Identify each of the six trigonometric functions of θ with exactly
one of the line segments in Figure 5.1.10, keeping in mind that the
radius of the circle is 1. To get you started, we have sin θ = MP
N
θ
x
(why?).
O
M
1
14. For Exercise 13, how would you draw the line segments in Fig-
ure 5.1.10 if θ was in QII? Recall that some of the trigonometric
functions are negative in QII, so you will have to come up with a
convention for how to treat some of the line segment lengths as
Figure 5.1.10
negative.
15. For any point ( x, y) on the unit circle and any angle α, show that the point Rα( x, y) defined by Rα( x, y) = ( x cos α − y sin α, x sin α + y cos α) is also on the unit circle. What is the geometric interpretation of Rα( x, y)? Also, show that R− α( Rα( x, y)) = ( x, y) and Rβ( Rα( x, y)) = Rα+ β( x, y).
Properties of Graphs of Trigonometric Functions • Section 5.2
109
5.2 Properties of Graphs of Trigonometric Functions
We saw in Section 5.1 how the graphs of the trigonometric functions repeat every 2 π radi-
ans. In this section we will discuss this and other properties of graphs, especially for the
sinusoidal functions (sine and cosine).
First, recall that the domain of a function f ( x) is the set of all numbers x for which the
function is defined. For example, the domain of f ( x) = sin x is the set of all real numbers,
whereas the domain of f ( x) = tan x is the set of all real numbers except x = ± π ,
,
,
2
± 3 π
2
± 5 π
2
.... The range of a function f ( x) is the set of all values that f ( x) can take over its domain.
For example, the range of f ( x) = sin x is the set of all real numbers between −1 and 1 (i.e.
the interval [−1,1]), whereas the range of f ( x) = tan x is the set of all real numbers, as we
can see from their graphs.
A function f ( x) is periodic if there exists a number p > 0 such that x + p is in the domain of f ( x) whenever x is, and if the following relation holds:
f ( x + p) = f ( x) for all x
(5.1)
There could be many numbers p that satisfy the above requirements. If there is a smallest
such number p, then we call that number the period of the function f ( x).
Example 5.3
The functions sin x, cos x, csc x, and sec x all have the same period: 2 π radians. We saw in Section 5.1 that the graphs of y = tan x and y = cot x repeat every 2 π radians but they also repeat every π
radians. Thus, the functions tan x and cot x have a period of π radians.
Example 5.4
What is the period of f ( x) = sin 2 x ?
Solution: The graph of y = sin 2 x is shown in Figure 5.2.1, along with the graph of y = sin x for comparison, over the interval [0, 2 π]. Note that sin 2 x “goes twice as fast” as sin x.
y
y = sin 2 x
1
y = sin x
0
x
π
π
3 π
π
5 π
3 π
7 π
2 π
4
2
4
4
2
4
−1
Figure 5.2.1
Graph of y = sin 2 x
For example, for x from 0 to π , sin x goes from 0 to 1, but sin 2 x is able to go from 0 to 1 quicker,
2
just over the interval [0, π ]. While sin x takes a full 2 π radians to go through an entire cycle (the
4
largest part of the graph that does not repeat), sin 2 x goes through an entire cycle in just π radians.
So the period of sin 2 x is π radians.
110
Chapter 5 • Graphing and Inverse Functions
§5.2
The above example made use of the graph of sin 2 x, but the period can be found analyt-
ically. Since sin x has period 2 π,1 we know that sin ( x + 2 π) = sin x for all x. Since 2 x is a number for all x, this means in particular that sin (2 x + 2 π) = sin 2 x for all x. Now define
f ( x) = sin 2 x. Then
f ( x + π) = sin 2( x + π)
= sin (2 x + 2 π)
= sin 2 x (as we showed above)
= f ( x)
for all x, so the period p of sin 2 x is at most π, by our definition of period. We have to show
that p > 0 can not be smaller than π. To do this, we will use a proof by contradiction. That
is, assume that 0 < p < π, then show that this leads to some contradiction, and hence can not
be true. So suppose 0 < p < π. Then 0 < 2 p < 2 π, and hence
sin 2 x = f ( x)
= f ( x + p) (since p is the period of f ( x))
= sin 2( x + p)
= sin (2 x + 2 p)
for all x. Since any number u can be written as 2 x for some x (i.e u = 2( u/2)), this means that sin u = sin ( u + 2 p) for all real numbers u, and hence the period of sin x is as most 2 p.
