Euler’s formula

Proof

Here is a beautiful proof of this formula. For a real number $x$, let \begin{align} f(x)&=e^{-ix}(\cos x+i\sin x)\\ \frac{d}{dx}f(x)&=\frac{d}{dx}\left[e^{-ix}(\cos x+i\sin x)\right]\notag\\ &=\frac{d}{dx}\left(e^{-ix}\cos x\right)+i\frac{d}{dx}\left(e^{-ix}\sin x\right)\notag\\ &=\cos x\frac{d}{dx}e^{-ix}+e^{-ix}\frac{d}{dx}\cos x+i\sin x\frac{d}{dx}e^{-ix}+ie^{-ix}\frac{d}{dx}\sin x\notag\\ &=-ie^{-ix}\cos x-e^{-ix}\sin x+e^{-ix}\sin x+ie^{-ix}\cos x\notag\\ &=0. \end{align} Since $\frac{d}{dx}f(x)=0$, $f(x)$ is a constant for all $x$. For $x=0$, \begin{equation} f(0)=e^{-i0}(\cos 0+i\sin 0)=1. \end{equation} Now, \begin{align} f(x)&=e^{-ix}(\cos x+i\sin x)=1\\ e^{ix}&=\cos x+i\sin x. \end{align}

From Eq. \eqref{eq:euler}, we can also obtain \begin{align} \cos x&=\frac{e^{ix}+e^{-ix}}{2} \label{eq:cos}\\ \sin x&=\frac{e^{-ix}-e^{ix}}{2}i. \label{eq:sin} \end{align}