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

Question

Here $k$ is the vector along the direction of propagation $k = k_xi +k_y j + k_z k$ and $r$ is $xi +yj +zk$. I couldn't find a proof that $a\sin(kr - wt)$ or $a\cos(kr - \omega t)$ is equal to $ae^{(±i(k\cdot r ± \omega t))}$.

Hope some one can give me a simple mathematical explanation.

Answer 1

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.

Answer 2

$e^{i\phi}=\cos(\phi)+i\sin(\phi)$ the ±part is strange? but sin is just the imaginary part of $e^{i\phi}$ or $e^{i\phi}-e^{-i\phi}=i\sin(\phi)$ and sometimes it is just easier to do calculations with the exponential function