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De Moivre's theorem

  1. May 4, 2014 #1
    1. The problem statement, all variables and given/known data

    Use De Moivre's Theorem with n = 3 to express cos(3θ) and sin(3θ) in terms of cos(θ) and sin(θ)
    cos(3θ) =
    sin(3θ) =

    3. The attempt at a solution

    To me this seems like a very vague problem. I am not entirely sure what they are asking me to do here.
  2. jcsd
  3. May 4, 2014 #2


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    Your first step should probably be to write down De Moivre's Theorem for the case [itex]n = 3[/itex].
  4. May 4, 2014 #3
    [itex] r^3(cos(3θ)+isin(3θ)) [/itex]

    I cant figure out how to go about solving this because I am not really sure what they are asking me to find.
  5. May 4, 2014 #4


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    \frac{\cos(3\theta)}{\cos(\theta)}=A\, \cos^2(\theta)+B\, \sin^2(\theta) \\
    \frac{\sin(3\theta)}{\sin(\theta)}=C\, \cos^2(\theta)+D\, \sin^2(\theta) $$

    What are A,B,C,and D?
    Last edited: May 4, 2014
  6. May 4, 2014 #5


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    That should be an equation, what is the other side?
  7. May 4, 2014 #6
    the other side would be r(cosθ+isinθ) ?
  8. May 4, 2014 #7


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    what does that tell you about how sin(3theta),cos(3theta),cos(theta), and sin(theta) are related?
  9. May 4, 2014 #8
    cos(3θ) = cos(θ)^3

    sin(3θ) = sin(θ)^3

    was it really that simple?
  10. May 4, 2014 #9


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    $$\cos(3\theta)+\imath \sin(3\theta)=[\cos(\theta)+\imath \sin(\theta)]^3$$
    expand the right hand side using the distributive property
    this will allow you to relate sin(3theta),cos(3theta),cos(theta), and sin(theta)
  11. May 4, 2014 #10
    No. You are wrong.
    Write down the full formula (De Moivre's).

    You can use a calculator to check your answer numerically.
    If cos(3x) = cos(x)^3, cos(3) - cos(1)^3 = 0.
    While cos(3) - cos(1)^3 is approximately -1.1477211018514388810043820516307.
  12. May 4, 2014 #11
    ok so far after expanding i have:

    [itex] r^3(cos(3θ)+isin(3θ)) = r^3(cos^3(θ)+3cos^2(θ)isin(θ)-3cos(θ)sin^2(θ)-sin^3(θ)i) [/itex]

    [itex] cos(3θ)+isin(3θ) = (cos^3(θ)+3cos^2(θ)isin(θ)-3cos(θ)sin^2(θ)-sin^3(θ)i) [/itex]

    [itex] cos(3θ)+isin^3(θ) = (cos^3(θ)+3cos^2(θ)isin(θ)-3cos(θ)sin^2(θ)-sin^3(θ)i) [/itex]

    [itex] cos(3θ)= (cos^3(θ)+3cos^2(θ)isin(θ)-3cos(θ)sin^2(θ)-2sin^3(θ)i) [/itex]

    and then i would do the same thing for sin?
  13. May 5, 2014 #12


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    actually you have done both at once
    equate the real and imaginary parts
    $$\cos(3\theta)+\imath \sin(3\theta)=[\cos^2(\theta)-3 \sin^2(\theta)]\cos(\theta)+\imath[-\sin^2(\theta)+3 \cos^2(\theta)]\sin(\theta) $$
  14. May 5, 2014 #13
  15. May 5, 2014 #14


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    The original problem said "Use DeMoivre's theorem" and, in the first response, pasmith asked you to write out what DeMoivre's theorem is. You still have not done that. Do you not know what "DeMoivre's theorem" is?
  16. May 5, 2014 #15
    $$e^{\jmath3 \theta}= (e^{\jmath \theta})^3$$
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