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Using e^ix to determine a trig identity

  1. Oct 17, 2004 #1
    Hi people, could someone help me with this

    Q. Write cosx and sinx in terms of e^ix and e^-ix respectively

    So I wrote that cosx=Re(e^ix)=Re(e^-ix)

    and sinx =Im(e^ix) and -Im(e^ix)

    I think the above identites are correct, now I must use this to show that

    16cos^3(x)sin^2(x) = 2cosx - cos3x - cos5x
  2. jcsd
  3. Oct 17, 2004 #2


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    [tex]e^{ix} = \cos x + i \sin x[/tex]

    [tex]e^{-ix} = \cos x - i \sin x[/tex]

    Look at the two above and think how you could rearrange them so you have one for sin(x) in terms of e^(ix) and e^(-ix) and same for cos(x).
  4. Oct 17, 2004 #3

    Dr Transport

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    also remember, [tex] e^{inx} = \cos(nx) + i\sin(nx) = (e^{ix})^n [/tex]. This is all you need to find any trig identity.......
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