This is a contradiction. Why? Because the period of sin x is 2 π > 2 p. Hence, the period p of
sin 2 x can not be less than π, so the period must equal π.
The above may seem like a lot of work to prove something that was visually obvious from
the graph (and intuitively obvious by the “twice as fast” idea). Luckily, we do not need to
go through all that work for each function, since a similar argument works when sin 2 x is
replaced by sin ωx for any positive real number ω: instead of dividing 2 π by 2 to get the
period, divide by ω. And the argument works for the other trigonometric functions as well.
Thus, we get:
For any number ω > 0:
2 π
2 π
sin ωx has period
csc ωx has period
ω
ω
2 π
2 π
cos ωx has period
sec ωx has period
ω
ω
π
π
tan ωx has period
cot ωx has period
ω
ω
If ω < 0, then use sin (− A) = −sin A and cos (− A) = cos A (e.g. sin (−3 x) = −sin 3 x).
1We will usually leave out the “radians” part when discussing periods from now on.
Properties of Graphs of Trigonometric Functions • Section 5.2
111
Example 5.5
The period of y = cos 3 x is 2 π and the period of y
x is 4 π. The graphs of both functions are
3
= cos 12
shown in Figure 5.2.2:
y = cos 1 x
y
2
y = cos 3 x
1
0
x
π
π
π
2 π
5 π
π 7 π 4 π 3 π 5 π 11 π 2 π 13 π 7 π 5 π 8 π 17 π 3 π 19 π 10 π 7 π 11 π 23 π 4 π
6
3
2
3
6
6
3
2
3
6
6
3
2
3
6
6
3
2
3
6
−1
Figure 5.2.2
Graph of y = cos 3 x and y = cos 1 x
2
We know that −1 ≤ sin x ≤ 1 and −1 ≤ cos x ≤ 1 for all x. Thus, for a constant A = 0,
−| A| ≤ A sin x ≤ | A| and
− | A| ≤ A cos x ≤ | A|
for all x. In this case, we call | A| the amplitude of the functions y = A sin x and y = A cos x.
In general, the amplitude of a periodic curve f ( x) is half the difference of the largest and
smallest values that f ( x) can take:
(maximum of f ( x)) − (minimum of f ( x))
Amplitude of f ( x) =
2
In other words, the amplitude is the distance from either the top or bottom of the curve to
the horizontal line that divides the curve in half, as in Figure 5.2.3.
y
| A|
| A|
2 | A|
0
x
π
π
3 π
π
5 π
3 π
7 π
2 π
4
2
4
4
2
4
| A|
−| A|
Figure 5.2.3
Amplitude = max−min
2
= | A|−(−| A|)
2
= | A|
Not all periodic curves have an amplitude. For example, tan x has neither a maximum
nor a minimum, so its amplitude is undefined. Likewise, cot x, csc x, and sec x do not have
an amplitude. Since the amplitude involves vertical distances, it has no effect on the period
of a function, and vice versa.
112
Chapter 5 • Graphing and Inverse Functions
§5.2
Example 5.6
Find the amplitude and period of y = 3 cos 2 x.
Solution: The amplitude is |3| = 3 and the period is 2 π
2 = π. The graph is shown in Figure 5.2.4:
y
3
2
3
1
6
0
x
π
π
π
2 π
5 π
π
7 π
4 π
3 π
5 π
11 π
2 π
−1
6
3
2
3
6
6
3
2
3
6
3
−2
−3
Figure 5.2.4
y = 3 cos 2 x
Example 5.7
Find the amplitude and period of y = 2 − 3 sin 2 π x.
