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Thread: De moivres theorem

  1. #1
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    De moivres theorem

    Simplify the following

    A) (cosx - isinx)^5
    B) (sinx - icosx)^4



    for A i get

    cos5x - isin5x
    I think this is right...?

    not sure if B is done the same or not. any help?
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by djmccabie View Post
    Simplify the following

    A) (cosx - isinx)^5
    B) (sinx - icosx)^4



    for A i get

    cos5x - isin5x
    I think this is right...?

    not sure if B is done the same or not. any help?
    Here is what you can do

    De Moivre's formula says : $\displaystyle (\cos(x)+i\sin(x))^n=\cos(nx)+i \sin(nx)$

    Remember that the sine function is odd and the cosine function is even.
    So $\displaystyle (\cos(x)-i\sin(x))^n=(\cos(-x)+i \sin(-x))^n=$ $\displaystyle \cos(-nx)+i \sin(-nx)=\cos(nx)-i \sin(nx)$
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    so did i do it right :S

    is it different when cosine and sine are switched round?

    thanks
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  4. #4
    Moo
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    Quote Originally Posted by djmccabie View Post
    so did i do it right :S

    is it different when cosine and sine are switched round?

    thanks
    Yes, it's the correct result.


    If they're switched... do you mean $\displaystyle (-\cos(x)+i \sin(x))^n$ ?
    Then it's just $\displaystyle (-1)^n (\cos(x)-i\sin(x))^n=(-1)^n(\cos(nx)-i\sin(nx))$

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    hi thanks for helping me. its actualyy (sinx - icosx) does this change things??
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  6. #6
    Moo
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    Quote Originally Posted by djmccabie View Post
    hi thanks for helping me. its actualyy (sinx - icosx) does this change things??
    Yes :

    $\displaystyle \sin(x)-i \cos(x)$
    Multiply and divide by i :

    $\displaystyle \frac 1i \left(i [\sin(x)-i\cos(x)]\right)=\frac 1i (\cos(x)+i \sin(x))$

    So $\displaystyle (\sin(x)-i \cos(x))^n=\frac{1}{i^n} (\cos(nx)+i \sin(nx))$
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