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Math Help - Eulers relation

  1. #1
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    Eulers relation

    I know that the Eulers relation:

    2 cos x = e^ ix + e^-ix
    2i sin x = e^ix - e^-ix


    I was just wondering how i can

    change 1. 4i*e^x(e^2ix-e^-2ix)

    2. 4i*e^x[(-1/2*e^2ix+1/2e^-2ix+ i *(1/2*e^2ix+1/2e^-2ix)]

    in Eulers relation

    Thanks
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  2. #2
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    Hello,
    Quote Originally Posted by 1234567 View Post
    I know that the Eulers relation:

    2 cos x = e^ ix + e^-ix
    2i sin x = e^ix - e^-ix


    I was just wondering how i can

    change 1. 4i*e^x(e^2ix-e^-2ix)
    =2e^x (2i[e^{2ix}-e^{-2ix}])=2e^x \sin(2x)

    2. 4i*e^x[(-1/2*e^2ix+1/2e^-2ix+ i *(1/2*e^2ix+1/2e^-2ix)]

    in Eulers relation

    Thanks
    is it 4ie^x \left[-\frac 12 e^{2ix}+\frac 12 e^{-2ix}+i \left(\frac 12 e^{2ix}+\frac 12 e^{-2ix}\right)\right] ?

    you can see that \cos(x)=\frac{e^{ix}+e^{-ix}}{2} and i \sin(x)=\frac{e^{ix}-e^{-ix}}{2}

    so we have :
    =4i e^x \left[-i \sin(2x)+i \cos(2x)\right]=4e^x \left[\sin(2x)-\cos(2x)\right]
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  3. #3
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    Thank you that helped alot!!
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