$. It follows that $$ \cos^6(\theta) = \left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^6 $$

Using the binomial expansion on the RHS yields: $$ \begin{align} \left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^6 &= \frac{1}{2^6}\left(e^{6i\theta} + 6e^{4i\theta} +15e^{2i\theta} + 20 + 15e^{-2i\theta} + 6e^{-4i\theta} + e^{-6i\theta}\right) \\ &= \frac{1}{2^6}\left(2\cos(6\theta) + 12\cos(4\theta) + 15\cos(2\theta) + 20\right) \end{align} $$

Euler's formula 2

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

We can proceed in a near identical way to the previous example; however, it is first worth noting that the right side is a function of cosines; this should be expected, because $\sin^6(\theta)$ is an even function, whereas $\sin(n\theta)$ is an odd function for all $n\in\mathbb{Z}$.

$$ \sin^6\theta = \left(\frac{e^{i\theta} - e^{-i\theta}}{2i}\right)^6 $$ Now apply the binomial expansion $$ \begin{align} \left(\frac{e^{i\theta} - e^{-i\theta}}{2i}\right)^6 &= \left(\frac{1}{2i}\right)^6\left(e^{6i\theta} - 6e^{4i\theta} + 15e^{2i\theta} -20 + 15e^{2i\theta} - 6e^{4i\theta} + e^{6i\theta}\right)\\ &= -\frac{1}{64}\left(2\cos(6\theta) - 12\cos(4\theta) + 30\cos(2\theta) - 20\right) \end{align} $$