3 Graphs of y = a sin (bx + c) and y = a cos (bx + c)
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304
CHAPTER 10 Graphs of the Trigonometric Functions
Note from Example 1 that the graph of y = sin 1 2x + p4 2 is precisely the same as
the graph of y = sin 2x, except that it is shifted p>8 units to the left. Fig. 10.15 shows
the graphs of y = sin 2x and y = sin 1 2x + p4 2 . We see that the shapes are the same
and that the graph of y = sin 1 2x + p4 2 is about 0.4 unit (p>8 ≈ 0.39) to the left of
the graph of y = sin 2x.
In general, the effect of c in the equation y = sin1bx + c2 can be understood
x
when we write bx + c = b1x + bc 2. This means that the function y = sin1 bx + c2
is obtained as a result of adding the constant c>b to x in the function y = sin bx. As we
discussed in Section 3.5, by adding a constant to x, the graph of y = a sin bx is shifted
to the left or to the right. In this case, the graph is shifted to the left if c 7 0 and to the
right if c 6 0. The direction and magnitude of the shift is called the displacement (or
phase shift), and it is given by –c>b. Note that the displacement can be obtained by
solving for x in the equation bx + c = 0. In Example 1, the displacement is -p>8.
We can verify that the displacement is -c>b by noting corresponding points on the graphs
of y = sin bx and y = sin1bx + c2. For y = sin bx, when x = 0, then y = 0. For
y = sin1bx + c2, when x = -c>b, then y = 0. The point 1 -c>b, 02 on the graph of
y = sin1bx + c2 is -c>b units to the left of the point 10, 02 on the graph of y = sin bx.
We can use the displacement combined with the amplitude and the period along
with the other information from the previous sections to sketch curves of the functions
y = a sin1bx + c2 and y = a cos 1bx + c2, where b 7 0.
y
Period = p
1
0
p
y = sin (2x + 4 )
p
4
3p
4
p
2
5p
4
p
y = sin 2x
−1
p
shift
8
Fig. 10.15
LEARNING TIP
Carefully note the difference
between y = sin1bx + c 2 and
y = sin bx + c. Writing sin1bx + c 2
means to find the sine of the quantity bx + c, whereas sin bx + c means
to find the sine of bx and then add
the value c.
Important Quantities for Sketching Graphs of y = a sin (bx + c) and
y = a cos (bx + c)
Amplitude = 0 a 0
2p
Period =
b
c
Displacement = b
LEARNING TIP
Note that the constant c and the displacement - c>b differ in sign:
t *Gc 7 0, the graph is shifted to the
left, and displacement is negative.
(10.1)
These quantities allow us to evaluate the function at key values each one-fourth
period. Table 10.3 summarizes these key values for the cycle that starts at x = -c>b
and is completed at x = -c>b + period. A general illustration of the graph of
y = a sin1bx + c2 is shown in Fig. 10.16.
t *Gc 6 0, the graph is shifted to the
right, and displacement is positive.
Table 10.3
Key values
(one cycle)
- bc
- bc +
- bc +
- bc +
y = a sin 1bx + c2
0
y = a cos 1 bx + c2
a
0
0
-a
-a
0
0
a
period
4
period
2
3 # period
4
- bc + period
y
a
c
−b
y
2p
b
Since c < 0,
−c/b is positive
c
0
−b +
−a
y = a sin (bx + c), c > 0
Fig. 10.16
a
(a)
2p
b
x
0
For each
a > 0, b > 0
2p
b
a
c
c
−b
−b +
−a
y = a sin (bx + c), c < 0
(b)
2p
b
x
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10.3 Graphs of y = a sin 1bx + c2 and y = a cos 1bx + c2
305
E X A M P L E 2 Sketching the graph of y = a sin (bx + c)
Sketch the graph of y = 2 sin13x - p2.
