Euler's Equation states that: \[e^{ix}=\cos x + i\sin x\] (where \(i=\sqrt{-1}\) )

Proof

If we consider the Teylor expansions of \(e^x\), \(\cos x\) and \(\sin x\), we see that the latter two almost combine to make the former, with some differences in the signs of the terms:

\[ e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+\cdots\]

\[\hspace{0.5cm} \cos x=1\hspace{1cm}-\frac{x^2}{2!}\hspace{1.2cm}+\frac{x^4}{4!}\hspace{1.2cm}-\frac{x^6}{6!}\hspace{1cm}+\cdots\]

\[\sin x=\hspace{1cm}x\hspace{1.2cm}-\frac{x^3}{3!}\hspace{1.2cm}+\frac{x^5}{5!}\hspace{1.2cm}+\cdots \hspace{0.7cm}\]

Notice however, that if we substitute \(ix\) for \(x\) in the Taylor expansion of \(e^x\), we get the correct signs:

\[ e^{ix}=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\frac{(ix)^4}{4!}+\frac{(ix)^5}{5!}+\frac{(ix)^6}{6!}+\cdots\]

so:

\[ e^{ix}=1+ix-\frac{x^2}{2!}-\frac{ix^3}{3!}+\frac{x^4}{4!}+\frac{ix^5}{5!}-\frac{x^6}{6!}+\cdots\]

\[\hspace{0.5cm} \cos x=1\hspace{1cm}-\frac{x^2}{2!}\hspace{1.2cm}+\frac{x^4}{4!}\hspace{1.2cm}-\frac{x^6}{6!}\hspace{1cm}+\cdots\]

\[\sin x=\hspace{1cm}x\hspace{1.2cm}-\frac{x^3}{3!}\hspace{1.2cm}+\frac{x^5}{5!}\hspace{1.2cm}+\cdots \hspace{0.7cm}\]

i.e.    `e^{ix}=cos x +i sin x`

\[Q.E.D.\]