How do you show that ${{e}^{-ix}}=\cos x-i\sin x$?

Hint:This equation is called Euler Formula. We will prove this equation by using Taylor’s / Maclaurin’s series. The Taylor’s series of a function is an infinite sum of terms that are expressed in terms of derivatives of a function at a single point . The Taylor’s series of a real or complex – valued function $f(x)$ that is infinitely differentiable at a real or complex number $a$ is the power series
$f(a)+\dfrac{f'(a)}{1!}(x-a)+\dfrac{f''(a)}{2!}{{(x-a)}^{2}}+...$
Where, \[n!\] denotes the factorial of \[n\].

Complete step by step solution:
We will first write the identities in Taylor’s Series for all the three i.e. $\sin x$ and $\cos x$ as well as ${{e}^{x}}$.
We can write ,
$\sin x=x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}-...$
$\cos x=1-\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}-...$
and ${{e}^{x}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{4}}}{4!}+...$
To prove Euler’s Formula , we usually multiply ${{e}^{x}}$by $i$ , but in this case we will multiply ${{e}^{x}}$ by $-i$.
And then on multiplication , it will give ${{e}^{-ix}}$, then it will become :
${{e}^{-ix}}=1+(-ix)+\dfrac{{{(-ix)}^{2}}}{2!}+\dfrac{{{(-ix)}^{3}}}{3!}+\dfrac{{{(-ix)}^{4}}}{4!}+...$
On expanding , we get
$\Rightarrow$\[{{e}^{-ix}}=1-ix-\dfrac{{{x}^{2}}}{2!}-i\dfrac{{{x}^{3}}}{3!}+\dfrac{{{(-ix)}^{4}}}{4!}...\]
On factoring it , we get
$\Rightarrow$\[{{e}^{-ix}}=(1-\dfrac{{{x}^{2}}}{2!}+\dfrac{{{(-ix)}^{4}}}{4!}...)-i(x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}...)\]
From here, we can clearly see that the first part of the equation is equal to $\cos x$ and the second part is equal to $\sin x$, and hence we can replace them in the above equation to get the desired result.
i.e. \[{{e}^{-ix}}=\cos x-i\sin x\]
Hence Proved.

Note: Trigonometric form of Complex number is also calculated by the formula
$z=r(\cos \theta -i\sin \theta )$, where $r=|z|$ and $\theta =Angle(z)$.