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 ( 2π, 0) (π, 0); (2π, 0) = Maximum: ( 2π, 4) 1 4 2 π ( , 4) = Minimum: ( 2π, – 4) 3 4 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 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

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: ( 1, 0) 1 4 = ( , 0); ( 1 , 0) 3 = ( , 0) Maximums: (0, ) ; 1 2 (1, ) Minimum: ( 1, – ) 1 2 = ( , – )

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 1 2000 = b 2 π 4000 = b 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 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

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

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