First, note that a = 2, b = 3, and c = -p. Therefore, the amplitude is 2, the
period is 2p>3, and the displacement is - 1 -p>32 = p>3. (We can also get the displacement from 3x - p = 0, x = p>3.)
p
One-fourth of the period is 14 1 2p
3 2 = 6 , so key values for one full cycle start at p>3, end
at p, and are found every p>6 units in between. This cycle is represented by the magenta
portion of the graph in Fig. 10.17, obtained from the following table of important values.
y
2
Amplitude
p
_
3
0
2__
p
3
p
x
−2
Displacement
Period
Fig. 10.17
x
0
p
6
p
3
p
2
2p
3
5p
6
p
y
0
-2
0
2
0
-2
0
y
■
E X A M P L E 3 Sketching the graph of y = a cos (bx + c)
1
_
− 12
Amplitude
p
− _6
p
_
3
p
2p
__
3
−1
Period
Sketch the graph of the function y = -cos 1 2x + p6 2 .
First, we determine that the amplitude is 1, the period is 2p
2 = p,
x
p
p
and the displacement is - p6 , 2 = - 12
. Moreover, one-fourth of
the period is p>4, so key values for one cycle are found every p>4
p
units beginning at - 12
and ending at 11p
12 . This cycle is represented
by the magenta portion of the graph in Fig. 10.18, sketched from the
following table:
Displacement
Fig. 10.18
x
p
- 12
p
6
5p
12
2p
3
11p
12
y
-1
0
1
0
-1
■
Practice Exercise
1. For the graph of y = 8 sin12x - p>32 ,
determine amplitude, period, and
displacement.
y
E X A M P L E 4 Graph with b < 0
Sketch the graph of the function y = sin1 -2x + p3 2.
We first rewrite the function with a positive angle. We use the fact that
sin1 -x2 = - sin x (see Eq. 8.7) to write
1
0
y = sin1 -2x +
p
6
5p
12
2p 11p
3 12
7p
6
x
= sin1 - 12x -
p
6
5p
12
2p
3
11p
12
7p
6
0
-1
0
1
0
From this table, we sketch the graph in Fig. 10.19.
Fig. 10.19
p
322
= -sin12x -
p
32
Therefore, we have a = -1, b = 2, c = - p3 . We determine that the amplitude is 1,
the period is 2p , 2 = p, and the displacement is - 1 - p3 2 , 2 = p6 . Also, onefourth of the period is p4 , so key values are p4 units apart, starting at p6 and ending at 7p
6.
We now make a table of important values:
x
y
−1
p
32
■
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306
CHAPTER 10 Graphs of the Trigonometric Functions
E X A M P L E 5 Using a trigonometric function to model daylight hours
At 48° N latitude, the number of daylight hours during the longest day of the year
(June 21) is 16.1 hours. During the shortest day of the year (December 21), the number of
daylight hours is 8.31 hours. Approximate the number of hours h of daylight each day during a year with a function of the form h = a sin 1bt + c2 + d, where time is measured
in days. Cities at 48° N latitude include Paris, Vienna, and Victoria, British Columbia.
We determine the values of a, b, c, and d as follows:
1. The amplitude is half the distance between the maximum and the minimum day16.1 - 8.31
light hours, so a =
= 3.90.
2
2p
2p
2. The period is 365 days. Therefore,
= 365 and b =
.
b
365
3. The maximum of the sine function occurs at one fourth of the period, which is
365
4 = 91.25. Since the maximum for the data at hand occurs at t = 172 (June 21),
we must shift the sine function to the right 172 - 91.25 = 80.75 days. This gives
2p
- bc = 80.75, so that c = -80.75 * 365
= -1.39.
4. Finally, the range is 3 8.31, 16.14 instead of 3 -3.90, 3.904 , so we must shift the
sine function 16.1 - 3.90 = 12.2 hours vertically by setting d = 12.2.
h
20
(172, 16.1)
15
12.2
10
(353, 8.31)
5
50 100 150 200 250 300 350
t
Our sine model is
h = 3.90 sin a
Fig. 10.20
The graph is shown in Fig. 10.20.
