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-bx + c

  1. Sep 30, 2013 #1
    Consider a trig function such as: y = A cos (bx - c)

    For the phase shift, we would use (-c/b); which aligns with the original function equation and makes sense to me.

    But in the case of a trig function such as: y = A cos (-bx + c)

    For the phase shift, we would use (+c/-b); which would be a negative phase shift instead of the positive phase shift. This is probably because my school just gave us a mechanism for phase shift of simply dividing c by b. But doing this:

    -bx + c = 0 -> x = -c/-b -> is this legal algebraically?

    Seems to be different answer, +c which would align with the function. (y = A cos (-bx + c))

    In the case of a negative B and a positive C, could I just put it together like y = [-b(x-(-c))] and use the standard c/b type approach?

    Any help in 'visualizing' this problem would be greatly appreciated.

    Thanks so much for your patience.

    mk
     
  2. jcsd
  3. Sep 30, 2013 #2

    Mark44

    Staff: Mentor

    Let's look at the equation y = cos(-bx + c) = cos(-b(x - c/b))

    Relative to the graph of y = cos(-bx) (which is exactly the same as y = cos(bx)), the -c/b term is a phase shift. The shift is to the right by c/b units of the graph of y = cos(bx).
     
  4. Sep 30, 2013 #3
    Thanks! Had to play around with it for a bit!
     
  5. Oct 1, 2013 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I think the simplest way of looking at it is this: the "basic" function, [itex]cos(\theta)[/itex], has period [itex]2\pi[/itex]. It starts one period at 0 and ends at [itex]2\pi[/itex]. So cos(bx+ c) starts a period where bx+ c= 0 and ends at [itex]bx+ c= 2\pi[/itex]. Solving x, a period starts at [itex]x= -c/b[/itex] and ends at [itex]x= (2\pi- c)/b[/itex]. The "phase shift" is -c/b and the period is [itex](2\pi- c)/b-(-c/b)= 2\pi/b[/itex].
     
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