Results 1 to 4 of 4

Math Help - de Moivre and Binomial Theorem

  1. #1
    PTL
    PTL is offline
    Junior Member
    Joined
    Aug 2009
    Posts
    29

    de Moivre and Binomial Theorem

    Use de Moivre's Theorem and the Binomial Theorem to find integers A,B,C,D,E,F such that
    \sin(5\theta)=A\sin^{5}\theta + B\sin^{3}\theta+ C\sin\theta
    16\sin^{5}\theta=D\sin\theta+E\sin(3\theta)+F\sin(  5\theta).

    Any help much appreciated
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Do you know the statement of De Moivre's theorem? Do you know how to expand (x+y)^5? If so just do it and set x=\cos\theta, \: y=i\sin\theta...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    PTL
    PTL is offline
    Junior Member
    Joined
    Aug 2009
    Posts
    29
    So I get
    \cos^{5}\theta+5\cos^{4}\theta\sin\theta + 10\cos^{3}\theta\sin^{2}\theta + 10\cos^{2}\theta\sin^{3}\theta+ 5\cos\theta\sin^{4}\theta +\sin^{5}\theta

    (\cos x + i \sin x)^{n}= \cos n\theta + i \sin n\theta

    I still don't see where I'm supposed to go with it...

    Is it something to do with matching up the real parts on both sides?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Quote Originally Posted by PTL View Post
    So I get
    \cos^{5}\theta+5\cos^{4}\theta\sin\theta + 10\cos^{3}\theta\sin^{2}\theta + 10\cos^{2}\theta\sin^{3}\theta+ 5\cos\theta\sin^{4}\theta +\sin^{5}\theta

    (\cos x + i \sin x)^{n}= \cos n\theta + i \sin n\theta

    I still don't see where I'm supposed to go with it...

    Is it something to do with matching up the real parts on both sides?
    You forgot i my friend!

    you should get

    (\cos x + i \sin x)^{5} =
    \cos^{5}\theta+5i\cos^{4}\theta\sin\theta - 10\cos^{3}\theta\sin^{2}\theta - 10i\cos^{2}\theta\sin^{3}\theta+ 5\cos\theta\sin^{4}\theta +i\sin^{5}\theta

    (just alternate 1,i,-1,-i,...)

    Then equate the imaginary parts because the real part is \cos 5\theta which is not what you want.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. de Moivre's Theorem
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: May 8th 2011, 03:28 AM
  2. [SOLVED] De Moivre's Theorem
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: November 18th 2010, 01:06 PM
  3. De Moivre's formula and Binomial theorem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 22nd 2009, 11:55 AM
  4. De Moivre's Theorem
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: August 31st 2009, 05:45 AM
  5. De Moivre's Theorem
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: April 15th 2009, 01:43 PM

Search Tags


/mathhelpforum @mathhelpforum