0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric...

121
56 56 x y 1 sin x 2 coordinates 0 y 1 sin 0 =− 2 1 0 = 0 =− 2 (0, 0) π 2 1 π y =− sin 2 2 1 1 = − 1 =− 2 2 π 1 , 2 2 π y 1 sin π =− 2 1 0 = 0 =− 2 (π , 0) 3 π 2 1 3 π y =− sin 2 2 1 ( 1) = 1 =− 2 2 3 π 1 , 2 2 2π y 1 sin 2π =− 2 1 =− 0 = 0 2 (2π , 0) x y = 3sin x coordinates 0 y = 3sin 0 = 3 0 = 0 (0, 0) π 2 y = 3sin π = 3 1 = 3 2 π , 3 2 π y = 3sin x = 3 0 = 0 (π , 0) 3 π 2 y = 3sin 3 π 2 = 3(1) = 3 3 π , 3 2 2π y = 3sin 2π= 3 0 = 0 (2 π , 0) x = 0 x = 0 + π = π 2 2 x = π + π = π adding quarter- periods. The five x-values are x = 0 x = 0 + π = π 2 2 x = π + π = π 2 2 2 2 x = π+ π = 3 π x = π+ π = 3 π 2 2 2 2 x = 3 π + π = 2π 2 2 Evaluate the function at each value of x. x = 3 π + π = 2 π 2 2 Evaluate the function at each value of x. =− Connect the five points with a smooth curve and graph one complete cycle of the given function with the graph of y = sin x . Full file at https://testbank123.eu/Solutions-Manual-for-Trigonometry-2nd-Edition-Blitzer Full file at https://testbank123.eu/Solutions-Manual-for-Trigonometry-2nd-Edition-Blitzer

Transcript of 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric...

Page 1: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

56 56

x y 1

sin x 2

coordinates

0 y 1

sin 0

= − 2 1

⋅ 0 = 0 = −

2

(0, 0)

π

2

1 π y = − sin

2 2 1

⋅1 = − 1 = −

2 2

π 1 , − 2 2

π y 1

sin π = −

2 1

⋅ 0 = 0 = −

2

(π , 0)

2

1 3π y = − sin

2 2 1

(−1) = 1 = −

2 2

3π 1 , 2 2

2π y 1

sin 2π = −

2 1

= − ⋅ 0 = 0 2

(2π , 0)

x y = 3sin x coordinates

0 y = 3sin 0 = 3 ⋅ 0 = 0 (0, 0)

π

2 y = 3sin

π = 3 ⋅1 = 3

2

π , 3

2

π y = 3sin x = 3 ⋅ 0 = 0 (π , 0)

2 y = 3sin

3π 2

= 3(−1) = −3

3π , −3

2

2π y = 3sin 2π = 3 ⋅ 0 = 0 (2π , 0)

x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π

adding quarter- periods. The five x-values are

x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

2 2

x = π + π

= 3π x = π +

π =

3π 2 2

2 2

x = 3π

+ π

= 2π 2 2

Evaluate the function at each value of x.

x = 3π

+ π

= 2π 2 2

Evaluate the function at each value of x.

= −

Connect the five points with a smooth curve and

graph one complete cycle of the given function with

the graph of y = sin x .

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Page 2: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

57 57

Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions Section 2.1 Graphs of Sine and Cosine Functions

3π 1 y = 2sin ⋅ 3π

32 π

= 2sin 2

= 2 ⋅ (−1) = −2

(3π , − 2)

4π y = 2sin 1

⋅ 4π 2

= 2sin 2π = 2 ⋅ 0 = 0

(4π , 0)

x y = 2sin 1

x 2

coordinates

0 y = 2sin 1

⋅ 0

2

= 2sin 0 = 2 ⋅ 0 = 0

(0, 0)

π y = 2sin 1

⋅π

= 2sin π

= 2 ⋅1 = 2 2

(π , 2)

2π y = 2sin 1

⋅ 2π

= 2 sin π = 2 ⋅ 0 = 0

(2π , 0)

3

Connect the five key points with a smooth curve and

graph one complete cycle of the given function with

the graph of y = sin x . Extend the pattern of each

graph to the left and right as desired. Connect the five key points with a smooth curve and

graph one complete cycle of the given function. Extend

the pattern of the graph another full period to the right.

3. The equation y = 2sin 1

x is of the form 2

y = A sin Bx with A = 2 and B = 1

. 2

The amplitude is A = 2 = 2 .

The period is 2π =

2π = 4π .

π B 1 4. The equation

2 y = 3sin 2 x −

is of the form

Find the x–values for the five key points by dividing

the period, 4π , by 4, period

= 4π

= π , then by y = A sin(Bx − C) with A = 3, B = 2, and C =

π . The

34 4

adding quarter-periods. amplitude is A = 3 = 3 .

The five x-values are The period is 2π

= 2π

= π .x = 0 B 2 x = 0 + π = π πx = π + π = 2π The phase shift is

C = 3 =

π ⋅

1 = π

.x = 2π + π = 3π B 2 3 2 6x = 3π + π = 4π Find the x-values for the five key points by dividingEvaluate the function at each value of x.

the period, π , by 4, period

= π

, then by adding4 4

quarter-periods to the value of x where the cycle

begins, x = π

. 6

The five x-values are

x = π 6 π π 2π 3π 5π

2

x = + = + = 6 4 12 12 12

x = 5π

+ π

= 5π

+ 3π

= 8π

= 2π

12 4 12 12 12 3

x = 2π

+ π

= 8π

+ 3π

= 11π

3 4 12 12 12 2

11π π

11π 3π

14π 7π

x = + = + = = 12 4 12 12 12 6

Evaluate the function at each value of x.

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Page 3: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

58 58

Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions Section 2.1 Graphs of Sine and Cosine Functions

x y = 3sin 2 x −

π

coordinates

π

6 y = 3sin

2 ⋅ π

− π

= 3sin 0 = 3 ⋅ 0 = 0

π , 0

12 y = 3sin

2 ⋅

5π − π

= 3sin 3π

= 3sin π

6 2 = 3 ⋅1 = 3

5π , 3

3 y = 3sin

2 ⋅

2π − π

= 3sin 3π

= 3sin π 3

= 3 ⋅ 0 = 0

2π , 0

11π

12 y = 3sin

2 ⋅

11π − π

= 3sin 9π

= 3sin 3π

6 2 = 3 (−1) = −3

11π , − 3

6 y = 3sin

2 ⋅

7π − π

= 3sin 6π

= 3sin 2π 3

= 3 ⋅ 0 = 0

6

x y = −4 cos π x coordinates

0 y = −4 cos (π ⋅ 0) = −4 cos 0 = −4

(0, –4)

1

2

1 y = −4 cos π ⋅

2

= −4 cos π

= 0 2

1 2

, 0

1 y = −4 cos(π ⋅1) = −4 cos π = 4

(1, 4)

3

2 y = −4 cos

π ⋅

3

= −4 cos = 0 2

3 , 0

2 y = −4 cos(π ⋅ 2) = −4 cos 2π = −4

(2, –4)

x = 0

1 1 3

6 3

12 3

6

12

x = 0 + = 2 2

x = 1

+ 1

= 1 2 2

x = 1 + 1

= 3

2 2

x = 3

+ 1

= 2 2 2

Evaluate the function at each value of x.

3 3

3

12 3

12

6 3

7π , 0

2

2

Connect the five key points with a smooth curve

and graph one complete cycle of the given graph.

Connect the five key points with a smooth curve and

graph one complete cycle of the given function. Extend

the pattern of the graph another full period to the left.

5. The equation

y = −4 cos π x is of the form

6. y = 3

cos(2 x + π ) = 3

cos(2 x − (−π )) 2 2

y = A cos Bx with A = −4, and B = π . The equation is of the form y = A cos(Bx − C ) with

Thus, the amplitude is A = −4 = 4 . A = 3

, B = 2 , and C = −π .

The period is 2π

= 2π

= 2 . 2

B π Thus, the amplitude is A = 3 =

3 .

Find the x-values for the five key points by dividing

the period, 2, by 4, period

= 2

= 1

, then by adding

2 2

The period is 2π

= 2π

= π .4 4 2 B 2

quarter periods to the value of x where the cycle

begins. The five x-values are The phase shift is

C =

−π = − π

. B 2 2

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Page 4: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

59 59

Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions Section 2.1 Graphs of Sine and Cosine Functions

x y = 3

cos(2x + π ) 2

coordinates

− π 2

y = 3

cos(−π + π ) 2

= 3

⋅1 = 3

2 2

− π

, 3

− π 4

y = 3

cos − π

+ π 2

2

= 3

⋅ 0 = 0 2

− π

, 0

0 y =

3 cos(0 + π )

2

= 3

⋅ −1 = − 3

2 2

0, −

3

2

π

4 y =

3 cos

π + π

2 2

= 3

⋅ 0 = 0 2

π , 0

4

π

2

3 y = cos(π + π )

2

= 3

⋅1 = 3

2 2

π 3 , 2 2

x y = 2 cos x + 1 coordinates

0 y = 2 cos 0 + 1 = 2 ⋅1 + 1 = 3

(0, 3)

π

2

π y = 2 cos + 1

2 = 2 ⋅ 0 + 1 = 1

π 2

π y = 2 cos π + 1 = 2 ⋅ (−1) + 1 = −1

(π , − 1)

2

3π y = 2 cos + 1

= 2 ⋅ 0 + 2

1

1 =

3π , 1 2

2π y = 2 cos 2π + 1 = 2 ⋅1 + 1 = 3

(2π , 3)

= −

Find the x-values for the five key points by dividing

the period, π , by 4, period

= π

, then by adding 4 4

quarter-periods to the value of x where the cycle

begins, x π

. 2

The five x-values are

π x = −

2 π π π

x = − + = − 2 4 4

7. The graph of

y = 2 cos x + 1 is the graph ofπ π

x = − + = 0 4 4

x = 0 + π

= π

y = 2 cos x shifted one unit upwards. The period for

both functions is 2π . The quarter-period is

4 4 2π or

π . The cycle begins at x = 0. Add quarter-

x = π

+ π

= π

4 4 2 Evaluate the function at each value of x.

4 2 periods to generate x-values for the key points.

x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

2 2

x = π + π

= 3π

4

2 2

x = 3π

+ π

= 2π 2 2

Evaluate the function at each value of x.

, 1

Connect the five key points with a smooth curve and

graph one complete cycle of the given graph.

By connecting the points with a smooth curve, we

obtain one period of the graph.

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Page 5: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

60 60

Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions Section 2.1 Graphs of Sine and Cosine Functions

6 2

8. Select several values of x over the interval.

x

0 π

4

π

2

4 π

4

2

4 2π

y1 = 2 sin x 0 1.4 2 1.4 0 −1.4 −2 −1.4 0

y2 = cos x 1 0.7 0 −0.7 −1 −0.7 0 0.7 1

y = 2 sin x + cos x 1 2.1 2 0.7 −1 −2.1 −2 −0.7 1

9. A, the amplitude, is the maximum value of y. The graph shows that this maximum value is 4, Thus,

π , and period =

2π .

A = 4 . The period is

2

Thus,

B

π =

2 Bπ B = 4π

B = 4

Substitute these values into

y = A sin Bx . The graph is modeled by

y = 4sin 4 x .

10. Because the hours of daylight ranges from a minimum of 10 hours to a maximum of 14 hours, the curve oscillates about

the middle value, 12 hours. Thus, D = 12. The maximum number of hours is 2 hours above 12 hours. Thus, A = 2. The

graph shows that one complete cycle occurs in 12–0, or 12 months. The period is 12.

Thus, 12 = 2π B

12B = 2π

B = 2π

= π

12 6

The graph shows that the starting point of the cycle is shifted from 0 to 3. The phase shift, C

, is 3. B

3 = C

B

3 = C

π 6

π = C

2

Substitute these values into y = A sin(Bx − C ) + D . The number of hours of daylight is modeled by

y = 2sin π

x − π

+ 12 .

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Page 6: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

61 61

Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions Section 2.1 Graphs of Sine and Cosine Functions

2

3π y = 4sin

2 = 4(−1) = −4

3π , − 4 2

2π y = 4sin 2π = 4 ⋅ 0 = 0 (2π , 0)

x y = 5sin x coordinates

0 y = 5sin 0 = 5 ⋅ 0 = 0 (0, 0)

π

2

π y = 5sin

2 = 5 ⋅1 = 5

π 2

, 5

π y = 5sin π = 5 ⋅ 0 = 0 (π , 0)

2 y = 5sin

3π = 5(−1) = −5

2

3π , − 5

2π y = 5sin 2π = 5 ⋅ 0 = 0 (2π , 0)

x y = 4sin x coordinates

0 y = 4sin 0 = 4 ⋅ 0 = 0 (0, 0)

π

2 y = 4sin

π = 4 ⋅1 = 4

2

π , 4

2

π y = 4sin π = 4 ⋅ 0 = 0 (π , 0)

Concept and Vocabulary Check 2.1

1. A ; 2π B

2. 3; 4π

3. π ; 0; π

;

π ;

3π ; π

Connect the five key points with a smooth curve and

graph one complete cycle of the given function with4 2 4 the graph of y = sin x .

4. C

; right; left B

5. A ; 2π B

6. 1

; 2π

2 3

7. false

8. true

2. The equation

y = 5sin x is of the form

y = A sin x

9. true with A = 5. Thus, the amplitude is A = 5 = 5 .

10. true The period is 2π . The quarter-period is 2π or

π .

Exercise Set 2.1

4 2 The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points.

x = 0

1. The equation

y = 4sin x

is of the form

y = A sin x x = 0 + π

= π

2 2with A = 4. Thus, the amplitude is A = 4 = 4 . x =

π + π

= π 2 2

The period is 2π . The quarter-period is 2π or

π . x = π +

π =

3π4 2

The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points.

x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2

Evaluate the function at each value of x.

2 2

x = 3π

+ π

= 2π 2 2

Evaluate the function at each value of x.

2

Connect the five key points with a smooth curve and

graph one complete cycle of the given function with

the graph of y = sin x .

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Page 7: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

62 62

Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions Section 2.1 Graphs of Sine and Cosine Functions

x y = 1

sin x 4

coordinates

0 y = 1

sin 0 = 1

⋅ 0 = 0 4 4

(0, 0)

π

2 y =

1 sin π

= 1

⋅1 = 1

4 2 4 4

π ,

1

π y = 1

sin π = 1

⋅ 0 = 0 4 4

(π , 0)

2 y =

1 sin

3π =

1 (−1) = −

1 4 2 4 4

3π , −

1 2 4

2π 1 1

y = sin 2π = ⋅ 0 = 0 4 4

(2π , 0)

x y =

1 sin x

3

coordinates

0 y =

1 sin 0 =

1 ⋅ 0 = 0

3 3

(0, 0)

π

2 y =

1 sin π

= 1

⋅1 = 1

3 2 3 3

π ,

1 2 3

π y =

1 sin π =

1 ⋅ 0 = 0

3 3

(π , 0)

2 y =

1 sin

3π 3 2

= 1

(−1) = − 1

3 3

3π , −

1

2 3

2π y =

1 sin 2π =

1 ⋅ 0 = 0

3 3

(2π , 0)

3. The equation

y = 1

sin x is of the form 3

y = A sin x

4. The equation

1

y = 1

sin x is of the form 4

y = A sin x

1 1

with A = 1

. Thus, the amplitude is A = 1 =

1 .

with A = . Thus, the amplitude is 4

A = = . 4 4

3 3 3 2π π

The period is 2π . The quarter-period is 2π

or π

. The period is 2π . The quarter-period is or .

4 24 2 The cycle begins at x = 0. Add quarter-periods to

The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points.

x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π

generate x-values for the key points. x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

2 2

x = π + π

= 3π

x = π + π

= 3π

2 22 2

x = 3π

+ π

= 2π 2 2

Evaluate the function at each value of x.

x = 3π

+ π

= 2π 2 2

Evaluate the function at each value of x.

2 4

Connect the five key points with a smooth curve and

graph one complete cycle of the given function withConnect the five key points with a smooth curve and graph one complete cycle of the given function with

the graph of y = sin x .

the graph of y = sin x .

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Page 8: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

63 63

Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions Section 2.1 Graphs of Sine and Cosine Functions

x y = −4sin x coordinates

0 y = −4sin 0 = −4 ⋅ 0 = 0 (0, 0)

π

2 y = −4sin

π = −4 ⋅1 = −4

2

π , − 4

π y = −4sin π = −4 ⋅ 0 = 0 (π , 0)

2

3π y = −4 sin = −4(−1) = 4

2

3π , 4 2

2π y = −4sin 2π = −4 ⋅ 0 = 0 (2π , 0)

x y = −3sin x coordinates

0 y = −3sin x = −3 ⋅ 0 = 0

(0, 0)

π

2 y = −3sin

π 2

= −3 ⋅1 = −3

π , − 3

π y = −3sin π = −3 ⋅ 0 = 0

(π , 0)

2 y = −3sin

3π 2

= −3(−1) = 3

3π , 3

2

2π y = −3sin 2π = −3 ⋅ 0 = 0

(2π , 0)

6. The equation y = −4sin x is of the form y = A sin x

with A = –4. Thus, the amplitude is A = −4 = 4 .

The period is 2π . The quarter-period is 2π or

π .

4 2 The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x = 0

x = 0 + π

= π

5. The equation y = −3sin x is of the form y = A sin x 2 2 π π

with A = –3. Thus, the amplitude is A = −3 = 3 . x = + = π

2 2

The period is 2π . The quarter-period is 2π

or π

. x = π +

π =

3π 2 2

4 2 3π πThe cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2

Evaluate the function at each value of x.

x = + = 2π 2 2

Evaluate the function at each value of x.

2

2

Connect the five key points with a smooth curve and

graph one complete cycle of the given function with

the graph of y = sin x .

Connect the five key points with a smooth curve and graph one complete cycle of the given function with

the graph of y = sin x .

Full file at https://testbank123.eu/Solutions-Manual-for-Trigonometry-2nd-Edition-Blitzer

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Page 9: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

64 64

Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions Section 2.1 Graphs of Sine and Cosine Functions

x y = sin 2 x coordinates

0 y = sin 2 ⋅ 0 = sin 0 = 0 (0, 0)

π

4 y = sin

2 ⋅ π

= sin π

= 1 2

π ,1

π

2 y = sin

2 ⋅ π

= sin π = 0

π , 0

2

4 y = sin

2 ⋅

= sin 3π

= −1 2

3π , −1

π y = sin(2 ⋅π ) = sin 2π = 0

(π , 0)

x y = sin 4 x coordinates

0 y = sin(4 ⋅ 0) = sin 0 = 0 (0, 0)

π

8

π π y = sin 4 ⋅ = sin = 1

8 2

π

8

π

4 y = sin

4 ⋅ π

= sin π = 0

4

π , 0

4

8 y = sin

4 ⋅

= sin 3π

= −1 2

3π , −1

8

π

2

y = sin 2π = 0 π 2

, 0

7. The equation y = sin 2 x is of the form y = A sin Bx 8. The equation y = sin 4 x is of the form y = A sin Bx

with A = 1 and B = 2. The amplitude is with A = 1 and B = 4. Thus, the amplitude is

A = 1 = 1 . The period is 2π

= 2π

= π . The

A = 1 = 1 . The period is 2π

= 2π

= π

. TheB 2 B 4 2

πquarter-period is

π . The cycle begins at x = 0. Add

quarter-period is 2 π 1 π

4 quarter-periods to generate x-values for the key points.

x = 0

x = 0 + π 4

x = π

+ π

= π

4 4 2

x = π

+ π

= 3π

= ⋅ = . The cycle begins at 4 2 4 8

x = 0. Add quarter-periods to generate x-values for

the key points. x = 0

x = 0 + π

= π

8 8

x = π

+ π

= π

8 8 4

2 4 4 x =

π + π

= 3π

x = 3π

+ π

= π 4 8 8

3π π π4 4 x = + =

8 8 2Evaluate the function at each value of x. Evaluate the function at each value of x.

4

4 , 1

2

4

4

8

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

Full file at https://testbank123.eu/Solutions-Manual-for-Trigonometry-2nd-Edition-Blitzer

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Page 10: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

65 65

Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions Section 2.1 Graphs of Sine and Cosine Functions

x 1 y = 3sin x

2

coordinates

0 1 y = 3sin

⋅ 0

2 = 3sin 0 = 3 ⋅ 0 = 0

(0, 0)

π 1 y = 3sin ⋅π

2

= 3sin π

= 3 ⋅1 = 3 2

(π , 3)

2π 1 y = 3sin ⋅ 2π

2 = 3sin π = 3 ⋅ 0 = 0

(2π , 0)

3π 1 y = 3sin ⋅ 3π

2

= 3sin 3π 2

= 3(−1) = −3

(3π , − 3)

4π 1 y = 3sin ⋅ 4π

2 = 3sin 2π = 3 ⋅ 0 = 0

(4π , 0)

π

9. The equation y = 3sin 1

x is of the form 2

y = A sin Bx

10. The equation y = 2sin 1

x is of the form 4

with A = 3 and B = 1

. The amplitude is 2

A = 3

= 3. y = A sin Bx with A = 2 and B = 1

. Thus, the 4

The period is 2π =

2π = 2π ⋅ 2 = 4π . The quarter-

amplitude is A = 2 = 2 . The period is

B 1 2 2

B =

2π = 2π ⋅ 4 = 8π . The quarter-period is

1

period is 4π

= π . The cycle begins at x = 0. Add 4

quarter-periods to generate x-values for the key points. x = 0 x = 0 + π = π x = π + π = 2π x = 2π + π = 3π x = 3π + π = 4π

Evaluate the function at each value of x.

4

8π = 2π . The cycle begins at x = 0. Add quarter-

4 periods to generate x-values for the key points. x = 0 x = 0 + 2π = 2π x = 2π + 2π = 4π x = 4π + 2π = 6π x = 6π + 2π = 8π Evaluate the function at each value of x.

x y = 2sin 1

x 4

coordinates

0 y = 2sin 1

⋅ 0

4

= 2sin 0 = 2 ⋅ 0 = 0

(0, 0)

2π 1 y = 2sin

4 ⋅ 2π

= 2sin π

= 2 ⋅1 = 2 2

(2π , 2)

4π y = 2sin π = 2 ⋅ 0 = 0 (4π , 0)

6π 3π

y = 2sin = 2(−1) = −2 2

(6π , − 2)

8π y = 2sin 2π = 2 ⋅ 0 = 0 (8π , 0)

Connect the five points with a smooth curve and

graph one complete cycle of the given function.

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

Full file at https://testbank123.eu/Solutions-Manual-for-Trigonometry-2nd-Edition-Blitzer

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Page 11: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

66 66

Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

x y = 4sin π x coordinates

0 y = 4sin(π ⋅ 0) = 4sin 0 = 4 ⋅ 0 = 0

(0, 0)

1

2 y = 4sin

π ⋅

1

= 4sin π

= 4(1) = 4 2

1 , 4

1 y = 4sin(π ⋅1) = 4sin π = 4 ⋅ 0 = 0

(1, 0)

3

2 y = 4sin

π ⋅

3

= 4sin 3π 2

= 4(−1) = −4

3 , − 4

2 y = 4sin(π ⋅ 2) = 4sin 2π = 4 ⋅ 0 = 0

(2, 0)

x y = 3sin 2π x coordinates

0 y = 3sin(2π ⋅ 0) = 3sin 0 = 3 ⋅ 0 = 0

(0, 0)

1

4 y = 3sin

2π ⋅

1

= 3sin π

= 3 ⋅1 = 3 2

1 , 3

1

2 y = 3sin

2π ⋅

1

= 3sin π = 3 ⋅ 0 = 0

1 , 0

3

4

3 y = 3sin 2π ⋅

4

= 3sin 3π

= 3(−1) = −3 2

3 , − 3 4

1 y = 3sin(2π ⋅1) = 3sin 2π = 3 ⋅ 0 = 0

(1, 0)

11. The equation y = 4sin π x is of the form 12. The equation y = 3sin 2π x is of the form

y = A sin Bx with A = 4 and B = π . The amplitude is y = A sin Bx with A = 3 and B = 2π . The amplitude

A = 4 = 4 . The period is 2π

= 2π

= 2 . The

is A = 3 = 3 . The period is 2π

= 2π

= 1 . TheB π B 2π

quarter-period is 2

= 1

. The cycle begins at x = 0. 4 2

Add quarter-periods to generate x-values for the key points.

x = 0

x = 0 + 1

= 1

2 2

x = 1

+ 1

= 1 2 2

x = 1 + 1

= 3

2 2

x = 3

+ 1

= 2 2 2

Evaluate the function at each value of x.

quarter-period is 1

. The cycle begins at x = 0. Add 4

quarter-periods to generate x-values for the key points. x = 0

x = 0 + 1

= 1

4 4

x = 1

+ 1

= 1

4 4 2

x = 1

+ 1

= 3

2 4 4

x = 3

+ 1

= 1 4 4

Evaluate the function at each value of x.

2

2

4

4

2

2

2

2

Connect the five points with a smooth curve and

graph one complete cycle of the given function. Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

Full file at https://testbank123.eu/Solutions-Manual-for-Trigonometry-2nd-Edition-Blitzer

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Page 12: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

66 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 67

x y = −3sin 2π x coordinates

0 y = −3sin(2π ⋅ 0) = −3sin 0 = −3 ⋅ 0 = 0

(0, 0)

1

4 y = −3sin

2π ⋅

1

= −3sin π 2

= −3 ⋅1 = −3

1 , − 3 4

1

2 y = −3sin

2π ⋅

1

= −3sin π = −3 ⋅ 0 = 0

1 , 0 2

3

4 y = −3sin

2π ⋅

3

= −3sin 3π 2

= −3(−1) = 3

3 , 3 4

1 y = −3sin(2π ⋅1) = −3sin 2π = −3 ⋅ 0 = 0

(1, 0)

x y = −2sin π x coordinates

0 y = −2sin(π ⋅ 0) = −2sin 0 = −2 ⋅ 0 = 0

(0, 0)

1

2

1 y = −2sin π ⋅

2

= −2sin π

= −2 ⋅1 = −2 2

1 , − 2 2

1 y = −2sin(π ⋅1) = −2sin π = −2 ⋅ 0 = 0

(1, 0)

3

2 y = −2sin

π ⋅

3

= −2sin = −2(−1) = 2 2

3 , 2

2 y = −2sin(π ⋅ 2) = −2sin 2π = −2 ⋅ 0 = 0

(2, 0)

13. The equation y = −3sin 2π x is of the form 14. The equation y = −2sin π x is of the form

y = A sin Bx with A = –3 and B = 2π . The amplitude y = A sin Bx with A = –2 and B = π . The amplitude

is A = −3 = 3 . The period is 2π

= 2π

= 1 . The

is A = −2 = 2 . The period is 2π

= 2π

= 2 . TheB 2π B π

quarter-period is 1

. The cycle begins at x = 0. Add 4

quarter-periods to generate x-values for the key points.

x = 0

x = 0 + 1

= 1

4 4

x = 1

+ 1

= 1

4 4 2

x = 1

+ 1

= 3

2 4 4

x = 3

+ 1

= 1 4 4

Evaluate the function at each value of x.

quarter-period is 2

= 1

. 4 2

The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x = 0

x = 0 + 1

= 1

2 2

x = 1

+ 1

= 1 2 2

x = 1 + 1

= 3

2 2

x = 3

+ 1

= 2 2 2

Evaluate the function at each value of x.

4

2

4

2

2

Connect the five points with a smooth curve and graph one complete cycle of the given function.

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

Full file at https://testbank123.eu/Solutions-Manual-for-Trigonometry-2nd-Edition-Blitzer

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Page 13: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

66 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 68

x 2 y = − sin x

3

coordinate s

0 2 y = − sin

⋅ 0 3

= − sin 0 = 0

(0, 0)

4 y = − sin

2 ⋅

= − sin π

= −1 2

3π , −1

2 y = − sin

2 ⋅

= − sin π = 0

3π , 0

4 y = − sin

2 ⋅

9π 3 4

= − sin 3π

= −(−1) = 1 2

9π , 1

3π 2 y = − sin ⋅ 3π

3 = − sin 2π = 0

(3π , 0)

x 4

y = − sin x 3

coordinates

0 4

y = − sin ⋅ 0 = − sin 0 = 0 3

(0, 0)

8 y = − sin

4 ⋅

3π = − sin

π = −1

3 8 2

3π , −1

8

4

4 3π y = − sin ⋅ = − sin π = 0

3 4

3π , 0 4

8

4 9π y = − sin ⋅

3 8

= − sin 3π

= −(−1) = 1 2

9π , 1 8

2

4 3π y = − sin ⋅ = − sin 2π = 0

3 2

3π , 0 2

15. The equation y = − sin 2

x is of the form 3

y = A sin Bx

16. The equation y = − sin 4

x is of the form 3

with A = –1 and B = 2

. 3

y = A sin Bx with A = –1 and B = 4

. 3

The amplitude is A = −1 = 1 . The amplitude is A = −1 = 1 .

The period is 2π =

2π = 2π ⋅

3 = 3π .

The period is 2π =

2π = 2π ⋅

3 =

3π .

B 2 2 3

B 4 4 2 3

The quarter-period is 3π

. The cycle begins at x = 0. Add 3π

1 3π4 The quarter-period is 2 = ⋅ = .

quarter-periods to generate x-values for the key points.

x = 0

x = 0 + 3π

= 3π

4 2 4 8

The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x = 0

4 4

x = 3π

+ 3π

= 3π x = 0 +

3π 8

= 3π 8

4 4 2

x = 3π

+ 3π

= 9π

2 4 4

x = 9π

+ 3π

= 3π 4 4

Evaluate the function at each value of x.

x = 3π 8

x = 3π 4

x = 9π 8

+ 3π 8

+ 3π 8

+ 3π 8

= 3π 4

= 9π 8

= 3π 2

Evaluate the function at each value of x.

3 4

4

3 2

2

4

Connect the five points with a smooth curve and

graph one complete cycle of the given function.

