1. EXAMPLE 1
Graph sine and cosine functions
Graph (a) y = 4 sin x and (b) y = cos 4x.
SOLUTION 2 b π = 1
2π.
a. The amplitude is a = 4 and the period is
Intercepts:
(0, 0); 1 2
( π, 0)
(π, 0); (2π, 0) =
Maximum:
( π, 4) 1 4 2 π
( , 4) =
Minimum:
( π, – 4) 3 4 2 3π
( , – 4) =

2. 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 is 2 b π = 4
= .
Intercepts:
( , 0) 1 4 π 2
= ( , 0); 8 3
= ( , 0) 3π
Maximums:
(0, 1); 2 π
( , 1)
Minimum:
( , –1) 1 2 π
= ( , –1) 4

3. EXAMPLE 2
Graph a cosine function
Graph y = cos 2
π x. 1 2
SOLUTION
The amplitude is a = 1 2
and the period is b π = 2π 1.
Intercepts:
( , 0) 1 4
= ( , 0);
( , 0) 3
= ( , 0)
Maximums:
(0, ) ; 1 2
(1, )
Minimum:
( , – ) 1 2
= ( , – )

4. 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

5. 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 1
2000 = b 2 π
= b
The pressure P as a function of time t is given by P = 2 sin 4000πt.

6. EXAMPLE 3
Model with a sine function
STEP 2
Graph the model. The amplitude is a = 2 and the period is
2000 1 f =
Intercepts:
(0 , 0);
( , 0) 1 2
2000
= ( , 0) ;
4000
( , 0)
Maximum:
( , 2) 1 4
2000
= ( , 2)
8000
Minimum:
( , –2) 3 4
2000 1
= ( , –2)
8000

7. EXAMPLE 4
Graph a tangent function
Graph one period of the function y = 2 tan 3x.
SOLUTION b π = 3 .
The period is
Intercepts:
(0, 0)
Asymptotes:
x = 2b π =
2 3
, or x = ; 6
x = 2b π – =
2 3
, or x = ; 6

8. EXAMPLE 4
Graph a tangent function
Halfway points:
( , a) 4b π =
4 3
( , 2) 12
( , 2);
( , – a) 4b π – =
4 3
( , – 2) 12
( , – 2)