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Use the Euler identity to prove sin^2x+cos^2x=1

  1. Aug 28, 2014 #1
    1. The problem statement, all variables and given/known data

    Just like my title says, we are to prove the trig identity sin^2x+cos^2x=1 using the Euler identity.

    2. Relevant equations

    Euler - e^(ix) = cosx + isinx
    trig identity - sin^2x + cos^2x = 1

    3. The attempt at a solution

    I tried solving the Euler for sinx and cosx, then plugging it into the trig identity. I just ended up with something really sloppy with multiple i's left over. I also tried squaring Euler and manipulating values that way to get an answer, but that just ended up really sloppy as well.

    If anyone has any tips, I'd really appreciate it. I spent about 30mins on this now and am just at a loss.
     
  2. jcsd
  3. Aug 28, 2014 #2

    HallsofIvy

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    I presume, then, that you got [tex]sin(x)= \frac{e^{ix}- e^{-ix}}{2i}[/tex]
    and [tex]cos(x)= \frac{e^{ix}+ e^{-ix}}{2}[/tex]

    So that [tex]sin^2(x)= \frac{e^{2ix}- 2+ e^{-2ix}}{-4}[/tex]
    and [tex]cos^2(x)= \frac{e^{2ix}+ 2+ e^{-2ix}}{4}[/tex]

    That doesn't look very "sloppy" to me. What happens when you add them? (Be careful of that "-" in the denominator of [itex]sin^2(x)[/itex].)
     
  4. Aug 28, 2014 #3
    Hmm, that is much different than what I got.

    When I solved with Euler for cosx, I got:

    cosx = e^(ix)-isinx, then I squared that and that is when it got sloppy.

    I'm not sure how you got rid of the "isinx" and got the rest of what you did in your equations?
     
  5. Aug 28, 2014 #4

    Orodruin

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    I suggest an alternative route. If
    $$
    e^{ix} = \cos x + i \sin x
    $$
    what is then ##e^{-ix}##?
     
  6. Aug 28, 2014 #5
    I'm sorry, but I'm not sure where that is going? Do you mean just 1/(cosx +isinx)? I don't see what to do with that? My problem is I keep ending up with "i" somewhere in my equations that I can't get rid of.
     
  7. Aug 28, 2014 #6

    HallsofIvy

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    No, he means replace "x" with "-x". If [itex]e^{ix}= cos(x)+ isin(x)[/itex] then [itex]e^{-ix}= cos(-x)+ isin(-x)= cos(x)- isin(x)[/itex]. Now you can solve for cos(x) in terms of [itex]e^{ix}[/itex] and [itex]e^{-ix}[/tex] by adding those two equation and solve for sin(x) by subtracting.
     
  8. Aug 28, 2014 #7

    olivermsun

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    He means if ##e^{ix} = \cos x + i \sin x,##

    then ##e^{-ix} = \cos (-x) + i \sin (-x) = \cos (x) - i \sin (x).##

    Then you have enough pieces to do some creative assembly and get the identity in question.

    Edit: Never mind, Halls just posted.
     
    Last edited: Aug 28, 2014
  9. Aug 28, 2014 #8
    Oh, thank you guys. I didn't know I was able to do that. Clearly I have some things to learn this semester. Now it's a piece of cake. I appreciate the help.
     
  10. Aug 29, 2014 #9

    Orodruin

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    No, I intended for him to multiply them. Solving for the cos and sin seems like an unnecessary detour.
     
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