10.1 Graph Sine, Cosine, and Tangent Functions

What graphs are periodic?What graphs have maximum and

minimum values?What values do you need to graph

the or the ?

Vocabulary

• Amplitude of each functions graph: half the difference of the maximum and the minimum , or ½ () = ½[1−(−1)] = 1.

• Periodic: the graph has a repeating pattern.• Cycle: the shortest repeating portion of the

graph.• Period: the horizontal length of each cycle.

Parent Graph of

D: all real numbers R: This graph has a horizontal length and a period of 2.The x-intercepts for occur when

Parent Graph of

D: all real numbers R: This graph has a horizontal length and a period of 2.The x-intercepts for occur when

p. 612

𝑦=𝑎 sin𝑏𝑥 𝑦=𝑎cos 𝑏𝑥

Each graph shows five key x-values on the interval when .Maximum x values occur at the blue points.Minimum x values occur at the green points.X-intercepts occur at the red points.

Graph (a) y = 4 sin x

SOLUTION2bπ= 2

1π= 2π.a. The amplitude is a = 4 and the period is

Intercepts: (0, 0); 12 ( 2π, 0) (π, 0); (2π, 0)=

Maximum: ( 2π, 4)14 2

π( , 4)=

Minimum: ( 2π, – 4)34 2

3π( , – 4)=

Graph (b) y = cos 4x.

SOLUTION

b. The amplitude is a = 1 and the period is 2bπ= 2

4π= .

π2

Intercepts: ( , 0)14

π2 = ( , 0);

π8 ( , 0)3

4π2 = ( , 0)3π

8Maximums: (0, 1); 2

π( , 1)

Minimum: ( , –1) 12

π2 = ( , –1)

π4

Graph the function.1. y = 2 cos x

The amplitude is a = 2 and the period is 2bπ= 2

1π= 2π

Intercepts: ( 2π, 0)14 ( 2π, 0)= ( , 0);

π2

34 = ( , 0)3π

2 ( , 2)Maximums: (0, 2); 2π

Minimum: ( 2π, –2) 12 = ( , –2)

π4

SOLUTION

Graph the function.2. y = 5 sin x

The amplitude is a = 5 and the period is 2bπ= 2

1π= 2π

Intercepts: (0, 0) (π, 0)= ( 2π, 0) =12 = ( , 0)2π

Maximums: ( 2π, 5) =14 ( , 5)π

2

SOLUTION

Changes in a and b

Notice how changes in a and b affect the graphs of and . • When the value of a increases, the amplitude

increases. • When the value of b increases, the period

decreases.

Graph the function.3. f (x) = sin πx

The amplitude is a = 1 and the period is 2bπ= 2

ππ= 2

Intercepts: (0, 0) (1, 0); (2, 0)= ( 2, 0) =12

Minimum: ( 2, –1) 34 = ( , –1)

32

Maximums: ( 2, 1 = ( , 1)12

14

SOLUTION

Graph the function.4. g(x) = cos 4πx

The amplitude is a = 1 and the period is 2bπ= 2

4ππ =

12

Intercepts: ( , 0) = ( , 0); ( , 0) = ( , 0)14

12

18

34

12

18

Maximums: 12( , 1)( 0, 1);

Minimum: = ( , –1) 14 ( , –1) 1

212

SOLUTION

• What graphs are periodic?The graphs of and are periodic.• What graphs have maximum and minimum

values?The graphs of and have maximum and minimum values.• What values do you need to graph the or the ?You can use the maximum and minimum values and the x-intercepts to graph and .

10.1 Assignment, day 1Page 617, 3 – 15

No graphing calculators are to used for this assignment. Graphing solutions need to include:• intercepts information• maximum information• minimum information• graphsLearn where this information comes from!

10.1 Graph Sine, Cosine, and Tangent Functions, day 2

What graphs have asymptotes?What is frequency?

What values do you need to graph tangents?

SOLUTION

Graph y = cos 2 π x.12

The amplitude is a = 12 and the period is 2

bπ= 2π =2π 1.

Intercepts: ( 1, 0)14 = ( , 0);

14 ( 1 , 0)3

4 = ( , 0)34

Maximums: (0, ) ;12 2

1(1, )

Minimum: ( 1, – )12

12 = ( , – )

12

12

Frequency

• Frequency: the reciprocal of the period which gives the number of cycles per unit of time.

