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

1 Answer

no need

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#

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Hi. I’m a junior in high school and this is a challenge problem that I was assigned for my Analysis class. We are allowed to use absolutely any sources to solve and understand it. Here it is:

Given: eix = cosx + isinx

  1. substitute -x for x to find e-ix, simplifying your answer

  2. use the given and part a to find an identity for cosx making no reference to trig functions

  3. find an identity for sinx

  4. find an identity for tanx. Then put it in a form where you are not “stacking fractions.”

  5. use your new “definitions” to confirm that cos2x + sin2x = 1 and tan2x + 1 = sec2x

  6. check that your definitions are consistent with cos2x = cos2x – sin2x and two other identities of your choice.

  7. now repeat a through e for the hyperbolic cos (coshx) and the hyperbolic sin (sinhx) defined as follows: ex = coshx + sinhx where coshx is an even function and sinhx is an odd function. (By the way, tanhx = (sinhx)/(coshx) and sechx = 1/(coshx). Your identities in part e will not be identical to those for the equivalent trig functions.

Thank you very much for your time. I look forward to hearing from you.

Thanks again,
Peter

Hi Peter,

Here is a start.

  1. substitute -x for x to find e-ix, simplifying your answer

    e-ix = cos(-x) + isin(-x) = cosx – isinx
  2. use the given and part a to find an identity for cosx making no reference to trig functions

    Add the equations

    eix = cosx + isinx
    e-ix = cosx – isinx

Cheers,
Penny
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