Chapter 2 Graphs of the Trigonometric Functions; …...Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctionsSec tion 2.1 Graphs of Sine and Cosine Functions

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Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

Section 2.1 Connect the five points with a smooth curve and

graph one complete cycle of the given function withCheckpoint Exercises the graph of y = sin x .

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 + π

= π

2. The equation

with 1

y 1

sin x is of the form 2

y = A sin x

2 2

x = π

+ π

= π A = − . Thus, the amplitude is

2

1 1 1

2 2 A = − = . The period for both

2 2 y = − sin x

2

x = π + π

= 3π

2 2 and y = sin x is 2π .

x = 3π

+ π

= 2π 2 2

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

period 2π π

Evaluate the function at each value of x. the period, 2π , by 4, = = , then by 4 4 2

adding quarter- periods. The five x-values are

x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π

2

2

π 3π




x y = 3sin x coordinates

0 y = 3sin 0 = 3 ⋅ 0 = 0 (0, 0)

π

2 y = 3sin

π = 3 ⋅1 = 3

2

π , 3

2

π y = 3sin x = 3 ⋅ 0 = 0 (π , 0)

3π

2 y = 3sin

3π 2

= 3(−1) = −3

3π , −3

2π y = 3sin 2π = 3 ⋅ 0 = 0 (2π , 0)

2

x = π + = 2 2

x = 3π

+ π

= 2π 2 2

Evaluate the function at each value of x.


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

x y 1

sin x 2

coordinates

0 y 1

sin 0 = −

2

1 ⋅ 0 = 0

= − 2

(0, 0)

π

2 y

1 sin

π 2 2

1 ⋅1 = −

1 = − 2 2

π , −

1

2 2

π y 1

sin π = −

2

1 ⋅ 0 = 0

= − 2

(π , 0)

3π

2 y

1 sin

3π 2 2

1 ( −1) =

1 = − 2 2

3π 1 , 2 2

2π y 1

sin 2π = −

2

1 ⋅ 0 = 0

= − 2

(2π , 0)

x y = 2 sin 1

x 2

coordinates

0

1 y = 2 sin

⋅ 0 2

= 2 sin 0

= 2 ⋅ 0 = 0

(0, 0)

π y = 2 sin 1

⋅ π 2

= 2 sin π

= 2 ⋅1 = 2 2

(π , 2)

2π y = 2sin 1

⋅ 2π 2

= 2 sin π = 2 ⋅ 0 = 0

(2π , 0)

3π

1 y = 2 sin ⋅ 3π

2

= 2 sin 3π 2

= 2 ⋅ ( −1) = −2

(3π , − 2)

4π y = 2sin 1

⋅ 4π 2

= 2sin 2π = 2 ⋅ 0 = 0

(4π , 0)

B

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

= − period 4π

= −

the period, 4π , by 4,

adding quarter-periods.

The five x-values are x = 0

x = 0 + π = π

x = π + π = 2π

x = 2π + π = 3π

x = 3π + π = 4π

= = π , then by 4 4


= −

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. The equation

y = 2 sin 1

x 2

is of the form


graph one complete cycle of the given function. Extend

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

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

. 2

The amplitude is A = 2 = 2 .

The period is



2π =

2π = 4π .

1 2



11π

12

11π π y = 3sin 2 ⋅

12 −

3

= 3sin 9π

= 3sin 3π

6 2

= 3 ( −1) = −3

11π , − 3 12

7π

6

7π π y = 3sin 2 ⋅ −

6 3

6π = 3sin = 3sin 2π

3

= 3 ⋅ 0 = 0

7π

6 , 0

x y = 3sin 2 x −

π

coordinates

π

6

π π y = 3sin 2 ⋅ −

6 3

= 3sin 0 = 3 ⋅ 0 = 0

π , 0 6

5π

12

5π π y = 3sin 2 ⋅ −

12 3

= 3sin 3π

= 3sin π

6 2

= 3 ⋅1 = 3

5π , 3 12

2π

3

2π π y = 3sin 2 ⋅ −

3 3

= 3sin 3π

= 3sin π 3

= 3 ⋅ 0 = 0

2π , 0 3

4. The equation

y = 3sin 2 x −

π is of the form

3

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

. 3


The period is 2π

= 2π

= π .B

The phase shift is

2

C π

π 1 π

= 3 = ⋅ = .

B 2 3 2 6

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

the period, π , by 4, period

= π

, then by adding 4 4

quarter-periods to the value of x where the cycle

begins, x = π

. 6

The five x-values are

x = π 6

x = π

+ π

= 2π

+ 3π

= 5π

6 4 12 12 12

x = 5π

+ π

= 5π

+ 3π

= 8π

= 2π

12 4 12 12 12 3

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

x = 2π

+ π

= 8π

+ 3π

= 11π

3 4 12 12 12

5. The equation

y = −4 cos π x is of the form

x = 11π

+ π

= 11π

+ 3π

= 14π

= 7π y = A cos Bx with A = −4, and B = π .

12 4 12 12 12 6 Thus, the amplitude is A = −4 = 4 .

Evaluate the function at each value of x. The period is

2π =

2π = 2 .

B π

3


period 2 1

the period, 2, by 4, = = , then by adding 4 4 2

quarter periods to the value of x where the cycle begins. The five x-values are

x = 0

x = 0 + 1

= 1

2 2

1 1 x = + = 1

2 2

x = 1 + 1

= 3

2 2

x = 3

+ 1

= 2



2 2




x y = −4 cos π x coordinates

0 y = −4 cos (π ⋅ 0) = −4 cos 0 = −4

(0, –4)

1

2

1 y = −4 cos π ⋅

2

= −4 cos π

= 0 2

1 2

, 0

1 y = −4 cos(π ⋅1)

= −4 cos π = 4

(1, 4)

3

2

3 y = −4 cos π ⋅

2

= −4 cos 3π

= 0 2

3 2

, 0

2 y = −4 cos(π ⋅ 2)

= −4 cos 2π = −4

(2, –4)

x 3

y = cos(2 x + π ) 2

coordinates

π −

2

3 y =

2 cos(−π + π )

3 3 =

2 ⋅1 =

2

π 3 −

2 ,

2

− π 4

y = 3

cos −

π + π

2 2

= 3

⋅ 0 = 0 2

−

π , 0

4

0

3 y = cos(0 + π )

2

= 3

⋅ −1 = − 3

2 2

3 0, − 2

π

4

3 π y = cos + π

2 2

3 = ⋅ 0 = 0

2

π , 0 4

π

2

y = 3

cos(π + π ) 2

= 3

⋅1 = 3

2 2

π ,

3

x

The five x-values are

π = −

2

π π π x = − + = −

2 4 4

π π x = − + = 0

4 4

x = 0 + π

= π

4 4

x = π

+ π

= π

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


graph one complete cycle of the given function. Extend

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

6. y = 3

cos(2 x + π ) = 3

cos(2 x − (−π )) 2 2

The equation is of the form y = A cos(Bx − C ) with

A = 3

, B = 2 , and C = −π . 2

2 2

Thus, the amplitude is A = 3 =

3 .

2 2

The period is 2π

= 2π

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

B

The phase shift is

2

C =

−π = −

π .

graph one complete cycle of the given graph.

B 2 2



= −


the period, π , by 4, period

= π

, then by adding 4 4

quarter-periods to the value of x where the cycle

begins, x π

. 2



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 or

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

4 2

x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2


x y = 2 cos x + 1 coordinates

0 y = 2 cos 0 + 1

= 2 ⋅1 + 1 = 3

(0, 3)

π

2 y = 2 cos

π + 1

2 = 2 ⋅ 0 + 1 = 1

π , 1

2

π y = 2 cos π + 1

= 2 ⋅ (−1) + 1 = −1

(π , −1)

3π

2 y = 2 cos

3π + 1

2 = 2 ⋅ 0 + 1 = 1

3π , 1

2

2π y = 2 cos 2π + 1

= 2 ⋅1 + 1 = 3

(2π , 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.

x

0 π

4

π

2

3π

4 π

5π

4

3π

2

7π

4 2π

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

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

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



x y = 4 sin x coordinates

0 y = 4sin 0 = 4 ⋅ 0 = 0 (0, 0)

π

2 y = 4 sin

π = 4 ⋅1 = 4

2

π , 4

π y = 4 sin π = 4 ⋅ 0 = 0 (π , 0)

3π

2 y = 4 sin

3π 2

= 4(−1) = −4

3π , − 4

2π y = 4sin 2π = 4 ⋅ 0 = 0 (2π , 0)

6 2

9. A, the amplitude, is the maximum value of y. The

graph shows that this maximum value is 4, Thus,

A = 4 . The period is π

, and period = 2π

. Thus, 2 B

π =

2π

7. false

8. true

9. true

10. true

2 B

π B = 4π

B = 4

Substitute these values into

y = A sin Bx .

Exercise Set 2.1 1. The equation

y = 4 sin x

is of the form

y = A sin x

The graph is modeled by y = 4sin 4 x . with A = 4. Thus, the amplitude is A = 4 = 4 .

2π π .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

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

x = 0 + π

= π

months. The period is 12. Thus, 12 = 2π B

12B = 2π

B = 2π

= π

12 6

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

The graph shows that the starting point of the cycle is

shifted from 0 to 3. The phase shift, C

, is 3. B

x = 3π

+ π

= 2π 2 2


3 = C

B

3 = C π 6

π = C

2

Substitute these values into

y = A sin(Bx − C ) + D .

2

The number of hours of daylight is modeled by

y = 2 sin π

x − π

+ 12 .

2

Concept and Vocabulary Check 2.1

1. A ; 2π

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

B

2. 3; 4π

3. π ; 0; π

;

π ;

3π ; π

the graph of y = sin x .



4 2 4

4. C

; right; left B

5. A ; 2π B

6. 1

; 2π

2 3



x y = 5sin x coordinates

0 y = 5sin 0 = 5 ⋅ 0 = 0 (0, 0)

π

2 y = 5sin

π = 5 ⋅1 = 5

2

π , 5

2

π y = 5sin π = 5 ⋅ 0 = 0 (π , 0)

3π

2 y = 5sin

3π = 5(−1) = −5

2

3π , − 5

2π y = 5sin 2π = 5 ⋅ 0 = 0 (2π , 0)

x 1

y = sin x 3

coordinates

0 1 1

y = sin 0 = ⋅ 0 = 0 3 3

(0, 0)

π

2 y =

1 sin

π =

1 ⋅1 =

1 3 2 3 3

π ,

1 2 3

π y = 1

sin π = 1

⋅ 0 = 0 3 3

(π , 0)

3π

2

1 3π y = sin

3 2

= 1

(−1) = − 1

3 3

3π 1 , − 2 3

2π y = 1

sin 2π = 1

⋅ 0 = 0 3 3

(2π , 0)

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

3. The equation y = 1

sin x is of the form

y = A sin x

with A = 5. Thus, the amplitude is A = 5 = 5 . 3

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

or π

.

with A = 1

. Thus, the amplitude is A = 1 =

1 .

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

3

The period is 2π . The quarter-period is

3 3

2π or

π .

4 2

x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2


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

x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2


2









x y = 1

sin x 4

coordinates

0 y = 1

sin 0 = 1

⋅ 0 = 0 4 4

(0, 0)

π

2 y =

1 sin

π =

1 ⋅1 =

1 4 2 4 4

π ,

1

π y = 1

sin π = 1

⋅ 0 = 0 4 4

(π , 0)

3π

2 y =

1 sin

3π =

1 (−1) = −

1 4 2 4 4

3π , −

1 2 4

2π y = 1

sin 2π = 1

⋅ 0 = 0 4 4

(2π , 0)

x y = −3sin x coordinates

0 y = −3sin x

= −3 ⋅ 0 = 0

(0, 0)

π

2 y = −3sin

π 2

= −3 ⋅1 = −3

π , − 3

π y = −3sin π

= −3 ⋅ 0 = 0

(π , 0)

3π

2

3π y = −3sin

2

= −3(−1) = 3

3π , 3 2

2π y = −3sin 2π

= −3 ⋅ 0 = 0

(2π , 0)

4. The equation y = 1

sin x

is of the form

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

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

with A = 1

. Thus, the amplitude is A = 1 =

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

2π or

π .

4 4 4 4 2

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

π .

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

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

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

generate x-values for the key points.

x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x 3π π

2π

2 2 =

2 +

2 =

x = 3π

+ π

= 2π 2 2



2 4

2



the graph of y = sin x . Connect the five key points with a smooth curve and







x y = −4 sin x coordinates

0 y = −4sin 0 = −4 ⋅ 0 = 0 (0, 0)

π

2 y = −4sin

π = −4 ⋅1 = −4

2

π , − 4

2

π y = −4sin π = −4 ⋅ 0 = 0 (π , 0)

3π

2 y = −4sin

3π = −4(−1) = 4

2

3π , 4

2π y = −4sin 2π = −4 ⋅ 0 = 0 (2π , 0)

x y = sin 2 x coordinates

0 y = sin 2 ⋅ 0 = sin 0 = 0 (0, 0)

π

4

π y = sin 2 ⋅

4

= sin π

= 1 2

π 4

,1

π

2 y = sin

2 ⋅

π

= sin π = 0

π , 0

3π

4

3π y = sin 2 ⋅

4

= sin 3π

= −1 2

3π

4 , −1

π y = sin(2 ⋅ π )

= sin 2π = 0

(π , 0)

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

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


or π

.

A = 1 = 1 . The period is 2π

= 2π

= π . The

4 2 B 2


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

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2


quarter-period is π

. The cycle begins at x = 0. Add 4

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

x = 0

x = 0 + π 4

x = π

+ π

= π

4 4 2

x = π

+ π

= 3π

2 4 4

x = 3π

+ π

= π 4 4


2



2

2


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





x 1 y = 3sin

2 x

coordinates

0 1 y = 3sin ⋅ 0

2

= 3sin 0 = 3 ⋅ 0 = 0

(0, 0)

π 1 y = 3sin

2 ⋅π

= 3sin π

= 3 ⋅1 = 3 2

(π , 3)

2π 1 y = 3sin

2 ⋅ 2π

= 3sin π = 3 ⋅ 0 = 0

(2π , 0)

3π 1 y = 3sin ⋅ 3π

2

3π = 3sin

2 = 3(−1) = −3

(3π , − 3)

4π 1 y = 3sin ⋅ 4π

2

= 3sin 2π = 3 ⋅ 0 = 0

(4π , 0)

x y = sin 4 x coordinates

0 y = sin(4 ⋅ 0) = sin 0 = 0 (0, 0)

π

8 y = sin

4 ⋅

π = sin

π = 1

π , 1 8

π

4

π y = sin 4 ⋅ = sin π = 0

4

π 4

, 0

3π

8

3π y = sin 4 ⋅

8

= sin 3π

= −1 2

3π

8 , −1

π

2

y = sin 2π = 0 π , 0 2

B 2

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

9. The equation y = 3sin 1

x is of the form

y = A sin Bx

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


= 2π

= π

. The with A = 3 and B = 1

. The amplitude is

A = 3

= 3.

