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Simplify complex number expression into acos(wt+x)

  1. Sep 6, 2011 #1

    sharks

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    1. The problem statement, all variables and given/known data
    I could not type it here, so i made a screenshot and posted it below.
    d4fcb5f5e44c282602483486b9542237.jpg

    2. Relevant equations
    it's about complex numbers undergraduate level. i'm currently doing Euler's formula and De Moivre's theorem, although i'm not sure that the solution lies there.


    3. The attempt at a solution
    i tried doing it by adding the values inside the cos() but when verifying via calculator with a dummy data, it didn't add up.
     
  2. jcsd
  3. Sep 6, 2011 #2

    tiny-tim

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    hi sharks! :smile:

    try expaning each cos into the form Acos100t + Bsin100t :wink:
     
  4. Sep 6, 2011 #3

    sharks

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    Hi tiny-tim, thanks for your time to help me out. :smile:

    OK, i did the expansion and got this:
    (5cos25-3cos38+7cos40)cos100t + (-5sin25+3sin38-7sin40)sin100t.

    It's in the form that you said above. But then i'm stuck again. How to get the answer in the final form?
     
  5. Sep 6, 2011 #4

    tiny-tim

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    use the same Acos100t + Bsin100t formula backwards! :wink:

    (btw, i think your 3sin38° may have the wrong sign)
     
  6. Sep 6, 2011 #5

    sharks

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    I tripled-checked and the expansion is correct, as:
    -3cos(100t-38) = -3(cos100tcos38 - sin100tsin38)

    So, (5cos25-3cos38+7cos40)cos100t + (-5sin25+3sin38-7sin40)sin100t becomes:

    7.5298cos100t - 4.7656sin100t
    I'm not sure if i did right by simplifying it with a calculator. Or maybe i should have kept the answer in cosine form?

    By dividing each term:
    1.580(cos100t - sin100t)
    But i can't figure out about cos (alpha) for the final form.

    I expanded the term Acos(wt+x) (i'm using x as i can't figure out how to type the alpha symbol here).
    And got this:
    Acosxcoswt - Asinxsinwt
     
  7. Sep 6, 2011 #6

    tiny-tim

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    no, you do need it in this form
    that's not true :confused:

    you need to write 7.5298cos100t - 4.7656sin100t in the form (Acos100t - Bsin100t)C :wink:
     
  8. Sep 6, 2011 #7

    LCKurtz

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    But doing that makes it no longer equal to the original expression.
    Think about this:
    [tex]A\cos(t) + B\sin(t) =\sqrt{A^2+B^2}\left(\frac A {\sqrt{A^2+B^2}}\cos(t)
    + \frac B {\sqrt{A^2+B^2}}\sin(t)\right)[/tex]

    where those two fractions can be a sine and cosine of some angle since the sum of their squares is 1.
     
  9. Sep 6, 2011 #8

    sharks

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    OK... Then, using LCKurtz's equation above, i get:

    8.9112(0.8449cos100t - 0.5348sin100t)

    I need the values 0.8449 and 0.5348 to be cosx and sinx respectively.

    So, i used arc cos and arc sin and got the answer, with alpha = 32.33 degrees in both.

    The answer is: 8.91cos(100t+32.33)

    I think it's finally solved now.

    Thanks to both of you. :smile:

    On another note, is there any other way to solve this? By using complex numbers? I mean, Euler, DeMoivre, etc? Just wondering.

    OK, i figured out an easier way of doing it:
    By comparing coefficients of coswt and sinwt.
     
    Last edited: Sep 6, 2011
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