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de moivre's theorem watch

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    Show using de moivres theorem

    sin^4x = 1/8 cos(4x) - 1/2 cos(2x) + 3/8

    hint given: let z = e^ix, then sinx = 1/(2i)(z-1/z) and cos(nx) = 1/2(z^n + 1/z^n)

    I have replaced the character theta in the question with x because i dont have a theta symbol. If anyone could help me understand why sinx = 1/(2i)(z-1/z) when z = e^ix it would be much appreciated.

    Sorry for the sloppy layout. Pls help, thnx
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    If  sinx = \frac{z - z^{-1}}{2i}

    what is   sin^4x ? (Use binomial expansion) Noting that  cosnx = \frac{z^n + z^{-n}}{2}
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    I dont understand how if let z = e^ix, then sinx = 1/(2i)(z-1/z)

    also, are u sure that sinx = (z-1/z)/2i, because i got confused by the mathematical expression, i thought it meant 1/[2i(z-1/z)]. thnx for helping me
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    you know e^ix = cosx + isinx, and e^-ix = cosx - isinx. so work it out...
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    I'm quite sure

     sinx = \frac{e^{ix}-e^{-ix}}{2i}

    I thought that was knowledge you knew?

    Also

     cosx = \frac{e^{ix}+e^{-ix}}{2}

    Do you know where to go from what ive said to you ?
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    I see it now, thanks so much guys. Im very rusty with the matsh right now, but im getting there
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    No probs if you have any other problems dont hesistate to ask.
 
 
 
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Updated: August 7, 2007

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