B 4 2 2 π

quarter-period is 2 = π

⋅ 1

= π

. The cycle begins at

The period is 2π =

2π = 2π ⋅ 2 = 4π . The quarter-

4 2 4 8 1

x = 0. Add quarter-periods to generate x-values for 4πthe key points. period is

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

x = 0

x = 0 + π

= π

8 8

x = π

+ π

= π

8 8 4

x = π

+ π

= 3π

4 8 8

x = 3π

+ π

= π


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

x = 0 + π = π

x = π + π = 2π

x = 2π + π = 3π

x = 3π + π = 4π


8

2


graph one complete cycle of the given function.

Connect the five points with a smooth curve and






x y = 2 sin 1

x 4

coordinates

0 y = 2sin 1

⋅ 0

= 2sin 0 = 2 ⋅ 0 = 0

(0, 0)

2π y = 2 sin 1

⋅ 2π

= 2 sin π

= 2 ⋅1 = 2 2

(2π , 2)

4π y = 2 sin π = 2 ⋅ 0 = 0 (4π , 0)

6π y = 2 sin 3π

= 2(−1) = −2 2

(6π , − 2)

8π y = 2sin 2π = 2 ⋅ 0 = 0 (8π , 0)

x y = 4 sin π x coordinates

0 y = 4sin(π ⋅ 0)

= 4sin 0 = 4 ⋅ 0 = 0

(0, 0)

1

2

1 y = 4 sin π ⋅

2

= 4 sin π

= 4(1) = 4 2

1 2

, 4

1 y = 4sin(π ⋅1)

= 4 sin π = 4 ⋅ 0 = 0

(1, 0)

3

2

3 y = 4sin π ⋅

2

= 4sin 3π 2

= 4(−1) = −4

3 , − 4

2

2 y = 4sin(π ⋅ 2)

= 4sin 2π = 4 ⋅ 0 = 0

(2, 0)

B 4

10. The equation y = 2 sin 1

x

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

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

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

. Thus, the 4

The period is 2π B

= 2π

π

= 2 . The quarter-period is

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

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

2π =

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

1 generate x-values for the key points.

x = 08π = 2π . The cycle begins at x = 0. Add quarter-

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

x = 0 + 2π = 2π

x = 2π + 2π = 4π

x = 4π + 2π = 6π

x = 6π + 2π = 8π Evaluate the function at each value of x.

x = 0 + 1

= 1

2 2

x = 1

+ 1

= 1 2 2

x = 1 + 1

= 3

2 2

x = 3

+ 1

= 2 2 2


4

4









x y = −3sin 2π x coordinates

0 y = −3sin(2π ⋅ 0)

= −3sin 0

= −3 ⋅ 0 = 0

(0, 0)

1

4 y = −3sin

2π ⋅

1

= −3sin π 2

= −3 ⋅1 = −3

1 , − 3 4

1

2

1 y = −3sin 2π ⋅

2

= −3sin π

= −3 ⋅ 0 = 0

1 2

, 0

3

4

3 y = −3sin 2π ⋅

4

= −3sin 3π 2

= −3(−1) = 3

3 , 3 4

1 y = −3sin(2π ⋅1)

= −3sin 2π

= −3 ⋅ 0 = 0

(1, 0)

x y = 3sin 2π x coordinates

0 y = 3sin(2π ⋅ 0)

= 3sin 0 = 3 ⋅ 0 = 0

(0, 0)

1

4

1 y = 3sin 2π ⋅

4

= 3sin π

= 3 ⋅1 = 3 2

1

1

2 y = 3sin

2π ⋅

1

= 3sin π = 3 ⋅ 0 = 0

1 , 0

3

4

3 y = 3sin 2π ⋅

4

= 3sin 3π

= 3(−1) = −3 2

3 4

, − 3

1 y = 3sin(2π ⋅1)

= 3sin 2π = 3 ⋅ 0 = 0

(1, 0)

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

is A = 3 = 3 . The period is 2π

= 2π

= 1 . The

is A = −3 = 3 . The period is 2π

= 2π

= 1 . The

B 2π B 2π

quarter-period is 1


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

x = 0 + 1

= 1

4 4

x = 1

+ 1

= 1

4 4 2

x = 1

+ 1

= 3

2 4 4

x = 3

+ 1

= 1 4 4


quarter-period is 1



x = 0

x = 0 + 1

= 1

4 4

x = 1

+ 1

= 1

4 4 2

x = 1

+ 1

= 3

2 4 4

x = 3

+ 1

= 1 4 4


4

, 3

4

2

2







x y = −2sin π x coordinates

0 y = −2sin(π ⋅ 0)

= −2sin 0 = −2 ⋅ 0 = 0

(0, 0)

1

2

1 y = −2sin π ⋅

2

= −2sin π

= −2 ⋅1 = −2 2

1 2

, − 2

1 y = −2sin(π ⋅1)

= −2sin π = −2 ⋅ 0 = 0

(1, 0)

3

2

3 y = −2sin π ⋅

2

= −2sin 3π

= −2(−1) = 2 2

3 2

, 2

2 y = −2sin(π ⋅ 2)

= −2sin 2π = −2 ⋅ 0 = 0

(2, 0)

x y = − sin 2

x 3

coordinates

0 y = − sin 2

⋅ 0

3

= − sin 0 = 0

(0, 0)

3π

4 y = − sin

2 ⋅

3π

= − sin π

= −1

3π , −1

3π

2

2 3π y = − sin

⋅ 3 2

= − sin π = 0

3π

2 , 0

B

2





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

15. The equation y = − sin 2

x is of the form

y = A sin Bx

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

is A = −2 = 2 . The period is 2π

= 2π

= 2 . The with A = –1 and B =

2 .

B π

quarter-period is 2

= 1

.

The amplitude is

3

A = −1

= 1 .

4 2 The period is

2π =

2π = 2π ⋅

3 = 3π .

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

to generate x-values for the key points. 3

x = 0

x = 0 + 1

= 1

2 2

x = 1

+ 1

= 1

The quarter-period is 3π

. The cycle begins at x = 0. 4

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

x = 0

2 2

x = 1 + 1

= 3

x = 0 + 3π 4

= 3π 4

2 2

x = 3

+ 1

= 2 2 2

x = 3π 4

3π

+ 3π 4

3π

= 3π 2

9π

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

9π 3π x = + = 3π

4 4 Evaluate the function at each value of x.

3 4

2

4



9π

4

2 9π y = − sin ⋅

3 4

= − sin 3π 2

= −(−1) = 1

9π , 1 4

3π y = − sin 2

⋅ 3π

= − sin 2π = 0

(3π , 0)

x y = − sin 4

x 3

coordinates

0 y = − sin 4

⋅ 0

3

= − sin 0 = 0

(0, 0)

3π

8

4 3π y = − sin ⋅ 3 8

π = − sin

2 = −1

3π 8

3π

4

4 3π y = − sin ⋅

3 4

= − sin π = 0

3π , 0 4

9π

8

4 9π y = − sin ⋅

3 8

= − sin 3π

= −(−1) = 1 2

9π , 1 8

3π

2

4 3π y = − sin

3 ⋅

2

= − sin 2π = 0

3π , 0

B

3

3 , −1



16. The equation

y = − sin 4

x is of the form 3

2

y = A sin Bx with A = –1 and B = 4

. 3


graph one complete cycle of the given function.The amplitude is A = −1 = 1 .

The period is 2π =

2π = 2π ⋅

3 =

3π .

4 4 2

3π

The quarter-period is 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

x = 0 + 3π

= 3π

8 8

x = 3π

+ 3π

= 3π

8 8 4

x = 3π

+ 3π

= 9π

4 8 8

x = 9π

+ 3π

= 3π




x y = sin( x − π ) coordinates

π y = sin(π − π )

= sin 0 = 0

(π , 0)

3π

2

3π y = sin − π

2

= sin π

= 1 2

3π

2 , 1

2π y = sin(2π − π )

= sin π = 0

(2π , 0)

5π

2

5π y = sin − π

2

= sin 3π

= −1 2

5π

2 , − 1

3π y = sin(3π − π )

= sin 2π = 0

(3π , 0)

x π y = sin x −

2

coordinates

π

2

y = sin

π −

π = sin 0 = 0

2 2

π 2

, 0

π y = sin π −

π = sin

π = 1

2 2

(π , 1)

3π

2

3π π y = sin −

2 2

= sin π = 0

3π

2 , 0

2π y = sin 2π −

π 2

3π = sin

2 = −1

(2π , − 1)

5π

2

5π π y = sin −

2 2

= sin 2π = 0

5π , 0 2

π

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

18. The equation

y = sin x −

π is of the form

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

2

amplitude is

A = 1

= 1 . The period is


.

2π =

2π = 2π . The phase shift is

C =

π = π . The

2

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

quarter-period is 2π

= π

. The cycle begins at 2π

= 2π

= 2π . The phase shift is C π

2

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

B 1

2π π

= = . The B 1 2

the key points.

x = π

quarter-period is

π

= . The cycle begins at 4 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2

x = 2π + π

= 5π

2 2

x = . Add quarter-periods to generate 2

x-values for the key points.

x = π 2

x = π

+ π

= π 2 2

x = 5π

+ π

= 3π 2 2


x = π + π

= 3π

2 2

3π π

x = + = 2π 2 2

x = 2π + π

= 5π








5π

4

5π y = sin 2 ⋅ − π

4

= sin 5π

− π 2

= sin 3π

= −1 2

5π

4 , − 1

3π

2

3π y = sin 2 ⋅ − π

2

= sin(3π − π )

= sin 2π = 0

3π

2 , 0

x y = sin(2 x − π ) coordinates

π

2

π y = sin 2 ⋅ − π

2

= sin(π − π )

= sin 0 = 0

π 2

, 0

3π

4

3π y = sin 2 ⋅ − π

4

= sin 3π

− π 2

= sin π

= 1 2

3π

4 , 1

π y = sin(2 ⋅ π − π ) (π , 0)

2

2



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 graph one complete cycle of the given function.

2π =

2π = π . The phase shift is C

= π

. The

B 2 B 2


. The cycle begins at x = π

. Add

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

x = π 2

x = π

+ π

= 3π

2 4 4

π

x = 3π

+ π

= π 4 4

20. The equation y = sin 2 x −

is of the form

x = π + π

= 5π

4 4 y = A sin( Bx − C ) with A = 1, B = 2, and C =

π .

2

x = 5π

+ π

= 3π The amplitude is A = 1 = 1 .

4 4 2 The period is

2π =

2π = π .

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

C π 1 π

= = ⋅ = .

The phase shift is B

2 2 2 4

π

The quarter-period is . 4

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


x = π 4

x = π

+ π

= π



4 4 2

x = π

+ π

= 3π

2 4 4

x = 3π

+ π

= π 4 4

x = π + π

= 5π

4 4




x y = sin 2 x −

π

coordinates

π

4

π π y = sin 2 ⋅ −

4 2

= sin π

− π

= sin 0 = 0

π , 0 4

π

2

π π y = sin 2 ⋅ −

2 2

= sin π −

π = sin

π = 1

π , 1 2

3π

4

3π π y = sin 2 ⋅ −

4 2

= sin 3π

− π

2 2

= sin π = 0

3π , 0 4

π y = sin 2 ⋅π −

π

= sin 2π −

π

= sin 3π

= −1 2

(π , −1)

5π

4

5π π y = sin 2 ⋅ −

4 2

= sin 5π

− π

= sin 2π = 0

5π , 0 4

x y = 3sin(2 x − π ) coordinates

π

2

π y = 3sin 2 ⋅ − π

2

= 3sin(π − π )

= 3sin 0 = 3 ⋅ 0 = 0

π 2

, 0

3π

4 y = 3sin

2 ⋅

3π − π

3π = 3sin − π

2

= 3sin π

= 3 ⋅1 = 3 2

3π , 3

π y = 3sin(2 ⋅ π − π )

= 3sin(2π − π )

= 3sin π = 3 ⋅ 0 = 0

(π , 0)

5π

4

5π y = 3sin 2 ⋅ − π

4

= 3sin 5π

− π 2

= 3sin 3π 2

= 3(−1) = −3

5π

4 , − 3

3π

2

3π y = 3sin 2 ⋅ − π

2

= 3sin(3π − π )

= 3sin 2π = 3 ⋅ 0 = 0

3π

2 , 0

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

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

amplitude is A = 3 = 3 . The period is

2π =

2π = π . The phase shift is C

= π

. The

B 2 B 2

π π

2 2

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

4 2


x = π 2

x = π

+ π

= 3π

2

2 2 4 4

x = 3π

+ π

= π 4 4

x = π + π

= 5π

4 4

5π π 3π x = + =


2

2

4

4

2 2







x y = 3sin 2 x −

π 2

coordinates

π

4

π π y = 3sin 2 ⋅ −

4 2

= sin π

− π

2 2

= 3sin 0 = 3 ⋅ 0 = 0

π , 0 4

π

2

π π y = 3sin 2 ⋅ −

2 2

π = 3sin π −

2

π = 3sin

2 = 3 ⋅1 = 3

π , 3 2

3π

4

3π π y = 3sin 2 ⋅ −

4 2

= 3sin 3π

− π

2 2

= 3sin π = 3 ⋅ 0 = 0

3π , 0

4

π π

= 3sin 2π −

π

2

= 3sin 3π

= 3 ⋅ (−1) = −3 2

(π , − 3)

5π

4

5π π y = 3sin 2 ⋅

4 −

2

= 3sin 5π

− π

2 2

= 3sin 2π = 3 ⋅ 0 = 0

5π

4 , 0



22. The equation

y = 3sin 2 x −

π

is of the form

2


. 2


The period is 2π

= 2π

= π . B 2

π

The phase shift is C

= 2 = π

⋅ 1

= π

.

B 2 2 2 4

The quarter-period is π

. 4

The cycle begins at x = π

. Add quarter-periods to 4


x = π 4

x = π

+ π

= π

4 4 2

x = π

+ π

= 3π

2 4 4

x = 3π

+ π

= π 4 4

x = π + π

= 5π

y = 3sin 2 ⋅ π − 2







π 1 π y = sin π +

2 2

= 1

sin 3π

2 2

= 1

⋅ (−1) = − 1

2 2

1 π , − 2

3π

2 y =

1 sin

3π +

π 2 2 2

= 1

sin 2π 2

= 1

⋅ 0 = 0 2

3π , 0

x y = 1

sin x +

π 2 2

coordinates

− π 2

y = 1

sin −

π +

π 2 2 2

= 1

sin 0 = 1

⋅ 0 = 0 2 2

2

0 y = 1

sin 0 +

π 2 2

= 1

sin π

= 1

⋅1 = 1

2 2 2 2

0,

1

π

2

1 π π y = sin +

2 2 2

= 1

sin π = 1

⋅ 0 = 0 2 2

π , 0 2

= −

x

23.

y = 1

sin x +

π =

1 sin

x −

−

π

2

2

2

2

The equation

y =

1 sin

x −

−

π is of the form

2

2

y = A sin( Bx − C ) with

A = 1

, B = 1, and C π

.

The amplitude is

A = 1

2 2

= 1

. The period is

2 2

2π =

2π = 2π . The phase shift is

C − π π

= 2 = − .

2

B 1 B 1 2


= π

. The cycle begins at 4 2

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

2 for the key points.



