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How to solve for theta for this trig question?

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data

    cos(theta) = -3/4, pi <= theta <= 2pi

    2. Relevant equations



    3. The attempt at a solution

    I forgot how to do this. How do I use the special triangles to do this question? Do I need to square the sqrt(3) / 2 one or something? Thanks.
     
  2. jcsd
  3. Nov 10, 2012 #2

    SteamKing

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    Try the Pythagorean theorem.
     
  4. Nov 10, 2012 #3
    So I use Pythagorean to get the opposite length? (sqrt(7) / 4) ?
     
  5. Nov 10, 2012 #4

    haruspex

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    It depends what you're trying to find. That will help if you want some other trig function of theta, but not much use if you want theta itself.
    If it's a numerical value for theta that you want, at some point you will have to use a calculator or whatever to compute an arccos. But the more interesting aspect of the question is getting an answer in the right quadrant. Do you know the relationships between cos(x), cos(-x), cos(pi+x), etc?
     
  6. Nov 10, 2012 #5

    SteamKing

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    Not all trig functions require the use of a calculator to evaluate.
     
  7. Nov 10, 2012 #6

    haruspex

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    True, but certainly to find arccos in this case.
     
  8. Nov 11, 2012 #7

    SteamKing

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    Not really. If you know basic trig values for 45-45-90 and 30-60-90 triangles, solving this problem should be possible without a calculator.
     
  9. Nov 11, 2012 #8

    haruspex

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    For |cos(theta)| = 3/4? Are you sure about that? If 3/5 I'd agree.
     
  10. Nov 11, 2012 #9

    SammyS

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    I'm pretty sure that zeion has left the building ...

    At any rate, given a value for cos(θ), for π ≤ θ ≤ 2π, one can determine in which quadrant θ lies, and thus determine the values of all the other trig functions at angle θ.

    Of course, we're still not sure about just what zeion was supposed to do in regards to this problem.
     
  11. Nov 11, 2012 #10
    I'm supposed to

    i) Sketch the angle and state the related acute angle
    ii) Determine the exact value of each trigonometric ratio

    The previous 3 questions I could do because they were all values I could find on the special triangle so they were easy. But I'm not sure how to relate 3/4 to the special triangles?
     
  12. Nov 11, 2012 #11

    HallsofIvy

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    Since [itex]\theta[/itex] lies between [itex]\pi[/itex] and [itex]2\pi[/itex], and [itex]cos(\theta)[/itex] is negative, [itex]\theta[/itex] is in the fourth quadrant. Use a calculator to find [itex]cos^{-1}[/itex] of -4/3 and add [itex]\pi[/itex]. Equivalently, find [itex]cos^{1}[/itex] of 3/4 and subtract from [itex]2\pi[/itex].
     
  13. Nov 11, 2012 #12

    SammyS

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    That should be, θ is in the third quadrant.

    cos(θ) is positive for θ in the first & fourth quadrants.

    cos(θ) is negative for θ in the second & third quadrants.
     
    Last edited: Nov 11, 2012
  14. Nov 11, 2012 #13

    haruspex

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    It is possible that a numerical answer is not required. An expression involving 'arccos(3/4)' may be acceptable. The trick is to get the expression right. arccos is defined to return a value in [0, pi), but the required answer is in [pi, 2pi].
     
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