Euler's Identities

In the expansion of $e^{j\theta}$ if one replaces $\theta$ by $j \theta$ and $-j \theta$, we get the following two equations, respectively.

(1)\begin{eqnarray} e^{j \theta} & = & 1 + \frac{j \theta}{1!} + \frac{(j\theta)^2}{2!} + \frac{(j\theta)^3}{3!} + \frac{(j\theta)^4}{4!} + \frac{(j \theta)^5}{5!} + \ldots \\ e^{-j \theta} & = & 1 - \frac{j \theta}{1!} + \frac{(j\theta)^2}{2!} - \frac{(j\theta)^3}{3!} + \frac{(j\theta)^4}{4!} - \frac{(j \theta)^5}{5!} + \ldots \end{eqnarray}

From the above equations and along with the expansions of $cos \theta$ and $sin \theta$ the following relationship can be seen to be true:

(2)\begin{eqnarray} \nonumber e^{j\theta} &=& \cos\theta + j\sin\theta\\ \nonumber e^{-j\theta} &=& \cos\theta - j\sin\theta \\ \nonumber \cos \theta & = & \frac{1}{2}\left( e^{j \theta} + e^{-j \theta} \right) \\ \nonumber \sin \theta & = & \frac{1}{2j}\left( e^{j \theta} - e^{-j \theta} \right) \end{eqnarray}

page revision: 2, last edited: 22 Aug 2016 20:02