1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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


    User Avatar
    Science Advisor

    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
    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


    User Avatar
    Science Advisor

    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)?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook