Question

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

Answer 1

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`