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Consider a trig function such as: y = A cos (bx - c)

  • Thread starter majinkenji
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  • #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
 

Answers and Replies

  • #2
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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
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).
 
  • #3
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).
Thanks! Had to play around with it for a bit!
 
  • #4
HallsofIvy
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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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