Euler's Formula

Q1: How can $e^{j\theta}$ be expressed in terms of trigonometric functions?
A1: $e^{j\theta}$ can be simplified using a Maclaurin series:
Let $\theta \to x$:
$\Rightarrow e^{j\theta} \to e^{jx}$
$e^{jx}=\sum\limits_{n=0}^n \frac {j^nx^n}{n!}$
Expanding the Maclaurin series:
 $e^{jx}=1+jx-\frac{x^2}{2}-j\frac{x^3}{3!}+\frac{x^4}{4!}+j\frac{x^5}{5!}-\frac{x^6}{6!}-j\frac{x^7}{7!}+...$
Collecting real and imaginary coefficients:
 $e^{jx}=[1-\frac{x^2}{2}+\frac{x^4}{4!}-\frac{x^6}{6!}+...]+[j(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...)]$
 Note that the expressions in the square brackets are the Maclaurin expansions of $cosx$ and $sinx$

$e^{jx}=cosx+jsinx$
$e^{j\theta}=cos\theta+jsin\theta$

Q2: What's the expression for $e^{-j\theta}$?
A2: Let $j \to -j$:
$e^{-j\theta}=cos\theta+jsin\theta$

Q3: Is there one equation that encapsulates the results from A1 and A2?
A3: $e^{\pm j\theta}=cos\theta\pm jsin\theta$