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Math Help - De Moivre's Theorem

  1. #1
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    Post De Moivre's Theorem

    Q: Prove that,
    (1 + sin 0 + i cos 0)^n
    -------------------- = cos n (pi/2 - 0) + i sin n(pi/2 - 0)
    (1 + sin 0 - i cos 0)^n

    where, 0 = theta (Cant type it)

    Thanks !

    I've thought about multipying the conjugate (sin 0 + i cos 0) but it didnt work out nicely.. so I need some help with this! Thanks!
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  2. #2
    ynj
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    \frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\frac{1+e^{(\frac{\pi}{2}-\theta)i}}{1+e^{(\theta-\frac{\pi}{2})i}}=e^{(\frac{\pi}{2}-\theta)i}=\cos (\frac{\pi}{2}-\theta)+i\sin (\frac{\pi}{2}-\theta)
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  3. #3
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    Smile

    Argh.. my textbook never went into the Eulers rule..
    so I dont really know how you got that..
    I did a search on Wiki
    it says,
    e^i(nx) = cos (nx) + i sin (nx)
    But this never appears in my textbook! Argh~

    Is there any other ways to do it?

    Thanks anyway though!!!
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  4. #4
    ynj
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    Quote Originally Posted by Modernized View Post
    Argh.. my textbook never went into the Eulers rule..
    so I dont really know how you got that..
    I did a search on Wiki
    it says,
    e^i(nx) = cos (nx) + i sin (nx)
    But this never appears in my textbook! Argh~

    Is there any other ways to do it?

    Thanks anyway though!!!
    \frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\frac{(1+\sin\theta+i\cos\theta)(\sin  \theta+i\cos\theta)}{(1+\sin\theta-i\cos\theta)(\sin\theta+i\cos\theta)}
    =\frac{(1+\sin\theta+i\cos\theta)(\sin\theta+i\cos  \theta)}{\sin\theta+i\cos\theta+\sin\theta^2+\cos\  theta^2}=\frac{(1+\sin\theta+i\cos\theta)(\sin\the  ta+i\cos\theta)}{\sin\theta+i\cos\theta+1}
    =\sin\theta+i\cos\theta
    understand now?
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  5. #5
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    Smile

    Yes. I got till that step.. because I said I tried to times the conjugate like what you've done.. and we got the same answer, but I just dont understand how to reduce it to the final answer that they want me to show.
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  6. #6
    ynj
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    Quote Originally Posted by Modernized View Post
    Yes. I got till that step.. because I said I tried to times the conjugate like what you've done.. and we got the same answer, but I just dont understand how to reduce it to the final answer that they want me to show.
    oh...there may be some mistake in the problem??
    I think it shoud be:
    (\sin\theta+i\cos\theta)^n=\sin n\theta+i\cos n\theta=\cos (\frac{\pi}{2}-n\theta)+i\sin (\frac{\pi}{2}-n\theta)
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  7. #7
    Senior Member pacman's Avatar
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    ynj, i think your observation is correct, the 1 should not be there.
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