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

1 Answer
Jul 9, 2018

Answer:

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#