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

Hint: Given identity is one of the trigonometric forms of a complex number. And it is another form of Euler’s identity and we can prove this identity by making use of Taylor’s series or maclaurin’s series expansions of \[\sin x\] , \[\cos x\] , \[{e^x}\]

Complete step by step solution:
To prove the given identity let us first write out the identities in Taylor’s series for \[\sin x\] , \[\cos x\] , \[{e^x}\] \[\sin x = x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}}...\]
 \[\cos x = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}}...\]
 \[{e^x} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}}...\]
so we will have,
 \[{e^{ - ix}} = 1 + \left( { - ix} \right) + \dfrac{{{{\left( { - ix} \right)}^2}}}{{2!}} + \dfrac{{{{\left( { - ix} \right)}^3}}}{{3!}} + \dfrac{{{{\left( { - ix} \right)}^4}}}{{4!}}...\]
We know that the value of \[{i^2} = - 1\] , substituting it in the above expansion we get
 \[{e^{ - ix}} = 1 - ix - \dfrac{{{x^2}}}{{2!}} - i\dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}}...\]
Rearranging the terms in the above expansion, we get
 \[{e^{ - ix}} = \left( {1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}}...} \right) - i\left( {x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}}...} \right)\]
And now we can see that the first part of the equation is the expansion of \[\sin x\]
And the second part of the equation is the expansion of \[\cos x\]
So, the above equation can be written as
 \[{e^{ - ix}} = (\cos x) - i(\sin x)\]
Hence the required proof.

Note: The Taylor Series, or Taylor Polynomial, is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. A Maclaurin Polynomial, is a special case of the Taylor Polynomial, that uses zero as our single point. And in the above problem we used \[i\] it is an imaginary number.