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Summing sines/cosines (Harmonic Addition Theorem)

  1. Jun 5, 2012 #1
    I have a mathematical model which, in part, does a calculation based on the location of certain points on a unit circle.
    I am just working in 2d so for some arbitrary values of a,b,c,d,e,f
    In the case of the unit circle equally divided so that I have three points on the unit circle (120° apart) this would look like this:
    x=a cos (θ) + b cos (θ) + c cos (θ)
    y=d sin (θ) + e sin (θ) + f sin (θ)
    Now, I want to examine what happens if any two of these points are "free". That is, only one of the points is fixed and the others may individually take on any value from 0° to 360°.
    Here is my question:
    In this case I believe the model must use the Harmonic Addition Theorem, yes?
    I believe this is the case since the two free positions on the circle have the same period but are out of phase with each other since they are moving independent of each other and can take on any arbitrary value. Is that right?
    In this case my model would then look like this (where A and B represent the differences in phase)
    x=a cos (θ) + (b cos (θ-A) + c cos (θ-B) )
    y=d sin (θ) + (e sin (θ-A) + f sin (θ-B))
    Is this correct?
     
  2. jcsd
  3. Jun 5, 2012 #2
    If you are using different points on the circle, you need different arguments i.e. ##\theta_1, \theta_2, \theta_3##.

    In any case I believe the formulas you are asking for are the sum-to-product formulas. Look at the bottom of this link: trig formulas, or if you really want the harmonic addition theorem, it's here harmonic addition
     
  4. Jun 6, 2012 #3
    I am not asking for the formulas. I already know those! What am I asking is if this is an appropriate application of the Harmonic Addition Theorem. Is it?
     
  5. Jun 6, 2012 #4
    In that case I'm not sure what you're asking. I'm not sure what your model is suppose to be or what symmetry you're trying to take advantage of. The harmonic addition theorem is just a formula, so there's not really a wrong application of it.
     
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