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Homework Help: Solving Trigonometric Equations

  1. Dec 11, 2005 #1
    I have a question when solving trigonometric equations.

    For example:

    Find all the solutions in the interval [0,2pi)

    [tex]\sin \theta \tan \theta = \sin \theta \][/tex]

    If you choose to divide through by [tex]\sin \theta\][/tex] we get,

    [tex]\tan \theta = 1\][/tex] such that [tex]\sin \theta \ne 0\][/tex]
    otherwise we are essentially dividing both sides by zero, which we
    cannot do.

    Do we need to be careful when solving trigonometric equations using
    multiplication and division?

    We're dividing by a term that can take on the value of zero. Does
    this have any special name? How can I learn more about this?
    Are there any techniques to use when solving trig equations
    so this doesn't happen?
     
  2. jcsd
  3. Dec 11, 2005 #2
    you can devide through even if it has the posibility of being = to zero.
    the reason for this is because trig functions are FUNCTIONS. when you take calculus you'll see that you can examin how a function behaves close to zero.
    as φ -> 0 for sinφ/sinφ , that ratio actually -> 1.

    when it is = 0 try not deviding it out.
    also, recall tanφ = sinφ/cosφ, so tan(0) = 0/1 = 0.
     
  4. Dec 11, 2005 #3
    When you devide, you must make sure that the quantity that you use in the denominator is not zero. But in this case you can bypass division like this :

    [tex]\sin \theta \tan \theta = \sin \theta[/tex]
    [tex]\sin \theta \tan \theta - \sin \theta = 0[/tex]
    [tex]\sin \theta ( \tan \theta -1) = 0[/tex]

    Then you solve both

    [tex]\sin \theta = 0[/tex]
    and
    [tex] \tan \theta -1 = 0[/tex]

    regards
    marlon
     
  5. Dec 11, 2005 #4
    Squaring the terms is another thing which you have to be careful. You get extraneous roots out there.
     
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