Page 1

Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions Connect the five points with a smooth curve and graph one complete cycle of the given function with the graph of y = sin x .

Section 2.1 Checkpoint Exercises 1.

The equation y = 3sin x is of the form

y = A sin x with A = 3. Thus, the amplitude is

A = 3 = 3 The period for both y = 3sin x and y = sin x is 2π . We find the three x–intercepts, the maximum point, and the minimum point on the interval [0, 2π ] by dividing the period, 2π , by 4, period 2π π = = , then by adding quarter-periods to 4 4 2 generate x-values for each of the key points. The five x-values are x=0

x = 0+ x=

π 2

π

+

x =π +

=

2

π 2

π 2

π 2

=π =

3π 2

3π π + = 2π 2 2 Evaluate the function at each value of x. x

y = 3sin x

0

y = 3sin 0 = 3 ⋅ 0 = 0

56

π

coordinates (0, 0)

= 3 ⋅1 = 3

π   2 , 3  

π

y = 3sin x = 3 ⋅ 0 = 0

(π , 0)

3π 2

y = 3sin

y = 3sin 2π = 3 ⋅ 0 = 0

2

y = 3sin

1 The equation y = − sin x is of the form y = A sin x 2 1 with A = − . Thus, the amplitude is 2 1 1 1 A = − = . The period for both y = − sin x 2 2 2 and y = sin x is 2π .

x=

π

2.

2

3π 2 = 3(−1) = −3

 3π   2 , −3   

Find the x–values for the five key points by dividing period 2π π = = , then by the period, 2π , by 4, 4 4 2 adding quarter- periods. The five x-values are x=0

x = 0+ x=

π 2

π

+

x =π +

2

π 2

π

=

π 2

=π =

3π 2

2 3π π x= + = 2π 2 2 Evaluate the function at each value of x.

(2π , 0)

Copyright © 2014 Pearson Education, Inc.


Section 2.1 Graphs of Sine and Cosine Functions

x 0

π 2

π

3π 2

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

coordinates (0, 0)

x = 0 +π = π 1 π  2 , − 2  

(π , 0)

x = π + π = 2π x = 2π + π = 3π x = 3π + π = 4π Evaluate the function at each value of x. x 0

 3π 1   2 , 2  

(2π , 0)

1 x is of the form 2 1 y = A sin Bx with A = 2 and B = . 2 The amplitude is A = 2 = 2 .

y = 2 sin

1 x 2

1  y = 2 sin  ⋅ 0  2  = 2 sin 0

coordinates (0, 0)

= 2⋅0 = 0

π

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.

3.

Find the x–values for the five key points by dividing period 4π = = π , then by the period, 4π , by 4, 4 4 adding quarter-periods. The five x-values are x=0

1  y = 2 sin  ⋅ π  2  = 2 sin

π 2

(π , 2)

= 2 ⋅1 = 2

1  y = 2sin  ⋅ 2π  2  = 2sin π = 2 ⋅ 0 = 0

1  y = 2 sin  ⋅ 3π  2  3π = 2 sin 2 = 2 ⋅ ( −1) = −2

1  y = 2sin  ⋅ 4π  2   = 2sin 2π = 2 ⋅ 0 = 0

(2π , 0)

(3π , − 2)

(4π , 0)

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.

The equation y = 2 sin

The period is

2π 2π = 1 = 4π . B 2

Copyright © 2014 Pearson Education, Inc.

57


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

4.

π  The equation y = 3sin  2 x −  is of the form 3  y = A sin( Bx − C ) with A = 3, B = 2, and C = The amplitude is

π 3

  11π π   11π −   , − 3 y = 3sin  2 ⋅   12 3   12 9π 3π = 3sin = 3sin 6 2 = 3 ( −1) = −3

7π 6

 7π π  − y = 3sin  2 ⋅ 6 3   6π = 3sin = 3sin 2π 3 = 3⋅ 0 = 0

.

A = 3 = 3.

2π 2π = =π . B 2 π C 3 π 1 π The phase shift is = = ⋅ = . B 2 3 2 6 Find the x-values for the five key points by dividing period π the period, π , by 4, = , then by adding 4 4 quarter-periods to the value of x where the cycle The period is

begins, x =

11π 12

 7π   6 , 0  

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

π

. 6 The five x-values are

x=

π 6

2π 3π 5π + = 6 4 12 12 12 5π π 5π 3π 8π 2π + = + = = x= 12 4 12 12 12 3 2π π 8π 3π 11π + = + = x= 3 4 12 12 12 11π π 11π 3π 14π 7π + = + = = x= 12 4 12 12 12 6 Evaluate the function at each value of x. x=

π

x

π

π

=

π  y = 3sin  2 x −  3 

coordinates

 π π y = 3sin  2 ⋅ −   6 3 = 3sin 0 = 3 ⋅ 0 = 0

π   6 , 0  

5π 12

 5π π  −  y = 3sin  2 ⋅  12 3  π 3π = 3sin = 3sin 6 2 = 3 ⋅1 = 3

 5π   12 , 3   

2π 3

 2π π  − y = 3sin  2 ⋅ 3 3   3π = 3sin = 3sin π 3 = 3⋅ 0 = 0

 2π   3 , 0  

6

58

+

5.

The equation y = −4 cos π x is of the form y = A cos Bx with A = −4, and B = π . Thus, the amplitude is

A = −4 = 4 .

2π 2π = =2. π B Find the x-values for the five key points by dividing period 2 1 = = , then by adding the period, 2, by 4, 4 4 2 quarter periods to the value of x where the cycle begins. The five x-values are x=0 The period is

1 1 = 2 2 1 1 x = + =1 2 2 1 3 x = 1+ = 2 2 3 1 x= + =2 2 2 Evaluate the function at each value of x. x = 0+

Copyright © 2014 Pearson Education, Inc.


Section 2.1 Graphs of Sine and Cosine Functions

x

y = −4 cos π x

0

y = −4 cos (π ⋅ 0 )

coordinates (0, –4)

= −4 cos 0 = −4 1 2

 1 y = −4 cos  π ⋅   2 = −4 cos

1

π 2

1   2 , 0  

=0

y = −4 cos(π ⋅1)

The five x-values are

x=− x=− x=−

π 2

π 2

π 4

x = 0+ (1, 4)

= −4 cos π = 4 3 2

 3 y = −4 cos  π ⋅   2 3π = −4 cos =0 2

3 2, 

 0 

2

y = −4 cos(π ⋅ 2)

(2, –4)

= −4 cos 2π = −4 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.

x=

π

+ +

π

3 , B = 2 , and C = −π . 2

Thus, the amplitude is

A =

x

π 2

π 4

4

2π 2π = =π . B 2 π C −π The phase shift is = =− . B 2 2 Find the x-values for the five key points by dividing period π = , then by adding the period, π , by 4, 4 4 quarter-periods to the value of x where the cycle The period is

begins, x = −

π 2

4

=0

4

π 4

π

y=

3 cos(2 x + π ) 2

coordinates

3 cos(−π + π ) 2 3 3 = ⋅1 = 2 2

 π 3 − 2 , 2   

3  π  cos  − + π  2  2  3 = ⋅0 = 0 2

 π   − 4 , 0  

3 cos(0 + π ) 2 3 3 = ⋅ −1 = − 2 2

3   0, − 2   

3 π  cos  + π  2 2  3 = ⋅0 = 0 2

π   4 , 0  

y=

y=

y=

y=

3 π 3 cos(π + π )  2, 2 2   2 3 3 = ⋅1 = 2 2 Connect the five key points with a smooth curve and graph one complete cycle of the given graph.

π

3 3 = . 2 2

π

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

y=

A=

π

π

π 3 3 cos(2 x + π ) = cos(2 x − ( −π )) 2 2 The equation is of the form y = A cos( Bx − C ) with

=−

4

=

4

0

6.

π

y=

.

Copyright © 2014 Pearson Education, Inc.

59


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

7.

The graph of y = 2 cos x + 1 is the graph of y = 2 cos x shifted one unit upwards. The period for both functions is 2π . The quarter-period is

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

x=0 x = 0+ x=

π 2

π

+

x =π +

=

2

π

2

2

π

π

=

3π 2

2 3π π x= + = 2π 2 2 Evaluate the function at each value of x. y = 2 cos x + 1 coordinates x 0

π 2

π 3π 2

y = 2 cos 0 + 1 = 2 ⋅1 + 1 = 3

(0, 3)

π

π   2 , 1  

y = 2 cos π + 1 = 2 ⋅ (−1) + 1 = −1

(π , − 1)

3π +1 2 = 2⋅ 0 +1 = 1

 3π   2 , 1  

y = 2 cos

+1 2 = 2 ⋅ 0 +1 = 1

y = 2 cos

y = 2 cos 2π + 1 ( 2π , 3) = 2 ⋅1 + 1 = 3 By connecting the points with a smooth curve, we obtain one period of the graph.

8.

Select several values of x over the interval.

4 0 1.4 1 0.7

2 2 0

3π 5π 3π π 4 4 2 1.4 0 −1.4 −2 −0.7 −1 −0.7 0

y = 2sin x + cos x 1 2.1

2

0.7

x y1 = 2sin x y2 = cos x

60

0

π

π

−1 −2.1

−2

7π 2π 4 −1.4 0 0.7 1 −0.7

1

Copyright © 2014 Pearson Education, Inc.


Section 2.1 Graphs of Sine and Cosine Functions

9.

A, the amplitude, is the maximum value of y. The graph shows that this maximum value is 4, Thus, π 2π A = 4 . The period is , and period = . Thus, B 2 π 2π = 2 B π 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 2π months. The period is 12. Thus, 12 = B 12 B = 2π

2π π = 12 6 The graph shows that the starting point of the cycle is C , is 3. shifted from 0 to 3. The phase shift, B C 3= B C 3= π B=

6

π

=C 2 Substitute these values into y = A sin( Bx − C ) + D . The number of hours of daylight is modeled by π π y = 2 sin  x −  + 12 . 6 2  Concept and Vocabulary Check 2.1

A;

2.

3; 4π

3.

π ; 0;

4.

C ; right; left B

5. 6.

4

false

8.

true

9.

true

10. true Exercise Set 2.1 1.

The equation y = 4 sin x is of the form y = A sin x with A = 4. Thus, the amplitude is

A = 4 = 4.

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

x = 0+ x=

π 2

π

+

x =π +

=

2

π 2

π

π 2

=π =

3π 2

2 3π π + = 2π x= 2 2 Evaluate the function at each value of x. x

y = 4 sin x

0

y = 4sin 0 = 4 ⋅ 0 = 0

(0, 0)

π

= 4 ⋅1 = 4

π   2 , 4  

π

y = 4sin π = 4 ⋅ 0 = 0

(π , 0)

3π 2

y = 4 sin

y = 4sin 2π = 4 ⋅ 0 = 0

π 2

y = 4 sin

2

coordinates

3π 2 = 4(−1) = −4

 3π   2 , − 4  

(2π , 0)

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 .

2π B

1.

π

7.

;

π 2

;

3π ; π 4

2π B 1 2π ; 2 3

A;

Copyright © 2014 Pearson Education, Inc.

61


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

2.

The equation y = 5sin x is of the form y = A sin x with A = 5. Thus, the amplitude is

A = 5 = 5.

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

x = 0+ x=

π 2

π

+

=

2

π 2

π 2

3π x =π + = 2 2 3π π x= + = 2π 2 2 Evaluate the function at each value of x. x

y = 5sin x

0

y = 5sin 0 = 5 ⋅ 0 = 0

2

y = 5sin

1 The equation y = sin x is of the form y = A sin x 3 1 1 1 with A = . Thus, the amplitude is A = = . 3 3 3 2π π The period is 2π . The quarter-period is or . 2 4 The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x=0 x = 0+

π

π

3.

π 2

x=

π

y = 5sin π = 5 ⋅ 0 = 0

3π 2

y = 5sin

y = 5sin 2π = 5 ⋅ 0 = 0

3π = 5(−1) = −5 2

2

+

x =π +

coordinates

= 5 ⋅1 = 5

π

π 2

π 2

π 2

(2π , 0)

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 .

2

=π =

3π 2

3π π + = 2π 2 2 Evaluate the function at each value of x. x

1 y = sin x 3

0

1 1 y = sin 0 = ⋅ 0 = 0 3 3

(0, 0)

1 π 1 1 y = sin = ⋅1 = 3 2 3 3

π 1  2 , 3  

π

1 1 y = sin π = ⋅ 0 = 0 3 3

(π , 0)

3π 2

1 3π y = sin 3 2 1 1 = (−1) = − 3 3

1 1 y = sin 2π = ⋅ 0 = 0 3 3

(π , 0)  3π   2 , − 5  

π

x=

(0, 0)

π   2 , 5  

=

π 2

coordinates

1  3π  2 , − 3  

(2π , 0)

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 .

62

Copyright © 2014 Pearson Education, Inc.


Section 2.1 Graphs of Sine and Cosine Functions

4.

1 sin x is of the form y = A sin x 4 1 1 1 with A = . Thus, the amplitude is A = = . 4 4 4 2π π The period is 2π . The quarter-period is or . 4 2 The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x=0 The equation y =

x = 0+ x=

π 2

π

+

=

2

π 2

π

1 sin x 4

0

π

coordinates

1 π 1 1 sin = ⋅1 = 4 2 4 4

π 1   2, 4  

π

y=

1 1 sin π = ⋅ 0 = 0 4 4

(π , 0)

3π 2

y=

1 3π 1 1 sin = (−1) = − 4 2 4 4

y=

1 1 sin 2π = ⋅ 0 = 0 4 4

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

x = 0+

π 2

π

+

=

2

π 2

π

π 2

=π =

3π 2

x

y = −3sin x

coordinates

0

y = −3sin x

(0, 0)

(0, 0)

y=

2

A = −3 = 3 .

2 3π π x= + = 2π 2 2 Evaluate the function at each value of x.

3π x =π + = 2 2 3π π x= + = 2π 2 2 Evaluate the function at each value of x.

1 1 y = sin 0 = ⋅ 0 = 0 4 4

with A = –3. Thus, the amplitude is

x =π +

π

y=

The equation y = −3sin x is of the form y = A sin x

x=

2

x

5.

1  3π  2 , − 4   (2π , 0)

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 .

= −3 ⋅ 0 = 0

π 2

π

y = −3sin

π

2 = −3 ⋅1 = −3

y = −3sin π

π   2 , − 3  

(π , 0)

= −3 ⋅ 0 = 0 3π 2

y = −3sin

3π 2 = −3(−1) = 3

 3π   2 , 3  

y = −3sin 2π

(2π , 0)

= −3 ⋅ 0 = 0 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 .

Copyright © 2014 Pearson Education, Inc.

63


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

6.

The equation y = −4 sin x is of the form y = A sin x with A = –4. Thus, the amplitude is

A = −4 = 4 .

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

x = 0+ x=

π 2

π

+

x =π +

=

2

π 2

π 2

π 2

3π 2

x

y = −4 sin x

0

y = −4sin 0 = −4 ⋅ 0 = 0

(0, 0)

π

π   2 , − 4  

2

2

coordinates

= −4 ⋅ 1 = −4

π

y = −4sin π = −4 ⋅ 0 = 0

3π 2

y = −4sin

y = −4sin 2π = −4 ⋅ 0 = 0

(π , 0)

3π = −4(−1) = 4 2

π 4

π

+

4

π 4

π

2 3π x= + = 2 4 4 3π π x= + =π 4 4 Evaluate the function at each value of x. x

y = sin 2 x

0

y = sin 2 ⋅ 0 = sin 0 = 0

(0, 0)

π

 π y = sin  2 ⋅   4

π   4 ,1  

4

= sin

π 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 .

=

π

 3π   2 , 4  

(2π , 0)

π

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

π

3π π + = 2π 2 2 Evaluate the function at each value of x.

y = −4sin

quarter-period is

x=

x=

π

The equation y = sin 2 x is of the form y = A sin Bx with A = 1 and B = 2. The amplitude is 2π 2π A = 1 = 1 . The period is = = π . The B 2

x = 0+

=π =

7.

π 2

coordinates

=1

 π y = sin  2 ⋅   2 = sin π = 0

3π 4

 3π  y = sin  2 ⋅ 4   3π = sin = −1 2

π

y = sin(2 ⋅ π )

π 2, 

 0 

 3π   4 , − 1  

(π , 0)

= sin 2π = 0 Connect the five key points with a smooth curve and graph one complete cycle of the given function.

64

Copyright © 2014 Pearson Education, Inc.


Section 2.1 Graphs of Sine and Cosine Functions

8.

The equation y = sin 4 x is of the form y = A sin Bx with A = 1 and B = 4. Thus, the amplitude is 2π 2π π A = 1 = 1 . The period is = = . The B 4 2 π π 1 π quarter-period is 2 = ⋅ = . 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+ x=

π 8

π

=

8

+

π 8

x = 0 +π = π

=

π

x = 2π + π = 3π x = 3π + π = 4π Evaluate the function at each value of x. x

y = sin(4 ⋅ 0) = sin 0 = 0

π

π  π y = sin  4 ⋅  = sin = 1 8 2  

π   8 , 1  

 π y = sin  4 ⋅  = sin π = 0  4

π   4 , 0  

π 2

 3π  y = sin  4 ⋅ 8   3π = sin = −1 2

y = sin 2π = 0

period is

x = π + π = 2π

0

3π 8

1 . The amplitude is A = 3 = 3. 2 2π 2π The period is = 1 = 2π ⋅ 2 = 4π . The quarterB 2 with A = 3 and B =

8

y = sin 4 x

4

1 x is of the form y = A sin Bx 2

π

x

π

The equation y = 3sin

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

4 π π 3π x= + = 4 8 8 3π π π x= + = 8 8 2 Evaluate the function at each value of x.

8

9.

coordinates

π 2, 

1  y = 3sin  ⋅ 0  2  = 3sin 0 = 3 ⋅ 0 = 0

π

1  y = 3sin  ⋅ π  2 

= 3sin

π 2

coordinates (0, 0)

(π , 3)

= 3 ⋅1 = 3

1  y = 3sin  ⋅ 2π  2   = 3sin π = 3 ⋅ 0 = 0

1  y = 3sin  ⋅ 3π  2  3π = 3sin 2 = 3(−1) = −3

 0 

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

1 x 2

0

(0, 0)

 3π   8 , − 1  

y = 3sin

(2π , 0)

(3π , − 3)

1  (4π , 0) y = 3sin  ⋅ 4π  2  = 3sin 2π = 3 ⋅ 0 = 0 Connect the five points with a smooth curve and graph one complete cycle of the given function.

Copyright © 2014 Pearson Education, Inc.

65


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

1 x is of the form 4 1 y = A sin Bx with A = 2 and B = . Thus, the 4 amplitude is A = 2 = 2 . The period is

10. The equation y = 2 sin

2π 2π = 1 = 2π ⋅ 4 = 8π . The quarter-period is B 4 8π = 2π . The cycle begins at x = 0. Add quarter4 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 0

y = 2 sin

1 x 4

1  y = 2sin  ⋅ 0  4  = 2sin 0 = 2 ⋅ 0 = 0 1  y = 2 sin  ⋅ 2π  4 

= 2 sin

π 2

coordinates

y = 2sin π = 2 ⋅ 0 = 0

y = 2 sin

y = 2sin 2π = 2 ⋅ 0 = 0

3π = 2(−1) = −2 2

with A = 4 and B = π . The amplitude is The period is

(2π , 2)

1 1 = 2 2 1 1 x = + =1 2 2 1 3 x = 1+ = 2 2 3 1 x= + =2 2 2 Evaluate the function at each value of x. x

y = 4 sin π x

0

y = 4sin(π ⋅ 0)

coordinates (0, 0)

= 4sin 0 = 4 ⋅ 0 = 0 1 2

 1 y = 4 sin  π ⋅   2

= 4 sin

π 2

1 2, 

 4 

= 4(1) = 4

y = 4sin(π ⋅1)

(1, 0)

= 4sin π = 4 ⋅ 0 = 0

(4π , 0) 3 2

 3 y = 4sin  π ⋅   2 3π = 4sin 2 = 4(−1) = −4

2

y = 4sin(π ⋅ 2)

(8π , 0)

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

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

x = 0+

1

(6π , − 2)

A = 4 = 4.

2 1 = . The cycle begins at x = 0. Add quarter-periods to 4 2 generate x-values for the key points. x=0

(0, 0)

= 2 ⋅1 = 2

11. The equation y = 4 sin π x is of the form y = A sin Bx

3   2 , − 4  

(2, 0)

= 4sin 2π = 4 ⋅ 0 = 0 Connect the five points with a smooth curve and graph one complete cycle of the given function.

66

Copyright © 2014 Pearson Education, Inc.


Section 2.1 Graphs of Sine and Cosine Functions 12. The equation y = 3sin 2π x is of the form

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

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

A = 3 = 3 . The period is

2π 2π = = 1 . The B 2π

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

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

A = −3 = 3 . The period is

2π 2π = = 1 . The B 2π

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

quarter-period is

quarter-period is

1 1 x = 0+ = 4 4 1 1 1 x= + = 4 4 2 1 1 3 x= + = 2 4 4 3 1 x = + =1 4 4 Evaluate the function at each value of x.

1 1 = 4 4 1 1 1 x= + = 4 4 2 1 1 3 x= + = 2 4 4 3 1 x = + =1 4 4 Evaluate the function at each value of x. x coordinates y = −3sin 2π x

x

y = 3sin 2π x

0

y = 3sin(2π ⋅ 0)

x = 0+

coordinates 0 (0, 0)

1  y = 3sin  2π ⋅  4  = 3sin

π 2

= −3 ⋅ 0 = 0 1   4 , 3  

1 4

1  y = 3sin  2π ⋅  2  = 3sin π = 3 ⋅ 0 = 0

3 4

3  y = 3sin  2π ⋅  4  3π = 3sin = 3(−1) = −3 2

1

y = 3sin(2π ⋅1)

1  y = −3sin  2π ⋅  4  = −3sin

= 3 ⋅1 = 3

1 2

(0, 0)

= −3sin 0

= 3sin 0 = 3 ⋅ 0 = 0 1 4

y = −3sin(2π ⋅ 0)

1 2, 

 0 

3   4 , − 3  

2 = −3 ⋅1 = −3 1 2

1  y = −3sin  2π ⋅  2  = −3sin π

1 2, 

 0 

= −3 ⋅ 0 = 0 3 4

3  y = −3sin  2π ⋅  4  3π = −3sin 2 = −3(−1) = 3

1

y = −3sin(2π ⋅1)

(1, 0)

= 3sin 2π = 3 ⋅ 0 = 0 Connect the five key points with a smooth curve and graph one complete cycle of the given function.

π

1   4 , − 3  

3   4 , 3  

(1, 0)

= −3sin 2π = −3 ⋅ 0 = 0

Copyright © 2014 Pearson Education, Inc.

67


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

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

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

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

A = −2 = 2 . The period is

2π 2π = = 2 . The π B

2 1 = . 4 2 The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x=0 quarter-period is

1 1 = 2 2 1 1 x = + =1 2 2 1 3 x = 1+ = 2 2 3 1 x= + =2 2 2 Evaluate the function at each value of x. x = 0+

x

y = −2sin π x

0

y = −2sin(π ⋅ 0)

coordinates (0, 0)

= −2sin 1

1   2 , − 2  

 1 y = −2sin  π ⋅   2

π 2

y = −2sin(π ⋅1)

(1, 0)

2

 3 y = −2sin  π ⋅   2 3π = −2sin = −2(−1) = 2 2

y = −2sin(π ⋅ 2)

The amplitude is The period is

2 . 3 A = −1 = 1 .

2π 2π 3 = 2 = 2π ⋅ = 3π . B 2 3

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

3π 3π = 4 4 3π 3π 3π x= + = 4 4 2 3π 3π 9π x= + = 2 4 4 9π 3π x= + = 3π 4 4 Evaluate the function at each value of x. x

3 2, 

 2 

(2, 0)

3π 4

y = − sin

3π 2

Copyright © 2014 Pearson Education, Inc.

2 x 3

2  y = − sin  ⋅ 0  3  = − sin 0 = 0  2 3π  y = − sin  ⋅  3 4 

= − sin

= −2sin 2π = −2 ⋅ 0 = 0

68

2 x is of the form y = A sin Bx 3

with A = –1 and B =

0

= −2 ⋅ 1 = −2

= −2sin π = −2 ⋅ 0 = 0 3 2

15. The equation y = − sin

x = 0+

= −2sin 0 = −2 ⋅ 0 = 0 1 2

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

π 2

coordinates

(0, 0)

 3π   4 , − 1  

= −1

 2 3π  y = − sin  ⋅  3 2  = − sin π = 0

 3π   2 , 0  


Section 2.1 Graphs of Sine and Cosine Functions

9π 4

 2 9π  y = − sin  ⋅  3 4  3π = − sin 2 = −(−1) = 1

 9π   4 , 1  

2  y = − sin  ⋅ 3π  3   = − sin 2π = 0

(3π , 0)

0

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

4 x is of the form 3 4 y = A sin Bx with A = –1 and B = . 3 The amplitude is A = −1 = 1 .

