EXAMPLE 1 Graph sine and cosine functions Graph

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EXAMPLE 1 Graph sine and cosine functions Graph (a) y = 4 sin x

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

EXAMPLE 1 Graph sine and cosine functions Graph (a) y = 4 sin x

EXAMPLE 1 Graph sine and cosine functions Graph (a) y = 4 sin x and (b) y = cos 4 x. SOLUTION b. π The amplitude is a = 1 and the period is 2 π =. b 4 2 Intercepts: ( 1 π , 0) = ( π , 0); ( 3 π , 0) = ( 3π, 0) 4 2 8 8 π (0, 1); ( 2 , 1) Maximums: π 1 π Minimum: ( 2 , – 1) = ( , – 1) 2 4

EXAMPLE 2 Graph a cosine function 1 Graph y = 2 cos 2 π

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

EXAMPLE 3 Model with a sine function Audio Test A sound consisting of a

EXAMPLE 3 Model with a sine function Audio Test 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).

EXAMPLE 3 Model with a sine function SOLUTION STEP 1 Find the values of

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. 1 frequency = period b 2000 = 2 π 4000 π = 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

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

EXAMPLE 4 Graph a tangent function Graph one period of the function y =

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

EXAMPLE 4 Graph a tangent function Halfway points: ( π , a) = (

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