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% i sin \theta
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% cos \theta
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Leonhard Euler demonstrated a relationship between the base of the natural logarithm, sine, cosine, and the imaginary unit, which is now coined \emph{Euler's Formula}: 

\begin{nicebox}
$$ e^{i \theta} = \cos \theta + i \sin \theta $$
\end{nicebox}

%\begin{figure}[htp]
%\centering
%\includegraphics{images/euler/eulers.ps}
%\caption{ A visual look at Euler's Formula }
%\label{ fig:eulers }
%\end{figure}

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% y-axis
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% x-axis
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% i sin \theta
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% cos \theta
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% the circle
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% quadrants
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% angle line
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% angle omega
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% z = e^{i \omega}
\rput(3,2.8){$\mathbf{z} = e^{i \theta} = \cos\theta + i\sin\theta$}
		


\end{pspicture}
\end{center}

Euler's formula can be derived by looking at the complex exponential on the unit circle. The horizontal component of $e^{i \theta}$ is $ \cos \theta$, and the vertical component is $i\sin \theta$.

The formula can also be derived using infinite series.
\begin{align*}
e^\theta &= \sum \limits_{n = 0}^{\infty} \frac{\theta^n}{n!} = 1 + \theta + \frac{\theta^2}{2!} + \frac{\theta^3}{3!} + \frac{\theta^4}{4!} + \cdots \\
\sin \theta &= \sum \limits_{n = 0}^{\infty} \frac{(-1)^n\theta^{2n+1}}{(2n+1)!} = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} + \cdots \\
\cos \theta &= \sum \limits_{n = 0}^{\infty} \frac{(-1)^n\theta^{2n}}{(2n)!} = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} + \cdots
\end{align*}
Given the above formulae, we can derive $e^{i\theta}$ in terms of $\sin \theta$ and $\cos \theta$:
\begin{align*}
e^{i\theta} &= \sum \limits_{n = 0}^{\infty} \frac{(i\theta)^n}{n!} = 1 + i\theta + \frac{i^2\theta^2}{2!} + \frac{i^3\theta^3}{3!} + \frac{i^4\theta^4}{4!} + \cdots \\
e^{i\theta} &= \sum \limits_{n = 0}^{\infty} \frac{(i\theta)^n}{n!} = 1 + i\theta - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} + \frac{\theta^4}{4!} + \frac{i\theta^5}{5!} -\frac{\theta^6}{6!} - \cdots \\
e^{i\theta} &= \sum \limits_{n = 0}^{\infty} \frac{(i\theta)^n}{n!} = \left[ 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots \right] + \left[  i\theta - \frac{i\theta^3}{3!} + \frac{i\theta^5}{5!} - \cdots\right] \\
e^{i\theta} &= \sum \limits_{n = 0}^{\infty} \frac{(i\theta)^n}{n!} = \sum \limits_{n = 0}^{\infty} \frac{(-1)^n\theta^{2n}}{(2n)!} +  i\sum \limits_{n = 0}^{\infty} \frac{(-1)^n\theta^{2n+1}}{(2n+1)!} \\
e^{i \theta} &= \cos \theta + i \sin \theta
\end{align*}

We can also use this formula to come up with useful identities:

For $\cos \theta$:
\begin{align*}
e^{-i \theta} &= \cos (\theta) + i \sin(- \theta) \\
e^{-i \theta} &= \cos \theta - i \sin \theta \quad \mbox{(since cosine is even and sine is odd)} \\
e^{i \theta} + e^{-i \theta} &= (\cos \theta + i \sin \theta) +  (\cos \theta - i \sin \theta) \\
e^{i \theta} + e^{-i \theta} &= 2\cos \theta \\
\frac{e^{i \theta} + e^{-i \theta}}{2} &= \cos \theta
\end{align*}

This demonstrates that a cosine wave defined by $\cos \omega_0t = \frac{1}{2}e^{i\omega_0t} + \frac{1}{2}e^{-i\omega_0t}$ has two frequencies, namely $\omega_0$ and $-\omega_0$. The magnitude represents how much each frequency contributes. The same goes for the sine wave:

For $\sin \theta$:
\begin{align*}
e^{i \theta} - e^{-i \theta} &= (\cos \theta + i \sin \theta) -  (\cos \theta - i \sin \theta) \\
e^{i \theta} - e^{-i \theta} &= 2i\sin \theta \\
\frac{e^{i \theta} - e^{-i \theta}}{2i} &= \sin \theta
\end{align*}

For example, the A major triad defined by $x(t) = sin(\omega_0t) + sin(\omega_1t) + sin(\omega_2t)$, where
\begin{align*}
\omega_0 &= 2\pi\cdot 440 = \mbox{A} \\
\omega_1 &= 2\pi \cdot 554 = \mbox{C\#} \\
\omega_2 &= 2\pi \cdot 554 \cdot (\sqrt[12]{2})^3 = \mbox{E}
\end{align*}

This implies that there exists 6 frequencies in the A major triad!

Eulers Formula