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Trig graph problem

  1. Sep 22, 2005 #1
    Please Help,urgent

    i have a trigonometry question and i really dont know how to solve i checked couple of books but still no hope here it is:

    im supposed to draw the graph of y=cos¶/3(x-1) but i couldnt

    so please help..
     
  2. jcsd
  3. Sep 22, 2005 #2

    Tom Mattson

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    Your equation is showing up funny on my browser.

    Is it supposed to be this:

    [tex]
    y=cos(3x-1)
    [/tex]

    or this:

    [tex]
    y=cos(\frac{1}{3x-1})
    [/tex]

    ?
     
  4. Sep 22, 2005 #3

    arildno

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    y=cos¶/3(x-1)???????????????????????????????????????????????????
    Neither can I. Please type in the CORRECT formula.
     
  5. Sep 22, 2005 #4

    Lisa!

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    I think it's something else!
     
  6. Sep 22, 2005 #5
    y=cos pi/3(x-1)

    now ok?
     
  7. Sep 22, 2005 #6

    arildno

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    Ok, so it is:
    [tex]y=\cos(\frac{\pi}{3x-1})[/tex]

    Hints:
    1. Find the point of singularity, where y is not defined.
    2. Determine the set of points were y is -1, 0, 1 respectively.
    In particular, see what this implies for the function's behaviour very close to the singularity found in 1.
     
  8. Sep 22, 2005 #7
    ok it is not that y is equal to cos multiplied by pi/3 multiplied by (x-1)
     
  9. Sep 22, 2005 #8

    Tom Mattson

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    Still not OK, and I suspect that it's because you aren't placing your parentheses properly.

    What you have written is this:

    [tex]y=cos(\frac{\pi}{3x-1})[/tex].

    Now that's a perfectly legitimate function, but I strongly suspect that it is not the one you are supposed to graph, simply because it's too difficult for a trigonometry course. I am inclined to think that you really meant this:

    [tex]y=cos[\frac{\pi}{3}(x-1)][/tex].

    Am I right?
     
  10. Sep 22, 2005 #9
    (x-1) isnt near 3
     
  11. Sep 22, 2005 #10

    Tom Mattson

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    I don't know what that means.
     
  12. Sep 22, 2005 #11

    arildno

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    özge:
    If Tom Mattson is right, here's how you could have written this properly:
    y=cos(pi*(x-1)/3)
     
  13. Sep 22, 2005 #12
    u are right
     
  14. Sep 22, 2005 #13

    Tom Mattson

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    This can't be right. cos without an argument makes no sense.
     
  15. Sep 22, 2005 #14

    arildno

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    Do you know what the cosine function is??
    What is its argument?
     
  16. Sep 22, 2005 #15
    what u wrote :[tex]y=cos[\frac{\pi}{3}(x-1)][/tex]
    was right im sorry im new to these things and i diddnt know how to write
     
  17. Sep 22, 2005 #16

    Tom Mattson

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    Alrighty then!

    Here are a couple of tips to get you started.

    *Examine the graph of [itex]y=cos(x)[/itex].

    *Note that if you have the graph of some function [itex]y=f(x)[/itex] then you can obtain the graph of [itex]y=f(x+a)[/itex] by shifting the graph of [itex]y=f(x)[/itex] horizontally by [itex]a[/itex] units. You move to the right if [itex]a<0[/itex] and to the left if [itex]a>0[/itex].

    *The period [itex]T[/itex] of a cosine function [itex]y=A\cos[k(x+a)][/itex] is [itex]T=\frac{2\pi}{k}[/itex].

    Give it a shot and see how you do, OK?
     
  18. Sep 22, 2005 #17
    ok, thanks
     
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