How is $φ =a \sin(k\cdot r - \omega t)$ converted to $φ = ae^{(±i(k\cdot r ± \omega t) )}$ using Fourier transform? [closed]

The Euler constant is defined as: \begin{equation} e^{x}=\sum_{n=0}^\infty \frac{x^n}{n!}. \end{equation} Now substitute $x=i\theta$ ,into the equation, grouping real terms and imaginary terms: \begin{equation} e^{i\theta}=\sum_{n=0}^\infty\frac{(-1)^{n}\theta^{2n}}{(2n)}+i\sum_{n=0}^\infty\frac{(-1)^n\theta^{2n+1}}{(2n+1)!} \end{equation} You can recognise the real term and imaginary term is cosine and sine taylor series expansion correspondingly.

Therefore you arrive to the famous Euler-equation: \begin{equation} e^{i\theta}=\cos\theta+i\sin\theta \end{equation} performing Taylor expansion of $e^{-i\theta}$ yield similar result: \begin{equation} e^{-i\theta}=\cos\theta-i\sin\theta. \end{equation}

Thus, using these expression you obtain: \begin{equation} Ae^{i(\vec{k}\cdot\vec{r} \pm \omega t)}=A\cos(\vec{k}\cdot\vec{r} \pm \omega t)\pm iA\sin(\vec{k}\cdot\vec{r} \pm \omega t), \end{equation} which means taking the imaginary part of the complex exponential function convert to the sine wave function.