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De Moivres formula for multiple angles. Where does the sinus go?

  1. Aug 26, 2011 #1
    Hi!

    Yesterday i read and mostly understood that (cos([itex]\theta[/itex]) + isin([itex]\theta[/itex])[itex]^{n}[/itex] = (cos([itex]n\theta[/itex]) + isin([itex]n\theta[/itex])
     
  2. jcsd
  3. Aug 26, 2011 #2

    HallsofIvy

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    Okay, so what is your question? What about the multiple angles that you mention in your title?

    Note that the formula you give can be derived from the formula for multiplication of complex numbers in "polar form"- [itex]r_1(cos(\theta_1)+ i sin(\theta_1))r_2(cos(\theta_2)+ i sin(\theta_2))= (r_1r_2)(cos(\theta_1+ \theta_2)+ i sin(\theta_1+ \theta_2))[/itex].

    Also, we can use Euler's formula, [itex]e^{ix}= cos(x)+ i sin(x)[/itex] to write [itex](cos(\theta)+ i sin(\theta))^n= cos(n\theta)+ i sin(n\theta)[/itex] in the simpler form [itex](e^{i\theta})^n= e^{i n\theta}[/itex].
     
  4. Aug 26, 2011 #3
    Sorry, I'm not best friends with the post publisher today.

    My question was about the multiple angles: I understand that
    [itex](cos x +i sinx)^{n} = cos (nx) + i*sin(nx)[/itex]. Then i read that [itex]cos (nx) = cos^{n}(x) - (nC2)cos^{n-2}x*sin^{2}(x)- ...-(-1)^{k/2}(nCk)cos^{n-k}(k)*sin^{k}(x) [/itex]

    Now, that looks alot like what I would expect to get from [itex]cos (nx) + i*sin(nx) = (cos x +i sinx)^{n},[/itex] but i cannot figure out where the sin(nx) goes.

    My first thought is that [itex]cos (nx) = (cos x +i sinx)^{n} - sin(nx)[/itex]
     
  5. Aug 26, 2011 #4
    Hi, and thanks for such a quick response! I'm sorry about the first post. Seems like I hit the submit rather than the preview button, a bit prematurely..
     
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