# -bx + c

1. Sep 30, 2013

### majinkenji

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. Sep 30, 2013

### 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).

3. Sep 30, 2013

### majinkenji

Thanks! Had to play around with it for a bit!

4. Oct 1, 2013

### HallsofIvy

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