Why is $sinx$ the imaginary part of $e^{ix}$?

Question

Most of us who are studying mathematics are familiar with the famous $e^{ix}=cos(x)+isin(x)$. Why is it that we have $e^{ix}=cos(x)+isin(x)$ and not $e^{ix}=sin(x)+icos(x)$? I haven't studied Complex Analysis to know the answer to this question. It pops up in Linear Algebra, Differential Equations, Multivariable Equations and many other fields. But I feel like textbooks and teachers just expect us students to take it as given without explaining it to a certain extent. I also couldn't find any good article that explains this.

Answer 1

The way I make sense of it is to consider the McLaurin expansions of $e^{ix}$, $\cos(x)$, and $\sin(x)$.

Note:

$$e^{x} = \sum^{\infty}_{n=0} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$

$$\sin x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$

$$\cos x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} x^{2n} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$$

Answer 2

Let $y=ix$. Then the Maclaurin series of $e^{y}$ is: $$ \begin{eqnarray} e^y=e^{ix}&=&\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}\\ &=& 1+ix-\frac{x^2}{2!}-i\frac{x^3}{3!}+\frac{x^4}{4!}+i\frac{x^5}{5!}\ldots\\ &=& \Big(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\ldots\Big)+i\Big(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots\Big)\\ &=& \Big(\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}\Big)+i\Big(\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}\Big)\\ &=&\cos x+i\sin x \end{eqnarray} $$ Hence $\sin x$ is the imaginary part of $e^{ix}$.