How do you show that e^(-ix)=cosx-isinx?

1 Answer
Jul 9, 2018

You can prove this using Taylor's/Maclaurin's Series.

Explanation:

First write out the identities in Taylor's Series for sin x and cos x as well as e^x.

sin x = x-x^3/(3!)+x^5/(5!)...

cos x = 1-x^2/(2!)+x^4/(4!)...

e^x = 1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)...

Usually to prove Euler's Formula you multiply e^x by i, in this case we will multiply e^x by -i.

And we will end with e^(-ix) thus it will be equal to...
1+(-ix)+(-ix)^2/(2!)+(-ix)^3/(3!)+(-ix)^4/(4!)...

Expand...

1-ix-x^2/(2!)-ix^3/(3!)+x^4/(4!)...

Factorise it...

(1-x^2/(2!)+x^4/(4!)...) -i(x-x^3/(3!)+x^5/(5!)...)

And the first part of the equation is equal to cos x and the second part to sin x, now we can replace them.

(cos x) -i(sin x)

And expand to find...

cos x -isin x

Tada, proof...

e^(-ix) = cos x -isin x