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Desperate Trig help

  1. Dec 25, 2011 #1
    I have these practice trig problems for a test. I understand the simple graphs but its these more complicated ones.

    y=2sec(3x-pi)

    my approach:

    I first find the period and phase shift. Since the period of a normal secant function which is 2pi. Then the period of this one is 2pi/3. The phase shift is pi/3. There is gonna a be a vertical strech of 2.

    After that I am stuck because I dont know how to exactly graph this even though I have all the information. Its mostly because of the odd period, 2pi/3. Really confused.
     
  2. jcsd
  3. Dec 25, 2011 #2

    SammyS

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    I don't know about you, but I'd rather plot cosine rather than secant.

    In addition to the information you have deduced, which all looks good, I like to look at the y-intercept, and find some points where the cosine is 1, where it's -1, and where it's zero.

    For the y-intercept, set x=0 → y = cos(3(0)-π) = cos(-π) = ?

    To find where cos(θ) = 1, of course an obvious place is where θ = 0, so that's 3x-π = 0. But in general cos(θ) = 1, where θ is an integer multiple of 2π. Of course you can generate these in terms of x, by using the easy case of the x which satisfies 3x-π = 0, and then using your information on period.

    You can do similar for the the values of x which make cos(3x-π) equal to -1 and equal to zero. You can also generate these values from the information in the above paragraph, and then using the fact that the cosine is equal to -1 at 1/2 a period on either side of where it is +1. The zeros occur 1/4 of a period on either side of where it is +1.

    Once you have cos(3x-π), then sketch sec(3x-π), then stretch that vertically to get y=2sec(3x-π).
     
  4. Dec 25, 2011 #3
    A good way to check your graphs is this (after you've made an attempt of course!).
     
  5. Dec 26, 2011 #4
    Thanks SammyS. I orginally planned to graph 1/cos which was the secant but it made it more confusing lol. Now I see what you mean by graph cos first.

    My question is that can I also apply to secant and graph sine first or cotangent and graph tangent first? Im gonna have all three of those on my test as well.
     
  6. Dec 26, 2011 #5
    Naturally, yes.
     
  7. Dec 26, 2011 #6

    eumyang

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    I don't think you really want to "apply to secant and graph sine first." :wink:
     
  8. Dec 26, 2011 #7
    lol cosecant but thanks cause now im gonna ace this test.
     
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