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

Full file at https://testbank123.eu/Solutions-Manual-for-Trigonometry-2nd-Edition-Blitzer

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Page 14: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

68 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 68

x y = sin( x − π ) coordinates

π y = sin(π − π ) = sin 0 = 0

(π , 0)

2 y = sin

3π − π

= sin π

= 1 2

3π , 1

2π y = sin(2π − π ) = sin π = 0

(2π , 0)

2 y = sin

5π − π

= sin 3π

= −1 2

5π , − 1

3π y = sin(3π − π ) = sin 2π = 0

(3π , 0)

x π y = sin x −

2

coordinates

π

2

π π y = sin − = sin 0 = 0

2 2

π 2

, 0

π y = sin π − π

= sin π

= 1

2

2

(π , 1)

2 y = sin

3π − π

= sin π = 0 2 2

3π , 0

2

2π π 3π

y = sin 2π − = sin = −1 2 2

(2π , −1)

2

5π π y = sin − = sin 2π = 0

2 2

5π , 0 2

17. The equation y = sin( x − π ) is of the form

18. The equation y = sin x − π

is of the formy = A sin(Bx − C) with A = 1, B = 1, and C = π . The 2

amplitude is A = 1 = 1 . The period is y = A sin(Bx − C) with A = 1, B = 1, and C =

π . The

2π =

2π = 2π . The phase shift is

C = π

= π . The 2

B 1 B 1 amplitude is A = 1 = 1 . The period is

πquarter-period is

2π = π

. The cycle begins at 2π =

2π = 2π . The phase shift is

C = 2 =

π . The

4 2 x = π . Add quarter-periods to generate x-values for

B 1

2π π B 1 2

the key points.

x = π quarter-period is = . The cycle begins at

4 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2

x = 2π + π

= 5π

2 2

x = 5π

+ π

= 3π

x = π

. Add quarter-periods to generate 2

x-values for the key points.

x = π 2

x = π

+ π

= π 2 2

2 2 Evaluate the function at each value of x.

x = π + π

= 3π

2 2

2

2

x = 3π

+ π

= 2π 2 2

x = 2π + π

= 5π

2 2 Evaluate the function at each value of x.

2

2

Connect the five points with a smooth curve and

graph one complete cycle of the given function.

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

Full file at https://testbank123.eu/Solutions-Manual-for-Trigonometry-2nd-Edition-Blitzer

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Page 15: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

69 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 69

x y = sin(2 x − π ) coordinates

π

2 y = sin

2 ⋅ π

− π

= sin(π − π ) = sin 0 = 0

π , 0

4 y = sin

2 ⋅

3π − π

= sin 3π

− π

= sin π

= 1 2

3π , 1

π y = sin(2 ⋅ π − π ) = sin(2π − π ) = sin π = 0

(π , 0)

4 y = sin

2 ⋅

5π − π

= sin 5π

− π

= sin 3π

= −1 2

5π , − 1

2 y = sin

2 ⋅

3π − π

= sin(3π − π ) = sin 2π = 0

2

x y = sin 2 x −

π

coordinates

π

4 y = sin

2 ⋅ π

− π

4 2 π π

= sin − = sin 0 = 0 2 2

π , 0

4

π

2 y = sin

2 ⋅ π

− π

= sin π − π

= sin π

= 1

2

2

π , 1

2

2

19. The equation y = sin(2 x − π ) is of the form Connect the five points with a smooth curve and

y = A sin(Bx − C) with A = 1, B = 2, and C = π . The graph one complete cycle of the given function.

amplitude is A = 1 = 1 . The period is

2π =

2π = π . The phase shift is

C = π

. The

B 2 B 2

quarter-period is π

. The cycle begins at x = π

. Add

4 2quarter-periods to generate x-values for the key points.

x = π 2

x = π

+ π

= 3π

2 4 4

x = 3π

+ π

= π 4 4

20. The equation

y = sin 2 x −

π

is of the form

π

x = π + π

= 5π

y = A sin(Bx − C) with A = 1, B = 2, and C = . The 2

4 4

x = 5π

+ π

= 3π amplitude is A = 1 = 1 .

4 4 2 The period is 2π

= 2π

= π .B 2

Evaluate the function at each value of x. π C π 1 π

= = ⋅ = .The phase shift is

B

2 2 2 4 π

2

2

The quarter-period is . 4

The cycle begins at x = π

. Add quarter-periods to 4

generate x-values for the key points.

π 4

4 x =

4 π π π

2

x = + = 4 4 2

x = π

+ π

= 3π

2 4 4 3π π

x = + = π 4 4

x = π + π

= 5π

4 4

4

4

Evaluate the function at each value of x.

2

2

2

3π , 0

2 2

2

Full file at https://testbank123.eu/Solutions-Manual-for-Trigonometry-2nd-Edition-Blitzer

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Page 16: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

70 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 70

4 y = sin

2 ⋅

3π − π

= sin 3π

− π

2 2

= sin π = 0

3π , 0

π y = sin 2 ⋅ π −

π

= sin 2π −

π

= sin 3π

= −1 2

(π , − 1)

4 y = sin

2 ⋅

5π − π

= sin 5π

− π

2 2

= sin 2π = 0

5π , 0

x y = 3sin(2x − π ) coordinates

π

2

π y = 3sin 2 ⋅ − π

2 = 3sin(π − π ) = 3sin 0 = 3 ⋅ 0 = 0

π 2

, 0

4 y = 3sin

2 ⋅

3π − π

= 3sin 3π

− π 2

= 3sin π

= 3 ⋅1 = 3 2

3π , 3

4

π y = 3sin(2 ⋅π − π ) = 3sin(2π − π ) = 3sin π = 3 ⋅ 0 = 0

(π , 0)

4

5π y = 3sin 2 ⋅ − π

4

= 3sin 5π

− π

= 3sin 3π 2

= 3(−1) = −3

4 , − 3

2 y = 3sin

2 ⋅

3π − π

= 3sin(3π − π ) = 3sin 2π = 3 ⋅ 0 = 0

3π , 0

2

4 2

4

2

2

4

4 2

4

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

2

2

2

Connect the five points with a smooth curve and

graph one complete cycle of the given function.

21. The equation y = 3sin(2x − π ) is of the form

y = A sin(Bx − C ) with A = 3, B = 2, and C = π . The

amplitude is A = 3 = 3 . The period is

2π =

2π = π . The phase shift is

C = π

. The quarter-

B 2 B 2

period is π

. The cycle begins at x = π

. Add quarter-

4 2periods to generate x-values for the key points.

x = π

22. The equation

y = 3sin 2 x −

π

is of the form

2

2

x = π

+ π

= 3π

2 4 4

x = 3π

+ π

= π

y = A sin(Bx − C ) with A = 3, B = 2, and C = π

. 2

4 4

x = π + π

= 5π

The amplitude is

2π A = 3

2π = 3 .

4 4

x = 5π

+ π

= 3π

4 4 2

The period is = = π . B 2

C π

π 1 π

= = ⋅ = .Evaluate the function at each value of x.

The phase shift is B

2 2 2 4

Full file at https://testbank123.eu/Solutions-Manual-for-Trigonometry-2nd-Edition-Blitzer

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Page 17: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

71 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 71

x y = 3sin 2 x −

π

coordinates

π

4 y = 3sin

2 ⋅ π

− π

= sin π

− π

= 3sin 0 = 3 ⋅ 0 = 0

π , 0

π

2 y = 3sin

2 ⋅ π

− π

= 3sin π − π

= 3sin π

= 3 ⋅1 = 3 2

π , 3

4 y = 3sin

2 ⋅

3π − π

= 3sin 3π

− π

2 2

= 3sin π = 3 ⋅ 0 = 0

4

π y = 3sin 2 ⋅ π −

π

2

= 3sin 2π −

π

2

= 3sin 3π

= 3 ⋅ (−1) = −3 2

(π , − 3)

4 y = 3sin

2 ⋅

5π − π

= 3sin 5π

− π

2 2

= 3sin 2π = 3 ⋅ 0 = 0

5π , 0

x

= −

π

The quarter-period is π

. 4

The cycle begins at x = π

. Add quarter-periods to 4

generate x-values for the key points.

x = π 4

x = π

+ π

= π

4 4 2

x = π

+ π

= 3π

2 4 4

x = 3π

+ π

= π

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

4 4 1 π 1 π x = π +

π =

5π 23. y = sin x + = sin x − − 2 2 2 2

4 4

Evaluate the function at each value of x.

The equation

y = 1

sin x −

− π

is of the form2

2

2

1 πy = A sin(Bx − C ) with A = , B = 1, and C = − .

4 2

4

The amplitude is

2 2

A = 1

= 1

. The period is 2 2

2 2

2π 2π π

C 2= = 2π . The phase shift is B 1

2π π

= = − . B 1 2

2 2

2

The quarter-period is

π

= . The cycle begins at 4 2

2 x = − . Add quarter-periods to generate x-values

2 for the key points.

π = −

4 2

3π , 0

2

x π

+ π

= 0 2 2

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2 Evaluate the function at each value of x.

4 2

4

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

72 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 72

x y = 1

sin x + π

2

2

coordinates

− π 2

y = 1

sin − π

+ π

2

2 2

= 1

sin 0 = 1

⋅ 0 = 0 2 2

2

0 y = 1

sin 0 + π

2

2

= 1

sin π

= 1

⋅1 = 1

2 2 2 2

0,

1

π

2 y =

1 sin

π + π

2 2 2

= 1

sin π = 1

⋅ 0 = 0 2 2

2

π y = 1

sin π + π

2

2

= 1

sin 3π

2 2

= 1

⋅ (−1) = − 1

2 2

π , −

1

2 y =

1 sin

3π + π

2

2 2

= 1

sin 2π 2

= 1

⋅ 0 = 0 2

3π , 0

x 1

y = sin( x + π ) 2

coordinates

−π 1

y = sin(−π + π ) 2 1 1

= sin 0 = ⋅ 0 = 0 2 2

(−π , 0)

π −

2

1 π y = sin − + π

2 2

= 1

sin π

= 1

⋅1 = 1

2 2 2 2

π 1 −

2 ,

2

0 y = 1

sin(0 + π ) 2 1 1

= sin π = ⋅ 0 = 0 2 2

(0, 0)

π

2 y =

1 sin

π + π

2 2

= 1

sin 3π

= 1

⋅ (−1) = − 1

2 2 2 2

π , −

1 2 2

π y = 1

sin(π + π ) 2 1 1

= sin 2π = ⋅ 0 = 0 2 2

(π , 0)

= 2π

= 2π . The phase shift is C

= −π

= −π . B 1 B 1

− π

, 0

The quarter-period is 2π

= π

. The cycle begins at 4 2

x = −π . Add quarter-periods to generate x-values for the key points. x = −π

x = −π + π

= − π

2 2 π π

2

x = − + = 0 2 2

x = 0 + π

= π

2 2

x = π

+ π

= π π

, 0 2 2 Evaluate the function at each value of x.

2

2

Connect the five points with a smooth curve and

graph one complete cycle of the given function.

24.

y = 1

sin( x + π ) = 1

sin( x − (−π )) 2 2

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

The equation y = 1

sin( x − (−π )) 2

is of the form

y = A sin(Bx − C) with A = 1

, B = 1, and C = −π . 2

The amplitude is A = 1 =

1 . The period is

2 2

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

73 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 73

π

2 y = −2sin

2 ⋅ π

+ π

2 2 π

= −2sin π +

3π 2

= −2sin 2

= −2(−1) = 2

π , 2 2

4 y = −2sin

2 ⋅

3π + π

= −2sin 3π

+ π

2 2 = −2sin 2π = −2 ⋅ 0 = 0

3π , 0

4

x y = −2sin

2 x +

π

coordinates

− π 4

y = −2sin 2⋅

− π

+ π

= −2sin − π

+ π

= −2sin 0 = −2 ⋅ 0 = 0

− π

, 0

0 y = −2sin

2 ⋅ 0 +

π

= −2sin 0 + π

= −2sin π 2

= −2 ⋅1 = −2

(0, –2)

π

4 y = −2sin

2 ⋅ π

+ π

= −2sin π

+ π

2 2

= −2sin π = −2 ⋅ 0 = 0

4

= −

= −

= −

C

25.

y = −2sin 2 x +

π = −2 sin

2x −

− π

2

2

The equation

y = −2sin 2 x −

− π

is of the form 2

y = A sin(Bx − C) with A = –2,

B = 2, and C π

. The amplitude is

A = −2

2

= 2 . The period is 2π

= 2π

= π . The

4 2

B 2 π

phase shift is C

= −

2 = − π

⋅ 1

= − π

. The quarter- B 2 2 2 4

period is π

. The cycle begins at x π

. Add Connect the five points with a smooth curve and4 4 graph one complete cycle of the given function.

quarter-periods to generate x-values for the key

points.

π x = −

4 π π

x = − + = 0 4 4

x = 0 + π

= π

4 4

x = π

+ π

= π

4 4 2

x = π

+ π

= 3π

2 4 4

26.

y = −3sin 2x +

π = −3sin

2 x −

− π

Evaluate the function at each value of x.

2

2

π The equation y = −3sin 2 x − −

is of the form

2 2

π

4

2

4

y = A sin(Bx − C) with A = –3, B = 2, and C = − . 2

The amplitude is A = −3 = 3 . The period is

2 2

2π =

2π = π . The phase shift is

B 2

π−

2 π 1 π π

2 = = − ⋅ = − . The quarter-period is . B 2 2 2 4 4

π 2 The cycle begins at x = − . Add quarter-periods to

4 generate x-values for the key points.

π x = −

4 2 π

, 0

4

x π

+ π

= 0 4 4

x = 0 + π

= π

4 4

x = π

+ π

= π

4 4 2

x = π

+ π

= 3π

2 4 4

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

74 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 74

x y = −3sin 2 x +

π

coordinates

− π 4

y = −3sin 2 ⋅

− π

+ π

= −3sin − π

+ π

2 2

= −3sin 0 = −3 ⋅ 0 = 0

− π

, 0

0 y = −3sin 2 ⋅ 0 +

π

2

= −3sin 0 + π

= −3sin π

= −3 ⋅1 = −3 2

(0, –3)

π

4 y = −3sin

2 ⋅ π

+ π

= −3sin π

+ π

2 2

= −3sin π = −3 ⋅ 0 = 0

π , 0

π

2 y = −3sin

2 ⋅ π

+ π

= −3sin π + π

= −3sin 3π

= −3 ⋅ (−1) = 3 2

π , 3

4 y = −3sin

2 ⋅

3π + π

= −3sin 3π

+ π

= −3sin 2π = −3 ⋅ 0 = 0

3π , 0

x y = 3sin(π x + 2) coordinates

− 2 π

y = 3sin π

2 + 2

= 3sin(−2 + 2) = 3sin 0 = 3 ⋅ 0 = 0

2 − , 0 π

π − 4

π − 4 y = 3sin π + 2

= 3sin π − 4

+ 2 2

π = 3sin − 2 + 2

2

= 3sin π 2

= 3 ⋅1 = 3

π − 4 2π

π − 2

π

π − 2 y = 3sin π

π + 2

= 3sin(π − 2 + 2) = 3sin π = 3 ⋅ 0 = 0

π − 2 , 0

3π − 4

3π − 4 y = 3sin π + 2

= 3sin 3π − 4

+ 2 2

= 3sin 3π

− 2 + 2 2

= 3sin 3π 2

= 3(−1) = −3

5π , − 3 4

2π − 2

π y = 3sin

π

2π − 2 + 2

= 3sin(2π − 2 + 2) = 3sin 2π = 3 ⋅ 0 = 0

2π − 2 , 0

Evaluate the function at each value of x. 2π =

2π = 2 . The phase shift is C

= −2

= − 2

. The

2

B π B π π 2 1

quarter-period is = . The cycle begins at 4 2

4

2

4

x = − 2

. Add quarter-periods to generate x-values π

for the key points.

= − 2

x π

= − 2

+ 1

= π − 4

x π 2 2π

= π − 4

+ 1

= π − 2

x 2π 2 π

2 x = π − 2

+ 1

= 3π − 4

π 2 2π

x = 3π − 4

+ 1

= 2π − 2

2π 2 π

4 2

4

Evaluate the function at each value of x.

2 2

2

2

π

, 3

4 2

2 2

4

27.

Connect the five key points with a smooth curve and graph one complete cycle of the given function.

y = 3sin(π x + 2)

π

The equation y = 3sin(π x − (−2)) is of the form

y = A sin(Bx − C) with A = 3, B = π , and C = –2.

The amplitude is A = 3 = 3 . The period is

π π

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

75 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 75

π − 4

π − 4 y = 3sin 2π + 4

2π = 3sin(π − 4 + 4) = 3sin π = 3 ⋅ 0 = 0

π − 4

2π , 0

3π − 8

3π − 8 y = 3sin 2π + 4

= 3sin 3π − 8

+ 4 2

= 3sin 3π

− 4 + 4 2

= 3sin 3π

= 3(−1) = −3 2

3π − 8 , − 3

π − 2

π

π − 2 y = 3sin 2π

π + 4

= 3sin(2π 4 + 4)

= π = 3 ⋅ 0 = 0 3sin 2

π − 2 ,0

x y = 3sin(2π x + 4) coordinates

2 − π

2 y = 3sin 2π − + 4

π = 3sin(−4 + 4) = 3sin 0 = 3 ⋅ 0 = 0

2 − π

, 0

π − 8

π − 8 y = 3sin 2π + 4

= 3sin π − 8

+ 4

= 3sin π

− 4 + 4

= 3sin π

= 3 ⋅1 = 3 2

π − 8 , 3

= −

Connect the five points with a smooth curve and

graph one complete cycle of the given function.

28. y = 3sin(2π x + 4) = 3sin(2π x − (−4))

The equation y = 3sin(2π x − (−4)) is of the form

y = A sin(Bx − C) with A = 3, B = 2π , and

C = –4. The amplitude is

A = 3

= 3 . The period − π

is 2π

= 2π

= 1 . The phase shift is C

= −4

= − 2

.B 2π B 2π π Connect the five key points with a smooth curve and

The quarter-period is 1

. The cycle begins at 4

x = − 2

. Add quarter-periods to generate x-values π

for the key points.

x = − 2 π

x = − 2

+ 1

= π − 8

graph one complete cycle of the given function.

π 4 4π

x = π − 8

+ 1

= π − 4

4π 4 2π 29. y = −2 sin(2π x + 4π ) = −2 sin(2π x − (−4π ))x = π − 4

+ 1

= 3π − 8

The equation

y = −2sin(2π x − (−4π )) is of the form2π 4 4π

x = 3π − 8

+ 1

= π − 2 y = A sin(Bx − C) with A = –2, B = 2π , and

4π 4 π C = −4π . The amplitude is A = −2 = 2 . TheEvaluate the function at each value of x.

period is 2π

= 2π

= 1 . The phase shift is B 2π

C −4π 1 = = −2 . The quarter-period is . The cycle

B 2π 4

2

begins at x = −2 . Add quarter-periods to generate x-

values for the key points.

x = −2

x = −2 + 1

= − 7

4 4

x 7

+ 1

= − 3

4 4 2 3 1 5

2

x = − + = −

2 4 4 5 1

= − + = −x 1

4 4

Evaluate the function at each value of x.

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

76 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 76

x y = −2sin(2π x + 4π ) coordinates

–2 y = −2sin(2π (−2) + 4π ) = −2sin(−4π + 4π ) = −2sin 0 = −2 ⋅ 0 = 0

(–2, 0)

− 7 4

y = −2sin 2π

7 + 4π

= −2sin −

7π + 4π

2

= −2sin π

= −2 ⋅1 = −2 2

7 , − 2

− 3 2

y = −2sin 2π

3 + 4π

= −2sin(−3π + 4π ) = −2sin π = −2 ⋅ 0 = 0

3 , 0

− 5 4

y = −2sin 2π

5 + 4π

= −2sin −

5π + 4π

= −2sin 3π 2

= −2(−1) = 2

5 , 2

4

–1 y = −2sin(2π (−1) + 4π ) = −2sin(−2π + 4π ) = −2sin 2π = −2 ⋅ 0 = 0

(–1, 0)

x y = −3sin(2π x + 4π ) coordinates

–2 y = −3sin(2π (−2) + 4π ) = −3sin(−4π + 4π ) = −3sin 0 = −3 ⋅ 0 = 0

(–2, 0)

7 −

4

7 y = −3sin 2π − + 4π

4 π

= −3sin − 7

+ 4π 2

= −3sin π

= −3 ⋅1 = −3 2

7 −

4 , − 3

3 −

2

3 y = −3sin 2π − + 4π

2 = −3sin(−3π + 4π ) = −3sin π = −3 ⋅ 0 = 0

3 − , 0 2

5 −

4

5 y = −3sin 2π − 4

+ 4π

= −3sin −

5π + 4π

= −3sin 3π

= −3(−1) = 3 2

5 , 3

4

–1 y = −3sin(2π (−1) + 4π ) = −3sin(−2π + 4π ) = −3sin 2π = −3 ⋅ 0 = 0

(–1, 0)

= −

= −

x = −2

x = −2 + 1

= − 7

4 4 7

x = − + 1

= − 3

4 4 2

4

4

x 3

+ 1

= − 5

2 4 4

x 5

+ 1

= −1 4 4

Evaluate the function at each value of x.

2

2

4

2

Connect the five points with a smooth curve and

graph one complete cycle of the given function.

2

30.

y = −3sin(2π x + 4π ) = −3sin(2π x − (−4π ))

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

The equation y = −3sin(2π x − (−4π )) is of the form

y = A sin(Bx − C ) with A = –3, B = 2π , and

C = −4π . The amplitude is A = −3 = 3 . The

period is 2π

= 2π

= 1 . The phase shift is B 2π

C =

−4π = −2 . The quarter-period is

1 . The cycle

B 2π 4

begins at x = −2 . Add quarter-periods to generate x-

values for the key points.

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

77 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 77

x y = 2 cos x coordinates

0 y = 2 cos 0 = 2 ⋅1 = 2 (0, 2)

π

2 y = 2 cos

π = 2 ⋅ 0 = 0

2

2

π y = 2 cos π = 2 ⋅ (−1) = −2

(π , − 2)

2 y = 2 cos

3π 2

= 2 ⋅ 0 = 0

2

2π y = 2 cos 2π = 2 ⋅1 = 2

(2π , 2)

x y = 3cos x coordinates

0 y = 3cos 0 = 3 ⋅1 = 3 (0, 3)

π

2

π y = 3cos = 3 ⋅ 0 = 0

2

π

2

π y = 3cos π = 3 ⋅ (−1) = −3 (π , − 3)

2 y = 3cos

3π = 3 ⋅ 0 = 0

2

3π , 0

2

2π y = 3cos 2π = 3 ⋅1 = 3 (2π , 3)

31. The equation y = 2 cos x is of the form y = A cos x 32. The equation y = 3cos x is of the form y = A cos x

with A = 2. Thus, the amplitude is A = 2 = 2 . with A = 3. Thus, the amplitude is A = 3 = 3 .

The period is 2π . The quarter-period is 2π or

π . The period is 2π . The quarter-period is

2π or π

.4 2

The cycle begins at x = 0 . Add quarter-periods to

generate x-values for the key points.

x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

4 2 The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2 3π

2 2

x = 3π

+ π

= 2π 2 2

Evaluate the function at each value of x.

x = + π

= 2π 2 2

Evaluate the function at each value of x.

π , 0 , 0

3π , 0

Connect the five key points with a smooth curve

and graph one complete cycle of the given function

Connect the five points with a smooth curve and

graph one complete cycle of the given function with

with the graph of y = cos x .

the graph of y = 2 cos x .

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Page 24: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

78 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 78

x y = −3cos x coordinates

0 y = −3cos 0 = −3 ⋅1 = −3 (0, –3)

π

2

π y = −3cos

2 = −3 ⋅ 0 = 0

π 2

, 0

π y = −3cos π = −3 ⋅ (−1) = 3 (π , 3)

2

3π y = −3cos = −3 ⋅ 0 = 0

2

3π , 0 2

2π y = −3cos 2π = −3 ⋅1 = −3 (2π , − 3)

x y = −2 cos x coordinates

0 y = −2 cos 0 = −2 ⋅1 = −2

(0, –2)

π

2 y = −2 cos

π 2

= −2 ⋅ 0 = 0

π , 0

2

π y = −2 cos π = −2 ⋅ (−1) = 2

(π , 2)

2 y = −2 cos

3π 2

= −2 ⋅ 0 = 0

3π , 0

2

2π y = −2 cos 2π = −2 ⋅1 = −2

(2π , − 2)

33. The equation y = −2 cos x is of the form y = A cos x x = 0

with A = –2. Thus, the amplitude is x = 0 + π

= π

A = −2 = 2 . The period is 2π . The quarter-

2 2

π

period is 2π

or π

. The cycle begins at x = 0 . Add

x = + π

= π 2 2

4 2 quarter-periods to generate x-values for the key points.

x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2

Evaluate the function at each value of x.

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2

Evaluate the function at each value of x.

Connect the five key points with a smooth curve and

graph one complete cycle of the given function with

the graph of y = cos x .

Connect the five points with a smooth curve and 35. The equation y = cos 2 x is of the form y = A cos Bx

graph one complete cycle of the given function with with A = 1 and B = 2. Thus, the amplitude isthe graph of y = cos x .

A = 1

= 1 . The period is 2π

= 2π

= π . The B 2

34. The equation

y = −3cos x is of the form

y = A cos x

quarter-period is π

. The cycle begins at x = 0 . Add 4

quarter-periods to generate x-values for the key points.

x = 0

x = 0 + π

= π

4 4 π π π

with A = –3. Thus, the amplitude is A = −3 = 3 . x = + = 4 4 2

The period is 2π . The quarter-period is 2π or

π . x =

π + π

= 3π

4 2 The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points.

2 4 4

x = 3π

+ π

= π 4 4

Evaluate the function at each value of x.

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

79 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 79

x y = cos 2 x coordinates

0 y = cos(2 ⋅ 0) = cos 0 = 1

(0, 1)

π

4 y = cos

2 ⋅ π

= cos π

= 0 2

π , 0

π

2 y = cos

2 ⋅ π

= cos π = −1

π , −1

4 y = cos

2 ⋅

= cos 3π

= 0 2

3π , 0

π y = cos(2 ⋅ π ) = cos 2π = 1

(π , 1)

x y = cos 4 x coordinates

0 y = cos(4 ⋅ 0) = cos 0 = 1 (0, 1)

π

8

π π y = cos 4 ⋅

8 = cos

2 = 0

π 8

, 0

π

4 y = cos

4 ⋅ π

= cos π = −1

π , −1

8

3π y = cos 4 ⋅

8

= cos 3π

= 0 2

3π 8

, 0

π

2

π y = cos 4 ⋅

2 = cos 2π = 1

π 2

, 1

4

4

2

2

4

4

4

4

Connect the five points with a smooth curve and graph one complete cycle of the given function.

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

37. The equation y = 4 cos 2π x is of the form y = A cos Bx

with A = 4 and B = 2π . Thus, the amplitude is

A = 4 = 4 . The period is 2π

= 2π

= 1 . The quarter-B 2π

136. The equation y = cos 4 x is of the form y = A cos Bx period is . The cycle begins at x = 0 . Add quarter-

4with A = 1 and B = 4. Thus, the amplitude is periods to generate x-values for the key points.

A = 1

= 1 . The period is 2π

= 2π

= π

. The x = 0

B 4 2 π

x = 0 + 1

= 1

4 4

quarter-period is 2 = π

⋅ 1

= π

. The cycle begins at 4 2 4 8

x = 0. Add quarter-periods to generate x-values for

the key points. x = 0

x = 0 + π

= π

8 8

x = π

+ π

= π

8 8 4

x = π

+ π

= 3π

4 8 8

x = 3π

+ π

= π

8 8 2 Evaluate the function at each value of x.

x = 1

+ 1

= 1

4 4 2

x = 1

+ 1

= 3

2 4 4

x = 3

+ 1

= 1 4 4

Evaluate the function at each value of x.

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Page 26: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

80 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 80

x y = 4 cos 2π x coordinates

0 y = 4 cos(2π ⋅ 0) = 4 cos 0 = 4 ⋅1 = 4

(0, 4)

1

4 y = 4 cos

2π ⋅

1

= 4 cos π 2

= 4 ⋅ 0 = 0

1 , 0

1

2 y = 4 cos

2π ⋅

1

= 4 cos π = 4 ⋅ (−1) = −4

1 , − 4

3

4 y = 4 cos

2π ⋅

3

= 4 cos 3π 2

= 4 ⋅ 0 = 0

3 , 0

1 y = 4 cos(2π ⋅1) = 4 cos 2π = 4 ⋅1 = 4

(1, 4)

x y = 5cos 2π x coordinates

0 y = 5 cos(2π ⋅ 0) = 5cos 0 = 5 ⋅1 = 5

(0, 5)

1

4 y = 5cos

2π ⋅

1 4

= 5cos π

= 5 ⋅ 0 = 0 2

1 , 0

4

1

2 y = 5cos

2π ⋅

1 2

= 5cos π = 5 ⋅ (−1) = −5

1 , − 5

2

3

4 y = 5cos

2π ⋅

3π 4

= 5cos 3π

= 5 ⋅ 0 = 0 2

3 , 0

4

1 y = 5cos(2π ⋅1) = 5cos 2π = 5 ⋅1 = 5

(1, 5)

4

4

2

2

4

4

Connect the five points with a smooth curve and

graph one complete cycle of the given function.

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

38. The equation

y = 5cos 2π x

is of the form

39. The equation y = −4 cos 1

x is of the form 2

1y = A cos Bx with A = 5 and B = 2π . Thus, the y = A cos Bx with A = –4 and B = . Thus, the

2amplitude is A = 5 = 5 . The period is amplitude is A = −4 = 4 . The period is

2π =

2π = 1 . The quarter-period is

1 . The cycle 2π

= 2π

= 2π ⋅ 2 = 4π . The quarter-period isB 2π 4 B 1

begins at x = 0. Add quarter-periods to generate x-

values for the key points. x = 0

x = 0 + 1

= 1

4 4

x = 1

+ 1

= 1

4 4 2

x = 1

+ 1

= 3

2 4 4

x = 3

+ 1

= 1 4 4

Evaluate the function at each value of x.

