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Thread: trig and complex numbers

  1. #1
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    trig and complex numbers

    if n is an integer x is real, then show that :
    $\displaystyle (1+\cos x + i\sin x )^n + (1+ \cos x-i\sin x)^n = 2^{n+1}\cos ^n {\frac{x}{2}}\cos \frac{nx}{2} $

    I played around a bit with de moivre's theorem and other things...

    thanks in advance.
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  2. #2
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    Re: trig and complex numbers

    Quote Originally Posted by earthboy View Post
    if n is an integer x is real, then show that :
    $\displaystyle (1+\cos x + i\sin x )^n + (1+ \cos x-i\sin x)^n = 2^{n+1}\cos ^n {\frac{x}{2}}\cos \frac{nx}{2} $

    I played around a bit with de moivre's theorem and other things...

    thanks in advance.
    If you can't solve after a massive amount of calculations, there is always the mighty mathematical induction.
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  3. #3
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    Re: trig and complex numbers

    Quote Originally Posted by ChessTal View Post
    If you can't solve after a massive amount of calculations, there is always the mighty mathematical induction.
    yes there is.....but there got be a silly neat trick here...anyone???
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  4. #4
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    Re: trig and complex numbers

    Start by using the identities

    $\displaystyle \cos 2A = 2\cos^{2}A - 1$ and $\displaystyle \sin 2A = 2\sin A \cos A$ so as to get everything inside the brackets in terms of half angles $\displaystyle x/2.$

    Next, remove $\displaystyle 2\cos(x/2)$ as a common factor from each of the brackets.

    Then, write the two complex numbers in exponential form and finally use the exponential form for $\displaystyle \cos A.$
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  5. #5
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    Re: trig and complex numbers

    trig and complex numbers-14-aug-13.png
    Thanks from earthboy
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