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Math Help - identity help

  1. #1
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    identity help

    Given  sinx = \frac{e^{ix} - e^{-ix}}{2i} and  cosx = \frac{e^{ix}+ e^{-ix}}{2}

    where  i = \sqrt{-1}

    show that  sin^{4}x = \frac{3}{8} -\frac{1}{2}cos2x + \frac{1}{8} cos4x


    I have a solution for this, provided by my teacher, but i cannot seem to understand it.

    the solution is

     sin^{4}x =( \frac{e^{ix}-e^{-ix}}{2})^{4}  what happened to the i, in the denominator?

    this expression is than expanded out, and looks like this; which i dont know how to get to

     \frac{1}{16} ( e^{2ix} + 4e^{2ix} +6 - 4e^{-2ix} + e^{-4ix}


    any help appreciated, thank you.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: identity help

    Quote Originally Posted by Tweety View Post
    what happened to the i, in the denominator?
    i^4=1.

    this expression is than expanded out, and looks like this; which i dont know how to get to

     \frac{1}{16} ( e^{2ix} + 4e^{2ix} +6 - 4e^{-2ix} + e^{-4ix}
    Use the Binomial Theorem. (The first term should be e^{4ix}).
    Thanks from Tweety
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  3. #3
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    Re: identity help

    Thank you.
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  4. #4
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    Re: identity help

    Actually, just tried expanding using the binomial expansion, but couldn't quite do it.

    because in the binomial expansion, the first term has to be 1  (1+x)^{n}  = 1 + nx + \frac{n(n-1)}{2!}......
    how do i apply this formula to  e^{ix}-e^{-ix} ?
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Re: identity help

    Use for example the Pascal Triangle (see here: Binomial theorem - Wikipedia, the free encyclopedia),


    \sin^4 x=\dfrac{1}{16}(e^{ix}-e^{-ix})^4=\dfrac{1}{16}(e^{4ix}-4e^{3ix}e^{-ix}+6e^{2ix}e^{-2ix}-4e^{ix}e^{-3ix}+e^{-4ix})

    So,

    \sin^4 x=\dfrac{1}{16}(e^{4ix}-4e^{2ix}+6-4e^{-2ix}+e^{-4ix})
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