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Linear Combinations of Trig Functions - Finding Roots

  1. Mar 11, 2010 #1
    Hi there
    I was wondering if there is a simple way to solve for the roots of a complicated summation of trig functions that can't be combined with any simple identities.
    I have an equation of the form:

    0 = sin(8x-arctan(4/3))+3.2sin(16x+pi/2)

    where the two sines have different amplitudes, phases and a frequency that differs by a factor of 2. I would like to find the values of x on [0,pi/2] that satisfy this condition. I don't know very much about Fourier analysis but is there some method for solving this?
    I need an analytical solution - not an approximation
  2. jcsd
  3. Mar 11, 2010 #2


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    I don't think you need to go into anything complicated. What does [tex]Sin(\theta + \frac{\pi}{2})[/tex] equal?
  4. Mar 11, 2010 #3
    Yes, it equals cosx but the frequency is still different so you can't easily combine the two terms. I dont think there is any way of combining the two terms into one simple function. The solution must have an amplitude that oscillates as a function of x; like a form:

    Not sure how to solve for the parameters though
  5. Mar 11, 2010 #4


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    Oh right, I didn't see that, maybe this isn't as simple as it looks. I'm assuming you've exausted double angle and angle addition formulas?
  6. Mar 11, 2010 #5
    Yes, Ive tried almost every standard substitution. I dont think this will solve the problem since the amplitude varies with x. Plotting the function in maple reveals that it doesn't have the form of a simple cosine function
  7. Mar 12, 2010 #6
    you could still use the cosx if you treat it as sin(16x+pi/2) = cos(16x) = cos(2*8x) = 2cos^2(8x)-1
    whereas you could expand the first term sin(8x- ... getting it in terms of sin8x and cos8x, substituting sin8x=sqrt(1-cos^2(8x)) and squaring to eliminate the sqrt leaves you with a 4th degree polynomial of cos8x. which is technically solvable

    if i thought it through enough, didn't try in written
  8. Mar 12, 2010 #7
    No this wont work either, you will end up with a function that has non integer cos terms
    Unless I misread what you were saying
  9. Mar 12, 2010 #8


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    Why? What is your application?
  10. Mar 12, 2010 #9
    Its an assignment question,
    but i'm also curious if there is a generalized technique for approaching this type of problem
  11. Mar 12, 2010 #10
    you don't need integer terms to solve a 4th degree polynomial, though you can just multiply the eq with sth to make them integer

    Sin(8x-arctan(4/3))+3.2Sin(16x+\pi/2)= \pm 0.6 \sqrt{1-Cos^2(8x)}-0.8Cos(8x)+6.4Cos^2(8x)-3.2 = 0

    [tex] \pm 3 \sqrt{1-a^2}-4 a+32a^2-16=0 [/tex]
    where a=cos(8x)
    Last edited: Mar 12, 2010
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