16. The equation y = − sin

The period is

x

3π 8

y = − sin

4 x 3

4  y = − sin  ⋅ 0  3  = − sin 0 = 0  4 3π  y = − sin  ⋅  3 8 

= − sin

π 2

coordinates

(0, 0)

 3π   8 , − 1  

= −1

3π 4

 4 3π  y = − sin  ⋅  3 4  = − sin π = 0

 3π   4 , 0  

9π 8

 4 9π  y = − sin  ⋅  3 8  3π = − sin = −(−1) = 1 2

 9π   8 , 1  

3π 2

 4 3π   3π  y = − sin  ⋅ , 0   3 2   2  = − sin 2π = 0 Connect the five key points with a smooth curve and graph one complete cycle of the given function.

2π 2π 3 3π . = 4 = 2π ⋅ = B 4 2 3 3π 2

3π 1 3π = ⋅ = . 4 2 4 8 The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x=0 The quarter-period is

3π 3π = 8 8 3π 3π 3π x= + = 8 8 4 3π 3π 9π x= + = 4 8 8 9π 3π 3π x= + = 8 8 2 Evaluate the function at each value of x. x = 0+

Copyright © 2014 Pearson Education, Inc.

69


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions 17. The equation y = sin( x − π ) is of the form

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

A = 1 = 1 . The period is

C π 2π 2π = = 2π . The phase shift is = = π . The B 1 B 1 2π π quarter-period is = . The cycle begins at 4 2 x = π . Add quarter-periods to generate x-values for the key points. x =π

π

3π x =π + = 2 2 3π π + = 2π x= 2 2 π 5π x = 2π + = 2 2 5π π + = 3π x= 2 2 Evaluate the function at each value of x. x

y = sin( x − π )

coordinates

π

y = sin(π − π ) = sin 0 = 0

(π , 0)

3π 2

 3π  −π  y = sin   2 

= sin 2π

π 2

 3π   2 , 1  

=1

y = sin(2π − π ) = sin π = 0

5π 2

 5π  y = sin  −π   2  3π = sin = −1 2

y = sin(3π − π ) = sin 2π = 0

π  18. The equation y = sin  x −  is of the form 2  y = A sin( Bx − C ) with A = 1, B = 1, and C = The amplitude is

 5π   2 , − 1  

(3π , 0)

.

A = 1 = 1 . The period is π

x=

π

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

x= x=

π 2

π 2

+

x =π +

π 2

π

=π =

3π 2

2 3π π + = 2π x= 2 2 π 5π x = 2π + = 2 2 Evaluate the function at each value of x. x

2

π  y = sin  x −  2  π π  y = sin  −  = sin 0 = 0 2 2

π

π π  y = sin  π −  = sin = 1 2 2 

3π 2

 3π π  y = sin  −   2 2 = sin π = 0

π  y = sin  2π −  2  3π = sin = −1 2

5π 2

 5π π  y = sin  −   2 2 = sin 2π = 0

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

70

2

C 2 π 2π 2π = = 2π . The phase shift is = = . The B 1 2 B 1 2π π quarter-period is = . The cycle begins at 4 2

π (2π , 0)

π

Copyright © 2014 Pearson Education, Inc.

coordinates

π 2, 

 0 

(π , 1)  3π   2 , 0  

(2π , − 1)

 5π   2 , 0  


Section 2.1 Graphs of Sine and Cosine Functions

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

5π 4

 5π  y = sin  2 ⋅ −π  4    5π  = sin  −π  2   3π = sin = −1 2

 5π   4 , − 1  

3π 2

 3π  y = sin  2 ⋅ −π  2   = sin(3π − π ) = sin 2π = 0

 3π   2 , 0  

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

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

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

C π 2π 2π = = π . The phase shift is = . The B 2 B 2

π

π

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

x=

π 2

π 3π + = 2 4 4 3π π + =π x= 4 4 π 5π x =π + = 4 4 5π π 3π + = x= 4 4 2 x=

π

π  20. The equation y = sin  2 x −  is of the form 2  y = A sin( Bx − C ) with A = 1, B = 2, and C = The amplitude is

x

y = sin(2 x − π )

π

 π  y = sin  2 ⋅ − π   2  = sin(π − π )

coordinates

π 2, 

 0 

 3π  −π  y = sin  2 ⋅ 4    3π  = sin  −π   2 

= sin

π

.

A = 1 =1.

2π 2π = =π . B 2 C π2 π 1 π The phase shift is = = ⋅ = . B 2 2 2 4

π 2

 3π   4 , 1  

π 4

The cycle begins at x =

x=

.

π

x=

π 4

π 4

(π , 0)

+

π 4

=

π

2 3π x= + = 2 4 4 3π π + =π x= 4 4 π 5π x =π + = 4 4 Evaluate the function at each value of x.

π

=1

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

The quarter-period is

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

= sin 0 = 0 3π 4

2

The period is

Evaluate the function at each value of x.

2

π

π

Copyright © 2014 Pearson Education, Inc.

71


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

x

π  y = sin  2 x −  2 

coordinates

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

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

π 4

π 2

 π π y = sin  2 ⋅ −   4 2 π π  = sin  −  = sin 0 = 0 2 2

π   4 , 0  

 π π y = sin  2 ⋅ −   2 2

π   2 , 1  

π

 3π π  − y = sin  2 ⋅ 4 2    3π π  = sin  −   2 2 = sin π = 0

π  y = sin  2 ⋅ π −  2 

 5π π  − y = sin  2 ⋅ 4 2    5π π  = sin  −   2 2 = sin 2π = 0

π

. The cycle begins at x =

 3π   4 , 0  

(π , − 1)

x= x=

π 2

π

+

π

=

3π 4

2 4 3π π x= + =π 4 4 π 5π x =π + = 4 4 5π π 3π x= + = 4 4 2 Evaluate the function at each value of x. x

y = 3sin(2 x − π )

π

 π  y = 3sin  2 ⋅ − π   2  = 3sin(π − π )

2

coordinates

π 2, 

 0 

= 3sin 0 = 3 ⋅ 0 = 0 3π 4  5π   4 , 0  

 3π  −π  y = 3sin  2 ⋅ 4    3π  = 3sin  −π   2  = 3sin

π

 3π   4 , 3  

π

= 3 ⋅1 = 3 2 y = 3sin(2 ⋅ π − π )

(π , 0)

= 3sin(2π − π )

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

= 3sin π = 3 ⋅ 0 = 0 5π 4

 5π  −π  y = 3sin  2 ⋅ 4    5π  = 3sin  −π  2   3π = 3sin 2 = 3(−1) = −3

 5π   4 , − 3  

3π 2

 3π  −π  y = 3sin  2 ⋅ 2   = 3sin(3π − π )

 3π   2 , 0  

= 3sin 2π = 3 ⋅ 0 = 0

72

π

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

π  = sin  2π −  2  3π = sin = −1 2 5π 4

2π 2π C π = = π . The phase shift is = . The B 2 B 2 quarter-period is

π π  = sin  π −  = sin = 1 2 2   3π 4

A = 3 = 3 . The period is

Copyright © 2014 Pearson Education, Inc.


Section 2.1 Graphs of Sine and Cosine Functions

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

x

π 4

π  y = 3sin  2 x −  2   π π y = 3sin  2 ⋅ −   4 2

coordinates

π 4, 

 0 

π π  = sin  −  2 2 = 3sin 0 = 3 ⋅ 0 = 0

π 2

π  22. The equation y = 3sin  2 x −  is of the form 2  y = A sin( Bx − C ) with A = 3, B = 2, and C = The amplitude is

π 2

.

2π 2π = =π . B 2 C π2 π 1 π = = ⋅ = . The phase shift is B 2 2 2 4 The period is

π 4

The cycle begins at x =

= 3sin

. Add quarter-periods to

x=

π  y = 3sin  2 ⋅ π −  2 

+

π

=

π

2 3π x= + = 2 4 4 3π π x= + =π 4 4 π 5π x =π + = 4 4 Evaluate the function at each value of x.

π

4

π

 3π   4 , 0  

(π , − 3)

π  = 3sin  2π −  2  3π = 3sin = 3 ⋅ (−1) = −3 2

4 4

= 3 ⋅1 = 3

π

π π

2

 3π π  − y = 3sin  2 ⋅ 4 2    3π π  = 3sin  −   2 2 = 3sin π = 3 ⋅ 0 = 0

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

π

3π 4

.

π

π   2 , 3  

π  = 3sin  π −  2 

A = 3 = 3.

The quarter-period is

 π π y = 3sin  2 ⋅ −   2 2

5π 4

 5π π  − y = 3sin  2 ⋅ 4 2    5π π  = 3sin  −   2 2 = 3sin 2π = 3 ⋅ 0 = 0

 5π   4 , 0  

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

Copyright © 2014 Pearson Education, Inc.

73


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

23.

y=

1 π  1   π   sin  x +  = sin  x −  −   2 2 2   2  

The equation y =

1 π  sin  π +  2 2  1 3π = sin 2 2 1 1 = ⋅ (−1) = − 2 2

π

y=

3π 2

y=

 1  π  sin  x −  −   is of the form 2  2  

1 π , B = 1, and C = − . 2 2 1 1 = . The period is The amplitude is A = 2 2 π C − π2 2π 2π = = 2π . The phase shift is = =− . B 1 2 B 1 2π π The quarter-period is = . The cycle begins at 4 2 y = A sin( Bx − C ) with A =

x=−

1  3π   3π π  +  sin   2 , 0 2 2 2     1 = sin 2π 2 1 = ⋅0 = 0 2 Connect the five points with a smooth curve and graph one complete cycle of the given function.

π

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

x=− x=−

π 2

π

+

2

x = 0+ x=

π 2

π

+

π 2

=0

2 =

2

x

π 2

π

π

π 2

3π 2 2 Evaluate the function at each value of x. x =π +

=

y=

1 π  sin  x +  2 2 

1  π π sin  − +  2  2 2 1 1 = sin 0 = ⋅ 0 = 0 2 2

y=

1 π  sin  0 +  2 2  π 1 1 1 = sin = ⋅1 = 2 2 2 2

0

y=

π

1 π π  y = sin  +  2 2 2 1 1 = sin π = ⋅ 0 = 0 2 2

2

24.

coordinates

 π − 2 , 

 0 

 1  0, 2   

π

  2 , 0  

1 1 sin( x + π ) = sin( x − (−π )) 2 2 1 The equation y = sin( x − (−π )) is of the form 2 1 y = A sin( Bx − C ) with A = , B = 1, and C = −π . 2 1 1 = . The period is The amplitude is A = 2 2 2π 2π C −π = = 2π . The phase shift is = = −π . B 1 B 1 2π π The quarter-period is = . The cycle begins at 4 2 x = −π . Add quarter-periods to generate x-values for the key points. x = −π y=

x = −π + x=−

π

π 2

π

π

+

2

π 2

Copyright © 2014 Pearson Education, Inc.

=−

2

+

2

x = 0+ x=

74

1  π , − 2   

π 2 =

π

=0

π 2

2


Section 2.1 Graphs of Sine and Cosine Functions

C − π2 π 1 π = = − ⋅ = − . The quarterB 2 2 2 4

Evaluate the function at each value of x. phase shift is x −π

π 2

y=

1 sin( x + π ) 2

coordinates

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

(−π , 0)

y=

1  π  y = sin  − + π  2  2  π 1 1 1 = sin = ⋅1 = 2 2 2 2

 π 1 − 2 , 2   

(0, 0)

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

y=

π

1 π  y = sin  + π  2 2  1 3π 1 1 = sin = ⋅ (−1) = − 2 2 2 2

π

1 π  2 , − 2  

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

x=−

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

π

π 4

π

x=

+

4

x = 0+

π 4

π

+

4

π 4

π

=0

4 =

π

=

4

π

2 π π 3π x= + = 2 4 4 Evaluate the function at each value of x. x

π  y = −2sin  2 x +  2 

π

  π π   π y = −2sin  2⋅  −  +   − ,   4 2   4

− (π , 0)

y=

. The cycle begins at x = −

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

x=−

0

2

π

period is

4

coordinates

 0 

 π π = −2sin  − +   2 2 = −2sin 0 = −2 ⋅ 0 = 0 0

π  y = −2sin  2 ⋅ 0 +  2 

(0, –2)

π  = −2sin  0 +  2  = −2sin

π

2 = −2 ⋅ 1 = −2

π 4 25.

 π π y = −2sin  2 ⋅ +   4 2

 π   π  y = −2sin  2 x +  = −2sin  2 x −  −   2    2     π  The equation y = −2sin  2 x −  −   is of the form  2   y = A sin( Bx − C ) with A = –2, B = 2, and C = −

π 2

π 4, 

 0 

π π  = −2sin  +  2 2 = −2sin π = −2 ⋅ 0 = 0

. The amplitude is

A = −2 = 2 . The period is

2π 2π = = π . The B 2

Copyright © 2014 Pearson Education, Inc.

75


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

π

 π π y = −2sin  2 ⋅ +   2 2

2

π 2, 

 2 

x=−

π  = −2sin  π +  2  3π = −2sin 2 = −2(−1) = 2 3π 4

 3π π  y = −2sin  2 ⋅ + 4 2  

x=−

π 4

π

x = 0+ x=

+

4

π 4

π

+

π =

4

π

π

=

4

π

2 3π x= + = 2 4 4 Evaluate the function at each value of x.

π

 3π   4 , 0  

 3π π  = −2sin  +   2 2 = −2sin 2π

4

=0

4

π

x

= −2 ⋅ 0 = 0 Connect the five points with a smooth curve and graph one complete cycle of the given function.

π 4

π  y = −3sin  2 x +  2    π π y = −3sin  2 ⋅  −  +    4 2

coordinates

 π − 4 , 

 0 

 π π = −3sin  − +   2 2 = −3sin 0 = −3 ⋅ 0 = 0 0

π  y = −3sin  2 ⋅ 0 +  2  π  = −3sin  0 +  2  = −3sin

26.

 π   π  y = −3sin  2 x +  = −3sin  2 x −  −   2    2  

π

  π  The equation y = −3sin  2 x −  −   is of the form  2   y = A sin( Bx − C ) with A = –3, B = 2, and C = − The amplitude is

π 2

4

.

A = −3 = 3 . The period is

2π 2π = = π . The phase shift is B 2

π 2

π

− π 1 π C π = 2 = − ⋅ = − . The quarter-period is . 4 B 2 2 2 4 The cycle begins at x = −

π

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

76

3π 4

Copyright © 2014 Pearson Education, Inc.

π 2

(0, –3)

= −3 ⋅ 1 = −3

 π π y = −3sin  2 ⋅ +   4 2 π π  = −3sin  +  2 2 = −3sin π = −3 ⋅ 0 = 0

π 4, 

 π π y = −3sin  2 ⋅ +   2 2 π  = −3sin  π +  2  3π = −3sin = −3 ⋅ (−1) = 3 2

π   2 , 3  

 3π π  + y = −3sin  2 ⋅ 4 2    3π π  = −3sin  +   2 2 = −3sin 2π = −3 ⋅ 0 = 0

 3π   4 , 0  

 0 


Section 2.1 Graphs of Sine and Cosine Functions

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

x

2

π

y = 3sin(π x + 2)   2  y = 3sin  π  −  + 2  π     = 3sin(−2 + 2) = 3sin 0 = 3 ⋅ 0 = 0

π −4  π −4  + 2 y = 3sin  π   2π   2π   27.

= 3sin

π

2

π −4   2π , 3   

π

π −2  π −2  + 2 y = 3sin  π   π   π  

π −2  π , 

 0 

= 3sin(π − 2 + 2) = 3sin π = 3 ⋅ 0 = 0 3π − 4   3π − 4   + 2 y = 3sin  π   2π   2π  

1 π −4 = π 2 2π π −4 1 π −2 + = x= π 2π 2 π − 2 1 3π − 4 + = x= π 2 2π 3π − 4 1 2π − 2 + = x= π 2π 2 Evaluate the function at each value of x. x=−

 0 

2 = 3 ⋅1 = 3

π

for the key points. 2 x=−

 2 − π , 

π −4  = 3sin  + 2  2  π  = 3sin  − 2 + 2  2  

y = 3sin(π x + 2) 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 2π 2π C −2 2 = = 2 . The phase shift is = =− . π B π π B 2 1 The quarter-period is = . The cycle begins at 4 2 2 x = − . Add quarter-periods to generate x-values

coordinates

 5π   4 , − 3  

 3π − 4  = 3sin  + 2  2   3π  = 3sin  − 2 + 2 2   3π = 3sin 2 = 3(−1) = −3

+

2π − 2

  2π − 2    2π − 2  , 0 + 2  y = 3sin  π   π    π    π = 3sin(2π − 2 + 2) = 3sin 2π = 3 ⋅ 0 = 0 Connect the five points with a smooth curve and graph one complete cycle of the given function.

Copyright © 2014 Pearson Education, Inc.

77


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

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

3π − 8    3π − 8    3π − 8 , − 3 + 4  y = 3sin  2π   4π    4π    4π  3π − 8  = 3sin  + 4  2   3π  = 3sin  − 4 + 4 2   3π = 3sin = 3(−1) = −3 2

2π 2π C −4 2 = = 1 . The phase shift is = =− . π B 2π B 2π 1 The quarter-period is . The cycle begins at 4 2 x = − . Add quarter-periods to generate x-values

is

π

π − 2 y = 3sin  2π  π − 2  + 4    π   π = 3sin(2π − 4 + 4) = 3sin 2π = 3 ⋅ 0 = 0

for the key points. 2 x=−

π

1 π −8 = 4 4π π −8 1 π − 4 + = x= 4π 4 2π π − 4 1 3π − 8 + = x= 2π 4 4π 3π − 8 1 π − 2 + = x= π 4π 4 Evaluate the function at each value of x. x=−

2

π

x

2

π

+

y = 3sin(2π x + 4)

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

coordinates

 2 − π , 

  2  y = 3sin  2π  −  + 4    π  = 3sin(−4 + 4) = 3sin 0 = 3 ⋅ 0 = 0

π −8  π −8   + 4 y = 3sin  2π   4π   4π  

 0 

π −8   4π , 3   

 π −8  = 3sin  + 4  2  π  = 3sin  − 4 + 4  2  

= 3sin

π 2

= 3 ⋅1 = 3

π −4  π −4  + 4 y = 3sin  2π   2π   2π   = 3sin(π − 4 + 4) = 3sin π = 3 ⋅ 0 = 0

π −2   π ,0  

π −4  2π , 

 0 

29.

y = −2sin(2π x + 4π ) = −2sin(2π x − (−4π )) The equation y = −2sin(2π x − (−4π )) is of the form y = A sin( Bx − C ) with A = –2, B = 2π , and C = −4π . The amplitude is A = −2 = 2 . The 2π 2π = = 1 . The phase shift is B 2π 1 C −4π = = −2 . The quarter-period is . The cycle 4 B 2π begins at x = −2 . Add quarter-periods to generate xvalues for the key points. x = −2 1 7 x = −2 + = − 4 4 7 1 3 x=− + =− 4 4 2 3 1 5 x=− + =− 2 4 4 5 1 x = − + = −1 4 4

period is

Evaluate the function at each value of x.

78

Copyright © 2014 Pearson Education, Inc.


Section 2.1 Graphs of Sine and Cosine Functions

x

y = −2sin(2π x + 4π )

–2 y = −2sin(2π (−2) + 4π )

coordinates (–2, 0)

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

 7π  = −2sin  − + 4π  2   = −2sin

2

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

y = A sin( Bx − C ) with A = –3, B = 2π , and C = −4π . A = −3 = 3 . The period is

C −4π 2π 2π = = −2 . The = = 1 . The phase shift is B 2π B 2π 1 quarter-period is . The cycle begins at x = −2 . Add 4 quarter-periods to generate x-values for the key points. x = −2 1 7 =− 4 4 7 1 3 x=− + =− 4 4 2 3 1 5 x=− + =− 2 4 4 5 1 x = − + = −1 4 4 Evaluate the function at each value of x. x = −2 +

= −2 ⋅ 1 = −2

3   3   3  y = −2sin  2π  −  + 4π   − , 0  2   2   2  = −2sin(−3π + 4π ) = −2sin π = −2 ⋅ 0 = 0

5   5   5 − y = −2sin  2π  −  + 4π   − , 4   4   4

 2 

 5π  = −2sin  − + 4π  2   3π = −2sin 2 = −2(−1) = 2 –1 y = −2sin(2π (−1) + 4π )

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

The amplitude is

7    7   7 y = −2sin  2π  −  + 4π   − , − 2  4 4    4  

π

30.

x

y = −3sin(2π x + 4π )

–2

y = −3sin(2π (−2) + 4π )

coordinates (–2, 0)

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

7 4

  7  y = −3sin  2π  −  + 4π    4   7π  = −3sin  − + 4π   2 

(–1, 0)

= −3sin

= −2sin(−2π + 4π ) = −2sin 2π

= −2 ⋅ 0 = 0

3 2

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

π 2

 7   − 4 , − 3  

= −3 ⋅ 1 = −3

  3  y = −3sin  2π  −  + 4π    2  = −3sin(−3π + 4π )

 3   − 2 , 0  

= −3sin π = −3 ⋅ 0 = 0 −

5 4

–1

  5  y = −3sin  2π  −  + 4π  4      5π  = −3sin  − + 4π  2   3π = −3sin = −3(−1) = 3 2 y = −3sin(2π (−1) + 4π )

 5   − 4 , 3  

(–1, 0)

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

Copyright © 2014 Pearson Education, Inc.

79


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

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 the graph of y = 2 cos x .

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

A = 2 = 2.

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

x = 0+ x=

π 2

π

+

x =π +

=

2

π 2

π

π

y = 2 cos x

coordinates

0

y = 2 cos 0 = 2 ⋅1 = 2

(0, 2)

π 3π 2 2π

80

π 2

π

2 = 2⋅0 = 0

y = 2 cos π = 2 ⋅ (−1) = −2

π 2, 

 0 

(π , − 2)

3π 2 = 2⋅0 = 0

 3π   2 , 0  

y = 2 cos 2π = 2 ⋅1 = 2

(2π , 2)

y = 2 cos

π

+

2

π 2

π 2

=

π 2

=π =

3π 2

x=

x

2

x = 0+

3π π + = 2π 2 2 Evaluate the function at each value of x.

3π 2

y = 2cos

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

x =π +

2 3π π x= + = 2π 2 2 Evaluate the function at each value of x.

π

A = 3 = 3.

The period is 2π . The quarter-period is

x=

2

=π =

with A = 3. Thus, the amplitude is

x

y = 3cos x

0

y = 3cos 0 = 3 ⋅1 = 3

(0, 3)

π

π   2 , 0  

π 2

y = 3cos

2

coordinates

= 3⋅ 0 = 0

π

y = 3cos π = 3 ⋅ (−1) = −3

3π 2

y = 3cos

y = 3cos 2π = 3 ⋅1 = 3

3π = 3⋅ 0 = 0 2

(π , − 3)  3π   2 , 0  

(2π , 3)

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 .

Copyright © 2014 Pearson Education, Inc.


Section 2.1 Graphs of Sine and Cosine Functions 33. The equation y = −2 cos x is of the form y = A cos x with A = –2. Thus, the amplitude is A = −2 = 2 . The period is 2π . The quarter-

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

π 2

π

+

=

2

π 2

π 2

3π x =π + = 2 2 3π π x= + = 2π 2 2 Evaluate the function at each value of x. x

y = −2 cos x

0

y = −2 cos 0 = −2 ⋅ 1 = −2

2

π

y = −2 cos

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

x = 0+ x=

π

π

34. The equation y = −3cos x is of the form y = A cos x with A = –3. Thus, the amplitude is A = −3 = 3 .

π

2 = −2 ⋅ 0 = 0

y = −2 cos π = −2 ⋅ (−1) = 2

coordinates (0, –2)

π 2

 0 

(π , 2)

3π 2

3π y = −2 cos 2 = −2 ⋅ 0 = 0

 3π   2 , 0  

y = −2 cos 2π = −2 ⋅ 1 = −2

(2π , − 2)

+

x =π +

2

π 2

π 2

=

π 2

=π =

3π 2

3π π + = 2π 2 2 Evaluate the function at each value of x. x=

x

y = −3cos x

0

y = −3cos 0 = −3 ⋅1 = −3

π 2

π 2, 

π

y = −3cos

π 2

coordinates

= −3 ⋅ 0 = 0

(0, –3)

π   2 , 0  

π

y = −3cos π = −3 ⋅ (−1) = 3

3π 2

y = −3cos

3π = −3 ⋅ 0 = 0 2

 3π   2 , 0  

y = −3cos 2π = −3 ⋅1 = −3

(2π , − 3)

(π , 3)

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 graph one complete cycle of the given function with the graph of y = cos x .

