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3 Graphs of y = a sin (bx + c) and y = a cos (bx + c)

# 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 *Gc 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 *Gc 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

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

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 +

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.

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

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.

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