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Determining the Period of sin(ax)*cos(bx)

  1. Nov 3, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm trying to work out the period of a function of the form sin(ax)*cos(bx), where a =/= b.

    I'm trying solve how the values for a and b relate to the period. I've graphed a lot of functions, but I'm struggling to notice any patterns. The period always seems somewhat related to a and b, but I can't for the life of me work out how.

    I'd like to be able to show this algebraically for all values of a and b.
    2. Relevant equations
    Possibly the double angle identities:
    sin2x = 2sinxcosx
    cos2x = cos2x–sin2x = 2cos2x–1 = 1–2sin2x

    3. The attempt at a solution
    I've been trying to take out common values and use the double angle identities, but so far nothing seems to make sense. Am I just missing something obvious?
  2. jcsd
  3. Nov 3, 2014 #2


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    Hello deusy and welcome to PF. Nice mug shot :)

    You have something really good to start with, right? You know the periodicty of ##\sin x## is ##2\pi## and for ##\cos x## it is also ## 2\pi##.
    So ##\sin (x+P) = \sin(x) ## with ## P = 2\pi##. And ##\sin (x+nP) = \sin(x) ## with ## n## an integer number.

    Now can you figure out the periodicity for ##\sin(ax)## ? Definitely has something to do with a :)
    Same question for ##\cos(bx)##.

    If you have two periods (different because they tell us ##a \ne b\;##), then what is the period of the product ?
    Think of what conditions it must fulfil: it has to be an integer multiple of .., and also ..
  4. Nov 3, 2014 #3
    You can't generally be sure that your function is periodic. But you can always express it as the sum of two periodic functions, with generally different periods.
  5. Nov 4, 2014 #4
    I know the period is 2pi/a for sin(ax), and the same (2pi/b) for cos(bx).

    I'm not sure what you mean about the integer multiple bit? Would you mind explaining a bit more? :)
  6. Nov 4, 2014 #5


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    Hello again,

    Yes, good. So you know that ##\sin\left ( a (x + {2\pi\over a})\right) = \sin(ax)##. If you use this twice, you see that also ##\sin\left ( a (x + {4\pi\over a})\right) = \sin(ax)##. So :

    If a function repeats with a period of ##2\pi/a##, it also repeats after twice that, three times, etcetera. If one of these multiples coincides with a multiple of ##2\pi/b##, you have found an interval over which the product repeats !

    This only works in cases where the ratio a/b is rational; that's why it would be nice to know if there were limitations to a and b in the original problem description. If there aren't, then stating such conditions will be part of the answer.

    The other approach (writing the product as a sum) is also worth investigating. But my feeling is that it doesn't help all that much to make things easier. Chet ?
  7. Nov 4, 2014 #6
    I think both methods lead to the same conclusion.

  8. Nov 5, 2014 #7
    Eeugh. I think I know what you're getting at but I can't quite grasp it. If the multiples coincide, how do you determine what the period of the overall sin(ax)*cos(bx) function is? o:)

    Also, there's no limitations on the values of a and b. I'm trying to work out the period in general form for ALL values of a and b, if that's even possible.

  9. Nov 5, 2014 #8


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    A tried my best, but we can do step by step:
    I expect the first paragraph in my post #5 is understood. Right ?
    And I lose you when I write "If one of these multiples coincides with a multiple of ##\; 2\pi/b\ ##, you have found an interval over which the product repeats !", or not ? (because you ask "how do you determine the period of the over-all function" -- when that's what I had just written):

    In case of doubt, work out an example...a=4, b=6 is a simple one.

    But if I summarize: you got 2pi/a as period of the sine and 2pi/b for the cosine.
    Is it clear that n times 2pi/a (n has to be an integer) is also a period of the sine, since it fulfils $$ \sin\left ( a (x + n\;{2\pi\over a})\right) = \sin(ax)\ ?$$
    Likewise m times 2pi/b (m an integer) for the cosine factor ?

    Then the next bit is somewhat a spoiler:

    So that if P = n x 2pi/a = m x 2pi/b both the sin factor and the cos factor are the same for x+P as for x, which means P is the period of sin(ax) * cos(bx) ?

    And it's fine with me that there are no limitations on a and b, but if there are no n and m for which n/a = m/b exactly, then the product is aperiodic.
  10. Oct 30, 2015 #9
    I was facing same problem. I was looking for a generalized formula. After scratching my head for some time and found the period =2pi/a*m=2pi/b*n. where m and n are smallest possible integers such that a/b=m/n. For example if a=0.1 and b=0.25 then m=2 and n=5. So for given values of a and b the period will be 40 pi. However I have not plotted any graph. Please plot and check it. The formula should work.
  11. Oct 31, 2015 #10


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    I think you mean a/b=n/m. Yes, that's the right formula. I don't know how you propose to plot a graph, though.
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