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Graphing Trigonometric Functions Question

  1. Feb 21, 2005 #1
    Ok I can graph sin(x) , cos(x) , and tan(x) pretty easily, but I'm having a hard time graphing the csc, sec, and cot ones. For the first three I just found the values of pi/2 pi 3pi/2 and 2pi. So for example pi/2 for sin(x) would be sin 90 or sin pi/2 which is equal to 1. I then just put a dot at 1 and did the same thing for the rest of the angles in radian measure and then just draw a smooth curve through them, but how do I do this with csc, sec, and cot?
     
  2. jcsd
  3. Feb 21, 2005 #2

    Integral

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    Consider what happens to csc([itex] \Theta [/itex]) When Sin([itex] \Theta [/itex])=0 .
     
  4. Feb 22, 2005 #3

    HallsofIvy

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    In other words, you can't draw "smooth" graphs for sec(x) and csc(x) (or for that matter tan(x) and cot(x)). They aren't "smooth"- they aren't even continuous.
     
  5. Feb 22, 2005 #4

    dextercioby

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    You should have seen that when u graphed "tan".You said u graphed it,so u have noticed that is "blows up" in certain points (namely where the cosine is zero).It's discontinuous as well.


    Daniel.
     
  6. Feb 22, 2005 #5
    [tex] \tan{x} [/tex], 'breaks up' at [tex] \frac{\pi}{2}(2k-1)[/tex], and [tex] \cot{x} [/tex], 'breaks up' at [tex] k\pi[/tex].
     
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