Results 1 to 3 of 3

Math Help - Trigonometric Sum

  1. #1
    MHF Contributor
    Joined
    Jul 2008
    From
    NYC
    Posts
    1,489

    Trigonometric Sum

    I would like you to show me how to get the amplitude and phase shift of the sum of the trigonometric sum below.

    8sin(50pi*x+pi/4)+7sin(50pi*x-pi/3)

    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1

    Trigonometry

    Hello magentarita
    Quote Originally Posted by magentarita View Post
    I would like you to show me how to get the amplitude and phase shift of the sum of the trigonometric sum below.

    8sin(50pi*x+pi/4)+7sin(50pi*x-pi/3)

    8\sin(50\pi x +\pi/4) +7\sin(50\pi x -\pi/3)

    = 8\sin(50\pi x)\cos(\pi/4) +8\cos(50\pi x)\sin(\pi/4) + 7\sin(50\pi x)\cos(\pi/3) -7\cos(50\pi x)\sin(\pi/3)

    = \Big(\frac{8}{\sqrt2}+\frac{7}{2}\Big)\sin(50\pi x) + \Big(\frac{8}{\sqrt2}-\frac{7\sqrt3}{2}\Big)\cos(50\pi x)

    = \frac{1}{2\sqrt2}\Big((16+7\sqrt2)\sin(50\pi x)-(7\sqrt6-16)\cos(50\pi x)\Big)

     = r\sin(50\pi[x-\alpha]), where

    r^2 = \tfrac18\Big((16+7\sqrt2)^2+(7\sqrt6-16)^2\Big)

    and \tan(50\pi\alpha) = \frac{(7\sqrt6-16)}{(16+7\sqrt2)}

    Can you tidy up and simplify now?

    Grandad
    Last edited by Grandad; May 22nd 2009 at 11:47 PM. Reason: Corrected solution
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Jul 2008
    From
    NYC
    Posts
    1,489

    yes...

    Quote Originally Posted by Grandad View Post
    Hello magentarita 8\sin(50\pi x +\pi/4) +7\sin(50\pi x -\pi/3)

    = 8\sin(50\pi x)\cos(\pi/4) +8\cos(50\pi x)\sin(\pi/4) + 7\sin(50\pi x)\cos(\pi/3) -7\cos(50\pi x)\sin(\pi/3)

    = \Big(\frac{8}{\sqrt2}+\frac{7}{2}\Big)\sin(50\pi x) + \Big(\frac{8}{\sqrt2}-\frac{7\sqrt3}{2}\Big)\cos(50\pi x)

    = \frac{1}{2\sqrt2}\Big((16+7\sqrt2)\sin(50\pi x)-(7\sqrt6-16)\cos(50\pi x)\Big)

     = r\sin(50\pi[x-\alpha]), where

    r^2 = \tfrac18\Big((16+7\sqrt2)^2+(7\sqrt6-16)^2\Big)

    and \tan(50\pi\alpha) = \frac{(7\sqrt6-16)}{(16+7\sqrt2)}

    Can you tidy up and simplify now?

    Grandad
    Yes, I can. This question is not easy.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigonometric sum.
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: January 15th 2011, 09:31 PM
  2. Replies: 6
    Last Post: August 29th 2010, 06:23 PM
  3. Replies: 2
    Last Post: February 19th 2010, 11:37 AM
  4. Trigonometric Sum
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: May 26th 2009, 05:03 PM
  5. trigonometric
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: March 16th 2009, 10:33 PM

Search Tags


/mathhelpforum @mathhelpforum