5.5 Graphs of Other Trigonometric Functions

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5.5 Graphs of Other Trigonometric Functions

Objectives:•Understand the graph of y = sin x.•Graph variations of y = sin x.•Understand the graph of y = cos x.•Graph variations of y = cos x.•Use vertical shifts of sine and cosine curves.•Model periodic behavior.Dr .Hayk Melikyan

Department of Mathematics and CS

[email protected]


The Graph of y = sinx


Some Properties of the Graph of y = sinx

The domain is ( , ). The range is [–1, 1].

The period is 2 . The function is an odd function:sin( ) sin .x x


Graphing Variations of y = sinx


Example: Graphing a Variation of y = sinx

Determine the amplitude of Then graph

and for

Step 1 Identify the amplitude and the period.

The equation is of the form y = Asinx with A = 3. Thus, the amplitude is This means that the maximum value of y is 3 and the minimum value of y is – 3. The period is

3sin .y x siny x3siny x 2 .x

3siny x3.A

2 .


Example: Graphing a Variation of y = sinx (continued)


and for

Step 2 Find the values of x for the five key points. To generate x-values for each of the five key points, we begin by dividing the period, by 4. The cycle begins at x1 = 0. We add quarter periods to generate x-values for each of the key points.


2 ,

24 2 1 0x 2 0

2 2x

3 2 2x

4

32 2

x 5

32

2 2x



Step 3 Find the values of y for the five key points.

3siny x ( , )x y

3sin 0 3 0 0y (0,0)0

2

32

2

3sin 3 1 32

y

x

3sin 3 0 0y

33sin 3 ( 1) 3

2y

3sin 2 3 0 0y

,32

,0

3, 3

2

2 ,0




and for

Step 4 Connect the five keypoints with a smooth curve and graph one complete cycle of the given function.


3siny x

siny x

amplitude = 3

period 2


Amplitudes and Periods


Example: Graphing a Function of the Form y = AsinBx

Determine the amplitude and period of Then graph the function for


The equation is of the form y = A sinBx with A = 2 and

The maximum value of y is 2, the minimum value of y is –2. The period of tells us that the graph completes one cycle from 0 to

12sin .

2y x

0 8 .x

1.

2B

44 .

2A 2 2

412

B amplitude: period:


Example: Graphing a Function of the Form y = AsinBx (continued)

Step 2 Find the values of x for the five key points.

To generate x-values for each of the five key points, we begin by dividing the period, by 4. The cycle begins at x1 = 0. We add quarter periods to generate x-values for each of the key points.

4 ,

44 1 0x 2 0x 3 2x

4 2 3x 5 3 4x



Step 3 Find the values of y for the five key points.1

2sin2

y xx ( , )x y

0

2

3

4 12sin 4 2sin 2 2 0 0

2y

1 32sin 3 2sin 2 ( 1) 2

2 2y

12sin 2 2sin 2 0 0

2y

12sin 2sin 2 1 2

2 2y

12sin 0 2sin 0 0

2y

(0,0)

,2

2 ,0

3 , 2

4 ,0




Step 4 Connect the five key pointswith a smooth curve and graphone complete cycle of the given function.

12sin .

2y x

0 8 .x

amplitude = 2

period 4




Step 5 Extend the graph in step 4 to the left or right as desired. We will extend the graph to include

the interval

12sin .

2y x

0 8 .x

0 8 .x

amplitude = 2

period 4


The Graph of y = Asin(Bx – C)


Example: Graphing a Function of the Form y = Asin(Bx – C)

Determine the amplitude, period, and phase shift of

Then graph one period of the function.

Step 1 Identify the amplitude, the period, and the

phase shift.

3sin 2 .3

y x

3sin 23

y x

3, 2,

3A B C

amplitude: period:

phase shift:

3 3A 2 22B

132 3 2 6

CB


Example: Graphing a Function of the Form y = Asin(Bx – C) (continued)

Step 2 Find the values of x for the five key points.period

4 4 1 6

x 2

2 3 56 4 12 12 12

x

3

5 5 3 8 212 4 12 12 12 3

x

4

2 8 3 113 4 12 12 12

x

5

11 11 3 14 712 4 12 12 12 6

x




x 3sin 23

y x

( , )x y

6

512

3sin 2 3sin 3sin 0 3 0 06 3 3 3

y

5 5 2 33sin 2 3sin 3sin

12 3 6 6 6

3sin 3 1 32

y

23

2 4 33sin 2 3sin 3sin

3 3 3 3 3

3sin 3 0 0

y

,06

5,3

12

2,0

3



Step 3 (cont) Find the values of y for the five key points.

x 3sin 23

y x

( , )x y

1112

76

11 11 93sin 2 3sin 3sin

12 3 6 6 6

33sin 3 ( 1) 3

2

y

7 7 63sin 2 3sin 3sin

6 3 3 3 3

3sin 2 3 0 0

y

11, 3

12

7,0

6


Step 4 Connect the five key points with a smooth curve and graph one complete cycle of the function


amplitude = 3

period

phase shift6

3sin 2 .3

y x


The graph of y = cosx


Some Properties of the Graph of y = cosx

The domain is ( , ). The range is [–1, 1].

