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Homework Help: Trig pi question

  1. May 27, 2010 #1
    1. The problem statement, all variables and given/known data
    If pi ≤ Θ ≤ 2pi and cos Θ = cos 1, what is the value of Θ? Round to nearest hundredth.


    2. Relevant equations



    3. The attempt at a solution
    cos 1 = ~.54 then I didn't really understand how to interpret the "If pi ≤ Θ ≤ 2pi" Thanks for your help.
     
  2. jcsd
  3. May 27, 2010 #2

    Mentallic

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    Homework Helper

    Well since [itex]cos(\theta)=cos(1)[/itex] then this would instinctively mean [itex]\theta=1[/itex] but the restriction is [itex]\pi \leq \theta \leq 2\pi[/itex] or [itex]3.14 \leq \theta \leq 6.28[/itex] (approx). Obviously [itex]1<\pi[/itex] so we can't use the instinctive solution.

    What other values of [itex]\theta[/itex] make the same value cos(1)? There are infinite values of [itex]\theta[/itex] that do this. Take a look at the cosine graph and find where [itex]\theta=1[/itex] (or x=1). Now the y-value at that point is cos(1). Where else does the same y value occur between [itex]\pi < \theta< 2\pi[/itex]?
     
  4. Jun 12, 2010 #3
    Apologies for the rather crude diagram (follow the link below) but hopefully this will help along with the advice in the previous post.

    http://yfrog.com/6fpf1kj

    :smile:
     
  5. Jun 12, 2010 #4

    cronxeh

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    Gold Member

    180 deg <= Θ <= 360 deg

    cos Θ = cos 1

    cos 359 = cos 1

    Particularly,
    Θ= 2*pi*n - 1
    Θ=2*180*n - 1 = 360*n - 1
    Θ= 359, 719, 1079, 1439, etc degrees
    180 < 359 < 360 for this case
     
    Last edited: Jun 12, 2010
  6. Jun 12, 2010 #5
    Think about this.. the cos graph in the region [itex]0 \leq \pi \leq 2\pi[/itex] is symmetrical about [itex]\pi[/itex].

    At [itex]x=1[/itex] you have [itex]y=cos(1)[/itex], and you need to find the other point, i.e. at a different x value, where you get the same y value.

    Perhaps if you knew the difference from the lower limit to the first x value, you could say this was the same as from the upper limit to the value you require? :wink: if that makes sense, look at the graph again. :smile:

    Also, you should be able to see that [itex]cos(1)\neq cos(359)[/itex]
     
  7. Jun 12, 2010 #6

    cronxeh

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    heh. Ok enlighten us, for which value of 180<x<360 is cos(x)=cos(1) ?

    And dont tell me its cos 5.283185 = cos 1. Nobody said we doing this in radians
     
    Last edited: Jun 12, 2010
  8. Jun 12, 2010 #7
    .. hah yeah, what's happened there, is that I've done some bad maths! :blushing: How strange! :redface:

    Hopefully my method made some sense though. I've been working in radians not degrees as that's what was stated in the question. So what I did was:

    [tex]cos(1)=0.54=cos(2\pi -1)=cos(5.28)[/tex]

    So [itex]\theta=5.28 radians=302.7 degrees[/itex]

    .. Perhaps I need a refresher in Trig! :wink: ..never mind..
     
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