2

4π = π . The cycle begins at x = 0 . Add quarter-

4 periods to generate x-values for the key points. x = 0 x = 0 + π = π x = π + π = 2π x = 2π + π = 3π x = 3π + π = 4π

Evaluate the function at each value of x.

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

81 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 81

x 1 y = −4 cos x

2

coordinates

0 y = −4 cos

1 ⋅ 0

= −4 cos 0 = −4 ⋅1 = −4

(0, –4)

π 1 y = −4 cos ⋅ π

2

= −4 cos π 2

= −4 ⋅ 0 = 0

(π , 0)

2π 1 y = −4 cos ⋅ 2π

2 = −4 cos π = −4 ⋅ (−1) = 4

(2π , 4)

3π 1 y = −4 cos ⋅ 3π

2

= −4 cos 3π 2

= −4 ⋅ 0 = 0

(3π , 0)

4π 1 y = −4 cos ⋅ 4π

2 = −4 cos 2π = −4 ⋅1 = −4

(4π , – 4)

x y = −3cos

1 x

3

coordinates

0 y = −3cos

1 ⋅ 0

3 = −3cos 0 = −3 ⋅1 = −3

(0, –3)

2 y = −3cos

1 ⋅

= −3cos π

= −3 ⋅ 0 = 0 2

3π , 0

3π y = −3cos

1 ⋅ 3π

3 = −3cos π = −3 ⋅ (−1) = 3

(3π , 3)

2 y = −3cos

1 ⋅

3π = −3cos

2 = −3 ⋅ 0 = 0

9π , 0

6π y = −3cos

1 ⋅ 6π 3

= −3cos 2π = −3 ⋅1 = −3

(6π , − 3)

= − 3

x = 0

3π 3π

2

x = 0 + = 2 2

x = 3π

+ 3π

= 3π 2 2

x = 3π + 3π

= 9π

2 2

x = 9π

+ 3π

= 6π 2 2

Evaluate the function at each value of x.

3 2

2

Connect the five points with a smooth curve and

graph one complete cycle of the given function.

3 2

2

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

40. The equation

y = A cos Bx

y = −3cos 1

x is of the form 3

with A = –3 and B = 1

. Thus, the 3

amplitude is A = −3 = 3 . The period is

2π =

2π = 2π ⋅ 3 = 6π . The quarter-period is

41. The equation

y

1 cos π

x is of the formB 1

2 3

6π =

3π . The cycle begins at x = 0. Add quarter-

4 2

y = A cos Bx

with 1

A = − 2

and B = π

. Thus, the 3

periods to generate x-values for the key points.

amplitude is A = − 1

= 1

. The period is 2 2

2π =

2π = 2π ⋅

3 = 6 . The quarter-period is

6 =

3 .

B π π 4 2 3

The cycle begins at x = 0 . Add quarter-periods to

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Page 28: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

82 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 82

x y 1

cos π

x 2 3

coordinates

0 y 1

cos π

⋅ 0 2 3 1

cos 0 2 1

⋅1 = − 1

2 2

0, −

1 2

3

2 y

1 cos

π 3 2 3 2 1

cos π = −

2 2 1

⋅ 0 = 0 = −

2

3 , 0 2

3 y 1

cos π

⋅ 3 2 3 1

cos π = −

2 1

⋅ (−1) = 1

2 2

3,

1 2

9

2 y

1 cos

π 9 2 3 2 1

cos 3π

2 2 1

⋅ 0 = 0 2

9 , 0 2

6 y 1

cos π

⋅ 6 2 3 1

cos 2π 2 1

⋅1 = − 1

2 2

6, −

1

2

x y

1 cos π

x

= − 2 4

coordinates

0 1 π y

2 cos

4 ⋅ 0 = −

1

cos 0 1

⋅1 = − 1 = − = −

2 2 2

1 0, −

2

2 y

1 cos

π ⋅ 2

= − 2 4

1 cos π

= − 1

⋅ 0 = 0 2 2 2

(2, 0)

4 1 π y

2 cos

4 ⋅ 4 = −

1 1 1 = − cos π = − ⋅ (−1) =

2 2 2

1 4,

2

6 y

1 cos

π ⋅ 6

2 4 1

cos 3π

= − 1

⋅ 0 = 0

= − 2 2 2

(6, 0)

8 1 π y = − cos ⋅ 8

2 4 1

cos 2π = − 1

⋅1 = − 1 = −

2 2 2

1 8, − 2

= −

generate x-values for the key points.

x = 0

x = 0 + 3

= 3

2 2

x = 3

+ 3

= 3 2 2

x = 3 + 3

= 9

2 2

x = 9

+ 3

= 6 2 2

Evaluate the function at each value of x.

Connect the five points with a smooth curve and

graph one complete cycle of the given function.

= − 42. The equation y

1 cos π

x 2 4

1

is of the form

π

= −

y = A cos Bx with A = − and B = . Thus, the 2 4

1 1

= − amplitude is

A = − = . The period is

2 2

= − 2π

= 2π

B π 4

= 2π ⋅ 4

= 8 . The quarter-period is π

8 = 2 . 4

= − ⋅ The cycle begins at x = 0. Add quarter-periods to

generate x-values for the key points. x = 0 x = 0 + 2 = 2 x = 2 + 2 = 4 x = 4 + 2 = 6 x = 6 + 2 = 8

Evaluate the function at each value of x.

= −

= −

= − ⋅

= −

= − = −

= −

= −

= −

= −

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

83 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 83

x coordinates

− π 2

− π

, 1

0 (0, 0)

π

2

π , −1 2

π (π , 0)

2

3π , 1

x coordinates

π

2

π , 1

π (π , 0)

2

3π , −1

2

2π (2π , 0)

2

5π , 1 2

x

= −

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

Connect the five points with a smooth curve and

graph one complete cycle of the given function

π

43. The equation

y = cos x − π

is of the form 44. The equation y = cos x +

2 is of the form

2

y = A cos ( Bx − C ) with A = 1, and B = 1, and y = A cos ( Bx − C ) with A = 1, and B = 1, and

π

C = π

. Thus, the amplitude is A = 1 = 1 . The C = − . Thus, the amplitude is 2

A = 1 = 1 . The

2

period is 2π

= 2π

= 2π . The phase shift is

period is

2π =

2π B 1

= 2π . The phase shift is

B 1 π

π −C 2 π 2π π

C = 2 =

π . The quarter-period is

2π = π

. The = = − . The quarter-period is B 1 2

= . The 4 2

B 1 2 4 2 π

cycle begins at x = π

. Add quarter-periods to 2

generate x-values for the key points.

x = π 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π

cycle begins at x = − . Add quarter-periods to 2

generate x-values for the key points.

π = −

2

x π

+ π

= 0 2 2

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

2 2

x = 2π + π

= 5π x = π +

π =

3π 2 2

2 2 Evaluate the function at each value of x.

Evaluate the function at each value of x.

2

2

2

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Page 30: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

84 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 84

x coordinates

π

2

π 2

, 4

4

3π , 0

4

π (π , − 4)

5

4

5π , 0

4

2

3π , 4

2

x coordinates

π

2

π , 3

4

3π , 0

4

π (π , − 3)

4

4 , 0

2

3π , 3 2

Connect the five points with a smooth curve and

graph one complete cycle of the given function

Connect the five points with a smooth curve and

graph one complete cycle of the given function

45. The equation

y = 3cos(2 x − π ) is of the form 46. The equation y = 4 cos(2 x − π ) is of the form

y = A cos ( Bx − C ) with A = 3, and B = 2, and y = A cos(Bx − C ) with A = 4, and B = 2, and C = π .

C = π . Thus, the amplitude is

A = 3

= 3 . The Thus, the amplitude is A = 4 = 4 . The period is

2π =

2π = π . The phase shift is

C = π

. The

period is 2π

= 2π

= π . The phase shift is C

= π

. B 2 B 2 B 2 B 2 π π

The quarter-period is π

. The cycle begins at x = π

. quarter-period is . The cycle begins at x = . Add

4 24 2 quarter-periods to generate x-values for the key points.

Add quarter-periods to generate x-values for the key points. x =

π 2

x = π 2

x = π

+ π

= 3π

2 4 4

x = 3π

+ π

= π 4 4

x = π + π

= 5π

4 4

x = 5π

+ π

= 3π

4 4 2 Evaluate the function at each value of x.

x = π

+ π

= 3π

2 4 4

x = 3π

+ π

= π 4 4

x = π + π

= 5π

4 4

x = 5π

+ π

= 3π

4 4 2

Evaluate the function at each value of x.

2

π

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

85 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 85

x coordinates

− π 6

− π

, 1

6 2

0 (0, 0)

π

6

π , −

1 6 2

π

3

π , 0

3

π

2

π ,

1 2 2

= −

= −

2

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

47.

y = 1

cos 3x +

π =

1 cos

3x −

− π

2

2

2

2

Connect the five points with a smooth curve and

The equation

y = 1

cos 3x −

− π

is of the form graph one complete cycle of the given function

2

2

y = A cos(Bx − C ) with

C π

A = 1

, and B = 3, and 2

1 1= − . Thus, the amplitude is

2 A = = . The

2 2

period is 2π

= 2π

. The phase shift is

B 3 π

C =

− 2 = −

π ⋅ 1

= − π

. The quarter-period is 1 1B 3 2 3 6

48. y = cos(2 x + π ) = cos(2 x − (−π )) 2 2

3 = 2π

⋅ 1

= π

. The cycle begins at x π

. Add

The equation y = 1

cos(2x − (−π ))

is of the form4 3 4 6 6 2

quarter-periods to generate x-values for the key points.

y = A cos(Bx − C ) with

A = 1

, and B = 2, andπ

x = − 6 π π

x = − + = 0 6 6

C = −π . Thus, the amplitude is A = 1

= 1

. The 2 2

x = 0 + π

= π period is

2π =

2π = π . The phase shift is

6 6 B 2

x = π

+ π

= π C

= −π

= − π

. The quarter-period is π

. The cycle6 6 3

x = π

+ π

= π B 2 2 4

3 6 2

Evaluate the function at each value of x.

begins at x π

. Add quarter-periods to generate x- 2

values for the key points.

π x = −

2 π π π

x = − + = − 2 4 4 π π

x = − + = 0 4 4

x = 0 + π

= π

4 4

x = π

+ π

= π

4 4 2 Evaluate the function at each value of x.

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Page 32: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

86 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 86

x

coordinates

− π 2

− π

, 1

− π 4

− π

, 0

0 0, −

1

4

4

π

2

π , 1

x

coordinates

π

4

π , − 3 4

2

2

4

4 , 3

π

(π , 0)

4

5π , − 3 4

Evaluate the function at each value of x.

2 2

4

2

π π , 0

π π , 0

2 2

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

Connect the five points with a smooth curve and

graph one complete cycle of the given function

49. The equation

y = −3cos

2 x −

π is of the form

π

2 50. The equation y = −4 cos 2 x −

is of the form 2

y = A cos(Bx − C ) with A = –3, and y = A cos(Bx − C ) with A = –4, and B = 2, and

B = 2, and C = π

. Thus, the amplitude is 2

C = π

. Thus, the amplitude is 2

A = −4

= 4 . The

A = −3 = 3 . The period is 2π

= 2π

= π . The period is

2π =

2π = π . The phase shift is

B 2 B 2 π π

phase shift is C

= 2 = π

⋅ 1

= π

. C

= 2 = π

⋅ 1

= π

. The quarter-period is π

. TheB 2 2 2 4 B 2 2 2 4 4

The quarter-period is π

. The cycle begins at x = π

. cycle begins at x =

π . Add quarter-periods to

4 4 4

Add quarter-periods to generate x-values for the key

points.

x = π 4

x = π

+ π

= π

4 4 2

x = π

+ π

= 3π

2 4 4

generate x-values for the key points.

x = π 4

x = π

+ π

= π

4 4 2

x = π

+ π

= 3π

2 4 4

x = 3π

+ π

= π

x = 3π

+ π

= π 4 4 π 5π4 4

x = π + π

= 5π

4 4

x = π + = 4 4

Evaluate the function at each value of x.

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Page 33: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

87 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 87

x

coordinates

π

4

π , − 4

4

2

2

4

4

π

(π , 0)

4

5π , − 4

4

x coordinates

–4 (–4, 2)

15 −

4

15 − , 0 4

7 −

2

7 , − 2

2

− 13 4

13 , 0

4

–3 (–3, 2)

= − = −

= −

= −

= −

= −

π π

, 0

, 4

Connect the five key points with a smooth curve

and graph one complete cycle of the given function.

52.

Connect the five points with a smooth curve and

graph one complete cycle of the given function

y = 3 cos(2π x + 4π ) = 3 cos(2π x − (−4π ))51. y = 2 cos(2π x + 8π ) = 2 cos(2π x − (−8π ))

The equation y = 3cos(2π x − (−4π )) is of the formThe equation y = 2 cos(2π x − (−8π )) is of the form

y = A cos(Bx − C ) with A = 3, and B = 2π , andy = A cos(Bx − C ) with A = 2, B = 2π , and

C = −4π . Thus, the amplitude is A = 3 = 3 . TheC = −8π . Thus, the amplitude is A = 2 = 2 . The

2π 2π

period is 2π

= 2π

= 1 . The phase shift is period is = = 1 . The phase shift is

B 2πB 2π C −4π 1

C =

−8π = −4 . The quarter-period is

1 . The cycle

= B 2π

= −2 . The quarter-period is . The cycle 4

B 2π 4 begins at x = –2. Add quarter-periods to generate x-begins at x = –4. Add quarter-periods to generate x- values for the key points.

x = −4

x = −4 + 1

= − 15

values for the key points. x = −2

x = −2 + 1

= − 7

4 44 4

x 15

+ 1

= − 7

x 7

+ 1

= − 3

4 4 24 4 2

x 7

+ 1

= − 13

2 4 4

x 13

+ 1

= −3

x 3

+ 1

= − 5

2 4 4

x 5

+ 1

= −1 4 4

4 4

Evaluate the function at each value of x.

Evaluate the function at each value of x.

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Page 34: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

88 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 88

x coordinates

–2 (–2, 3)

− 7 4

7 , 0

4

− 3 2

3 , − 3

2

− 5 4

5π , 0

4

–1 (–1, 3)

x y = sin x − 2 coordinates

0 y = sin 0 − 2 = 0 − 2 = −2 (0, –2)

π

2 y = sin

π − 2 = 1 − 2 = −1

2

π , −1

π y = sin π − 2 = 0 − 2 = −2 (π , − 2)

2 y = sin

3π − 2 = −1 − 2 = −3

2

3π , − 3

2π y = sin 2π − 2 = 0 − 2 = −2 (2π , − 2)

x y = sin x + 2 coordinates

0 y = sin 0 + 2 = 0 + 2 = 2 (0, 2)

π

2 y = sin

π + 2 = 1 + 2 = 3

2

π , 3

2

π y = sin π + 2 = 0 + 2 = 2 (π , 2)

2 y = sin

3π + 2 = −1 + 2 = 1

2

3π , 1

2

2π y = sin 2π + 2 = 0 + 2 = 2 (2π , 2)

By connecting the points with a smooth curve we

obtain one period of the graph.

54. The graph of y = sin x − 2 is the graph of y = sin x

Connect the five key points with a smooth curve and

graph one complete cycle of the given function. shifted 2 units downward. The period for both

2π πfunctions is 2π . The quarter-period is

4 or .

2

The cycle begins at x = 0. Add quarter-periods to

generate x-values for the key points. x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

53. The graph of

y = sin x + 2 is the graph of

y = sin x x = π +

π =

3π 2 2

shifted up 2 units upward. The period for both

functions is 2π . The quarter-period is 2π

or π

.

x = 3π

+ π

= 2π 2 2

4 2 Evaluate the function at each value of x.

The cycle begins at x = 0. Add quarter-periods to

generate x-values for the key points.

x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π

4

2 2

Evaluate the function at each value of x.

2

By connecting the points with a smooth curve we

obtain one period of the graph.

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Page 35: 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctioSectins on 2.1 Graphs of Sine and Cosine Functions 3 2 3 y = 4sin 2 = 4( 1) = 4 3 ,

Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

89 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 89

x y = cos x − 3 coordinates

0 y = cos 0 − 3 = 1 − 3 = −2

(0, –2)

π

2 y = cos

π − 3

2 = 0 − 3 = −3

π , − 3

π y = cos π − 3 = −1 − 3 = −4

(π , − 4)

2

3π y = cos − 3

2 = 0 − 3 = −3

3π , − 3 2

2π y = cos 2π − 3 = 1 − 3 = −2

(2π , − 2)

x y = cos x + 3 coordinates

0 y = cos 0 + 3 = 1 + 3 = 4 (0, 4)

π

2 y = cos

π + 3 = 0 + 3 = 3

2

π , 3

π y = cos π + 3 = −1 + 3 = 2 (π , 2)

2

3π y = cos + 3 = 0 + 3 = 3

2

3π , 3 2

2π y = cos 2π + 3 = 1 + 3 = 4 (2π , 4)

55. The graph of y = cos x − 3 is the graph of y = cos x 56. The graph of y = cos x + 3 is the graph of y = cos x

shifted 3 units downward. The period for both

functions is 2π . The quarter-period is 2π

or π

.

shifted 3 units upward. The period for both functions

is 2π . The quarter-period is 2π

or π

. The cycle4 2 4 2

The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2

Evaluate the function at each value of x.

begins at x = 0. Add quarter-periods to generate x- values for the key points. x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2

Evaluate the function at each value of x.

2

2

By connecting the points with a smooth curve we

obtain one period of the graph.

By connecting the points with a smooth curve we obtain one period of the graph.

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

90 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 90

x 1 y = 2sin x + 1

2

coordinates

0 1 y = 2sin

⋅ 0

+ 1 2

= 2sin 0 + 1 = 2 ⋅ 0 + 1 = 0 + 1 = 1

(0, 1)

π 1 y = 2sin ⋅π + 1

2

= 2sin π

+ 1 2

= 2 ⋅1 + 1 = 2 + 1 = 3

(π , 3)

2π 1 y = 2sin ⋅ 2π + 1

2 = 2 sin π + 1 = 2 ⋅ 0 + 1 = 0 + 1 = 1

(2π , 1)

3π y = 2sin

1 ⋅ 3π

+ 1

2

= 2sin 3π

+ 1 2

= 2 ⋅ (−1) + 1 = −2 + 1 = −1

(3π , − 1)

4π 1 y = 2sin ⋅ 4π + 1

2 = 2sin 2π + 1 = 2 ⋅ 0 + 1 = 0 + 1 = 1

(4π , 1)

57. The graph of y = 2sin 1 x + 1 is the graph 2

58. The graph of y = 2 cos 1

x + 1 is the graph of

of y = 2sin 1 x shifted one unit upward. The 2

y = 2 cos 1

x

2

shifted one unit upward. The amplitudeamplitude for both functions is 2 = 2 . The period 2

for both functions is 2π

= 2π ⋅ 2 = 4π . The quarter- 1

for both functions is 2

2π = 2 . The period for both

2

period is 4π

= π . The cycle begins at x = 0. Add

functions is 1

= 2π ⋅ 2 = 4π . The quarter-period is 2

4 quarter-periods to generate x-values for the key points. x = 0 x = 0 + π = π x = π + π = 2π x = 2π + π = 3π x = 3π + π = 4π Evaluate the function at each value of x.

4π = π . The cycle begins at x = 0. Add quarter-

4 periods to generate x-values for the key points. x = 0 x = 0 + π = π x = π + π = 2π x = 2π + π = 3π x = 3π + π = 4π Evaluate the function at each value of x.

x 1

y = 2 cos 2

x + 1

coordinates

0 1

y = 2 cos ⋅ 0

+ 1

2 = 2 cos 0 + 1 = 2 ⋅1 + 1 = 2 + 1 = 3

(0, 3)

π 1 y = 2 cos

2 ⋅π + 1

= 2 cos π

+ 1 2

= 2 ⋅ 0 + 1 = 0 + 1 = 1

(π , 1)

2π 1 y = 2 cos

2 ⋅ 2π + 1

= 2 cos π + 1 = 2 ⋅ (−1) + 1 = −2 + 1 = −1

(2π , −1)

3π 1 y = 2 cos

2 ⋅ 3π + 1

= 2 ⋅ 0 + 1 = 0 + 1 = 1

(3π , 1)

4π 1 y = 2 cos

2 ⋅ 4π + 1

= 2 cos 2π + 1 = 2 ⋅1 + 1 = 2 + 1 = 3

(4π , 3)

By connecting the points with a smooth curve we

obtain one period of the graph.

By connecting the points with a smooth curve we

obtain one period of the graph.

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

91 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 91

x y = −3cos 2π x + 2 coordinates

0 y = −3cos(2π ⋅ 0) + 2 = −3cos 0 + 2 = −3 ⋅1 + 2 = −3 + 2 = −1

(0, –1)

1

4 y = −3cos

2π ⋅

1 + 2

= −3cos π

+ 2 2

= −3 ⋅ 0 + 2 = 0 + 2 = 2

1 , 2

1

2 y = −3cos

2π ⋅

1 + 2

= −3cos π + 2 = −3 ⋅ (−1) + 2 = 3 + 2 = 5

1 , 5

3

4 y = −3cos

2π ⋅

3 + 2

= −3cos 3π

+ 2 2

= −3 ⋅ 0 + 2 = 0 + 2 = 2

3 , 2

1 y = −3cos(2π ⋅1) + 2 = −3cos 2π + 2 = −3 ⋅1 + 2 = −3 + 2 = −1

(1, –1)

x y = −3sin 2π x + 2 coordinates

0 y = −3sin(2π ⋅ 0) + 2 = −3sin 0 + 2 = −3 ⋅ 0 + 2 = 0 + 2 = 2

(0, 2)

1

4

1 y = −3sin 2π ⋅

4 + 2

= −3sin π

+ 2 2

= −3 ⋅1 + 2 = −3 + 2 = −1

1 4

, − 1

1

2

1 y = −3sin 2π ⋅ + 2

2 = −3sin π + 2 = −3 ⋅ 0 + 2 = 0 + 2 = 2

1 , 2

2

3

4 y = −3sin 2π ⋅

3 + 2

4

= −3sin 3π

+ 2 2

= −3 ⋅ (−1) + 2 = 3 + 2 = 5

, 5

4

1 y = −3sin(2π ⋅1) + 2 = −3sin 2π + 2 = −3 ⋅ 0 + 2 = 0 + 2 = 2

(1, 2)

59. The graph of y = −3cos 2π x + 2 is the graph of 60. The graph of y = −3sin 2π x + 2 is the graph of

y = −3cos 2π x shifted 2 units upward. The y = −3sin 2π x shifted two units upward. The

amplitude for both functions is −3 = 3 . The period amplitude for both functions is A = −3 = 3 . The

for both functions is 2π

= 1 . The quarter-period is 2π

1 . The cycle begins at x = 0. Add quarter-periods to

4 generate x-values for the key points.

x = 0

x = 0 + 1

= 1

4 4

x = 1

+ 1

= 1

4 4 2

x = 1

+ 1

= 3

2 4 4

x = 3

+ 1

= 1 4 4

Evaluate the function at each value of x.

period for both functions is 2π

= 1 . The quarter- 2π

period is 1

. The cycle begins at x = 0. Add quarter– 4

periods to generate x-values for the key points. x = 0

x = 0 + 1

= 1

4 4

x = 1

+ 1

= 1

4 4 2

x = 1

+ 1

= 3

2 4 4

x = 3

+ 1

= 1 4 4

Evaluate the function at each value of x.

4

4

2

2

3

4 4

By connecting the points with a smooth curve we

obtain one period of the graph.

By connecting the points with a smooth curve we

obtain one period of the graph.

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

92 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 92

61. Select several values of x over the interval.

x

0 π

4

π

2

4 π

4

2

4 2π

y1 = 2 cos x 2 1.4 0 −1.4 −2 −1.4 0 1.4 2

y2 = sin x 0 0.7 1 0.7 0 −0.7 −1 −0.7 0

y = 2 cos x + sin x 2 2.1 1 −0.7 −2 −2.1 −1 0.7 2

62. Select several values of x over the interval.

x

0 π

4

π

2

4 π

4

2

4 2π

y1 = 3 cos x 3 2.1 0 −2.1 −3 −2.1 0 2.1 3

y2 = sin x 0 0.7 1 0.7 0 −0.7 −1 −0.7 0

y = 3 cos x + sin x 3 2.8 1 −1.4 −3 −2.8 −1 1.4 3

63. Select several values of x over the interval.

x

0 π

4

π

2

4 π

4

2

4 2π

y1 = sin x 0 0.7 1 0.7 0 −0.7 −1 −0.7 0

y2 = sin 2x 0 1 0 −1 0 1 0 −1 0

y = sin x + sin 2 x 0 1.7 1 −0.3 0 0.3 −1 −1.7 0

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

93 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 93

64. Select several values of x over the interval.

x

0 π

4

π

2

4 π

4

2

4 2π

y1 = cos x 1 0.7 0 −0.7 −1 −0.7 0 0.7 1

y2 = cos 2 x 1 0 −1 0 1 0 −1 0 1

y = cos x + cos 2x 2 0.7 −1 −0.7 0 −0.7 −1 0.7 2

65. Select several values of x over the interval.

x

0 π

4

π

2

4 π

4

2

4 2π

y1 = sin x 0 0.7 1 0.7 0 −0.7 −1 −0.7 0

y2 = cos 2 x 1 0 −1 0 1 0 −1 0 1

y = sin x + cos 2 x 1 0.7 0 0.7 1 −0.7 −2 −0.7 1

66. Select several values of x over the interval.

x

0 π

4

π

2

4 π

4

2

4 2π

y1 = cos x 1 0.7 0 −0.7 −1 −0.7 0 0.7 1

y2 = sin 2 x 0 1 0 −1 0 1 0 −1 0

y = cos x + sin 2x 1 1.7 0 −1.7 −1 0.3 0 −0.3 1

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

94 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 94

67. Select several values of x over the interval.

x

0 1

2

1 3

2

2 5

2

3 7

2

4

y1 = sin π x 0 1 0 −1 0 1 0 −1 0

π y2 = cos

2 x

1

0.7

0

−0.7

−1

−0.7

0

0.7

1

y = sin π x + cos π

x 2

1

1.7

0

−1.7

−1

0.3

0

−0.3

1

68. Select several values of x over the interval.

x

0 1

2

1 3

2

2 5

2

3 7

2

4

y1 = cos π x 1 0 −1 0 1 0 −1 0 1

π y2 = sin

2 x

0

0.7

1

0.7

0

−0.7

−1

−0.7

0

y = cos π x + sin π

x 2

1

0.7

0

0.7

1

−0.7

−2

−0.7

1

69. Using y = A cos Bx the amplitude is 3 and A = 3 , The period is 4π and thus

B = 2π

= 2π

= 1

period 4π 2

y = A cos Bx

y = 3cos 1

x 2

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

95 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 95

70. Using y = A sin Bx the amplitude is 3 and A = 3 , The 76.

period is 4π and thus

B = 2π

= 2π

= 1

period 4π 2

y = A sin Bx

y = 3sin 1

x 2

71. Using y = A sin Bx the amplitude is 2 and A = −2 ,

The period is π and thus

B = 2π

= 2π

= 2

77.

period π

y = A sin Bx y = −2sin 2 x

72. Using y = A cos Bx the amplitude is 2 and A = −2 ,

The period is 4π and thus

B = 2π

= 2π

= 2

78.

period π

y = A cos Bx y = −2 cos 2x

73. Using y = A sin Bx the amplitude is 2 and A = 2 , The

period is 4 and thus

B = 2π

= 2π

= π

period 4 2

y = A sin Bx

y = 2sin π

x

79.

2

74. Using y = A cos Bx the amplitude is 2 and A = 2 ,

The period is 4 and thus

B = 2π

= 2π

= π

period 4 2

y = A cos Bx

y = 2 cos π

x

80.

2

75.

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

96 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 96

81.

82.

92. The information gives the five key points of the graph.

(0, 23) corresponds to Noon,

(3, 38) corresponds to 3 P.M., (6, 53) corresponds to 6 P.M., (9, 38) corresponds to 9 P.M., (12, 23) corresponds to Midnight. By connecting the five key points with a smooth curve

we graph information from noon to midnight. Extend

the graph one cycle to the right to graph the information

for 0 ≤ x ≤ 24.

83. The period of the physical cycle is 33 days.

93. The function y = 3 sin 2π

( x − 79) + 12 is of the form 365

84. The period of the emotional cycle is 28 days. C 2π

85. The period of the intellectual cycle is 23 days. y = A sin B x −

B + D with A = 3 and B =

365 .

86. In the month of February, the physical cycle is at a

minimum on February 18. Thus, the author should

a. The amplitude is A = 3 = 3 .

not run in a marathon on February 18. b. The period is 2π =

2π = 2π ⋅

365 = 365 .

87. In the month of March, March 21 would be the best

day to meet an on-line friend for the first time,

because the emotional cycle is at a maximum.

88. In the month of February, the intellectual cycle is at a

maximum on February 11. Thus, the author should

begin writing the on February 11.

89. Answers may vary.

90. Answers may vary.

91. The information gives the five key point of the graph.

(0, 14) corresponds to June,

(3, 12) corresponds to September, (6, 10) corresponds to December, (9, 12) corresponds to March, (12, 14) corresponds to June By connecting the five key points with a smooth

curve we graph the information from June of one

year to June of the following year.

B 2π 2π 365

c. The longest day of the year will have the most

hours of daylight. This occurs when the sine

function equals 1.

y = 3 sin 2π

( x − 79) + 12 365

y = 3(1) + 12 y = 15

There will be 15 hours of daylight.

d. The shortest day of the year will have the least

hours of daylight. This occurs when the sine

function equals –1.

y = 3 sin 2π

( x − 79) + 12 365

y = 3(−1) + 12 y = 9

There will be 9 hours of daylight.

e. The amplitude is 3. The period is 365. The

phase shift is C

= 79 . The quarter-period is B

365 = 91.25 . The cycle begins at x = 79. Add

4 quarter-periods to find the x-values of the key points.

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

97 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 97

3

x = 79 x = 79 + 91.25 = 170.25 x = 170.25 + 91.25 = 261.5 x = 261.5 + 91.25 = 352.75 x = 352.75 + 91.25 = 444

Because we are graphing for 0 ≤ x ≤ 365 , we

will evaluate the function for the first four x-

values along with x = 0 and x = 365. Using a

calculator we have the following points.

(0, 9.1) (79, 12) (170.25, 15) (261.5, 12) (352.75, 9) (365, 9.1)

By connecting the points with a smooth curve

we obtain one period of the graph, starting on

January 1.

The highest average monthly temperature is 56° in

July.

95. Because the depth of the water ranges from a

minimum of 6 feet to a maximum of 12 feet, the

curve oscillates about the middle value, 9 feet. Thus,

D = 9. The maximum depth of the water is 3 feet

above 9 feet. Thus, A = 3. The graph shows that one

complete cycle occurs in 12-0, or 12 hours. The

period is 12. Thus,

12 = 2π

94. The function

y = 16 sin π

x − 2π

+ 40

is in the B 6 3

12B = 2π

form

y = A sin(Bx − C ) + D with A = 16,

B = π

, 6

B = 2π

= π

12 6

and C = 2π

. The amplitude is 3

A = 16

= 16 . The Substitute these values into y = A cos Bx + D . The

π x

period is 2π =

2π = 2π ⋅

6 = 12 . The phase shift is

depth of the water is modeled by y = 3cos

6 + 9 .

B π π 6

2πC 2π 6 12

= = ⋅ = 4 . The quarter-period is = 3 .

96. Because the depth of the water ranges from aB π 3 π 6

4 minimum of 3 feet to a maximum of 5 feet, the curve

The cycle begins at x = 4. Add quarter-periods to find the x-values for the key points.

x = 4 x = 4 + 3 = 7 x = 7 + 3 = 10 x = 10 + 3 = 13

oscillates about the middle value, 4 feet. Thus, D = 4.

The maximum depth of the water is 1 foot above 4

feet. Thus, A = 1. The graph shows that one complete

cycle occurs in 12–0, or 12 hours. The period is 12.

Thus,

2πx = 13 + 3 = 16 12 =

BBecause we are graphing for 1 ≤ x ≤ 12 , we will

evaluate the function for the three x-values between 1

and 12, along with x = 1 and x = 12. Using a

calculator we have the following points.

(1, 24) (4, 40) (7, 56) (10, 40) (12, 26.1) By connecting the points with a smooth curve we

12B = 2π

B = 2π

= π

12 6 Substitute these values into

y = A cos Bx + D . The

π xobtain the graph for 1 ≤ x ≤ 12 .

depth of the water is modeled by y = cos 6

+ 4 .

97. – 110. Answers may vary.

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

98 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 98

111. The function y = 3sin(2 x + π ) = 3 sin(2 x − (−π )) is of We choose −10 ≤ x ≤ 30 , and −1 ≤ y ≤ 1 for our

the form y = A sin(Bx − C ) with A = 3, B = 2, and graph.

C = −π . The amplitude is A = 3 = 3 . The

period is 2π

= 2π

= π . The cycle begins at

B 2

x = C

= −π

= − π

. We choose − π

≤ x ≤ 3π

, and

B 2 2 2 2 −4 ≤ y ≤ 4 for our graph.

114. The function y = 3sin(2 x − π ) + 5 is of the form

y = A cos( Bx − C ) + D with A = 3, B = 2, C = π , and

D = 5. The amplitude is A = 3 = 3 . The period is

2π =

2π = π . The cycle begins at x =

C = π

. Because

B 2 B 2 D = 5, the graph has a vertical shift 5 units upward. We

112. The function

y = −2 cos

2π x −

π is of the form

choose π ≤ x ≤

5π , and 0 ≤ y ≤ 10 for our graph.

2

2 2

y = A cos(Bx − C ) with A = –2, B = 2π , and C = π

. 2

The amplitude is A = −2 = 2 . The period is

2π =

2π = 1 . The cycle begins at

B 2π π

x = C

= 2 = π

⋅ 1

= 1

. We choose 1

≤ x ≤ 9

,B 2π 2 2π 4 4 4

and −3 ≤ y ≤ 3 for our graph.

115.

The graphs appear to be the same from − π

to π

.

113. The function

y = 0.2sin π

x + π = 0.2sin

π x − (−π )

is of the

2 2

116.

10

10

form

y = A sin(Bx − C ) with A = 0.2, B = π

, and 10

C = −π . The amplitude is A = 0.2 = 0.2 . The

period is 2π =

2π = 2π ⋅

10 = 20 . The cycle begins

B π π 10

π πat x =

C =

−π = −π ⋅

10 = −10 .

The graphs appear to be the same from − to . 2 2

B π π 10

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

99 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 99

117. 125. a. Since A = 3 and D = −2, the maximum will

occur at 3 − 2 = 1 and the minimum will occur

at −3 − 2 = −5 . Thus the range is [−5,1]

Viewing rectangle: − π

, 23π

, π

by [−5,1,1] 6 6 6

b. Since A = 1 and D = −2, the maximum will

The graph is similar to

y = sin x , except the occur at 1 − 2 = −1 and the minimum will occur

118.

amplitude is greater and the curve is less smooth. at −1 − 2 = −3 . Thus the range is [−3, −1]

Viewing rectangle: − π

, 7π

, π

by [−3, −1,1] 6 6 6

126. A = π

B = 2π

= 2π

= 2π period 1

C C = = −2

B 2π

The graph is very similar to

smooth.

119. a. see part c.

y = sin x , except not C = −4π

y = A cos( Bx − C ) y = π cos(2π x + 4π ) or y = π cos [2π ( x + 2)]

b. y = 22.61sin(0.50 x − 2.04) + 57.17

127.

y = sin 2 x = 1

− 1

cos 2 xc. graph for parts a and c: 2 2

120. Answers may vary.

121. makes sense

122. does not make sense; Explanations will vary. Sample

explanation: It may be easier to start at the highest

point.

123. makes sense

124. makes sense

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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

100 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 100

π

128. y = cos2

x = 1

+ 1

cos 2 x 2 2

132. 4π

lies in quadrant III. The reference angle is 3

θ ′ = 4π

− π = 4π

− 3π

= π

. 3 3 3 3

tan π

= 3 3

Because the tangent is positive in quadrant III,

tan 5π

= + tan π

= 3 4 4

133. − π

< x + π

< π

2 4 2

− π

− π

< x + π

− π

< π

− π

129. Answers may vary. 2 4

2 π 4 4 2 4

π− π

− < x < 2

− π

130. a. cos 47

0 sec 47

0 = cos 470 1

= 1 4 4 4 4

3π πcos 47

0

− < x < 4 4

b. sin 2 θ + cos2 θ = 1

x − 3π

< x < π

or −

3π , π

4 4

4 4

sin 2 π + cos2 π

= 1

5 5 3π π 2π π

131. Use the Pythagorean Theorem,

find b.

c2 = a2 + b2

, to

134.

− + − − 4 4 = 4 = 2 = −

2 2 2 4

a2 + b2 = c2

12 + b2 = 22

1 + b2 = 4

b2 = 3

135. a.

b = 3 = 3

Note that side a is opposite θ and side b is adjacent

to θ .

sin θ = opposite

= 1

hypotenuse 2

cosθ = adjacent

= 3

hypotenuse 2

tan θ = opposite

= 1

= 3

b. The reciprocal function is undefined.

adjacent 3 3

cscθ = hypotenuse

= 2

= 2 opposite 1

secθ = hypotenuse

= 2

= 2 3

adjacent 3 3

cot θ = adjacent

= 3

= 3 opposite 1

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101 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 101

Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

2

= −

= −

Section 2.2 y = tan

x −

π from 0 to π . Continue the pattern

Check Point Exercises

1. Solve the equations 2x π

and 2x = π

and extend the graph another full period to the right.

2 2

x= − π x =

π

4 4

Thus, two consecutive asymptotes occur at x π 4

and x = π

. Midway between these asymptotes is 4

x = 0. An x-intercept is 0 and the graph passesthrough (0, 0). Because the coefficient of the tangent

is 3, the points on the graph midway between an x-

intercept and the asymptotes have y-coordinates of

3. Solve the equations

π π–3 and 3. Use the two asymptotes, the x-intercept, and the points midway between to graph one period

x = 0 and 2

x = π 2

π

of y = 3tan 2x from − π 4

to π

. In order to graph for 4

x = 0 x = π 2

x = 2− π

< x < 3π

, Continue the pattern and extend the 4 4

graph another full period to the right.

Two consecutive asymptotes occur at x = 0 and x = 2.

Midway between x = 0 and x = 2 is x = 1. An x-

intercept is 1 and the graph passes through (1, 0).

Because the coefficient of the cotangent is 1

, the 2

points on the graph midway between an x-intercept

and the asymptotes have y-coordinates of − 1 2

and 1

. 2

Use the two consecutive asymptotes, x = 0 and x = 2,

to graph one full period of y = 1

cot π

x . The curve 2 2

2. Solve the equations is repeated along the x-axis one full period as shown.

x − π

= − π

and x − π

= π

2 2 2 2

x = π

− π x =

π + π

2 2 2 2

x = 0 x = π

Thus, two consecutive asymptotes occur at

x = 0 and x = π .

x-intercept = 0 + π

= π

2 2

An x-intercept is π

and the graph passes through 2

π , 0 . Because the coefficient of the tangent is 1,

2

the points on the graph midway between an x-

intercept and the asymptotes have y-coordinates of –1

and 1. Use the two consecutive asymptotes,

x = 0 and x = π , to graph one full period of

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102 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 102

Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

C

4. The x-intercepts of

y = sin x +

π correspond to

4.

− π

, 3π

;

− π

; 3π

4

vertical asymptotes of

y = csc x +

π .

4 4 4 4

4

5. 3sin 2 x

6. y = 2 cos π x

7. false

8. true

5. Graph the reciprocal cosine function,

y = 2 cos 2x .

Exercise Set 2.2

πThe equation is of the form y = A cos Bx with A = 2 1. The graph has an asymptote at x = − .

2and B = 2. C π πamplitude: A = 2 = 2 The phase shift, , from

B 2 to −

2 is −π units.

period: 2π

= 2π

= π Thus, C

= C

= −πB 2 B 1

Use quarter-periods, π

, to find x-values for the five 4

key points. Starting with x = 0, the x-values are

C = −π The function with C = −π

is y = tan( x + π ) .

2. The graph has an asymptote at x = 0 .

0, π

, π

, 3π

, and π . Evaluating the function at each 4 2 4

value of x, the key points are

The phase shift,

C , from

π B 2

to 0 is − π

units. Thus, 2

(0, 2), π

, 0 ,

π , −2

, 3π

, 0 , (π , 2) . In order to C C π

= = − 4

2

4

B 1 2 π

= − graph for −

3π ≤ x ≤

3π , Use the first four points

4 4 2

The function with C π

is y = tan x +

π .

and extend the graph − 3π

units to the left. Use the = −

2

2

4 graph to obtain the graph of the reciprocal function. Draw vertical asymptotes through the x-intercepts,

3. The graph has an asymptote at x = π .

πand use them as guides to graph y = 2sec 2x . π = + C

2

C = π 2

The function is

y = − tan x −

π . 2

4. The graph has an asymptote at π

. 2

There is no phase shift. Thus, C

= C

= 0

Concept and Vocabulary Check 2.2 B 1

1. − π

, π

;

− π

; π

The function with C = 0 is

C = 0 y = − tan x .

4 4

4 4

2. (0, π ) ; 0; π

3. (0, 2); 0; 2

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103 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 103

Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

= −

= −

= −

5. Solve the equations x π

and x = π Continue the pattern and extend the graph another

full period to the right.4 2 4 2

x = − π

4

x = π

4 2

2

x = −2π x = 2π

Thus, two consecutive asymptotes occur at x = −2π and x = 2π .

x-intercept = −2π + 2π

= 0

= 0 2 2

An x-intercept is 0 and the graph passes through (0,

0). Because the coefficient of the tangent is 3, the

7. Solve the equations 2 x

π and 2 x

π

points on the graph midway between an x-intercept

and the asymptotes have y-coordinates of –3 and 3.

= − = 2 2

− π π

Use the two consecutive asymptotes, x = −2π and x = 2 x = 2

x = 2π , to graph one full period of

y = 3tan x

2 2

from π π

−2π 4

to 2π . x = − x =

4 4

Continue the pattern and extend the graph another full period to the right.

Thus, two consecutive asymptotes occur at x π 4

and x = π

. 4

− π + π 0

x-intercept = 4 4 = = 0 2 2

An x-intercept is 0 and the graph passes through (0,

0). Because the coefficient of the tangent is 1

, the 2

points on the graph midway between an x-intercept

6. Solve the equations and the asymptotes have y-coordinates of − 1 and

1 .

x π and

x π 2 2= − =

4 2 4 2 Use the two consecutive asymptotes, x π

and

x = − π

4

x = π

4 4

2

2

1 x = π

, to graph one full period of y =

tan 2x fromx = −2π x = 2π 4 2

Thus, two consecutive asymptotes occur at x = −2π and x = 2π .

− π

to π

. Continue the pattern and extend the graph 4 4

x-intercept =

−2π + 2π =

0 = 0

2 2

another full period to the right.

An x-intercept is 0 and the graph passes through

(0, 0). Because the coefficient of the tangent is 2, the points on the graph midway between an x-intercept and the asymptotes have

y-coordinates of –2 and 2. Use the two consecutive

asymptotes, x = −2π and x = 2π ,

to graph one full period of

2π .

y = 2 tan x 4

from −2π to

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104 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 104

Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

= −

= −

8. Solve the equations graph another full period to the right.

π 2 x = − and 2 x =

π

2 2

− π π

x = 2 x = 2

2 2

x π

x π

= − = 4 4

Thus, two consecutive asymptotes occur at x π 4

10. Solve the equations

and x = π

. 1 π 1 π

4

− π + π

x = − 2 2

and x = 2 2

x-intercept = 4 4 = 0

= 0

x = − π

2

x = π

22 2

2

2

An x-intercept is 0 and the graph passes through (0, 0). Because the coefficient of the tangent is 2, the x = −π x = π

points on the graph midway between an x-intercept and the asymptotes have y-coordinates of –2 and 2.

Thus, two consecutive asymptotes occur at x = −π and x = π .

Use the two consecutive asymptotes, x π

and x-intercept =

−π + π =

0 = 0

x = π

, to graph one full period of 4

4

y = 2 tan 2x

from

2 2 An x-intercept is 0 and the graph passes through (0, 0). Because the coefficient of the tangent is –3, the

points on the graph midway between an x-intercept

− π

to π

. 4 4

and the asymptotes have y-coordinates of 3 and –3. Use the two consecutive

Continue the pattern and extend the graph another full period to the right.

asymptotes, x = −π and x = π , to graph one full

1period of y = −3 tan

2 x from −π to π . Continue the

9. Solve the equations

pattern and extend the graph another full period to the

right.

1 π x = −

and 1 x =

π

2 2 2 2

x = − π

2

x = π

2 2

2 11. Solve the equations

x = −π x = π π π

Thus, two consecutive asymptotes occur at x = −π x − π = − 2 and x − π =

2and x = π . π π

x = − + π x = + π

x-intercept = −π + π

= 0

= 0 2 2

x = π

2 2

x = 3π

An x-intercept is 0 and the graph passes through (0,

0). Because the coefficient of the tangent is –2, the points on the graph midway between an x-intercept and the asymptotes have y-coordinates of 2 and –2.

2 2

Thus, two consecutive asymptotes occur at x = π 2

3πUse the two consecutive asymptotes, x = −π and and x = .

2

x = π , to graph one full period of y = −2 tan

1 x

π + 3π 4π

from −π

2 to π . Continue the pattern and extend the

x-intercept = 2 2 = 2 = 4π

= π 2 2 4

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105 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 105

Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

= −

= −

An x-intercept is π and the graph passes through

(π , 0) . Because the coefficient of the tangent is 1,

the points on the graph midway between an x- intercept and the asymptotes have y-coordinates of –1

and 1. Use the two consecutive asymptotes, x = π

2

Continue the pattern and extend the graph another

full period to the right.

and x = 3π

, to graph one full period of 2

y = tan( x − π ) from π

to 3π

. Continue the pattern 2 2

and extend the graph another full period to the right.

13. There is no phase shift. Thus,

C =

C = 0

B 1 C = 0

Because the points on the graph midway between an x-intercept and the asymptotes have y-coordinates of

–1 and 1, A = –1. The function with C = 0 and A = –1 is y = − cot x .

12. Solve the equations 14. The graph has an asymptote at π

. The phase shift, 2

x − π

= − π

and x − π

= π

C π π4 2 4 2 , from 0 to is units.

x 2π π

x 2π π B 2 2= − + = +

4 4 4 4 C C π

x π

x 3π Thus, = =

= − = B 1 24 4

Thus, two consecutive asymptotes occur at x π

C = π 2

4 The function with C =

π is y = − cot x −

π .

and x = 3π

. 4

2 2

− π

+ 3π 2π

15. The graph has an asymptote at − π

. The phase shift,x-intercept = 4 4 = 4 =

π 2 2 4

C π π

2

C C π, from 0 to − is − units. Thus,

= = −

An x-intercept is π

and the graph passes through 4

π , 0 . Because the coefficient of the tangent is 1,

B 2 2 B 1 2 π

C = − 2

4

= − is .

The function with C π y = cot

x +

π

the points on the graph midway between an x- intercept and the asymptotes have y-coordinates of –1

and 1. Use the two consecutive asymptotes, x π

2

2

16. The graph has an asymptote at −π . The phase shift,

4

and x = 3π

, to graph one full period of

C , from 0 to −π

B is −π units.

4 Thus, C

= C

= −π

y = tan x −

π from 0 to π .

B 1

4 C = −π

The function with C = −π is y = cot( x + π ) .

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106 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 106

Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

π

17. Solve the equations x = 0 and x = π . Two

consecutive asymptotes occur at x = 0 and x = π .

x-intercept = 0 + π

= π

2 2

An x-intercept is π

and the graph passes through 2

π , 0 . Because the coefficient of the cotangent is

The curve is repeated along the x-axis one full period

as shown.

2

2, the points on the graph midway between an x- intercept and the asymptotes have y-coordinates of 2 and –2. Use the two consecutive asymptotes, x = 0and x = π , to graph one full period of y = 2 cot x .

19. Solve the equations 2 x = 0 and 2 x = π

The curve is repeated along the x-axis one full period x = 0

x = π 2

as shown. Two consecutive asymptotes occur at x = 0 and

x = π

. 2

0 + π π

x-intercept = 2 = 2 = 2 2 4

An x-intercept is π

and the graph passes through 4

π , 0 . Because the coefficient of the cotangent is

4

18. Solve the equations

1 , the points on the graph midway between an x-

x = 0 and x = π 2

1Two consecutive asymptotes occur at x = 0

and x = π .

x-intercept = 0 + π

= π

intercept and the asymptotes have y-coordinates of 2

and − 1

. Use the two consecutive asymptotes, x = 0 2

2 2

An x-intercept is π

and the graph passes through and x = π

, to graph one full period of 2

y = 1

cot 2x . 2

2

π , 0 . Because the coefficient of the cotangent is

The curve is repeated along the x-axis one full period

as shown. 2

1 , the points on the graph midway between an x-

2

intercept and the asymptotes have y-coordinates of 1 2

and − 1

. Use the two consecutive asymptotes, x = 0 2

and x = π , to graph one full period of y = 1

cot x . 2

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107 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 107

Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

π

= −

= −

20. Solve the equations 2 x = 0 and 2 x = π 22. Solve the equations π

x = 0 and π

x = πx = 0 x =

π 2

4 4 x = 0

x = π

π

Two consecutive asymptotes occur at x = 0 and x = π

. 2

0 + π π

x-intercept = 2 = 2 = 2 2 4

4 x = 4

Two consecutive asymptotes occur at x = 0 and

x = 4 .

0 + 4 4

An x-intercept is π

and the graph passes through x-intercept = = = 2 2 2

4

π , 0 . Because the coefficient of the cotangent is 2,

An x-intercept is 2 and the graph passes through

(2, 0 ) . Because the coefficient of the cotangent is – 4

the points on the graph midway between an x-intercept and the asymptotes have y-coordinates of 2 and –2. Use

the two consecutive asymptotes, x = 0 and x = π

, to 2

2, the points on the graph midway between an x- intercept and the asymptotes have y-coordinates of –2

and 2. Use the two consecutive asymptotes, x = 0

and x = 4 , to graph one full period of

graph one full period of

y = 2 cot 2x . The curve is y = −2 cot π

x . The curve is repeated along the x- 4

repeated along the x-axis one full period as shown. axis one full period as shown.

21. Solve the equations π x = 0 and

π x = π

2 2 x = 0

x = π

23. Solve the equations

π ππ x + = 0 and x + = π2 2 2

x = 2 x = 0 −

π x = π − π

Two consecutive asymptotes occur at x = 0 and x = 2. 2 2 π π

x-intercept = 0 + 2

= 2

= 1 2 2

x = − x = 2 2

An x-intercept is 1 and the graph passes through (1, 0).

Because the coefficient of the cotangent is –3, the

points on the graph midway between an x-intercept

and the asymptotes have y-coordinates of –3 and 3.

Use the two consecutive asymptotes, x = 0 and x = 2, to graph one full period of

y = −3cot π

x . The curve is repeated along the x-axis 2

one full period as shown.

Two consecutive asymptotes occur at x π

and 2

x = π

. 2

− π + π 0

x-intercept = 2 2 = = 0 2 2

An x-intercept is 0 and the graph passes through (0, 0). Because the coefficient of the cotangent is 3, the points on the graph midway between an x-intercept and the asymptotes have y-coordinates of 3 and –3.

Use the two consecutive asymptotes, x π

and

x = π

, to graph one full period of

2

y = 3cot x +

π .

2

2

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108 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 108

Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

= −

= −

= −

= −

The curve is repeated along the x-axis one full period 1 xas shown. 25. The x-intercepts of y = − sin

2 2 corresponds to

vertical asymptotes of y 1

csc x

. Draw the 2 2

vertical asymptotes, and use them as a guide to

sketch the graph of y 1

csc x

. 2 2

24. Solve the equations

x + π

= 0 and

x + π

= π4 4

x = 0 − π x = π −

π

4 4

x π

x 3π

= − = 4 4

26. The x-intercepts of y = 3sin 4x correspond to

Two consecutive asymptotes occur at x π

and vertical asymptotes of y = 3csc 4x . Draw the vertical

4 asymptotes, and use them as a guide to sketch the

x = 3π

. 4

− π + 3π

π

graph of y = 3csc 4x .

x-intercept = 4 4 = 4 = 2 2 4

An x-intercept is π

and the graph passes through 4

π , 0 . Because the coefficient of the cotangent is

4

3, the points on the graph midway between an x- intercept and the asymptotes have y-coordinates of 3

and –3. Use the two consecutive asymptotes, x π

27. The x-intercepts of

y = 1

cos 2π x corresponds to 2

4

and x = 3π

, to graph one full period of

vertical asymptotes of y =

1 sec 2π x . Draw the

24

y = 3cot x +

π . The curve is repeated along the x-

vertical asymptotes, and use them as a guide to

1 4

sketch the graph of y =

sec 2π x . 2

axis one full period as shown.

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109 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 109

Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

28. The x-intercepts of y = −3cos π

x correspond to 30. Graph the reciprocal sine function, y = 2sin x . The

vertical asymptotes of

2

y = −3sec π

x . Draw the

equation is of the form

B = 1 .

y = A sin Bx with A = 2 and

2 vertical asymptotes, and use them as a guide to

amplitude:

2π A = 2 = 2

sketch the graph of y = −3sec π

x . 2

period: = = 2π B 1

πUse quarter-periods, , to find x-values for the five

2key points. Starting with x = 0, the x-values are

0, π

, π , 3π

, and 2π . Evaluating the function at 2 2

each value of x, the key points are

(0, 0), π

, 2 , (π , 0),

3π , − 2

, and (2π , 0) .

2

2

29. Graph the reciprocal sine function,

y = 3sin x . The Use these key points to graph y = 2sin x from 0 to

equation is of the form

B = 1.

y = A sin Bx with A = 3 and 2π . Extend the graph one cycle to the right. Use the graph to obtain the graph of the reciprocal function.

Draw vertical asymptotes through the x-intercepts,amplitude: A = 3 = 3 and use them as guides to graph y = 2cscx.

period: 2π

= 2π

= 2π B 1

Use quarter-periods, π

, to find x-values for the 2

five key points. Starting with x = 0, the x-values are

0, π

, 2 π ,

3π , and 2π . Evaluating the function at

2each value of x, the key points are (0, 0),

π

, 3 , (π , 0),

3π , − 3

, and (2π , 0) . Use

31. Graph the reciprocal sine function, y = 1

sin x

. The

2

2

2 2

these key points to graph y = 3sin x from 0 to 2π .

equation is of the form y = A sin Bx with A = 1

andExtend the graph one cycle to the right.

Use the graph to obtain the graph of the reciprocal

function. Draw vertical asymptotes through the x-

intercepts, and use them as guides to graph

2

B = 1

. 2

1 1

y = 3csc x . amplitude: A = = 2 2

period: 2π

= 2π

= 2π ⋅ 2 = 4π B 1

2

Use quarter-periods, π , to find x-values for the five

key points. Starting with x = 0, the x-values are 0,

π , 2π , 3π , and 4π . Evaluating the function at each

value of x, the key points are (0, 0),

π ,

1 , (2π , 0),

3π , −

1 , and (4π , 0) . Use these

2

2

key points to graph y = 1

sin x

2 2

from 0 to 4π .

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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

110 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 110

Extend the graph one cycle to the right. Use the graph 33. Graph the reciprocal cosine function, y = 2 cos x .

to obtain the graph of the reciprocal function. Draw vertical asymptotes through the x-intercepts, and use

The equation is of the form

and B = 1.

y = A cos Bx with A = 2

them as guides to graph y = 1

csc x

. 2 2

amplitude: A = 2 = 2

period: 2π

= 2π

= 2π B 1

Use quarter-periods, π

, to find x-values for the five 2

key points. Starting with x = 0, the x-values are 0,

π , π ,

3π , 2π . Evaluating the function at each value

2 2

of x, the key points are (0, 2), π

, 0 , (π , − 2),

2

32. Graph the reciprocal sine function,

y = 3

sin x

. The

3π 2 4

, 0 , and (2π , 2) . Use these key points to 2

equation is of the form y = A sin Bx with A = 3 and graph y = 2 cos x from 0 to 2π . Extend the graph

B = 1

. 4

2 one cycle to the right. Use the graph to obtain the

graph of the reciprocal function. Draw vertical asymptotes through the x-intercepts, and use them as

amplitude: A = 3

= 3 guides to graph y = 2sec x .

2 2

period: 2π

= 2π

= 2π ⋅ 4 = 8π B 1

4

Use quarter-periods, 2π , to find x-values for the five

key points. Starting with x = 0, the x-values are

0, 2π , 4π , 6π , and 8π . Evaluating the function at

each value of x, the key points are

(0, 0), 2π ,

3 , (4π , 0),

6π , −

3 , and (8π , 0) .

34. Graph the reciprocal cosine function,

y = 3cos x .

2

2 The equation is of the form y = A cos Bx with A = 3

Use these key points to graph y = 3

sin x

2 4

from 0 to and B = 1 .

amplitude:

A = 3 = 3

8π . Extend the graph one cycle to the right.

period: 2π

= 2π

= 2πUse the graph to obtain the graph of the reciprocal B 1

function. Draw vertical asymptotes through the x-

intercepts, and use them as guides to graph

y = 3

csc x

. 2 4

Use quarter-periods, π

, to find x-values for the five 2

key points. Starting with x = 0, the x-values are

0, π

, π , 3π

, and 2π . Evaluating the function at 2 2

each value of x, the key points are

(0, 3), π

, 0 , (π , − 3), 3π

, 0 , (2π , 3) .

2

2

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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

111 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 111

Use these key points to graph y = 3cos x from 0 to 36. Graph the reciprocal cosine function, y = cos

x . The

2π . Extend the graph one cycle to the right. Use the graph to obtain the graph of the reciprocal function. Draw vertical asymptotes through the x-intercepts,

equation is of the form

1

y = A cos Bx

with

2 A = 1 and

and use them as guides to graph y = 3sec x . B = . 2

amplitude:

A = 1 = 1

period: 2π

= 2π

= 2π ⋅ 2 = 4π B 1

2

35. Graph the reciprocal cosine function,

y = cos x

. The 3

Use quarter-periods, π , to find x-values for the five

key points. Starting with x = 0, the x-values are

0, π , 2π , 3π , and 4π . Evaluating the function at

each value of x, the key points are

(0, 1), (π , 0 ) , (2π , − 1), (3π , 0) , and (4π , 1) .

equation is of the form y = A cos Bx with A = 1 and

Use these key points to graph y = cos x

2

from 0 to

B = 1

. 3

amplitude:

A = 1 = 1

4π . Extend the graph one cycle to the right. Use the

graph to obtain the graph of the reciprocal function.

Draw vertical asymptotes through the x-intercepts,

period: 2π

= 2π

= 2π ⋅ 3 = 6π

and use them as guides to graph y = sec x

.B 1 2

3

Use quarter-periods, 6π

= 3π

, to find x-values for 4 2

the five key points. Starting with x = 0, the x-values

are 0, 3π

, 3π , 9π

, and 6π . Evaluating the function 2 2

at each value of x, the key points are (0, 1), 3π

, 0 ,

2

(3π , −1),

9π , 0 , and (6π , 1) . Use these key

2 37. Graph the reciprocal sine function, y = −2sin π x .

The equation is of the form

y = A sin Bx with A = –2points to graph y = cos

x

3 from 0 to 6π . Extend the

and B = π .

graph one cycle to the right. Use the graph to obtain the graph of the reciprocal function. Draw vertical

amplitude:

2π A = −2 = 2

2πasymptotes through the x-intercepts, and use them as

period: = = 2 B π

guides to graph y = sec x

. 3

Use quarter-periods, 2

= 1

, to find 4 2

x-values for the five key points. Starting with x = 0,

the x-values are 0, 1

, 1, 3

, and 2. Evaluating the 2 2

function at each value of x, the key points are (0, 0),

1 , − 2

, (1, 0),

3 , 2

, and (2, 0) . Use these key

2

2

points to graph

y = −2sin π x from 0 to 2. Extend the

graph one cycle to the right. Use the graph to obtain

the graph of the reciprocal function.

