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Solving simultaneous trigonometric equations

  1. Dec 17, 2006 #1
    Hi, all I am new to this site and I was wonder if anyone could help.

    I took linear algear algebra last semster and am currently taking statics. I want to know if I can use some of the techniques I learned in linear algebra to solve simtultaneous equations which involve sine or cosines

    For example how can I solve this use linear algebra (In know how to solve by methods of subsition but I would like to learn a method which is faster).

  2. jcsd
  3. Dec 17, 2006 #2
    I only know how to solve system of equations when Ax=B and x=[x;y;z]. I am not sure what to do when x=[T,T:sin(x),cos(x)] like in the example above
  4. Dec 17, 2006 #3
    Maybe it would help to think about them geometrically
  5. Dec 17, 2006 #4

    matt grime

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    These aren't linear equations. So why would linear algebra solve them? It doesn't. You can linearize them, but that is an approximate answer (replace a function by the first two (possibly zero) terms of its Taylor series - then you'd have a linear system).
  6. Dec 25, 2006 #5

    Chris Hillman

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    Equations involving trig polynomials?

    Well, your example may suggest that this is not what you have in mind, but if you are interesting in the (practical) solution of equations involving trig polynomials (e.g. from problems in robotic motion), these can be attacked using methods from commutative algebra. The basis of these methods is Buchberger's algorithm, a common generalization of Gaussian reduction and Euclid's algorithm, which allows us to compute with ideals in polynomial rings. See for example the superb undergraduate textbook Ideals, Varieties and Algorithms by Cox, Little, and O'Shea.
  7. Dec 25, 2006 #6


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    Well, you're going to give up the notion that you can do this problem using linear algebra and nothing else. As the others noted, this is not a linear problem, so you're going to have to use some other tools.

    But there's no reason you can't use linear algebra as one of your tools. For example, if I desired to eliminate T from this system of equations, I tend to do it quicker if I do it via (a form of) Gaussian elimination, rather than solve one equation for T and substitute into the other.

    Or... you seem to know how to write this system in the form

    [tex]A \left( \begin{array}{c} T \\ \sin x \\ \cos x \end{array} \right) = b.[/tex]

    Well, if you have a solution to the original system of equations, then

    [tex]\left( \begin{array}{c} T \\ \sin x \\ \cos x \end{array} \right)[/tex]

    must be a solution to the linear system of equations

    (*) Av = b.
    So... you could first find the general solution for v, and then try and solve

    [tex]\left( \begin{array}{c} T \\ \sin x \\ \cos x \end{array} \right) = v.[/tex]

    However, I would prefer to eliminate x, rather than eliminate T. You can do it purely algebraically, by invoking sinĀ² x + cosĀ² x = 1. Hint: you don't have to take square roots or inverse trig functions to eliminate x.
    Last edited: Dec 25, 2006
  8. Dec 26, 2006 #7


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    As you have been told, these equations are not linear and so "Linear Algebra" methods will not work. Relatively simple algebra will, however. From the first equation sin(x)= -(T/20)cos(60)= -T/40 (since cos(60 degrees)= 1/2). The second equation is cos(x)= T/40- 1= sin(x)- 1. That's a single trigonometric equation for x and can be solved- though not trivially. For example, replace sin(x) by [itex]\sqrt{1- cos^2(x)} and replace cos(x) by y: you need to solve [itex]y= -\sqrt{1- y^2}- 1[/itex].
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