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(e^ix)^n=(e^ixn) & Trig identities

  1. Aug 22, 2005 #1
    ok...this was meant to be a fun problem :grumpy: but looks like I don't deserve to have fun!!!
    How am I meant to derive trig identities like sin(x)cos^3(x) from some complex **** like [tex]\left( {e}^{{\it ix}} \right) ^{n}={e}^{{\it ixn}}[/tex]!!! I just don't get the idea!!! :cry:
  2. jcsd
  3. Aug 22, 2005 #2
  4. Aug 22, 2005 #3


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    Homework Helper

    It might be more clear in the trigonometric form where the equation is
    [cos(x)+i sin(x)]^n=cos(n x)+i sin(n x)
    so if you wanted to know
    cos(3 x)=[cos(x)]^3-3cos(x)[sin(x)]^2
    you could consider
    [cos(x)+i sin(x)]^3=cos(3 x)+i sin(3 x)
    so expand the left side find its real part and you have the identitiy
    the binomial theorem can be helpful here
    just be aware that using lots of odd identities at intermediate steps will mess things up
  5. Aug 22, 2005 #4
    thanks for the links inha. :smile:
  6. Aug 24, 2005 #5
    Thanks guys :)
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