Results 1 to 1 of 1

Thread: find expression for an angle

  1. #1
    Junior Member
    Joined
    Apr 2012
    From
    planet earth
    Posts
    31
    Thanks
    1

    find expression for one line segment

    Attachment 25106


    Link to image:

    In this geometric setup, I'm trying to find an expression for A dependent on $\displaystyle \phi$, given that $\displaystyle R$ and $\displaystyle \theta$ are known.
    My take on this is to write the two expressions for the two sides $\displaystyle DL$ and $\displaystyle DR$

    1) $\displaystyle DL^2 = A^2 + R^2 - 2 A R \cos\left(\frac{1}{2}\pi - \phi\right)$
    2) $\displaystyle DR^2 = A^2 + R^2 - 2 A R \cos\left(\frac{1}{2}\pi + \phi\right) = A^2 + R^2 + 2 A R \cos\left(\frac{1}{2}\pi - \phi\right) $

    and an expression for $\displaystyle DS = 2A$

    3) $\displaystyle DS^2 = DL^2 + DR^2 - 2 DL \cdot DR \cos\left(\theta\right)$

    Then it seems that I could achieve my goal by eliminating $\displaystyle DL$ and $\displaystyle DR$ in the equation for $\displaystyle DS$.

    Summation of 1) and 2) gives

    4) $\displaystyle DL^2 + DR^2= 2A^2 + 2R^2$

    and multiplication of 1) and 2) gives

    5) $\displaystyle DL^2\cdot DR^2 = \left(A^2 + R^2 - 2 A R \cos\left(\frac{1}{2}\pi - \phi\right)\right)\left(A^2 + R^2 + 2 A R \cos\left(\frac{1}{2}\pi - \phi\right) \right)$
    $\displaystyle = \left(A^2 + R^2 \right)^2 - 4 A^2 R^2 \cos^2\left(\frac{1}{2}\pi - \phi\right)$

    Inserting 4) into 3) and squaring provides

    6) $\displaystyle DL^2 \cdot DR^2 = \frac{(R^2 - A^2)^2}{\cos^2\left(\theta\right)} $

    Equating 5) and 6) gives an equation dependent on only $\displaystyle A$ and $\displaystyle \phi$

    7) $\displaystyle \left(A^2 + R^2 \right)^2 - 4 A^2 R^2 \cos^2\left(\frac{1}{2}\pi - \phi\right) = \frac{(R^2 - A^2)^2}{\cos^2\left(\theta\right)}$

    Rearranging

    8) $\displaystyle \cos^2\left(\theta\right)\left(A^4 + R^4 + 2 A^2R^2\right) - 4 A^2 R^2 \cos^2\left(\frac{1}{2}\pi - \phi\right)\cos^2\left(\theta\right) = A^4 + R^4 - 2 A^2R^2$

    9) $\displaystyle A^4(1-\cos^2\left(\theta\right)) - A^2(2 R^2 + 2 R^2\cos^2\left(\theta\right) - 4 R^2 \cos^2\left(\frac{1}{2}\pi - \phi\right)\cos^2\left(\theta\right)) + R^4(1-\cos^2\left(\theta\right)) = 0 $

    10) $\displaystyle A^4 - A^2 \frac{2 R^2(1 + \cos^2\left(\theta\right) - 2 \cos^2\left(\frac{1}{2}\pi - \phi\right)\cos^2\left(\theta\right)^2)}{1-\cos^2\left(\theta\right)} + R^4 = 0 $

    I have a hard time believing this expression to be true because of the following reasoning:
    For $\displaystyle \phi=0$ then $\displaystyle A = R\tan\left(\frac{1}{2}\theta\right)$
    As $\displaystyle \phi$ is increased I would assume that $\displaystyle A$ should also increase (relative to $\displaystyle R\tan\left(\frac{1}{2}\theta\right)$). However, for the numerical values that I have used to test the expression in 10) I find that $\displaystyle A$ decreases.

    Looking back, I can't find out where I go wrong.
    Last edited by niaren; Oct 8th 2012 at 06:32 AM. Reason: Found possible error in derivation
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Apr 4th 2011, 11:36 AM
  2. Replies: 1
    Last Post: Feb 4th 2010, 11:07 PM
  3. Replies: 7
    Last Post: Nov 27th 2009, 06:35 PM
  4. Simplify the Expression (Double-Angle or Half-Angle)
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Nov 3rd 2009, 03:38 PM
  5. Replies: 3
    Last Post: Apr 16th 2009, 08:45 AM

Search Tags


/mathhelpforum @mathhelpforum