6.4 Graphing Sine and Cosine Functions
Referring back to The Unit Circle, let’s create a table of angles and values of sine and
cosine functions. We will consider θ in radian measure.
angle θ
0
π/6
sin θ
0
½
cos θ
1
3 /2
π/2
2π/3
2 /2
3 /2
1
3 /2
3π/4
5π/6
2 /2
½
π/4
π/3
π
0
angle θ
π
7π/6
sin θ
0
-½
cos θ
-1
- 3 /2
2 /2
½
5π/4
4π/3
- 2 /2
-½
0
-½
3π/2
5π/3
- 2 /2
- 3 /2
-1
- 3 /2
- 2 /2
- 3 /2
-1
7π/4
11π/6
- 2 /2
-½
2π
0
2 /2
3 /2
1
0
½
By plotting (θ, sin θ), we create the graph representing one cycle of y = sin x.
By plotting (θ, cos θ), we create the graph representing one cycle of y = cos x.
To graph these functions on your calculator, make sure you are in radian mode and once
you type in the equation be sure to select ZOOM-TRIG. This allows the x interval to be
partitioned every π/2.
Vertical stretching can be determined by a coefficient on the trigonometric function. This
coefficient is multiplied to the y-values in the previous chart.
Example Graph y = 3sin x on the interval [-2π, 2π].
Notice that the y-values are multiplied by 3. The important angles to note when graphing
are 0, π/2, π, 3π/2, and 2π.
Try the following:
Graph on [-2π, 2π].
y = 2cos x
y = 4sin x
y = 5cos x
y = ½ sin x
Answers:
Notice the pattern in these graphs. What is the length of one period? Sine and Cosine
both have a period of 2π. Examine the graphs above, notice the absolute values of the
minimum and maximum y-values correlate with the coefficient of the functions.
What do you think a negative coefficient would do to the graphs of sine and cosine
functions?
Example Graph y = -2cos x on [0, 2π].
Notice this graph is a refection in the x-axis of the previous problem y = 2cos x.
Try the following:
Graph on [0, 2π].
y = -4sin x
y = - ½ sin x
y = -3/2 cos x
Answers:
How could we affect the length of the period of these graphs? Is there a way to have 2
cycles of sine on the interval [0, 2π]? We can establish this by having a multiplier to the
x-value. For instance, y = sin 2x means we will graph sine function containing 2 cycles
on the interval [0, 2π].
So what would happen to the graph of y = sin ( ½ x)? Only half of a cycle will occur on
the interval [0, 2π]. So that means that one cycle will be on the interval [0, 4π].
Example Graph y = cos (2x) on [0, 2π]. State the minimum and maximum values of the
graph.
The minimum is at –1 when x = π/2, 3π/2 and the maximum is at 1 when x = 0, π, 2π.
We could algorithmically calculate the length of one cycle, or period.
2
Period =
, where b is the coefficient on x.
b
Try the following:
Graph one period of the functions below. In addition, determine the length of a period
and state the minimum and maximum values.
y = 3sin ( ½ x)
y = -2cos x
y = -sin (3x)
y = 3cos (2x)
Answers:
Min: -1 when
x = 3π
Max: 1 when
x=π
Min: -1 when
x = 0 and 2π
Max: 1 when
x=π
Min: -1 when
x = π/6, 5π/6, & 3π/2
Max: 1 when
x = π/2, 7π/6, & 11 π/6
Min: -1 when
x = π/2
Max: 1 when
x = 0 and π
Note: The coefficient of the function is called the amplitude of the function.