Copyright © 2014 Pearson Education, Inc.

81


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions 35. The equation y = cos 2 x is of the form y = A cos Bx with A = 1 and B = 2. Thus, the amplitude is 2π 2π A = 1 = 1 . The period is = = π . The B 2

π

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

x = 0+ x=

π 4

π

+

=

4

π 4

π

=

x = 0+

4

π

2 π π 3π x= + = 2 4 4 3π π x= + =π 4 4 Evaluate the function at each value of x. x

y = cos 2 x

0

y = cos(2 ⋅ 0) = cos 0 = 1

π

 π y = cos  2 ⋅   4

4

= cos

π 2 3π 4

π

36. The equation y = cos 4 x is of the form y = A cos Bx with A = 1 and B = 4. Thus, the amplitude is 2π 2π π A = 1 = 1 . The period is = = . The B 4 2 π π 1 π quarter-period is 2 = ⋅ = . The cycle begins at 4 2 4 8 x = 0. Add quarter-periods to generate x-values for the key points. x=0

π 2

x=

π 8

π   4 , 0  

π   2 , − 1  

 3π  y = cos  2 ⋅ 4   3π = cos =0 2

 3π   4 , 0  

y = cos(2 ⋅ π ) = cos 2π = 1

π

y = cos 4 x

0

y = cos(4 ⋅ 0) = cos 0 = 1

π

π  π y = cos  4 ⋅  = cos = 0 8 2  

8

coordinates (0, 1)

π 8, 

 0 

 π y = cos  4 ⋅  = cos π = −1  4

π   4 , − 1  

3π 8

 3π  y = cos  4 ⋅ 8   3π = cos =0 2

 3π   8 , 0  

π

 π y = cos  4 ⋅  = cos 2π = 1  2

π   2 , 1  

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.

82

=

x

2 (π , 1)

8

8

4 3π x= + = 4 8 8 3π π π x= + = 8 8 2 Evaluate the function at each value of x.

4

 π y = cos  2 ⋅   2 = cos π = −1

+

π

π

π

π

=0

8

=

π

coordinates (0, 1)

π

Copyright © 2014 Pearson Education, Inc.


Section 2.1 Graphs of Sine and Cosine Functions 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 2π 2π A = 4 = 4 . The period is = = 1 . The B 2π 1 quarter-period is . The cycle begins at x = 0 . Add 4 quarter-periods to generate x-values for the key points. x=0

1 1 = 4 4 1 1 1 x= + = 4 4 2 1 1 3 x= + = 2 4 4 3 1 x = + =1 4 4 Evaluate the function at each value of x. x y = 4 cos 2π x coordinates (0, 4) 0 y = 4 cos(2π ⋅ 0) x = 0+

1 4

= 4 cos 0 = 4 ⋅1 = 4 1  y = 4 cos  2π ⋅  4  = 4 cos

1 2

1   4 , 0  

amplitude is

2 = 4⋅0 = 0 1  y = 4 cos  2π ⋅  2  = 4 cos π = 4 ⋅ (−1) = −4

A = 5 = 5 . The period is

2π 2π 1 = = 1 . The quarter-period is . The cycle 4 B 2π begins at x = 0. Add quarter-periods to generate xvalues for the key points. x=0 1 1 x = 0+ = 4 4 1 1 1 x= + = 4 4 2 1 1 3 x= + = 2 4 4 3 1 x = + =1 4 4 Evaluate the function at each value of x. x

y = 5 cos 2π x

0

y = 5cos(2π ⋅ 0) = 5cos 0 = 5 ⋅1 = 5

1 4

1  y = 5cos  2π ⋅  4 

π

= 5cos 1   2 , − 4  

3 3   y = 4 cos  2π ⋅   4 , 0 4    3π = 4 cos 2 = 4⋅0 = 0 (1, 4) 1 y = 4 cos(2π ⋅1) = 4 cos 2π = 4 ⋅1 = 4 Connect the five points with a smooth curve and graph one complete cycle of the given function. 3 4

38. The equation y = 5 cos 2π x is of the form y = A cos Bx with A = 5 and B = 2π . Thus, the

π 2

coordinates (0, 5)

1 4, 

 0 

= 5⋅0 = 0 1   2 , − 5  

1 2

1  y = 5cos  2π ⋅  2  = 5cos π = 5 ⋅ (−1) = −5

3 4

3π   y = 5cos  2π ⋅ 4   3π = 5cos = 5⋅0 = 0 2

3   4 , 0  

1

y = 5cos(2π ⋅1) = 5cos 2π = 5 ⋅1 = 5

(1, 5)

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

Copyright © 2014 Pearson Education, Inc.

83


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

1 x is of the form 2 1 y = A cos Bx with A = –4 and B = . Thus, the 2 amplitude is A = −4 = 4 . The period is

39. The equation y = −4cos

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

2π 2π = 1 = 2π ⋅ 2 = 4π . The quarter-period is B 2 4π = π . The cycle begins at x = 0 . Add quarter4 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 0

π

y = −4cos

1 x 2

1  y = −4 cos  ⋅ 0  2  = −4 cos 0 = −4 ⋅1 = −4 1  y = −4 cos  ⋅ π  2 

= −4 cos

coordinates (0, –4)

(π , 0)

π

2 = −4 ⋅ 0 = 0 2π

1  y = −4 cos  ⋅ 2π  2   = −4 cos π = −4 ⋅ (−1) = 4

(2π , 4)

1  y = −4cos  ⋅ 3π  2  3π = −4cos 2 = −4 ⋅ 0 = 0

(3π , 0)

1  y = −4 cos  ⋅ 4π  2  = −4 cos 2π

= −4 ⋅ 1 = −4

84

1 40. The equation y = −3cos x is of the form 3 1 y = A cos Bx with A = –3 and B = . Thus, the 3 amplitude is A = −3 = 3 . The period is 2π 2π = 1 = 2π ⋅ 3 = 6π . The quarter-period is B 3 6π 3π = . The cycle begins at x = 0. Add quarter4 2 periods to generate x-values for the key points. x=0 3π 3π x = 0+ = 2 2 3π 3π x= + = 3π 2 2 3π 9π x = 3π + = 2 2 9π 3π x= + = 6π 2 2 Evaluate the function at each value of x. x

0

3π 2

1 y = −3cos x 3 1  y = −3cos  ⋅ 0  3  = −3cos 0 = −3 ⋅1 = −3  1 3π  y = −3cos  ⋅  3 2 

= −3cos (4π , – 4)

π 2

(0, –3)

 3π   2 , 0  

= −3 ⋅ 0 = 0

1  y = −3cos  ⋅ 3π  3   = −3cos π = −3 ⋅ (−1) = 3

Copyright © 2014 Pearson Education, Inc.

coordinates

(3π , 3)


Section 2.1 Graphs of Sine and Cosine Functions

9π 2

 1 9π  y = −3cos  ⋅  3 2  3π = −3cos = −3 ⋅ 0 = 0 2

 9π   2 , 0  

1  y = −3cos  ⋅ 6π  3  = −3cos 2π = −3 ⋅1 = −3

(6π , − 3)

x

0

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

3 2

3

1 π 41. The equation y = − cos x is of the form 2 3 1 π y = A cos Bx with A = − and B = . Thus, the 2 3 1 1 = . The period is amplitude is A = − 2 2 2π 2π 3 6 3 = π = 2π ⋅ = 6 . The quarter-period is = . 4 2 B π 3

9 2

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

3 3 = 2 2 3 3 x= + =3 2 2 3 9 x = 3+ = 2 2 9 3 x= + =6 2 2 Evaluate the function at each value of x. x = 0+

6

1 π y = − cos x 2 3 1 π  y = − cos  ⋅ 0  2 3  1 = − cos 0 2 1 1 = − ⋅1 = − 2 2

coordinates

1   0, − 2   

1 π 3 y = − cos  ⋅  2  3 2 1 π = − cos 2 2 1 = − ⋅0 = 0 2

3 2, 

1 π  y = − cos  ⋅ 3  2 3  1 = − cos π 2 1 1 = − ⋅ (−1) = 2 2

 1  3, 2   

1 π 9  y = − cos  ⋅  2  3 2 1 3π = − cos 2 2 1 = − ⋅0 = 0 2

9 2, 

1 π  y = − cos  ⋅ 6  2 3  1 = − cos 2π 2 1 1 = − ⋅1 = − 2 2

 0 

 0 

1   6, − 2   

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

Copyright © 2014 Pearson Education, Inc.

85


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

1 π 42. The equation y = − cos x is of the form 2 4 1 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 4 = π = 2π ⋅ = 8 . The quarter-period is 84 = 2 . B π 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. x 0

2

4

6

8

86

1 π y = − cos x 2 4

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

π  43. The equation y = cos  x −  is of the form 2  y = A cos ( Bx − C ) with A = 1, and B = 1, and C=

coordinates

π

. Thus, the amplitude is

2

A = 1 = 1 . The

2π 2π = = 2π . The phase shift is B 1

period is

π

1 π  y = − cos  ⋅ 0  2 4  1 1 1 = − cos 0 = − ⋅1 = − 2 2 2

1   0, − 2   

C 2 π 2π π = = . The quarter-period is = . The 4 2 B 1 2 cycle begins at x =

π

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

1 π  y = − cos  ⋅ 2  2 4  1 1 π = − cos = − ⋅ 0 = 0 2 2 2

(2, 0)

x= x=

1 π  y = − cos  ⋅ 4  2 4  1 1 1 = − cos π = − ⋅ (−1) = 2 2 2 1 π  y = − cos  ⋅ 6  2 4  1 1  3π  = − cos  = − ⋅0 = 0  2 2  2  1 π  y = − cos  ⋅ 8  2 4  1 1 1 = − cos 2π = − ⋅1 = − 2 2 2

 1  4, 2   

(6, 0)

1   8, − 2   

π 2

π 2

+

x =π +

π 2

π 2

=π =

3π 2

3π π + = 2π 2 2 π 5π x = 2π + = 2 2 Evaluate the function at each value of x. x=

x

coordinates

π 2

π   2 , 1  

π

(π , 0 )

3π 2

 3π   2 , − 1  

( 2π , 0 )

5π 2

 5π   2 , 1  

Copyright © 2014 Pearson Education, Inc.


Section 2.1 Graphs of Sine and Cosine Functions

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

π  44. The equation y = cos  x +  is of the form 2  y = A cos ( Bx − C ) with A = 1, and B = 1, and C=−

π

. Thus, the amplitude is

2

A = 1 = 1 . The

2π 2π = = 2π . The phase shift is B 1

period is

π

− π C 2π π = . The = 2 = − . The quarter-period is 4 2 B 1 2 cycle begins at x = −

π

. Add quarter-periods to

2 generate x-values for the key points. x=− x=−

π 2

π

x = 0+ x=

+

2

π 2

π

+

2

π 2

2

=0

π

=

2

3π 2 2 Evaluate the function at each value of x. x =π +

π

π

=

x

coordinates

π

 π   − 2 , 1  

2

0

( 0, 0 )

π 2

π   2 , − 1  

π

(π , 0 )

3π 2

 3π   2 , 1  

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

y = A cos ( Bx − C ) with A = 3, and B = 2, and C = π . Thus, the amplitude is

A = 3 = 3 . The

2π 2π C π = = π . The phase shift is = . period is B 2 B 2 The quarter-period is

π

. The cycle begins at x =

π

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

.

π 2

π

+

π

=

3π 4

2 4 3π π + =π x= 4 4 π 5π x =π + = 4 4 5π π 3π + = x= 4 4 2 Evaluate the function at each value of x. x

coordinates

π 2

π   2 , 3  

3π 4

 3π   4 , 0  

π

(π , − 3)

5π 4

 5π   4 , 0  

3π 2

 3π   2 , 3  

Copyright © 2014 Pearson Education, Inc.

87


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

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.

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

y = A cos( Bx − C ) with A = 4, and B = 2, and C = π . Thus, the amplitude is A = 4 = 4 . The period is 2π 2π C π = = π . The phase shift is = . The B 2 B 2

π

π

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

x=

π 2

π 3π + = 2 4 4 3π π + =π x= 4 4 π 5π x =π + = 4 4 5π π 3π + = x= 4 4 2 x=

π

Evaluate the function at each value of x. x

π 2

88

47.

y=

 1 π 1   π  cos  3x +  = cos  3x −  −   2 2 2    2  

The equation y =

1 , and B = 3, and 2 π 1 1 = . C = − . Thus, the amplitude is A = 2 2 2 2π 2π = . The phase shift is The period is B 3 C − π2 π 1 π = = − ⋅ = − . The quarter-period is B 3 2 3 6 y = A cos( Bx − C ) with A =

2π 3

2π 1 π π ⋅ = . The cycle begins at x = − . Add 4 3 4 6 6 quarter-periods to generate x-values for the key points.

=

x=− x=−

coordinates

π 2, 

 4 

3π 4

 3π   4 , 0  

π

(π , − 4)

5π 4

 5π   4 , 0  

3π 2

 3π   2 , 4  

 1  π  cos  3x −  −   is of the form 2  2  

π 6

π

x = 0+ x= x=

π 6

π

+

6

π

+

6

π 6

π

π

=0

6

=

π

=

6

π 3

π

+ = 3 6 2 Evaluate the function at each value of x.

Copyright © 2014 Pearson Education, Inc.


Section 2.1 Graphs of Sine and Cosine Functions

x

coordinates

π

 π 1

6

(0, 0)

π

1 π  6 , − 2  

π 3

π

x=−

− 6 , 2   

0

6

x=−

π 3, 

x=−

π 2

π π

x=

π

4

π

π 1   2, 2  

π

=

4

π

x

coordinates

π

 π 1 − 2 , 2   

− −

2

π 4

0

π 4

π 1 1 cos(2 x + π ) = cos(2 x − (−π )) 2 2 1 The equation y = cos(2 x − (−π )) is of the form 2 1 y = A cos( Bx − C ) with A = , and B = 2, and 2 1 1 C = −π . Thus, the amplitude is A = = . The 2 2 2π 2π period is = = π . The phase shift is B 2 π π C −π = = − . The quarter-period is . The cycle B 2 2 4 y=

begins at x = −

4

=0

4

=

π

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

 0 

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

48.

π

π

+

=−

4

+

4

x = 0+

π

+

2

2

 π   − 4 , 0  

1   0, − 2    π 4, 

 0 

π 1   2, 2  

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

π

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

Copyright © 2014 Pearson Education, Inc.

89


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

π  49. The equation y = −3cos  2 x −  is of the form 2  y = A cos( Bx − C ) with A = –3, and B = 2, and C =

π 2

C=

. Thus, the amplitude is

A = −3 = 3 . The period is phase shift is

π  50. The equation y = −4cos  2 x −  is of the form 2  y = A cos( Bx − C ) with A = –4, and B = 2, and

2π 2π = = π . The B 2

π

. The cycle begins at x =

π

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

π 4

π 4

+

π

2 3π x= + = 2 4 4 3π π + =π x= 4 4 π 5π x =π + = 4 4 Evaluate the function at each value of x. x coordinates

π

π 4

π

4

=

π

π   4 , − 3  

2

π   2 , 0  

3π 4

 3π   4 , 3  

π

(π , 0 )

5π 4

.

cycle begins at x =

A = −4 = 4 . The

π

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

x=

π 4

π 4

+

π

=

π

2 3π x= + = 2 4 4 3π π + =π x= 4 4 π 5π x =π + = 4 4 Evaluate the function at each value of x.

π

4

π

x

coordinates

π

π   4 , − 4  

4

π 2

 5π   4 , − 3   Connect the five points with a smooth curve and graph one complete cycle of the given function

90

. Thus, the amplitude is

2π 2π = = π . The phase shift is B 2 C π2 π 1 π π = = ⋅ = . The quarter-period is . The B 2 2 2 4 4

x=

π

2

period is

C π2 π 1 π = = ⋅ = . B 2 2 2 4

The quarter-period is

π

π 2, 

 0 

3π 4

 3π   4 , 4  

π

(π , 0)

5π 4

 5π   4 , − 4  

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

Copyright © 2014 Pearson Education, Inc.


Section 2.1 Graphs of Sine and Cosine Functions

51.

y = 2 cos(2π x + 8π ) = 2cos(2π x − (−8π )) The equation y = 2 cos(2π x − (−8π )) is of the form y = A cos( Bx − C ) with A = 2, B = 2π , and C = −8π . Thus, the amplitude is A = 2 = 2 . The 2π 2π = = 1 . The phase shift is B 2π C −8π 1 = = −4 . The quarter-period is . The cycle 4 B 2π begins at x = –4. Add quarter-periods to generate xvalues for the key points. x = −4 1 15 x = −4 + = − 4 4 15 1 7 x=− + =− 4 4 2 7 1 13 x=− + =− 2 4 4 13 1 x = − + = −3 4 4 Evaluate the function at each value of x. period is

x

coordinates

–4

(–4, 2)

15  15  − − , 0 4  4  7  7  − − , − 2 2  2  13  13 − − , 4  4 –3

 0 

52.

y = 3cos(2π x + 4π ) = 3cos(2π x − (−4π )) The equation y = 3cos(2π x − (−4π )) is of the form y = A cos( Bx − C ) with A = 3, and B = 2π , and C = −4π . Thus, the amplitude is A = 3 = 3 . The 2π 2π = = 1 . The phase shift is B 2π C −4π 1 = = −2 . The quarter-period is . The cycle 4 B 2π begins at x = –2. Add quarter-periods to generate xvalues for the key points. x = −2 1 7 x = −2 + = − 4 4 7 1 3 x=− + =− 4 4 2 3 1 5 x=− + =− 2 4 4 5 1 x = − + = −1 4 4 Evaluate the function at each value of x. period is

x

coordinates

–2

(–2, 3)

7 4

 7   − 4 , 0  

3 2

 3   − 2 , − 3  

5 4

 5π   − 4 , 0  

–1 (–1, 3) Connect the five key points with a smooth curve and graph one complete cycle of the given function.

(–3, 2)

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

Copyright © 2014 Pearson Education, Inc.

91


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

53. The graph of y = sin x + 2 is the graph of y = sin x shifted up 2 units upward. The period for both 2π π or . functions is 2π . The quarter-period is 4 2 The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x=0

x = 0+ x=

π 2

π

+

x =π +

=

2

π 2

π

π

x = 0+

2

x=

=π =

54. The graph of y = sin x − 2 is the graph of y = sin x shifted 2 units downward. The period for both 2π π or . functions is 2π . The quarter-period is 4 2 The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x=0

3π 2

2 3π π x= + = 2π 2 2 Evaluate the function at each value of x.

π 2

π

+

x =π +

2

π 2

π

π 2

=π =

3π 2

2 3π π + = 2π x= 2 2 Evaluate the function at each value of x.

x

y = sin x + 2

coordinates

x

y = sin x − 2

0

y = sin 0 + 2 = 0+2 = 2

(0, 2)

0

y = sin 0 − 2 = 0 − 2 = −2

π 2

y = sin

π

+2 2 = 1+ 2 = 3

π

y = sin π + 2 = 0+2 = 2

3π 2

y = sin

y = sin 2π + 2 = 0+2 = 2

3π +2 2 = −1 + 2 = 1

π π

  2 , 3  

(π , 2 )  3π   2 , 1  

( 2π , 2 )

π

coordinates (0, –2)

− 2 = 1 − 2 = −1

π   4 , − 1  

π

y = sin π − 2 = 0 − 2 = −2

(π , − 2 )

3π 2

y = sin

y = sin 2π − 2 = 0 − 2 = −2

2

y = sin

2

3π 3π  − 2 = −1 − 2 = −3  , − 3 2  2  (2π , − 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.

92

=

Copyright © 2014 Pearson Education, Inc.


Section 2.1 Graphs of Sine and Cosine Functions 55. The graph of y = cos x − 3 is the graph of y = cos x shifted 3 units downward. The period for both 2π π or . functions is 2π . The quarter-period is 4 2 The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x=0

x = 0+ x=

π 2

π

+

x =π +

=

2

π 2

π

π

x = 0+

2

x=

=π =

3π 2

2 3π π x= + = 2π 2 2 Evaluate the function at each value of x. x

y = cos x − 3

coordinates

0

y = cos 0 − 3 = 1 − 3 = −2

(0, –2)

π 2

π

y = cos

56. The graph of y = cos x + 3 is the graph of y = cos x shifted 3 units upward. The period for both functions 2π π or . The cycle is 2π . The quarter-period is 4 2 begins at x = 0. Add quarter-periods to generate xvalues for the key points. x=0

π

−3

2 = 0 − 3 = −3

y = cos π − 3 = −1 − 3 = −4 3π −3 2 = 0 − 3 = −3

π   2 , − 3  

(π , − 4 )

3π 2

y = cos

 3π   2 , − 3  

y = cos 2π − 3 = 1 − 3 = −2

( 2π , − 2 )

π 2

2

π

+

x =π + x=

π

2

π 2

=

π 2

=π =

3π 2

3π π + = 2π 2 2

Evaluate the function at each value of x. x

y = cos x + 3

0

y = cos 0 + 3 = 1 + 3 = 4

π 2

y = cos

π 2

coordinates

+3 = 0+3 = 3

(0, 4)

π   2 , 3  

π

y = cos π + 3 = −1 + 3 = 2

3π 2

y = cos

3π +3 = 0+3= 3 2

 3π   2 , 3  

y = cos 2π + 3 = 1 + 3 = 4

(2π , 4)

(π , 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.

57. The graph of y = 2sin 12 x + 1 is the graph of y = 2sin 12 x shifted one unit upward. The amplitude for both functions is 2 = 2 . The period for both functions is

2π 1 2

= 2π ⋅ 2 = 4π . The quarter-

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

period is

Copyright © 2014 Pearson Education, Inc.

93


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

points. x=0

58. The graph of y = 2 cos

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 0

π

y = 2sin

1 x +1 2

1  y = 2sin  ⋅ 0  + 1 2  = 2sin 0 + 1 = 2 ⋅ 0 +1 = 0 +1 = 1 1  y = 2sin  ⋅ π  + 1 2 

= 2sin

π

coordinates (0, 1)

(π , 3)

+1

1 x shifted one unit upward. The amplitude 2 for both functions is 2 = 2 . The period for both y = 2 cos

functions is

1  y = 2sin  ⋅ 2π  + 1 2  = 2sin π + 1

x

( 2π , 1)

= 2 ⋅ 0 +1 = 0 +1 = 1 3π

1  y = 2sin  ⋅ 3π  + 1 2  3π = 2sin +1 2 = 2 ⋅ (−1) + 1

0

π (3π , − 1)

1 2

= 2π ⋅ 2 = 4π . The quarter-period is

y = 2 cos

1 x +1 2

1  y = 2 cos  ⋅ 0  + 1 2  = 2 cos 0 + 1 = 2 ⋅1 + 1 = 2 + 1 = 3 1  y = 2 cos  ⋅ π  + 1 2 

= 2 cos

coordinates (0, 3)

(π , 1)

π

+1 2 = 2 ⋅ 0 +1 = 0 +1 = 1

= −2 + 1 = −1 1  y = 2sin  ⋅ 4π  + 1 ( 4π , 1) 2  = 2sin 2π + 1 = 2 ⋅ 0 +1 = 0 +1 = 1 By connecting the points with a smooth curve we obtain one period of the graph.

1  y = 2 cos  ⋅ 2π  + 1 2   = 2 cos π + 1 = 2 ⋅ (−1) + 1 = −2 + 1 = −1

1  y = 2 cos  ⋅ 3π  + 1 2  = 2 ⋅ 0 +1 = 0 +1 = 1

(3π , 1)

1  y = 2 cos  ⋅ 4π  + 1 2   = 2 cos 2π + 1 = 2 ⋅1 + 1 = 2 + 1 = 3

(4π , 3)

94

4π = π . The cycle begins at x = 0. Add quarter4 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.

2 = 2 ⋅1 + 1 = 2 + 1 = 3 2π

1 x + 1 is the graph of 2

Copyright © 2014 Pearson Education, Inc.

(2π , − 1)


Section 2.1 Graphs of Sine and Cosine Functions

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

3 4

1

59. The graph of y = −3cos 2π x + 2 is the graph of y = −3cos 2π x shifted 2 units upward. The amplitude for both functions is −3 = 3 . The period for both functions is

x

y = −3cos 2π x + 2

0

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

(0, –1)

1 4

1  y = −3cos  2π ⋅  + 2 4 

1   4 , 2  

π

2 = −3 ⋅ 0 + 2 = 0+2 = 2 1 2

3 4, 

 2 

(1, –1)

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

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 1 1 x = 0+ = 4 4 1 1 1 x= + = 4 4 2 1 1 3 x= + = 2 4 4 3 1 x = + =1 4 4 Evaluate the function at each value of x.

