1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    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


    User Avatar
    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 :)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Trig equations with multiple angles
  1. Trig equation (Replies: 1)

  2. Trig equation (Replies: 4)

  3. Trig. Equation (Replies: 8)

  4. Trig equation (Replies: 10)