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  1. #1
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    complex

    express in trig form

    (-2 Ix)/(e^-Ix - e^(Ix)  - (8I) (e^(-Ix) + e^(Ix) (I/2) (e^(-Ix) - e^(Ix) - (e^(-Ix) + (e^(Ix)/2 /(e^(-Ix) - e^(I x)^3  = 4/(e^(2 Ix) (e^(-Ix) - e^(Ix)^3 - (4 e^(2 Ix) / (e^(-Ix) - e^(Ix)^3 + (12Ix) / (e^(-Ix) - e^(I x)^3 + (2Ix) / (e^(2Ix) (e^(-Ix) - e^(I x)^3 + (2I) e^(2 Ix) x)/(e^(-Ix) - e^(Ix)^3

    sorry first time using latex. i dont know how to show the numerator over the denominator and the powers keep ending up looking larger than its supposed to be.

    nvm figured it out
    Last edited by purebladeknight; November 7th 2009 at 11:42 PM.
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  2. #2
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    Quote Originally Posted by purebladeknight View Post
    express in trig trig form

    ((-2 I) x)/(E^((-I) x) - E^(I x)) - ((8 I) (E^((-I) x) + E^(I x)) ((I/2) (E^((-I) x) - E^(I x)) - ((E^((-I) x) + E^(I x)) x)/2))/(E^((-I) x) - E^(I x))^3 == 4/(E^((2 I) x) (E^((-I) x) - E^(I x))^3) - (4 E^((2 I) x))/(E^((-I) x) - E^(I x))^3 + ((12 I) x)/(E^((-I) x) - E^(I x))^3 + ((2 I) x)/(E^((2 I) x) (E^((-I) x) - E^(I x))^3) + ((2 I) E^((2 I) x) x)/(E^((-I) x) - E^(I x))^3
    Recall that e^{i\theta} = \cos{\theta} + i\sin{\theta}.

    You should be able to convert them all to sines and cosines, and then use trigonometric identities to simplify.
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  3. #3
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    Quote Originally Posted by purebladeknight View Post
    express in trig form

    (-2 Ix)/(e^-Ix - e^(Ix)  - (8I) (e^(-Ix) + e^(Ix) (I/2) (e^(-Ix) - e^(Ix) - (e^(-Ix) + (e^(Ix)/2 /(e^(-Ix) - e^(I x)^3  = 4/(e^(2 Ix) (e^(-Ix) - e^(Ix)^3 - (4 e^(2 Ix) / (e^(-Ix) - e^(Ix)^3 + (12Ix) / (e^(-Ix) - e^(I x)^3 + (2Ix) / (e^(2Ix) (e^(-Ix) - e^(I x)^3 + (2I) e^(2 Ix) x)/(e^(-Ix) - e^(Ix)^3

    sorry first time using latex. i dont know how to show the numerator over the denominator and the powers keep ending up looking large
    Please revise this and make sure that your brackets match.

    Also an exponent should be written e^{ix} in LaTeX.

    Also^2 try asking WolframAlpha to simplify this.

    CB
    Last edited by CaptainBlack; November 8th 2009 at 02:25 AM.
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