= −3cos

3  y = −3cos  2π ⋅  + 2 4  3π = −3cos +2 2 = −3 ⋅ 0 + 2 = 0+2 = 2 y = −3cos(2π ⋅1) + 2 = −3cos 2π + 2 = −3 ⋅ 1 + 2 = −3 + 2 = −1

coordinates

+2

1  y = −3cos  2π ⋅  + 2 2  = −3cos π + 2 = −3 ⋅ (−1) + 2 = 3+ 2 = 5

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

y = −3sin 2π x shifted two units upward. The amplitude for both functions is A = −3 = 3 . The period for both functions is

2π = 1 . The quarter2π

1 . The cycle begins at x = 0. Add quarter– 4 periods to generate x-values for the key points. x=0 1 1 x = 0+ = 4 4 1 1 1 x= + = 4 4 2 1 1 3 x= + = 2 4 4 3 1 x = + =1 4 4 Evaluate the function at each value of x.

period is

1   2 , 5  

Copyright © 2014 Pearson Education, Inc.

95


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

x

y = −3sin 2π x + 2

0

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

1 4

1  y = −3sin  2π ⋅  + 2 4 

= −3sin

coordinates (0, 2)

1   4 , − 1  

π

+2 2 = −3 ⋅1 + 2 = −3 + 2 = −1 1 2

1  y = −3sin  2π ⋅  + 2 2  = −3sin π + 2 = −3 ⋅ 0 + 2 = 0 + 2 = 2

1 2, 

3 4

3  y = −3sin  2π ⋅  + 2 4  3π = −3sin +2 2 = −3 ⋅ (−1) + 2 = 3 + 2 = 5

3   4 , 5  

1

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

 2 

(1, 2)

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

61. Select several values of x over the interval.

4 2 1.4 0 0.7

2 0 1

3π 5π 3π 7π π 2π 4 4 2 4 −1.4 −2 −1.4 0 1.4 2 0.7 0 −0.7 −1 −0.7 0

y = 2cos x + sin x 2 2.1

1

−0.7 −2 −2.1

x y1 = 2 cos x y2 = sin x

96

0

π

π

−1

0.7

2

Copyright © 2014 Pearson Education, Inc.


Section 2.1 Graphs of Sine and Cosine Functions

62. Select several values of x over the interval.

4 3 2.1 0 0.7

2 0 1

3π 5π 3π 7π π 2π 4 4 2 4 −2.1 −3 −2.1 0 2.1 3 0.7 0 −0.7 −1 −0.7 0

y = 3cos x + sin x 3 2.8

1

−1.4 −3 −2.8

x y1 = 3cos x y2 = sin x

0

π

π

−1

1.4

3

63. Select several values of x over the interval.

4 0 0.7 0 1

2 1 0

3π 4 0.7 −1

y = sin x + sin 2 x 0 1.7

1

−0.3 0

x y1 = sin x y2 = sin 2 x

0

π

π

5π 3π 7π 2π 4 2 4 −0.7 −1 −0.7 0 −1 1 0 0

π 0 0

0.3

−1 −1.7

0

64. Select several values of x over the interval.

x y1 = cos x y2 = cos 2 x

3π 5π 3π π 4 2 4 4 2 1 0.7 0 −0.7 −1 −0.7 0 −1 1 0 −1 0 1 0 0

π

π

y = cos x + cos 2 x 2 0.7 −1 −0.7

0

−0.7

−1

7π 2π 4 0.7 1 0 1 0.7

2

Copyright © 2014 Pearson Education, Inc.

97


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

65. Select several values of x over the interval.

π

π 3π π 4 2 4 0 0.7 1 0.7 0 1 0 −1 0 1 0

x y1 = sin x y2 = cos 2 x

y = sin x + cos 2 x 1 0.7

0

0.7

5π 3π 7π 2π 4 2 4 −0.7 −1 −0.7 0 −1 0 0 1

−0.7

1

−2

−0.7

1

66. Select several values of x over the interval.

4 1 0.7 0 1

2 0 0

3π 5π 3π π 4 4 2 −0.7 −1 −0.7 0 −1 0 1 0

y = cos x + sin 2 x 1 1.7

0

−1.7 −1

x

π

0

y1 = cos x y2 = sin 2 x

π

0.3

0

7π 4 0.7 −1

−0.3

2π 1 0 1

67. Select several values of x over the interval.

x y1 = sin π x y2 = cos

π 2

1 0.7 0 −0.7 −1 −0.7 0

0.7

0

x

y = sin π x + cos

98

0

7 2 −1

0

π 2

1 2 1

1 0

3 2 −1

2 0

x 1 1.7 0 −1.7 −1

5 2 1

0.3

3

4 0 1

0 −0.3 1

Copyright © 2014 Pearson Education, Inc.


Section 2.1 Graphs of Sine and Cosine Functions

68. Select several values of x over the interval.

x

0

y1 = cos π x

1

y2 = sin

π 2

π 2

1 −1

3 2 0

2 1

5 2 0

3 −1

7 2 0

4 1

0 0.7

1

0.7 0 −0.7 −1 −0.7 0

x 1 0.7

0

0.7 1 −0.7 −2 −0.7 1

x

y = cos π x + sin

1 2 0

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

2π 2π 1 = = period 4π 2 y = A cos Bx 1 y = 3cos x 2

B=

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

2π 2π 1 = = period 4π 2 y = A sin Bx

B=

y = 3sin

1 x 2

71. Using y = A sin Bx the amplitude is 2 and A = −2 , The period is π and thus

2π 2π = =2 period π y = A sin Bx y = −2sin 2 x

B=

72. Using y = A cos Bx the amplitude is 2 and A = −2 , The period is 4π and thus

2π 2π = =2 period π y = A cos Bx y = −2cos 2 x

B=

Copyright © 2014 Pearson Education, Inc.

99


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

73. Using y = A sin Bx the amplitude is 2 and A = 2 , The period is 4 and thus

2π 2π π = = period 4 2 y = A sin Bx

B=

π y = 2sin  2

 x 

74. Using y = A cos Bx the amplitude is 2 and A = 2 , The period is 4 and thus

2π 2π π = = period 4 2 y = A cos Bx

B=

π  y = 2 cos  x  2  75.

76.

77.

78.

100

Copyright © 2014 Pearson Education, Inc.


Section 2.1 Graphs of Sine and Cosine Functions

79. Select several values of x over the interval.

π 4 y1 = cos x 1 0.7 y2 = sin 2 x 0 1 y = cos x − sin 2 x 1 −0.3 x

0

π 2 0 0 0

3π 5π 3π π 4 4 2 −0.7 −1 −0.7 0 −1 0 1 0 0.3

−1 −1.7

0

7π 2π 4 0.7 1 −1 0 1.7

1

80. Select several values of x over the interval.

π π 3π 4 2 4 −1 y1 = cos 2 x 1 0 0 y2 = sin x 0 0.7 1 0.7 y = cos 2 x − sin x 1 −0.7 −2 −0.7 x

0

1 0

5π 3π 7π 2π 4 2 4 −1 0 0 1 −0.7 −1 −0.7 0

1

0.7

π

0

0.7

1

81. Select several values of x over the interval.

x y1 = x y2 = sin x

π π 3π 5π 3π 7π 2π π 4 2 4 4 2 4 0 0.9 1.3 1.5 1.8 2.0 2.2 2.3 2.5 0 0.7 1 0.7 0 −0.7 −1 −0.7 0 0

y = x + sin x 0 1.6 2.3 2.2 1.8

1.3

1.2

1.6

2.5

Copyright © 2014 Pearson Education, Inc.

101


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

82. Select several values of x over the interval.

π π 3π 5π 3π π 4 2 4 4 2 y1 = x 0 0.8 1.6 2.4 3.1 3.9 4.7 y2 = cos x 1 0.7 0 −0.7 −1 −0.7 0 y = x + cos x 1 1.5 1.6 1.6 2.1 3.2 4.7 x

0

7π 2π 4 5.5 6.3 0.7 1 6.2 7.3

83. The period of the physical cycle is 33 days. 84. The period of the emotional cycle is 28 days. 85. The period of the intellectual cycle is 23 days. 86. In the month of February, the physical cycle is at a minimum on February 18. Thus, the author should not run in a marathon on February 18. 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.

102

Copyright © 2014 Pearson Education, Inc.


Section 2.1 Graphs of Sine and Cosine Functions

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.

e.

The amplitude is 3. The period is 365. The C = 79 . The quarter-period is phase shift is B 365 = 91.25 . The cycle begins at x = 79. Add 4 quarter-periods to find the x-values of the key points.

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 xvalues 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)

2π 93. The function y = 3sin ( x − 79) + 12 is of the form 365 C  y = A sin B  x −  + D with B  A = 3 and B =

2π . 365 A = 3 = 3.

a.

The amplitude is

b.

The period is

c.

The longest day of the year will have the most hours of daylight. This occurs when the sine function equals 1. 2π y = 3sin ( x − 79) + 12 365 y = 3(1) + 12

2π 2π 365 = 2π = 2π ⋅ = 365 . B 2π 365

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.

2π ( x − 79) + 12 365 y = 3(−1) + 12 y=9 There will be 9 hours of daylight. y = 3sin

By connecting the points with a smooth curve we obtain one period of the graph, starting on January 1.

2π π 94. The function y = 16sin  x − 3 6

  + 40 is in the 

form y = A sin( Bx − C ) + D with A = 16, B =

π , 6

2π . The amplitude is A = 16 = 16 . 3 2π 2π 6 The period is = π = 2π ⋅ = 12 . The phase B π 6

and C =

shift is

C = B

2π 3

π

6

=

2π 6 ⋅ = 4 . The quarter-period is 3 π

12 = 3 . The cycle begins at x = 4. Add quarter4 periods to find the x-values for the key points.

Copyright © 2014 Pearson Education, Inc.

103


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

x=4 x = 4+3 = 7 x = 7 + 3 = 10 x = 10 + 3 = 13 x = 13 + 3 = 16 Because 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 obtain the graph for 1 ≤ x ≤ 12 .

96. Because the depth of the water ranges from a minimum of 3 feet to a maximum of 5 feet, the curve 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, 2π 12 = B 12 B = 2π

B=

2π π = 12 6

Substitute these values into y = A cos Bx + D . The depth of the water is modeled by y = cos

πx +4. 6

97. – 110. Answers may vary. 111. The function y = 3sin(2 x + π ) = 3sin(2 x − (−π )) is of the form y = A sin( Bx − C ) with A = 3, B = 2, and

C = −π . The amplitude is 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, 2π 12 = B 12 B = 2π

B=

2π π = 12 6

A = 3 = 3 . The

2π 2π = = π . The cycle begins at B 2 π π 3π C −π , and = − . We choose − ≤ x ≤ x= = 2 2 B 2 2 −4 ≤ y ≤ 4 for our graph.

period is

π  112. The function y = −2cos  2π x −  is of the form 2  y = A cos( Bx − C ) with A = –2, B = 2π , and π . The amplitude is A = −2 = 2 . The 2 2π 2π period is = = 1 . The cycle begins at B 2π π C π 1 1 1 9 = . We choose ≤ x ≤ , x= = 2 = ⋅ 4 4 B 2π 2 2π 4 C=

Substitute these values into y = A cos Bx + D . The depth of the water is modeled by y = 3cos

πx +9. 6

and −3 ≤ y ≤ 3 for our graph.

104

Copyright © 2014 Pearson Education, Inc.


Section 2.1 Graphs of Sine and Cosine Functions

113. The function π  π  y = 0.2sin  x + π  = 0.2sin  x − (−π )  is of the  10   10  form y = A sin( Bx − C ) with A = 0.2, B =

C = −π . The amplitude is period is at x =

116.

π

, and 10 A = 0.2 = 0.2 . The

2π 2π 10 = π = 2π ⋅ = 20 . The cycle begins B π 10

The graphs appear to be the same from −

π π to . 2 2

117.

C −π 10 = π = −π ⋅ = −10 . We choose B 10 π

−10 ≤ x ≤ 30 , and −1 ≤ y ≤ 1 for our graph. The graph is similar to y = sin x , except the amplitude is greater and the curve is less smooth. 118. 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 C π = = π . The cycle begins at x = = . Because B 2 B 2 D = 5, the graph has a vertical shift 5 units upward. We π 5π choose ≤x≤ , and 0 ≤ y ≤ 10 for our graph. 2 2

The graph is very similar to y = sin x , except not smooth. 119. a.

b.

y = 22.61sin(0.50 x − 2.04) + 57.17

115.

120. Answers may vary. The graphs appear to be the same from −

π π to . 2 2

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

Copyright © 2014 Pearson Education, Inc.

105


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

125. a.

Since A = 3 and D = −2, the maximum will occur at 3 − 2 = 1 and the minimum will occur

128. y = cos 2 x =

1 1 + cos 2 x 2 2

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

 π 23π π  Viewing rectangle:  − , ,  by [ −5,1,1]  6 6 6 b.

Since A = 1 and D = −2, the maximum will occur at 1 − 2 = −1 and the minimum will occur at −1 − 2 = −3 . Thus the range is [ −3, −1]

 π 7π π  Viewing rectangle:  − , ,  by [ −3, −1,1]  6 6 6 126. A = π

B=

129. Answers may vary.

2π 2π = = 2π period 1

C C = = −2 B 2π C = −4π y = A cos( Bx − C )

π

y = π cos(2π x + 4π ) y = π cos [ 2π ( x + 2) ] 1 1 − cos 2 x 2 2

− 131.

2

π

π 2

< x+

< x+

π

π 4 −

<

π

π 2 <

π

π

4 4 4 2 4 2π π 2π π − − <x< − 4 4 4 4 π 3π − <x< 4 4  3π π  3π π  < x <  or  − ,  x − 4 4  4 4 

or

127. y = sin 2 x =

130.

3π π 2π π + − − 4 4 = 4 = 2 = −π 2 2 2 4

132. a.

b.

106

The reciprocal function is undefined.

Copyright © 2014 Pearson Education, Inc.


Section 2.2 Graphs of Other Trigonometric Functions

π  y  tan  x   from 0 to π . Continue the pattern  2 and extend the graph another full period to the right.

Section 2.2 Check Point Exercises 1.

Solve the equations 2 x  

x= 

π

and

2

π

2x  x

4

π 2

π 4

Thus, two consecutive asymptotes occur at x   and x 

π 4

π

. Midway between these asymptotes is 4 x = 0. 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 and the asymptotes have y-coordinates of –3 and 3. Use the two asymptotes, the x-intercept, and the points midway between to graph one period of y  3 tan 2 x from 

π 4

to

π 4

2.

π

Solve the equations

x

π 2



x

π

π

and

2 

π

x

π 2

x

Solve the equations

π 2

x  0 and x0

. In order to graph

3π x , Continue the pattern and extend 4 4 the graph another full period to the right. for 

3.

π 2

π

π 2

xπ x

π π

2

x2 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). 1 Because the coefficient of the cotangent is , the 2 points on the graph midway between an x-intercept 1 and the asymptotes have y-coordinates of  and 2 1 . Use the two consecutive asymptotes, x = 0 and 2 1 π x = 2, to graph one full period of y  cot x . The 2 2 curve is repeated along the x-axis one full period as shown.

π

2 2 2 2 x0 xπ Thus, two consecutive asymptotes occur at x = 0 and x  π . 0π π x-intercept   2 2 An x-intercept is

π 2

and the graph passes through

π   , 0  . Because the coefficient of the tangent is 1, 2 the points on the graph midway between an xintercept 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

Copyright © 2014 Pearson Education, Inc.

107


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

4.

π  The x-intercepts of y  sin  x   correspond to  4

4.

π 3π  π 3π    ,  ;  ; 4 4 4 4

π  vertical asymptotes of y  csc  x   .  4

5.

3sin 2x

6.

y  2 cos π x

7.

false

8.

true

Exercise Set 2.2 5.

Graph the reciprocal cosine function, y  2 cos 2 x . The equation is of the form y  A cos Bx with A = 2 and B = 2. amplitude: A  2  2 period:

1.

The phase shift,

2π 2π  π B 2

Use quarter-periods,

The graph has an asymptote at x  

π 2

.

C π π , from to  is π units. 2 2 B

C C   π B 1 C  π The function with C  π is y  tan( x  π ) .

Thus,

π

, to find x-values for the five 4 key points. Starting with x = 0, the x-values are π π 3π 0, , , , and π . Evaluating the function at each 4 2 4 value of x, the key points are π  π   3π  (0, 2),  ,0  ,  , 2  ,  , 0  , (π , 2) . In order to 4  2   4 

3π 3π x , Use the first four points 4 4 3π and extend the graph  units to the left. Use the 4 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 2 x .

2.

The graph has an asymptote at x  0 . C π π , from to 0 is  units. Thus, The phase shift, B 2 2 π C C   B 1 2

C

graph for 

π 2

The function with C   3.

π 2

π  is y  tan  x   .  2

The graph has an asymptote at x  π .

π C

π

2

C

π 2

π  The function is y   tan  x   .  2 4. Concept and Vocabulary Check 2.2 1.

π π  π π   ,  ;  ; 4 4 4 4

2.

(0, π ) ; 0; π

3.

(0, 2); 0; 2

108

The graph has an asymptote at

π

. 2 C C There is no phase shift. Thus,  0 B 1 C0 The function with C = 0 is y   tan x .

Copyright © 2014 Pearson Education, Inc.


Section 2.2 Graphs of Other Trigonometric Functions

5.

π x x π  and  4 2 4 2  π π  x    4 x   4  2 2 x  2π x  2π Thus, two consecutive asymptotes occur at x  2π and x  2π . 2π  2π 0  0 x-intercept = 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 and the asymptotes have y-coordinates of –3 and 3. Use the two consecutive asymptotes, x  2π and x x  2π , to graph one full period of y  3 tan from 4 2π to 2π .

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

Solve the equations

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

7.

Solve the equations 2 x  

x

Solve the equations π x x π  and  4 2 4 2  π π  x    4 x   4  2 2 x  2π x  2π Thus, two consecutive asymptotes occur at x  2π and x  2π . 2π  2π 0  0 x-intercept = 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  2π and x  2π , x to graph one full period of y  2 tan from 2π to 4 2π .

2  π2 2

x

π 4

and 2 x  x x

π 2 π

2

2

π 4

Thus, two consecutive asymptotes occur at x   and x 

π 4

π 4

.

 π4  π4

0 0 2 2 An x-intercept is 0 and the graph passes through (0, 1 0). Because the coefficient of the tangent is , the 2 points on the graph midway between an x-intercept 1 and the asymptotes have y-coordinates of  and 2 1 π . Use the two consecutive asymptotes, x   2 4 π 1 and x  , to graph one full period of y  tan 2 x 2 4 x-intercept =

6.

π

from 

π

π

to . Continue the pattern and extend the 4 4 graph another full period to the right.

Copyright © 2014 Pearson Education, Inc.

109


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

8.

Use the two consecutive asymptotes, x  π and 1 x  π , to graph one full period of y  2 tan x 2 from π to π . Continue the pattern and extend the graph another full period to the right.

Solve the equations

2x   

x

π

and 2 x 

2 π

x

2

2

x

π

x

4

π 2 π

2

2

π 4

Thus, two consecutive asymptotes occur at x   and x 

π 4

π 4

.

 π4  π4

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. x-intercept =

Use the two consecutive asymptotes, x  

x

π

π 4

π 4

and

, to graph one full period of y  2 tan 2 x from

π

. 4 4 Continue the pattern and extend the graph another full period to the right.

9.

to

Solve the equations 1 π and x 2 2  π x    2  2

10. Solve the equations 1 π and x 2 2  π x    2  2

x  π xπ Thus, two consecutive asymptotes occur at x  π and x  π . π  π 0  0 x-intercept = 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 and the asymptotes have y-coordinates of 3 and –3. Use the two consecutive asymptotes, x  π and x  π , to graph one full 1 period of y  3tan x from π to π . Continue 2 the pattern and extend the graph another full period to the right.

1 π x 2 2 π  x   2 2

x  π xπ Thus, two consecutive asymptotes occur at x  π and x  π . π  π 0  0 x-intercept = 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.

110

1 π x 2 2 π  x   2 2

Copyright © 2014 Pearson Education, Inc.


Section 2.2 Graphs of Other Trigonometric Functions

11. Solve the equations

xπ   x x

π 2

π 2

xπ 

and π

x

13. An x-intercept is

π

2

Thus, two consecutive asymptotes occur at x  and x 

π

2

3π 2

4π 2

4π π 2 2 4 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 xintercept and the asymptotes have y-coordinates of –1 2

and 1. Use the two consecutive asymptotes, x 

and the graph passes through

and 1. Use the two consecutive asymptotes, x  

π

3π . 2

x-intercept =

4

π   , 0  . Because the coefficient of the tangent is 1, 4 the points on the graph midway between an xintercept and the asymptotes have y-coordinates of –1

2

π

π 2 3π x 2

π

π

π 4

3π and x  , to graph one full period of 4 π  y  tan  x   from 0 to π .  4 Continue the pattern and extend the graph another full period to the right.

π

2 3π and x  , to graph one full period of 2 π 3π to . Continue the pattern y  tan( x  π ) from 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

x

π 4



π

and

2 2π π  x 4 4 x

π 4

x

π 4

 x-intercept =

π

π 2

. The phase shift,

C π π , from 0 to is units. 2 2 B C C π Thus,   B 1 2

2 2π π  x 4 4 3π x 4

Thus, two consecutive asymptotes occur at x  

3π and x  . 4

14. The graph has an asymptote at

π

π 4

C

π 2

The function with C 

π

π  is y   cot  x   .  2 2

3π 2π 4 4  4 π 2 2 4 

Copyright © 2014 Pearson Education, Inc.

111


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

15. The graph has an asymptote at 

π

. The phase shift,

2 C π π , from 0 to  is  units. Thus, 2 2 B C C π   2 B 1 C

π

An x-intercept is

2

The function with C  

π

π  is y  cot  x   .  2 2

16. The graph has an asymptote at π . The phase shift, C , from 0 to π is π units. B C C Thus,   π B 1 C  π The function with C  π is y  cot( x  π ) . 17. Solve the equations x  0 and x  π . Two consecutive asymptotes occur at x  0 and x  π . x-intercept 

18. Solve the equations x  0 and x  π Two consecutive asymptotes occur at x  0 and x  π . 0π π x-intercept =  2 2

0π π  2 2

An x-intercept is

π 2

π 2

and the graph passes through

π   , 0  . Because the coefficient of the cotangent is 2 1 , the points on the graph midway between an x2 intercept and the asymptotes have y-coordinates of 1 1 and  . Use the two consecutive asymptotes, 2 2 x  0 and x  π , to graph one full period of 1 y  cot x . 2 The curve is repeated along the x-axis one full period as shown.

and the graph passes through

π   , 0  . Because the coefficient of the cotangent is 2 2, the points on the graph midway between an xintercept and the asymptotes have y-coordinates of 2 and –2. Use the two consecutive asymptotes, x = 0 and x  π , to graph one full period of y  2 cot x . The curve is repeated along the x-axis one full period as shown.

19. Solve the equations 2 x  0

and

x0

2x  π x

π

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

π 2

.

x-intercept =

0  π2 2

An x-intercept is

π 4

π

2

2

π 4

and the graph passes through

π   , 0  . Because the coefficient of the cotangent is 4 1 , the points on the graph midway between an x2 intercept and the asymptotes have y-coordinates of 1 1 and  . Use the two consecutive asymptotes, 2 2

112

Copyright © 2014 Pearson Education, Inc.


Section 2.2 Graphs of Other Trigonometric Functions

x = 0 and x 

π

, to graph one full period of

2 1 y  cot 2 x . The curve is repeated along the x-axis 2 one full period as shown.

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.

20. Solve the equations 2 x  0

2x  π

and

x0

x

π

2 Two consecutive asymptotes occur at x  0 and x

π 2

22. Solve the equations

.

x-intercept =

0  π2 2

An x-intercept is

π 4

π

2

2

π

4

x  0 and x0

π

π 4

xπ x

π π

4

4

and the graph passes through

π   , 0  . Because the coefficient of the cotangent is 2, 4 the points on the graph midway between an xintercept and the asymptotes have y-coordinates of 2 and –2. Use the two consecutive asymptotes, x  0 and x 

π

, to graph one full period of y  2 cot 2 x .

2 The curve is repeated along the x-axis one full period as shown.

x4 Two consecutive asymptotes occur at x  0 and x  4. 04 4 x-intercept =  2 2 2 An x-intercept is 2 and the graph passes through  2, 0 . Because the coefficient of the cotangent is –2, the points on the graph midway between an xintercept 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

y  2 cot

π

x . The curve is repeated along the x4 axis one full period as shown.

21. Solve the equations

π 2

x0 x0

and

π 2

xπ x

π π

2

x2 Two consecutive asymptotes occur at x = 0 and x = 2. 02 2  1 x-intercept = 2 2 An x-intercept is 1 and the graph passes through (1, 0).

Copyright © 2014 Pearson Education, Inc.

113


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

23. Solve the equations

x

π 2

0

x  0 x

x

and

π 2

π

An x-intercept is

π

xπ

2

π

x

2

π 2

π 2

Two consecutive asymptotes occur at x  

x

π 2

π 2

and

.

 π2  π2

0 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. x-intercept =

Use the two consecutive asymptotes, x  

π 4

and the graph passes through

π   , 0  . Because the coefficient of the cotangent is 4 3, the points on the graph midway between an xintercept and the asymptotes have y-coordinates of 3 and –3. Use the two consecutive asymptotes, π 3π x   and x  , to graph one full period of 4 4 π  y  3cot  x   . The curve is repeated along the x 4 axis one full period as shown.

π

and 2 π π  x  , to graph one full period of y  3cot  x   .  2 2

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

1 x 25. The x-intercepts of y   sin corresponds to 2 2 1 x vertical asymptotes of y   csc . Draw the 2 2 vertical asymptotes, and use them as a guide to 1 x sketch the graph of y   csc . 2 2

24. Solve the equations

x

π 4

0

and

x  0 x

x

π

π

4

π

xπ

4

π

x

4

π

3π 4

Two consecutive asymptotes occur at x  

x

π 4

and

3π . 4

x-intercept =

114

26. The x-intercepts of y  3sin 4 x correspond to vertical asymptotes of y  3csc 4 x . Draw the vertical asymptotes, and use them as a guide to sketch the graph of y  3csc 4 x .

4

 π4 

2

3π 4

2π 4

2

π 4

Copyright © 2014 Pearson Education, Inc.


Section 2.2 Graphs of Other Trigonometric Functions

1 cos 2π x corresponds to 2 1 vertical asymptotes of y  sec 2π x . Draw the 2 vertical asymptotes, and use them as a guide to 1 sketch the graph of y  sec 2π x . 2

27. The x-intercepts of y 

intercepts, and use them as guides to graph y  3csc x .

30. Graph the reciprocal sine function, y  2 sin x . The equation is of the form y  A sin Bx with A  2 and B 1. amplitude: A  2  2 period: 28. The x-intercepts of y  3cos

π 2

x correspond to

vertical asymptotes of y  3sec

π

x . Draw the 2 vertical asymptotes, and use them as a guide to sketch the graph of y  3sec

29.

π 2

x.

2π 2π   2π 1 B

Use quarter-periods,

π , to find x-values for the five 2

key points. Starting with x = 0, the x-values are π 3π 0, , π , , and 2π . Evaluating the function at 2 2 each value of x, the key points are π   3π  (0, 0),  , 2  , (π , 0),  ,  2  , and (2π , 0) . 2   2  Use these key points to graph y  2 sin x from 0 to 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, and use them as guides to graph y = 2cscx.

Graph the reciprocal sine function, y  3sin x . The equation is of the form y  A sin Bx with A = 3 and B = 1. amplitude: A  3  3 period:

2π 2π   2π 1 B

Use quarter-periods,

π

, to find x-values for the 2 five key points. Starting with x = 0, the x-values are π 3π , and 2π . Evaluating the function at 0, , π , 2 2 each value of x, the key points are (0, 0), π   3π   , 3 , (π , 0),  ,  3 , and (2π , 0) . Use 2 2 these key points to graph y  3sin x from 0 to 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-

Copyright © 2014 Pearson Education, Inc.

115


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

1 x sin . The 2 2 1 equation is of the form y  A sin Bx with A  and 2 1 B . 2 1 1 amplitude: A   2 2 2π 2π period:  1  2π  2  4π B 2

31. Graph the reciprocal sine function, y 

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

3  (0, 0),  2π ,  , (4π , 0),  2

3   6π ,   , and (8π , 0) . 2 3 x Use these key points to graph y  sin from 0 to 2 4 8π . Extend the graph one cycle to the right. Use the graph to obtain the graph of the reciprocal function. Draw vertical asymptotes through the xintercepts, and use them as guides to graph 3 x y  csc . 2 4

 1 value of x, the key points are (0, 0),  π ,  , (2π , 0),  2 1   3π ,   , and (4π , 0) . Use these key points to 2 1 x graph y  sin from 0 to 4π . 2 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 1 x them as guides to graph y  csc . 2 2

33. Graph the reciprocal cosine function, y  2 cos x . The equation is of the form y  A cos Bx with A = 2 and B = 1. amplitude: A  2  2 period:

2π 2π   2π B 1

Use quarter-periods,

3 x sin . The 2 4 3 equation is of the form y  A sin Bx with A  and 2 1 B . 4 3 3 amplitude: A   2 2 2π 2π period:  1  2π  4  8π B 4

32. Graph the reciprocal sine function, y 

, to find x-values for the five 2 key points. Starting with x = 0, the x-values are 0, π 3π , π, , 2π . Evaluating the function at each value 2 2 π  of x, the key points are (0, 2),  , 0  , (π ,  2), 2 

 3π   , 0  , and (2π , 2) . Use these key points to graph 2 y  2 cos x from 0 to 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, and use them as guides to graph y  2 sec x .

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

116

π

Copyright © 2014 Pearson Education, Inc.


Section 2.2 Graphs of Other Trigonometric Functions 34. Graph the reciprocal cosine function, y  3cos x . The equation is of the form y  A cos Bx with A  3 and B  1 . amplitude: A  3  3 period:

2π 2π   2π B 1

Use quarter-periods,

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 x guides to graph y  sec . 3

π

, to find x-values for the five 2 key points. Starting with x = 0, the x-values are π 3π 0, , π , , and 2π . Evaluating the function at 2 2 each value of x, the key points are π   3π  (0, 3),  , 0  , (π ,  3),  , 0  , (2π , 3) . 2   2 

Use these key points to graph y  3cos x from 0 to 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, and use them as guides to graph y  3sec x .

x . The 2 equation is of the form y  A cos Bx with A  1 and

36. Graph the reciprocal cosine function, y  cos

1 . 2 amplitude: A  1  1 B

period:

2π 2π  1  2π  2  4π B 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, 1), π , 0 , (2π ,  1),  3π , 0 , and (4π , 1) .

x from 0 to 2 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, x and use them as guides to graph y  sec . 2 Use these key points to graph y  cos

x . The 3 equation is of the form y  A cos Bx with A = 1 and

35. Graph the reciprocal cosine function, y  cos

1 . 3 amplitude: B

A  1 1

2π 2π period:  1  2π  3  6π B 3 6π 3π  , to find x-values for 4 2 the five key points. Starting with x = 0, the x-values 3π 9π are 0, , 3π , , and 6π . Evaluating the function 2 2  3π  at each value of x, the key points are (0, 1),  , 0  ,  2  Use quarter-periods,

 9π  (3π ,  1),  , 0  , and (6π , 1) . Use these key  2  points to graph y  cos

x from 0 to 6π . Extend the 3

Copyright © 2014 Pearson Education, Inc.

117


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions 37. Graph the reciprocal sine function, y  2 sin π x . The equation is of the form y  A sin Bx with A = –2 and B  π . amplitude: A  2  2 period:

Draw vertical asymptotes through the x-intercepts, 1 and use them as guides to graph y   csc π x . 2

2π 2π  2 π B

2 1  , to find 4 2 x-values for the five key points. Starting with x = 0, 1 3 the x-values are 0, , 1, , and 2. Evaluating the 2 2 function at each value of x, the key points are (0, 0), 1  3   ,  2  , (1, 0),  , 2  , and (2, 0) . Use these key 2 2 points to graph y  2 sin 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. Use quarter-periods,

1 39. Graph the reciprocal cosine function, y   cos π x . 2 The equation is of the form y  A cos Bx with A

1 and B  π . 2

amplitude: Draw vertical asymptotes through the x-intercepts, and use them as guides to graph y  2 csc π x .

period:

A  

1 1  2 2

2π 2π  2 π B

2 1  , to find x-values for the 4 2 five key points. Starting with x = 0, the x-values are 1 3 0, , 1, , and 2. Evaluating the function at each 2 2 1  value of x, the key points are  0,   ,  2 Use quarter-periods,

1 38. Graph the reciprocal sine function, y   sin π x . 2 The equation is of the form y  A sin Bx with 1 and B  π . 2 1 1 amplitude: A    2 2 2π 2π  2 period: π B 2 1 Use quarter-periods,  , to find x-values for the 4 2 five key points. Starting with x = 0, the x-values are 1 3 0, , 1, , and 2 . Evaluating the function at each 2 2 value of x, the key points are 1 1 3 1 (0, 0),  ,   , (1, 0),  ,  , and (2, 0) . 2 2 2 2 A

1   , 0  , 2

1  1 3   1,  ,  , 0  ,  2,   . Use these key 2 2 2 1 points to graph y   cos π x from 0 to 2. Extend 2 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 y   sec π x . 2

1 Use these key points to graph y   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.

118

Copyright © 2014 Pearson Education, Inc.


Section 2.2 Graphs of Other Trigonometric Functions

3 40. Graph the reciprocal cosine function, y   cos π x . 2 The equation is of the form y  A cos Bx with 3 and B  π . 2 3 3 amplitude: A    2 2 2π 2π  2 period: π B 2 1 Use quarter-periods,  , to find x-values for the 4 2 five key points. Starting with x = 0, the x-values are 1 3 0, , 1, , and 2 . Evaluating the function at each 2 2 value of x, the key points are 3 1   3  3   3   0,   ,  , 0  , 1,  ,  , 0  ,  2,   . 2 2 2 2 2 A

3 Use these key points to graph y   cos 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, 3 and use them as guides to graph y   sec π x . 2

these key points to graph 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. Draw vertical asymptotes through the x-intercepts, and use them as guides to graph y  csc( x  π ) .

π  42. Graph the reciprocal sine function, y  sin  x   .  2 The equation is of the form y  A sin( Bx  C ) with π

A  1 , B  1 , and C 

2

.

amplitude: A  1  1

2π 2π   2π B 1 π C 2 π   phase shift: B 1 2 period:

Use quarter-periods,

π 2

, to find x-values for the five

key points. Starting with x 

π 2

, the x-values are

π

41. Graph the reciprocal sine function, y  sin( x  π ) . The equation is of the form y  A sin( Bx  C ) with A = 1, and B = 1, and C  π . amplitude: A  1  1

2π 2π   2π period: B 1 C π phase shift:  π B 1 2π π Use quarter-periods,  , to find 4 2 x-values for the five key points. Starting with x  π , 3π 5π the x-values are π , , 2π , , and 3π . Evaluating 2 2 the function at each value of x, the key points are  3π   5π  (π , 0),  , 1 , (2π , 0),  ,  1 , (3π , 0) . Use  2   2 

3π 5π . Evaluating the function at , π, , 2π , and 2 2 2 each value of x, the key points are π   3π   5π   , 0  , π , 1 ,  , 0  ,  2π ,  1 , and  , 0  . 2 2 2

π  Use these key points to graph y  sin  x   from  2 π 2

to

5π . Extend the graph one cycle to the right. 2

Copyright © 2014 Pearson Education, Inc.

119


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

Use the graph to obtain the graph of the reciprocal function. Draw vertical asymptotes through the xintercepts, and use them as guides to graph π  y  csc  x   .  2

44. Graph the reciprocal cosine function, π  y  2 cos  x   . The equation is of the form  2 y  A cos( Bx  C ) with A  2 and B  1 , and

C

π

. 2 amplitude: A  2  2

2π 2π   2π B 1 π C 2 π   phase shift: B 1 2 period:

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 . amplitude:

2π 2π   2π B 1 C π phase shift:   π B 1 2π π Use quarter-periods,  , to find x-values for 4 2 the five key points. Starting with x  π , the xperiod:

π

π

, and π . Evaluating the 2 2 function at each value of x, the key points are  π  π  (π , 2),   , 0  ,  0,  2  ,  , 0  , and (π , 2) . 2   2  Use these 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 xintercepts, and use them as guides to graph y  2 sec( x  π ) .

120

, 0,

π 2

, to find x-values for the five

key points. Starting with x  

A  2 2

values are π , 

Use quarter-periods,

π 2

, 0,

π 2

, π , and

π 2

, the x-values are

3π . 2

Evaluating the function at each value of x, the key points are  π  π   3π    , 2  ,  0, 0 ,  ,  2  , π , 0 ,  , 2  . 2 2 2

π  Use these key points to graph y  2 cos  x    2 3π . Extend the graph one cycle to the 2 2 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 π  graph y  2 sec  x   .  2

from 

π

Copyright © 2014 Pearson Education, Inc.

to


Section 2.2 Graphs of Other Trigonometric Functions

45.

50.

51. 46.

52. 47.

53.

π  y   f  h  ( x )  f (h( x ))  2 sec  2 x    2

54.

π  y   g  h  ( x )  g ( h( x ))  2 tan  2 x    2

48.

49.

Copyright © 2014 Pearson Education, Inc.

121


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions 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 3π 7π ,  , , x 4 4 4 4 x  3.93,  0.79, 2.36, 5.50

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

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 3π 7π ,  , , x 4 4 4 4 x  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 58. Use a graphing utility with y1  1 / cos x and y2  1 . For the window use Xmin  2π , Xmax  2π , Ymin  2 , and Ymax  2 . x  2π , 0, 2π

x  6.28, 0, 6.28 59.

d  12 tan 2π t a. Solve the equations 2π t   t

π

2  π2

and

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. d cot x  2 d  2 cot x Use the equations x  0 and x  π . Two consecutive asymptotes occur at x  0 and x  π . Midway between x  0 and x  π is x  An x-intercept is

2π t  t

π 2 π

2

2π 2π 1 1 t t 4 4 Thus, two consecutive asymptotes occur at 1 1 x   and x  . 4 4  14  14 0 x-intercept   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 ycoordinates of –12 and 12. Use the two 1 1 consecutive asymptotes, x   and x  , to 4 4 graph one full period of d  12 tan 2π t . To

122

b.

π 2

π 2

.

and the graph passes through

π   , 0  . Because the coefficient of the cotangent is 2 2, the points on the graph midway between an xintercept and the asymptotes have y-coordinates of –2 and 2. Use the two consecutive asymptotes, x  0 and x  π , to graph y  2 cot x for 0  x  π .

Copyright © 2014 Pearson Education, Inc.


Section 2.2 Graphs of Other Trigonometric Functions

61. Use the function that relates the acute angle with the hypotenuse and the adjacent leg, the secant function. d sec x  10 d  10sec x Graph the reciprocal cosine function, y  10 cos x . The equation is of the form y  A cos Bx with A  10 and B  1. amplitude: A  10  10 period: For 

π 2

2π 2π   2π B 1 x

π 2

65. – 76. Answers may vary.

, use the x-values 

 π  find the key points   , 0  ,  2 

64. Graphs will vary.

π

, 0, and

π

2 π  (0, 10), and  , 0  . 2  2

to

77.

π

period:

B

π 1 4

Graph y  tan

 π  4  4π x for 0  x  8π . 4

Connect these points with a smooth curve, then draw vertical asymptotes through the x-intercepts, and use  π π them as guides to graph d  10sec x on   ,  .  2 2

78. period:

π B

π 4

Graph y  tan 4 x for 0  x 

π 2

.

62. Graphs will vary.

79. period:

π

80. period:

π

π

B 2 Graph y  cot 2 x for 0  x  π .

63.

B

π 1 2

Graph y  cot

Copyright © 2014 Pearson Education, Inc.

 π  2  2π x for 0  x  4π . 2

123


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

81. period:

π B

π 1 π

2π 2π  π B 2 π C 6 π   phase shift: B 2 12

85. period:

1 tan π x for 0  x  2 . 2

Graph y 

Thus, we include

π

x

12 5π graph for 0  x  . 2

25π in our graph, and 12

82. Solve the equations

π x 1   πx   x

π

and π x  1 

2

π

1

2

π 2

1

π

π  2 x 2π x  0.82

π

πx  x

π 2

π

1

2 π

2

1

2π 2π  2 B π π C 6 π 1 1     phase shift: B π 6 π 6 1 25 Thus, we include  x  in our graph, and 6 6 9 graph for 0  x  . 2

86. period:

π π 2 x 2π x  0.18

π

 1 B π Thus, we include 0.82  x  1.18 in our graph of 1 y  tan(π x  1) , and graph for 0.85  x  1.2 . 2

period:

87. 83. period:

2π 2π  1  2π  2  4π B 2

Graph the functions for 0  x  8π . The graph shows that carbon dioxide concentration rises and falls each year, but over all the concentration increased from 1990 to 2008. 88. 84. period:

y  sin

1 x

2π 2π 3  π  2π   6 B π 3

Graph the functions for 0  x  12 . The graph is oscillating between 1 and –1. The oscillation is faster as x gets closer to 0. Explanations may vary. 89.

124

makes sense

Copyright © 2014 Pearson Education, Inc.


Section 2.2 Graphs of Other Trigonometric Functions

90.

makes sense

91.

does not make sense; Explanations will vary. 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 consecutive asymptotes at 2π 2π 2π x = 0 and x  . The period is 0 . Thus, 3 3 3 π 2π  B 3 2π B  3π

3π 3  B 2π 2 The points on the graph midway between an xintercept and the asymptotes have y-coordinates of 1 and –1. Thus, A = 1. There is no phase shift. Thus, 3 C = 0. An equation for this graph is y  cot x . 2 94. The graph has the shape of a secant function. The reciprocal function has amplitude A  1 . The

8π 2π 8π . Thus,  3 B 3 8π B  6π 6π 3  B 8π 4 There is no phase shift. Thus, C  0 . An equation for 3 the reciprocal function is y  cos x . Thus, an 4 3 equation for this graph is y  sec x . 4

96. The range shows that A  π . Since the period is 2, the coefficient of x is given by 2π B where 2 B 2π 2 B 2 B  2π

Bπ Thus, y  π csc π x 97. a.

Since A=1, the range is  , 1  1,  

7π   π Viewing rectangle:   , π ,  by  3, 3,1 6   6 b.

Since A=3, the range is  , 3   3,  

 1 7  Viewing rectangle:   , ,1 by  6,6,1  2 2  98.

y  2 x sin x 2  x decreases the amplitude as x gets larger. Examples may vary.

99. a.

period is

95. The range shows that A  2. Since the period is 3π , the coefficient of x is given 2π by B where  3π B 2π  3π B 3Bπ  2π

B

2 3

Thus, y  2 csc

2x 3

b.

yes; Explanations will vary.

c.

The angle is 

π 6

.

 π 1 This is represented by the point   ,   .  6 2 100. a.

b.

yes; Explanations will vary.

c.

The angle is

5π . 6

 5π 3 This is represented by the point  ,  . 2   6

Copyright © 2014 Pearson Education, Inc.

125


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

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

101. a.

b.

yes; Explanations will vary.

c.

The angle is 

π 3

.

 π  This is represented by the point   ,  3  .  3 

Mid-Chapter 2 Check Point 1.

The equation y  4 sin 2 x is of the form y  A sin Bx with A = 4 and B  2 . Thus, the amplitude is

A  4  4 . The period is

2π 2π π   π . The quarter-period is . The cycle 4 B 2 begins at x = 0. Add quarter-periods to generate xvalues for the key points and evaluate the function at each value of x. x

y  4 sin 2 x

π 4

π 2

3π 4

π

126

1 π cos x is of the form 2 3 1 π y  A cos Bx with A  and B  . The 2 3 1 1 amplitude is A   . The period is 2 2 6 3 2π 2π  π  6 . The quarter-period is  . Add B 4 2 3 The equation y 

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

0  π 4 sin  2    4  1  4 4  π 4 sin  2    4  0  2 0  3π  4 sin  2    4(1)  4   4 4 sin(2  π )  4  0 0

(0, 0)

π   , 4  4 π   ,0  2

 3π   , 4  4

y

1 π cos x 2 3

coordinates

0

1 1 π  1 cos   0   1  3  2 2 2

 1  0,  2

3 2

1 π  3 1 cos     0  0 2 3 2 2

3   , 0  2

3

1 1 π  1 cos   3   1   3  2 2 2

1   3,   2

9 2

1 π 9 1 cos      0  0  3 2 2 2

9   , 0  2

6

1 1 π  1 cos   6   1  3  2 2 2

 1  6,  2

coordinates

4 sin(2  0)  4  0 0

2.

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

(π , 0)

Copyright © 2014 Pearson Education, Inc.


Mid-Chapter 2 Check Point

3.

The equation y  3sin( x  π ) is of the form y  A sin( Bx  C ) with A = 3, B = 1, and C  π . The amplitude is

A  3  3 . The period is

2π 2π C π   2π . The phase shift is   π . The 1 B B 1 2π π quarter-period is  . The cycle begins at x  π . 4 2 Add quarter-periods to generate x-values for the key points and evaluate the function at each value of x. x

π

3π 2

y  3sin( x  π )

coordinates

3sin π  π   3sin(0)  3 0 0  3π  π   π   3sin   3sin   2  2  31

π , 0   3π   , 3 2

3

3sin  2π  π   3sin(π )  3 0

(2π , 0)

0 5π 2

 5π   3π   π   3sin   3sin   2   2   3(1)

 5π   ,  3 2

 3

3sin  3π  π   3sin(2π )  3 0

4.

π  The equation y  2 cos  2 x   is of the form  4 y  A cos( Bx  C ) with A = 2, and B = 2, and C

π 4

. Thus, the amplitude is

period is

A  2  2 . The

2π 2π   π . The phase shift is 2 B

π

C 4 π   . B 2 8

π

π

. The cycle begins at x  . 4 4 Add quarter-periods to generate x-values for the key points and evaluate the function at each value of x. The quarter-period is

x

coordinates

π 8

π   , 2  8

3π 8

 3π   , 0  8

5π 8

 5π   ,  2  8

7π 8

 7π   , 0  8

9π 8

 9π   , 2  8

3π , 0 Connect the five points with a smooth curve and graph one complete cycle of the given function

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

Copyright © 2014 Pearson Education, Inc.

127


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

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

2π π  π . The quarter-period is . The cycle begins at x = 0. Add quarter2 4 periods to generate x-values for the key points and evaluate the function at each value of x. 1  1 . The period for both functions is

x

y  cos 2 x  1

coordinates

0

y  cos  2  0  1  2

(0, 2)

 π y  cos  2    1  1  4

π   ,1 4

2

 π y  cos  2    1  0  2

π   ,0  2

3π 4

 3π  y  cos  2    1  1  4 

 3π   ,1 4

π

y  cos  2  π   1  2

(π , 2)

π 4

π

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 y1  2sin x 0 1.4 2 1.4 y2  2 cos x y  2sin x  2 cos x 2 2.8

128

π 2 2 0 2

3π 5π π 4 4 1.4 0 1.4 1.4 2 1.4 2 2.8 0

3π 2 2 0 2

7π 2π 4 1.4 0 1.4 2 0 2

Copyright © 2014 Pearson Education, Inc.


Mid-Chapter 2 Check Point

7.

Solve the equations

π 4

x

π

π

and

2

4

 π4 x     2π

x

π

2 π  4 x  2π

x  2 x2 Thus, two consecutive asymptotes occur at x  2 and x  2 . 2  2 0 x-intercept =  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 to 2 . 4 Continue the pattern and extend the graph another full period to the right.

8.

Solve the equations 2 x  0

and

x0

2x  π x

π 2

Two consecutive asymptotes occur at x  0 and x  x-intercept =

0  π2 2

π

2

2

π 2

.

π 4

π  and the graph passes through  , 0  . Because the coefficient of the cotangent is 4, the points on the 4  4 graph midway between an x-intercept and the asymptotes have y-coordinates of 4 and –4. Use the two consecutive An x-intercept is

π

asymptotes, x  0 and x 

π 2

, to graph one full period of y  4 cot 2 x . The curve is repeated along the x-axis one full period

as shown.

Copyright © 2014 Pearson Education, Inc.

129


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

9.

Graph the reciprocal cosine function, y  2 cos π x . The equation is of the form y  A cos Bx with A  2 and B  π . amplitude: A   2  2 period:

2π 2π  2 π B

Use quarter-periods,

2 1  , to find x-values for the five key points. Starting with x = 0, the x-values are 4 2

1 3 , 1, , and 2 . Evaluating the function at each value of x, the key points are 2 2 1 3 0,  2 ,  , 0  , 1, 2 ,  , 0  ,  2,  2 . 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, and use them as guides to graph y  2 sec π x . 0,

10. Graph the reciprocal sine function, y  3sin 2π x . The equation is of the form y  A sin Bx with A  3 and B  2π . amplitude: A  3  3 period:

2π 2π  1 B 2π

1 1 1 3 , to find x-values for the five key points. Starting with x = 0, the x-values are 0, , , , and 1 . 4 4 2 4 1  1  3  Evaluating the function at each value of x, the key points are (0, 0),  , 3 ,  , 0  ,  , 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 . Use quarter-periods,

130

Copyright © 2014 Pearson Education, Inc.


Section 2.3 Inverse Trigonometric Functions

Section 2.3 Check Point Exercises 1.

Let θ  sin 1

3 3 , then sin θ  . 2 2

3 π π 3 π  π π The only angle in the interval   ,  that satisfies sin θ  is . Thus, θ  , or sin 1  . 3 3 2 2 3  2 2 2.

 2 2 Let θ  sin 1   , then sin θ   .  2  2   2 π π 2 π  π π is  . Thus θ   , or sin 1   The only angle in the interval   ,  that satisfies cos θ   4 . 4 4 2  2 2  2 

3.

1 1 2π  1 . Thus, Let θ  cos1    , then cos θ   . The only angle in the interval [0, π ] that satisfies cos θ   is  2 2 2 3

θ 4.

2π  1  2π , or cos1     .  2 3 3

π  π π Let θ  tan 1 ( 1) , then tan θ  1 . The only angle in the interval   ,  that satisfies tan θ  1 is  . Thus  2 2 4 θ

π 4

or tan 1 θ  

π 4

.

5.

Scientific Calculator Solution Function

Mode

a.

1 cos1    3

Radian

1  3 

b.

tan 1 ( 35.85)

Radian

35.85

Display (rounded to four places)

Keystrokes

COS1

1.2310

TAN 1

–1.5429

Graphing Calculator Solution

6.

Function

Mode

a.

1 cos1    3

Radian

b.

tan 1 ( 35.85)

Radian

a.

cos cos1 0.7

Display (rounded to four places)

Keystrokes

COS1

TAN 1

( 1  3 )

ENTER

 35.85 ENTER

1.2310

–1.5429

x  0.7 , x is in [–1,1] so cos(cos 1 0.7)  0.7

Copyright © 2014 Pearson Education, Inc.

131


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

b.

sin 1 (sin π )  π π x  π , x is not in   ,  . x is in the domain  2 2 1 of sin x, so sin (sin π )  sin 1 (0)  0

c.

cos cos1 π

9.

x  π , x is not in [–1,1] so cos cos1 π is not

x . 1

Use the Pythagorean Theorem to find the third side, a. a 2  x 2  12

defined. 7.

Let θ  tan 1 x , then tan θ =x =

a  x2  1 Use the right triangle to write the algebraic expression.

3 3 Let θ  tan   , then tan θ  . Because tan θ is 4 4 positive, θ is in the first quadrant. 1

sec tan 1 x  sec θ 

x2  1  1

Concept and Vocabulary Check 2.3

8.

π

Use the Pythagorean Theorem to find r. r 2  32  42  9  16  25

2.

0  x  π ; cos 1 x

r  25  5 Use the right triangle to find the exact value. 3 side opposite θ 3  sin  tan 1   sin θ    4 hypotenuse 5

3.

4.

 π π [1,1] ;   ,   2 2

5.

[1,1] ; [0, π ]

6.

 π π (,  ) ;   ,   2 2

7.

 π π  2 , 2   

8.

[0, π ]

9.

 π π   ,  2 2

1  1 Let θ  sin 1    , then sin θ   . Because sin θ  2 2 is negative, θ is in quadrant IV.

Use the Pythagorean Theorem to find x. x 2  (1)2  22

x2  1  4

π 2

x

10. false

x 3 2

x 3 Use values for x and r to find the exact value.  x 3  1  cos sin 1      cos θ     2 r 2  

132

; sin 1 x

2

x

π

1.

Copyright © 2014 Pearson Education, Inc.

2

π 2

; tan 1 x

x2  1


Section 2.3 Inverse Trigonometric Functions

Exercise Set 2.3 7. 1.

2.

1 1 , then sin θ  . The only angle in 2 2 1 π  π π the interval   ,  that satisfies sin θ  is . 6 2 2 2   π 1 π Thus, θ  , or sin 1  . 2 6 6 Let θ  sin 1

1

Let θ  sin 0 , then sin θ  0 . The only angle in the

8.

 π π interval   ,  that satisfies sin θ  0 is 0. Thus  2 2

θ  0 , or sin 1 0  0 . 3.

2 2 , then sin θ  . The only angle 2 2 2  π π in the interval   ,  that satisfies sin θ  is 2 2 2   Let θ  sin 1

π 4 4.

π 4

, or sin 1

2 π  . 2 4

3 3 , then sin θ  . The only angle 2 2 3  π π in the interval   ,  that satisfies sin θ  is 2 2 2   Let θ  sin 1

π 3 5.

. Thus θ 

. Thus θ 

π 3

, or sin 1

3 π  . 2 3

1  1 Let θ  sin 1    , then sin θ   . The only  2 2  π π angle in the interval   ,  that satisfies  2 2 1 π π sin θ   is  . Thus θ   , or 2 6 6 π  1 sin 1      .  2 6

6.

 3 3 Let θ  sin 1   , then sin θ   .  2  2   π π The only angle in the interval   ,  that satisfies  2 2

9.

3 3 , then cos θ  . The only angle 2 2 3 π in the interval [0, π ] that satisfies cos θ  is . 2 6 π 3 π  . Thus θ  , or cos1 2 6 6 Let θ  cos1

2 2 , then cos θ  . The only angle 2 2 2 π is . in the interval  0, π  that satisfies cos θ  2 4 π 2 π  . Thus θ  , or cos1 2 4 4 Let θ  cos1

 2 2 Let θ  cos 1    , then cos θ   2 . The only  2  angle in the interval [0, π ] that satisfies 2 3π 3π . Thus θ  , or is 2 4 4  2  3π cos1    4 . 2   cos θ  

 3 3 , then cos θ   . 10. Let θ  cos 1    2  2  The only angle in the interval  0, π  that satisfies cos θ  

3 5π 5π is . Thus θ  , or 2 6 6

 3  5π . cos1     2  6 11. Let θ  cos 1 0 , then cos θ  0 . The only angle in the interval [0, π ] that satisfies cos θ  0 is Thus θ 

π 2

, or cos1 0 

π 2

π

2

.

.

12. Let θ  cos1 1 , then cos θ  1 . The only angle in the interval  0, π  that satisfies cos θ  1 is 0 . Thus θ  0 , or cos1 1  0 .

3 π π is  . Thus θ   , 2 3 3   3 π or sin 1   3.  2  sin θ  

Copyright © 2014 Pearson Education, Inc.

133


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

3 3 , then tan θ  . The only angle in the interval 3 3

13. Let θ  tan 1

θ

π 6

, or tan 1

3 π  π π is . Thus   ,  that satisfies tan θ  2 2 3 6

3 π  . 3 6

π  π π 14. Let θ  tan 1 1 , then tan θ  1 . The only angle in the interval   ,  that satisfies tan θ  1 is .  2 2 4 Thus θ 

π

, or tan 1 1 

4

π 4

.

 π π 15. Let θ  tan 1 0 , then tan θ  0 . The only angle in the interval   ,  that satisfies tan θ  0 is 0. Thus θ  0 , or  2 2 tan 1 0  0 .

π  π π 16. Let θ  tan 1 ( 1) , then tan θ  1 . The only angle in the interval   ,  that satisfies tan θ  1 is  . Thus  2 2 4 θ

π 4

, or tan 1 (1)  

π 4

.

π  π π 17. Let θ  tan 1  3 , then tan θ   3 . The only angle in the interval   ,  that satisfies tan θ   3 is  .  2 2 3 Thus θ  

π

, or tan 1  3  

3

π 3

.

 3 3 π 3  π π , then tan θ   . The only angle in the interval   ,  that satisfies tan θ   is  . 18. Let θ  tan 1      2 2 3 3 6  3  Thus θ  

 3 π , or tan 1   6 . 6  3 

π

19.

Scientific Calculator Solution Function

Mode

Keystrokes

Display (rounded to two places)

sin 1 0.3

Radian

0.3 SIN 1

0.30

Graphing Calculator Solution Function

Mode

Keystrokes

Display (rounded to two places)

sin 1 0.3

Radian

SIN 1 0.3 ENTER

0.30

20.

Scientific Calculator Solution Function

Mode

sin 1 0.47

Radian

Keystrokes

Display (rounded to two places)

0.47 SIN 1

0.49

Graphing Calculator Solution

134

Function

Mode

Keystrokes

Display (rounded to two places)

sin 1 0.47

Radian

SIN 1 0.47 ENTER

0.49

Copyright © 2014 Pearson Education, Inc.


Section 2.3 Inverse Trigonometric Functions

21.

Scientific Calculator Solution Function

Mode

sin 1 ( 0.32)

Radian

Display (rounded to two places)

Keystrokes 

0.32

SIN 1

–0.33

Graphing Calculator Solution Function

Mode

sin 1 ( 0.32)

Radian

22.

Display (rounded to two places)

Keystrokes

SIN 1

 0.32 ENTER

–0.33

Scientific Calculator Solution Function

Mode

sin 1 (0.625)

Radian

Display (rounded to two places)

Keystrokes

0.625

SIN 1

–0.68

Graphing Calculator Solution Function

Mode

sin 1 (0.625)

Radian

23.

Display (rounded to two places)

Keystrokes

SIN 1

 0.625 ENTER

–0.68

Scientific Calculator Solution Function

Mode

 3 cos1   8

Radian

Display (rounded to two places)

Keystrokes

3  8 

COS1

1.19

Graphing Calculator Solution Function

Mode

 3 cos1   8

Radian

24.

Display (rounded to two places)

Keystrokes

COS1

( 3  8 )

1.19

ENTER

Scientific Calculator Solution Function

Mode

4 cos1   9

Radian

Display (rounded to two places)

Keystrokes

4  9 =

COS1

1.11

Graphing Calculator Solution Function

Mode

4 cos1   9

Radian

Display (rounded to two places)

Keystrokes

COS1

( 4  9 )

ENTER

Copyright © 2014 Pearson Education, Inc.

1.11

135


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

25.

Scientific Calculator Solution Function

Mode

5 7

Radian

cos1

Display (rounded to two places)

Keystrokes

COS1

 7 

5

1.25

Graphing Calculator Solution Function

Mode

5 7

Radian

cos1

26.

Display (rounded to two places)

Keystrokes

COS1

5  7 )

(

1.25

ENTER

Scientific Calculator Solution Function

Mode

7 10

Radian

cos1

Display (rounded to two places)

Keystrokes

 10 =

7

COS1

1.30

Graphing Calculator Solution Function

Mode

7 10

Radian

cos1

27.

Display (rounded to two places)

Keystrokes

COS1

7  10 )

(

1.30

ENTER

Scientific Calculator Solution Function

Mode

tan 1 ( 20)

Radian

Keystrokes 

20

Display (rounded to two places)

TAN 1

–1.52

Graphing Calculator Solution Function

Mode

tan 1 ( 20)

Radian

28.

Keystrokes

TAN 1

Display (rounded to two places)

 20 ENTER

–1.52

Scientific Calculator Solution Function

Mode

tan 1 ( 30)

Radian

Keystrokes 

30

Display (rounded to two places)

TAN 1

–1.54

Graphing Calculator Solution

136

Function

Mode

tan 1 ( 30)

Radian

Keystrokes

TAN 1

Display (rounded to two places)

 30 ENTER

Copyright © 2014 Pearson Education, Inc.

–1.54


Section 2.3 Inverse Trigonometric Functions

29.

Scientific Calculator Solution Function

tan 1  473

Mode

Radian

Display (rounded to two places)

Keystrokes 

473

TAN 1

–1.52

Graphing Calculator Solution Function

tan 1  473

Mode

Radian

30.

Display (rounded to two places)

Keystrokes

TAN 1

(

473 )

–1.52

ENTER

Scientific Calculator Solution Function

tan 1  5061

Mode

Radian

Display (rounded to two places)

Keystrokes 

5061

TAN 1

–1.56

Graphing Calculator Solution Function

tan 1  5061

31.

sin sin 1 0.9

Mode

Radian

Display (rounded to two places)

Keystrokes

TAN 1

(

5061 )

ENTER

–1.56

x  0.9, x is in [1, 1], so sin(sin 1 0.9)  0.9 32.

cos(cos1 0.57) x  0.57, x is in [1, 1], so cos(cos1 0.57)  0.57

33.

 π sin 1  sin   3 x

34.

π

 π π  π π , x is in   ,  , so sin 1  sin    3 3 3  2 2

2π   cos1  cos   3  2π , x is in [0, π ], 3 2π  2π  so cos1  cos   3  3 x

35.

 5π  sin 1  sin   6  x

5π , x is not in 6

5π  π  π π 1  1  1    2 , 2  , x is in the domain of sin x, so sin  sin 6   sin  2   6  

Copyright © 2014 Pearson Education, Inc.

137


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

36.

4π   cos1  cos   3 

43.

 π π x  π , x is not in   ,  ,  2 2 x is in the domain of sin x, so sin 1 (sin π )  sin 1 0  0

4π , x is not in [0, π ], x 3 x is in the domain of cos x, 4π    1  2π  cos1     so cos1  cos    2 3  3 37.

tan tan 1 125

44.

cos1 (cos 2π ) x  2π , x is not in [0, π ], x is in the domain of cos x, so cos1 (cos 2π )  cos 1 1  0

45.

sin sin 1 π

x  125, x is a real number, so tan tan 1 125  125

38.

tan(tan 1 380) 1

so tan(tan 380)  380 39.

tan

46.

  π   tan   6    

4 4 , then sin θ  . Because sin θ is 5 5 positive, θ is in the first quadrant.

47. Let θ  sin 1

π

 π π , x is in   ,  ,  2 2 3

  π  π so tan 1  tan         3 3  

x2  y2  r2 x 2  42  52

2π   tan 1  tan   3  2π , x is not in 3

 π π   ,  , x is in the domain of 2 2

2π  π  1 tan x, so tan 1  tan   tan  3    3 3 42.

cos(cos 1 3π )

  π  tan 1  tan       3 

x

x 2  25  16  9 x3 x 3 4  cos  sin 1   cos θ    5 r 5

3π   tan 1  tan   4  3π  π π , x is not in   ,  ,  2 2 4 x is in the domain of tan x 3π  π  so tan 1  tan   tan 1 ( 1)     4 4 x

138

so cos(cos1 3π ) is not defined.

  π  π tan 1  tan         6 6  

x

x  3π , x is not in [1, 1]

 π π x   , x is in   ,  , so  2 2 6

41.

defined.

π

40.

x  π , x is not in [1, 1] , so sin sin 1 π is not

x  380, x is a real number,

1

sin 1 (sin π )

Copyright © 2014 Pearson Education, Inc.


Section 2.3 Inverse Trigonometric Functions

7 7 , then tan θ  . 24 24 Because tan θ is positive, θ is in the first quadrant.

48. Let θ  tan 1

x2  y 2  r 2 x 2  52  132 x 2  144 x  12 x 12 5  cot  sin 1   cot θ    13  y 5 3  3 51. Let θ  sin 1    , then sin θ   . Because sin θ  5 5 is negative, θ is in quadrant IV.

r2  x2  y 2 r 2  72  242 r 2  625 r  25 y 7  7  sin  tan 1   sin θ    24  r 25 5 5 , then cos θ  . Because cos θ is 13 13 positive, θ is in the first quadrant.

49. Let θ  cos 1

x2  y2  r2 x 2  (3)2  52 x 2  16 x4  y 3  3  tan sin 1      tan θ      5 x 4   4  4 52. Let θ  sin 1    , then sin θ   .  5 5 Because sin θ is negative, θ is in quadrant IV.

x2  y 2  r2 52  y 2  132 y 2  169  25 y 2  144 y  12 y 12 5  tan  cos1   tan θ     13 x 5 5 5 50. Let θ  sin then sin θ  . 13 13 because sin θ is positive, θ is in the first quadrant. 1

x2  y2  r2 x 2  (4)2  52 x2  9 x3  x 3  4  cos sin 1      cos θ     5 r 5  

Copyright © 2014 Pearson Education, Inc.

139


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

2 2 , then cos θ  . Because cos θ 2 2 is positive, θ is in the first quadrant.

53. Let, θ  cos1

x2  y 2  r2 x 2  (1)2  42 x 2  15 x  15  r 4 4 15  1  sec sin 1      sec θ     4  x 15 15 

x y r 2

 2

2

2

1  1 56. Let θ  sin 1    , then sin θ   .  2 2 Because sin θ is negative, θ is in quadrant IV.

2

 y 2  22 y2  2 y 2

 2 y 2  sin θ   sin  cos1 2  r 2  1 1 , then sin θ  . 2 2 Because sin θ is positive, θ is in the first quadrant.

54. Let θ  sin 1

x2  y2  r2 x 2  (1)2  22 x2  3 x 3  r 2 2 3  1  sec sin 1      sec θ     2  x 3 3  1  1 57. Let θ  cos1    , then cos θ   . Because  3 3 cos θ is negative, θ is in quadrant II.

x2  y 2  r2 x 2  12  22 x2  3 x 3 1 x 3  cos  sin 1   cos θ    2 r 2 1  1 55. Let θ  sin 1    , then sin θ   . Because sin θ  4 4 is negative, θ is in quadrant IV.

x2  y2  r2

(1)2  y 2  32 y2  8 y 8 y2 2 Use the right triangle to find the exact value.  y 2 2  1  tan  cos1      tan θ    2 2   x 3 1  

140

Copyright © 2014 Pearson Education, Inc.


Section 2.3 Inverse Trigonometric Functions

1  1 58. Let θ  cos1    , then cos θ   .  4 4 Because cos θ is negative, θ is in quadrant II.

x2  y2  r2

x2   2

2

 22

x2  2 x 2

  2  r 2 sec sin 1    2   sec θ    x 2  2    2  2 61. Let θ  tan 1    , then tan θ   .  3 3 Because tan θ is negative, θ is in quadrant IV.

x2  y 2  r2

(1)2  y 2  42 y 2  15 y  15

 y 15  1  tan  cos1      tan θ     15   4  x 1   3 3 59. Let θ  cos    , then cos θ   2 . Because 2   cos θ is negative, θ is in quadrant II. 1

r2  x2  y 2 r 2  32   2 

2

r2  9  4 r 2  13 r  13

 x 3 3 13  2  cos  tan 1      cos θ     3  13 r 13  3  3 62. Let θ  tan 1    , then tan θ   .  4 4 Because tan θ is negative, θ is in quadrant IV.

x2  y2  r2

 3 

2

 y 2  22 y2  1 y 1

  3  r 2 csc  cos1     csc θ    2  y 1  2     2 2 60. Let θ  sin 1    , then sin θ   2 .  2  Because sin θ is negative, θ is in quadrant IV.

r2  x2  y 2 r 2  42   3

2

r 2  16  9 r 2  25 r5  y 3 3  3   sin  tan 1      sin θ     r 4  5 5 

Copyright © 2014 Pearson Education, Inc.

141


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

63. Let θ  cos1 x , then cos θ  x 

x . 1

Use the Pythagorean Theorem to find the third side, b. x 2  b2  12

b2  1  x 2 b  1  x2 Use the right triangle to write the algebraic expression.

tan cos 1 x  tan θ 

1  x2 x

64. Let θ  tan 1 x , then tan θ  x 

x . 1

Use the Pythagorean Theorem to find the third side, c. c 2  x 2  12

c  x2  1 Use the right triangle to write the algebraic expression.

sin(tan 1 )  sin θ x  x2  1  

x x2  1

66. Let θ  cos1 2 x. Use the Pythagorean Theorem to find the third side, b.

(2 x )2  b2  12 b2  1  4 x 2 b  1  4 x2

sin cos1 2 x  67. Let θ  sin 1

1 1 , then sin θ  . x x

Use the Pythagorean Theorem to find the third side, a. a 2  12  x 2

a 2  x2  1 a  x2  1 Use the right triangle to write the algebraic expression.

1  cos  sin 1   cos θ   x

x2  1 x2  1

x x2  1 x2  1

65. Let θ  sin 1 2x , then sin θ  2x y = 2x, r = 1 Use the Pythagorean Theorem to find x. x 2  (2 x )2  12

x2  1  4 x2 x  1  4 x2 cos(sin 1 2 x )  1  4 x 2

142

1  4 x2  1  4 x2 1

Copyright © 2014 Pearson Education, Inc.

x2  1 x


Section 2.3 Inverse Trigonometric Functions

68. Let θ  cos1

1 1 , then cos θ  . x x

Use the Pythagorean Theorem to find the third side, b. 12  b2  x 2

b  x 1 2

2

x 4

a2 

 2 x  70. cot  tan 1    x 2 x

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

73. a. , then sin θ 

x x 4 2

x2  9

2

 x2

a3 Use the right triangle to write the algebraic expression.  x2  9  3 cot  sin 1  2   x x 9

 3 x  69. cot  tan 1   x  3

2

x2  9 . x

a 2  x2  x2  9  9

b  x2  1 Use the right triangle to write the algebraic expression. 1 x  sec  cos1   sec θ   x  1 x

71. Let θ  sin 1

x2  9 , then sin θ  x

72. Let θ  sin 1

.

3 x2  9

x2  9 x2  9

3 x2  9 x2  9

y  sec x is the reciprocal of y  cos x . The xvalues for the key points in the interval [0, π ] are 0,

π π 3π

, , , 4 2 4 π 2 (0, 1),  , , 4 2 

and π . The key points are

2  π   3π  2 , 0  ,  4 ,  2  , and  

(π ,  1) , Draw a vertical asymptote at x 

x2  4

2

.

Now draw our graph from (0, 1) through π   , 2  to  on the left side of the 4 asymptote. From  on the right side of the  3π  asymptote through  ,  2  to (π ,  1) .  4 

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

a2  x2 

π

2

a2  x2  4  x2  4 a2 Use the right triangle to write the algebraic expression.   x x2  4 sec  sin 1  sec θ   2  x2  4 

b.

With this restricted domain, no horizontal line intersects the graph of y  sec x more than

Copyright © 2014 Pearson Education, Inc.

143


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

once, so the function is one-to-one and has an inverse function. c.

75.

Reflecting the graph of the restricted secant function about the line y = x, we get the graph of y  sec1 x .

domain: [−1, 1]; range: [0, π] 76. 74. a.

Two consecutive asymptotes occur at x  0 and x  π . Midway between x  0 and x  π is x  0 and x  π . An x-intercept for the graph

π  is  , 0  . The graph goes through the points 2  π   3π   , 1 and  ,  1 . Now graph the 4 4 function through these points and using the asymptotes.

domain: [−1, 1];  π 3π  range:  ,  2 2  77.

b.

With this restricted domain no horizontal line 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 y  cot 1 x .

domain: [−2, 0]; range: [0, π] 78.

c.

domain: [−2, 0];  π π range:   ,   2 2

144

Copyright © 2014 Pearson Education, Inc.


Section 2.3 Inverse Trigonometric Functions

83.

79.

domain: (−∞, ∞); range: (−π, π) 80.

domain: [−2, 2]; range: [0, π] 84.

domain: (−∞, ∞);  3π 3π  range:   ,  2 2  81.

domain: [−2, 2];  π π range:   ,   2 2 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 domain: (1, 3]; range: [−π, 0] 82.

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 domain: [1, 3];  π π range:   ,   2 2

interval  ,   . The outer function returns values

 π π on the interval   ,   2 2 domain:

Copyright © 2014 Pearson Education, Inc.

 ,   ;

 π π range:   ,   2 2

145


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

89. The inner function, cos x, accepts values on the

interval  ,   . The outer function returns values

95.

 π π on the interval   ,   2 2

 ,   ;

domain:

 π π range:   ,   2 2

θ  2 tan 1

97.

tan 1 b  tan 1 a  tan 1 2  tan 1 0  1.1071 square units

98.

tan 1 b  tan 1 a  tan 1 1  tan 1 (2)  1.8925 square units

interval  ,   . The outer function returns values

 ,   ;

domain:

range:  0, π 

91. The functions sin 1 x and cos1 x accept values on the interval  1,1 . The sum of these values is always

π 2

21.634  0.1440 radians; 300  180   8.2 0.1440   π 

96.

90. The inner function, sin x, accepts values on the on the interval  0, π 

21.634  1.3157 radians; 28  180   75.4 1.3157   π 

θ  2 tan 1

99. – 109. Answers may vary.

.

 1,1 ;

domain:

110. y  sin 1 x

π  range:   2

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 

θ  tan

1

10 15 20 25

to

3π 2

 π 3π  range:   ,   2 2 

33 8  tan 1 x x

θ

x 5

2

 1,1 ;

domain:

93.

π

33 8  tan 1  0.408 radians 5 5 33 8  tan 1  0.602 radians tan 1 10 10 33 8  tan 1  0.654 radians tan 1 15 15 33 8  tan 1  0.645 radians tan 1 20 20 33 8  tan 1  0.613 radians tan 1 25 25 tan 1

The graph of the second equation is the graph of the first equation shifted up 2 units. 111. The domain of y  cos1 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] 4  2 viewing rectangle.

The graph of the second equation is the graph of the first equation shifted right 1 unit.

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.

146

Copyright © 2014 Pearson Education, Inc.


Section 2.3 Inverse Trigonometric Functions

112. y  tan 1 x

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.

y  2 tan 1 x

118. makes sense The graph of the second equation is the graph of the first equation reversed and stretched. 113. The domain of y  sin 1 x is the interval

 π π [–1, 1], and the range is   ,  . 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 , 2 , 2  by   [–2, 3, 1] viewing rectangle.

The graph of the second equation is the graph of the first equation shifted left 2 units and up 1 unit. 114. y  tan 1 x

119. does not make sense; Explanations will vary. Sample explanation:  2 π  5π  sin 1  sin   sin 1   4  4  2   120. y  2sin 1 ( x  5)

y  sin 1 ( x  5) 2 y sin  x  5 2 y x  sin  5 2 121. 2sin 1 x 

π

sin 1 x 

π

4 8

x  sin

π 8

122. Prove: If x > 0, tan 1 x  tan 1

1 π  x 2

Since x > 0, there is an angle θ with 0  θ 

π 2

as

shown in the figure.

Observations may vary.

π  1 tan θ  x and tan   θ   thus 2  x

115.

It seems sin 1 x  cos1 x 

π 2

for 1  x  1 .

1 π tan 1 x  θ and tan 1     θ so x 2 1 π π tan 1 x  tan 1  θ   θ  x 2 2

116. does not make sense; Explanations will vary. Sample explanation: The cosine’s inverse is not a function over that interval.

Copyright © 2014 Pearson Education, Inc.

147


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions 123. Let α equal the acute angle in the smaller right triangle. 8 tan α  x 8 so tan 1  α x 33 tan(α  θ )  x 33 so tan 1  α θ x 33 8 θ  α  θ  α  tan 1  tan 1 x x 124.

tan A 

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. 8.4 tan 62.7  b 8.4 b  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. 8.4 sin 62.7  c 8.4 c  9.45 sin 62.7 In summary, B = 27.3°, b ≈ 4.34, and c ≈ 9.45.

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. a tan 85.4  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. 6.7 sin A  13.8 6.7 A  sin 1  29.0 13.8 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 triangle. Therefore, use the tangent function.

a b

a 12.1 a  12.1tan 22.3 a  4.96

tan 22.3 

b c 12.1 cos 22.3  c 12.1 c cos 22.3 c  13.08 cos A 

opposite adjacent 18 tan θ  25  18  θ  tan 1    25 

125. tan θ 

θ  35.8 π  126. 10cos  x  6  amplitude: 10  10 period:

2π π

6

148

 2π 

6

π

 12

Copyright © 2014 Pearson Education, Inc.


Section 2.4 Applications of Trigonometric Functions

a 800 a  800 tan 32  499.9 b tan 35  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. tan 32 

5.

a.

b.

6.

a.

7.

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. 2π period  4

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 northsouth line. Thus, the bearing from O to D is S 25°E. 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 northsouth line. Thus the bearing from O to C is S 15° W.

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 equation with the cosine function, y  a cos ω t , to

ω

2π  4ω 2π π  4 2 Substitute these values into d  a cos wt. The equation for the ball’s simple harmonic motion is

ω

d  6 cos 8.

To find your bearing from the entrance to the trail system, consider a north-south line passing through the entrance. The acute angle from this line to the ray on which you lie is 31  θ . Because we are measuring the angle from the south side of the line and you are west of the entrance, your bearing from the entrance is S (31  θ ) W. To find θ , Use a right triangle and the tangent function. 3.5 tan θ  2.3 3.5 θ  tan 1  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.

π

4

t , a  12 and ω 

π

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

f 

c  17.54  4.2 You are approximately 4.2 miles from the entrance to the trail system. b.

t.

2

We begin by identifying values for a and ω .

d  12 cos

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 c 2  a 2  b2  (2.3)2  (3.5)2  17.54

π

π ω π 1 1  4    2π 2π 4 2π 8

The frequency is c.

1 cm per second. 8

The time required for one cycle is the period. 2π 2π 4  π  2π   8 period 

ω

π

4

The time required for one cycle is 8 seconds. Concept and Vocabulary Check 2.4 1.

sides; angles

2.

north; south

3.

simple harmonic; a ;

Copyright © 2014 Pearson Education, Inc.

ω

;

ω 2π

149


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

Exercise Set 2.4 1.

2.

3.

150

Find the measure of angle B. Because 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. a tan 23.5  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. 10 cos 23.5  c 10 c  10.90 cos 23.5 In summary, B = 66.5°, a ≈ 4.35, and c ≈ 10.90. Find the measure of angle B. Because C = 90°, A + B = 90°. Thus, B  90  A  90  41.5  48.5. Because we have a known angle, a known adjacent side, and an unknown opposite side, use the tangent function. a tan 41.5  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. 20 cos 41.5  c 20 c  26.70 cos 41.5 In summary, B  48.5, a  17.69 , and c  26.70 . Find the measure of angle B. Because C = 90°, A + B = 90°. 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. a sin 52.6  54 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. b cos52.6  54 b  54 cos52.6  32.80 In summary, B = 37.4°, a ≈ 42.90, and b ≈ 32.80.

4.

Find the measure of angle B. Because C = 90°, A + B = 90°. 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. a sin 54.8  80 a  80sin 54.8  65.37 Because we have a known angle, a known hypotenuse, and an unknown adjacent side, use the cosine function. b cos54.8  80 b  80 cos54.8  46.11 In summary, B  35.2, a  65.37 , and c  46.11 .

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 opposite side and an unknown adjacent side, use the tangent function. 30.5 tan16.8  a 30.5 a  101.02 tan16.8 Because we have a known angle, a known opposite side, and an unknown hypotenuse, use the sine function. 30.5 sin16.8  c 30.5 c  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°. Thus, A  90  B  90  23.8  66.2 . Because we have a known angle, a known opposite side, and an unknown adjacent side, use the tangent function. 40.5 tan 23.8  a 40.5 a  91.83 tan 23.8 Because we have a known angle, a known opposite side, and an unknown hypotenuse, use the sine function. 40.5 sin 23.8  c 40.5 c  100.36 sin 23.8 In summary, A  66.2, a  91.83 , and c  100.36 .

Copyright © 2014 Pearson Education, Inc.


Section 2.4 Applications of Trigonometric Functions

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. 30.4 sin A  50.2  30.4  A  sin 1   37.3  50.2  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. a 2  b2  c 2

(30.4)2  b2  (50.2)2 b2  (50.2)2  (30.4)2  1595.88 b  1595.88  39.95 In summary, A ≈ 37.3°, B ≈ 52.7°, and b ≈ 39.95. 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. 11.2 sin A  65.8  11.2  A  sin 1   9.8  65.8  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. a 2  b2  c 2

(11.2)2  b2  (65.8)2 b  (65.8)  (11.2)  4204.2 2

2

2

b  4204.2  64.84 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. 10.8 tan A  24.7  10.8  A  tan 1   23.6  24.7  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. c 2  a 2  b2  (10.8)2  (24.7)2  726.73

c  726.73  26.96 In summary, A ≈ 23.6°, B ≈ 66.4°, and c ≈ 26.96.

10. 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. 15.3 tan A  17.6  15.3  A  tan 1   41.0  17.6  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. c 2  a 2  b2  (15.3)2  (17.6)2  543.85

c  543.85  23.32 In summary, A  41.0, B  49.0 , and c  23.32 . 11. Find the measure of angle A. Because we have a known hypotenuse, a known adjacent side, and unknown angle, use the cosine function. 2 cos A  7 2 A  cos 1    73.4 7 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. a 2  b2  c2

a 2  (2) 2  (7)2 a 2  (7)2  (2)2  45 a  45  6.71 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. 4 cos A  9 4 A  cos 1    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. a 2  b2  c2

a 2  (4) 2  (9)2 a 2  (9)2  (4)2  65 a  65  8.06 In summary, A  63.6, B  26.4 , and a  8.06 .

Copyright © 2014 Pearson Education, Inc.

151


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

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 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. 16. We need the acute angle between ray OD and the north-south line through O. This angle measures 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 t  0, d  6. Notice the greatest distance from rest position occurs at t  0. Thus, we will use the equation with the cosine function, y  a cos ω t to model 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. 2π period  4

ω

2π  4ω 2π π  4 2 Substitute these values into d  a cos ω t. The equation for the object’s simple harmonic motion is

ω

d  6 cos

π 2

t.

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 equation with the cosine function, y  a cos ω t , to model 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. 2π 2 period =

ω

2π  2ω 2π π 2 Substitute these values into d  a cos ω t . The equation for the object’s simple harmonic motion is d  8cos π t .

ω

19. When t = 0, d = 0. Therefore, we will use the equation with the sine function, y  a sin ω t , to 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. 2π period   1.5

ω

2π  1.5ω 2π 4π  1.5 3 Substitute these values into d  a sin ω t. The equation for the object’s simple harmonic motion is 4π d  3sin t. 3 20. When t = 0, d = 0. Therefore, we will use the equation with the sine function, y  a sin ω t , to model the

ω

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. 2π period =  2.5

ω

2π  2.5ω 2π 4π  2.5 5 Substitute these values into d  a sin ω t . The equation for the object’s simple harmonic motion is 4π d  5sin t. 5

ω

152

Copyright © 2014 Pearson Education, Inc.


Section 2.4 Applications of Trigonometric Functions 21. We begin by identifying values for a and ω .

d  5cos a.

b.

t , a  5 and ω 

π

2 2 The maximum displacement from the rest position is the amplitude. Because a = 5, the maximum displacement is 5 inches. The frequency, f, is

f 

c.

π

24. We begin by identifying values for a and ω .

d  8cos

ω

2

π

The maximum displacement from the rest position is the amplitude. Because a = –6, the maximum displacement is 6 inches.

b.

The frequency, f, is ω 2π f    1. 2π 2π The frequency is 1 inch per second.

c.

The time required for one cycle is the period. 2π 2π  1 period  ω 2π The time required for one cycle is 1 second.

2

b.

The frequency, f, is f  The frequency is

c.

π ω 1  2  . 2π 2π 4

1 inch per second. 4

The time required for one cycle is the period. 2π 2π 2 period =  π  2π   4

ω

2

π

The time required for one cycle is 4 seconds. 25. We begin by identifying values for a and ω . 1 1 d  sin 2t, a  and ω  2 2 2 a.

The maximum displacement from the rest position is the amplitude. 1 Because a  , the maximum displacement is 2 1 inch. 2

b.

The frequency, f, is ω 2 1 f     0.32. 2π 2π π The frequency is approximately 0.32 cycle per second.

c.

The time required for one cycle is the period. 2π 2π   π  3.14 period  ω 2 The time required for one cycle is approximately 3.14 seconds.

23. We begin by identifying values for a and ω . d  6 cos 2π t , a  6 and ω  2π a.

π

The maximum displacement from the rest position is the amplitude. Because a = –8, the maximum displacement is 8 inches.

The time required for one cycle is 4 seconds. 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. ω 2π b. The frequency, f, is f    1. 2π 2π The frequency is 1 inch per second. c. The time required for one cycle is the period. 2π 2π  1 period = ω 2π The time required for one cycle is 1 second.

2

t , a  8and ω 

a.

π ω π 1 1  2    . 2π 2π 2 2π 4

1 The frequency is inch per second. 4 The time required for one cycle is the period. 2π 2π 2  π  2π   4 period 

π

26. We begin by identifying values for a and ω . 1 1 d  sin 2t , a  and ω  2 3 3 a.

The maximum displacement from the rest position is the amplitude. 1 Because a  , the maximum displacement is 3 1 inch . 3

Copyright © 2014 Pearson Education, Inc.

153


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

b.

The frequency, f, is f 

ω 2 1    0.32 . 2π 2π π

The frequency is approximately 0.32 cycle per second. c.

The time required for one cycle is the period. 2π 2π   π  3.14 period = ω 2 The time required for one cycle is approximately 3.14 seconds.

27. We begin by identifying values for a and ω . 2π 2π d  5sin t , a  5and ω  3 3 a.

b.

The maximum displacement from the rest position is the amplitude. Because a = –5, the maximum displacement is 5 inches. 2π ω 2π 1 1  3    . 2π 2π 3 2π 3

The frequency is

The time required for one cycle is the period. 2π 2π 3 period =   2π  3 2π 2π ω 3 The time required for one cycle is 3 seconds.

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

f 

3π ω 3π 1 3  2    . 2π 2π 2 2π 4

The frequency is

3 cycle per second. 4

The time required for one cycle is the period. 2π 2π 2 4 period =  3π  2π   3 3 ω π 2 The required time for one cycle is

154

30.

x  100 tan 20  100 tan 8 x  50

31.

x  600 tan 28  600 tan 25 x  39

32.

x  400 tan 40  400 tan 28 x  123

33.

x

34.

x

35.

x

36.

x

37.

π  d  4 cos  π t    2

1 cycle per second. 3

28. We begin by identifying values for a and ω . 3π 3π d  4 sin t , a  4 and ω  2 2

c.

x  500 tan 40  500 tan 25 x  653

The frequency, f, is

f 

c.

29.

4 seconds. 3

300 300  tan 34 tan 64 x  298 500 500  tan 20 tan 48 x  924 400 tan 40 tan 20 tan 40  tan 20 x  257 100 tan 43 tan 38 tan 43  tan 38 x  482

a.

4 in.

b.

1 in. per sec 2

c.

2 sec

d.

1 2

Copyright © 2014 Pearson Education, Inc.


Section 2.4 Applications of Trigonometric Functions

38.

π  d  3cos  π t    2

a.

3 in.

b.

1 in. per sec 2

c.

2 sec

d.

39.

40.

1 2

 πt π  d  2 sin     4 2

a.

2 in.

b.

1 in. per sec 8

c.

8 sec

d.

−2

1  πt π  d   sin     4 2 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. a tan 21.3  5280 a  5280 tan 21.3  2059 The height of the tower is approximately 2059 feet.

3 ft  90 ft 1 yd Using a right triangle, we have a known angle, an unknown opposite side, a, and a known adjacent side. Therefore, use the tangent function. a tan 38.7  90 a  90 tan 38.7  72 The height of the building is approximately 72 feet.

42. 30 yd 

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. 305 tan 23.7  a 305 a  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. 200 tan 22.3  a 200 a  488 tan 22.3 The ship is about 488 feet offshore.

Copyright © 2014 Pearson Education, Inc.

155


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

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. 1000 tan 36  d 1000 d  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. 800 tan 72  d 800 d  260 tan 72 The stolen car is approximately 260 feet from a point directly below the helicopter. 47. Using a right triangle, we have an unknown angle, A, a known opposite side, and a known hypotenuse. Therefore, use the sine function. 6 sin A  23  6 A  sin 1    15.1  23  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. 250 tan A  40  250  A  tan 1   80.9  40  The angle of elevation of the sun is approximately 80.9°.

156

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. a tan19.2  125 a  125 tan19.2  43.5

b 125 b  125 tan 31.7  77.2 The balloon rises approximately 77.2 – 43.5 or 33.7 feet. tan 31.7 

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. a tan 53  330 a  330 tan 53  437.9

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). tan 63 

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. a sin 53  150 a  150sin 53  120 To find the adjacent side, b, use the cosine function. b cos53  150 b  150 cos 53  90 The boat has traveled approximately 90 miles north and 120 miles east.

Copyright © 2014 Pearson Education, Inc.


Section 2.4 Applications of Trigonometric Functions

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. a sin 64  40 a  40sin 64  36 To find the adjacent side, b, use the cosine function. b cos 64  40 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. 7 tan 28  b 7 b  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. 9 tan 34  b 9 b  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. 1.5 tan A  2  1.5  A  tan    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.

56. Using a right triangle, we have a known adjacent side, a known opposite side, and an unknown angle, A. Therefore, use the tangent function. 6 tan A  9 6 A  tan 1    34 9 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 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 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. 7 tan θ  5 7 θ  tan 1    54.5 5 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 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 port, the ship’s bearing from the port is S (40  θ ) W . To find θ , use a right triangle and the tangent function. 11 tan θ  7  11  θ  tan 1    57.5 7 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 northsouth line. The bearing of the ship from the port is N 82.5° W.

Copyright © 2014 Pearson Education, Inc.

157


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

59. The frequency, f, is f 

ω , so 2π

71.

1 ω  2 2π 1 ω   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.

72.

makes sense

ω

73.

does not make sense; Explanations will vary. Sample explanation: When using bearings, the angle must be less than 90.

74.

does not make sense; Explanations will vary. Sample explanation: When using bearings, north and south are listed before east and west.

75.

does not make sense; Explanations will vary. Sample explanation: Frequency and Period are inverses of each other. If the period is 10 seconds 1 then the frequency is  0.1 oscillations per 10 second.

60. The frequency, f, is f 

, so

1 ω  4 2π 1 π ω   2π  4 2 Because the amplitude is 8 feet, a = 8. Thus, the equation for the object’s simple harmonic motion is d  8sin

π 2

t.

61. The frequency, f, is f 

264 

ω

ω 2π

, so

2π ω  264  2π  528π Thus, the equation for the tuning fork’s simple harmonic motion is d  sin 528π t. 62. The frequency, f, is f 

98,100,000 

ω 2π

ω 2π

, so

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, we can find r by using the sine function. r sin 76.6  r  112 sin 76.6( r  112)  r

r  r sin 76.6  112 sin 76.6

Thus, the equation for the radio waves’ simple harmonic motion is d  sin196, 200,000π t . 63. – 69. Answers may vary.

10 complete oscillations occur.

r sin 76.6  112 sin 76.6  r

ω  98,100,000  2π  196, 200,000π

70.

y  6e 0.09 x cos 2π x

r 1  sin 76.6  112 sin 76.6

112 sin 76.6  4002 1  sin 76.6 The Earth's radius is approximately 4002 miles.

y  4e 0.1x cos 2 x

3 complete oscillations occur.

158

Copyright © 2014 Pearson Education, Inc.

r


Chapter 2 Review Exercises

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. h tan 20  75  d h  (75  d ) tan 20

h Also, tan 40  d h  d tan 40 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 75 tan 20 d tan 40  tan 20 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. 79.

80.

81.

sec x cot x 

1 cos x 1   or csc x cos x sin x sin x

tan x csc x cos x  sec x  tan x 

sin x 1 cos x   1 cos x sin x 1

1 sin x 1  sin x   cos x cos x cos x

Chapter 2 Review Exercises 1.

The equation y  3sin 4 x is of the form y  A sin Bx with A = 3 and B = 4. The amplitude is

A  3  3. 2π 2π π   . The quarter-period is B 4 2

The period is π

2

π 1 

π

. The cycle begins at x = 0. Add 4 2 4 8 quarter-periods to generate x-values for the key points. x0

x  0 x

π 8

π

π

8

π 8

8

π

4 π π 3π x   4 8 8 3π π π   x 8 8 2 Evaluate the function at each value of x. x

coordinates

0

(0, 0)

π

π   , 3 8

8

π 4

π   , 0  4

3π 8

 3π   ,  3 8

π

(2π , 0)

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

Copyright © 2014 Pearson Education, Inc.

149


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

2.

The equation y  2 cos 2 x is of the form y  A cos Bx with A = –2 and B = 2. The amplitude is A  2  2. The period is

2π 2π   π . The B 2

π

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

quarter-period is

x  0 x

π 4

π

π

4

π 4

4

π

3.

1 x is of the form 2 1 y  A cos Bx with A = 2 and B  . The amplitude 2 2π 2π is A  2  2. The period is  1  2π  2  4π . B 2 The equation y  2 cos

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

x  0π  π

2 π π 3π x   2 4 4 3π π x  π 4 4 Evaluate the function at each value of x.

x  π  π  2π x  2π  π  3π x  3π  π  4π Evaluate the function at each value of x. x

coordinates

0

(0, 2)

π

(π , 0)

(2π ,  2)

 2 

(3π , 0)

3π 4

 3π   , 0  4

(4π , 2)

π

(π ,  2)

x

coordinates

0

(0, –2)

π

π   , 0  4

4

π 2

π  , 2

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.

150

Copyright © 2014 Pearson Education, Inc.


Chapter 2 Review Exercises

4.

1 π sin x is of the form 2 3 1 π and B = . The amplitude y  A sin Bx with A = 3 2 1 1 is A   . The period is 2 2 2π 2π 3  π  2π   6. The quarter-period is B π 3 The equation y 

6 3  . The cycle begins at x = 0. Add quarter4 2 periods to generate x-values for the key points. x0 3 3  2 2 3 3 x  3 2 2 3 9 x  3  2 2 9 3 x  6 2 2 Evaluate the function at each value of x. x  0

5.

The equation y   sin π x is of the form y  A sin Bx with A = –1 and B = π . The amplitude is A  1  1. The period is

2π 2π   2. The B π

2 1  . The cycle begins at x = 0. 4 2 Add quarter-periods to generate x-values for the key points. x0 quarter-period is

1 1  2 2 1 1 x   1 2 2 1 3 x  1  2 2 3 1 x  2 2 2 Evaluate the function at each value of x. x  0

x

coordinates

0

(0, 0)

x

coordinates

1 2

1   ,  1 2

0

(0, 0)

1

(1, 0)

3 2

3 1  ,  2 2

3 2

3   , 1 2

3

(3, 0)

2

(2, 0)

9 2

1 9  ,   2 2

6

(6, 0)

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.

Copyright © 2014 Pearson Education, Inc.

151


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

6.

x is of the form 3 1 y  A cos Bx with A = 3 and B = . The amplitude 3 is A  3  3. The equation y  3cos

The period is

2π 2π  1  2π  3  6π . The quarterB 3

6π 3π . The cycle begins at x = 0. Add  4 2 quarter-periods to generate x-values for the key points. x0 period is

3π 3π x  0  2 2 3π 3π   3π x 2 2 3π 9π  x  3π  2 2 9π 3π   6π x 2 2 Evaluate the function at each value of x. x

coordinates

0

(0, 3)

3π 2

 3π   , 0  2

(3π ,  3)

9π 2

 9π   , 0  2

(6π , 3)

7.

The equation y  2 sin( x  π ) is of the form y  A sin( Bx  C ) with A = 2, B = 1, and C = π . The amplitude is A  2  2. The period is

2π 2π C π   2π . The phase shift is   π . The B 1 B 1 2π π quarter-period is  . 4 2 The cycle begins at x  π . Add quarter-periods to generate x-values for the key points. xπ xπ

π

3π 2

2 3π π   2π x 2 2 π 5π x  2π   2 2 5π π   3π x 2 2 Evaluate the function at each value of x. x

coordinates

π

(π , 0)

3π 2

 3π   , 2  2

(2π , 0)

5π 2

 5π   ,  2  2

(3π , 0)

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.

152

Copyright © 2014 Pearson Education, Inc.


Chapter 2 Review Exercises

8.

y  3cos( x  π )  3cos( x  ( π )) The equation y  3cos( x  ( π )) is of the form y  A cos( Bx  C ) with A = –3, B = 1, and C = π .

The amplitude is A  3  3.

2π 2π   2π . The phase shift is B 1 C π 2π π   π . The quarter-period is  . The 4 2 B 1 cycle begins at x  π . Add quarter-periods to generate x-values for the key points. x  π The period is

x  π  x

π 2

x  0 x

π

π

π

π 2 

2

2

π

 π    , 0  2

0

(0, 3)

π

π   , 0  2

π

π 4

. The amplitude is

A

π

. Add quarter-periods to 8 generate x-values for the key points.

x x

π 8

π

π

π

π

x

coordinates

π

 π 3   ,  8 2

8

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

4

π

(π ,  3)

8

8 3π x   8 4 8 3π π 5π x   8 4 8 5π π 7π x   8 4 8 Evaluate the function at each value of x.

coordinates

π

2

B = 2, and C = 

3 , 2

The cycle begins at x  

(π ,  3)

2

the form y  A cos( Bx  C ) with A 

π

2

π

 3  π  cos  2 x      is of  4  2 

The equation y 

0

π

 π 2 2 Evaluate the function at each value of x. x

 3 π 3   π  cos  2 x    cos  2 x        4  2 4 2 

3 3  . 2 2 2π 2π The period is   π . The phase shift is B 2 π C π 1 π π  4      . The quarter-period is . B 2 4 2 8 4



2

9. y 

8

π   , 0  8

3π 8

3  3π  ,   8 2

5π 8

 5π   , 0  8

7π 8

 7π 3   ,  8 2

Copyright © 2014 Pearson Education, Inc.

153


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

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

x

coordinates

π

(

4

π π 5 π 5    π  sin  2 x    sin  2 x        2  2 2 2 

The equation y 

 5  π  sin  2 x      is of  2  2 

5 the form y  A sin( Bx  C ) with A  , 2 B = 2, and C = 

π 2

, 0)

π   , 0  4

4

y

4

 5  0,  2

0

10.

π

2

5 π  ,   2 2

3π 4

 3π   , 0  4

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

. The amplitude is

5 5  . 2 2 2π 2π The period is   π . The phase shift is B 2 π C 2 π 1 π π       . The quarter-period is . B 2 2 2 4 4 A

The cycle begins at x  

π

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

π 4

π 4

x  0 x

π 4

π

4

π

4  

B=

0

154

π

π 3

, and C = 3π . The amplitude is A  3  3.

The period is

π 4

π

is

2 3π x   2 4 4 Evaluate the function at each value of x.

π

4

π

π  11. The equation y  3sin  x  3π  is of 3  the form y  A sin( Bx  C ) with A = –3,

2π 2π 3  π  2π   6. The phase shift B π 3

C 3π 3  π  3π   9. The quarter-period is B π 3

6 3  . The cycle begins at x = 9. Add quarter4 2 periods to generate x-values for the key points.

Copyright © 2014 Pearson Education, Inc.


Chapter 2 Review Exercises

x9 3 21 x  9  2 2 21 3   12 x 2 2 3 27 x  12   2 2 27 3   15 x 2 2 Evaluate the function at each value of x.

12. The graph of y  sin 2 x  1 is the graph of y  sin 2 x shifted one unit upward. The period for

2π π  π . The quarter-period is . 2 4 The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x0 both functions is

x  0 x

π 4

π

4

π 4

π

4

π

2 3π x   2 4 4 3π π x  π 4 4 Evaluate the function at each value of x.

π

π

x

coordinates (0, 1)

x

coordinates

9

(9, 0)

21 2

 21   ,  3 2

12

(12, 0)

0

27 2

 27   , 3 2

π

15

(15, 0)

π

4

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

π  , 4

 2 

2

π   , 1 2

3π 4

 3π   , 0  4

π

(π , 1)

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

Copyright © 2014 Pearson Education, Inc.

155


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

1 1 13. The graph of y  2 cos x  2 is the graph of y  2 cos x shifted two units downward. The period for both functions is 3 3 6π 3π 2π . The cycle begins at x = 0. Add quarter-periods to generate x-values   2π  3  6π . The quarter-period is 1 4 2 3 for the key points. x0

3π 3π  2 2 3π 3π x   3π 2 2 3π 9π x  3π   2 2 9π 3π x   6π 2 2 Evaluate the function at each value of x. x  0

x

coordinates

0

(0, 0)

3π 2

 3π   ,  2  2

(3π ,  4)

9π 2

 9π   ,  2  2

(6π , 0)

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

156

Copyright © 2014 Pearson Education, Inc.


Chapter 2 Review Exercises

14. Select several values of x over the interval.

x

0

π

4 y1  3sin x 0 2.1 y2  cos x 1 0.7 y  3sin x  cos x 1 2.8

π

3π 5π 3π π 4 4 2 2.1 0 2.1 3 0.7 1 0.7 0 1.4 1 2.8 3

2 3 0 3

7π 2π 4 2.1 0 0.7 1 1.4 1

15. Select several values of x over the interval.

x

0

y1  sin x

π

4 0 0.7

π 2 1

3π 5π π 4 4 0.7 0 0.7

1 1 0.9 0.7 0.4 x 2 1 y  sin x  cos x 1 1.6 1.7 1.1 2 y2  cos

16. a.

3π 2 1

7π 2π 4 0.7 0

0 0.4 0.7 0.9

1

1.1 1.7 1.6

1

0

11π  π At midnight x = 0. Thus, y  98.6  0.3sin   0    12 12   11π   98.6  0.3sin    12   98.6  0.3( 0.2588)  98.52 The body temperature is about 98.52°F.

b.

period:

2π 2π 12  π  2π   24 hours B π 12

Copyright © 2014 Pearson Education, Inc.

157


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

c.

x  11

Solve the equation π 11π π  x 12 12 2 π π 11π 6π 11π 17π    x  12 2 12 12 12 12 17π 12   17 x 12 π The body temperature is highest for x = 17.

x  11  6  17 x  17  6  23 x  23  6  29 x  29  6  35 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.

11π  π y  98.6  0.3sin   17    12 12   98.6  0.3sin

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 11π 3π x  12 12 2 3π 11π 18π 11π 29π π x     12 2 12 12 12 12 29π 12 x   29 12 π The body temperature is lowest for x = 29. 11π  π y  98.6  0.3sin   29    12 12 

 3π   98.6  0.3sin    2   98.6  0.3( 1)  98.3 29 hours after midnight or 5 hours after midnight, at 5 A.M., the body temperature is 98.3°F. e.

11π  π The graph of y  98.6  0.3sin  x   is  12 12  of the form y  D  A sin( Bx  C ) with A = 0.3, 11π , and D = 98.6. The 12 12 amplitude is A  0.3  0.3. The period from part (b) is 24. The quarter-period is 24  6. The phase shift is 4 11π C 11π 12  12    11. The cycle begins at x π B 12 π 12 B

π

,C

= 11. Add quarter-periods to generate x-values for the key points.

158

17. Blue: This is a sine wave with a period of 480. Since the amplitude is 1, A  1. 2π 2π π   B period 480 240 The equation is y  sin

π 240

x.

Red: This is a sine wave with a period of 640. Since the amplitude is 1, A  1. 2π 2π π   B period 640 320 The equation is y  sin

π 320

x.

18. Solve the equations

2x   x

π

and 2 x 

2  π2

2 π

x

2

x

π

π

x

2

2

π

4 4 Thus, two consecutive asymptotes occur at x

π 4

and x 

π

. 4  π4  π4

0 0 2 2 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 and the asymptotes have y-coordinates of –4 and 4. x-intercept 

Copyright © 2014 Pearson Education, Inc.


Chapter 2 Review Exercises

Use the two consecutive asymptotes. x  

x 

π 4

π 4

and

, to graph one full period of y  4 tan 2 x from

π

π

to . 4 4 Continue the pattern and extend the graph another full period to the right.

19. Solve the equations

π 4

x

π 2

π 4 x  2 π

and

π 4

x

π 2

π 4 x  2 π

x  2 x2 Thus, two consecutive asymptotes occur at x = –2 and x = 2. 2  2 0 x-intercept   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.

20. Solve the equations

x π   x

π

and

2

xπ 

π

π 2

π

π x  π 2 2 3π π x x 2 2 Thus, two consecutive asymptotes occur at 3π π and x   . x 2 2 3π π   2π x-intercept  2 2   π 2 2 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 xintercept and the asymptotes have y-coordinates of –1 and 1. Use the two consecutive asymptotes, 3π π and x   , to graph one full period of x 2 2 3π π to  . y  tan( x  π ) from  2 2 Continue the pattern and extend the graph another full period to the right.

Use the two consecutive asymptotes, x = –2 and x = 2, to graph one full period of y  2 tan

π

x from –2 4 to 2. Continue the pattern and extend the graph another full period to the right.

21. Solve the equations

x

π 4



x

π

and

2

π

π 4

4

An x-intercept is

Copyright © 2014 Pearson Education, Inc.

x

π 2

π

π

 2 4 3π π x x 4 4 Thus, two consecutive asymptotes occur at π 3π x   and x   . 4 4 π  π4  34π π  2  x-intercept  2 2 4 2

π

x

π 4

and the graph passes through

159


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

π   , 0  . Because the coefficient of the tangent is –1, 4 the points on the graph midway between an xintercept and the asymptotes have y-coordinates of 1 and –1. Use the two consecutive asymptotes, π 3π , to graph one full period of x   and x  4 4 π π 3π  y   tan  x   from  to . Continue the  4 4 4 pattern and extend the graph another full period to the right.

full period to the right.

23. Solve the equations

π 2

x  0 and x0

x

π

3 Thus, two consecutive asymptotes occur at x = 0 and x  x-intercept 

π

. 3 0  π3 2

An x-intercept is

π 6

π

3

2

6

and the graph passes through

π   , 0  . 6 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

π 3

, to

graph one full period of y  2 cot 3 x from 0 to

π

3 Continue the pattern and extend the graph another

160

xπ xπ

2

π

x = 0 and x = 2, to graph one full period of 1 π y   cot x from 0 to 2. Continue the pattern and 2 2 extend the graph another full period to the right.

π

two consecutive asymptotes, x = 0 and x 

2

x2 Thus, two consecutive asymptotes occur at x = 0 and x = 2. 02 2 x-intercept   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 x2 intercept and the asymptotes have y-coordinates of 1 1  and . Use the two consecutive asymptotes, 2 2

22. Solve the equations 3x  0 and 3x  π

x0

π

.

Copyright © 2014 Pearson Education, Inc.


Chapter 2 Review Exercises Use these key points to graph y  3cos 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  3sec 2π x.

24. Solve the equations

x

π 2

0

and

x  0 x

x

π

π 2

π

xπ

2

π

π 2

π

x 2 2 Thus, two consecutive asymptotes occur at x

π

and x 

2

π

. 2  π2  π2

0 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. x-intercept 

Use the two consecutive asymptotes, x  

π

and 2 π π  x  , to graph one full period of y  2 cot  x    2 2

from 

π

π

to . Continue the pattern and extend the 2 2 graph another full period to the right.

25. Graph the reciprocal cosine function, y  3cos 2π x . The equation is of the form y  A cos Bx with A = 3 and B  2π . amplitude: A  3  3 period:

26. Graph the reciprocal sine function, y  2 sin π x . The equation is of the form y  A sin Bx with A = –2 and B  π . amplitude: A  2  2 period:

2π 2π  2 B π

2 1  , to find 4 2 x-values for the five key points. Starting with 1 3 x = 0, the x-values are 0, , 1, , 2 . Evaluating the 2 2 function at each value of x, the key points are (0, 0), 1  3   ,  2  , 1, 0 ,  , 2  , (2,0) . Use these key points 2 2 to graph y  2 sin 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, and use them as guides to graph y  2 csc π x. Use quarter-periods,

2π 2π  1 B 2π

1 , to find x-values for the five 4 1 key points. Starting with x = 0, the x-values are 0, , 4 1 3 , , 1 . Evaluating the function at each value of x, 2 4 the key points are (0, 3), 1  1  3   , 0  ,  ,  3 ,  , 0  , (1, 3) . 4 2 4

Use quarter-periods,

Copyright © 2014 Pearson Education, Inc.

161


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

27. Graph the reciprocal cosine function, y  3cos( x  π ) . The equation is of the form y  A cos( Bx  C ) with A = 3, B = 1, and C  π . amplitude:

function at each value of x, the key points 5  3π 5   5π are (π , 0),  ,  , (2π , 0) ,  ,   , (3π , 0).  2 2  2 2

A  3 3

2π 2π   2π B 1 C π phase shift:   π B 1 2π π Use quarter-periods,  , to find 4 2 x-values for the five key points. Starting with period:

x  π , the x-values are π , 

π

π

5 sin( x  π ) from 2 π to 3π . Extend the graph one cycle to the right. Use the graph to obtain the graph of the reciprocal function. Draw vertical asymptotes through the xintercepts, and use them as guides to graph 5 y  csc( x  π ). 2 Use these key points to graph y 

,π. 2 2 Evaluating the function at each value of x, the key  π  points are  π , 3 ,   , 0  ,  0,  3 ,  2  , 0,

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 graph of the reciprocal function. Draw vertical asymptotes through the x-intercepts, and use them as guides to graph y  3sec( x  π ).

29. Let θ  sin 1 1 , then sin θ  1 .

 π π The only angle in the interval   ,  that satisfies  2 2 sin θ  1 is

π 2

. Thus θ 

π 2

, or sin 1 1 

π 2

.

30. Let θ  cos1 1 , then cos θ  1 . The only angle in the interval  0, π  that satisfies

cos θ  1 is 0 . Thus θ  0 , or cos1 1  0 . 5 28. Graph the reciprocal sine function, y  sin( x  π ) . 2 The equation is of the form y  A sin( Bx  C ) with 5 , B = 1, and C  π . 2 5 5 amplitude: A   2 2 2π 2π period:   2π B 1 C π phase shift:  π B 1 2π π Use quarter-periods,  , to find 4 2 x-values for the five key points. Starting with x  π , 3π 5π the x-values are π , , 2π , , 3π . Evaluating the 2 2 A

162

31. Let θ  tan 1 1 , then tan θ  1 .

 π π The only angle in the interval   ,  that satisfies  2 2 tan θ  1 is

π 4

. Thus θ 

π 4

, or tan 1 1 

π 4

.

 3 3 32. Let θ  sin 1    , then sin θ   2 .  2   π π The only angle in the interval   ,  that satisfies  2 2 3 π π is  . Thus θ   , or 3 3 2   3 π sin 1   3. 2  

sin θ  

Copyright © 2014 Pearson Education, Inc.


Chapter 2 Review Exercises

1  1 33. Let θ  cos1    , then cos θ    2 2

. The only angle in the interval  0, π  that satisfies

1 2π 2π . Thus θ  , or is 2 3 3  1  2π . cos1      2 3

cos θ  

 3 3 34. Let θ  tan 1    , then tan θ   3 . 3  

3 π is  . 6 3

π 6

3 π is . 6 3  3 π Thus csc  tan 1   csc 6  2 . 3   tan θ 

 3 π , or tan 1   6 . 3  

2 2 , then sin θ  . The only angle 2 2 2  π π in the interval   ,  that satisfies sin θ  is 2  2 2

35. Let θ  sin 1

π 4

3 3 , then tan θ  . 3 3  π π The only angle in the interval   ,  that satisfies  2 2

39. Let θ  tan 1

 π π The only angle in the interval   ,  that satisfies  2 2

Thus θ  

angle in the interval  0, π  that satisfies

3 5π is . 2 6   3  5π 3 . Thus, tan  cos1      tan  6 3  2   

cos θ  

tan θ  

 3 3 38. Let θ  cos 1    , then cos θ   2 . The only 2  

3 3 , then tan θ  4 4 Because tan θ is positive, θ is in the first quadrant.

40. Let θ  tan 1

.

 2 π 2 Thus, cos  sin 1 .  cos   2  4 2  36. Let θ  cos 1 0 , then cos θ  0 . The only angle in the interval  0, π  that satisfies cos θ  0 is

Thus, sin cos1 0  sin

π 2

π

2

.

 1.

1  1 37. Let θ  sin 1    , then sin θ   . The only  2 2

r2  x2  y 2 r 2  42  32 r 2  25 r5 3 x 4  cos  tan 1   cos θ    4 r 5

 π π angle in the interval   ,  that satisfies  2 2 1 π sin θ   is  . 6 2  1  1   3  π . Thus, tan sin      tan       2   6 3 

Copyright © 2014 Pearson Education, Inc.

163


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

x2  y 2  r2

3 3 , then cos θ  . 5 5 Because cos θ is positive, θ is in the first quadrant.

41. Let θ  cos1

(4)2  y 2  52 y 2  25  16  9 y 9 3 Use the right triangle to find the exact value.  3  4  tan  cos1      tan θ    5  4   1 44. Let θ  tan 1    ,  3 Because tan θ is negative, θ is in quadrant IV and x  3 and y  1 .

x2  y2  r2 32  y 2  52 y 2  25  9  16 y  16  4

r2  x2  y 2

3 y 4  sin  cos1   sin θ    5 r 5

r 2  32  ( 1) 2 r 2  10 r  10

3  3 42. Let θ  sin 1    , then sin θ   .  5 5 Because sin θ is negative, θ is in quadrant IV.

 1 y 10  4   sin  tan 1      sin θ     5 r 10 10   45.

46.

x 2  (3)2  52 x y r 2

2

47.

x  16  4  y 3  3  tan sin 1      tan θ      5 x 4   4  4 43. Let θ  cos    , then cos θ   .  5 5 Because cos θ is negative, θ is in quadrant II.

164

π

 π π  π π , x is in   ,  , so sin 1  sin    3 3 3  2 2

2π , x is not in 3 sin x , so x

 π π   2 , 2  . x is in the domain of  

π  2π  1 3  sin 1  sin   sin  3  2 3

2

x 2  25  9  16

1

x

2π   1  1  sin 1  cos   sin     3 2 1  1 Let θ  sin 1    , then sin θ   . The only  2 2  π π angle in the interval   ,  that satisfies  2 2 1 π π sin θ   is  . Thus, θ   , or 6 6 2 2 π 1 π      sin 1      . sin 1  cos    2 3  6

Copyright © 2014 Pearson Education, Inc.


Chapter 2 Review Exercises

48. Let θ  tan 1

50. Find the measure of angle B. Because C  90 , A  B  90 . Thus, B  90  A  90º 22.3  67.7 We have a known angle, a known hypotenuse, and an unknown opposite side. Use the sine function. a sin 22.3  10 a  10sin 22.3  3.79 We have a known angle, a known hypotenuse, and an unknown adjacent side. Use the cosine function. b cos 22.3  10 b  10cos 22.3  9.25 In summary, B  67.7, a  3.79 , and b  9.25 .

x x , then tan θ  . 2 2

r 2  x 2  22 r2  x2  y 2 r

x2  4

Use the right triangle to write the algebraic expression. x 2 2 x2  4  cos  tan 1   cos θ    2 x2  4 x2  4 49. Let θ  sin 1

1 1 , then sin θ  . x x

Use the Pythagorean theorem to find the third side, b. 12  b2  x 2

b2  x 2  1 b  x2  1 Use the right triangle to write the algebraic expression. 1 x x x2  1  sec  sin 1   sec θ    x x2  1 x2  1

51. Find the measure of angle A. Because C  90 , A  B  90 . Thus, A  90  B  90  37.4  52.6 We have a known angle, a known opposite side, and an unknown adjacent side. Use the tangent function. 6 tan 37.4  a 6 a  7.85 tan 37.4

We have a known angle, a known opposite side, and an unknown hypotenuse. Use the sine function. 6 sin 37.4  c 6 c  9.88 sin 37.4 In summary, A  52.6, a  7.85 , and c  9.88 .

Copyright © 2014 Pearson Education, Inc.

165


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

52. Find the measure of angle A. We have a known hypotenuse, a known opposite side, and an unknown angle. Use the sine function. 2 sin A  7 2 A  sin 1    16.6 7 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. a 2  b2  c 2

22  b 2  7 2 b2  72  22  45 b  45  6.71 In summary, A  16.6, B  73.4 , and b  6.71 . 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. 1.4 tan A  3.6  1.4  A  tan 1   21.3  3.6  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. c 2  a 2  b2  (1.4)2  (3.6)2  14.92

c  14.92  3.86 In summary, A  21.3, B  68.7 , and c  3.86 . 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. h tan 25.6  80 h  80 tan 25.6

 38.3 The building is about 38 feet high.

166

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. h tan 40  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. 25 tan 68  d 125 d  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. h tan 71  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 side of the north-south line and lies east of the northsouth line. Thus the bearing from O to A is N35°E. 58. 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 south side of the north-south line and lies west of the northsouth line. Thus the bearing from O to A is S35°W. 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. d tan 64  12 d  12 tan 64  24.6 The ship is about 24.6 miles from the lighthouse.

Copyright © 2014 Pearson Education, Inc.


Chapter 2 Review Exercises

60.

62.

1 sin 4t 2 1 a  and ω  4 2

d

a.

a.

Using the figure, B  58  32  90 Thus, use the Pythagorean Theorem to find the distance from city A to city C. 8502  9602  b2

b.

c.

b2  1,644,100

b.

Using the figure, opposite 960   1.1294 tan A  adjacent 850 1

A  tan (1.1294)  48 180  58  48  74 The bearing from city A to city C is S74°E. 61.

d  20 cos

π 4

a.

b.

c.

π

π ω π 1 1  4    2π 2π 4 2π 8

63. Because the distance of the object from the rest position at t  0 is a maximum, use the form 2π d  a cos ω t . The period is so,

2

ω

ω

2π π ω 2 Because the amplitude is 30 inches, a  30 .

5

1 frequency: cm per second 8 2π 2π 4 period:  π  2π   8

ω

2π π   1.57 4 2 ω The time required for one cycle is about 1.57 seconds. period:

64. Because the distance of the object from the rest position at t  0 is 0, use the form d  a sin ω t . The 2π period is so

4 maximum displacement: a  20  20 cm f 

ω 4 2    0.64 2π 2π π

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 .

t

a  20 and ω 

f 

frequency: 0.64 cm per second

b2  722,500  921,600 b  1,644,100  1282.2 The distance from city A to city B is about 1282.2 miles.

maximum displacement: 1 1 a   cm 2 2

4

π

The time required for one cycle is 8 seconds.

ω

ω

2π ω 5 1 1 inch, a  . a is 4 4 negative since the object begins pulled down. The equation for the object’s simple harmonic motion is 1 2π d   sin t. 4 5

Because the amplitude is

Copyright © 2014 Pearson Education, Inc.

167


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

Chapter 2 Test 1.

The equation y  3sin 2 x is of the form y  A sin Bx with A = 3 and B = 2. The amplitude is A  3  3.

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

x  0 x

π 4

π

π

4

π

π

x

coordinates

0

(0, 0)

π

π   , 3 4

4

π 2

 3π   ,  3 4

π

(π , 0)

C=

π 2

. The amplitude is A  2  2.

2π 2π   2π . The phase shift is B 1 π C 2 π 2π π   . The quarter-period is  . B 1 2 4 2

The period is

π

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

π 2

π 2

xπ

π 2

π 2

π 

3π 2

3π π   2π 2 2 π 5π x  2π   2 2 Evaluate the function at each value of x. x

π   , 0  2

3π 4

π  The equation y  2 cos  x   is of the form  2 y  A cos( Bx  C ) with A = –2, B = 1, and

The cycle begins at x =

π

2 3π x   2 4 4 3π π x  π 4 4 Evaluate the function at each value of x.

π

4

4

2.

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

x

coordinates

π 2

π   ,  2  2

π

π , 0 

3π 2

 3π   , 2  2

 2π , 0

5π 2

 5π   , 2  2

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

168

Copyright © 2014 Pearson Education, Inc.


Chapter 2 Test

3.

Solve the equations x x π π  and  2 2 2 2

x

π 2

2

x

π 2

2

x  π xπ Thus, two consecutive asymptotes occur at x  π and x  π . π  π 0 x-intercept   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  π and x  π , to graph one x full period of y  2 tan from π to π . 2

4.

1 1 Graph the reciprocal sine function, y   sin π x. The equation is of the form y  A sin Bx with A =  and B  π . 2 2 1 1 amplitude: A    2 2 2π 2π period:  2 B π 2 1 Use quarter-periods,  , to find x-values for the five key points. Starting with x = 0, the 4 2 1 3 x-values are 0, , 1, , 2. Evaluating the function at each value of x, the key points are 2 2 1 1   3 1 (0, 3),  ,   , 1, 0 ,  ,  , (2, 0) . 2 2 2 2 1 Use these key points to graph y   sin π x from 0 to 2. Use the graph to obtain the graph of the reciprocal function. 2 1 Draw vertical asymptotes through the x-intercepts, and use them as guides to graph y   csc π x. 2

Copyright © 2014 Pearson Education, Inc.

169


Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

5.

Select several values of x over the interval.

x

0

π

π

4

2

3π 4

π

5π 4

3π 2

7π 4

1 sin x 0 0.4 0.5 0.4 0 0.4 0.5 0.4 2 y2  2 cos x 2 1.4 0 1.4 2 1.4 0 1.4 1 y  sin x  2 cos x 2 1.8 0.5 1.1 2 1.8 0.5 1.1 2 y1 

6.

2π 0 2 2

1  1 Let θ  cos1    , then cos θ   .  2 2 Because cos θ is negative, θ is in quadrant II. x2  y2  r2 (1)2  y 2  22 y2  4  1  3 y 3  3 y  1   3 tan  cos1      tan θ      2 x 1  

7.

x x Let θ  cos1   , then cos θ  .  3 3 Because cos θ is positive, θ is in quadrant I. x2  y 2  r 2 x 2  y 2  32 y2  9  x2 y  9  x2  y  x  sin cos 1     sin θ     3  r 

170

9  x2 3

Copyright © 2014 Pearson Education, Inc.


Chapter 2 Test

8.

Find the measure of angle B. Because C = 90°, A + B = 90°. Thus, B = 90° – A = 90° – 21° = 69°. We have a known angle, a known hypotenuse, and an unknown opposite side. Use the sine function. a sin 21  13 a  13sin 21  4.7 We have a known angle, a known hypotenuse, and an unknown adjacent side. Use the cosine function. b cos 21  13 b  13cos 21  12.1 In summary, B = 69°, a  4.7, and b  12.1.

9.

x  1000 tan 56  1000 tan 51 x  247.7 ft

10. We need the acute angle between ray OP and the north-south line through O. This angle measures 90° – 10°. 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 P is N80°W. 11.

d  6cos π t a  6 and ω  π a. b.

c.

maximum displacement: a  6  6 in.

π 1  2π 2 1 in. per second frequency: 2 f 

ω

period =

ω

π

2

The time required for one cycle is 2 seconds.

Copyright © 2014 Pearson Education, Inc.

171

Download Trigonometry 1st Edition by Blitzer solution manual  

Download Trigonometry first Edition by Blitzer arrangement manual Connection FULL DOWNLOAD : https://bit.ly/2EgG098 Dialect: English I...

Download Trigonometry 1st Edition by Blitzer solution manual  

Download Trigonometry first Edition by Blitzer arrangement manual Connection FULL DOWNLOAD : https://bit.ly/2EgG098 Dialect: English I...

Advertisement