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Deriving DeMoivre's Formula?

  1. Feb 2, 2009 #1
    1. The problem statement, all variables and given/known data
    Derive from Demoivre's theorem that
    cos2theta=(costheta)^2-(sintheta)^2

    sin2theta=2sin theta costheta

    2. Relevant equations



    3. The attempt at a solution
    I really am clueless on this.
    I considered using (costheta +isintheta)^n, but don't know where to go with this idea.
     
    Last edited: Feb 2, 2009
  2. jcsd
  3. Feb 2, 2009 #2

    Dick

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    Sure. e^(i*2*theta)=(e^(i*theta))^2. Write down both sides and equate the imaginary parts.
     
  4. Feb 2, 2009 #3
    cos2theta+isin2theta=(costheta)^2+2isintheta-(sintheta)^2
    cos2theta=costheta^2-sintheta^2




    cos2theta+isin2theta=(costheta)^2+2isintheta-(sintheta)^2
    sin2theta=2sintheta-costheta-sintheta^2+costheta^2
    I get stuck here
     
  5. Feb 2, 2009 #4

    Dick

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    Let's just call theta t, ok? On one side you have cos(2t)+i*sin(2t). On the other side you have (cos(t)+i*sin(t))*(cos(t)+i*sin(t)). Multiply that out carefully. You are getting things all mushed up.
     
  6. Feb 2, 2009 #5
    cos(2t)+isin(2t)=(cost+isint)^2
    cos(2t)+isin(2t)=(cost)^2+2isint-(sint)^2
    isin2t=(cost)^2+2isint-(sint)^2-cos(2t)
    Is there something I'm missing on what to do here?
     
  7. Feb 2, 2009 #6

    Dick

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    On your second line the imaginary part is wrong. (a+bi)*(a+bi)=a^2-b^2+2abi. Now if the two sides are equal then the real part of one side equals the real part of the other side. Same for the imaginary parts.
     
  8. Feb 2, 2009 #7

    cos(2t)+isin(2t)=(cost+isint)^2
    cos(2t)+isin(2t)=(cost)^2+2isintcost-(sint)^2
    sin(2t)=2sintcost
     
  9. Feb 2, 2009 #8

    Dick

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    Yes. And cos(2t)=cos^2(t)-sin^2(t). Right?
     
  10. Feb 2, 2009 #9
    Yep, that makes sense. i was having a brain fart
     
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