2p
t - 1.39 b + 12.2
365
■
E XE R C I S E S 1 0 .3
In Exercises 1 and 2, graph the function if the given changes are made
in the indicated examples of this section.
1. In Example 3, if the sign before p>6 is changed, sketch the graph
of the resulting function.
2. In Example 4, if the sign before p>6 is changed, sketch the graph
of the resulting function.
In Exercises 3–26, determine the amplitude, period, and displacement
for each function. Then sketch the graphs of the functions.
p
3. y = sinax - b
6
p
4. y = 3 sinax + b
4
p
5. y = cos ax + b
6
p
6. y = 2 cos ax - b
8
p
7. y = 0.2 sina2x + b
2
p
8. y = -sina3x - b
2
9. y = -cos 12x - p2
11. y =
10. y = 0.4 cos a3x +
1
1
p
sina x - b
2
2
4
1
p
12. y = 2 sina x + b
4
2
1
p
13. y = 30 cos a x + b
3
3
15. y = sinapx +
p
b
8
17. y = 0.08 cos a4px -
p
b
3
14. y =
p
b
5
1
1
p
cos a x - b
3
2
8
1
b
3
19. y = -0.6 sin12px - 12
20. y = 1.8 sinapx +
21. y = 40 cos 13px + 22
23. y = sin1p2x - p2
22. y = 360 cos 16px - 12
3
p2
25. y = - cos apx +
b
2
6
1
1
26. y = p cos a x + b
p
3
1
1
24. y = - sina2x - b
p
2
In Exercises 27–30, write the equation for the given function with the
given amplitude, period, and displacement, respectively 1a 7 02.
27. sine, 4, 3p, - p>4
28. cosine, 8, 2p>3, p>3
29. cosine, 12, 1>2, 1>8
30. sine, 18, 4, - 1
In Exercises 31–34, explain why the given equations are correct. The
method using a graphing calculator is indicated in Exercise 31.
31. By viewing the graphs of y1 = sin x and y2 = cos 1x - p>22 ,
show that cos 1x - p>22 = sin x.
32. Show that cos 12x - 3p>82 = cos 13p>8 - 2x2 .
33. Show that sin1x>2 - 3p>42 = - sin13p>4 - x>22 .
34. Show that 2 ͉ sin1x - p>32 ͉ = sin1p>3 - x2 - sin1x - p>32
for -2p>3 6 x 6 p>3.
16. y = -2 sin12px - p2
In Exercises 35–40, solve the given problems. In Exercises 39 and 40,
use a graphing calculator to view the indicated curves.
p
b
2
35. Find the function and graph it for a function of the form
y = 2 sin12x + c2 that passes through 1 - p>8, 02 and for
which c has the smallest possible positive value.
18. y = 25 cos a3px +
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307
10.4 Graphs of y = tan x, y = cot x, y = sec x, y = csc x
36. Find the function and graph it for a function of the form
y = 2 cos 12x - c2 that passes through 1p>6, 22 and for which
c has the smallest possible positive value.
In Exercises 41–44, give the specific form of the equation by evaluating
a, b, and c through an inspection of the given curve. Explain how a, b,
and c are found.
37. A wave travelling in a string may be represented by the equation
x
t
y = A sin 2pa - b. Here, A is the amplitude, t is the time the
T
l
wave has travelled, x is the distance from the origin, and T is the
time required for the wave to travel one wavelength l (the Greek
letter lambda). Sketch three cycles of the wave for which
A = 2.00 cm, T = 0.100 s, l = 20.0 cm, and x = 5.00 cm.
42. y = a cos 1bx + c2
Fig. 10.21
41. y = a sin1bx + c2
Fig. 10.21
43. y = a cos 1bx + c2
Fig. 10.22
44. y = a sin1bx + c2
Fig. 10.22
y
y
0.8
5
38. The electric current i (in mA) in a certain circuit is given by
i = 3.8 cos 2p1t + 0.202, where t is the time in seconds. Sketch
three cycles of this function.
