Proof of Euler's theorem in complex numbers I am an engineer and not a mathematician, but I have used the formula stated below many number of times.$$e^{i\theta} = \cos{\theta} + i\sin{\theta}$$ Just out of curiosity, what is the proof to this theorem? I know that adding the Taylor series of the sines and the cosines and comparing it to the Taylor series of $e^{i\theta}$ is a proof, but is there some other way to go about this?

It can also be seen with series.  Robert mentioned this but here are the details.  It needs to be noted that the series expansions for $e^z$, $\sin x$, and $\cos x$, with $z \in \Bbb C$ and $x \in \Bbb R$, each have an infinite radius of convergence.  Also, these series all converge absolutely.  Absolute convergence makes it safe to add the sine and cosine series like we do in the end.
Recall:
\begin{align*}
  e^x &= \sum_{n=0}^{+\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}2 + \frac{x^3}{3!} + \cdots\\[0.3cm]
  \cos x &= \sum_{n=0}^{+\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}2 + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\\[0.3cm]
  \sin x &= \sum_{n=0}^{+\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots
\end{align*}
Therefore:
\begin{align*}
  e^{ix} &= 1 + ix + \frac{(ix)^2}2 + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \frac{(ix)^6}{6!} + \frac{(ix)^7}{7!} + \cdots\\[0.3cm]
    &= 1 + ix - \frac{x^2}2 - i\frac{x^3}{3!} + \frac{x^4}{4!} + i\frac{x^5}{5!} - \frac{x^6}{6!} - i\frac{x^7}{7!} + \cdots\\[0.3cm]
  i\sin x &= ix - i\frac{x^3}{3!} + i\frac{x^5}{5!} - i\frac{x^7}{7!} + \cdots
\end{align*}
And so:
\begin{align*}
  \cos x + i\sin x &= \left(1 - \frac{x^2}2 + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\right) + \left(ix - i\frac{x^3}{3!} + i\frac{x^5}{5!} - i\frac{x^7}{7!} + \cdots\right)\\[0.3cm]
    &= 1 + ix - \frac{x^2}2 - i\frac{x^3}{3!} + \frac{x^4}{4!} + i\frac{x^5}{5!} - \frac{x^6}{6!} - i\frac{x^7}{7!} + \cdots\\[0.3cm]
    &= e^{ix}
\end{align*}
