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Solving simultaneous trig equations

  1. Nov 20, 2009 #1
    a step by step explanation would be appreciated =]]
    as part of the static equilibrium unit, im asked to solve the following:

    -11.7cosA + 11.4cosB = 0 .......I
    11.7sinA + 11.4sinB = 4.9 ......II

    how do i get the angles A and B??
    (btw this problem bugged me for quite a while already)
     
  2. jcsd
  3. Nov 20, 2009 #2
    Have you tried squaring both equations and then adding them..
     
  4. Nov 20, 2009 #3
    well that would still leave me with 2 variables wouldn't it?
     
  5. Nov 20, 2009 #4
    You try this yourself: get a relation between cos A and cos B from I, use the identity sin^2 x + cos^2 x = 1, then use the relation to eliminate (or better substitute) one of the variables in II . From that you should be able to solve one of the variables and so the other.
     
  6. Nov 20, 2009 #5
    Hello!
    Although I didn't post this question, I am intrigued as to what the answer is. My equation when I substitute in is always too long, or at least it looks that way. Can someone show me what the equation whould look like when you substitute in the value? If it is the same, I'll try again from there. If you don;t want to post it here (to allow the other person to make an attmept), can someone IM me? Please! This is annoying me!:mad::frown:
     
  7. Nov 20, 2009 #6
    Yes where it can have multiple solutions.

    I was envisioning something like cos(A.B) or similar. Once you know what values A.B are equal to you can pick any value for A and B as far the meet with the requirement.
     
  8. Nov 21, 2009 #7
    The system indeed has multiple solutions due to the fact that the sine and cosine functions are periodic functions. Furthermore, the two equations intersect only once in a period, the point of intersection of which is the solution to the system at that certain period.
     
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