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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

112 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 112

= −

= −

= −

2 2

2 2

= −

= −

= −

= −

Draw vertical asymptotes through the x-intercepts, 1and use them as guides to graph y = −2 csc π x . 39. Graph the reciprocal cosine function, y = − cos π x .

2The equation is of the form y = A cos Bx with

A 1

and B = π . 2

amplitude: A = − 1

= 1

2 2

period: 2π

= 2π

= 2

B π

38. Graph the reciprocal sine function, y 1

sin π x . 2

Use quarter-periods, 2

= 1

, to find x-values for the 4 2

The equation is of the form y = A sin Bx with five key points. Starting with x = 0, the x-values are

1 3

A 1

and B = π . 0, , 1, , and 2. Evaluating the function at each

2 22

1

amplitude:

A = − 1

= 1 value of x, the key points are 0, −

2 ,

2 2

3

1

1 , 0 , 1,

1 ,

, 0 , 2, − . Use these key

period: 2π

= 2π

= 2

2

2

2 2

B π

1

Use quarter-periods, 2

= 1

, to find x-values for the 4 2

points to graph y = − cos π x from 0 to 2. Extend 2

five key points. Starting with x = 0, the x-values are

0, 1

, 1, 3

, and 2 . Evaluating the function at each 2 2

value of x, the key points are

the graph one cycle to the right. Use the graph to obtain the graph of the reciprocal function. Draw vertical asymptotes through the x-intercepts, and use them as guides to graph

1

(0, 0), 1

, − 1

, (1, 0), 3

, 1

, and (2, 0) .

y = − sec π x . 2

Use these key points to graph y 1

sin π x from 0 2

to 2 . Extend the graph one cycle to the right. Use the

graph to obtain the graph of the reciprocal function.

Draw vertical asymptotes through the x-intercepts,

and use them as guides to graph y 1

csc π x . 2

40. Graph the reciprocal cosine function,

y 3

cos π x . 2

The equation is of the form y = A cos Bx with

A 3

and B = π . 2

amplitude: A = − 3

= 3

2 2

period: 2π

= 2π

= 2

B π

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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

113 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 113

= −

= −

.

Use quarter-periods, 2

= 1

, to find x-values for the Draw vertical asymptotes through the x-intercepts,

4 2 five key points. Starting with x = 0, the x-values are

0, 1

, 1, 3

, and 2 . Evaluating the function at each 2 2

value of x, the key points are

0, −

3 ,

1 , 0

, 1,

3 ,

3 , 0

,

2, −

3 .

and use them as guides to graph y = csc( x − π ) .

2

2

2

2

2

Use these key points to graph y 3

cos π x from 0 2

to 2 . Extend the graph one cycle to the right. Use the 42. Graph the reciprocal sine function,

y = sin

x −

π .

graph to obtain the graph of the reciprocal function.

2

Draw vertical asymptotes through the x-intercepts, The equation is of the form

y = Asin(Bx − C) with

and use them as guides to graph y 3

sec π x . 2

π A = 1 , B = 1 , and C = .

2

amplitude: A = 1 = 1

period: 2π

= 2π

= 2π B 1

π

phase shift: C

= 2 = π

B 1 2

41. Graph the reciprocal sine function,

y = sin(x − π ) .

Use quarter-periods, π

, to find x-values for the five 2

key points. Starting with x = π

, the x-values areThe equation is of the form

= 1, and B = 1, and C = π .

y = Asin(Bx − C) with A 2

π , π ,

3π , 2π , and

5π . Evaluating the function at

amplitude: A = 1 = 1 2 2 2 each value of x, the key points are

period: 2π

= 2π

= 2π

π 3π

5π B 1

, 0 , (π , 1) , , 0 , ( 2π , − 1) , and , 0 . 2

2

2

phase shift: C

= π

= π

B 1 Use these key points to graph

y = sin x −

π from

2 Use quarter-periods,

2π = π

, to find 4 2

π to

5π . Extend the graph one cycle to the right.

x-values for the five key points. Starting with x = π , 2 2

the x-values are π , 3π

, 2π , 2

5π , and 3π .

2

Use the graph to obtain the graph of the reciprocal function. Draw vertical asymptotes through the x-

Evaluating the function at each value of x, the key intercepts, and use them as guides to graph

points are (π , 0), 3π

, 1 , (2π , 0),

y = csc x −

π

2

2

5π , − 1

, (3π , 0) . Use these key points to graph

2

y = sin(x − π ) from π to 3π . Extend the graph one

cycle to the right. Use the graph to obtain the graph

of the reciprocal function.

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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

114 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 114

2

− π

C

= −

43. Graph the reciprocal cosine function,

y = 2 cos( x + π ) . The equation is of the form

y = A cos(Bx + C ) with A = 2, B = 1, and C = −π .

Evaluating the function at each value of x, the key

points are

− π

, 2 , (0, 0 ) ,

π , − 2

, (π , 0 ) ,

3π , 2

.

amplitude: A = 2 = 2 2 2 2

π

period: 2π

= 2π

= 2π Use these key points to graph y = 2 cos x +

2 from

B 1

− π

to 3π

. Extend the graph one cycle to the right.

phase shift: C

= −π

= −π

B 1 2 2

Use quarter-periods, 2π

= π

, to find x-values for the 4 2

Use the graph to obtain the graph of the reciprocal function. Draw vertical asymptotes through the x- intercepts, and use them as guides to graph

five key points. Starting with x = −π , the x-values π .

are −π , − π

, 2

0, π

, and π . Evaluating the function 2

y = 2sec x +

at each value of x, the key points are (−π , 2),

, 0 , (0, − 2 ) , π

, 0 , and (π , 2) . Use these

2

2

key points to graph

y = 2 cos( x + π ) from −π

to π .

Extend the graph one cycle to the right. Use the graph

to obtain the graph of the reciprocal function. Draw

vertical asymptotes through the x-intercepts, and use

them as guides to graph y = 2 sec( x + π ) . 45.

44. Graph the reciprocal cosine function,

y = 2 cos x +

π . The equation is of the form

46.

2

y = A cos(Bx + C ) with

π = − .

2

A = 2

and B = 1 , and

amplitude: A = 2 = 2

period: 2π

= 2π

= 2π B 1

π

phase shift: C

= −

2 = − π

B 1 2

Use quarter-periods, π

, to find x-values for the five 2

key points. Starting with x π

, the x-values are 2

− π

, 0, π

, π , and 3π

. 2 2 2

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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

115 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 115

2

, −

, −

,

47.

52.

53.

y = ( f h ) ( x) = f (h( x)) = 2 sec 2 x −

π

2

48.

π

49.

50.

54. y = ( g h ) ( x) = g (h( x)) = −2 tan 2 x −

55. Use a graphing utility with y1 = tan x and y2 = −1 .

For the window use Xmin = −2π , Xmax = 2π ,

Ymin = −2 , and Ymax = 2 .

5π π x = − ,

3π ,

4 4 4 4

51. x ≈ −3.93, − 0.79, 2.36, 5.50

56. Use a graphing utility with y1 = 1/ tan x and

y2 = −1 .

For the window use Xmin = −2π , Xmax = 2π ,

Ymin = −2 , and Ymax = 2 .

5π π x = − ,

3π ,

4 4 4 4x ≈ −3.93, − 0.79, 2.36, 5.50

57. Use a graphing utility with y1 = 1/ sin x and y2 = 1 .

For the window use Xmin = −2π , Xmax = 2π ,

Ymin = −2 , and Ymax = 2 .

3π π x = −

2 2 x ≈ −4.71, 1.57

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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

116 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 116

0

= −

= −

58. Use a graphing utility with y1 = 1/ cos x and y2 = 1 . Two consecutive asymptotes occur at x = 0 and

For the window use Xmin = −2π , Xmax = 2π , Ymin = −2 , and Ymax = 2 .

x = −2π , 0, 2π

x = π . Midway between x = 0 and x = π is x = π

. 2

59.

x ≈ −6.28, 0, 6.28

d = 12 tan 2π t

An x-intercept is π

and the graph passes through 2

π , 0 . Because the coefficient of the cotangent is

a. Solve the equations

2

2π t π

and 2π t = π

2, the points on the graph midway between an x-

2 2 − π π

intercept and the asymptotes have y-coordinates of –2 and 2. Use the two consecutive asymptotes, x = 0

t = 2

2π t = 2

2π and x = π , to graph y = 2 cot x for 0 < x < π .1

t = − t = 1

4 4

Thus, two consecutive asymptotes occur at

1 x = −

4 and x =

1 .

4 1 1− +

x-intercept = 4 4 = = 0 2 2

An x-intercept is 0 and the graph passes through (0, 0). Because the coefficient of the tangent is 12, the points on the graph midway between an

x-intercept and the asymptotes have y-

coordinates of –12 and 12. Use the two

61. Use the function that relates the acute angle with the

hypotenuse and the adjacent leg, the secant function.

sec x = d

10 d = 10 sec x

consecutive asymptotes, x 1 and x =

1 , to Graph the reciprocal cosine function, y = 10 cos x .

4 4

graph one full period of d = 12 tan 2π t . To

The equation is of the form

A = 10 and B = 1.

y = A cos Bx with

graph on [0, 2], continue the pattern and extend the graph to 2. (Do not use the left hand side of

amplitude: A = 10 = 10

the first period of the graph on [0, 2].) period: 2π

= 2π

= 2πB 1

For − π

< x < π

, use the x-values − π

, 0, and π

to2 2 2

2

π

find the key points − π

, 0 , (0, 10), and , 0 . 2

2

Connect these points with a smooth curve, then draw

vertical asymptotes through the x-intercepts, and use

them as guides to graph d = 10 sec x on − π

, π

.

2 2

b. The function is undefined for t = 0.25, 0.75,

1.25, and 1.75.

The beam is shining parallel to the wall at these times.

60. In a right triangle the angle of elevation is one of the

acute angles, the adjacent leg is the distance d, and

the opposite leg is 2 mi. Use the cotangent function.

cot x = d 2

d = 2 cot x

Use the equations x = 0 and x = π .

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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

117 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 117

62. Graphs will vary.

78. period: π

= π

B 4

Graph y = tan 4x for 0 ≤ x ≤ π

. 2

63.

79. period: π

= π

B 2Graph y = cot 2x for 0 ≤ x ≤ π .

64. Graphs will vary.

80. period: π

= π

= π ⋅ 2 = 2πB 1

2

y = x

for 0 ≤ x ≤ 4π .65. – 76. Answers may vary.

77. period: π

= π

= π ⋅ 4 = 4π

Graph cot 2

B 1 4

Graph y = tan x 4

for 0 ≤ x ≤ 8π .

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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

118 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 118

−1

2

81. period: π

= π

= 1 84. period:

2π =

2π = 2π ⋅

3 = 6

B π B π π 3

Graph y = 1

tan π x for 0 ≤ x ≤ 2 . 2

Graph the functions for 0 ≤ x ≤ 12 .

82. Solve the equations 85. period:

2π =

2π = π

π x + 1 = − π

and π x + 1 = π B 2

2 2 C π

π π

π x = − π x = π

− 1 phase shift: = 6 =

B 2 122 2

−π π

Thus, we include π

≤ x ≤ 25π

in our graph, and− 1 x = 2

− 1 x = 2 12 12

π π graph for 0 ≤ x ≤

5π .

x = −π − 2

x ≈ −0.82

period: π

= π

= 1

x = π − 2 2

2π x ≈ 0.18

B π

Thus, we include −0.82 ≤ x ≤ 1.18 in our graph of

y = 1

tan(π x + 1) , and graph for −0.85 ≤ x ≤ 1.2 . 2

86. period: 2π

= 2π

= 2B π

π

phase shift: C

= 6 = π

⋅ 1

= 1

B π 6 π 6

Thus, we include 1

≤ x ≤ 25

in our graph, and graph 6 6

83. period: 2π

= 2π

= 2π ⋅ 2 = 4π for 0 ≤ x ≤ 9

.B 1

2

Graph the functions for 0 ≤ x ≤ 8π .

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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

119 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 119

2 2

87. 1 and –1. Thus, A = 1. There is no phase shift. Thus,

C = 0. An equation for this graph is y = cot 3

x . 2

94. The graph has the shape of a secant function.

The reciprocal function has amplitude

period is 8π

. Thus, 2π

= 8π

A = 1 . The

3 B 3The graph shows that carbon dioxide concentration

rises and falls each year, but over all the

concentration increased from 1990 to 2008.

8π B = 6π

B = 6π

= 3

8π 4

There is no phase shift. Thus, C = 0 . An equation for

88. y = sin 1 x

the reciprocal function is

equation for this graph is

y = cos 3

x . Thus, an 4

y = sec 3

x . 4

95. The range shows that A = 2.

Since the period is 3π , the coefficient of x is given

by B where 2π

= 3π B

The graph is oscillating between 1 and –1.

The oscillation is faster as x gets closer to 0.

Explanations may vary.

89. makes sense

2π = 3π

B 3Bπ = 2π

B = 2

3

2 x

90. makes sense Thus, y = 2 csc

3

91. does not make sense; Explanations will vary.

96. The range shows that

A = π .

Sample explanation: To obtain a cosecant graph,

you can use a sine graph.

92. does not make sense; Explanations will vary.

Sample explanation: To model a cyclical temperature, use sine or cosine.

93. The graph has the shape of a cotangent function with

Since the period is 2, the coefficient of x is given by

B where 2π

= 2 B

2π = 2

B 2B = 2π

B = π

consecutive asymptotes at Thus, y = π cscπ x

x = 0 and x = 2π

. The period is 2π

− 0 = 2π

. Thus,

97. a. Since A=1, the range is ( ] [ )3 3 3 −∞, −1 1, ∞

π =

2π Viewing rectangle: −

π ,π ,

7π by [−3, 3,1]

B 3

6 6

2π B = 3π

B = 3π

= 3

b. Since A=3, the range is (−∞, −3] [3, ∞ )2π 2

1 7 The points on the graph midway between an

x-intercept and the asymptotes have y-coordinates of

Viewing rectangle: − , ,1 by [−6, 6,1]

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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

120 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 120

o

98. y = 2− x sin x

2− x

decreases the amplitude as x gets larger.

Examples may vary.

102. a.

99. The formula s = rθ can only be used when θ is

expressed in radians. Thus, we begin by converting

150° to radians. Multiply by π radians

. 180°

150° = 150° ⋅ π radians

= 150

π

radians

b. yes; Explanations will vary.180°

= 5 π radians

6

180

c. The angle is − π

. 6

Now we can use the formula s = rθ to find the This is represented by the point

− π

, − 1

.length of the arc. The circle’s radius is 8 inches : r = 8 inches. The measure of the central angle in

radians is 5 π : θ =

5 π . The length of the arc

6 6

intercepted by this central angle is s = rθ

= (8 inches) 5 π

6 2

103. a.

6

= 20π 3

inches

b. yes; Explanations will vary.

≈ 20.94 inches.

100. 25π

− 2π ⋅ 4 = 25π

− 8π

c. The angle is 5π

. 6

3

101.

3 3

25π 24π π = − =

3 3 3

tan 35o

= a 120

a = 120 tan 35 a ≈ 120(0.7002) ≈ 84 m

104. a.

This is represented by the point 6

, − 2

.

b. yes; Explanations will vary.

c. The angle is − π

. 3

This is represented by the point − π

, −

3 . 3

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121 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 121

Mid-Chapter 2 Check Point Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

x y 4sin 2x

coordinates

0

4sin(2 0) 4 0

0

(0, 0)

π

4

π 4sin

2

4 4 1

4

π ,4

π

2

4 sin 2 π

4 0

0

π ,0

4

4 sin 2

3π 4(1)

4

, 4

4

π 4sin(2 π ) 4 0

0

(π , 0)

x

1 π y

2 cos

3 x

coordinates

0

1 cos

π 0

1

1 1

2

0, 1

2

3

2

cos 0 0 2 3

3

2 , 0

3

1 cos

π 3

1

1 1

2

3,

1

9

2

1 cos

π 9

1

0 0 2

9 , 0

6

1 π

1

1

2 cos

3 6

2 1

2

6, 1

2

Mid-Chapter 2 Check Point 2. The equation y

1 cos π

x is of the form

1. The equation y 4sin 2x is of the form y Asin Bx 2 3

1 πwith A = 4 and B 2 . Thus, the amplitude is y A cos Bx with A and B . The amplitude

A 4 4 . The period is 2π

π . The

2 3

1 1 2π 2πB 2 is A . The period is

2 2 6 . The B π

quarter-period is π

. The cycle begins at x = 0. Add 4

quarter-periods to generate x-values for the key

points and evaluate the function at each value of x.

3

quarter-period is 6

3

. Add quarter-periods to 4 2

generate x-values for the key points and evaluate the function at each value of x.

4 1 π 3 1

2 2

2

4 2

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

Connect the five points with a smooth curve and

graph one complete cycle of the given function.

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122 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 122

Mid-Chapter 2 Check Point Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

x

y 3sin( x π ) coordinates

π

3sin π π 3sin(0)

3 0

0

π , 0

2

3sin 3π

π 3sin

π

3 1

3

, 3

3sin 2π π 3sin(π )

3 0

0

(2π , 0)

2

3sin 5π

π 3sin

3(1)

3

5π 2

, 3

3sin 3π π 3sin(2π )

3 0

0

3π , 0

x

coordinates

π

8

π , 2

8

8 , 0

8

5π , 2

8

8

8 , 0

8

8 , 2

3. The equation y 3sin( x π ) is of the form π

y Asin( Bx C) with A = 3, B = 1, and C π . The 4. The equation y 2cos

2 x

4 is of the form

amplitude is A 3 3 . The period is y Acos( Bx C) with A = 2, and B = 2, and

2π 2π

2π . The phase shift is C

π

π . The C π

. Thus, the amplitude is

A 2

2 . TheB 1 B 1 4

quarter-period is 2π

π

. The cycle begins at x π . period is 2π

π . The phase shift is4 2

Add quarter-periods to generate x-values for the key

points and evaluate the function at each value of x.

B 2

C π

π 4 .

B 2 8

The quarter-period is π

. The cycle begins at x π

.

4 4 Add quarter-periods to generate x-values for the key points and evaluate the function at each value of x.

2 2 8

2

2 2

Connect the five points with a smooth curve and

graph one complete cycle of the given function.

Connect the five points with a smooth curve and graph one complete cycle of the given function

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123 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 123

Mid-Chapter 2 Check Point Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

x y cos 2 x 1 coordinates

0 y cos 2 0 1 2 (0, 2)

π

4 y cos

2 π

1 1 π

,1

π

2 y cos

2 π

1 0 π

,0

4 y cos

2

3π 1 1

3π ,1

π y cos 2 π 1 2 (π , 2)

5. The graph of y cos 2 x 1 is the graph of y cos 2 x shifted one unit upward. The amplitude for both functions is

1 1 . The period for both functions is 2π

π . The quarter-period is π

. The cycle begins at x = 0. Add quarter-

2 4

periods to generate x-values for the key points and evaluate the function at each value of x.

4 4

2 2

4 4

By connecting the points with a smooth curve we obtain one period of the graph.

6. Select several values of x over the interval.

x

0 π

4

π

2

4 π

4

2

4 2π

y1 2 sin x 0 1.4 2 1.4 0 1.4 2 1.4 0

y2 2 cos x 2 1.4 0 1.4 2 1.4 0 1.4 2

y 2 sin x 2 cos x 2 2.8 2 0 2 2.8 2 0 2

7. Solve the equations

π π x

and π

x π

4 2 4 2

x π 4

x π 4

x 2

2 π 2 π

x 2

Thus, two consecutive asymptotes occur at x 2 and x 2 .

x-intercept = 2 2

0

0 2 2

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124 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 124

Mid-Chapter 2 Check Point Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

An x-intercept is 0 and the graph passes through (0, 0).

Because the coefficient of the tangent is 2, the points

Use quarter-periods, 2 1

, to find x-values for the 4 2

on the graph midway between an x-intercept and the asymptotes have y-coordinates of –2 and 2. Use the two consecutive asymptotes, x 2 and x 2 , to

five key points. Starting with x = 0, the x-values are

0, 1

, 1, 3

, and 2 . Evaluating the function at each 2 2

graph one full period of y 2 tan π

x 4

from 2 to 2 . value of x, the key points are

Continue the pattern and extend the graph another full 0, 2 , 1

, 0 , 1, 2 ,

3 , 0

, 2, 2 .

period to the right. 2 2

Use these key points to graph y 2 cos π x from 0

to 2 . Extend the graph one cycle to the right. Use the

graph to obtain the graph of the reciprocal function.

Draw vertical asymptotes through the x-intercepts,

8. Solve the equations 2 x 0 and 2 x π

and use them as guides to graph y 2sec π x .

x 0 x

π 2

Two consecutive asymptotes occur at x 0 and x π

. 2

0 π π π x-intercept = 2 2

10. Graph the reciprocal sine function,

y 3sin 2π x . The2 2 4

An x-intercept is π

and the graph passes through equation is of the form

B 2π .

y Asin Bx with A 3 and

4

π , 0 . Because the coefficient of the cotangent is 4,

amplitude: A 3 3

4 period: 2π

1

the points on the graph midway between an x-intercept and the asymptotes have y-coordinates of 4 and –4. Use

the two consecutive asymptotes, x 0 and x π

, to 2

B 2π

Use quarter-periods, 1

, to find x-values for the five key 4

points. Starting with x = 0, the x-values are 0, 1

, 1

, 3

, andgraph one full period of y 4 cot 2 x . The curve is 4 2 4

repeated along the x-axis one full period as shown. 1. Evaluating the function at each value of x, the key

points are (0, 0), 1

, 3 ,

1 , 0

,

3 , 3 , and (1, 0) .

4 2 4

Use these key points to graph y 3sin 2π x from 0 to 1 .

Extend the graph one cycle to the right. Use the graph to

obtain the graph of the reciprocal function. Draw vertical

asymptotes through the x-intercepts, and use them as

guides to graph y 3csc 2π x .

9. Graph the reciprocal cosine function, y 2 cos π x .

The equation is of the form

A 2 and B π .

y A cos Bx with

amplitude: A 2 2

period: 2π

2 B π

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125 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 125

Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

Scientific Calculator Solution

Function

Mode

Keystrokes Display

(rounded to four places)

cos−1 1

3

Radian

1 ÷ 3 = COS−1

1.2310

tan−1

(−35.85)

Radian

35.85 + TAN−1

–1.5429

Graphing Calculator Solution

Function

Mode

Keystrokes Display

(rounded to four places)

cos−1 1

Radian

COS−1

( 1 ÷ 3 ) ENTER

1.2310

tan−1

(−35.85)

Radian

TAN−1 − 35.85 ENTER

–1.5429

2

= −

.

= −

= −

Section 2.3

Check Point Exercises

1. Let θ = sin −1 3 , then sin θ =

3 .

2 2

The only angle in the interval − π

, π

that satisfies sin θ = 3

is π

. Thus, θ = π

, or sin−1 3 = π

. 2 2

2 3 3 2 3

2. Let θ = sin−1 − 2

, then sin θ = − .

2 2

The only angle in the interval − π

, π

that satisfies cosθ = − 2

is − π

. Thus θ π

, or sin−1 − 2

= − π

. 2 2 = −

2 4 4 2 4

3. Let θ = cos−1 − 1

, then cosθ 1

. The only angle in the interval [0, π ] that satisfies cosθ 1

is 2π

. Thus, 2

2 2 3

θ = 2π

, or cos−1 − 1

= 2π

.

3

2

3

4. Let θ = tan−1

(−1) , then tan θ = −1 . The only angle in the interval − π

, π

that satisfies tan θ = −1 is − π

. Thus

2 2

4

π

θ = − 4

or tan−1 θ

π 4

5.

a.

b.

a.

3

b.

6. a. cos (cos−1 0.7) x = 0.7 , x is in [–1,1] so cos(cos

−1 0.7) = 0.7

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126 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 126

Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

4 x + 1 4

b. sin −1

(sin π )

x = π , x is not in − π

, π

. x is in the domain

9. Let θ = tan −1 x , then tan θ =x= x

. 1

2 2

of sin x, so sin−1

(sin π ) = sin −1

(0) = 0

c. cos (cos−1 π )

x = π , x is not in [–1,1] so cos (cos−1 π ) is not

defined.

7. Let θ = tan −1 3

, then tan θ = 3

. Because tan θ is

Use the Pythagorean Theorem to find the third side, a.

a2

= x2

+ 12

a = x2

+ 1

Use the right triangle to write the algebraic expression. 2

sec ( tan −1 x ) = secθ = =

x2

+ 1positive, θ is in the first quadrant. 1

Concept and Vocabulary Check 2.3

1. − π

≤ x ≤ π

; 2 2

sin −1

x

Use the Pythagorean Theorem to find r.

r 2

= 32

+ 42

= 9 + 16 = 25

2. 0 ≤ x ≤ π ; cos−1

x

r = 25 = 5 3. − π

≤ x ≤ π

; tan −1 x

Use the right triangle to find the exact value. 2 2

sin tan

−1 3 = sin θ =

side opposite θ =

3 π π

4

hypotenuse 5 4. [−1,1] ;

2 ,

2

8. Let θ = sin −1

− 1

, then sin θ

1

= − .

Because sin θ

5. [−1,1] ; [0,π ]

2

2

is negative, θ is in quadrant IV. 6. (−∞, ∞) ;

− π

, π

2 2

7. − π

, π

2 2

8. [0,π ]

Use the Pythagorean Theorem to find x.

9. − π

, π

2 2

x2 + (−1)2 = 22

x2 + 1 = 4

x2 = 3

10. false

x = 3

Use values for x and r to find the exact value.

cos sin −1

− 1

= cos θ = x

= 3

2

r 2

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127 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 127

Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

= −

is

3

,

= −

= −

Exercise Set 2.3

7. Let θ = cos−1 3

, then cos θ = 3

. The only angle

1. Let θ = sin −1 1 , then sin θ =

1 . The only angle in

2 2

2 2 in the interval [0, π ] that satisfies cos θ = 3 is

π .

the interval − π

, π

that satisfies sin θ = 1

is π

. 2 6

2 2 2 6

π − 3 π

Thus,

θ = π

, or sin −1 1

= π

. Thus θ = , or cos

1 = . 6 2 6

6 2 6

8. Let θ = cos−1 2

, then cos θ = 2

. The only angle2. Let θ = sin

−1 0 , then sin θ = 0 . The only angle in the 2 2

interval − π

, π

that satisfies sin θ = 0 is 0. Thus

in the interval [0, π ] that satisfies cosθ = 2

is π

. 2 2 2 4

θ = 0 , or sin −1

0 = 0 .

3. Let θ = sin −1 2

, then sin θ = 2

. The only angle

Thus θ = π

, or cos−1 2

= π

. 4 2 4

2 2 9. Let θ = cos−1 −

2 , then cosθ = −

2 . The only

in the interval − π

, π

that satisfies sin θ = 2

is 2 2

2 2

2 angle in the interval [0, π ] that satisfies

π . Thus

θ = π

, or sin −1 2

= π

.

cosθ = − 2

is 3π

. Thus θ = 3π

, or4 4 2 4 2 4 4

cos−1 − 2

= 3π

.4. Let θ = sin

−1 3 , then sin θ =

3 . The only angle in

2 42 2

the interval − π

, π

that satisfies sin θ = 3

is

−1

3 3 2 2

2 10. Let θ = cos

2

, then cosθ = −

2 .

π . Thus θ =

π , or sin−1 3

= π

.

The only angle in the interval [0, π ] that3 3 2 3

satisfies cosθ = − 3

is 5π

. Thus θ = 5π

, or

5. Let θ = sin −1

− 1

, then sin θ 1

. The only angle 2 6 6

2

2

cos−1 −

3 =

5π .

in the interval − π

, π

that satisfies sin θ 1 2 6

2 2 2

− π

. Thus θ

π , or sin −1 −

1 = −

π .

11. Let θ = cos−1

0 , then cosθ = 0 . The only angle in= −

6 6 2 6 the interval [0, π ] that satisfies cosθ = 0 is

π .

2

6. Let θ = sin −1 − 3

, then sin θ = − . Thus θ = π

, or cos−1 0 = π

. 2 2 2 2

The only angle in the interval − π

, π

that satisfies −1 2 2

12. Let θ = cos 1 , then cosθ = 1 . The only angle in the

3 sin θ = −

2

is − π

. Thus θ π

3 3

interval [0, π ] that satisfies cosθ = 1 is 0 .

Thus θ = 0 , or cos−1

1 = 0 .

or sin−1 − 3

= − π

.

2 3

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128 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 128

Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

Scientific Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

sin −1

0.3

Radian

0.30 0.3 SIN−1

= −

= −

13. Let θ = tan −1 3

, then tan θ = 3

. The only angle in the interval − π

, π

that satisfies tan θ = 3

is π

. Thus3 3

2 2

3 6

θ = π

, or tan−1 3 = π

.

6 3 6

14. Let θ = tan −1

1 , then tanθ = 1 . The only angle in the interval − π

, π

that satisfies tanθ = 1 is π

.

2 2

4

Thus θ = π

, or tan−1 1 = π

.

4 4

15. Let θ = tan −1

0 , then tan θ = 0 . The only angle in the interval − π

, π

that satisfies tan θ = 0 is 0. Thus θ = 0 , or

2 2

tan −1

0 = 0 .

16. Let θ = tan−1

(−1) , then tan θ = −1 . The only angle in the interval − π

, π

that satisfies tan θ = −1 is − π

. Thus

2 2

4

θ π

, or tan−1 (−1) = − π

.4

17. Let θ = tan−1 (−

4

3 ) , then tan θ = − 3 . The only angle in the interval − π

, π

that satisfies tan θ = − 3 is

− π

. 2 2

3

Thus θ

π , or tan

−1 (−

3 ) = − π

.3 3

18. Let θ = tan−1

− 3

, then tan θ = − 3

. The only angle in the interval − π

, π

that satisfies tan θ = − 3

is − π

. 3 3 2 2 3 6

Thus θ π

, or tan−1 − 3

= − π

.= −

6 3 6

19.

