Question

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

Answer 1

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.