The period is 2 . The function is an even function:cos( ) cos .x x


Sinusoidal Graphs

The graph of is the graph ofwith a phase shiftof

The graphs of sinefunctions and cosinefunctions are calledsinusoidal graphs.

siny xcosy x

.2

cos sin2

x x


Graphing Variations of y = cosx


Example: Graphing a Function of the Form y = AcosBx

Determine the amplitude and period of

Then graph the function for


4cos .y x2 2.x

amplitude: period:

cos 4cosy A Bx x

4 4A 2 22

B


Example: Graphing a Function of the Form y = AcosBx (continued)




4cos .y x2 2.x

period 2 14 4 2

1 0x 2

1 10

2 2x

3

1 11

2 2x 4

1 31

2 2x 5

3 12

2 2x



Step 3 Find the values of y for the five key points.x ( , )x y4cosy x

0

12

1

4cos( 0) 4cos0 4 1 4y

14cos 4cos 4 0 0

2 2y

4cos 1 4cos 4 ( 1) 4y

(0, 4)

1,0

2

(1,4)



Step 3 Find the values of y for the five key points.x ( , )x y4cosy x

32

2

3 34cos 4cos 4 0 0

2 2y

4cos 2 4cos(2 ) 4 1 4y

3,0

2

2, 4





Step 4 Connect the five key

points with a smooth curve

and graph one complete

cycle of the given function.

4cos .y x2 2.x

1 2 3

-5

-4

-3

-2

-1

1

2

3

4

5

x

y

amplitude = 4

period = 2





Step 5 Extend the graph

to the left or right

as desired.

4cos .y x2 2.x

-2 -1 1 2

-5

-4

-3

-2

-1

1

2

3

4

5

x

y

amplitude = 4

period = 2


The Graph of y = Acos(Bx – C)


Example: Graphing a Function of the Form y = Acos(Bx – C)



Step 1 Identify the amplitude, the period, and the

phase shift.

3cos 2 .

2y x

3cos( ) cos(2 )

2y A Bx C x

amplitude: period:

phase shift:

3 32 2

2 22B

2 2CB


Example: Graphing a Function of the Form y = Acos(Bx – C) (continued)



Step 2 Find the x-values for the five key points.

3cos 2 .

2y x

period4 4

1 2x

2 2 4 4x

3 04 4

x 4 0

4 4x

5

24 4 4 2

x




x 3cos 2

2y x ( , )x y

2

4

3 3 3cos 2 cos( ) cos0

2 2 2 2

3 31

2 2

y

3 3 3cos 2 cos cos

2 4 2 2 2 2

30 0

2

y

3,

2 2

,04




x 3cos 2

2y x ( , )x y

0

4

3 3 3 3cos 2 0 cos( ) ( 1)

2 2 2 2y

3 3 3 3cos 2 cos cos

2 4 2 2 2 2

30 0

2

y

30,

2

,04




x 3cos 2

2y x ( , )x y

2

3 3 3cos 2 cos( ) cos(2 )

2 2 2 2

3 31

2 2

y

3,

2 2





Step 4 Connect the five

key points with

a smooth curve

and graph one

complete cycle of the

given function.

3cos 2 .

2y x

amplitude32

period


Vertical Shifts of Sinusoidal Graphs

For sinusoidal graphs of the form

and

the constant D causes a vertical shift in the graph.

These vertical shifts result in sinusoidal graphs oscillating about the horizontal line y = D rather than about the x-axis.

The maximum value of y is

The minimum value of y is

sin( )y A Bx C D cos( ) y A Bx C D

.D A.D A


Example: A Vertical Shift

Graph one period of the function

Step 1 Identify the amplitude, period, phase shift, and vertical shift.

2cos 1.y x

cos( )y A Bx C D 2cos 1y x

amplitude: period:

phase shift:

2 2A 2 22

1B

0 vertical shift: one unit upward





2cos 1.y x

period 24 4 2

1 0x 2 02 2

x

3 2 2x

4

32 2

x 5

32

2 2x




x ( , )x y2cos 1y x

0

2

2cos0 1 2 1 1 3y

2cos 1 2 0 1 12

y

2cos 1 2 ( 1) 1 1y

32 3

2cos 1 2 0 1 12

y

0,3

,12

, 1

3,1

2




x ( , )x y2cos 1y x

2 2cos 2 1 2 1 1 3y 2 ,3




Step 4 Connect the five

key points with

a smooth curve

and graph one

complete cycle of

the given function.

2cos 1.y x


Example: Modeling Periodic Behavior

A region that is 30° north of the Equator averages a minimum of 10 hours of daylight in December. Hours of daylight are at a maximum of 14 hours in June. Let x represent the month of the year, with 1 for January, 2 for February, 3 for March, and 12 for December. If y represents the number of hours of daylight in month x, use a sine function of the form y = Asin(Bx – C) + D to model the hours of daylight.

Because the hours of daylight range from a minimum of 10 to a maximum of 14, the curve oscillates about the middle value, 12 hours. Thus, D = 12.


Example: Modeling Periodic Behavior (continued)


The maximum number of hours of daylight is 14, which is 2 hours more than 12 hours. Thus, A, the amplitude, is 2; A = 2.




One complete cycle occurs over a period of 12 months.2

12B 12 2B

212 6

B



The starting point of the cycle is March, x = 3.

The phase shift is

3CB

3

6

C 3

36 6 2

C

.2




2, , , 126 2

A B C D

The equation that models the hours of daylight is

2sin 12.6 2

y x

5.5 Graphs of Other Trigonometric Functions

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