Phase, Period, and Vertical Shifts of Trigonometric Functions

We begin by considering
the general formula for the graphs of sine and cosine. The general formulas for
sine and cosine are *y* = *A*sin(*Bx*
*C*) +
*D* and *y* = *A*cos(*Bx* *C*) +
*D*.

We shall begin by labeling the various components.

* A* amplitude of the function = (max-min)/2
(vertical stretch of the graph)

* B* stretch/shrink on the *x*-axis. (It compresses or expands the graph.)

*B* has the following relationship with
the period:

,
where *P* is the period. Solving for *P*, we have .

* C*/*B*
the phase shift of the graph (the shift left
(if *C*/*B* is neg.) or right (if *C*/*B* is pos.))

* D* the vertical shift of the graph.

Let us examine
what effect these different values have on the graph of sin(*x*). Below is the graph of *y* = sin(*x*) on the interval [0, 2π].

**Figure 1**: Graph of *y* =
sin(*x*).

Let us consider
what the graph of *y* = *A*sin(*Bx*)
on the interval [0, 2π/*B*] looks like.

**Figure 2**: Graph of *y* = *A*sin(*Bx*).

Notice that the
graph looks precisely the same. The only difference is the range of the
function is now [-*A*, *A*] instead of [-1, 1] and the period is
2π/*B*
instead of 2π. Let us
look at some examples of the two graphs imposed on top of each other, to see
exactly what the changes described above do to the graphs.

**Example 1:**

Consider the
graphs of *f*(*x*) = sin(*x*) and *g*(*x*)
= sin(2*x*).

**Figure 3**: Graphs of *f*(*x*) = sin(*x*) and *g*(*x*) = sin(2*x*).

The two graphs
are the same, except *g*(*x*) has period equal to 2π/2 = π. Thus, on
the interval [0, 2π], the
graph of *g*(*x*) will repeat twice while *f*(*x*) will only show one period.

**Example 2:**

Consider the
graphs of *f*(*x*) = sin(*x*) and *g*(*x*)
= 2sin(*x*).

**Figure 4**: Graphs of *f*(*x*) = sin(*x*) and *g*(*x*) = 2sin(*x*).

Here, we notice
that the two graphs are the same, except for the fact that *g*(*x*) has an amplitude
twice as large as *f*(*x*). That translates to a vertical
stretch by a factor of 2.

**Example 3:**

Consider the
graphs of *f*(*x*) = sin(*x*) and .

**Figure 5**: Graphs of *f*(*x*) = sin(*x*) and .

Here, *g*(*x*)
has period π and an amplitude half that of *f*(*x*).
This means that the graph of *g*(*x*) repeats twice for every period of *f*(*x*)
and its range is [-1/2, 1/2] instead of [-1, 1].

Things get a bit more complicated once we introduce shifts into the graph.

** **

** **

**Example 4:**

** **

** **How do the graphs of and *g*(*x*) = cos(*x*) compare?

**Solution:**

We begin by
graphing *f*(*x*). Notice that this is the same graph as sin(*x*), except that we have shifted every point to the *left* by π/2.

The following
graph has *f*(*x*) (solid blue) and the original graph sin(*x*) (dashed red).

**Figure 6**: Graphs of and sin(*x*).

Next, we graph *g*(*x*)
= cos(*x*).

**Figure 7**: Graph of *g*(*x*) = cos(*x*).

Observe that *f*(*x*)
and *g*(*x*) are the same graph. That means that sin(*x*) and cos(*x*) are
essentially the same graph, and differ only by a phase shift of π/2.

Now, let us look at a more complicated graph.

**Example 5:**

Graph *y* = 3cos(2*x* π) + 1.

**Solution:**

We start by
noting what *A*, *B*, *C* and *D* are in the above equation. Observe
that *A* = 3, *B* = 2, *C* = π and *D* = 1. Our graph will look like cosine,
but with some modifications.

(i) The graph is stretched vertically by a factor of 3.

(ii) The period is now 2π/2 = π instead of 2π.

(iii) There is a phase shift of π/2 to the right.

(iv) There is a vertical shift of 1 up.

Let us handle
these in pieces. We begin by graphing 3cos(2*x*),
since it will look the same as cos(*x*),
only with some changes to the labeling of the *x-* and *y*-axes.

