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**10.1 Graph Sine, Cosine, and Tangent Functions**

What graphs are periodic? What graphs have maximum and minimum values? What values do you need to graph the sinβ‘π₯ or the cosβ‘π₯?

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Vocabulary Amplitude of each functions graph: half the difference of the maximum π and the minimum π, or Β½ (πβπ) = Β½[1β(β1)] = 1. Periodic: the graph has a repeating pattern. Cycle: the shortest repeating portion of the graph. Period: the horizontal length of each cycle.

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**Parent Graph of π¦ = sinβ‘π₯**

D: all real numbers R: β1β€π¦β€1. This graph has a horizontal length and a period of 2. The x-intercepts for π¦=sinβ‘π₯ occur when π₯=0, Β±π, Β±2π, Β±3π,β¦

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**Parent Graph of π¦ = cosβ‘π₯**

D: all real numbers R: β1β€π¦β€1. This graph has a horizontal length and a period of 2. The x-intercepts for π¦=cosβ‘π₯ occur when π₯= Β± π 2 , Β± 3π 2 , Β± 5π 2 ,Β± 7π 2 ,β¦

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p. 612

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π¦=π sin ππ₯ π¦=π cosβ‘ππ₯ Each graph shows five key x-values on the interval 0β€π₯β€ 2π π when π>0 and π>0. Maximum x values occur at the blue points. Minimum x values occur at the green points. X-intercepts occur at the red points.

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Graph (a) y = 4 sin x SOLUTION 2 b Ο = 1 2Ο. a. The amplitude is a = 4 and the period is Intercepts: (0, 0); 1 2 ( Ο, 0) (Ο, 0); (2Ο, 0) = Maximum: ( Ο, 4) 1 4 2 Ο ( , 4) = Minimum: ( Ο, β 4) 3 4 2 3Ο ( , β 4) =

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Graph (b) y = cos 4x. SOLUTION b. The amplitude is a = 1 and the period is 2 b Ο = 4 = . Intercepts: ( , 0) 1 4 Ο 2 = ( , 0); 8 3 = ( , 0) 3Ο Maximums: (0, 1); 2 Ο ( , 1) Minimum: ( , β1) 1 2 Ο = ( , β1) 4

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Graph the function. 1. y = 2 cos x SOLUTION The amplitude is a = 2 and the period is 2 b Ο = 1 = 2Ο Intercepts: ( Ο, 0) 1 4 ( Ο, 0) = ( , 0); Ο 2 3 = ( , 0) 3Ο ( , 2) Maximums: (0, 2); 2Ο Minimum: ( Ο, β2) 1 2 = ( , β2) Ο 4

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

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Changes in a and b Notice how changes in a and b affect the graphs of π¦=π sinβ‘ππ₯ and π¦=π cosβ‘ππ₯. When the value of a increases, the amplitude increases. When the value of b increases, the period decreases.

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Graph the function. 3. f (x) = sin Οx SOLUTION The amplitude is a = 1 and the period is 2 b Ο = = 2 Intercepts: (0, 0) (1, 0); (2, 0) = ( , 0) = 1 2 Maximums: ( , 1 = ( , 1) 1 2 4 Minimum: ( , β1) 3 4 = ( , β1) 2

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Graph the function. 4. g(x) = cos 4Οx SOLUTION The amplitude is a = 1 and the period is 2 b Ο = 4Ο 1 Intercepts: ( , 0) = ( , 0); ( , 0) = ( , 0) 1 4 2 8 3 Maximums: 1 2 ( , 1) ( 0, 1); Minimum: = ( , β1) 1 4 ( , β1) 2

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**What graphs are periodic?**

The graphs of π¦=sinβ‘π₯ and π¦=cosβ‘π₯ are periodic. What graphs have maximum and minimum values? The graphs of π¦=sinβ‘π₯ and π¦=cosβ‘π₯ have maximum and minimum values. What values do you need to graph the sinβ‘π₯ or the cosβ‘π₯? You can use the maximum and minimum values and the x-intercepts to graph π¦=sinβ‘π₯ and π¦=cosβ‘π₯ .

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10.1 Assignment, day 1 Page 617, 3 β 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!

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**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?

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

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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,β¦)

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**A sound consisting of a single frequency is called a pure tone**

A sound consisting of a single frequency is called a pure tone. An audiometer produces pure tones to test a personβs 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

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**The pressure P as a function of time t is given by P = 2 sin 4000Οt.**

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 Ο = b The pressure P as a function of time t is given by P = 2 sin 4000Οt. 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: ( , β2) 3 4 2000 1 = ( , β2) 8000

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

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

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

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

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Graph of π¦= tan π₯ The domain is all real numbers except odd multiples of π 2 . At these x-values, the graph has vertical asymptotes. The range is all real numbers. The function π¦= tan π₯ does not have a maximum or minimum value. The graph of π¦= tan π₯ does not have an amplitude. The graph has a period of π. The x intercepts of the graph occur when π₯=0, Β±π, Β±2π, Β±3π,β¦

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Parent Graph π¦= tan π₯

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**Graphing Key Points Graph the x-intercept**

Graph the x-values where the asymptotes occur. Graph the x-values half-way between the x-intercept and the asymptotes. At each halfway point, the functionβs value is either π or βπ.

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**Graph one period of the function y = 2 tan 3x.**

SOLUTION b Ο = 3 . The period is Intercepts: (0, 0) Asymptotes: x = 2b Ο = 2 3 , or x = ; 6 x = 2b Ο β = 2 3 , or x = 6 Halfway points: ( , a) 4b Ο = 4 3 ( , 2) 12 ( , 2); ( , β a) 4b Ο β = 4 3 ( , β 2) 12 ( , β 2)

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**Graph one period of the function.**

f (x) = 2 tan 4x SOLUTION π π The period is Intercepts: π 2 π Asymptotes: π¦= a tan ππ₯ Β± π 4π Halfway points:

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**Graph one period of the function.**

g(x) = 5 tan Οx SOLUTION The period is π π Intercepts: π 2 π Asymptotes: π¦= a tan ππ₯ Β± π 4π Halfway points:

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**What graphs have asymptotes?**

π¦= a tan ππ₯ What is frequency? The reciprocal of the period which gives the number of cycles per unit of time. What values do you need to graph tangents? You can use the asymptotes, the x-intercepts, and the x-values halfway between the x-intercepts and the asymptotes to graph π¦= tan π₯ .

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**10.1 Assignment, day 2 Page 617, 16-24 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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