3
Solution: The amplitude of −3 sin 2 π x is
3
|−3| = 3. Adding 2 to that function to get the function
y = 2 − 3 sin 2 π x does not change the amplitude, even though it does change the maximum and
3
minimum. It just shifts the entire graph upward by 2. So in this case, we have
max − min
5 − (−1)
6
Amplitude =
=
=
= 3 .
2
2
2
2 π
The period is
= 3. The graph is shown in Figure 5.2.5:
2 π
3
y
5
4
3
3
6
2
1
3
0
x
3
3
9
3
4
2
4
−1
Figure 5.2.5
y = 2 − 3 sin 2 π x
3
Properties of Graphs of Trigonometric Functions • Section 5.2
113
Example 5.8
Find the amplitude and period of y = 2 sin ( x 2).
Solution: This is not a periodic function, since the angle that we are taking the sine of, x 2, is not a
linear function of x, i.e. is not of the form ax + b for some constants a and b. Recall how we argued that sin 2 x was “twice as fast” as sin x, so that its period was π instead of 2 π. Can we say that
sin ( x 2) is some constant times as fast as sin x ? No. In fact, we see that the “speed” of the curve keeps
increasing as x gets larger, since x 2 grows at a variable rate, not a constant rate. This can be seen in
the graph of y = 2 sin ( x 2), shown in Figure 5.2.6:2
2
1.5
1
0.5
y
0
-0.5
-1
-1.5
-2
0
π
π
3 π
2 π
2
2
x
Figure 5.2.6
y = 2 sin ( x 2)
Notice how the curve “speeds up” as x gets larger, making the “waves” narrower and narrower.
Thus, y = 2 sin ( x 2) has no period. Despite this, it appears that the function does have an amplitude,
namely 2. To see why, note that since |sin θ| ≤ 1 for all θ, we have
|2 sin ( x 2)| = |2| · |sin ( x 2)| ≤ 2 · 1 = 2 .
In the exercises you will be asked to find values of x such that 2 sin ( x 2) reaches the maximum value
2 and the minimum value −2. Thus, the amplitude is indeed 2.
Note: This curve is still sinusoidal despite not being periodic, since the general shape is still that of
a “sine wave”, albeit one with variable cycles.
So far in our examples we have been able to determine the amplitudes of sinusoidal curves
fairly easily. This will not always be the case.
2This graph was created using Gnuplot, an open-source graphing program which is freely available at http://
gnuplot.info. See Appendix B for a brief tutorial on how to use Gnuplot.
114
Chapter 5 • Graphing and Inverse Functions
§5.2
Example 5.9
Find the amplitude and period of y = 3 sin x + 4 cos x.
Solution: This is sometimes called a combination sinusoidal curve, since it is the sum of two such
curves. The period is still simple to determine: since sin x and cos x each repeat every 2 π radians,
then so does the combination 3 sin x + 4 cos x. Thus, y = 3 sin x + 4 cos x has period 2 π. We can see this in the graph, shown in Figure 5.2.7:
5
4
3
2
1
y
0
-1
-2
-3
-4
-5
0
π
π
3 π
2 π
5 π
3 π
7 π
4 π
2
2
2
2
x
Figure 5.2.7
y = 3 sin x + 4 cos x
The graph suggests that the amplitude is 5, which may not be immediately obvious just by looking
at how the function is defined. In fact, the definition y = 3 sin x +4 cos x may tempt you to think that
the amplitude is 7, since the largest that 3 sin x could be is 3 and the largest that 4 cos x could be is
4, so that the largest their sum could be is 3 + 4 = 7. However, 3 sin x can never equal 3 for the same
x that makes 4 cos x equal to 4 (why?).
There is a useful technique (which we will discuss further in Chapter 6) for showing
that the amplitude of y = 3 sin x + 4 cos x is 5. Let θ be the angle shown in the right
5
triangle in Figure 5.