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10.1 Graph Sine, Cosine, and Tangent Functions. Vocabulary.

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Presentation on theme: "10.1 Graph Sine, Cosine, and Tangent Functions. Vocabulary."— Presentation transcript:

1 10.1 Graph Sine, Cosine, and Tangent Functions

2 Vocabulary

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8 Graph (a) y = 4 sin x SOLUTION 2 b π = 2 1 π = 2π.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) π (, 4) = Minimum: ( 2π, – 4) π3π (, – 4) =

9 Graph (b) y = cos 4x. SOLUTION b. The amplitude is a = 1 and the period is 2 b π = 2 4 π =. π 2 Intercepts: (, 0) 1 4 π 2 = (, 0); π 8 (, 0) 3 4 π 2 = (, 0) 3π3π 8 Maximums: (0, 1); 2 π (, 1) Minimum: (, –1) 1 2 π 2 = (, –1) π 4

10 Graph the function. 1. y = 2 cos x The amplitude is a = 2 and the period is 2 b π = 2 1 π = 2π Intercepts: ( 2π, 0) 1 4 = (, 0); π = (, 0) 3π3π 2 (, 2) Maximums: (0, 2); 2π2π Minimum: ( 2π, –2) 1 2 = (, –2) π 4 SOLUTION

11 Graph the function. 2. y = 5 sin x The amplitude is a = 5 and the period is 2 b π = 2 1 π = 2π Intercepts: (0, 0) (π, 0) = ( 2π, 0) = 1 2 = (, 0) 2π2π Maximums: ( 2π, 5) = 1 4 (, 5) π 2 SOLUTION

12 Changes in a and b

13 Graph the function. 3. f (x) = sin πx The amplitude is a = 1 and the period is 2 b π = 2 π π = 2 Intercepts: (0, 0) (1, 0); (2, 0) = ( 2, 0) = 1 2 Minimum: ( 2, –1) 3 4 = (, –1) 3 2 Maximums: ( 2, 1 = (, 1) SOLUTION

14 Graph the function. 4. g(x) = cos 4πx The amplitude is a = 1 and the period is 2 b π = 2 4π4π π = 1 2 Intercepts: (, 0) = (, 0); (, 0) = (, 0) Maximums: 1 2 (, 1) ( 0, 1); Minimum: = (, –1) 1 4 (, –1) SOLUTION

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16 10.1 Assignment, day 1 Page 617, 3 – 15 No graphing calculators are to used for this assignment. Graphing solutions need to include: intercepts information maximum information minimum information graphs Learn where this information comes from!

17 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?

18 SOLUTION Graph y = cos 2 π x. 1 2 The amplitude is a = 1 2 and the period is 2 b π = 2 π = 2π2π 1. Intercepts: ( 1, 0) 1 4 = (, 0); 1 4 ( 1, 0) 3 4 = (, 0) 3 4 Maximums: (0, ) ; (1, ) Minimum: ( 1, – ) = (, – )

19 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,…) appears-frozen-in.html

20 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

21 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 = b 2 π 4000 = b π 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 f = Intercepts: (0, 0); (, 0) = (, 0) ; (, 0) Maximum: (, 2) = (, 2) Minimum: (, –2) = (, –2)

22 Graph the function. 5. y = sin πx 1 4 SOLUTION The amplitude is a = 1 4 and the period is 2 b π = 2 π = π 2. Intercepts: = (1, 0) ;(2, 0) (0, 0); 1 2 2, 0 ( ) Maximums: 1 4 2, 1 4 () 1, = () Minimum: 3 4 2, 1 4 () – = 3, () –

23 Graph the function. 6. y = cos πx 1 3 SOLUTION The amplitude is a = and the period is 2 π = π 2 b π = = 1 2, 0 (); 3 4 2, 0 ( ) Intercepts: 1 4 2, 0 ( ) 1 2, 0 ( ) = Maximums: 1 3 2,) 1 3 ( 0,);); ( Minimum: 1 2 2, 1 3 () – = 1 3 () – 1,

24 Graph the function. 7. f (x) = 2 sin 3x SOLUTION The amplitude is a = 2 and the period is 2 π 3 2 b π = Intercepts: (0, 0); 1 2 2π2π 3, 0 = π 3, 0 2π2π 3, 0 ( ) ( );( ) Maximums: 1 4 2π2π 3, 2 = π 6, 2() ( ) Minimum: 3 4 2π2π 3 – 2 ( ) π 2 – 2 = ()

25 Graph the function. 8. g(x) = 3 cos 4x SOLUTION The amplitude is a = 3 and the period is 2 π = 4 2 b π = π 2 Intercepts: 1 4 π 2, 2 = () π π 2, 0 ( );, 0 = () 3π3π 0, 0 ( ) Maximums: (0, 3); π 2, 3 ( ) Minimum: 1 2 π 2 – 3 ( ) π 4 – 2 = ()

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28 Graphing Key Points

29 Graph one period of the function y = 2 tan 3x. SOLUTION b π = 3 π. The period is Intercepts: (0, 0) Asymptotes: x = 2b2b π = 2 3 π, or x = ; 6 π x = 2b2b π – = 2 3 π –, or x = 6 π – Halfway points: (, a) 4b4b π = 4 3 π (, 2) = 12 π (, 2); (, – a) 4b4b π – = 4 3 π (, – 2) – = 12 π (, – 2) –

30 Graph one period of the function. 12. f (x) = 2 tan 4x SOLUTION The period is Intercepts: Asymptotes: Halfway points:

31 The period is Intercepts: Asymptotes: Halfway points: Graph one period of the function. 13. g(x) = 5 tan πx SOLUTION

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33 10.1 Assignment, day 2 Page 617, all No graphing calculators are to used for this assignment. Graphing solutions need to include: intercepts information maximum information minimum information asymptotes graphs Learn where this information comes from!


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