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Homework Help: Trig equations with multiple angles

  1. Jul 9, 2007 #1
    1. The problem statement, all variables and given/known data
    What I'm scratching my head with is trying to find mechanically, a way of finding all the solutions of the expression [tex]sin(a \theta ) - sin(b \theta) = 0[/tex] (where a and b are integers (for now) with a > b)

    2. Relevant equations
    I can get the roots that are regular rotations from a specific starting angle by first solving [tex]a \theta = \pi - b \theta[/tex] and then by letting all subsequent values of [tex] \theta = \frac{(1+2k) \pi}{a+b}[/tex] (where k is any integer)...it is also easy to spot that [tex]\theta = n \pi[/tex] (where n is an integer)

    3. The attempt at a solution
    what I'm having trouble with is finding quickly the sneaky roots such as [tex]\theta = \frac{2 \pi}{3} or \frac{4 \pi}{3} [/tex] in the expression [tex] sin4 \theta - sin \theta = 0[/tex](amongst others, though this is the question that I started with) that aren't found with the above method.

    I have tried to expand [tex] sin4 \theta - sin \theta = 0[/tex] but after:

    [tex] 2sin2\theta cos2\theta - sin\theta = 0[/tex]
    [tex] 4sin\theta cos\theta (cos^2\theta - sin^2\theta ) - sin\theta = 0[/tex]
    [tex] 4cos\theta (2cos^2\theta - 1 ) -1 = 0[/tex]
    [tex] 8cos^3\theta - 4cos\theta - 1 = 0[/tex] I can't see a simple way of solving it without using either newton's method, the cubic formula (there must surely be a simpler way than that though), or getting the protractor out and searching for them.

    I'm betting there is a simple method and either I just can't remember it or have not come across it yet...can anyone throw me some hints?
    Last edited: Jul 9, 2007
  2. jcsd
  3. Jul 9, 2007 #2


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    Homework Helper

    You could try for something a little less fancy:

    means that
    [tex]\theta=\phi + 2 \pi n[/tex]
    [tex]\theta=\pi - \phi + 2 \pi n[/tex]

  4. Jul 9, 2007 #3


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    Homework Helper

    You forget the family (?, wonder if this is used correctly) of solution:
    [tex]a \theta = b\theta + 2k' \pi \Rightarrow \theta = \frac{2k' \pi}{a - b}[/tex]
    No, that's not correct. [tex]\frac{1 + 2k}{a + b}[/tex] does not mean that it's a natural number, or even an integer. And, nor does [tex]\frac{2k'}{a - b}[/tex]. Say, we have k = 1, and a = 3, b = 1? [tex]\frac{1 + 2k}{a + b} = \frac{1 + 2}{3 + 1} = \frac{3}{4} \notin \mathbb{Z}[/tex]

    You've dropped the solution [tex]\sin \theta = 0[/tex], when moving from the former line to the later one.

    Well, you can see NateTG's post, if you want to have a neater version of it. :)
    Last edited: Jul 9, 2007
  5. Jul 10, 2007 #4
    agh, I knew it was something simple!...I forgot the family: [tex]a \theta = b\theta + 2k' \pi \Rightarrow \theta = \frac{2k' \pi}{a - b}[/tex] :grumpy::smile::smile: cheers!

    hmm...that wasn't what I was saying...I meant just that you fling integers into k and get an answer, not that the resulting fraction is an integer.

    agh...true! :redface:

    Thankyou Viet Dao
    Last edited: Jul 10, 2007
  6. Jul 10, 2007 #5
    Cheers for the reply but I was looking for a way to find specific elements of the solution set namely [tex]\frac {2n\pi}{3}[/tex] but it has been pointed out to me what I didn't do :)
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