Lesson 14.1, For use with pages 908-914
Find the sine, cosine, and tangent.
2. 2π
ANSWER sin: 1, cos: 0, tan: undefined
ANSWER sin: 0, cos: 1, tan: 0
1.π2
Lesson 14.1, For use with pages 908-914
Find the sine, cosine, and tangent.
3. The diameter of a wheel is 27 inches. Through how many radians does a point on the wheel move when the wheel moves 15 feet?
ANSWER about 13.3 radians
Graphing Trigonometry Functions 14.11
EXAMPLE 1 Graph sine and cosine functions
Graph (a) y = 4 sin x and (b) y = cos 4x.
SOLUTION
2bπ
=21π
= 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) =
EXAMPLE 1 Graph sine and cosine functions
Graph (a) y = 4 sin x and (b) y = cos 4x.
SOLUTION
b. The amplitude is a = 1 and the period is2bπ
=24π
= .π
2
Intercepts: ( , 0)14
π
2= ( , 0) ;
π
8 ( , 0 )
34
π2
= ( , 0)3π8
Maximums: (0, 1); 2π
( , 1)
Minimum: ( – , 1 ) 12
π
2= ( , –1)
π
4
GUIDED PRACTICE for Example 1
Graph the function.
1. y = 2 cos x
The amplitude is a = 2 and the period is2bπ
=21π
= 2π
Intercepts: ( 2π, 0)14
( 2π, 0)= ( , 0 ) π
234
= ( , 0 )3π2
( , 2)Maximums: (0, 2); 2π
Minimum: ( 2π, –2) 12
= ( , –2)π
4
SOLUTION
GUIDED PRACTICE for Example 1
Graph the function.
2. y = 5 sin x
The amplitude is a = 5 and the period is2bπ
=21π
= 2π
Intercepts: (0, 0) (π, 0)= ( 2π, 0) =1
2= ( , 0)2π
Minimum: ( 2π, –5) 34
= ( , –5)3π
2
Maximums: ( 2π, 5) =14 ( , 5)
π
2
SOLUTION
GUIDED PRACTICE for Example 1
Graph the function.
3. f (x) = sin πx
The amplitude is a = 1 and the period is2bπ
=2π
π= 2
Intercepts: (0, 0) (1, 0); (2, 0)= ( 2, 0) =1
2
Minimum: ( 2, –1) 34
= ( , –1) 3
2
Maximums: ( 2, 1 = ( , 1)1
214
SOLUTION
GUIDED PRACTICE for Example 1
Graph the function.
4. g(x) = cos 4πx
The amplitude is a = 1 and the period is2bπ
=24ππ
=1
2
Intercepts: ( , 0) = ( , 0); ( , 0) = ( , 0)1
4
1
2
1
8
3
4
1
2
1
8
Maximums:1
2( , 1)( 0, 1);
Minimum: = ( , –1) 1
4 ( , –1)
12
1
2
SOLUTION
EXAMPLE 2 Graph a cosine function
SOLUTION
Graph y = cos 2 π x.12
The amplitude is a =12
and the period is2bπ
=2 π
=2π
1.
Intercepts: ( 1, 0)14
= ( , 0);1
4 ( 1 , 0)34 = ( , 0)
34
Maximums: (0, ) ;12 2
1 (1, )
Minimum: ( 1, – )12
12
= ( , – )1
212
EXAMPLE 3 Model with a sine function
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
EXAMPLE 3 Model with a sine function
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
12000 =
b2 π 4000 = b Π
The pressure P as a function of time t is given by P = 2 sin 4000πt.
EXAMPLE 3 Model with a sine function
STEP 2 Graph the model. The amplitude is a = 2 and the period is
2000
11f
=
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
GUIDED PRACTICE for Examples 2 and 3
Graph the function.
5. y = sin πx14
SOLUTION
The amplitude is a =14
and the period is2bπ
=2 π
= π
2.
Intercepts: = (1, 0) ; (2, 0)(0 , 0);12
2, 0( )
Maximums:14
2,14
( )1 ,2
14
= ( )
Minimum:34
2,14
( )– =3 ,2
14
( )–
GUIDED PRACTICE for Examples 2 and 3
Graph the function.
6. y = cos πx13
SOLUTION
The amplitude is a = and the period is2 π
= π
2bπ
= 2.13
=12
, 0( );34
2, 0( )Intercepts: 14
2, 0( )12
, 0( )=
Maximums:13
2, )13
(0, ); (
Minimum:12
2,13
( )– =13
( )–1,
GUIDED PRACTICE for Examples 2 and 3
Graph the function.
7. f (x) = 2 sin 3x
SOLUTION
The amplitude is a = 2 and the period is2 π 3
2bπ
=
Intercepts: (0 , 0); 12
2π3
, 0 =π3
, 02π3
, 0( ) ( ); ( )
Maximums:14
2π3
, 2 =π6
, 2( ) ( )
Minimum:34
2π3
– 2( ) π2
– 2= ( )
GUIDED PRACTICE for Examples 2 and 3
Graph the function.
8. g(x) = 3 cos 4x
SOLUTION
The amplitude is a = 3 and the period is2 π
= 4
2bπ
=π2
Intercepts:14
π2
, 2 =( ) π8
34
π2
, 0( ); , 0 =( ) 3π0
, 0( )
Maximums: (0 , 3);π2
, 3( )
Minimum:12
π2
– 3( ) π4
– 2= ( )
GUIDED PRACTICE for Examples 2 and 3
9. What If ? In Example 3, how would the function change if the audiometer produced a pure tone with a frequency of 1000 hertz?
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
11000 =
b2 π 2000 = bπ
The pressure P as a function of time t is given by P = 2 sin 2000πt.
GUIDED PRACTICE for Examples 2 and 3
STEP 2
Graph the model. The amplitude is a = 2 and the period is
2000
f1f
=
Intercepts: (0 , 0); ( , 0)12 1000
1= ( , 0) ;
2000
1( , 0)
10001
Maximum: ( , 2)14 1000
1= ( , 2)
40001
Minimum: ( , –2)34 1000
1= ( , –2)
40003
The period would increase because the frequency is decreased p = 2 sin 2000πt
EXAMPLE 4 Graph a tangent function
Graph one period of the function y = 2 tan 3x.
SOLUTION
b
π=
3
π.The period is
Intercepts: (0, 0)
Asymptotes: x =2b
π=
2 3
π, or x = ;
6
π
x =2b
π– =
2 3
π– , or x = ;
6
π–
EXAMPLE 4 Graph a tangent function
Halfway points: ( , a)4b
π=
4 3
π( , 2) =12
π( , 2);
( , – a)4b
π– =
4 3
π( , – 2)– =12
π( , – 2)–
GUIDED PRACTICE for Example 4
Graph one period of the function.
10. y = 3 tan x
SOLUTION
GUIDED PRACTICE for Example 4
Graph one period of the function.
11. y = tan 2x
SOLUTION
GUIDED PRACTICE for Example 4
Graph one period of the function.
12. f (x) = 2 tan 4x
SOLUTION
GUIDED PRACTICE for Example 4
Graph one period of the function.
13. g(x) = 5 tan πx
SOLUTION