Results 1 to 5 of 5

Math Help - Trigonometric equation

  1. #1
    Member
    Joined
    Dec 2010
    Posts
    82
    Thanks
    2

    Trigonometric equation

    If \cos x = \cos y and \sin x = -\sin y

    Then find value of \sin (2006\; x)+\sin (2006\; y)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4

    Re: Trigonometric equation

    Quote Originally Posted by jacks View Post
    If \cos x = \cos y and \sin x = -\sin y

    Then find value of \sin (2006\; x)+\sin (2006\; y)
    So what have you tried?

    Clearly y=-x satisfies the condition, but does anything else?

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,504
    Thanks
    1400

    Re: Trigonometric equation

    Quote Originally Posted by jacks View Post
    If \cos x = \cos y and \sin x = -\sin y

    Then find value of \sin (2006\; x)+\sin (2006\; y)
    Are x and y mutually exclusive?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2010
    Posts
    82
    Thanks
    2

    Re: Trigonometric equation

    Thanks Got it

    Let z=\cos x+i\sin x and \omega = \cos y +i\sin y

    Then from Given equation z=\bar{\omega}

    So \cos (2006 \;x)+i\sin (2006\; x) = z^{2006}

    OR Im(z^{2006}) = \sin (2006\; x)

    and Im(\omega^{2006}) = \sin (2006\; y)

    So \sin (2006\; x)+\sin (2006\; y) = Im\left(z^{2006}+\omega^{2006}\right)=Im\left(\bar  {\omega})^{2006}+\omega^{2006}\right) = 0
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4

    Re: Trigonometric equation

    Quote Originally Posted by jacks View Post
    Thanks Got it

    Let z=\cos x+i\sin x and \omega = \cos y +i\sin y

    Then from Given equation z=\bar{\omega}

    So \cos (2006 \;x)+i\sin (2006\; x) = z^{2006}

    OR Im(z^{2006}) = \sin (2006\; x)

    and Im(\omega^{2006}) = \sin (2006\; y)

    So \sin (2006\; x)+\sin (2006\; y) = Im\left(z^{2006}+\omega^{2006}\right)=Im\left(\bar  {\omega})^{2006}+\omega^{2006}\right) = 0
    If you had indicated that you were familiar with complex analysis we could have pointed out that the conditions give:

    e^{ix}=e^{-iy + 2n\pi i}, \ \ n=...-1, 0, 1, ...

    and so:

    e^{2006 ix}=e^{- 2006 iy} e^{4012 \pi i}=e^{- 2006 iy}

    and the result follows immediately.

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigonometric Equation
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: September 28th 2009, 07:25 AM
  2. trigonometric equation
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: July 22nd 2009, 12:59 AM
  3. Replies: 2
    Last Post: April 28th 2009, 06:42 AM
  4. Trigonometric equation for x and y
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: March 19th 2009, 10:41 AM
  5. Trigonometric Equation
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: September 20th 2008, 02:22 AM

Search Tags


/mathhelpforum @mathhelpforum