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Solution to a system of transcendental equations

  1. Feb 12, 2015 #1
    I got stuck while solving a problem in statics unable to solve a system of equations of the form obtained as below,

    a cosθ - b sinθ = c1
    a sinθ + b cosθ = c2

    I wish to obtain a and b in terms of c1 and c2.Squaring and adding I obtained,

    a2 + b2 = c12 + c22

    I need another relation in a,b,c1 and c2 to achieve the result.How do I proceed next?
     
  2. jcsd
  3. Feb 12, 2015 #2

    PeroK

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    a and b will depend not only on c_1 and c_2 but on θ.

    Think about a rotation of θ about the origin.
     
  4. Feb 12, 2015 #3
    Well I did forget to mention c1 and c2 are functions in θ.How do I solve for a and b in terms of c1, c2 and θ?
     
  5. Feb 12, 2015 #4

    PeroK

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    If you know a little linear algebra, you can express the system of equations in matrix form, then multiple by the inverse rotation matrix.
     
  6. Feb 12, 2015 #5
    Well I am not quite used to linear algebra in practice but I will have a look if things appear easier in comparison to conventional algebra.
     
  7. Feb 12, 2015 #6

    PeroK

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    3
    If you recognise what you have as vector rotation, you can write down the answer without any calculations!

    The equation you derived shows that the magnitude of a vector is invariant under rotation.
     
  8. Feb 12, 2015 #7
    I apologize for using boldface type alphabets but the equation is a simple transcendental equation I used bold alphabets to highlight the quantities primarily involved in my problem.Noting that it's a simple scalar equation how can I solve for a and b in terms of c1, c2 and θ?
     
  9. Feb 12, 2015 #8
    These are not transcedental equations. If just Θ is unknown It is not even system of equations since you have only one indepedent variable in each equation. Such trigonometric equations can be solved by transforming sum of two trigonometric function in one trigonometric function. See method http://www.sinclair.edu/centers/mathlab/pub/findyourcourse/worksheets/DiffEq/ConvertingTheForm_asinx.pdf [Broken].
    Or one can use theory of complex numbers.
     
    Last edited by a moderator: May 7, 2017
  10. Feb 12, 2015 #9

    PeroK

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    Why not

    ##a = c_1 cos(\theta) + c_2 sin(\theta)##
    ##b = -c_1 sin(\theta) + c_2 cos(\theta)##
     
  11. Feb 12, 2015 #10
    Yes!!!

    Thank You :)
     
  12. Feb 12, 2015 #11
    Could you briefly outline the procedure that was used to obtain the solution?
     
  13. Feb 12, 2015 #12

    PeroK

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    I just wrote it down.

    ##(c_1, c_2)## is the vector obtained by rotating the vector ##(a, b)## by ##-\theta## about the origin.

    So, to obtain ##(a, b)## you rotate ##(c_1, c2)## by ##+\theta##
     
  14. Feb 12, 2015 #13
    Is there an easy analytical way to arrive to the solution using ordinary algebra rather than using linear algebra? :confused:
     
  15. Feb 12, 2015 #14

    PeroK

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    Multiply the first equation by ##cos\theta## and the second by ##sin\theta## and add to get ##a##.
     
  16. Feb 12, 2015 #15
    Thank You once again :)
     
  17. Feb 12, 2015 #16

    Mark44

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    In future posts, please do not decorate what you write with all of the font features. This is not permitted under PF rules (https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/). What I have emphasized below applies to specific items in your post.
     
  18. Feb 14, 2015 #17
  19. Feb 14, 2015 #18

    HallsofIvy

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    Multiply the first equation by [itex] cos(\theta)[/itex]: [itex]a cos^2(\theta)- b cos(\theta)sin(\theta)= c_1 cos(\theta)[/itex]
    Multiply the second equation by [itex]sin(\theta)[/itex]: [itex]a sin^2(\theta)+ b sin(\theta)cos(\theta)= c_2 sin(\theta)[/itex]
    Add those two equations: [itex]a= c_1 cos(\theta)a+ c_2 sin(\theta)[/itex].

    Multiply the first equation by [itex]sin(\theta)[/itex], the second by [itex]cos(\theta)[/itex], and subtract to get b.
     
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