<s>[INST] Taylor series expansion for $e^{-x}$ could anyone show me the Taylor series expansion for $e^{-x}$.I was trying to find out how
$e^{-i\theta}$=$\cos\theta-i\sin\theta$.
More specifically could you show me how $e^{-i\theta}$=$\cos\theta-i\sin\theta$ is obtained from Taylor series.
 [/INST] $$ e^{-x} = \sum\limits_{k=0}^{\infty}\frac{(-x)^k}{k!} = 1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}+\dots $$
So plugging in $x=i\theta$ we have that
\begin{align}
e^{-i\theta} &=1-i\theta+\frac{(i\theta)^2}{2!}-\frac{(i\theta)^3}{3!}+\frac{(i\theta)^4}{4!}+\dots\\
&=1-i\theta-\frac{\theta^2}{2!}+i\frac{\theta^3}{3!}+\frac{\theta^4}{4!}+\dots
\end{align}
using the fact that $i^2=-1$, $i^3=-i$ and $i^4=1$, etc.
The taylor expansions of $\sin$ and $\cos$ are 
$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots\hspace{10px}\text{and}\hspace{10px}\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots $$
So
\begin{align}
e^{-i\theta} &=\left(1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\dots\right)-i\left(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\dots\right)\\
&=\cos\theta-i\sin\theta.
\end{align}
As a side-note if you already know that $e^{i\theta}=\cos\theta+i\sin\theta$, then it is easy to show that $e^{-i\theta}$ without using taylor-series using the fact that cosine is even and sine is odd. That is
$$e^{-i\theta} = \cos(-\theta)+i\sin(-\theta)=\cos\theta-i\sin\theta, $$
because $\cos(x)=\cos(-x)$ and $\sin(-x)=-\sin(x).$
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