π x = −

2

π π x = − + = 0

2 2

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π


24.

y = 1

sin( x + π ) = 1

sin( x − (−π )) 2 2

1The equation y = sin( x − (−π ))

2

1

is of the form

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

1 1 −

π , 0 The amplitude is A = = . The period is

2 2

2π =

2π = 2π . The phase shift is C

= −π

= −π .

B 1 B 1

2


= π


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

the key points. x = −π



x = −π + π

= − π

2 2

π π x = − + = 0

2 2

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2



x y = 1

sin( x + π ) 2

coordinates

−π y = 1

sin(−π + π ) 2

= 1

sin 0 = 1

⋅ 0 = 0 2 2

(−π , 0)

− π 2

y = 1

sin −

π + π

2 2

= 1

sin π

= 1

⋅1 = 1

2 2 2 2

−

π ,

1 2 2

0 y = 1

sin(0 + π ) 2

= 1

sin π = 1

⋅ 0 = 0 2 2

(0, 0)

π

2

1 π y = sin + π

2 2

= 1

sin 3π

= 1

⋅ (−1) = − 1

2 2 2 2

π 1 , − 2 2

π y = 1

sin(π + π ) 2

= 1

sin 2π = 1

⋅ 0 = 0 2 2

(π , 0)

x y = −2sin

2 x +

π 2

coordinates

− π 4

y = −2sin 2⋅

−

π +

π

= −2sin −

π +

π 2 2

= −2sin 0 = −2 ⋅ 0 = 0

−

π , 0

0 y = −2sin

2 ⋅ 0 +

π 2

= −2sin 0 +

π 2

= −2sin π 2

= −2 ⋅1 = −2

(0, –2)

π

4

π π y = −2sin 2 ⋅ +

4 2

= −2sin π

+ π

= −2sin π

= −2 ⋅ 0 = 0

π , 0 4

= −

= −

= −

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

C − π π 1 π

= 2 = − ⋅ = − . The quarter-

B 2 2 2 4

period is π

. The cycle begins at x π

. Add

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

π x = −

4

x π

+ π

= 0 4 4

x = 0 + π

= π

4 4

x = π

+ π

= π

4 4 2

x = π

+ π

= 3π

2 4 4


4 2

4



25.

y = −2sin 2 x +

π = −2 sin

2 x −

−

π

2

2

2 2

The equation

π

y = −2sin 2 x − − is of the form

2

y = A sin( Bx − C ) with A = –2,

B = 2, and C π

. The amplitude is



A = −2 2

= 2 . The period is 2π

= 2π

= π . The

B 2



π

2

π π y = −2sin 2 ⋅ +

2 2

= −2sin π +

π

2

= −2sin 3π 2

= −2(−1) = 2

π , 2 2

3π

4

3π π y = −2sin 2 ⋅ +

4 2

= −2sin 3π

+ π

2 2

= −2sin 2π

= −2 ⋅ 0 = 0

3π , 0 4

x y = −3sin 2 x +

π 2

coordinates

π −

4 y = −3sin 2 ⋅

−

π +

π

= −3sin −

π +

π 2 2

= −3sin 0 = −3 ⋅ 0 = 0

4

0 y = −3sin 2 ⋅ 0 +

π 2

= −3sin 0 +

π

2

= −3sin π

= −3 ⋅1 = −3 2

(0, –3)

π

4

π π y = −3sin 2 ⋅ +

4 2

= −3sin π

+ π

= −3sin π = −3 ⋅ 0 = 0

π , 0 4

π

2 y = −3sin

2 ⋅

π +

π 2 2

π = −3sin π +

2

= −3sin 3π

= −3 ⋅ (−1) = 3 2

π , 3

2

3π

4 y = −3sin

2 ⋅

3π +

π

= −3sin 3π

+ π

2 2

= −3sin 2π = −3 ⋅ 0 = 0

4

= −

−

= −

π

x = − 4

π π x = − + = 0

4 4

x = 0 + π

= π

4 4

x = π

+ π

= π

4 4 2

x = π

+ π

= 3π


26.



y = −3sin 2 x +

π = −3sin

2 x −

−

π

4

2

− π

, 0

2

2

The equation

π

y = −3sin 2 x − −

is of the form

2

2 2

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

.

The amplitude is

A = −3

2

= 3 . The period is

2π

= 2π

= π . The phase shift is B 2

π C 2 π 1 π π

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

The cycle begins at x π

. Add quarter-periods to4


4 2 3π

, 0



x y = 3sin(π x + 2) coordinates

− 2 π

y = 3sin π

−

2 + 2

= 3sin(−2 + 2)

= 3sin 0 = 3 ⋅ 0 = 0

−

2 , 0

π − 4

2π

π − 4 y = 3sin π

2π + 2

= 3sin π − 4

+ 2

2

= 3sin π

− 2 + 2

2

= 3sin π 2

= 3 ⋅1 = 3

π − 4

2π , 3

π − 2

π

π − 2 y = 3sin π + 2

π

= 3sin(π − 2 + 2)

= 3sin π = 3 ⋅ 0 = 0

π − 2 , 0

π

3π − 4

2π

3π − 4 y = 3sin π + 2

2π

= 3sin 3π − 4

+ 2

3π = 3sin

2 − 2 + 2

3π = 3sin

2

= 3(−1) = −3

5π , − 3 4

2π − 2

π

2π − 2 y = 3sin π

π + 2

= 3sin(2π − 2 + 2)

= 3sin 2π = 3 ⋅ 0 = 0

2π − 2

π , 0



π

π

27.

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 = 2 . The phase shift is C

= −2

= − 2

.

B π B π π

The quarter-period is 2

= 1


x = − 2

. Add quarter-periods to generate x-values π

for the key points.

x = − 2 π

x = − 2

+ 1

= π − 4

π 2 2π 2

x = π − 4

+ 1

= π − 2

2π 2 π

x = π − 2

+ 1

= 3π − 4

π 2 2π

x = 3π − 4

+ 1

= 2π − 2

2π 2 πEvaluate the function at each value of x.







3π − 8

4π

3π − 8 y = 3sin 2π

4π + 4

= 3sin 3π − 8

+ 4

3π = 3sin

2 − 4 + 4

3π = 3sin = 3(−1) = −3

2

3π − 8

4π , − 3

π − 2

π

π − 2 y = 3sin 2π

π + 4

= 3sin(2π − 4 + 4)

= 3sin 2π = 3 ⋅ 0 = 0

π − 2 ,0

x y = 3sin(2π x + 4) coordinates

− 2 π

y = 3sin 2π

−

2 + 4

= 3sin(−4 + 4)

= 3sin 0 = 3 ⋅ 0 = 0

−

2 , 0

π − 8

4π

π − 8 y = 3sin 2π

4π + 4

= 3sin π − 8

+ 4

= 3sin π

− 4 + 4

2

= 3sin π

= 3 ⋅1 = 3 2

π − 8

4π , 3

π − 4

2π

π − 4 y = 3sin 2π

2π + 4

= 3sin(π − 4 + 4)

= 3sin π = 3 ⋅ 0 = 0

π − 4

2π , 0

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 2

is 2π

= 2π

= 1 . The phase shift is C =

−4 = −

2 .

B 2π B 2π π

The quarter-period is 1

. The cycle begins at 4

x = − 2

. Add quarter-periods to generate x-values π

for the key points.

x = − 2 π

x = − 2

+ 1

= π − 8

π

π 4 4π

x = π − 8

+ 1

= π − 4


4π 4 2π

x = π − 4

+ 1

= 3π − 8

2π 4 4π

x = 3π − 8

+ 1

= π − 2

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

π

π

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

period is 2π

= 2π

= 1 . The phase shift is

B 2π

C =

−4π = −2 . The quarter-period is

1 . The cycle

2

B 2π 4

begins at x = −2 . Add quarter-periods to generate x- values for the key points.

x = −2

x = −2 + 1

= − 7

4 4



= −

= −

x

7

+

1

=

−

3

4 4 2

x

3

+

1

=

−

5

2 4 4

5 1

x = − + = −1



4 4




x y = −2sin(2π x + 4π ) coordinates

–2 y = −2sin(2π (−2) + 4π )

= −2sin(−4π + 4π )

= −2sin 0

= −2 ⋅ 0 = 0

(–2, 0)

− 7 4

y = −2sin 2π

−

7 + 4π

= −2sin −

7π + 4π

2

= −2sin π

= −2 ⋅1 = −2 2

−

7 , − 2

− 3 2

y = −2sin 2π

−

3 + 4π

= −2sin(−3π + 4π )

= −2sin π = −2 ⋅ 0 = 0

2

− 5 4

y = −2sin 2π

−

5 + 4π

= −2sin −

5π + 4π

= −2sin 3π 2

= −2(−1) = 2

−

5 , 2

–1 y = −2sin(2π (−1) + 4π )

= −2sin(−2π + 4π )

= −2sin 2π

= −2 ⋅ 0 = 0

(–1, 0)

x y = −3sin(2π x + 4π ) coordinates

–2 y = −3sin(2π (−2) + 4π )

= −3sin(−4π + 4π )

= −3sin 0 = −3 ⋅ 0 = 0

(–2, 0)

− 7 4

y = −3sin 2π

−

7 + 4π

= −3sin −

7π + 4π

2

= −3sin π

= −3 ⋅1 = −3 2

−

7 , − 3

3 −

2

3 y = −3sin 2π −

2 + 4π

= −3sin(−3π + 4π )

= −3sin π = −3 ⋅ 0 = 0

3 −

2 , 0

− 5 4

y = −3sin 2π

−

5 + 4π

= −3sin −

5π + 4π

2

= −3sin 3π

= −3(−1) = 3 2

−

5 , 3

–1 y = −3sin(2π (−1) + 4π )

= −3sin(−2π + 4π )

= −3sin 2π = −3 ⋅ 0 = 0

(–1, 0)

= −

= −

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

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

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

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

2π =

2π = 1 . The phase shift is

B 2π

C =

−4π B 2π

= −2 . The

4

4 quarter-period is

1 . The cycle begins at x = −2 . Add

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

2

−

3 , 0

x = −2 + 1

= − 7

4 4

x 7

+ 1

= − 3

4 4 2

x 3

+ 1

= − 5

2 4 4

5 1 x = − + = −1


4

4

2

4

4

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

4

4



x y = 3 cos x coordinates

0 y = 3 cos 0 = 3 ⋅1 = 3 (0, 3)

π

2

π y = 3cos = 3 ⋅ 0 = 0

2

π , 0 2

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

3π

2 y = 3cos

3π = 3 ⋅ 0 = 0

2

2

2π y = 3 cos 2π = 3 ⋅1 = 3 (2π , 3)

x y = 2 cos x coordinates

0 y = 2 cos 0

= 2 ⋅1 = 2

(0, 2)

π

2 y = 2cos

π 2

= 2 ⋅ 0 = 0

π , 0

2

π y = 2 cos π

= 2 ⋅ (−1) = −2

(π , − 2)

3π

2 y = 2 cos

3π 2

= 2 ⋅ 0 = 0

2

2π y = 2 cos 2π

= 2 ⋅1 = 2

(2π , 2)





the graph of y = 2 cos x .

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

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

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

with A = 2. Thus, the amplitude is A = 2 = 2 . The period is 2π . The quarter-period is or . 4 2


or π

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

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


x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2



x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2


3π , 0

3π , 0

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 .





x y = −3cos x coordinates

0 y = −3cos 0 = −3 ⋅1 = −3 (0, –3)

π

2

π y = −3cos = −3 ⋅ 0 = 0

2

π , 0 2

π y = −3cos π = −3 ⋅ (−1) = 3 (π , 3)

3π

2

3π y = −3cos

2 = −3 ⋅ 0 = 0

3π

2 , 0

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

x y = −2 cos x coordinates

0 y = −2 cos 0

= −2 ⋅1 = −2

(0, –2)

π

2 y = −2 cos

π 2

= −2 ⋅ 0 = 0

π , 0

2

π y = −2 cos π

= −2 ⋅ (−1) = 2

(π , 2)

3π

2 y = −2 cos

3π 2

= −2 ⋅ 0 = 0

3π , 0

2π y = −2 cos 2π

= −2 ⋅1 = −2

(2π , − 2)

33. The equation y = −2 cos x is of the form y = A cos x 34. The equation y = −3cos x is of the form y = A cos x

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

A = −2

period is 2π

= 2 . The period is 2π . The quarter-

or π

. The cycle begins at x = 0 . Add


or π

. 4 2

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

x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2



x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2


2



the graph of y = cos x .


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



x y = cos 2 x coordinates

0 y = cos(2 ⋅ 0)

= cos 0 = 1

(0, 1)

π

4

π y = cos 2 ⋅

4

= cos π

= 0 2

π 4

, 0

π

2

π y = cos 2 ⋅

2

= cos π = −1

π 2

, −1

3π

4

3π y = cos 2 ⋅

4

= cos 3π

= 0 2

3π

4 , 0

π y = cos(2 ⋅π )

= cos 2π = 1

(π , 1)

x y = cos 4 x coordinates

0 y = cos(4 ⋅ 0) = cos 0 = 1 (0, 1)

π

8 y = cos

4 ⋅

π = cos

π = 0

π , 0

π

4 y = cos

4 ⋅

π = cos π = −1

π , −1

3π

8 y = cos

4 ⋅

3π

= cos 3π

= 0 2

3π , 0

π

2

π y = cos 4 ⋅

2 = cos 2π = 1

π 2

, 1

35. The equation y = cos 2 x is of the form y = A cos Bx 36. The equation y = cos 4 x is of the form y = A cos Bx

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


= 2π

= π . The


= 2π

= π

. The

B 2 B 4 2 π


. The cycle begins at x = 0 . Add 4


x = 0

x = 0 + π

= π

4 4

x = π

+ π

= π

4 4 2

x = π

+ π

= 3π

2 4 4

x = 3π

+ π

= π 4 4


quarter-period is 2 = π

⋅ 1

= π

. The cycle begins at 4 2 4 8

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

the key points. x = 0

x = 0 + π

= π

8 8

x = π

+ π

= π

8 8 4

x = π

+ π

= 3π

4 8 8

x = 3π

+ π

= π


8

2

8

4

4

8

8








x y = 4 cos 2π x coordinates

0 y = 4 cos(2π ⋅ 0)

= 4 cos 0

= 4 ⋅1 = 4

(0, 4)

1

4

1 y = 4 cos 2π ⋅

4

= 4 cos π 2

= 4 ⋅ 0 = 0

1 4

, 0

1

2

1 y = 4 cos 2π ⋅

2

= 4 cos π

= 4 ⋅ (−1) = −4

1 2

, − 4

3

4

3 y = 4 cos 2π ⋅

4

= 4 cos 3π 2

= 4 ⋅ 0 = 0

3 4

, 0

1 y = 4 cos(2π ⋅1)

= 4 cos 2π

= 4 ⋅1 = 4

(1, 4)

x y = 5 cos 2π x coordinates

0 y = 5 cos(2π ⋅ 0)

= 5 cos 0 = 5 ⋅1 = 5

(0, 5)

