Results 1 to 3 of 3

Math Help - Trig. - Elimination parameters

  1. #1
    Senior Member
    Joined
    Jul 2008
    Posts
    347

    Exclamation Trig. - Elimination parameters

    Hey guys

    How do i eliminate x in the following case: x = a sinx and y = b tan(x/2)?

    I could do the basic questions but I have no clue how to do this one.

    Thanx
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by xwrathbringerx View Post
    Hey guys

    How do i eliminate x in the following case: x = a sinx and y = b tan(x/2)?

    I could do the basic questions but I have no clue how to do this one.

    Thanx
    I assume that x as an angle is not meant to be the same as x as a coordinate??

    x = a \sin t = 2a \sin \left( \frac{t}{2}\right) \cos \left( \frac{t}{2}\right) .... (1)

    y = b \tan \left( \frac{t}{2}\right) \Rightarrow \frac{y}{b} = \left( \frac{t}{2}\right) .... (2)


    From equation (2): \sin \left( \frac{t}{2} \right) = \frac{y}{\sqrt{y^2 + b^2}} and \cos \left( \frac{t}{2}\right) = \frac{b}{\sqrt{y^2 + b^2}}

    assuming  0 \leq t \leq \pi (why?)

    Substitute these two expressions into equation (1).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,803
    Thanks
    692
    Hello, xwrathbringerx!


    Eliminate the parameter \theta: . \begin{array}{cccc}x &=& a\sin\theta & {\color{red}(a)} \\ y &=& b\tan\frac{\theta}{2} & {\color{red}(b)} \end{array}

    From (b) we have: . y \;=\;b\,\frac{\sin\frac{\theta}{2}}{\cos\frac{\the  ta}{2}} \quad\Rightarrow\quad y\;=\;b\sqrt{\frac{1-\cos x}{1 + \cos x}}

    Square both sides: . y^2 \;=\;b^2\,\frac{1-\cos x}{1+\cos x} \quad\Rightarrow\quad y^2 + y^2\cos x \;=\;b^2 - b^2\cos x

    . . b^2\cos x + y^2\cos x \;=\;b^2-y^2 \quad\Rightarrow\quad (b^2+y^2)\cos x \;=\;b^2-y^2

    . . \cos x \;=\;\frac{b^2-y^2}{b^2+y^2} .[1]


    From (a), we have: . x \;=\;a\sin x \quad\Rightarrow\quad x\;=\;a\sqrt{1-\cos^2\!x} .[2]

    Substitute [1] into [2]: . x \;=\;a\sqrt{1-\left(\frac{b^2-y^2}{b^2+y^2}\right)^2} \;=\;a\sqrt{\frac{(b^2+y^2)^2 - (b^2-y^2)^2}{(b^2+y^2)^2}}

    . . which simplifies to: . x \;=\;a\sqrt{\frac{4b^2y^2}{(b^2+y^2)^2}} \;=\;a\,\frac{2by}{b^2+y^2}


    Therefore: . x \;=\;\frac{2ab\,y}{b^2+y^2}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: September 12th 2011, 09:03 PM
  2. Random Parameters
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: June 5th 2011, 11:23 PM
  3. Replies: 1
    Last Post: February 16th 2011, 02:06 PM
  4. Replies: 2
    Last Post: January 25th 2010, 10:29 AM
  5. Using Parameters
    Posted in the Algebra Forum
    Replies: 5
    Last Post: July 10th 2009, 08:11 PM

Search Tags


/mathhelpforum @mathhelpforum