Ϫ1
0
39. A certain satellite circles the earth such that its distance y, in kilometres north or south (altitude is not considered) from the equator, is y = 7200 cos 10.025t - 0.252, where t is the time (in
min) after launch. View two cycles of the graph.
x
15
0
−0.8
Ϫ5
Fig. 10.22
Fig. 10.21
40. In performing a test on a patient, a medical technician used an
ultrasonic signal given by the equation I = A sin1vt + u 2 .
View two cycles of the graph of I vs. t if A = 5 nW>m2,
v = 2 * 105 rad>s, and u = 0.4.
x
5p
4
p
4
45. In Sydney, Australia, the number of daylight hours during the shortest day of the year (June 21) is 9.90 hours. During the longest day of
the year (December 21), the number of daylight hours is 14.4 hours.
Approximate the number of hours h of daylight each day during a
year with a function of the form h = a sin 1bt + c2 + d, where
time is measured in days.
Answers to Practice Exercise
1. amp. = 8, per. = p, disp. = p>6
10.4
Graphs of y = tan x, y = cot x, y = sec x, y = csc x
Graph of y = tan x t 3FDJQSPDBM
'VODUJPOT t "TZNQUPUFT t (SBQIT
of y = cot x, y = sec x, y = csc x
The graph of y = tan x is shown in Fig. 10.23. Then, knowing that csc x = 1>sin x,
sec x = 1>cos x, and cot x = 1>tan x (see Eq. 4.1 in Section 4.2), we are able to find
the values of y = csc x, y = sec x, and y = cot x from the indicated corresponding
reciprocal functions. Using these values, we can graph these functions, and in
Figs. 10.24–10.26 we show the resulting graphs.
y
y
Period
Period
4
4
Asymptote
2
2
0
p
−2
p
2
3p
2
2p
x
p
−2
−2
y = tan x
Fig. 10.23
p
2
p
3p
2
2p
x
−2
y = cot x
−4
0
Asymptote
−4
Fig. 10.24
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308
CHAPTER 10 Graphs of the Trigonometric Functions
■ Repeating from above the reciprocal
relationships that are the basis for graphing
y = cot x, y = sec x, and y = csc x, we have
csc x =
1
sin x
cot x =
1
tan x
sec x =
y
y
Period
Period
1
cos x
4
(10.2)
2
4
2
3p
2
p
p
0
−2
p
2
p
3p
2
2p
x
−2
0
−2
y = csc x
−4
p
2p
x
−2
Asymptote
y = sec x
p
2
Fig. 10.25
Asymptote
−4
Fig. 10.26
From these graphs, note that y = tan x and y = cot x have period p and have all
real numbers as their range. The functions y = sec x and y = csc x have period 2p, but
their ranges do not include the real numbers between –1 and 1.
The vertical dashed lines on the graphs are vertical asymptotes (see Section 3.4).
The curves approach these lines but never actually reach them. The values of x for
which the curve has an asymptote are not included in the domain of the function.
As we can see from the graphs, the functions have asymptotes, zeros, and maximum
or minimum values when x is a multiple of p>2, that is to say, at the same key values
which allowed us to sketch the graphs of the sine and cosine functions easily. The characteristics of the trigonometric functions at the key values from 0 to 2p are compared
in Table 10.4. For example, note that when the sine is zero, its reciprocal (the cosecant)
has an asymptote.
Table 10.4
Key
values
0
y = sin x
y = cos x
y = tan x
0
1
0
asymptote
p
2
1
0
p
0
-1
3p
2
-1
0
asymptote
0
1
0
0
2p
Key
values
0
y = csc x
y = sec x
y = cot x
asymptote
1
asymptote
p
2
1
asymptote
0
p
asymptote
3p
2
2p
-1
asymptote
-1
asymptote
0
asymptote
1
asymptote
To sketch functions such as y = a sec x, first sketch y = sec x and then multiply
the y-values by a. Here, a is not an amplitude, since the ranges of these functions are
not limited in the same way they are for the sine and cosine functions.