1

4

1 y = 5 cos 2π ⋅

4

= 5 cos π

= 5 ⋅ 0 = 0 2

1 , 0 4

1

2 y = 5 cos

2π ⋅

1

2

= 5 cos π = 5 ⋅ (−1) = −5

1 , − 5

2

3

4

3π y = 5 cos 2π ⋅

4

= 5 cos 3π

= 5 ⋅ 0 = 0 2

3 4

, 0

1 y = 5 cos(2π ⋅1)

= 5 cos 2π = 5 ⋅1 = 5

(1, 5)

37. The equation y = 4 cos 2π x is of the form y = A cos Bx 38. The equation y = 5 cos 2π x is of the form

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


= 2π

= 1 . The amplitude is A = 5 = 5 . The period is

B 2π 2π =

2π = 1 . The quarter-period is

1 . The cycle

quarter-period is 1

. The cycle begins at x = 0 . Add B 2π 4

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

x = 0

x = 0 + 1

= 1

4 4

x = 1

+ 1

= 1

4 4 2

x = 1

+ 1

= 3

2 4 4

x = 3

+ 1

= 1 4 4


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

x = 0 + 1

= 1

4 4

x = 1

+ 1

= 1

4 4 2

x = 1

+ 1

= 3

2 4 4

x = 3

+ 1

= 1 4 4









x y = −4cos 1

x 2

coordinates

0 y = −4 cos 1

⋅ 0

2

= −4 cos 0

= −4 ⋅1 = −4

(0, –4)

π y = −4 cos 1

⋅π

= −4 cos π 2

= −4 ⋅ 0 = 0

(π , 0)

2π y = −4 cos 1

⋅ 2π

= −4 cos π

= −4 ⋅ (−1) = 4

(2π , 4)

3π y = −4cos 1

⋅ 3π 2

= −4cos 3π 2

= −4 ⋅ 0 = 0

(3π , 0)

4π y = −4 cos 1

⋅ 4π

= −4 cos 2π

= −4 ⋅1 = −4

(4π , – 4)

x 1

y = −3cos x 3

coordinates

0 y = −3cos 1

⋅ 0

3

= −3cos 0 = −3 ⋅1 = −3

(0, –3)

3π

2 y = −3cos

1 ⋅

3π

= −3cos π

= −3 ⋅ 0 = 0 2

3π , 0

3π

1 y = −3cos ⋅ 3π

3

= −3cos π = −3 ⋅ (−1) = 3

(3π , 3)

B

= B

39. The equation y = −4cos 1

x 2

is of the form Connect the five points with a smooth curve and


y = A cos Bx with A = –4 and B = 1

. Thus, the 2

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

2π =

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

1 2

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


x = 0 + π = π

40. The equation

y = −3cos 1

x 3

is of the form

1

x = π + π = 2π y = A cos Bx with A = –3 and B = . Thus, the 3

x = 2π + π = 3π amplitude is A = −3 = 3 . The period is

x = 3π + π = 4π


2π 2π 1 3

= 2π ⋅ 3 = 6π . The quarter-period is

6π =

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

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

x = 0 + 3π 2

= 3π 2

x = 3π

+ 3π

= 3π 2 2

2

x = 3π + 3π

= 9π

2 2

9π 3πx = +

2 2 = 6π


2

3 2

2



2



9π

2

1 9π y = −3cos ⋅

3 2

= −3cos 3π

= −3 ⋅ 0 = 0 2

9π , 0 2

6π y = −3cos 1

⋅ 6π

= −3cos 2π = −3 ⋅1 = −3

(6π , − 3)

x y 1

cos π

x 2 3

coordinates

0 1 π y

2 cos

3 ⋅ 0 = −

1 cos 0

= −

2

1 ⋅1 = −

1 = − 2 2

0, −

1 2

3

2

1 π 3 y = − cos ⋅

2 3 2

1 cos

π = −

2 2

1 ⋅ 0 = 0

= − 2

3 , 0 2

3 y 1

cos π

⋅ 3 2 3

1 cos π

2

1 1

2 ⋅ (−1) =

2

= −

3,

1 2

9

2 y

1 cos

π 9 2 3 2

1 cos

3π 2 2

1 ⋅ 0 = 0

= − 2

9 2

, 0

6 y 1

cos π

⋅ 6

= − 2 3

1 cos 2π

= − 2

1 ⋅1 = −

1 = − 2 2

6, −

1

= −

π

= −

= −

3



41. The equation

y 1

cos π

x is of the form 2 3

= −

= −

y = A cos Bx

with 1

A = − 2

and B = π

. Thus, the 3

amplitude is A = − 1 =

1 . The period is

= − ⋅

2 2

2π =

2π = 2 ⋅

3 = 6 . The quarter-period is

6 =

3 .

π B 3

π 4 2

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


x = 0

x = 0 + 3

= 3

2 2

x = 3

+ 3

= 3 2 2

x = 3 + 3

= 9

2 2

2

x = 9

+ 3

= 6 2 2






x y 1

cos π

x

= − 2 4

coordinates

0 y 1

cos π

⋅ 0 2 4

1 cos 0

1 ⋅1 = −

1 = − = − 2 2 2

0, −

1

2

2 y 1

cos π

⋅ 2 2 4

1 cos

π 1 ⋅ 0 = 0

= − = −

2 2 2

(2, 0)

4 y 1

cos π

⋅ 4 2 4

1 cos π = −

1 ⋅ (−1) =

1 = − 2 2 2

4,

1 2

6 y 1

cos π

⋅ 6

= − 2 4

1 cos

3π 1 ⋅ 0 = 0

= − = −

2 2 2

(6, 0)

8 y 1

cos π

⋅ 8 2 4

1 cos 2π

1 ⋅1 = −

1 = − = − 2 2 2

8, −

1

2

x coordinates

π

2

π , 1 2

π (π , 0)

3π

2

3π , −1

2

2π (2π , 0)

5π

2

5π , 1

2

= −

2

2

B π

42. The equation y 1

cos π




y = A cos Bx

with 1

A = − 2

and B = π

. Thus, the 4

amplitude is A = − 1 =

1 . The period is

2 2

2π =

2π = 2π ⋅

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

π 4 4



π

x = 0 + 2 = 2 43. The equation y = cos x − is of the form

x = 2 + 2 = 4

x = 4 + 2 = 6 y = A cos ( Bx − C ) with A = 1, and B = 1, and

x = 6 + 2 = 8

Evaluate the function at each value of x. C =

π . Thus, the amplitude is

2 A = 1 = 1 . The

2π =

2π =

π . The phase shift isperiod is 2

B 1

π C π 2π π

= = . The quarter-period is = . The= − B 1 2 4 2

π

= −

= −

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


x = π 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2

x = 2π + π

= 5π


= −



x coordinates

π

2

π , 3 2

3π

4

3π , 0

π (π , − 3)

5π

4

5π , 0 4

3π 3π

x coordinates

− π 2

−

π , 1

2

0 (0, 0)

π

2

π , −1 2

π (π , 0)

C

−

= −


graph one complete cycle of the given function



44. The equation

y = cos x +

π is of the form

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

2

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

y = A cos ( Bx − C ) with A = 1, and B = 1, and C = π . Thus, the amplitude is A = 3 = 3 . The

π = − . Thus, the amplitude is

2

A = 1

= 1 . The

period is 2π

= 2π


C =

π .

B 2

period is 2π

= 2π

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


.

B 1 4 2

π C 2 π

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

= = − . The quarter-period is B 1 2

= . The 4 2 x =

π

cycle begins at x π



x = π

+ π

= 3π

2 4 4

π x = −

2 x =

3π +

π = π

4 4

π π x = − + = 0 x = π +

π =

5π

2 2

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π


4 4

x = 5π

+ π

= 3π


4


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

x coordinates

π

2

π , 4 2

3π

4

3π , 0 4

π (π , − 4)

5π

4

5π , 0

C

= −





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 .

47.

y = 1

cos 3x +

π =

1 cos

3x −

−

π

Thus, the amplitude is

A = 4

= 4 . The period is

2

2

2

2

2π =

2π = π . The phase shift is

C =

π . The

The equation

y =

1 cos

3x −

−

π is of the form

B 2 B 2

2

2



.

1

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

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

x = π

π = − . Thus, the amplitude is

2

A = 1

= 1

. 2 2

2 The period is 2π

= 2π

. The phase shift is

x = π

+ π

= 3π B 3

π

2 4 4 C =

− 2 = −

π ⋅

1 = −

π . The quarter-period is

x = 3π

+ π

= π 4 4

B 2π

3 2 3 6

3 = 2π

⋅ 1

= π


. Addx = π +

π =

5π 4 3 4 6 6

4 4

x = 5π

+ π

= 3π


4 4 2 π x = −


6

π π x = − + = 0

6 6

π πx = 0 +

6 =

6

π π π x = + =

6 6 3



x = + = π π π

3 6 2




x

coordinates

− π 6

−

π ,

1

0 (0, 0)

π

6

π 1 , − 6 2

π

3

π , 0 3

π

2

π ,

1 2 2

x

coordinates

− π 2

−

π ,

1

π −

4 −

π , 0

4

0 0, −

1 2

π

4

π , 0 4

π

2

π 1 , 2 2

x

= −

π

= − 2

6 2

π π π x = − + = −

2 4 4

π πx = − + = 0

4 4

x = 0 + π

= π

4 4

x = π

+ π

= π




2 2

48. y = 1

cos(2 x + π ) = 1

cos(2 x − (−π )) 2 2


The equation y = 1

cos(2x − (−π )) 2

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

y = A cos(Bx − C ) with A = 1

, and B = 2, and 2

C = −π . Thus, the amplitude is A = 1 =

1 . The

2 2

period is 2π

= 2π


C =

−π = −

π . The quarter-period is

π . The cycle

B 2 2 4

begins at x π

. Add quarter-periods to generate 2

x-values for the key points.



x coordinates

π

4

π 4

, − 4

π

2

π 2

, 0

3π

4

3π

4 , 4

π

(π , 0)

5π

4

5π

4 , − 4

x coordinates

π

4

π , − 3 4

π

2

π , 0 2

3π

4

3π

4

π (π , 0)

5π

4

5π , − 3 4

2 2

49. The equation

y = −3cos 2 x −

π is of the form

50. The equation

y = −4cos 2 x −

π

is of the form

2

2

y = A cos(Bx − C ) with A = –3, and

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

B = 2, and C = π

. Thus, the amplitude is 2

C = π

. Thus, the amplitude is 2

A = −4

= 4 . The

A = −3 = 3 . The period is 2π

= 2π

= π . The period is 2π

= 2π

= π . The phase shift is

B 2 B 2 π π

phase shift is C π 1 π

= = ⋅ = . C π 1 π π

= = ⋅ = . The quarter-period is . The

B 2 2 2 4 B 2 2 2 4 4



. cycle begins at x =

π . Add quarter-periods to

4 4 4

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

points.

x = π 4

x = π

+ π

= π

4 4 2

x = π

+ π

= 3π

2 4 4

x = 3π

+ π

= π 4 4

x = π + π

= 5π

4 4



x = π 4

x = π

+ π

= π

4 4 2

x = π

+ π

= 3π

2 4 4

x = 3π

+ π

= π 4 4

x = π + π

= 5π


, 3








x coordinates

–2 (–2, 3)

− 7 4

−

7 , 0

4

− 3 2

−

3 , − 3

− 5 4

−

5π , 0

4

–1 (–1, 3)

x coordinates

–4 (–4, 2)

− 15 4

−

15 , 0

4

− 7 2

−

7 , − 2

2

− 13 4

−

13 , 0

4

–3 (–3, 2)

= −

= −

= −

= −

= −

51. y = 2 cos(2π x + 8π ) = 2 cos(2π x − (−8π )) 52. y = 3 cos(2π x + 4π ) = 3 cos(2π x − (−4π ))

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

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

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

period is 2π

= 2π

= 1 . The phase shift is period is 2π

= 2π

= 1 . The phase shift is

B 2π B 2π

C =

−8π = −4 . The quarter-period is

1 . The cycle C

= −4π

= −2 . The quarter-period is 1

. The cycle

B 2π 4 B 2π 4

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

x = −4

x = −4 + 1

= − 15

4 4

x 15

+ 1

= − 7

4 4 2

x 7

+ 1

= − 13

2 4 4

x 13

+ 1

= −3 4 4

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

x = −2 + 1

= − 7

4 4

x 7

+ 1

= − 3

4 4 2

x 3

+ 1

= − 5

2 4 4

x 5

+ 1

= −1 = −

4 4

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

2








x y = sin x + 2 coordinates

0 y = sin 0 + 2

= 0 + 2 = 2

(0, 2)

π

2 y = sin

π + 2

2 = 1 + 2 = 3

π , 3

2

π y = sin π + 2

= 0 + 2 = 2

(π , 2)

3π

2

3π y = sin + 2

2 = −1 + 2 = 1

3π , 1 2

2π y = sin 2π + 2

= 0 + 2 = 2

(2π , 2)

x y = sin x − 2 coordinates

0 y = sin 0 − 2 = 0 − 2 = −2 (0, –2)

π

2

π y = sin − 2 = 1 − 2 = −1

2

π , −1 4

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

3π

2 y = sin

3π − 2 = −1 − 2 = −3

2

3π , − 3

2

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

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

shifted up 2 units upward. The period for both

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

or π

.

shifted 2 units downward. The period for both


or π

.

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

x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2



x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2


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

By connecting the points with a smooth curve we

obtain one period of the graph.



x y = cos x − 3 coordinates

0 y = cos 0 − 3

= 1 − 3 = −2

(0, –2)

π

2 y = cos

π − 3

2 = 0 − 3 = −3

π , − 3

π y = cos π − 3

= −1 − 3 = −4

(π , − 4)

3π

2

3π y = cos − 3

2 = 0 − 3 = −3

3π , − 3 2

2π y = cos 2π − 3

= 1 − 3 = −2

(2π , − 2)

x y = cos x + 3 coordinates

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

π

2

π y = cos + 3 = 0 + 3 = 3

2

π , 3 2

π y = cos π + 3 = −1 + 3 = 2 (π , 2)

3π

2 y = cos

3π + 3 = 0 + 3 = 3

2

3π , 3

2

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

2

2

1

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

shifted 3 units downward. The period for both


or π

.

shifted 3 units upward. The period for both functions

is 2π . The quarter-period is 2π

or π

. The cycle

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

x = 0

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2


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

x = 0 + π

= π

2 2

x = π

+ π

= π 2 2

x = π + π

= 3π

2 2

x = 3π

+ π

= 2π 2 2


2




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

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

amplitude for both functions is 2 = 2 . The period

for both functions is

2π = 2π ⋅ 2 = 4π . The quarter-

2



period is 4π

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

quarter-periods to generate x-values for the key



x y = 2 sin 1

x + 1 2

coordinates

0 y = 2sin 1

⋅ 0 + 1

2

= 2sin 0 + 1

= 2 ⋅ 0 + 1 = 0 + 1 = 1

(0, 1)

π y = 2sin 1

⋅ π + 1

= 2sin π

+ 1 2

= 2 ⋅1 + 1 = 2 + 1 = 3

(π , 3)

2π

1 y = 2sin ⋅ 2π + 1

2

= 2sin π + 1

= 2 ⋅ 0 + 1 = 0 + 1 = 1

(2π , 1)

3π y = 2 sin 1

⋅ 3π + 1

2

= 2 sin 3π

+ 1 2

= 2 ⋅ (−1) + 1

= −2 + 1 = −1

(3π , −1)

4π y = 2sin 1

⋅ 4π + 1

2

= 2sin 2π + 1

= 2 ⋅ 0 + 1 = 0 + 1 = 1

(4π , 1)

x y = 2 cos 1

x + 1 2

coordinates

0 y = 2 cos 1

⋅ 0 + 1

2

= 2 cos 0 + 1

= 2 ⋅1 + 1 = 2 + 1 = 3

(0, 3)

π y = 2 cos 1

⋅π + 1

2

= 2 cos π

+ 1 2

= 2 ⋅ 0 + 1 = 0 + 1 = 1

(π , 1)

2π y = 2 cos 1

⋅ 2π + 1

2

= 2 cos π + 1

= 2 ⋅ (−1) + 1 = −2 + 1 = −1

(2π , − 1)

3π y = 2 cos 1

⋅ 3π + 1

2

= 2 ⋅ 0 + 1 = 0 + 1 = 1

(3π , 1)

4π y = 2 cos 1

⋅ 4π + 1

2

= 2 cos 2π + 1

= 2 ⋅1 + 1 = 2 + 1 = 3

(4π , 3)

1

points. x = 0

x = 0 + π = π

58. The graph of

1

y = 2 cos 1

x + 1 is the graph of 2

x = π + π = 2π y = 2 cos x shifted one unit upward. The amplitude

2

x = 2π + π = 3π for both functions is 2 = 2 . The period for both

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

functions is 2π

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

2

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


x = 0 + π = π

x = π + π = 2π

x = 2π + π = 3π

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







3

4

3 y = −3cos 2π ⋅

4 + 2

= −3cos 3π

+ 2 2

= −3 ⋅ 0 + 2

= 0 + 2 = 2

3 4

, 2

1 y = −3cos(2π ⋅1) + 2

= −3cos 2π + 2

= −3 ⋅1 + 2

= −3 + 2 = −1

(1, –1)

x y = −3cos 2π x + 2 coordinates

0 y = −3 cos(2π ⋅ 0) + 2

= −3cos 0 + 2

= −3 ⋅1 + 2

= −3 + 2 = −1

(0, –1)

1

4

1 y = −3cos 2π ⋅ + 2

4

= −3cos π

+ 2 2

= −3 ⋅ 0 + 2

= 0 + 2 = 2

1 4

, 2

1

2

1 y = −3cos 2π ⋅ + 2

2

= −3cos π + 2

= −3 ⋅ (−1) + 2

= 3 + 2 = 5

1 2

, 5

x = + = 1



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


y = −3cos 2π x shifted 2 units upward. The obtain one period of the graph.

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

for both functions is 2π

= 1 . The quarter-period is 2π

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


x = 0

x = 0 + 1

= 1

4 4

x = 1

+ 1

= 1

60. The graph of

y = −3sin 2π x + 2 is the graph of

4 4 2 y = −3sin 2π x shifted two units upward. The

x = 1

+ 1

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

2 4 4

x = 3

+ 1

= 1 4 4

period for both functions is 2π

= 1 . The quarter- 2π

1

Evaluate the function at each value of x. period is . 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

x = 1

+ 1

= 1

4 4 2

x = 1

+ 1

= 3

2 4 4

3 1

4 4




x y = −3sin 2π x + 2 coordinates

0 y = −3sin(2π ⋅ 0) + 2

= −3sin 0 + 2

= −3 ⋅ 0 + 2 = 0 + 2 = 2

(0, 2)

1

4

1 y = −3sin 2π ⋅ + 2

4

= −3sin π

+ 2 2

= −3 ⋅1 + 2 = −3 + 2 = −1

1 , − 1 4

1

2

1 y = −3sin 2π ⋅ + 2

2

= −3sin π + 2

= −3 ⋅ 0 + 2 = 0 + 2 = 2

1 , 2 2

3

4

3 y = −3sin 2π ⋅ + 2

4

= −3sin 3π

+ 2 2

= −3 ⋅ (−1) + 2 = 3 + 2 = 5

3 , 5 4

1 y = −3sin(2π ⋅1) + 2

= −3sin 2π + 2

= −3 ⋅ 0 + 2 = 0 + 2 = 2

(1, 2)



x

0 π

4

π

2

3π

4 π

5π

4

3π

2

7π

4 2π

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

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

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






x

0 π

4

π

2

3π

4 π

5π

4

3π

2

7π

4 2π

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

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

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


x

0 π

4

π

2

3π

4 π

5π

4

3π

2

7π

4 2π

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

y2 = sin 2 x 0 1 0 −1 0 1 0 −1 0

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


x

0 π

4

π

2

3π

4 π

5π

4

3π

2

7π

4 2π

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

y2 = cos 2 x 1 0 −1 0 1 0 −1 0 1

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




x

0 π

4

π

2

3π

4 π

5π

4

3π

2

7π

4 2π

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

y2 = cos 2 x 1 0 −1 0 1 0 −1 0 1

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


x

0 π

4

π

2

3π

4 π

5π

4

3π

2

7π

4 2π

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

y2 = sin 2 x 0 1 0 −1 0 1 0 −1 0

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


x

0 1

2

1 3

2

2 5

2

3 7

2

4

y1 = sin π x 0 1 0 −1 0 1 0 −1 0

π y2 = cos

2 x

1

0.7

0

−0.7

−1

−0.7

0

0.7

1

y = sin π x + cos π

x 2

1

1.7

0

−1.7

−1

0.3

0

−0.3

1






x

0 1

2

1 3

2

2 5

2

3 7

2

4

y1 = cos π x 1 0 −1 0 1 0 −1 0 1

π y2 = sin

2 x

0

0.7

1

0.7

0

−0.7

−1

−0.7

0

y = cos π x + sin π

x 2

1

0.7

0

0.7

1

−0.7

−2

−0.7

1

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

B = 2π

= 2π

= 1

period 4π 2

y = A cos Bx

y = 3 cos 1

x 2

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

B = 2π

= 2π

= 1

period 4π 2

y = A sin Bx

y = 3sin 1

x 2

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

B = 2π

= 2π

= 2 period π

y = A sin Bx

y = −2 sin 2x

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

B = 2π

= 2π

= 2 period π

y = A cos Bx

y = −2cos 2x



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

B = 2π

= 2π

= π

period 4 2

y = A sin Bx

y = 2sin π

x

2

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

B = 2π

= 2π

= π

period 4 2

y = A cos Bx

y = 2 cos π

x

2

75.

76.

77.

78.




x

0 π

4

π

2

3π

4 π

5π

4

3π

2

7π

4 2π

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

y2 = sin 2x 0 1 0 −1 0 1 0 −1 0

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


x

0 π

4

π

2

3π

4 π

5π

4

3π

2

7π

4 2π

y1 = cos 2x 1 0 −1 0 1 0 −1 0 1

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

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


x

0 π

4

π

2

3π

4 π

5π

4

3π

2

7π

4 2π

y1 = x 0 0.9 1.3 1.5 1.8 2.0 2.2 2.3 2.5

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

y = x + sin x 0 1.6 2.3 2.2 1.8 1.3 1.2 1.6 2.5




x

0 π

4

π

2

3π

4 π

5π

4

3π

2

7π

4 2π

y1 = x 0 0.8 1.6 2.4 3.1 3.9 4.7 5.5 6.3

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

y = x + cos x 1 1.5 1.6 1.6 2.1 3.2 4.7 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.


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.



π

B 2π

B 3 π

92. The information gives the five key points of the

graph.

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

C

(0, 23) corresponds to Noon, (3, 38) corresponds to 3 P.M.,

phase shift is

365

= 79 . The quarter-period is B

(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.

= 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 x-

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

calculator we have the following points. (0, 9.1) (79, 12) (170.25, 15)

(261.5, 12) (352.75, 9) (365, 9.1)

93. The function

y = 3sin 2π

( x − 79) + 12 is of the form 365

By connecting the points with a smooth curve

we obtain one period of the graph, starting on

January 1.

y = A sin B x −

C + D

with

B

A = 3 and B = 2π

. 365

a. The amplitude is A = 3 = 3 .

b. The period is

2π =

2π = 2π ⋅

365 = 365 .

2π 365

94. The function y = 16 sin π

x − 2π

+ 40 is in the

c. The longest day of the year will have the most 6 3

hours of daylight. This occurs when the sine function equals 1.

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

, 6

y = 3sin 2π

( x − 79) + 12 365

and C = 2π

. The amplitude is 3

A = 16

= 16 .

y = 3(1) + 12

The period is 2π =

2π = 2 ⋅

6 = 12 . The phase

π

y = 15

There will be 15 hours of daylight.

B

C 2π

6 π

2π 6

d. The shortest day of the year will have the least shift is = 3 = ⋅ = 4 . The quarter-period is



π

hours of daylight. This occurs when the sine 6

function equals –1.

y = 3sin 2π

( x − 79) + 12 365

y = 3(−1) + 12

y = 9

There will be 9 hours of daylight.

12 = 3 . The cycle begins at x = 4. Add quarter-

4 periods to find the x-values for the key points.



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,

12 = 2π B

12B = 2π

B = 2π

= π

12 6

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

depth of the water is modeled by

97. – 110. Answers may vary.

y = cos π x

+ 4 . 6

111. The function y = 3sin(2x + π ) = 3sin(2x − (−π )) is of

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

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

period is 2π

= 2π

= π . The cycle begins at

The highest average monthly temperature is 56° in B 2July. C −π π π 3π

95. Because the depth of the water ranges from a x = = B

= − . We choose − ≤ x ≤ 2 2 2

, and 2

minimum of 6 feet to a maximum of 12 feet, the curve oscillates about the middle value, 9 feet. Thus,

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

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

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

period is 12.

Thus,

12 = 2π

−4 ≤ y ≤ 4 for our graph.

B 112. The function

y = −2cos 2π x −

π is of the form

12B = 2π

2

B = 2π

= π

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

12 6 C =

π . The amplitude is

2 A = −2 = 2 . The

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



2π =

2π = 1 . The cycle begins at

depth of the water is modeled by y = 3cos π x

+ 9 . 6

B 2π

π

x = C

= 2 = π

⋅ 1

= 1

. We choose 1

≤ x ≤ 9

,B 2π 2 2π 4 4 4

and −3 ≤ y ≤ 3 for our graph.



π

π

2 2

113. The function

y = 0.2 sin π

x + π = 0.2 sin

π x − (−π )

is of the

116.

10

10

form

y = A sin(Bx − C) with A = 0.2, B = π

, and 10

C = −π . The amplitude is A = 0.2 = 0.2 . The The graphs appear to be the same from − π to

π .

period is 2π =

2π = 2 ⋅

10 = 20 . The cycle begins

π

B 10

π 117.

at x = C

= −π

= − ⋅ 10

= −10 . We choose π B 10

π −10 ≤ x ≤ 30 , and −1 ≤ y ≤ 1 for our graph.

The graph is similar to y = sin x , except the

114. The function

y = 3sin(2x − π ) + 5 is of the form

118.

amplitude is greater and the curve is less smooth.

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 The graph is very similar to

y = sin x , except not

2π =

2π = π . The cycle begins at x =

C =

π . Because

smooth.

B 2 B 2

115.

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

choose π

≤ x ≤ 5π

, and 0 ≤ y ≤ 10 for our graph. 2 2

119. a.

b.

y = 22.61sin(0.50x − 2.04) + 57.17

The graphs appear to be the same from − π

to π

.


121. makes sense

2 2 122. 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



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

128.

y = cos 2 x = 1

+ 1

cos 2 x

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

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

Viewing rectangle: − π

, 23π

, π

by [−5,1,1]

6 6 6

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

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

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

Viewing rectangle: − π

, 7π

, π

by [−3, −1,1] 6 6 6

126. A = π

B = 2π

= 2π

= 2π


π π π

period 1

C =

C = −2

130. − < x + < 2 4 2

π π π π π π

B 2π − − < x + − < − 2 4 4 4 2 4

C = −4π 2π π 2π π

y = A cos( Bx − C ) − − < x < − 4 4 4 4

y = π cos(2π x + 4π ) 3π π

or − 4

< x < 4

y = π cos [2π ( x + 2)] 3π π 3π π

x − < x < or − ,

127.

y = sin 2 x = 1

− 1

cos 2 x 4 4 4 4

2 2 3π π 2π π

− + − −

131. 4 4 = 4 = 2 = − π

132. a.

2 2 2 4

b. The reciprocal function is undefined.


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

Section 2.2 y tan

x

π from 0 to π . Continue the pattern

Check Point Exercises 2

1. Solve the equations 2 x π

and 2 x π

and extend the graph another full period to the right.

2 2

x= π x

π

4 4

Thus, two consecutive asymptotes occur at x π 4

and x π

. Midway between these asymptotes is 4

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

is 3, the points on the graph midway between an

3. Solve the equations

x-intercept and the asymptotes have y-coordinates of π x 0 and π

x π

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

2 2

x 0

x π

πof y 3 tan 2 x from

π

4 to

π . In order to graph

4 2

x 2

for π

x 3π

, Continue the pattern and extend 4 4

the graph another full period to the right.

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

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

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

Because the coefficient of the cotangent is 1

, the 2

points on the graph midway between an x-intercept

and the asymptotes have y-coordinates of 1 2

and

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

2


x = 2, to graph one full period of y 1

cot π

x . The 2 2

x π

π

and x

π π

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

2 2 2 2

x π

π x

π π

2 2 2 2

x 0 x π

Thus, two consecutive asymptotes occur at x = 0 and x π .

x-intercept 0 π

π

2 2



An x-intercept is π and the graph passes through

2

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

2

the points on the graph midway between an x-

intercept and the asymptotes have y-coordinates of –1

and 1. Use the two consecutive asymptotes, x = 0 and x π , to graph one full period of



x

4. The x-intercepts of y sin x

π correspond to 4.

π

, 3π

; π

; 3π

4 4 4 4 4

vertical asymptotes of y csc x

π .

5. 3sin 2 x

4

6. y 2 cos π x

7. false

8. true

5. Graph the reciprocal cosine function,

y 2 cos 2 x .

Exercise Set 2.2

1. The graph has an asymptote at

π

.

The equation is of the form

and B = 2.

y A cos Bx with A = 2

The phase shift, C

, from π

to π

2

is π

units.

amplitude: A 2 2 B 2 2

period: 2π

2π

π

Thus, C C

π

B 2 B 1

Use quarter-periods, π

, to find x-values for the five 4

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

C π The function with C π

is y tan( x π ) .

2. The graph has an asymptote at

x 0 .

0, π

, π

, 3π

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

value of x, the key points are

The phase shift,

C , from

π B 2

to 0 is π

units. Thus, 2

(0, 2), π

, 0 ,

π , 2

, 3π

, 0 , (π , 2) . In order to C C π

4 2 4 B 1 2

π

graph for 3π

x 3π

, Use the first four points 4 4

C 2

and extend the graph 3π 4

units to the left. Use the The function with C π

is 2

y tan x

π

2 .

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

3. The graph has an asymptote at x π .

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

C 2

C π

2

The function is

y tan x

π .



Concept and Vocabulary Check 2.2 4. The graph has an asymptote at

There is no phase shift. Thus,

2

π

. 2

C C 0

B 1

1. π

, π

; 4 4

π

; π

4 4

The function with C = 0 is

C 0 y tan x .

2. (0, π ) ; 0; π

3. (0, 2); 0; 2



x

5. Solve the equations x π

and x π

Continue the pattern and extend the graph another

full period to the right.4 2 4 2

x π

4 x π

4

2 2

x 2π x 2π

Thus, two consecutive asymptotes occur at x 2π

and x 2π .

x-intercept = 2π 2π

0

0 2 2

An x-intercept is 0 and the graph passes through (0, 0). Because the coefficient of the tangent is 3, the points on the graph midway between an x-intercept

7. Solve the equations 2 x π 2

and 2 x π

2

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

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

2

x 2π , to graph one full period of y 3 tan x

2 2

from π π

2π

4 to 2π .

x x 4 4


full period to the right.


and x π

. 4

π π 0

x-intercept = 4 4 0 2 2

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

0). Because the coefficient of the tangent is 1

, the 2


6. Solve the equations and the asymptotes have y-coordinates of 1 2

and

x π and

x π

4 2 4 2

1 . Use the two consecutive asymptotes,

π 2 4

x π

4

x π

4

2 2 and x π

, to graph one full period of 4

y 1

tan 2 x 2

x 2π x 2π π π

Thus, two consecutive asymptotes occur at x 2π from 4 to . Continue the pattern and extend the

4

and x 2π .

x-intercept =

2π 2π

0

0 2 2

graph another full period to the right.



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

y-coordinates of –2 and 2. Use the two consecutiveasymptotes, x 2π and x 2π ,

to graph one full period of

2π .

y 2 tan x 4

from 2π to



x

8. Solve the equations Use the two consecutive asymptotes, x π and

π 2 x and 2 x

π

x π , to graph one full period of y 2 tan 1

x

2 2

π π

from π 2

to π . Continue the pattern and extend the

x 2 x

2 graph another full period to the right.

2 2

x π

x π

4 4

Thus, two consecutive asymptotes occur at x π

and x π

. 4

4

π π

x-intercept = 4 4 0

0 2 2

10. Solve the equationsAn x-intercept is 0 and the graph passes through (0, 0). Because the coefficient of the tangent is 2, the

1 π and 1

x π

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

and the asymptotes have y-coordinates of –2 and 2. x π

2 x π

2

Use the two consecutive asymptotes, x π 2 2

and

4 x π x π

x π


y 2 tan 2 x from Thus, two consecutive asymptotes occur at x π

and x π .

π

to π

. 4 4

x-intercept = π π

0

0 2 2



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

y-coordinates of 3 and –3. Use the two consecutiveasymptotes, x π and x π , to graph one full

period of y 3 tan 1

x from π 2

to π . Continue


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

1 π x

and 1 x

π

2 2 2 2

x π

2

x π

2



2

x π

2

x π


and x π .

x-intercept = π π

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


and the asymptotes have y-coordinates of 2 and –2.



π

2

11. Solve the equations 13. An x-intercept is

π

and the graph passes through

x π π

and x π π 4

2 2 π , 0 . Because the coefficient of the tangent is 1,

π x x

π π 4

x π

2 2

x 3π


intercept and the asymptotes have y-coordinates of –1

2 2


and x 3π

.

and 1. Use the two consecutive asymptotes, x π 4

and x 3π


2 π 3π

4π

y tan

x

π from 0 to π .

4

x-intercept = 2

2 4π

π2 2 4


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

points on the graph midway between an x- intercept

and the asymptotes have y-coordinates of –1

and 1. Use the two consecutive asymptotes, x π

2



and x 3π


y tan( x π ) from π

to 3π

. Continue the pattern 2 2

and extend the graph another full period to the right.

13. There is no phase shift. Thus,

C C 0

B 1

C 0

Because the points on the graph midway between an

x-intercept and the asymptotes have y-coordinates of

–1 and 1, A = –1. The function with C = 0 and A = –1

is y cot x .


x π

π

and

x π

π

14. The graph has an asymptote at π

. The phase shift, 2

4 2 4 2 C , from 0 to

π is

π units.

x 2π π

x 2π π

B 2 2

4 4 4 4 Thus,

C C

π

x π

x 3π

B 1 2



4 4


C π

2

4 The function with C

π is y cot

x

π .

and x 3π

. 4

2 2

π

3π 2π

x-intercept = 4 4 4 π

2 2 4



15. The graph has an asymptote at π

. The phase shift, 18. Solve the equations

2 x 0 and x π

C , from 0 to

π is

π

units. Thus, Two consecutive asymptotes occur at x 0

B 2 2

C C π

and x π .

x-intercept = 0 π

π

B 1 2 2 2

π C


2

The function with C π

is y cot x

π .

2

π , 0

. Because the coefficient of the cotangent is

2 2 2

116. The graph has an asymptote at π . The phase shift,

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

C , from 0 to π

is π

units. intercept and the asymptotes have y-coordinates of

B

Thus,

C C

π

1 and


2 2

B 1 x 0 and x π , to graph one full period of

C π The function with C π

is y cot( x π ) .

y 1

cot x . 2

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

consecutive asymptotes occur at x 0 and x π .

x-intercept 0 π

π

2 2

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

as shown.

An x-intercept is π

and the graph passes through 2

π , 0 . Because the coefficient of the cotangent is

2

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

intercept and the asymptotes have y-coordinates of 2

and –2. Use the two consecutive asymptotes, x = 0and x π , to graph one full period of y 2 cot x .

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

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

x 0 x

π 2

Two consecutive asymptotes occur at x = 0 and

x π

. 2

0 π π π






π , 0 . Because the coefficient of the cotangent is

4

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

2 intercept and the asymptotes have y-coordinates of

1 and


2 2



x = 0 and x π


y 1

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


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.

x 0 x

π 2

Two consecutive asymptotes occur at x 0 and π π

x π

. 2

0 π π π

22. Solve the equations x 0 and 4

x 0

x π 4

x π

πx-intercept = 2 2 4

2 2 4 x 4



Two consecutive asymptotes occur at x 0 and

x 4 .

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

4 x-intercept = 0 4

4

2 2 2

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

An x-intercept is 2 and the graph passes through

and –2. Use the two consecutive asymptotes,

x 0 2, 0 . Because the coefficient of the cotangent is –2,


and x π


y 2 cot 2 x .

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

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

and x 4 , to graph one full period of

y 2 cot π

x . The curve is repeated along the x- 4

axis one full period as shown.

21. Solve the equations π

x 0 and π x π

2 2 x 0

x π

π 2



x 2

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

x-intercept = 0 2

2

1 2 2

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



y


x π

0 and

x π

π An x-intercept is

π 4

and the graph passes through

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

x 0 π x π

π 4

2 2 3, the points on the graph midway between an x-

x π

x π

2 2



Two consecutive asymptotes occur at x π

and

x π 3π

and x , to graph one full period of 2 4 4

x π

. y 3cot x

π . The curve is repeated along the x-

2 4

π π 0


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



axis one full period as shown.

Use the two consecutive asymptotes, x π

and

x π

, to graph one full period of

2

y 3cot x

π .

2 2 25. The x-intercepts of

1 sin

x

corresponds to

2 2 The curve is repeated along the x-axis one full period 1 xas shown. vertical asymptotes of y csc . Draw the

2 2 vertical asymptotes, and use them as a guide to

sketch the graph of y 1

csc x

. 2 2


x π

0 and

x π

π

4 4

x 0 π x π

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

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



x π

x 3π vertical asymptotes, and use them as a guide to

4 4

Two consecutive asymptotes occur at x π

and

sketch the graph of

y 3csc 4 x .

x 3π

. 4

π 3π

4

2π

π

x-intercept = 4 4 4 2 2 4



27. The x-intercepts of y 1

cos 2π x 2

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

vertical asymptotes of y 1

sec 2π x . Draw the 2

vertical asymptotes, and use them as a guide to

sketch the graph of y 1

sec 2π x . 2

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

equation is of the form

B 1 .

y A sin Bx with A 2 and

amplitude: A 2 2

period: 2π

2π

2π B 1

28. The x-intercepts of y 3cos π

x correspond to 2 Use quarter-periods,

π , to find x-values for the five

2

vertical asymptotes of y 3sec π

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

2 π π

3π

π . Evaluating the function atvertical asymptotes, and use them as a guide to 0, , ,

2 , and 2

2

sketch the graph of y 3sec π

x . each value of x, the key points are

2 (0, 0),

π 2

, 2 , (π , 0),

3π 2

, 2 , and (2π , 0) .

Use these key points to graph y 2sin 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.

29. Graph the reciprocal sine function, y 3sin x . The


B = 1.

y A sin Bx with A = 3 and

amplitude: A 3 3

period: 2π

2π

2π B 1




, to find x-values for the 2

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

0, π

, 2

π , 3π

, and 2π . Evaluating the function at 2

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

π , 3 , (π , 0),

3π , 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-



B

2

2

31. Graph the reciprocal sine function, y 1

sin x

. The (0, 0),

2π ,

3 , (4π , 0), 6π ,

3 , and (8π , 0) .

2 2 2 2


y A sin Bx

with A 1 2

and Use these key points to graph y

3 sin

x 2 4

from 0 to

B 1

. 2

amplitude:

A 1

1

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

Use the graph to obtain the graph of the reciprocal

2 2 function. Draw vertical asymptotes through the x-

intercepts, and use them as guides to graph

period: 2π

2π

2π 2 4π 3 x

1 y csc . 2 4

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

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

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

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

, (2π , 0), 2

3π , 1

, and (4π , 0) . Use these key points to 2

graph y

1 sin

x 2 2

from 0 to 4π .

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

to obtain the graph of the reciprocal function. Draw


y 2 cos x .

vertical asymptotes through the x-intercepts, and use The equation is of the form y A cos Bx with A = 2

them as guides to graph y 1

csc x

. 2 2

and B = 1.

amplitude:

A 2 2

period: 2π

2π

2π B 1



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

π , π ,

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

2 2

of x, the key points are (0, 2), π

, 0 , (π , 2),


sin x

. The

2 4 3π , 0 , and (2π , 2) . Use these key points to graph

equation is of the form y A sin Bx with A

3

2 and 2

y 2 cos x from 0 to 2π . Extend the graph one

B 1

. 4



B

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

amplitude: A 3

3

graph y 2 sec x .

period:

2 2

2π

2π 2π 4 8π

1 4

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


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

each value of x, the key points are



B

B

34. Graph the reciprocal cosine function, y 3cos x . graph one cycle to the right. Use the graph to obtain


and B 1 .

y A cos Bx with A 3 the graph of the reciprocal function. Draw vertical asymptotes through the x-intercepts, and use them as

amplitude: A 3 3 guides to graph y sec x

. 3

period: 2π

2π

2π B 1




0, π

, π , 3π

, and 2π . Evaluating the function at 2 2


(0, 3), π

, 0 , (π , 3), 3π

, 0 , (2π , 3) . x

2

2

36.

Graph

the reciprocal cosine function,

y cos

. The


y A cos Bx with

2 A 1 and

Use these key points to graph

y 3cos x from 0 to B

1 .




2

amplitude:

2π

A 1 1

2π

and use them as guides to graph y 3sec x . period: 1

2

2π 2 4π

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


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


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

Use these key points to graph y cos

x 2

from 0 to

35. Graph the reciprocal cosine function, y cos x

. The 3



equation is of the form y A cos Bx with A = 1 and Draw vertical asymptotes through the x-intercepts,

x

B 1

. 3

amplitude:

A 1 1

and use them as guides to graph y sec . 2

period:

2π

2π 2π 3 6π

1 3

Use quarter-periods, 6π

3π

, to find x-values for 4 2



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

are 0, 3π

, 3π , 9π

, and 6π . Evaluating the function 2 2

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

, 0 , 2

9π (3π , 1),

2 , 0

, and (6π , 1) . Use these key

points to graph y cos

x 3

from 0 to 6π . Extend the



37. Graph the reciprocal sine function, y 2 sin π x . Draw vertical asymptotes through the x-intercepts,


and B π .

y A sin Bx with A = –2

and use them as guides to graph y 1

csc π x . 2

amplitude: A 2 2

period: 2π

2π

2 B π

Use quarter-periods, 2

1

, to find 4 2

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

the x-values are 0, 1

, 1, 3

, and 2. Evaluating the 2 2

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

, 2 , (1, 0),

3 , 2

, and (2, 0) . Use these key

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

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

points to graph y 2 sin π x from 0 to 2. Extend the 1

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

the graph of the reciprocal function.

A 2

and B π .

1 1

Draw vertical asymptotes through the x-intercepts, amplitude: A

2 2

and use them as guides to graph y 2 csc π x . period:

2π

2π 2

B π


1

, to find x-values for the 4 2


0, 1

, 1, 3

, and 2. Evaluating the function at each 2 2

value of x, the key points are 0, 1

,

38. Graph the reciprocal sine function,

y 1

sin π x . 1

2

1 3 1

2 , 0 , 1, , 2

, 0 ,

2, 2

. Use these key

The equation is of the form y A sin Bx with 2 2

1

1 A and B π .

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

2

amplitude:

A 1

1

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

obtain the graph of the reciprocal function. Draw

2 2

period: 2π

2π

2

vertical asymptotes through the x-intercepts, and use them as guides to graph

B π y 1 sec π x .




1

, to find x-values for the 2

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

0, 1

, 1, 3

, and 2 . Evaluating the function at each 2 2


(0, 0), 1

, 1

, (1, 0), 3

, 1

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

Use these key points to graph y

1 sin π x from 0

2

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




40. Graph the reciprocal cosine function, y 3

cos π x . these key points to graph y sin( x π ) from π to


2 y A cos Bx with



3 A

and B π .


2

amplitude:

A 3

3

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

2 2

period: 2π

2π

2 B π


1

, to find x-values for the 4 2


0, 1

, 1, 3

, and 2 . Evaluating the function at each 2 2



y sin

x π

. 2

0, 3

, 1

, 0 , 1,

3 ,

3 , 0

,

2,

3 . The equation is of the form y A sin( Bx C ) with

2 2 2 2 2 A 1 , B 1 , and C

π .

Use these key points to graph y 3

cos π x 2

from 0

amplitude:

2

A 1 1

to 2 . Extend the graph one cycle to the right. Use the graph to obtain the graph of the reciprocal function.

period: 2π

2π

2π

Draw vertical asymptotes through the x-intercepts, B 1 π

and use them as guides to graph y 3

sec π x .

phase shift: C 2

π

2 B 1 2



key points. Starting with x π

, the x-values are 2

π , π ,

3π , 2π , and

5π . Evaluating the function at

2 2 2 each value of x, the key points are

π π

3π

π

5π .


y sin( x π ) . , 0 ,

2 , 1, , 0 , 2 , 1, and , 0

2 2

The equation is of the form y A sin( Bx C ) with π

A = 1, and B = 1, and C π . Use these key points to graph y sin x from 2

amplitude: A 1 1 π to

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

period: 2π

2π

2π 2 2



1

B

phase shift:

1

C π π

B 1


π

, to find 4 2

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

the x-values are π , 3π

, 2π , 2

5π , and 3π . Evaluating

2

the function at each value of x, the key points are

(π , 0), 3π

2 , 1

, (2π , 0),

5π

2 ,

, (3π , 0) . Use



C

π

Use the graph to obtain the graph of the reciprocal 44. Graph the reciprocal cosine function,

function. Draw vertical asymptotes through the x-

intercepts, and use them as guides to graph y 2 cos

x π

. The equation is of the form 2

y csc x

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

2 π

. 2

amplitude:

A 2 2

period: 2π

2π

2π B 1

π

phase shift: C 2 π

B 1 2


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:

A 2 2 key points. Starting with x π

, the x-values are

period: 2π

2π

2π

2

π π 3π

B

phase shift:

1

C π π

, 0, , π , and . 2 2 2

B 1 Evaluating the function at each value of x, the key


π

, to find x-values for 4 2

points are

π

, 2 , 0, 0, π

, 2 , π , 0 ,

3π , 2

.

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

values are π , π

, 0, π

, and π . Evaluating the Use these key points to graph y 2 cos

x

π

2 2 2function at each value of x, the key points are

π 3π

(π , 2), , 0 , 0, 2 , π

, 0 , and (π , 2) . from to

2 . Extend the graph one cycle to the

2

2 2 right. Use the graph to obtain the graph of theUse these key points to graph y 2 cos( x π ) from reciprocal function. Draw vertical asymptotes

π to π . Extend the graph one cycle to the right. through the x-intercepts, and use them as guides to


function. Draw vertical asymptotes through the x-

intercepts, and use them as guides to graph y 2 sec( x π ) .

graph y 2 sec

x π

. 2





45.

46.

47.

50. 51. 52.

53.

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

π

48.

54.

2

y g h ( x) g (h( x)) 2 tan 2 x

π

2

49.



,

,

,

55. Use a graphing utility with y1 tan x and y2 1 . graph on [0, 2], continue the pattern and extend

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

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

5π π x ,

3π ,

7π

4 4 4 4

x 3.93, 0.79, 2.36, 5.50

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

y2 1 . For the window use Xmin 2π ,

Xmax 2π , Ymin 2 , and Ymax 2 .

5π π x ,

3π ,

7π

4 4 4 4x 3.93, 0.79, 2.36, 5.50 b. The function is undefined for t = 0.25, 0.75,

1.25, and 1.75. The beam is shining parallel to the wall at these

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

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

x 3π π

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

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

2 2 x 4.71, 1.57

58. Use a graphing utility with y1 1 / cos x and

y2 1 .

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

cot x d

2 d 2 cot x

For the window use Xmin 2π , Xmax 2π , Use the equations x 0 and x π .

Ymin 2 , and Ymax 2 . Two consecutive asymptotes occur at x 0 and

x 2π , 0, 2π

x 6.28, 0, 6.28

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

. 2

59.

d 12 tan 2π t

a. Solve the equations

An x-intercept is

π and the graph passes through

2

π 2π t

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

2

2 2

π π

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

t 2

2π t

2

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

1 t t

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

4 4



Thus, two consecutive asymptotes occur at

1 x 4

and x 1

. 4

1 1

0 x-intercept 4 4 0

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

12, the points on the graph midway between an

x-intercept and the asymptotes have y-

coordinates of –12 and 12. Use the two

consecutive asymptotes, 1

x 4

and x 1

, to 4

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



B 1

B

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

hypotenuse and the adjacent leg, the secant function.

sec x d

10 d 10 sec x

64. Graphs will vary.

Graph the reciprocal cosine function, y 10 cos x .


and B 1.

y A cos Bx with A 10

amplitude: A 10 10

period: 2π

2π

2π

B 1 65. – 76. Answers may vary.

For π

x π

, use the x-values π

, 0, and π

to π π2 2 2

2

π

77. period: π 4 4π 4

find the key points π

, 0 , (0, 10), and , 0 . x

2

2 Graph y tan 4

for 0 x 8π .

Connect these points with a smooth curve, then draw

vertical asymptotes through the x-intercepts, and use

them as guides to graph d 10 sec x on π

, π

.

2 2

78. period: π

π

B 4

Graph y tan 4 x for 0 x π

. 2

62. Graphs will vary.

79. period: π

π

B 2Graph y cot 2 x for 0 x π .

63.

80. period: π

π

π 2 2π 1 2

Graph y cot

x



2

for 0 x

4π .



1

B

81. period: π

π

1 85. period: 2π

2π

π

B π B 2 π

Graph y 1

tan π x for 0 x 2 .

phase shift: C 6 π

2 B 2 12


Thus, we include π

x 25π

12 12

graph for 0 x 5π

. 2

in our graph, and

π x 1 π

and π x 1 π

2 2

π π x π x

π 1

2 2 π π

2π 2π

x 2 1

x 2 1 86. period: 2

π π B π

π

x π 2

x π 2

phase shift: C 6 π 1

1

2π 2π B π 6 π 6

x 0.82

period: π

π

1

x 0.18 Thus, we include 1

x 25

6 6

9

in our graph, and

B π graph for 0 x .

Thus, we include 0.82 x 1.18 in our graph of 2

y 1

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

83. period:

2π

2π

2π 2 4π

87.

1 2

Graph the functions for 0 x 8π .

88.

The graph shows that carbon dioxide concentration

rises and falls each year, but over all the

concentration increased from 1990 to 2008.

y sin 1

x

84. period:



π

2π 2π

2π 3

6B

3 π

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. makes sense



90. makes sense 96. The range shows that A π .

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

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

B where 2π

2 B

2π 2

B

2B 2π

B π

consecutive asymptotes at Thus, y π csc π x

x = 0 and x 2π

. The period is 2π

0 2π

. Thus, 97. a. Since A=1, the range is , 1 1,

3 3 3

π 7π π

2π Viewing rectangle:

6 , π ,

6 by 3, 3,1

B 3

2π B 3π b. Since A=3, the range is , 3 3,

B 3π

3

1 7

2π 2 Viewing rectangle: 2

, 2

,1 by 6, 6,1

The points on the graph midway between an x-


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

98.

y 2 x

sin x

x

C = 0. An equation for this graph is

y cot 3

x . 2 decreases the amplitude as x gets larger.

2

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

Examples may vary.

99. a.

The reciprocal function has amplitude

period is 8π

. Thus, 2π

8π

A 1 . The

3 B 3

8π B 6π

B 6π

3

8π 4There is no phase shift. Thus, C 0 . An equation for

b. yes; Explanations will vary.

the reciprocal function is

equation for this graph is

y cos 3

x . Thus, an 4

y sec 3

x . 4

c. The angle is π

. 6

This is represented by the point π

, 1

.

95. The range shows that A 2. 6 2

Since the period is 3π , the coefficient of x is given by B where

2π 3π



B 100. a.

2π 3π

B

3Bπ 2π

B 2 3


Thus, y 2 csc 2 x 3

c. The angle is 5π

. 6

5π 3 This is represented by the point , .

6 2



x

1 π y

2 cos

3 x

coordinates

0

1 π

1

1

2 cos

3 0

2 1

2

0, 1

3

2

1 π 3 1

2 cos

3 2 2 0 0

3 , 0

3

1 π 1 1

2 cos

3 3

2

1 2

1 3,

2

9

2

1 cos

π 9

1

0 0 2 3 2 2

9 , 0

2

6

1 π 1 1 cos 6 1

1 6,

x y 4 sin 2 x

coordinates

0

4 sin(2 0) 4 0

0

(0, 0)

π

4

π

4

π

4 ,4

π

2

4 sin 2

π 4 0

0

π

2 ,0

3π

4

4 sin 2

3π 4(1)

4

3π

, 4

π 4 sin(2 π ) 4 0

0

(π , 0)

B

101. a. Connect the five key points with a smooth curve and



c. The angle is π

. 3

This is represented by the point π

,

3 .

2. The equation

y 1

cos π

x is of the form

3

y A cos Bx with

2 3

A 1 2

and B π

. The 3

Mid-Chapter 2 Check Point

amplitude is A 1 1

. The period is

1. The equation

y A sin Bx

y 4 sin 2 x is of the form

with A = 4 and B 2 . Thus, the

2π 2π

π 3

2 2

6 . The quarter-period is

6 3

. Add 4 2

amplitude is A 4 4 . The period is quarter-periods to generate x-values for the key

2π

2π π . The quarter-period is

π . The cycle points and evaluate the function at each value of x.

B 2 4 begins at x = 0. Add quarter-periods to generate x-

values for the key points and evaluate the function at

each value of x.

2

2

4 sin 2

4 4 1

2

2 3 2 2 2

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


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

x

y 3sin( x π )

coordinates

π

3sin π π 3sin(0)

3 0

0

π , 0

3π

2

3sin 3π

π 3sin

π

3 1

3

3π

, 3

2π

3sin 2π π 3sin(π )

3 0

0

(2π , 0)

5π

2

3sin 5π

π 3sin

3π

3(1)

3

5π 2

3π

3sin 3π π 3sin(2π )

3 0

0

3π , 0

x

coordinates

π

8

π , 2

3π

8

3π

8 , 0

5π

8

5π , 2

8

7π

8

7π

8 , 0

9π

8

9π , 2

8

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

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

2 x

4 is of the form

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

2π

2π 2π . The phase shift is C π

π . The C π

. Thus, the amplitude is

A 2

2 . The

B 1 B 1 4

quarter-period is 2π

π

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

2π

π . The phase shift is

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

points and evaluate the function at each value of x.

B 2

C π

π 4 .

B 2 8



. 4 4

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

points and evaluate the function at each value of x.

2 2 8

2

2 2







x y cos 2 x 1

coordinates

0 y cos 2 0 1 2

(0, 2)

π

4

y cos 2

π 1 1

π

4 ,1

π

2

y cos 2

π 1 0

π

2 ,0

3π

4

y cos 2

3π 1 1

3π

,1

π y cos 2 π 1 2

(π , 2)

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

1 1 . The period for both functions is 2π

π . The quarter-period is π

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

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

4

2

4 4



x

0 π

4

π

2

3π

4

π 5π

4

3π

2

7π

4

2π

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

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

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




π π x

and π x

π

4 2 4 2

x π 4

x π 4

2 π

x 2

2 π

x 2

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

x-intercept = 2 2

0

0 2 2

An x-intercept is 0 and the graph passes through (0, 0). Because the coefficient of the tangent is 2, the points on the graph midway between an x-intercept and the asymptotes have y-coordinates of –2 and 2. Use the two consecutive asymptotes,

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

x 4

from 2 to 2 .

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


x 0 x

π 2

Two consecutive asymptotes occur at x 0 and x π

. 2

0 π π π



π , 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

asymptotes,

as shown.

x 0 and x π


y 4 cot 2 x . The curve is repeated along the x-axis one full period



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 B π


1

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

0, 1

, 1, 3

, and 2 . Evaluating the function at each value of x, the key points are 2 2

0, 2, 1

, 0 , 1, 2 ,

3 , 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 .

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


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

, 1

, 3

, and 1 .4

Evaluating the function at each value of x, the key points are (0, 0), 1

, 3 ,

4 2 4

1 , 0

,

3 , 3 , and (1, 0) .

4 2 4

Use these key points to graph y 3sin 2π x from 0 to 1 . Extend the graph one cycle to the right. Use the graph to obtain

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

y 3csc 2π x .


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

Scientific Calculator Solution

Function

Mode

Keystrokes Display

(rounded to four places)

cos1 1

3

Radian

1 3 COS1

1.2310

tan1 (35.85)

Radian

35.85 TAN1

–1.5429

Graphing Calculator Solution

Function

Mode

Keystrokes Display

(rounded to four places)

cos1 1

3

Radian

COS1

( 1 3 ) ENTER

1.2310

tan1 (35.85)

Radian

TAN1

35.85 ENTER

–1.5429

.

Section 2.3

Check Point Exercises

1. Let θ sin1 3 , then sin θ

3 .

2 2

The only angle in the interval π

, π

that satisfies sin θ 3

is π

. Thus, θ π

, or sin1 3 π

.

2 2

2 3 3 2 3

2. Let θ sin1 2

, then sin θ 2

.

2

2


, π

that satisfies cosθ 2

is π

. Thus θ

π

, or sin1 2

π

.

2 2

2 4 4

2

4

3. Let θ cos1 1

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

1 is 2π

. Thus,

2 2 2 3

θ 2π

, or cos1 1

2π

. 3 2 3

4. Let θ tan1 (1) , then tan θ 1 . The only angle in the interval π

, π

that satisfies tan θ 1 is π

. Thus

π

θ 4

or tan1 θ

π 4

2 2 4

5.

a.

b.

a.

b.



6. a. cos cos1 0.7x 0.7 , x is in [–1,1] so cos(cos

1 0.7) 0.7



1

b. sin1 (sin π )

x π , x is not in π

, π

. x is in the domain

9. Let θ tan1

x , then tan θ =x= x

. 1

2 2

of sin x, so sin1 (sin π ) sin

1 (0) 0

c. cos cos1 π

x π , x is not in [–1,1] so cos cos1 π is not

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

2 2 2defined.

7. Let θ tan1 3 , then tan θ

3 . Because tan θ is

a x

a

1

x2

1

4 4 Use the right triangle to write the algebraic expression. 2positive, θ is in the first quadrant.

sec tan1 x secθ x 1


x2

1

1. π

x π

; 2 2

sin1

x

Use the Pythagorean Theorem to find r.

r2

32

42

9 16 25

2. 0 x π ;

π π

cos1

x

1

r 25 5 3. x ; 2 2

tan x

Use the right triangle to find the exact value.

sin tan

1 3 sin θ

side opposite θ 3

4. [1,1] ;

π

, π

4 hypotenuse 5 2 2

8. Let θ sin1 1

, then sin θ

1

.

Because sin θ

5. [1,1] ; [0, π ]

2 2

is negative, θ is in quadrant IV. π π

6. (, ) ; ,

7. π

, π

2 2

2 2

8. [0, π ]

Use the Pythagorean Theorem to find x.

x2

(1)2

22

x2

1 4

x2

3

x 3



9. π

, π

2 2

10. false

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

cos sin1

1

cosθ x

3

2

r 2



Exercise Set 2.3 7. Let θ cos1 3

, then cosθ 3

. The only angle

1. Let θ sin1 1

, then sin θ 1

. The only angle in 2 2

2

the interval π

, π

2


is π

.

in the interval [0, π ] that satisfies cosθ 3

2

is π

. 6

2 2 2 6

π 3 π

Thus,

θ π

, or sin1 1 π

. Thus θ , or cos

1 .

6 2 6

6 2 6

8. Let θ cos1 2

, then cosθ 2

. The only angle2. Let θ sin1 0 , then sin θ 0 . The only angle in the 2 2

interval π

, π

that satisfies sin θ 0 is 0. Thus

in the interval 0, π that satisfies cosθ 2

is π

.

2 2 2 4

θ 0 , or sin1

0 0 .