**Figure 8**: Graph of *y* = 3cos(2*x*).

Now, we take into account the phase shift, which moves the graph to the right by π/2.

**Figure 9**: Graph of *y* = 3cos(2*x* π).

Lastly, we take into account the vertical shift, which moves the graph up 1, which gives us our final answer.

**Figure 10**: Graph of *y* =
3cos(2*x* π) + 1.

**Example 6:**

Match the following equations with their graphs. Give reasons for your answers.

(i)
*y* = sin(*x*/2)

(ii)
*y* = cos(*x* π/2)

(iii)

(iv)
*y* = 3sin(*x*) 1

(v)

(vi)
*y* = sin(2*x* 2π)

(a)

(b)

(c)

(d)

(e)

(f)

**Solution:**

(a) corresponds with (i). The graph starts at the origin, has period equal to 2π/(1/2) = 4π, and has no amplitude shift, phase shift, or vertical shift.

(b) corresponds with (v). We see that the graph starts at the origin, the period is equal to 2π/2 = π, and the amplitude is 1/2.

(c) corresponds
with (ii). This looks like the graph of sin(*x*),
but that is not an option. Notice, though, that since sin(*x*) and cos(*x*) are related
by a phase shift of π/2, (ii)
is the only likely choice, since there are no amplitude, vertical or period
shifts.

(d) corresponds with (iii). The graph starts at 1/2 (which is the amplitude) and has period equal to 2π/2 = π.

(e) corresponds
with (vi). *y* = sin(2*x* 2π) is the
graph of *y* = sin(2*x*), shifted to the right by 2π/2 = π. But the
period of *y* = sin(2*x*) is 2π/2 = π, so, the
phase shift does not affect the appearance of t he graph. Hence, we look for
the graph that resembles sin(2*x*).

(f) corresponds
with (iv). Notice that the graph is not symmetric about the *x*-axis, so there is a vertical shift.
Since more of the graph lies *below*
the *x*-axis, we have that it is a
vertical shift down. The period is unaffected, but the amplitude is equal to
(2-(-4))/2 = 3.

Now, let us
consider the graph of the function: *y*
= *A*tan(*Bx* *C*) +
*D*.

As was
illustrated in Figure 2, the graphs of *y*
= *A*tan(*Bx*) and *y* = tan(*x*) look the same. The ranges are still
(-∞, ∞), so *A* does not have much effect on the
graph. The period is π/*B* instead of π. Also, the vertical asymptotes will occur at π/2*B* ±
*k*π/*B*,
instead of at π/2 ±
*k*π.

And as before,* C*/*B*
is the phase shift of the graph. The shift is to the left if *C*/*B*
is negative or to the right if *C*/*B* is positive. The phase shift also
moves the vertical asymptotes of the graph by *C*/*B*.* *Finally, *D* is the
vertical shift of the graph.

**Example 7:**

Graph *y* = 2tan(3*x* π/2) 1.

**Solution:**

We start by
noting what *A*, *B*, *C* and *D* are in the above equation. Observe
that *A* = 2, *B* = 3, *C* = π/2 and *D*
= -1. Our graph will look like cosine, but with some modifications.

(i) The graph is stretched vertically by a factor of 2.

(ii) The period is now π/3 instead of π.

(iii) There is a phase shift of π/6 to the right.

(iv) There is a vertical shift of 1 down.

Let us handle
these in pieces. We begin by graphing 2tan(3*x*),
since it will look the same as tan(*x*),
only with some changes to the labeling of the *x-* and *y*-axes.

**Figure 11**: Graph of *y* =
2tan(3*x*).

Now, we take into account the phase shift, which moves the graph to the right by π/6.

**Figure 12**: Graph of *y* =
2tan(3*x* π/2).

Lastly, we take into account the vertical shift, which moves the graph down 1, which gives us our final answer.

**Figure 13**: Graph of *y* =
2tan(3*x* π/2) 1.

The graphs of *y* = *A*csc(*Bx* *C*) +
*D* and *y* = *A*sec(*Bx* *C*) +
*D* follow exactly the same rules as
those of *y* = *A*sin(*Bx* *C*) +
*D* and *y* = *A*cos(*Bx* *C*) +
*D*, respectively, since they are
constructed by first drawing the corresponding sine/cosine graph.