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Trigometric functions

  1. Mar 19, 2007 #1
    State the period and phase shift of the function y= -4 tan (1/2x + 3pie/8).

    I know the answer is pie; -3pie/8, but I don't understand the process could someone explain how the period and phase shift are found in these functions.
     
  2. jcsd
  3. Mar 19, 2007 #2

    Integral

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    Could you share with us the general forumla given in your text or notes for this type of function?
     
  4. Mar 19, 2007 #3

    HallsofIvy

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    Do you know the definitions of "phase shift" and period? What is the period of tan(x)?
     
  5. Mar 20, 2007 #4
    The period of function y = tan k0 is pie/k where k>o. I'm not sure about the phase shift.
     
  6. Mar 20, 2007 #5
    Maybe this'll help:
    Consider y=x^2

    y=(x-2.5)^2 is the same function, shifted to the right 2.5 units.

    y=(2x-6)^2 would have to first be written as
    y=(2(x-3))^2
    This is the function y=x^2 shifted 3 units to the right. The 2 does something else to the function (stretches it vertically in this case.)
    Can you get (x-#) in your problem?
     
  7. Mar 20, 2007 #6

    HallsofIvy

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    tan(x) has period [itex]\pi[/itex]. In particular, [itex]tan(0)= tan(\pi)[/itex].
    One period starts at x= 0 and ends at [itex]x= \pi[/itex].

    Okay, one period of [itex]-4tan((1/2)x- 3\pi/8)[/itex] "starts" when [itex](1/2)x- 3\pi/8= 0[/itex] and ends when [itex](1/2)x- 3\pi/8= \pi[/itex]. What is the "starting" value of x (the phase shift) and what is the difference between the two values of x (the period)?
     
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