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Solving systems of equations with trig functions

  1. Sep 3, 2013 #1
    I've stumbled upon a system of equations that involves trig functions...

    100cos(θ) + 200cos(ω) = 250
    100sin(θ) + 200sin(ω) = 0

    How do you go about solving a system like this?

    It's nonlinear, so linear algebra won't work...
     
  2. jcsd
  3. Sep 3, 2013 #2

    chiro

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  4. Sep 3, 2013 #3
  5. Sep 3, 2013 #4
    I solved the system numerically, but I was wondering if any analytical solution exists.
     
  6. Sep 3, 2013 #5

    chiro

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    Sorry I thought your angles were on the same line.

    In this case yes the numerical method is the best method. You are better off using a matrix method iteration scheme and you iterate until you get a change that is small enough from your last iteration.
     
  7. Sep 3, 2013 #6
    An analytic solution is possible. Here are two potential approaches. One uses algebra only. The other uses geometry.

    Algebra

    We might try simplifying things by getting rid of the trig functions. In this case we can let ##\sin \theta = x## and ##\sin \omega = y##. Then ##\cos \theta = \sqrt{1 - x^2}## and ##\cos \omega = \sqrt{1 - y^2}##. With a little bit of massaging I believe you should get a quartic equation in one variable (or at any rate something like that) and can solve the whole thing.

    Geometry

    View the equations as components of an equation in vectors:

    ##100\begin{bmatrix}\cos \theta \\ \sin \theta\end{bmatrix} + 200\begin{bmatrix}\cos \omega \\ \sin \omega\end{bmatrix} = \begin{bmatrix}250 \\ 0 \end{bmatrix}.## Since the two vectors on the left hand side are unit vectors, this is really a geometry problem: you are asked to find the angles of a triangle whose side lengths are 100, 200, and 250 units. (You'll have to do a little work to figure out the signs of θ and ω.) Use your favorite techniques from analytic geometry (Law of Cosines).
     
  8. Sep 4, 2013 #7
    Just isolate 100 cos θ in one equation, and 100 sin θ in the other equation.
    Square both equations.
    Add the two and, after using sin2+cos2=1, you are left with an equation involving cos ω

    Solve for ω. and return to the original to solve for θ.

    [I get cos ω = 92500 / 100000 = 0.925]
     
  9. Sep 5, 2013 #8
    cyto, how would I go about solving the general case? Can you explain that to me please? I'd prefer a nice, short, well-written explanation. I mean the case:

    [tex]F=a[/tex]
    [tex]G=b[/tex]

    where F and G are some type of sine and cosine expressions. Well, what I'm getting at is that you really haven't solved this problem until you can. That's all.
     
    Last edited: Sep 5, 2013
  10. Sep 5, 2013 #9

    verty

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    There is no general solution. Each problem is unique.
     
  11. Dec 18, 2014 #10
    Hello, jumping in a tad (sic...) late on this post but I'd like to ask what numerical method you would have used to solve this set of equation.

    I have derived the equations of equilibrium and geometric compatibility of a statically indeterminate system. I am left with a system of 5 equations and 5 unknown. The equations use trig functions and 3 different unknown angles.
     
  12. Dec 18, 2014 #11
     
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