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Math Help - problem in solving equation..

  1. #1
    Newbie
    Joined
    Apr 2010
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    3
    hi all,



    x=L1cos q1+L2 cos(q1+ q2);
    y=L1sin q1+L2 sin(q1+q2);

    x,y and L1 L2 values are know..
    how to find q1 and q2...


    In other words, From the given diagram...
    x,y, L1,L2 are known...
    how to find q1& q2...
    Attached Thumbnails Attached Thumbnails problem in solving equation..-problem.png  
    Last edited by mr fantastic; April 30th 2010 at 04:46 AM. Reason: Merged posts and edited.
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, nani!

    I have part of the solution.

    Those subscripts are messy; I'll modify the problem.


    Solve for p and q.

    . . \begin{array}{ccc}x &=& L\cos(p) + M\cos(p+q) \\<br />
y &=& L\sin(p) + M\sin(p+q) \end{array}

    x,y,L,M are constants.

    Square the equations: . \begin{array}{ccc}<br />
x^2 &=& L^2\cos^2(p) + 2LM\cos(p)\cos(p+q) + M^2\cos^2(p+q) \\<br />
y^2 &=& L^2\sin^2(p) + 2LM\sin(p)\sin(p+q) + M^2\sin^2(p+q)<br />
\end{array}


    \text{Add: }\;x^2+y^2 \;=\;L^2\underbrace{\bigg[\sin^2(p) + \cos^2(p)\bigg]}_{\text{This is 1}} + \;2LM\underbrace{\bigg[\cos(p)\cos(p+q) + \sin(p)\sin(p+q)\bigg]}_{\text{This is }\cos(q)} + \;M^2\underbrace{\bigg[\sin^2(p+q) + \cos^2(p+q)\bigg]}_{\text{This is 1}}


    We have: . x^2+y^2 \;=\;L^2 + 2LM\cos(q) + M^2  \qquad\Rightarrow\qquad \cos(q) \;=\;\frac{(x^2+y^2) - (L^2+M^2)}{2LM}


    Therefore: . q \;=\;\cos^{-1}\left[\frac{(x^2+y^2) - (L^2+M^2)}{2LM}\right]

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  3. #3
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    thanx soroban...
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  4. #4
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    Apr 2010
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    pal am unable to get the value of p(using ur solution for q).. help me out with this pls......
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