Graphing Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

sin −1

0.3

Radian

0.30 SIN−1 0.3 ENTER

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129 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 129

Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

Scientific Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

sin −1

0.47

Radian 0.47 SIN−1

0.49

Scientific Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

sin −1

(−0.32)

Radian

–0.33 SIN−1 0.32 + −

Scientific Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

sin−1

(−0.625)

Radian 0.625 + SIN−1

–0.68

Scientific Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

cos−1 3

Radian

−1

1.19

Graphing Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

cos−1 3

Radian

−1

1.19

20.

Graphing Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

sin −1

0.47

Radian SIN−1

0.47 ENTER

0.49

21.

Graphing Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

sin −1

(−0.32)

Radian

–0.33 SIN−1 − 0.32 ENTER

22.

Graphing Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

sin−1

(−0.625)

Radian SIN−1 − 0.625 ENTER

–0.68

23.

8

3 ÷ 8 = COS

8

COS ( 3 ÷ 8 ) ENTER

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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

130 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 130

Scientific Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

cos−1 4

Radian

−1

1.11

Graphing Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

cos−1 4

Radian

−1

1.11

Scientific Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

cos−1 5

7

Radian

5 ÷ 7 = COS−1

1.25

Scientific Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

tan −1

(−20)

Radian

–1.52 TAN−1 20 + −

24.

9

4 ÷ 9 = COS

9

COS ( 4 ÷ 9 ) ENTER

25.

Graphing Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

cos−1 5

7

Radian

COS−1

( 5 ÷ 7 ) ENTER

1.25

26. Scientific Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

cos−1 7

10

Radian

7 ÷ 10 = COS−1

1.30

Graphing Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

cos−1 7

10

Radian

COS−1

( 7 ÷ 10 ) ENTER

1.30

27.

Graphing Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

tan −1

(−20)

Radian

–1.52 TAN−1 − 20 ENTER

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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

131 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 131

Scientific Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

tan −1

(−30)

Radian 30 + TAN−1

–1.54

Scientific Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

tan −1 (− 473 )

Radian

–1.52 473 + −

TAN−1

Scientific Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

tan−1 (− 5061) Radian

−1

–1.56

Graphing Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

tan−1 (− 5061) Radian

−1

–1.56

28.

Graphing Calculator Solution

Function

Mode

Keystrokes Display

(rounded to two places)

tan −1

(−30)

Radian TAN−1 − 30 ENTER

–1.54

29.

Graphing Calculator Solution

Function Mode Keystrokes Display

(rounded to two places)

tan −1 (− 473 )

Radian TAN−1

( −

473 ) ENTER

–1.52

30.

5061 + − TAN

TAN ( − 5061 ) ENTER

31. sin (sin−1 0.9) x = 0.9, x is in [−1, 1], so sin(sin −1 0.9) = 0.9

32.

33.

cos(cos−1

0.57)

x = 0.57, x is in [−1, 1],

so cos(cos−1 0.57) = 0.57

sin −1

sin π

3

x = π

, x is in − π

, π

, so sin −1

sin π

= π

3 2 2

3

3

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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

132 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 132

2 2

3

4

34.

cos−1

cos 2π

41.

tan −1 tan 2π

3

3

x = 2π

, x is in [0, π ],

π π 3

so cos−1

cos 2π

= 2π

x = , x is not in − , , x is in the domain of 3

3

3

−1

2π −1 π

35.

sin −1

sin 5π

tan x, so tan tan

= tan

(− 3 ) = − 3

6

−1

x = 5π

, x is not in − π

, π

, x is in the domain of

42. tan tan

6 2 2

x = 3π

, x is not in − π

, π

,

4 2 2 sin x, so sin

−1 sin

5π = sin

−1 1 =

π

x is in the domain of tan x 6

2

6

so tan −1

tan 3π

= tan −1

(−1) = − π

4

4

36.

cos−1

cos 4π

3

x = 4π

, x is not in [0, π ], 3

43. sin −1

(sin π )

π π x is in the domain of cos x, x = π , x is not in − , ,

so cos−1

cos 4π

= cos−1

− 1

= 2π 2 2

3

2

3

x is in the domain of sin x, so

37.

tan ( tan−1

125) x = 125, x is a real number, so tan ( tan−1 125) = 125

44.

sin−1

(sin π ) = sin−1

0 = 0

cos−1

(cos 2π ) x = 2π , x is not in [0, π ],

x is in the domain of cos x,−1 −1

38.

39.

tan(tan−1

380)

x = 380, x is a real number,

so tan(tan−1 380) = 380

tan −1 tan − π

45.

so cos (cos 2π ) = cos 1 = 0

sin (sin−1 π )

x = π , x is not in [−1, 1] , so sin (sin−1 π ) is not

6 defined.

x π

, x is in − π

, π

, so −1= −

46.

cos(cos 3π )6 2 2 x = 3π , x is not in [−1, 1]

tan −1 tan − π

= − π

so cos(cos−1

3π ) is not defined.

6

6

40.

tan −1

tan − π

3

x π

, x is in − π

, π

,

= −

3 2 2

so tan −1

tan − π

= − π

3

3

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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

133 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 133

= −

5

47. Let θ = sin −1 4

, then sin θ = 4

. Because sin θ is 50. Let θ = sin −1 5 then sin θ =

5 .

5 5 13 13

positive, θ is in the first quadrant.

x2 + y2 = r 2

x2 + 42 = 52

x2 = 25 −16 = 9 x = 3

cos sin

−1 4 = cosθ =

x =

3

because sin θ is positive, θ is in the first quadrant.

x2 + y2 = r 2

x2 + 52 = 13

2

x2 = 144

x = 12

cot sin

−1 5 = cot θ =

x =

12

5

r 5 13 y 5

48. Let θ = tan−1 7

, then tan θ = 7

.

51. Let θ = sin −1

− 3

, then sin θ

3

. Because sin θ24 24

5

5

Because tan θ is positive, θ is in the first quadrant.

r 2 = x2 + y2

r 2 = 72 + 242

r 2 = 625 r = 25

is negative, θ is in quadrant IV.

x2 + y2 = r

2

x2 + (−3)

2 = 52

x2 = 16 x = 4

sin tan

−1 7 = sin θ =

y =

7 −1

3

y 3 24

r 25

tan sin

− = tan θ = = −

5 x 4

49. Let θ = cos−1 5 , then cosθ =

5

. Because cosθ is

−1 4 413 13 52. Let θ = sin − , then sin θ

= − .

5positive, θ is in the first quadrant.

x2 + y 2 = r 2

52 + y 2 = 132

y 2 = 169 − 25

y 2 = 144 y = 12

Because sin θ is negative, θ is in quadrant IV.

x2 + y2 = r 2

x2 + (−4)2 = 52

x2 = 9 x = 3

cos sin −1

− 4

= cosθ = x

= 3

tan cos−1 5

= tan θ = y

= 12 5 r 5

13

x 5

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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

134 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 134

.

= −

= −

4

53. Let, θ = cos−1 2 , then cosθ =

2 . Because cosθ

56. Let θ = sin

−1 −

1 , then sin θ

1

2 2

2

2

is positive, θ is in the first quadrant.

x2 + y2 = r 2 2 2 2

Because sin θ is negative, θ is in quadrant IV.

x2 + y2 = r 2

2 2 2( 2 ) + y = 2 x + (−1) = 2

y2 = 2

y = 2

x2 = 3

x = 3

sec sin −1 −

1 = secθ =

r =

2 =

2 3

sin cos−1 2

= sin θ = y

= 2

2 x 3 3

2 r 2

54. Let θ = sin −1 1

, then sin θ = 1

.

57. Let θ = cos−1

− 1

, then cosθ

1 . Because

2 2

3

3

Because sin θ is positive, θ is in the first quadrant.

x2 + y 2 = r 2

x2 + 12 = 22

x2 = 3

x = 3

cosθ is negative, θ is in quadrant II.

x2 + y2 = r

2

(−1)2 + y2 = 32

y2 = 8

y = 8

cos sin

−1 1 = cosθ =

x =

3 y = 2 2

2

r 2

Use the right triangle to find the exact value.

tan cos−1 − 1

= tan θ = y

= 2 2

= −2 255. Let θ = sin

−1 − 1

, then sin θ 1

. Because sin θ

3

x −1 4

= −

is negative, θ is in quadrant IV.

x2 + y

2 = r 2

x2 + (−1)

2 = 42

x2 = 15 x = 15

sec sin −1

− 1

= secθ = r

= 4

= 4 15

4

x 15 15

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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

135 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 135

. 2

−1

.

.

−1

2

= −

= −

= −

58. Let θ = cos−1

− 1

, then cosθ 1 x

2 + y

2 = r

2

4

4

x2 + (−

2 ) = 22

Because cosθ is negative, θ is in quadrant II. x

2 = 2

x = 2

2 r 2

sec sin − = secθ = = = 2

2 x 2

61. Let θ = tan −1

− 2

, then tan θ 2

3

3

x2

+ y2

= r 2

(−1)2

+ y2

= 42

y2

= 15

Because tan θ is negative, θ is in quadrant IV.

y = 15 θ

tan cos−1

− 1

= tan θ = y

= 15

= − 15 4

x −1

r 13

59. Let θ = cos−1 −

3 , then cosθ = −

3 . Because

r 2 = x2 + y2

r 2 = 32 + (−2)2

2 2 r 2 = 9 + 4cosθ is negative, θ is in quadrant II. r 2 = 13

r = 13

cos tan−1

− 2

= cosθ = x

= 3

= 3 13

3

r 13 13

62. Let θ = tan −1

− 3

, then tan θ 3

4

4

x2 + y2 = r

2

2 2 2

Because tan θ is negative, θ is in quadrant IV.

(− 3 ) + y = 2

y2 = 1 y = 1

3 r 2

csc cos − = cscθ = = = 2

2 y 1

r 2 = x2 + y2

60. Let θ = sin−1 − 2

, then sin θ = − .

r 2 = 42 + ( −3)2

2 2 r

2 = 16 + 9Because sin θ is negative, θ is in quadrant IV.

r 2 = 25 r = 5

sin tan

−1 −

3 = sin θ =

y =

−3 = −

3

4

r 5 5

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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

136 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 136

( ) 1 − x

−1

63. Let θ = cos−1 x , then cosθ = x = x

. 1

65. Let θ = sin −1

2 x , then sin θ = 2 x

y = 2x, r = 1

Use the Pythagorean Theorem to find x.

x2

+ (2x)2

= 12

x2 = 1 − 4 x

2

x = 1 − 4 x2

Use the Pythagorean Theorem to find the third

side, b.

x2 + b2 = 12

cos(sin −1

2 x) =

66. Let θ = cos−1

2 x.

1 − 4 x2

b2 = 1 − x2

Use the Pythagorean Theorem to find the third side, b.

b = 1 − x2

Use the right triangle to write the algebraic expression.

2

tan cos−1 x = tan θ = x

64. Let θ = tan −1

x , then tan θ = x = x

.

2 2 2

1 (2 x) + b = 1

b2 = 1 − 4 x2

b = 1 − 4 x2

−1 1 − 4 x 2

2sin (cos 2 x ) = 1

= 1 − 4 x

Use the Pythagorean Theorem to find the third side, c. 67. Let θ = sin −1 1

, then sin θ = 1

.

c2

= x2

+ 12 x x

c = x2 + 1

Use the right triangle to write the algebraic expression.

sin(tan −1 ) = sin θ

= x

x2 + 1

x x2 + 1

= ⋅

x2 + 1

x x2 + 1

= x2 + 1

x2 + 1 Use the Pythagorean Theorem to find the third side,

a.

a2 +12 = x2

2 2a = x

a =

−1

x2 −1

Use the right triangle to write the algebraic

expression.

1 x 2

− 1

cos sin = cosθ =

x

x

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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

137 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 137

−1

68. Let θ = cos−1 1 , then cosθ =

1 .

−1 x2 − 9 x2 − 9

x x 72. Let θ = sin , then sin θ = . x x

Use the Pythagorean Theorem to find the third

side, b.

12 + b2 = x2

Use the Pythagorean Theorem to find the third side,

a.

b2 = x2 − 1

a2 +

2

x2 − 9

= x2

b = x2 − 1

a2 = x2 − x2 + 9 = 9

Use the right triangle to write the algebraic expression.

sec cos−1 1

= secθ = x

= x

a = 3

Use the right triangle to write the algebraic expression.

x

1

2 −

cot sin −1 x 9

= 3

x x2 − 9

69.

cot tan

−1 x =

3

3

x

3

x2 − 9 3 x

2 − 9 = ⋅ =

70.

cot tan

−1 x =

2 x

2 − 9 x

2 − 9 x2 − 9

2

x

73. a.

y = sec x is the reciprocal of

y = cos x . The x-

71. Let θ = sin −1 x

, then sin θ = x

.

values for the key points in the interval [0, π ]

are 0, π

, π

, 3π

, and π . The key points arex2 + 4 x2 + 4 4 2 4

π 2 π 3π 2

(0, 1), , , , 0 , , − , and

4 2 2 4 2

(π , − 1) , Draw a vertical asymptote at x = π

. 2

Use the Pythagorean Theorem to find the third side,

Now draw our graph from (0, 1) through

π , 2

to ∞ on the left side of the

a. 4

a2 + x2 =

2

x2 + 4

asymptote. From −∞ on the right side of the

asymptote through

3π , − 2

to (π , − 1) .

a2 = x2 + 4 − x2 = 4 a = 2

Use the right triangle to write the algebraic

expression.

4

x

sec sin

secθ x

2 + 4

= =

x2

+ 4 2

b. With this restricted domain, no horizontal lineintersects the graph of y = sec x more than

once, so the function is one-to-one and has an

inverse function.

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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

138 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 138

2 2 ,

c. Reflecting the graph of the restricted secant

function about the line y = x, we get the graph 75.

of y = sec−1 x .

74. a. Two consecutive asymptotes occur at x = 0 and

76.

domain: [−1, 1];

range: [0, π]

x = π . Midway between x = 0 and x = π is

x = 0 and x = π . An x-intercept for the graph

is π

, 0 . The graph goes through the points

2

3π π , 1

and

, − 1

. Now graph the function

4

4

through these points and using the asymptotes. domain: [−1, 1];

range: π 3π

b. With this restricted domain no horizontal line

77.

intersects the graph of y = cot x more than

once, so the function is one-to-one and has an

inverse function. Reflecting the graph of the

restricted cotangent function about the line y =

x, we get the graph of

c.

y = cot −1

x .

78.

domain: [−2, 0];

range: [0, π]

domain: [−2, 0];

range:

− π

, π

2 2

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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

139 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 139

79.

80.

domain: (−∞, ∞);

range: (−π, π)

domain: (−∞, ∞);

83. 84.

domain: [−2, 2];

range: [0, π]

domain: [−2, 2];

range:

− 3π

, 3π

π π 2 2

range: − , 2 2

81.

82.

domain: (1, 3];

range: [−π, 0]

domain: [1, 3];

85. The inner function, sin −1

x, accepts values on the

interval [−1,1] . Since the inner and outer functions

are inverses of each other, the domain and range are

as follows.

domain: [−1,1] ; range: [−1,1]

86. The inner function, cos−1

x, accepts values on the

interval [−1,1] . Since the inner and outer functions

are inverses of each other, the domain and range are

as follows.

domain: [−1,1] ; range: [−1,1] 87. The inner function, cos x, accepts values on the

interval (−∞, ∞ ) . The outer function returns values

on the interval [0, π ] domain: (−∞, ∞ ) ; range: [0, π ]

88. The inner function, sin x, accepts values on the

interval (−∞, ∞ ) . The outer function returns values

range:

− π

, π

π π 2 2

on the interval − ,

2 2

domain: (−∞, ∞ ) ; range:

− π

, π

2 2

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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

140 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 140

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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

140 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 140

x θ

5 tan −1 33

− tan −1 8

≈ 0.408 radians 5 5

10 tan −1 33

− tan −1 8

≈ 0.602 radians 10 10

15 tan −1 33

− tan −1 8

≈ 0.654 radians 15 15

20 tan −1 33

− tan −1 8

≈ 0.645 radians 20 20

25 tan −1 33

− tan −1 8

≈ 0.613 radians 25 25

2

89. The inner function, cos x, accepts values on the

interval (−∞, ∞ ) . The outer function returns values

95.

θ = 2 tan −1 21.634

28

≈ 1.3157 radians;

1.3157 180

≈ 75.4°on the interval −

π , π

π 2 2

domain: (−∞, ∞ ) ; range:

− π

, π

−1 21.634 2 2

96. θ = 2 tan 300

≈ 0.1440 radians;

0.1440 180

≈ 8.2°90. The inner function, sin x, accepts values on the π

interval (−∞, ∞ ) . The outer function returns values

on the interval [0, π ]

domain: (−∞, ∞ ) ; range: [0, π ]

97.

tan −1

b − tan −1

a = tan −1

2 − tan −1

0

≈ 1.1071 square units

−1 −1

−1 −1

91. The functions sin−1

x and cos−1

x

accept values on 98. tan b − tan a = tan 1 − tan (−2)

≈ 1.8925 square unitsthe interval [−1,1] . The sum of these values is always

π .

99. – 109. Answers may vary.

2 110.

y = sin −1 x

domain: [−1,1] ; range: π

y = sin −1

x + 2

92. The functions sin−1 x and cos−1 x accept values on

the interval [−1,1] . The difference of these values

range from − π 2

to 3π 2

domain: [−1,1] ; range:

− π

, 3π

2 2

93.

θ = tan −1 33

− tan −1 8

x x

The graph of the second equation is the graph of the first equation shifted up 2 units.

−1111. The domain of y = cos x is the interval

[–1, 1], and the range is the interval [0, π ] . Because

the second equation is the first equation with 1

subtracted from the variable, we will move our x

π π max to π , and graph in a − , π , by [0, 4, 1]

viewing rectangle.

2 4

94. The viewing angle increases rapidly up to about 16 feet, then it decreases less rapidly; about 16 feet;

when x = 15, θ = 0.6542 radians; when x = 17,

θ = 0.6553 radians.

The graph of the second equation is the graph of the first equation shifted right 1 unit.

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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

141 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 141

−1

112. y = tan −1

x

y = −2 tan −1

x

115.

The graph of the second equation is the graph of the

It seems sin−1

x + cos−1

x = π 2

for −1 ≤ x ≤ 1 .

first equation reversed and stretched. 116. does not make sense; Explanations will vary. Sample explanation: The cosine’s inverse is not a

113. The domain of y = sin−1 x is the interval function over that interval.

[–1, 1], and the range is − π

, π

. Because the 2 2

second equation is the first equation plus 1, and with 2 added to the variable, we will move our y max to 3, and move our x min to −π , and graph in a

−π , π

, π

by

117. does not make sense; Explanations will vary.

Sample explanation: Though this restriction works

for tangent, it is not selected simply because it is

easier to remember. Rather the restrictions are based

on which intervals will have inverses.

118. makes sense 2 2

[–2, 3, 1] viewing rectangle. 119. does not make sense; Explanations will vary.

Sample explanation:

sin −1 sin 5π

= sin −1 − 2

= − π

4 2 4

114.

The graph of the second equation is the graph of the first equation shifted left 2 units and up 1 unit.

y = tan −1

x

120.

121.

y = 2sin−1( x − 5) y

= sin ( x − 5) 2

sin y

= x − 5 2

x = sin y

+ 5 2

2sin −1

x = π

Observations may vary.

4

sin −1

x = π

8

x = sin π 8

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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

142 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 142

x y = 3cos 2π x coordinates

0 y = 3 cos(2π ⋅ 0) = 3cos 0 = 3 ⋅1 = 3

(0, 3)

1

4 y = 3cos

2π ⋅

1

π

= 3cos 2

= 3 ⋅ 0 = 0

1 , 0

4

1

2

1 y = 3cos 2π ⋅

2 = 3cos π = 3 ⋅ (−1) = −3

1 2

, − 3

3

4 y = 3cos

2π ⋅

3

3π 4

= 3cos 2

= 3 ⋅ 0 = 0

3 , 0

4

1 y = 3cos(2π ⋅1) = 3cos 2π

= 3 ⋅1 = 3

(1, 3)

x

= =

2

122. Prove: If x > 0, tan −1

x + tan −1 1

= π 126. The equation y = 3cos 2π x is of the form y = A cos Bx

x 2 with A = 3 and B = 2π . Thus, the amplitude is

Since x > 0, there is an angle θ with 0 < θ < π

as

A = 3 = 3 . The period is 2π

= 2π

= 1 . The quarter-2

shown in the figure. B 2π

1period is . The cycle begins at x = 0 . Add quarter-

4periods to generate x-values for the key points.

x = 0

x = 0 + 1

= 1

tan θ = x and tan π

− θ =

1 thus 1

4 4 2

x

x = + 1

= 1

tan −1

x = θ and tan −1 1

= π

− θ so 4 4 2

1 1 3 x = + = 2 4 4

tan −1

x + tan −1 1

= θ + π

− θ = π

x = 3

+ 1

= 1x 2 2 4 4

Evaluate the function at each value of x.123. Let α equal the acute angle in the smaller right

triangle.

tan α = 8 x

so tan −1 8 = α

x

tan(α + θ ) = 33

x

so tan −1 33 = α + θ

x

θ = α + θ − α = tan −1 33 − tan −1 8

x x

124. Because the tangent is negative and the cosine is

negative, θ lies in quadrant II. In quadrant II, x is negative and y is positive. Thus,

4

tan θ 2 y 2 = −

3 x −3 x = −3, y = 2

Furthermore,

r = x2 + y2 = (−3)2 + 22 = 9 + 4 = 13

Now that we know x, y, and r, we can find

sin θ and secθ .

sin θ = y

= 2

= 2

⋅ 13

= 2 13

Connect the five points with a smooth curve and

graph one complete cycle of the given function.

r 13 13 13 13

secθ = r

= 13

= − 13

x −3 3

125. 210° lies in quadrant III. The reference angle is

θ ′ = 210° − 180° = 30° .

sin 30° = 1 2

Because the sine is negative in quadrant III,

sin 210° = − sin 30° = − 1

. 2

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Section 2.4 Applications of Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

25

127. tan A = a

b

tan 22.3 =

a

12.1 a = 12.1 tan 22.3

a ≈ 4.96

cos A = b

c

cos 22.3 =

12.1

c

c = 12.1 cos 22.3

c ≈ 13.08

2. Using a right triangle, we have a known angle, an

unknown opposite side, a, and a known adjacent side.

Therefore, use the tangent function.

tan 85.4° = a 80

a = 80 tan 85.4° ≈ 994 The Eiffel tower is approximately 994 feet high.

3. Using a right triangle, we have an unknown angle, A,

a known opposite side, and a known hypotenuse.

Therefore, use the sine function.

sin A = 6.7

13.8

A = sin −1 6.7

≈ 29.0° 13.8

128.

tan θ = opposite

adjacent

tan θ = 18

25

θ = tan −1 18

θ ≈ 35.8

The wire makes an angle of approximately 29.0° with the ground.

4. Using two right triangles, a smaller right triangle

corresponding to the smaller angle of elevation drawn

inside a larger right triangle corresponding to the

larger angle of elevation, we have a known angle, an

unknown opposite side, a in the smaller triangle, b in

the larger triangle, and a known adjacent side in each

129. 10 cos π

x triangle. Therefore, use the tangent function. 6

a

amplitude: 10 = 10

tan 32° = 800

period: 2π

= 2π ⋅ 6

= 12

a = 800 tan 32° ≈ 499.9

tan 35° = b

π π 6

Section 2.4

Check Point Exercises

1. We begin by finding the measure of angle B. Because

C = 90° and the sum of a triangle’s angles is 180°, we

see that A + B = 90°. Thus, B = 90° – A = 90° – 62.7° = 27.3°. Now we find b. Because we have a known angle, a

known opposite side, and an unknown adjacent side,

use the tangent function.

tan 62.7° = 8.4 b

b = 8.4

≈ 4.34 tan 62.7°

Finally, we need to find c. Because we have a known

angle, a known opposite side and an unknown

hypotenuse, use the sine function.

sin 62.7° = 8.4 c

c = 8.4

≈ 9.45 sin 62.7

In summary, B = 27.3°, b ≈ 4.34, and c ≈ 9.45.

800 b = 800 tan 35° ≈ 560.2

The height of the sculpture of Lincoln’s face is 560.2 – 499.9, or approximately 60.3 feet.

5. a. We need the acute angle between ray OD and the north-south line through O. The measurement of this angle is given to be 25°. The angle is measured from the south side

of the north-south line and lies east of the north-

south line. Thus, the bearing from O to D is

S 25°E.

b. We need the acute angle between ray OC and

the north-south line through O. This angle measures 90° − 75° = 15°.

This angle is measured from the south side of

the north-south line and lies west of the north-

south line. Thus the bearing from O to C is S

15° W.

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Section 2.4 Applications of Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

4

° =

6. a. Your distance from the entrance to the trail

system is represented by the hypotenuse, c, of a

right triangle. Because we know the length of

the two sides of the right triangle, we find c

using the Pythagorean Theorem. We have

c2 = a2 + b2 = (2.3)

2 + (3.5)2 = 17.54

c = 17.54 ≈ 4.2

You are approximately 4.2 miles from the

entrance to the trail system.

8. We begin by identifying values for a and ω.

d = 12 cos π

t, a = 12 and ω = π

. 4 4

a. The maximum displacement from the rest

position is the amplitude. Because

a = 12, the maximum displacement is 12 centimeters.

b. The frequency, f, is

ω π

π 1 1 f = = = ⋅ =

b. To find your bearing from the entrance to the 2π 2π 4 2π 8

trail system, consider a north-south line passing

through the entrance. The acute angle from this The frequency is

1

8

cm per second.

line to the ray on which you lie is 31° + θ . Because we are measuring the angle from the

c. The time required for one cycle is the period.

south side of the line and you are west of the

period = 2π

= 2π

= 2π ⋅ 4

= 8

πentrance, your bearing from the entrance is ω 4

π

S (31° + θ ) W. To find θ , Use a right triangle

and the tangent function.

tan θ = 3.5 2.3

θ = tan −1 3.5

≈ 56.7° 2.3

Thus, 31° + θ = 31° + 56.7° = 87.7°. Your

bearing from the entrance to the trail system is S

87.7° W.

The time required for one cycle is 8 seconds.

Concept and Vocabulary Check 2.4

1. sides; angles 2. north; south

7. When the object is released (t = 0), the ball’s

distance, d, from its rest position is 6 inches down. Because it is down, d is negative: when t = 0, d = –6. Notice the greatest distance from

rest position occurs at t = 0. Thus, we will use the

3. simple harmonic; a ; 2π

; ω

Exercise Set 2.4

ω

equation with the cosine function, y = a cos ωt, to 1. Find the measure of angle B. Because

model the ball’s motion. Recall that a is the

maximum distance. Because the ball initially moves

down, a = –6. The value of ω can be found using the formula for the period.

period = 2π

= 4 ω

2π = 4ω

ω = 2π

= π

4 2

Substitute these values into d = a cos wt. The equation for the ball’s simple harmonic motion is

d = −6 cos π

t. 2

C = 90°, A + B = 90°. Thus, B = 90° − A = 90° − 23.5° = 66.5°.

Because we have a known angle, a known adjacent

side, and an unknown opposite side, use the tangent

function.

tan 23.5° = a 10

a = 10 tan 23.5° ≈ 4.35 Because we have a known angle, a known adjacent

side, and an unknown hypotenuse, use the cosine

function.

cos 23.5 10 c

c = 10

≈ 10.90 cos 23.5°

In summary, B = 66.5°, a ≈ 4.35, and c ≈ 10.90.

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Section 2.4 Applications of Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

50.2

2. Find the measure of angle B. Because C = 90°,

A + B = 90°. Thus, B = 90° − A = 90° − 41.5° = 48.5°.

5. Find the measure of angle A. Because

C = 90°, A + B = 90°. Thus, A = 90° − B = 90° −16.8° = 73.2° .

Because we have a known angle, a known adjacent side, and an unknown opposite side, use the tangent

Because we have a known angle, a known opposite side and an unknown adjacent side, use the tangent function.

function.

tan 41.5° = a

tan 16.8° = 30.5

a

20 a = 20 tan 41.5° ≈ 17.69

Because we have a known angle, a known adjacent side, and an unknown hypotenuse, use the cosine function.

a = 30.5

≈ 101.02 tan 16.8°

Because we have a known angle, a known opposite

side, and an unknown hypotenuse, use the sine function.

30.5

cos 41.5° = 20

sin 16.8° = c

c

c = 20

≈ 26.70 cos 41.5°

In summary, B = 48.5°, a ≈ 17.69 , and c ≈ 26.70 .

3. Find the measure of angle B. Because

c = 30.5

≈ 105.52 sin16.8°

In summary, A = 73.2°, a ≈ 101.02, and c ≈ 105.52.

6. Find the measure of angle A. Because C = 90°,

A + B = 90°.

C = 90°, A + B = 90°. Thus, A = 90° − B = 90° − 23.8° = 66.2° .

Thus, B = 90° − A = 90° − 52.6° = 37.4° .

Because we have a known angle, a known

hypotenuse, and an unknown opposite side, use the

sine function.

sin 52.6 = a

54

Because we have a known angle, a known opposite side, and an unknown adjacent side, use the tangent function.

tan 23.8° = 40.5

a

a 40.5

≈ 91.83

a = 54 sin 52.6° ≈ 42.90

Because we have a known angle, a known

hypotenuse, and an unknown adjacent side, use the

cosine function.

cos 52.6° = b 54

b = 54 cos 52.6° ≈ 32.80 In summary, B = 37.4°, a ≈ 42.90, and b ≈ 32.80.

= tan 23.8°

Because we have a known angle, a known opposite

side, and an unknown hypotenuse, use the sine

function.

sin 23.8° = 40.5

c

c = 40.5

≈ 100.36 sin 23.8°

4. Find the measure of angle B. Because C = 90°, A + B = 90°.

In summary, A = 66.2°, a ≈ 91.83 , and c ≈ 100.36 .

Thus, B = 90° − A = 90° − 54.8° = 35.2° .

