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

no need

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…

$\cos x - i \sin x$

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

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