Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By fkf

Math Help - Maclaurin series help

  1. #1
    Super Member
    Joined
    Sep 2008
    Posts
    619

    Maclaurin series help

     sin^{3}x = \frac{3}{4} sinx - \frac{1}{4}sin(3x) Use this identity and the Maclaurin series for sin x to
    show that
     sin^{3} x = x^{3} - \frac{1}{2}x^{5} + \frac{13}{120}x^{7}

    I am quite stuck,

    I know that sinx =  x-\frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!}

    and I know sin3x = sin(2x+x)

    so I have

     \frac{3}{4}( x-\frac{x^{3}}{3!} + \frac{x^{5}}{5!}) - \frac{1}{4} (sin2x +x)

    however when I expand sin2x +x I just it just gets more complicated, is this right so far? what should I do next ?

    Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    fkf
    fkf is offline
    Junior Member fkf's Avatar
    Joined
    Nov 2012
    From
    Sweden
    Posts
    70
    Thanks
    9

    Re: Maclaurin series help

    Quote Originally Posted by Tweety View Post
     sin^{3}x = \frac{3}{4} sinx - \frac{1}{4}sin(3x) Use this identity and the Maclaurin series for sin x to
    show that
     sin^{3} x = x^{3} - \frac{1}{2}x^{5} + \frac{13}{120}x^{7}

    I am quite stuck,

    I know that sinx =  x-\frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!}

    and I know sin3x = sin(2x+x)

    so I have

     \frac{3}{4}( x-\frac{x^{3}}{3!} + \frac{x^{5}}{5!}) - \frac{1}{4} (sin2x +x)

    however when I expand sin2x +x I just it just gets more complicated, is this right so far? what should I do next ?

    Thank you.
    We have the identity
    sin^3(x) = 3/4sin(x)-1/4sin(3x)

    And want to use Maclaurin expansion to show that
    sin^3(x) = x^3-1/2x^5+13/120x^7

    Since we know that the expansion of sin(x) is
    sin(x) = x-x^3/3!+x^5/5!-x^7/7! (1)

    We can raise both sides to the power or three hence we have
    sin^3(x) = (x-x^3/3!+x^5/5!-x^7/7!)^3

    But since we also know that
    sin^3(x) = 3/4sin(x)-1/4sin(3x) (2)

    By using the relationship (1) into (2) we get
    sin^3(x) = 3/4*(x-x^3/3!+x^5/5!-x^7/7!)-1/4*((3x)-(3x)^3/3!+(3x)^5/5!-(3x)^7/7!) = x^3-1/2x^5+13/120x^7
    Thanks from Tweety
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Sep 2008
    Posts
    619

    Re: Maclaurin series help

    Thank you^^
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: January 26th 2010, 09:06 AM
  2. Replies: 2
    Last Post: September 16th 2009, 08:56 AM
  3. Binomial Series to find a Maclaurin Series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 21st 2009, 08:15 AM
  4. Multiplying power series - Maclaurin series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 8th 2009, 12:24 AM
  5. Replies: 1
    Last Post: May 5th 2008, 10:44 PM

Search Tags


/mathhelpforum @mathhelpforum