1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    \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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper


    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


    User Avatar
    Homework Helper



    $$\cos(3\theta)+\imath \sin(3\theta)=[\cos(\theta)+\imath \sin(\theta)]^3$$
    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Science Advisor

    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
    $$e^{\jmath3 \theta}= (e^{\jmath \theta})^3$$
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted