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

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 ${e}^{- i x}$ 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.

$\left(\cos x\right) - i \left(\sin x\right)$

And expand to find...

$\cos x - i \sin x$

${e}^{- i x} = \cos x - i \sin x$