Because we have a known angle, a known hypotenuse,

and an unknown opposite side, use the sine function.

sin 54.8° = a 80

a = 80 sin 54.8° ≈ 65.37

Because we have a known angle, a known hypotenuse, and an unknown adjacent side, use the cosine function.

cos 54.8 = b 80

b = 80 cos 54.8° ≈ 46.11

In summary, B = 35.2°, a ≈ 65.37 , and c ≈ 46.11 .

7. Find the measure of angle A. Because we have a known hypotenuse, a known opposite side, and an unknown angle, use the sine function.

sin A = 30.4 50.2

A = sin −1 30.4

≈ 37.3°

Find the measure of angle B. Because C = 90°, A + B = 90°. Thus,

B = 90° − A ≈ 90° − 37.3° = 52.7° .

Use the Pythagorean Theorem.

a2 + b2 = c2

(30.4)2 + b2 = (50.2)2

b2 = (50.2)2 − (30.4)2 = 1595.88

b = 1595.88 ≈ 39.95

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Section 2.4 Applications of Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

In summary, A ≈ 37.3°, B ≈ 52.7°, and b ≈ 39.95.

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147 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 147

Section 2.4 Applications of Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

65.8

7

24.7

17.6

8. Find the measure of angle A. Because we have a

known hypotenuse, a known opposite side, and an

unknown angle, use the sine function.

sin A = 11.2 65.8

A = sin −1 11.2

≈ 9.8°

Find the measure of angle B. Because C = 90°,

A + B = 90°. Thus, B = 90° − A ≈ 90° − 9.8° = 80.2° .

Use the Pythagorean Theorem.

a2 + b2 = c2

(11.2)2 + b2 = (65.8)2

b2 = (65.8)

2 − (11.2)2 = 4204.2

b = 4204.2 ≈ 64.84

11. Find the measure of angle A. Because we have a

known hypotenuse, a known adjacent side, and

unknown angle, use the cosine function.

cos A = 2 7

A = cos−1 2

≈ 73.4°

Find the measure of angle B. Because

C = 90°, A + B = 90°. Thus, B = 90° − A ≈ 90° − 73.4° = 16.6° .

Use the Pythagorean Theorem.

a2 + b2 = c2

a2 + (2)2 = (7)2

a2 = (7)

2 − (2)2 = 45

a = 45 ≈ 6.71

In summary, A ≈ 9.8°, B ≈ 80.2°, and b ≈ 64.84.

9. Find the measure of angle A. Because we have a

known opposite side, a known adjacent side, and an

unknown angle, use the tangent function.

tan A = 10.8 24.7

A = tan −1 10.8

≈ 23.6°

Find the measure of angle B. Because

C = 90°, A + B = 90°. Thus, B = 90° − A ≈ 90° − 23.6° = 66.4° . Use the Pythagorean Theorem.

c2 = a2 + b2 = (10.8)

2 + (24.7)2 = 726.73

c = 726.73 ≈ 26.96

In summary, A ≈ 73.4°, B ≈ 16.6°, and a ≈ 6.71.

12. Find the measure of angle A. Because we have a

known hypotenuse, a known adjacent side, and an

unknown angle, use the cosine function.

cos A = 4 9

A = cos−1 4

≈ 63.6° 9

Find the measure of angle B. Because C = 90°,

A + B = 90°.

Thus, B = 90° − A ≈ 90° − 63.6° = 26.4° .

Use the Pythagorean Theorem.

a2 + b2 = c2

a2 + (4)2 = (9)2

In summary, A ≈ 23.6°, B ≈ 66.4°, and a2 = (9)

2 − (4)2 = 65

c ≈ 26.96.

10. Find the measure of angle A. Because we have a known opposite side, a known adjacent side, and an

a =

In summary,

65 ≈ 8.06

A ≈ 63.6°, B ≈ 26.4° , and a ≈ 8.06 .

unknown angle, use the tangent function.

tan A = 15.3 17.6

A = tan −1 15.3

≈ 41.0°

Find the measure of angle B. Because C = 90°, A + B = 90°.

Thus, B = 90° − A ≈ 90° − 41.0° = 49.0° .

Use the Pythagorean Theorem.

c2 = a2 + b2 = (15.3)

2 + (17.6)2 = 543.85

c = 543.85 ≈ 23.32

13. We need the acute angle between ray OA and the north-south line through O. This angle measure

90° − 75° = 15°. This angle is measured from the

north side of the north-south line and lies east of the

north-south line. Thus, the bearing from O and A is

N 15° E. 14. We need the acute angle between ray OB and the

north-south line through O. This angle measures

90° − 60° = 30° . This angle is measured from the

north side of the north-south line and lies west of the north-south line. Thus, the bearing from O to B is N

In summary, A ≈ 41.0°, B ≈ 49.0° , and c ≈ 23.32 . 30° W.

15. The measurement of this angle is given to be 80°.

The angle is measured from the south side of the

north-south line and lies west of the north-south line.

Thus, the bearing from O to C is S 80° W.

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Section 2.4 Applications of Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

16. We need the acute angle between ray OD and the 19. When t = 0, d = 0. Therefore, we will use the

north-south line through O. This angle measures equation with the sine function, y = a sin ωt, to

90° − 35° = 55° . This angle is measured from the

south side of the north-south line and lies east of the north-south line. Thus, the bearing from O to D is S 55° E.

17. When the object is released (t = 0), the object’s

distance, d, from its rest position is 6 centimeters

down. Because it is down, d is negative: When

model the object’s motion. Recall that a is the

maximum distance. Because the object initially

moves down, and has an amplitude of 3 inches,

a = –3. The value of ω can be found using the formula for the period.

period = 2π

= 1.5 ω

2π = 1.5ωt = 0, d = −6. Notice the greatest distance from rest

position occurs at t = 0. Thus, we will use the ω = 2π

= 4π

1.5 3equation with the cosine function, y = a cos ωt to Substitute these values into d = a sin ωt. Themodel the object’s motion. Recall that a is the

maximum distance. Because the object initially

moves down, a = –6. The value of ω can be found using the formula for the period.

equation for the object’s simple harmonic motion is

d = −3sin 4π

t. 3

20. When t = 0, d = 0. Therefore, we will use theperiod =

2π = 4

ω

equation with the sine function,

y = a sin ωt, to

2π = 4ω

ω = 2π

= π

4 2

Substitute these values into d = a cos ωt. The

equation for the object’s simple harmonic motion is

d = −6 cos π

t. 2

model the object’s motion. Recall that a is the

maximum distance. Because the object initially moves down, and has an amplitude of 5 centimeters,

a = –5. The value of ω can be found using the

formula for the period.

period = 2π

= 2.5 ω 2π = 2.5ω

18. When the object is released (t = 0), the object’s

distance, d, from its rest position is 8 inches down.

Because it is down, d, is negative: When

t = 0, d = –8. Notice the greatest distance from rest

position occurs at t = 0. Thus, we will use the

ω = 2π

= 4π

2.5 5

Substitute these values into d = a sin ωt . The

equation for the object’s simple harmonic motion is

4πequation with the cosine function, y = a cos ωt , to d = −5sin t .

5model the object’s motion. Recall that a is the

maximum distance. Because the object initially

moves down, a = –8. The value of ω can be found using the formula for the period.

period = 2π

= 2 ω 2π = 2ω

21. We begin by identifying values for a and ω.

d = 5 cos π

t, a = 5 and ω = π

2 2 a. The maximum displacement from the rest

position is the amplitude. Because a = 5, the

maximum displacement is 5 inches.

ω = 2π

= π ω

π 2

π 1 12

Substitute these values into d = a cos ωt .

The equation for the object’s simple harmonic motion

is d = −8 cos π t .

b. The frequency, f, is

The frequency is 1 4

f = = = ⋅ = . 2π 2π 2 2π 4

inch per second.

c. The time required for one cycle is the period.

period = 2π

= 2π

= 2π ⋅ 2

= 4

ω π π 2

The time required for one cycle is 4 seconds.

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Section 2.4 Applications of Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

22. We begin by identifying values for a and ω .

d = 10 cos 2π t, a = 10 and ω = 2π

a. The maximum displacement from the rest position is the amplitude. Because a = 10, the maximum displacement is 10 inches.

25. We begin by identifying values for a and ω.

d = 1

sin 2t, a = 1

and ω = 2 2 2

a. The maximum displacement from the rest

position is the amplitude.

1

b. The frequency, f, is f = ω

= 2π

= 1 . Because a = ,

2 the maximum displacement is

2π 2πThe frequency is 1 inch per second.

c. The time required for one cycle is the period.

period = 2π

= 2π

= 1

1 inch.

2

b. The frequency, f, is

ω 2π f = ω

= 2

= 1

≈ 0.32.The time required for one cycle is 1 second.

23. We begin by identifying values for a and ω.

d = −6 cos 2π t, a = −6 and ω = 2π

a. The maximum displacement from the rest

2π 2π π The frequency is approximately 0.32 cycle per

second.

c. The time required for one cycle is the period.

period = 2π

= 2π

= π ≈ 3.14position is the amplitude. ω 2Because a = –6, the maximum displacement is 6 inches.

b. The frequency, f, is

f = ω

= 2π

= 1.

The time required for one cycle is approximately 3.14 seconds.

26. We begin by identifying values for a and ω .

1 12π 2π d = sin 2t, a = and ω = 2The frequency is 1 inch per second.

c. The time required for one cycle is the period.

period = 2π

= 2π

= 1

3 3 a. The maximum displacement from the rest

position is the amplitude.

ω 2π 1The time required for one cycle is 1 second.

Because a = , the maximum displacement is 3

1

24. We begin by identifying values for a and ω .

d = −8cos π

t, a = −8 and ω = π

inch . 3

ω

2 1

2 2 b. The frequency, f, is f = = = ≈ 0.32 .2π 2π π

a. The maximum displacement from the rest

position is the amplitude.

Because a = –8, the maximum displacement is 8 inches.

The frequency is approximately 0.32 cycle per second.

c. The time required for one cycle is the period.

period = 2π

= 2π

= π ≈ 3.14

ω π ω 2

b. The frequency, f, is f = = 2 = 1

. The time required for one cycle is2π 2π 4

approximately 3.14 seconds.

The frequency is 1

4

inch per second.

c. The time required for one cycle is the period.

period = 2π =

2π = 2π ⋅

2 = 4

ω π π 2

The time required for one cycle is 4 seconds.

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Section 2.4 Applications of Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

3

2

27. We begin by identifying values for a and ω.

33. x = 300

− 300

d = −5sin 2π

t, a = −5 and ω = 2π

3 3

tan 34° x ≈ 298

tan 64°

a. The maximum displacement from the rest position is the amplitude. Because a = –5, the maximum displacement is 5 inches.

b. The frequency, f, is

ω 2π

2π 1 1 f = = = ⋅ = .

34.

35.

x = 500

− 500

tan 20° tan 48° x ≈ 924

x = 400 tan 40° tan 20° tan 40° − tan 20°

x ≈ 257

2π 2π 3 2π 3

100 tan 43° tan 38°The frequency is

1

cycle per second. 36. x =

tan 43° − tan 38°3

c. The time required for one cycle is the period.

x ≈ 482

π period =

2π =

2π = 2π ⋅

3 = 3

37. d = 4 cos π t − 2 ω 2π 2π

3

The time required for one cycle is 3 seconds.

28. We begin by identifying values for a and ω .

d = −4sin 3π

t, a = −4 and ω = 3π

2 2

a. The maximum displacement from the rest

position is the amplitude.

Because a = –4, the maximum displacement is 4 inches.

b. The frequency, f, is

a. 4 in.

1

ω 3π

3π 1 3 f = = = ⋅ = .

b. in. per sec 2

2π 2π 2 2π 4

c. 2 secThe frequency is

3 cycle per second.

4

c. The time required for one cycle is the period.

d. 1 2

period = 2π

= 2π

= 2π ⋅ 2

= 4

ω 3π 3π 3

38.

d = 3cos π t +

π 2 2

29.

30.

31.

32.

The required time for one cycle is 4

seconds. 3

x = 500 tan 40° + 500 tan 25° x ≈ 653

x = 100 tan 20° + 100 tan 8° x ≈ 50

x = 600 tan 28° − 600 tan 25° x ≈ 39

x = 400 tan 40° − 400 tan 28° x ≈ 123

a. 3 in.

b. 1

in. per sec 2

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Section 2.4 Applications of Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

150 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 150

4 2

c. 2 sec 42. 30 yd ⋅

3 ft = 90 ft

1 yd

d. − 1

2

Using a right triangle, we have a known angle, an

unknown opposite side, a, and a known adjacent

side. Therefore, use the tangent function.

39. d = −2sin π t

+ π

tan 38.7° = a

90a = 90 tan 38.7° ≈ 72

The height of the building is approximately 72 feet.

a. 2 in.

b. 1

in. per sec 8

c. 8 sec

d. −2

43. Using a right triangle, we have a known angle, a

known opposite side, and an unknown adjacent side,

a. Therefore, use the tangent function.

tan 23.7° = 305

a

a = 305

≈ 695 tan 23.7°

The ship is approximately 695 feet from the statue’s

base.

44. Using a right triangle, we have a known angle, a

known opposite side, and an unknown adjacent side,

a. Therefore, use the tangent function.

tan 22.3° = 200

a

a = 200

≈ 488

40. d 1

sin π t

− π

= − 2 4 2

a. 1

in. 2

b. 1

in. per sec 8

c. 8 sec

d. 2

41. Using a right triangle, we have a known angle, an

unknown opposite side, a, and a known adjacent side.

Therefore, use tangent function.

tan 21.3° = a 5280

a = 5280 tan 21.3° ≈ 2059 The height of the tower is approximately 2059 feet.

tan 22.3° The ship is about 488 feet offshore.

45. The angle of depression from the helicopter to point P

is equal to the angle of elevation from point P to the

helicopter. Using a right triangle, we have a known

angle, a known opposite side, and an unknown

adjacent side, d. Therefore, use the tangent function.

tan 36° = 1000

d

d = 1000

≈ 1376 tan 36°

The island is approximately 1376 feet off the coast. 46. The angle of depression from the helicopter to the stolen

car is equal to the angle of elevation from the stolen car to the helicopter. Using a right triangle, we have a known angle, a known opposite side, and an unknown adjacent side, d. Therefore, use the tangent function.

tan 72° = 800 d

d = 800

≈ 260 tan 72°

The stolen car is approximately 260 feet from a point

directly below the helicopter.

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23

40

47. Using a right triangle, we have an unknown angle, A,

a known opposite side, and a known hypotenuse.

Therefore, use the sine function.

sin A = 6 23

A = sin −1 6

≈ 15.1°

The ramp makes an angle of approximately 15.1°

with the ground.

48. Using a right triangle, we have an unknown angle, A,

a known opposite side, and a known adjacent side.

Therefore, use the tangent function.

cos 53° = b

150 b = 150 cos 53° ≈ 90

The boat has traveled approximately 90 miles north

and 120 miles east.

52. Using a right triangle, we have a known angle, a

known hypotenuse, and unknown sides. To find the

opposite side, a, use the sine function.

sin 64° = a 40

a = 40 sin 64° ≈ 36 To find the adjacent side, b, use the cosine function.

btan A =

250

40

cos 64° = 40

A = tan −1 250

≈ 80.9°

The angle of elevation of the sun is approximately 80.9°.

49. Using the two right triangles, we have a known angle,

an unknown opposite side, a in the smaller triangle, b

in the larger triangle, and a known adjacent side in

each triangle. Therefore, use the tangent function.

tan 19.2° = a 125

a = 125 tan 19.2° ≈ 43.5

tan 31.7° = b

125

b = 125 tan 31.7° ≈ 77.2 The balloon rises approximately 77.2 – 43.5 or 33.7 feet.

50. Using two right triangles, a smaller right triangle

corresponding to the smaller angle of elevation drawn

inside a larger right triangle corresponding to the

larger angle of elevation, we have a known angle, an

unknown opposite side, a in the smaller triangle, b in

the larger triangle, and a known adjacent side in each

triangle. Therefore, use the tangent function.

tan 53° = a 330

a = 330 tan 53° = 437.9

tan 63° = b

330

b = 330 tan 63° ≈ 647.7 The height of the flagpole is approximately 647.7 – 437.9, or 209.8 feet (or 209.7 feet).

51. Using a right triangle, we have a known angle, a

known hypotenuse, and unknown sides. To find the

opposite side, a, use the sine function.

sin 53° = a 150

a = 150 sin 53° ≈ 120 To find the adjacent side, b, use the cosine function.

b = 40 cos 64° ≈ 17.5 The boat has traveled about 17.5 mi south and 36 mi east.

53. The bearing from the fire to the second ranger is N

28° E. Using a right triangle, we have a known angle, a known opposite side, and an unknown adjacent side, b. Therefore, use the tangent function.

tan 28° = 7 b

b = 7

≈ 13.2 tan 28°

The first ranger is 13.2 miles from the fire, to the nearest tenth of a mile.

54. The bearing from the lighthouse to the second ship is

N 34° E. Using a right triangle, we have a known

angle, a known opposite side, and an unknown

adjacent side, b. Therefore, use the tangent function.

tan 34° = 9 b

b = 9

≈ 13.3 tan 34°

The first ship is about 13.3 miles from the lighthouse,

to the nearest tenth of a mile.

55. Using a right triangle, we have a known adjacent

side, a known opposite side, and an unknown angle,

A. Therefore, use the tangent function.

tan A = 1.5 2

A = tan 1.5

≈ 37° 2

We need the acute angle between the ray that runs

from your house through your location, and the

north-south line through your house. This angle

measures approximately 90° − 37° = 53°. This angle

is measured from the north side of the north-south

line and lies west of the north-south line. Thus, the

bearing from your house to you is N 53° W.

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152 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 152

9

7

56. Using a right triangle, we have a known adjacent

side, a known opposite side, and an unknown angle,

59. The frequency, f, is f = ω

, so 2π

A. Therefore, use the tangent function. 1 ω

tan A = 6

9

A = tan −1 6

≈ 34°

We need the acute angle between the ray that runs

from the ship through the harbor, and the north-south

line through the ship. This angle measures 90° − 34° = 56° . This angle is measured from the

= 2 2π

ω = 1

⋅ 2π = π 2

Because the amplitude is 6 feet, a = 6. Thus, the equation for the object’s simple harmonic motion is

d = 6sin π t.

north side of the north-south line and lies west of the

north-south line. Thus, the bearing from the ship to

the harbor is N 56° W. The ship should use a bearing

of N 56° W to sail directly to the harbor.

57. To find the jet’s bearing from the control tower, consider

60. The frequency, f, is

1 =

ω 4 2π

ω = 1

⋅ 2π = π

4 2

f = ω

, so 2π

a north-south line passing through the tower. The acute angle from this line to the ray on which the jet lies is

35° + θ . Because we are measuring the angle from the

north side of the line and the jet is east of the tower, the

jet’s bearing from the tower is N (35° + θ ) E. To find θ ,

use a right triangle and the tangent function.

Because the amplitude is 8 feet, a = 8. Thus, the equation for the object’s simple harmonic motion is

d = 8sin π

t . 2

ω

tan θ = 7 5

θ = tan −1 7

≈ 54.5°

5

61. The frequency, f, is

264 = ω 2π

f = , so 2π

Thus, 35° + θ = 35° + 54.5° = 89.5°.

The jet’s bearing from the control tower is N 89.5° E.

58. To find the ship’s bearing from the port, consider a

ω = 264 ⋅ 2π = 528π Thus, the equation for the tuning fork’s simple

harmonic motion is d = sin 528π t.

north-south line passing through the port. The acute

angle from this line to the ray on which the ship lies

is 40° + θ . Because we are measuring the angle from

the south side of the line and the ship is west of the

62. The frequency, f, is

98,100, 000 = ω

f = ω

, so 2π

port, the ship’s bearing from the port is

S (40° + θ ) W . To find θ , use a right triangle and

the tangent function.

tan θ = 11 7

θ = tan −1 11 ≈ 57.5°

ω = 98,100, 000 ⋅ 2π = 196, 200, 000π

Thus, the equation for the radio waves’ simple

harmonic motion is d = sin 196, 200, 000π t . 63. – 69. Answers may vary.

−0.1x

Thus, 40° + θ = 40° + 57.5 = 97.5° . Because this

angle is over 90° we subtract this angle from 180° to find the bearing from the north side of the north- south line. The bearing of the ship from the port is N 82.5° W.

70. y = 4e cos 2 x

3 complete oscillations occur.

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71. y = −6e−0.09 x

cos 2π x Using the transitive property we have

(75 + d ) tan 20° = d tan 40° 75 tan 20° + d tan 20° = d tan 40° d tan 40° − d tan 20° = 75 tan 20° d (tan 40° − tan 20°) = 75 tan 20°

d = 75 tan 20° tan 40° − tan 20°

10 complete oscillations occur.

72. makes sense

Thus, h = d tan 40° 75 tan 20°

= tan 40° ≈ 48 tan 40° − tan 20°

The height of the building is approximately 48 feet.

78. Answers may vary.

73. does not make sense; Explanations will vary.

Sample explanation: When using bearings, the angle

must be less than 90 .

79.

r = x2 + y2

r = (−3)2 + 42 =

9 + 16 =

25 = 5

74. does not make sense; Explanations will vary.

Sample explanation: When using bearings, north

Now that we know x, y, and r, we can find the six

trigonometric functions of θ .

sin θ = y

= 4

and south are listed before east and west. r 5

cosθ = x

= −3

= − 3

75. does not make sense; Explanations will vary. Sample explanation: Frequency and Period are inverses of each other. If the period is 10 seconds

r 5 5

tan θ = y

= 4

= − 4

x −3 3

cscθ = r

= 5

then the frequency is 1

10

= 0.1 oscillations per

y 4 r 5 5

second.

76. Using the right triangle, we have a known angle, an unknown opposite side, r, and an unknown hypotenuse, r + 112. Because both sides are in terms of the variable r,

secθ = = = − x −3 3

cot θ = x

= −3

= − 3

y 4 4

we can find r by using the sine function.

sin 76.6° = r

80. 13π

− 4π = 13π

− 12π

= π

3 3 3 3 lies in quadrant I.

r + 112

sin 76.6°(r + 112) = r Because the tangent is positive in quadrant I, π

r sin 76.6° + 112 sin 76.6° = r r − r sin 76.6° = 112 sin 76.6°

r (1 − sin 76.6°) = 112 sin 76.6°

tan = 3 . 3

o 46r =

112 sin 76.6° ≈ 4002 81. sin 26 =

c1 − sin 76.6° The Earth's radius is approximately 4002 miles.

c sin 26o = 46 46

77. Let d be the adjacent side to the 40° angle. Using the

right triangles, we have a known angle and unknown

sides in both triangles. Use the tangent function.

82.

c = ≈ 105 yd sin 26

o

sec x cot x = 1

⋅ cos x

= 1

or csc x

tan 20° = h 75 + d

cos x sin x sin x

h = (75 + d ) tan 20°

83. tan x csc x cos x = sin x

⋅ 1

⋅ cos x

= 1

Also, tan 40° = h d

cos x sin x 1

h = d tan 40°

84. sec x + tan x = 1

+ sin x

= 1 + sin x

cos x cos x cos x

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Section 2.4 Applications of Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

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x coordinates

0 (0, –2)

π

4

π , 0 4

π

2

π , 2 2

4

3π , 0 4

π (π , − 2)

x coordinates

0 (0, 0)

π

8

π , 3

8

π

4

π , 0

4

8

3π , − 3

8

π

2

(2π , 0)

Chapter 2 Review Exercises 2. The equation y = −2 cos 2x is of the form

1. The equation

y = 3sin 4x is of the form

y = A sin Bx y = A cos Bx with A = –2 and B = 2. The amplitude

2π =

2π = π

Thewith A = 3 and B = 4. The amplitude is A = 3 = 3. is A = −2 = 2. The period is . B 2

The period is 2π

= 2π

= π

. The quarter-period is quarter-period is

π . The cycle begins at x = 0. Add

B 4 2 π

π 1 π 2 = ⋅ = . The cycle begins at x = 0. Add 4 2 4 8

quarter-periods to generate x-values for the key

points.

4 quarter-periods to generate x-values for the key points.

x = 0

π πx = 0

x = 0 + π

= π

8 8

x = π

+ π

= π

8 8 4

x = π

+ π

= 3π

4 8 8

x = 3π

+ π

= π

8 8 2

Evaluate the function at each value of x.

x = 0 + = 4 4

x = π

+ π

= π

4 4 2

x = π

+ π

= 3π

2 4 4

x = 3π

+ π

= π 4 4

Evaluate the function at each value of x.

Connect the five key points with a smooth curve

and graph one complete cycle of the given function.

Connect the five key points with a smooth curve

and graph one complete cycle of the given function.

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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

x

coordinates

0

(0, 2)

π

(π , 0)

(2π , − 2)

(3π , 0)

(4π , 2)

x

coordinates

0

(0, 0)

3

2

3 , 1

2 2

3

(3, 0)

9

2

9 , −

1

6

(6, 0)

3. The equation y = 2 cos 1

x 2

is of the form

4. The equation y = 1

sin π

x is of the form 2 3

y = A cos Bx with A = 2 and B = 1

. The amplitude 2

y = A sin Bx with A = 1 2

and B = π

. The amplitude 3

2π 2π 1 1 2π =

2π =

π ⋅

3 =is A = 2 = 2. The period is

B =

1 2

= 2π ⋅ 2 = 4π . is A = = . The period is 2 2 B π

3

2 6. π

The quarter-period is 4π

= π . The cycle begins at 4

x = 0. Add quarter-periods to generate x-values for the key points. x = 0

x = 0 + π = π

x = π + π = 2π

x = 2π + π = 3π

x = 3π + π = 4π Evaluate the function at each value of x.

The quarter-period is 6

= 3

. The cycle begins at 4 2

x = 0. Add quarter-periods to generate x-values for the key points.

x = 0

x = 0 + 3

= 3

2 2

x = 3

+ 3

= 3 2 2

x = 3 + 3

= 9

2 2

x = 9

+ 3

= 6 2 2

Evaluate the function at each value of x.

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

2 2

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

x coordinates

0 (0, 0)

1

2

1 , − 1

2

1 (1, 0)

2

2

2 (2, 0)

x

coordinates

0

(0, 3)

2

3π , 0 2

(3π , −3)

2

9π , 0

(6π , 3)

5. The equation y = − sin π x is of the form

6. The equation y = 3cos x

is of the form

y = A cos Bxy = A sin Bx with A = –1 and B = π . The amplitude 3

is A = −1 = 1. The period is 2π

= 2π

= 2. The with A = 3 and B =

1 . The amplitude is

A = 3 = 3.B π 3

quarter-period is 2

= 1

. The cycle begins at x = 0.

The period is 2π =

2π = 2π ⋅ 3 = 6π . The quarter-

4 2 Add quarter-periods to generate x-values for the key points.

period is

B

6π =

1 3

. The cycle begins at x = 0. Add

x = 0

x = 0 + 1

= 1

2 2

x = 1

+ 1

= 1 2 2

x = 1 + 1

= 3

2 2

x = 3

+ 1

= 2 2 2

4 2 quarter-periods to generate x-values for the key points.

x = 0

x = 0 + 3π

= 3π

2 2

x = 3π

+ 3π

= 3π 2 2

x = 3π + 3π

= 9π

2 2Evaluate the function at each value of x.

9π 3πx = + = 6π

2 2 Evaluate the function at each value of x.

3 3

, 1

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

2

Connect the five key points with a smooth curve and graph one complete cycle of the given function.

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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

x coordinates

π (π , 0)

2

2 , 2

2π (2π , 0)

2

5π , − 2

2

3π (3π , 0)

x coordinates

−π (−π , − 3)

π −

2 − π

, 0

2

0 (0, 3)

π

2

π , 0

2

π (π , − 3)

= −

7. The equation y = 2sin( x − π ) is of the form 8. y = −3cos( x + π ) = −3cos( x − (−π ))

y = A sin(Bx − C) with A = 2, B = 1, and C = π . The The equation y = −3cos( x − (−π )) is of the form

amplitude is A = 2 = 2. The period is y = A cos(Bx − C ) with A = –3, B = 1, and C = −π .

2π =

2π = 2π . The phase shift is

C = π

= π . The The amplitude is A = −3 = 3.

B 1 B 1

The period is 2π =

2π = 2π . The phase shift is

quarter-period is 2π

= π

. B 1

4 2 C =

−π = −π

The quarter-period is

2π = π

. TheThe cycle begins at x = π . Add quarter-periods to B 1 .

4 2generate x-values for the key points. x = π

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2

x = 2π + π

= 5π

2 2

x = 5π

+ π

= 3π 2 2

Evaluate the function at each value of x.

cycle begins at x = −π . Add quarter-periods to

generate x-values for the key points. x = −π

x = −π + π

= − π

2 2

x π

+ π

= 0 2 2

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

Evaluate the function at each value of x.

Connect the five key points with a smooth curve

and graph one complete cycle of the given function.

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

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158 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 158

Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

x

coordinates

− π 8

− π

, 3

π

8

π , 0

8

8

3π , −

3

8

5π , 0

8

8

7π ,

3

8 2

2

= −

9. y = 3

cos 2 x +

π =

3 cos

2 x −

− π Connect the five key points with a smooth curve and

2

4

2

4 graph one complete cycle of the given function.

The equation

y = 3

cos 2 x −

− π

is of2

4

the form

y = A cos(Bx − C ) with A = 3

, 2

B = 2, and C = − π

. The amplitude is A = 3 =

3 .

4 2 2

The period is 2π

= 2π

= π . The phase shift is

B 2 −π

C = 4 = −

π ⋅

1 = − π

. The quarter-period is π

. 5 π 5 π

B 2 4 2 8 4 10. y = sin 2 x + = sin 2x − −

The cycle begins at x π

. Add quarter-periods to 2 2 2 2

= − 5

π 8 The equation y = sin 2x − −

is ofgenerate x-values for the key points. 2 2

π x = −

8

the form

y = A sin(Bx − C ) with

π

A = 5

, 2

5 5π π π x = − + =

8 4 8 B = 2, and C = − . The amplitude is A = = .

