Graphs of Trigonometric Functions

We begin by considering the general formula for the graphs of sine and cosine.

The general formula for sine: *y* = *A*sin(*Bx* *C*) +
*D*.

The general formula for cosine: *y* = *A*cos(*Bx* *C*) +
*D*.

Let’s 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’s look to see what these different values do to the graph
of sine. 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’s 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*).

Here, we notice that the two graphs are the same, except for
the fact that *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, 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.

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?

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 black).

**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’s look at a really complicated graph.

**Example 5**: Graph *y* = 3cos(2*x* π) + 1.

We start by noting what *A*,
*B*, *C* and *D* are in the above
equation.

*A* = 3, *B* = 2, *C* = π and *D*
= 1.

So, that means that we are going to graph cos(*x*), 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.

So, 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)

(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). Notice that *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 look of the graph. Thus, we look for the graph that
looks like *y* = 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.