The periodic nature of trig functions is useful to model oscillating motions or repeating patterns (sound waves, motion of a pendulum,…)

http://boingboing.net/2013/03/12/video-water-appears-frozen-in.html

A sound consisting of a single frequency is called a pure tone. An audiometer produces pure tones to test a person’s auditory functions. Suppose an audiometer produces a pure tone with a frequency f of 2000 hertz (cycles per second). The maximum pressure P produced from the pure tone is 2 millipascals. Write and graph a sine model that gives the pressure P as a function of the time t (in seconds).

Audio Test

SOLUTION

STEP 1 Find the values of a and b in the model P = a sin bt. The maximum pressure is 2, so a = 2. You can use the frequency f to find b.frequency = period

1 2000 = b

2π 4000 = bπ

The pressure P as a function of time t is given by P = 2 sin 4000πt.

STEP 2 Graph the model. The amplitude is a = 2 and the period is

200011

f =

Intercepts: (0 , 0); ( , 0)12 2000

1

= ( , 0) ; 40001 ( , 0)2000

1

Maximum: ( , 2)14 2000

1 = ( , 2) 80001

Minimum: ( , –2)34 2000

1 = ( , –2) 80003

Graph the function.5. y = sin πx1

4

SOLUTION

The amplitude is a = 14 and the period is 2

bπ= 2π = π 2.

Intercepts: = (1, 0) ; (2, 0)(0 , 0); 12 2, 0( )

Maximums: 14 2, 1

4( ) 1 ,2

14= ( )

Minimum: 34 2, 1

4( )– = 3 ,2

14( )–

Graph the function.6. y = cos πx1

3

SOLUTION

The amplitude is a = and the period is 2π = π2bπ= 2.1

3

= 12

, 0( ); 34 2, 0( )Intercepts: 1

4 2, 0( ) 12

, 0( )=

Maximums: 13 2, )1

3(0, ); (

Minimum: 12 2, 1

3( )– = 13( )–1,

Graph the function.7. f (x) = 2 sin 3x

SOLUTION

The amplitude is a = 2 and the period is 2π 3

2bπ=

Intercepts: (0 , 0); 12

2π3

, 0 = π3

, 0 2π3

, 0( ) ( ); ( )

Maximums: 14

2π3

, 2 = π6

, 2( ) ( )

Minimum: 34

2π3 – 2( ) π

2 – 2= ( )

Graph the function.8. g(x) = 3 cos 4x

SOLUTION

The amplitude is a = 3 and the period is 2π = 42bπ= π

2Intercepts: 1

4 π2

, 2 =( ) π8

34

π2

, 0( ); , 0 =( ) 3π0

, 0( )

Maximums: (0 , 3); π2

, 3( )

Minimum: 12

π2 – 3( ) π

4 – 2= ( )

Graph of• The domain is all real numbers except odd multiples

of . At these x-values, the graph has vertical asymptotes.

• The range is all real numbers. • The function does not have a maximum or minimum

value. The graph of does not have an amplitude.• The graph has a period of .• The x intercepts of the graph occur when

Parent Graph

Graphing Key Points

• Graph the x-intercept• Graph the x-values where

the asymptotes occur.• Graph the x-values half-way

between the x-intercept and the asymptotes.

• At each halfway point, the function’s value is either .

Graph one period of the function y = 2 tan 3x.

SOLUTION

bπ = 3

π .The period is

Intercepts: (0, 0)

Asymptotes: x = 2bπ = 2 3

π , or x = ; 6π

x = 2bπ– = 2 3

π– , or x =

6π–

Halfway points:( , a)4bπ = 4 3

π( , 2) = 12π( , 2);

( , – a)4bπ– = 4 3

π( , – 2)– = 12π( , – 2)–

Graph one period of the function.12. f (x) = 2 tan 4x

SOLUTION

The period is

Intercepts:

Asymptotes:

Halfway points:𝑦=a tan𝑏𝑥

𝜋𝑏

𝜋2|𝑏|± 𝜋4𝑏

The period is

Intercepts:

Asymptotes:

Halfway points:

Graph one period of the function.13. g(x) = 5 tan πx

SOLUTION

𝑦=a tan𝑏𝑥

𝜋𝑏

𝜋2|𝑏|

± 𝜋4𝑏

• What graphs have asymptotes?

• What is frequency?The reciprocal of the period which gives the number of cycles per unit of time.• What values do you need to graph tangents?You can use the asymptotes, the x-intercepts, and the x-values halfway between the x-intercepts and the asymptotes to graph

10.1 Assignment, day 2

Page 617, 16-24 all

No graphing calculators are to used for this assignment. Graphing solutions need to include:• intercepts information• maximum information• minimum information• asymptotes• graphsLearn where this information comes from!