2 2 2

x = π

+ π

= 3π

8 4 8

The period is

π

2π =

2π B 2

= π . The phase shift is

C − π 1 π π

= = − ⋅ = − . The quarter-period is .x =

3π + π

= 5π

8 4 8 B 2 2 2 4 4

x = 5π

+ π

= 7π

8 4 8

The cycle begins at x π

. Add quarter-periods to 4

generate x-values for the key points.

Evaluate the function at each value of x. π x = −

8 2

4

π π x = − + = 0

4 4

π π

8 2

x = 0 + = 4 4

x = π

+ π

= π

4 4 2

x = π

+ π

= 3π

2 4 4 Evaluate the function at each value of x.

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159 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 159

Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

x

coordinates

− π 4

(− π

, 0) 4

0 0,

5

2

π

4

π , 0

π

2

π , −

5 2 2

4

3π , 0

4

x

coordinates

9

(9, 0)

21

2

21 , − 3 2

12

(12, 0)

27

2

27 , 3

2

15

(15, 0)

4

x = 9

x = 9 + 3

= 21

2 2

x = 21

+ 3

= 12 2 2

x = 12 + 3

= 27

2 2

x = 27

+ 3

= 15 2 2

Evaluate the function at each value of x.

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

11. The equation

y = −3sin π

x − 3π

is of

Connect the five key points with a smooth curve and

graph one complete cycle of the given function.

3

the form

y = A sin(Bx − C ) with A = –3,

B = π

, and C = 3π . The amplitude is 3

A = −3 = 3.

The period is 2π =

2π = 2π ⋅

3 = 6. The phase shift

B π π 3

is C

= 3π

= 3π ⋅ 3

= 9. The quarter-period is 6

= 3

.

B π π 4 2 3

The cycle begins at x = 9. Add quarter-periods to

generate x-values for the key points.

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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

160 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 160

x coordinates

0 (0, 1)

π

4

π , 2

4

π

2

π , 1

2

4

3π , 0

π (π , 1)

x

coordinates

0

(0, 0)

2

3π , − 2

2

(3π , − 4)

2

9π , − 2

(6π , 0)

12. The graph of y = sin 2x + 1 is the graph of y = sin 2 x

13. The graph of y = 2 cos 1

x − 2 is the graph ofshifted one unit upward. The period for both

functions is 2π

= π . The quarter-period is π

. The

3

y = 2 cos 1

x shifted two units downward. The period2 4 3

cycle begins at x = 0. Add quarter-periods to generate x-values for the key points.

x = 0

for both functions is 2π 1 3

= 2π ⋅ 3 = 6π . The quarter-

x = 0 + π

= π

period is 6π

= 4

3π . The cycle begins at x = 0. Add

24 4

x = π

+ π

= π

4 4 2

x = π

+ π

= 3π

2 4 4

x = 3π

+ π

= π 4 4

quarter-periods to generate x-values for the key

points.

x = 0

x = 0 + 3π

= 3π

2 2

x = 3π

+ 3π

= 3π 2 2

Evaluate the function at each value of x.

x = 3π + 3π

= 9π

2 2

x = 9π

+ 3π

= 6π 2 2

Evaluate the function at each value of x.

4

By connecting the points with a smooth curve we obtain one period of the graph.

2

By connecting the points with a smooth curve we

obtain one period of the graph.

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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

161 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 161

12 12

14. Select several values of x over the interval.

x

0 π

4

π

2

4 π

4

2

4 2π

y1 = 3 sin x 0 2.1 3 2.1 0 −2.1 −3 −2.1 0

y2 = cos x 1 0.7 0 −0.7 −1 −0.7 0 0.7 1

y = 3 sin x + cos x 1 2.8 3 1.4 −1 −2.8 −3 −1.4 1

15. Select several values of x over the interval.

x

0 π

4

π

2

4 π

4

2

4 2π

y1 = sin x 0 0.7 1 0.7 0 −0.7 −1 −0.7 0

y = cos 1

x 2

2

1

0.9

0.7

0.4

0

−0.4

−0.7

−0.9

−1

y = sin x + cos 1

x 2

1

1.6

1.7

1.1

0

−1.1

−1.7

−1.6

−1

16. a. At midnight x = 0. Thus, y = 98.6 + 0.3 sin π

⋅ 0 − 11π

= 98.6 + 0.3 sin −

11π

12

≈ 98.6 + 0.3(−0.2588) ≈ 98.52

The body temperature is about 98.52°F.

b. period: 2π

= 2π

= 2π ⋅ 12

= 24 hoursB π π

12

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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

162 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 162

12 12

12 12

12

c. Solve the equation

π x −

11π = π

12 12 2

π x = π

+ 11π

= 6π

+ 11π

= 17π

12 2 12 12 12 12

x = 17π

⋅ 12

= 17

x = 11

x = 11 + 6 = 17

x = 17 + 6 = 23

x = 23 + 6 = 29

x = 29 + 6 = 3512 π

The body temperature is highest for x = 17.

y = 98.6 + 0.3 sin π

⋅17 − 11π

= 98.6 + 0.3 sin π

= 98.6 + 0.3 = 98.9 2

17 hours after midnight, which is 5 P.M., the body temperature is 98.9°F.

d. Solve the equation

π x −

11π =

3π 12 12 2

Evaluate the function at each value of x. The key points are (11, 98.6), (17, 98.9), (23, 98.6), (29, 98.3), (35, 98.6). Extend the pattern to the

left, and graph the function for 0 ≤ x ≤ 24.

17. Blue:

This is a sine wave with a period of 480.

π x =

3π +

11π =

18π +

11π =

29π 12 2 12 12 12 12

x = 29π

⋅ 12

= 29

Since the amplitude is 1,

B = 2π

= 2π

= π

period 480 240

A = 1.

12 π

The equation is y = sin π

x.The body temperature is lowest for x = 29.

y = 98.6 + 0.3 sin π

⋅ 29 − 11π

12 12

Red:

240

This is a sine wave with a period of 640.

= 98.6 + 0.3 sin 3π Since the amplitude is 1, A = 1.

2

2π π

= 98.6 + 0.3(−1) = 98.3°

B = = = period 640 320

29 hours after midnight or 5 hours after

midnight, at 5 A.M., the body temperature is 98.3°F.

The equation is y = sin π

x. 320

18. Solve the equations

e. The graph of y = 98.6 + 0.3 sin π

x − 11π

is

π 2 x = −

2 and 2 x =

π 2

π πof the form y = D + A sin(Bx − C ) with A = 0.3, − x = 2 x = 2

B = π

, C = 11π

, and D = 98.6. The amplitude 2 2

12 12

x π

x π

is A =

0.3 = 0.3. The period from part (b) = − =

4 4 Thus, two consecutive asymptotes occur at

is 24. The quarter-period is 24

= 6. The phase 4

π x = −

and x = π

.

C 11π

11π 12

shift is = = ⋅ = 11. The cycle

4 4

− π + π

B π 12

12 π x-intercept = 4 4 = 0

= 0 2 2

begins at x = 11. Add quarter-periods to generate x-values for the key points.

An x-intercept is 0 and the graph passes through (0, 0). Because the coefficient of the tangent is 4, the points on the graph midway between an x-intercept

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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

163 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 163

= −

− π

−2π

π

2 2

and the asymptotes have y-coordinates of –4 and 4. 20. Solve the equations

Use the two consecutive asymptotes. x π

and x + π = − π

and x + π = π

4 2 2

x = π

, to graph one full period of

y = 4 tan 2x

from π x = − x =

π − π

4 2 2

− π

to π

.

x 3π

x π

= − = − 4 4 2 2

Continue the pattern and extend the graph another full period to the right.

Thus, two consecutive asymptotes occur at

x 3π π

= − and x = − . 2 2

− 3π − π x-intercept = 2 2 = = −π

2 2An x-intercept is −π and the graph passes through

(−π , 0) . Because the coefficient of the tangent is 1,

the points on the graph midway between an x-

intercept and the asymptotes have y-coordinates of –1

and 1. Use the two consecutive asymptotes,

3π π19. Solve the equations x = − and x = − , to graph one full period of

π π x = −

and π x = π

2 2

3π π4 2 4 2 y = tan( x + π ) from − to − .

x π 4

x π 4

= − ⋅ = ⋅2 π

x = −2

2 π x = 2

Continue the pattern and extend the graph another full period to the right.

Thus, two consecutive asymptotes occur at

x = –2 and x = 2.

x-intercept = −2 + 2

= 0

= 0 2 2

An x-intercept is 0 and the graph passes through (0, 0). Because the coefficient of the tangent is –2, the points on the graph midway between an x-intercept and the asymptotes have y-coordinates of 2 and –2.

Use the two consecutive asymptotes, x = –2 and x =

2, to graph one full period of

y = −2 tan π

x from –2 21. Solve the equations

4 x − π

= − π

and x − π

= π

to 2. Continue the pattern and extend the graph 4 2 4 2

another full period to the right. x

π π x

π π= − + = +

2 4 2 4

x π

x 3π

= − = 4 4

Thus, two consecutive asymptotes occur at

x π 3π

= − and x = − . 4 4

− π − 3π π

x-intercept = 4 4 = 2 = 2 2 4

An x-intercept is π

and the graph passes through 4

π , 0 . Because the coefficient of the tangent is –1,

4

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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

164 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 164

= −

x

= −

the points on the graph midway between an x-

intercept and the asymptotes have y-coordinates of 1

23. Solve the equations

π πand –1. Use the two consecutive asymptotes,

x = 0 and 2

x = π 2

x π 3π

x = 0 x = π ⋅ 2

= − and x = , to graph one full period of 4 4 π

y = − tan x − π

from − π

to 3π

. Continue the x = 2

4

4 4

pattern and extend the graph another full period to the

right.

Thus, two consecutive asymptotes occur at x = 0 and x = 2.

x-intercept = 0 + 2

= 2

= 1 2 2

An x-intercept is 1 and the graph passes through (1, 0). Because the coefficient of the cotangent is

− 1

, the points on the graph midway between an x- 2

intercept and the asymptotes have y-coordinates of

− 1

and 1

. Use the two consecutive asymptotes, 2 2

22. Solve the equations x = 0 and x = 2, to graph one full period of

3x = 0 and 3x = π y 1

cot π

x from 0 to 2. Continue the pattern and 2 2

x = 0 x = π 3

extend the graph another full period to the right.

Thus, two consecutive asymptotes occur at

x = 0 and x = π

. 3

0 + π π π

x-intercept = 3 = 3 =

2 2 6

An x-intercept is π

and the graph passes through 6

π , 0 .

6

24. Solve the equations

Because the coefficient of the tangent is 2, the points x + π

= 0 and x + π

= πon the graph midway between an x-intercept and the 2 2

asymptotes have y-coordinates of 2 and –2. Use the x = 0 −

π x = π − π

two consecutive asymptotes, x = 0 and x = π

, to 2 2

3 x

π x

π

graph one full period of

y = 2 cot 3x

from 0 to π

. 3

= − = 2 2

Thus, two consecutive asymptotes occur at

Continue the pattern and extend the graph another full period to the right.

π π = − and x = .

2 2

− π + π 0

x-intercept = 2 2 = = 0 2 2

An x-intercept is 0 and the graph passes through (0, 0). Because the coefficient of the cotangent is 2, the points on the graph midway between an x-intercept and the asymptotes have y-coordinates of 2 and –2.

Use the two consecutive asymptotes, x π

and2

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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

165 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 165

, 0 , , − 3 ,

x = π

, to graph one full period of

y = 2 cot x + π 26. Graph the reciprocal sine function, y = −2sin π x .

2

2

The equation is of the form

y = A sin Bx with

from − π

to π

. Continue the pattern and extend the

A = –2 and B = π .

2 2 amplitude: A = −2 = 2

graph another full period to the right.

period: 2π

= 2π

= 2 B π

Use quarter-periods, 2

= 1

, to find 4 2

x-values for the five key points. Starting with

x = 0, the x-values are 0, 1

, 2

1, 3

, 2 . Evaluating the 2

function at each value of x, the key points are (0, 0),

1 , − 2

, (1, 0) ,

3 , 2

, (2, 0) . Use these key points

2

2

25. Graph the reciprocal cosine function,

y = 3cos 2π x . to graph y = −2sin π x from 0 to 2. Extend the graph

The equation is of the form

and B = 2π .

y = A cos Bx

with A = 3 one cycle to the right. Use the graph to obtain the graph of the reciprocal function. Draw vertical asymptotes through the x-intercepts, and use them as

amplitude: A = 3 = 3 guides to graph y = −2 csc π x.

period: 2π

= 2π

= 1 B 2π

Use quarter-periods, 1

, to find x-values for the five 4

key points. Starting with x = 0, the x-values are 0, 1

, 4

1 ,

3 , 1 . Evaluating the function at each value of x,

2 4 the key points are (0, 3),

1 1 3 , 0 , (1, 3) .

27. Graph the reciprocal cosine function,

y = 3cos( x + π ) . The equation is of the form 4

2

4 y = A cos(Bx − C ) with A = 3, B = 1, and C = −π .

Use these key points to graph y = 3cos 2π x from 0 amplitude: A = 3 = 3

to 1. Extend the graph one cycle to the right. Use the 2π =

2π = π

graph to obtain the graph of the reciprocal function.

Draw vertical asymptotes through the x-intercepts,

period: B

2 1

C −πand use them as guides to graph y = 3sec 2π x. phase shift: = = −π

B 1

Use quarter-periods, 2π

= π

, to find 4 2

x-values for the five key points. Starting with

x = −π , the x-values are −π , − π

, 2

0, π

, π . 2

Evaluating the function at each value of x, the key

points are (−π , 3) , − π

, 0 , (0, − 3) ,

2

π , 0 , (π , 3) . Use these key points to graph

2

y = 3cos( x + π ) from −π

to π . Extend the graph

one cycle to the right. Use the graph to obtain the

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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

166 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 166

3

2 2

= −

= −

graph of the reciprocal function. Draw vertical

asymptotes through the x-intercepts, and use them as 29. Let θ = sin

−1 1 , then sin θ = 1 .

guides to graph

y = 3 sec( x + π ).

The only angle in the interval − π

, π

that satisfies 2 2

sin θ = 1 is π

. Thus θ = π

, or sin −1 1 = π

.

2 2 2

28. Graph the reciprocal sine function,

y = 5

sin( x − π ) .

30. Let θ = cos−1

1 , then cosθ = 1 .

The only angle in the interval [0, π ] that satisfies

cosθ = 1 is 0 . Thus θ = 0 , or cos−1

1 = 0 .

31. Let θ = tan −1

1 , then tan θ = 1 .

The only angle in the interval − π

, π

that satisfies2 2 2

The equation is of the form

y = A sin(Bx − C ) with

π π −1 πtan θ = 1 is

. Thus θ = , or tan 1 = .A =

5 ,

2 B = 1, and C = π . 4 4 4

amplitude: A = 5

= 5 32. Let θ = sin −1 −

3 , then sin θ = − .

2 2

period: 2π

= 2π

= 2π

2 2

π π B 1 The only angle in the interval − , that satisfies

phase shift: C

= π

= π

3 π πB 1 sin θ = − is − . Thus θ = − , or

Use quarter-periods, 2π

= π

, to find 2 3 3

4 2 sin −1 − 3

= − π

.x-values for the five key points. Starting with x = π ,

the x-values are π , 3π

, 2π , 5π

, 3π . Evaluating the

2 3

2 2 33. Let θ = cos−1

− 1

, then cosθ 1

function at each value of x, the key points

2

2 .

are (π , 0), 3π

, 5

, (2π , 0) , 5π

, − 5

,

(3π , 0).

The only angle in the interval [0, π ] that satisfies 2 2

2 2

Use these key points to graph

y = 5

sin( x − π ) from

cosθ 1

is 2

2π . Thus θ =

2π , or

3 3

2 cos−1 −

1 =

2π .

π to 3π . Extend the graph one cycle to the right.

2

3 Use the graph to obtain the graph of the reciprocal function. Draw vertical asymptotes through the x-

−1

3 3intercepts, and use them as guides to graph 34. Let θ = tan

− 3

, then tan θ = − 3

.

y = 5

csc( x − π ).

π π 2 The only angle in the interval − , that satisfies

tan θ = − 3 3

is − π

. 6

2 2

Thus θ π

, or tan −1 − 3

= − π

.= −

6 3 6

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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

167 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 167

cos sin −1 2 = cos

π =

2 .

2

−1

Thus csc tan −1 3 = csc

π = 2 .

= −

35. Let θ = sin −1 2

, then sin θ = 2

. The only angle 40. Let θ = tan −1 3

, then tan θ = 3

2 2 4 4

in the interval − π

, π

that satisfies sin θ = 2

is

Because tan θ is positive, θ is in the first quadrant.

2 2 2

π .

4

Thus,

2 4 2

2 2 2

36. Let θ = cos−1 0 , then cosθ = 0 . The only angle in

the interval [0, π ] that satisfies cosθ = 0 is π

. 2

r = x + y

r 2 = 42 + 32

2

Thus, sin (cos−1 0) = sin π

= 1 . r = 25

r = 5

37. Let θ = sin −1 −

1 , then sin θ

1 . The only

cos

tan

−1 3 = cosθ =

x =

4= −

2

2

4 r 5

angle in the interval − π

, π

that satisfies 3 3 2 2

41. Let θ = cos−1 , then cosθ = . 5 5

sin θ 1

2

is − π

. 6

Because cosθ is positive, θ is in the first quadrant.

Thus, tan sin −1 −

1 = tan

− π

= − 3

. 2

6

3

38. Let θ = cos−1 −

3 , then cosθ = −

3 . The only

2 2 2

2 2 x + y = r

angle in the interval [0, π ] that satisfies

32 + y2 = 52

cosθ = − 3 2

is 5π

. 6

y2 = 25 − 9 = 16

3 5π 3

Thus, tan cos − = tan = − .

y = 16 = 4

2 6 3 sin

cos−1 3

= sin θ = y

= 4

5

r 5

39. Let θ = tan −1 3 , then tan θ =

3 .

3 3

The only angle in the interval − π

, π

that satisfies

2 2

tan θ = 3 3

is π

. 6

3 6

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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

168 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 168

.

.

= −

= −

= −

42. Let θ = sin −1 −

3 , then sin θ

3 44. Let θ = tan

−1 − 1

, 5

5

3

Because sin θ is negative, θ is in

quadrant IV.

Because tan θ is negative, θ is in quadrant IV and

x = 3 and y = −1 .

r 2 = x2 + y2

r 2 = 32 + (−1)

2

r 2 = 10

x2 + (−3)

2 = 52

r = 10

1

4

y −1 10sin tan

− −

= sin θ = = = −

x2 + y2 = r 2 5 r 10 10

x2 = 25 − 9 = 16

45.

x = π

, x is in − π

, π

, so sin −1

sin π

= π

3 2 2

3

3

x = 16 = 4

tan sin −1 −

3 = tan θ =

y = −

3

46.

x = 2π

, x is not in − π

, π

. x is in the domain of 5

x 4

3 2 2

sin x , so

43. Let θ = cos−1

− 4

, then cosθ 4

−1 2π −1 3 π

5

5 sin

sin

3 = sin

2 =

3

Because cosθ is negative, θ is in

quadrant II.

47.

sin −1

cos 2π

= sin −1 −

1

3

2

Let θ = sin −1 −

1 , then sin θ

1 . The only

2

2

angle in the interval − π

, π

that satisfies

2 2

sin θ

1 is − π

. Thus, θ

π , or

x2 + y2 = r

2

= − = − 2 6 6

(−4)2 + y2 = 52 sin

−1 cos

2π = sin

−1 − 1

= − π

. 3

2

6

y2 = 25 − 16 = 9

y = 9 = 3

48. Let θ = tan −1 x , then tan θ =

x .

2 2 Use the right triangle to find the exact value.

tan cos−1 −

4 = tan θ = −

3

5

4

r 2 = x2 + 22

r 2 = x2 + y2

r = x2 + 4

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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

169 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 169

7

3.6

Use the right triangle to write the algebraic

expression.

2 x2 + 4

We have a known angle, a known opposite side, and

an unknown hypotenuse. Use the sine function.

sin 37.4° = 6

cos tan −1 x

= cosθ = 2

= 2

x2

4

x2 + 4

c +

49. Let θ = sin −1 1 , then sin θ =

1 .

c =

6 ≈ 9.88

sin 37.4°

x x In summary, A = 52.6°, a ≈ 7.85 , and c ≈ 9.88 .

Use the Pythagorean theorem to find the third side, b.

12 + b2 = x2

b2 = x2 − 1

b = x2 − 1

52. Find the measure of angle A. We have a known

hypotenuse, a known opposite side, and an unknown

angle. Use the sine function.

sin A = 2 7

A = sin −1 2

≈ 16.6°

Find the measure of angle B. Because C = 90° ,

A + B = 90° . Thus, B = 90° − A ≈ 90° −16.6° = 73.4°

We have a known hypotenuse, a known opposite side,

and an unknown adjacent side. Use the Pythagorean

theorem.Use the right triangle to write the algebraic expression. a

2 + b2 = c2

−1 1 x x x2 − 1 2

2 + b2 = 72

sec sin x

= secθ = 2

= 2 x − 1 x − 1

b2 = 72 − 22 = 45

50. Find the measure of angle B. Because C = 90° ,

b = 45 ≈ 6.71A + B = 90° . Thus, B = 90° − A = 90º −22.3° = 67.7°

We have a known angle, a known hypotenuse, and an In summary, A ≈ 16.6°, B ≈ 73.4° , and b ≈ 6.71 .

unknown opposite side. Use the sine function.

sin 22.3° = a 10

53. Find the measure of angle A. We have a known

opposite side, a known adjacent side, and an

unknown angle. Use the tangent function.

1.4a = 10 sin 22.3° ≈ 3.79

We have a known angle, a known hypotenuse, and an

tan A =

3.6

unknown adjacent side. Use the cosine function.

cos 22.3° = b 10

b = 10 cos 22.3° ≈ 9.25

In summary, B = 67.7°, a ≈ 3.79 , and b ≈ 9.25 .

51. Find the measure of angle A. Because C = 90° ,

A = tan −1 1.4

≈ 21.3°

Find the measure of angle B. Because C = 90° ,

A + B = 90° . Thus, B = 90° − A ≈ 90° − 21.3° = 68.7°

We have a known opposite side, a known adjacent

side, and an unknown hypotenuse.

Use the Pythagorean theorem.

A + B = 90° . Thus, A = 90° − B = 90° − 37.4° = 52.6° c2 = a2 + b2 = (1.4)2 + (3.6)2 = 14.92We have a known angle, a known opposite side, and an unknown adjacent side. Use the tangent function.

c = 14.92 ≈ 3.86

tan 37.4° = 6 a

a = 6

≈ 7.85 tan 37.4°

In summary, A ≈ 21.3°, B ≈ 68.7° , and c ≈ 3.86 .

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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

170 170

54. Using a right triangle, we have a known angle, an

unknown opposite side, h, and a known adjacent side.

Therefore, use the tangent function.

tan 25.6° = h 80

h = 80 tan 25.6°

≈ 38.3

The building is about 38 feet high.

55. Using a right triangle, we have a known angle, an

unknown opposite side, h, and a known adjacent side.

Therefore, use the tangent function.

tan 40° = h 60

h = 60 tan 40° ≈ 50 yd

The second building is 50 yds taller than the first.

Total height = 40 + 50 = 90 yd .

56. Using two right triangles, a smaller right triangle

corresponding to the smaller angle of elevation drawn

inside a larger right triangle corresponding to the

larger angle of elevation, we have a known angle, a

known opposite side, and an unknown adjacent side,

d, in the smaller triangle. Therefore, use the tangent

function.

59. Using a right triangle, we have a known angle, a

known adjacent side, and an unknown opposite side,

d. Therefore, use the tangent function.

tan 64° = d 12

d = 12 tan 64° ≈ 24.6

The ship is about 24.6 miles from the lighthouse.

60.

a. Using the figure, B = 58° + 32° = 90° Thus, use the Pythagorean Theorem to find the

distance from city A to city C.

8502 + 960

2 = b2

b2 = 722, 500 + 921, 600

b2 = 1, 644,100

tan 68 125

b = 1, 644,100 ≈ 1282.2° = d

The distance from city A to city B is about

d = 125

≈ 50.5 tan 68°

We now have a known angle, a known adjacent side,

and an unknown opposite side, h, in the larger

triangle. Again, use the tangent function.

tan 71° = h 50.5

h = 50.5 tan 71° ≈ 146.7

The height of the antenna is 146.7 − 125 , or 21.7 ft,

to the nearest tenth of a foot.

57. We need the acute angle between ray OA and the

north-south line through O. This angle measures 90° − 55° = 35° . This angle measured from the north

61.

1282.2 miles.

b. Using the figure,

tan A = opposite

= 960

≈ 1.1294 adjacent 850

A ≈ tan −1

(1.1294) ≈ 48°

180° − 58° − 48° = 74°

The bearing from city A to city C is S74°E.

d = 20 cos π

t 4

a = 20 and ω = π 4

a. maximum displacement:side of the north-south line and lies east of the north- south line. Thus the bearing from O to A is N35°E.

a = 20 = 20 cm

ω π

π

58. We need the acute angle between ray OA and the b. f = = 4 = ⋅ 1

= 1

north-south line through O. This angle measures 90° − 55° = 35° . This angle measured from the south

side of the north-south line and lies west of the north-

2π 2π

frequency: 1 8

4 2π 8

cm per second

south line. Thus the bearing from O to A is S35°W.

c. period: 2π =

ω π 4

= 2π ⋅ 4

= 8 π

The time required for one cycle is 8 seconds.

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171 171

Chapter 2 Test Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

x coordinates

0 (0, 0)

π

4

π , 3

π

2

π 2

, 0

4

3π , − 3 4

π (π , 0)

= −

62. d = 1

sin 4t 2

Chapter 2 Test

1. The equation

y = 3sin 2x is of the form

y = A sin Bx

a = 1

and ω = 4 with A = 3 and B = 2. The amplitude is A = 3 = 3.2

2π 2π πThe period is = = π . The quarter-period is .

a. maximum displacement: B 2 4

a = 1 =

1 cm

The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points.

2 2

b. f = ω

= 4

= 2

≈ 0.64

x = 0

x = 0 + π

= π

2π 2π π 4 4frequency: 0.64 cm per second

c. period: 2π

= 2π

= π

≈ 1.57 ω 4 2

x = π

+ π

= π

4 4 2

π π 3πThe time required for one cycle is about 1.57

seconds.

63. Because the distance of the object from the rest position at t = 0 is a maximum, use the form

x = + = 2 4 4

x = 3π

+ π

= π 4 4

Evaluate the function at each value of x.

d = a cos ωt . The period is 2π ω

so,

2 = 2π

ω

ω = 2π

= π 2

Because the amplitude is 30 inches,

4

a = 30 .

because the object starts below its rest position

a = −30 . the equation for the object’s simple

harmonic motion is d = −30 cos π t .

64. Because the distance of the object from the rest

position at t = 0 is 0, use the form d = a sin ωt . The

period is 2π

so ω

Connect the five key points with a smooth curve

and graph one complete cycle of the given function.

5 = 2π

ω

ω = 2π

5

Because the amplitude is 1 4

inch,

a = 1

. a is 4

negative since the object begins pulled down. The equation for the object’s simple harmonic motion is

d 1

sin 2π

t. 4 5

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172 172

Chapter 2 Test Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

x coordinates

π

2

π , − 2

π (π , 0)

2

3π , 2

2

2π (2π , 0)

2

5π , −2

= −

2 2

2 2

2. The equation

y = −2 cos x − π

is of the form 3. Solve the equations

2

x π x π = − and =y = A cos(Bx − C ) with A = –2, B = 1, and 2 2 2 2

π πC =

π . The amplitude is

2 A = −2 = 2.

x = − ⋅ 2 2

x = −π

x = ⋅ 2 2

x = πThe period is

2π =

2π = 2π . The phase shift is

Thus, two consecutive asymptotes occur atB 1

x = −π and x = π . π

C = 2 =

π . The quarter-period is

2π = π

. x-intercept = −π + π

= 0

= 0B 1 2 4 2 2 2

The cycle begins at x = π

. Add quarter-periods to 2

generate x-values for the key points.

x = π

An x-intercept is 0 and the graph passes through (0,

0). Because the coefficient of the tangent is 2, the

points on the graph midway between an x-intercept

and the asymptotes have y-coordinates of –2 and 2.

2

x = π

+ π

= π

Use the two consecutive asymptotes, x = −π x = π , to graph one

x

and

2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2

x = 2π + π

= 5π

2 2

Evaluate the function at each value of x.

full period of y = 2 tan 2

from −π to π .

4. Graph the reciprocal sine function, y 1

sin π x. 2

2 The equation is of the form y = A sin Bx with A =

− 1

and B = π . 2

amplitude:

A = − 1

= 1

2 2

period: 2π

= 2π

= 2 B π

2

2 1

Connect the five key points with a smooth curve and Use quarter-periods, = , to find x-values for the

4 2

graph one complete cycle of the given function. five key points. Starting with x = 0, the

x-values are 0, 1

, 1, 3

, 2. Evaluating the function at 2 2

each value of x, the key points are

(0, 3), 1

, − 1

, (1, 0) , 3

, 1

, (2, 0) .

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173 173

Chapter 2 Test Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

= −

= −

. = −

Use these key points to graph y 1

sin π x 2

from 0 to 2. Use the graph to obtain the graph of the reciprocal function.

Draw vertical asymptotes through the x-intercepts, and use them as guides to graph

5. Select several values of x over the interval.

y 1

csc π x. 2

x

0 π

4

π

2

4 π

4

2

4 2π

y = 1

sin x 1

2

0

0.4

0.5

0.4

0

−0.4

−0.5

−0.4

0

y2 = 2 cos x 2 1.4 0 −1.4 −2 −1.4 0 1.4 2

y = 1

sin x + 2 cos x 2

2

1.8

0.5

−1.1

−2

−1.8

−0.5

1.1

2

6. Let θ = cos−1

− 1

, then cosθ 1

2

2

Because cosθ is negative, θ is in quadrant II.

x2 + y2 = r

2

(−1)2 + y2 = 22

y2 = 4 − 1 = 3

y = 3

tan cos−1 −

1 = tan θ =

y =

3 = − 3

2

x −1

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