Thus θ

π

, or

cos

1 2

π

.

3. Let θ sin1 2

, then sin θ 2

. The only angle

4 2 4

2 2 9. Let θ cos1

2

, then cosθ 2

. The only

in the interval π

, π


is 2 2

2 2

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

π . Thus

θ π

, or sin1 2

π

.

cosθ 2

is 3π

. Thus θ 3π

, or

4 4 2 4 2 4 4

cos1

2

3π

.4. Let θ sin1 3

, then sin θ 3

. The only angle

2

4

2

in the interval π

, π

2


is

3 3

2 2

2 10. Let θ cos1 , then cosθ . 2 2

π . Thus θ

π , or sin1 3

π

.

The only angle in the interval 0, π that

3 3 2 3

satisfies cosθ 3

is 5π

. Thus θ 5π

, or

5. Let θ sin1 1

, then sin θ 1 . The only 2 6 6

2 2 cos

1 3 5π

.

angle in the interval π

, π

that satisfies

2 2 2 6

sin θ

1

is π

. Thus θ π

, or 11. Let θ cos1 0 , then cosθ 0 . The only angle in

2 6 6



,

the interval [0, π ] that satisfies cosθ 0 is π

.

sin1

1

π

. 2

2 6 π 1 π

Thus θ , or cos 0 . 2 2

6. Let θ sin1 3

, then sin θ 3

.

12. Let θ cos1 1 , then cosθ 1 . The only angle in the

2

2


, π

that satisfies

interval 0, π that satisfies cosθ 1 is 0 .

Thus θ 0 , or cos1 1 0 .

2 2

sin θ 3 2

is π

. Thus θ π

3 3

or sin1 3

π

.

2

3




Function

Mode

Keystrokes Display

(rounded to two places)

sin1

0.3 Radian

0.30 0.3 SIN 1


Function

Mode

Keystrokes Display


sin1

0.47 Radian 0.47 SIN 1 0.49

1 3


3 . The only angle in the interval

π

, π

that satisfies tan θ 3 is

π . Thus

3 3

θ π

, or tan1 3 π

.

2 2 3 6

6 3 6

14. Let θ tan1 1 , then tan θ 1 . The only angle in the interval π

, π


.

Thus θ π

, or tan1 1 π

.

2 2 4

4 4

15. Let θ tan1 0 , then tan θ 0 . The only angle in the interval π

, π

that satisfies tan θ 0 is 0. Thus θ 0 , or 2 2

tan1

0 0 .

16. Let θ tan1 (1) , then tan θ 1 . The only angle in the interval π

, π


. Thus

θ π

, or tan1 (1) π

.

2 2 4

4

17. Let θ tan1

4

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

, π

that satisfies tan θ 3 is

π

.

Thus θ

π

, or tan1

3 π

.

2 2 3

3 3

3 π π 3 18. Let θ tan , then tan θ . The only angle in the interval , that satisfies tan θ is

π .

3

3

2 2 3 6

Thus θ

π , or tan1

3

π

.

6 3 6

19.


Function

Mode

Keystrokes Display


sin1

0.3 Radian

0.30 SIN 1 0.3 ENTER

20.




Function

Mode

Keystrokes Display


sin1

0.47 Radian SIN1

0.47 ENTER 0.49




Function

Mode

Keystrokes Display


sin1 (0.32) Radian

–0.33 0.32

SIN1


Function

Mode

Keystrokes Display


sin1 (0.625) Radian 0.625 SIN

1

–0.68


Function

Mode

Keystrokes Display


cos1 3

8

Radian

3 8 COS1

1.19


Function

Mode

Keystrokes Display


cos1 4

9

Radian

4 9 = COS1

1.11

21.


Function

Mode

Keystrokes Display


sin1 (0.32) Radian

SIN

1

–0.33 0.32 ENTER

22.


Function

Mode

Keystrokes Display


sin1 (0.625) Radian SIN

1 0.625 ENTER –0.68

23.


Function

Mode

Keystrokes Display


cos1 3

8

Radian

COS1

( 3 8 ) ENTER

1.19

24.


Function

Mode

Keystrokes Display


cos1 4

9

Radian

COS1

( 4 9 ) ENTER

1.11




Function

Mode

Keystrokes Display


cos1 5

7

Radian

5 7 COS1

1.25


Function

Mode

Keystrokes Display


tan1 (20) Radian

–1.52 20

TAN1


Function

Mode

Keystrokes Display


tan1 (30) Radian 30 TAN

1

–1.54

25.


Function

Mode

Keystrokes Display


cos1 5

7

Radian

COS1

( 5 7 ) ENTER

1.25

26. Scientific Calculator Solution

Function

Mode

Keystrokes Display


cos1 7

10

Radian

7 10 = COS1

1.30


Function

Mode

Keystrokes Display


cos1 7

10

Radian

COS1

( 7 10 ) ENTER

1.30

27.


Function

Mode

Keystrokes Display


tan1 (20) Radian

TAN

1

–1.52 20 ENTER

28.


Function

Mode

Keystrokes Display


tan1 (30) Radian TAN

1 30 ENTER –1.54




Function

Mode

Keystrokes Display


tan1 473 Radian

473

–1.52

TAN 1


Function

Mode

Keystrokes Display


tan1 5061 Radian

5061 TAN 1

–1.56

29.


Function

Mode

Keystrokes Display


tan1 473 Radian

TAN1

(

–1.52 473 ) ENTER

30.


Function

Mode

Keystrokes Display


tan1 5061 Radian

TAN1

( 5061 ) ENTER

–1.56

31. sin sin1 0.9x 0.9, x is in [1, 1], so sin(sin

1 0.9) 0.9

32.

33.

cos(cos1

0.57)

x 0.57, x is in [1, 1], so

cos(cos1

0.57) 0.57

sin1 sin π

3

x π

, x is in π

, π

, so sin1 sin π

π

3 2 2 3 3

34.

cos1 cos 2π

3

x 2π

, x is in [0, π ], 3

so cos1

cos 2π

2π

35.

3 3

sin1 sin 5π



6

x 5π

, x is not in π

, π

, x is in the domain of sin x, so sin1 sin 5π

sin1 1 π

6 2 2

6 2 6



36. cos1 cos 4π 43. sin

1 (sin π )

3 x π , x is not in

π , π

,

x 4π

, x is not in [0, π ], 2 2

3

x is in the domain of cos x,

so cos1

cos 4π

cos1

1

2π

x is in the domain of sin x, so

sin1 (sin π ) sin

1 0 0

37.

3 2 3

tan tan1 125

44. cos1 (cos 2π )

x 2π , x is not in [0, π ],

x is in the domain of cos x,

1 1

x 125, x is a real number, so tan tan1 125 125 so cos (cos 2π ) cos 1 0

38.

tan(tan

1 380)

x 380, x is a real number,

so tan(tan1

380) 380

45. sin sin1 π

x π , x is not in [1, 1] , so sin sin1 π is not

defined.

39.

tan1

tan π 46. cos(cos

1 3π )

6

x π

, x is in π

, π

, so

x 3π , x is not in [1, 1]

so cos(cos1 3π ) is not defined.

6 2 2

1 4 4

tan1

tan π

π 47. Let θ sin , then sin θ . Because sin θ is

5 5

6

6

40.

tan1

tan π

positive, θ is in the first quadrant.

3

x π

, x is in π

, π

,

3 2 2

so tan1

tan π

π

3

3

x

2 y

2 r

2

41.

tan1 tan 2π

x2

42

52

x2

25 16 9

3 x 3

x 2π

, x is not in π

, π

, x is in the domain of

4

x 3

3 2 2 cos sin1 cosθ

tan x, so tan1

tan 2π

tan1

3 π

5 r 5



42.

3 3

tan1 tan 3π

4

x 3π

, x is not in π

, π

, 4 2 2

x is in the domain of tan x

so tan1 tan 3π

tan1 (1) π

4 4



2 2 2

48. Let θ tan1 7

, then tan θ 7

. x y r

24 24 x2

52

132

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

x2

144

x 12

cot sin1 5 cot θ

x 12

13 y 5

51. Let θ sin1 3

, then sin θ 3

. Because sin θ

r2

x2

y 2 5 5

r2

72

242

r2

625

r 25

sin tan1 7

sin θ y

7

is negative, θ is in

quadrant IV.

24 r 25

49. Let θ cos1 5 , then cosθ

5 . Because cosθ is

13 13 x2

y2

r 2

positive, θ is in the first quadrant.

x2

(3)2

52

x2

16 x 4

tan sin1

3

tan θ y

3

5

x 4

1 4 4

x2

y 2

r2

52. Let θ sin 5

, then sin θ

. 5

52

y 2

132

y 2

169 25

y 2

144

y 12

tan cos1 5

tan θ y

12

Because sin θ is negative, θ is in quadrant IV.

13 x 5 x

2 y

2 r

2

50. Let θ sin1 5

13 then sin θ

5 .

13

x2

(4)2

52

because sin θ is positive, θ is in the first quadrant. x2 9

x 3

cos sin1

4

cosθ x

3

5

r 5



2 2 2

53. Let, θ cos1 2 , then cosθ

2 . Because cosθ x y r

2 2 x2

(1)2

42

is positive, θ is in the first quadrant.

x2

15

x 15

sec sin1

1

secθ r

4

4 15

4

x 15 15

56. Let θ sin1 1

, then sin θ 1

.

x2

y 2

r2 2 2

2 2 2

Because sin θ is negative, θ is in quadrant IV.

2 y 2

y 2

2

y 2

sin cos1 2

sin θ y

2

2

r 2


1 .

x

2 y

2 r

2

x2 (1)2 22

2 2Because sin θ is positive, θ is in the first quadrant. x2 3

x 3

sec sin1

1

secθ r

2

2 3

2

x 3 3

57. Let θ cos1 1

, then cosθ

1

. Because

3 3

x2 y 2 r2

x2

12

22

x2

3

cosθ is negative, θ is in

quadrant II.

x 3

cos sin1 1 cosθ

x 3

2 r 2

55. Let θ sin1 1

, then sin θ 1 . Because sin θ x2 y2 r 2

4 4 (1)

2 y

2 3

2

is negative, θ is in

quadrant IV.

y2

8

y 8

y 2 2




tan cos1

1

tan θ y

2 2

2 2 3

x 1



.

2

.

2 2 2

58. Let θ cos1 1

, then cosθ 1 x y r

4 4 2 2

Because cosθ is negative, θ is in quadrant II. x 2 2

x2

2

x 2

2 r 2

sec sin1 secθ 2

2 x 2

x2

y 2

r2 61. Let θ tan1

2 , then tan θ

2

3 3(1)2 y 2 42

y 2

15

Because tan θ is negative, θ is in quadrant IV.

y 15

tan cos1

1

tan θ y

15

15 4

x 1

59. Let θ cos1

3

, then cosθ 3

. Because

r2

x2

y 2

2 2 2

2

2

r 3

2

cosθ is negative, θ is in quadrant II.

r2

9 4

r2

13

r 13

cos tan1

2

cosθ x

3

3 13

3

r 13 13

1 3 3x

2 y

2 r

2

62. Let θ tan 4

, then tan θ

. 4

2 2 2

Because tan θ is negative, θ is in quadrant IV. 3 y 2

y 2

1

y 1

3 r 2

csc cos1 cscθ 2

2 y 1

2 2 2

60. Let θ sin1

2

, then sin θ 2

. r x y

2

2

2 2 2

r 4 3



Because sin θ is negative, θ is in quadrant IV. r2

16 9

r2

25

r 5

sin tan1

3

sin θ y

3

3

4

r 5 5



1 x

1

2

63. Let θ cos1

x , then cosθ x x

. 1

Use the Pythagorean Theorem to find the third side,

b.

x2

b2

12

b2

1 x2

66. Let θ cos1

2 x.

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

(2 x)2

b2

12

b2

1 4 x2

b 1 x2

b 1 4 x2

Use the right triangle to write the algebraic expression.

2

tan cos1 x tan θ x

1 4 x 2

sin cos 2 x 1

1 4 x2

67. Let θ sin1 1

, then sin θ 1

.

64. Let θ tan1

x , then tan θ x x

. x x

1

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

c2

x2

12

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

a 2 12 x2

c x2

1 2 2


sin(tan1 ) sin θ

a x

a

1

x2

1

x

x2

1

x x 2 1


cos sin1 1 cosθ

x 1

x

2 1

x x2 1

x x

x2 1

x

2 1

65. Let θ sin1

2 x , then sin θ 2 x

y = 2x, r = 1

Use the Pythagorean Theorem to find x.

x2 (2 x)2 12



x2

1 4 x2

x 1 4 x2

cos(sin1

2 x) 1 4 x2



2

68. Let θ cos1 1 , then cosθ

1 .

1 x 2 9 x 2 9

x x 72. Let θ sin , then sin θ . x x


b.

12

b2

x2


a. 2

b2

x2

1 a

x 9 x2 2 2

b x2

1 a 2

x2

x2

9 9

Use the right triangle to write the algebraic

expression.

sec cos

1 1 secθ

x x

a 3


expression.

x 1 cot sin

1 x

2 9

3

69.

cot tan1 x

3

x x2 9

3 x 3 x 2

9 3 x 2

9

70.

cot tan1 x

2

x2

9 x2

9 x2 9

2 x 73. a. y sec x is the reciprocal of y cos x . The x-

71. Let θ sin1 x

, then sin θ x

.

values for the key points in the interval [0, π ]

are 0, π

, π

, 3π

, and π . The key points are

x2 4 x2 4 4 2 4

π 2 π (0, 1), , , , 0 ,

3π

, 2

, and

4 2

2

4 2

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

. 2


Now draw our graph from (0, 1) through

π , 2

a. 4 to on the left side of the

a2 x2

x2 4 asymptote. From on the right side of the



asymptote through 3π

, 2 to (π , 1) .

a 2 x2 4 x2 4 4

a 2


expression.

x

sec sin1

secθ

x 2 4

x

2 4 2

b. With this restricted domain, no horizontal line

intersects the graph of y sec x more than



1

2 2

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

inverse function.

c. Reflecting the graph of the restricted secant

function about the line y = x, we get the graph

75.

of y sec1

x .

74. a. Two consecutive asymptotes occur at x 0 and

76.

domain: [−1, 1];

range: [0, π]

x π . Midway between x 0 and x π is

x 0 and x π . An x-intercept for the graph

is π

, 0 . The graph goes through the points 2

π , 1

and

3π ,

. Now graph the

domain: [−1, 1];

4

4

π 3π function through these points and using the asymptotes.

range: ,

b. With this restricted domain no horizontal line

77.

intersects the graph of y cot x more than

domain: [−2, 0];

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 =

78.

range: [0, π]

x, we get the graph of y cot1 x .

c.



domain: [−2, 0];

range:

π

, π

2 2



79.

80.

domain: (−∞, ∞);

range: (−π, π)

domain: (−∞, ∞);

83. 84.

domain: [−2, 2];

range: [0, π]

domain: [−2, 2];

range: 3π

, 3π

π π

2 2 range: ,

81.

2 2

85. The inner function, sin1

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

82.

domain: (1, 3];

range: [−π, 0]

domain: [1, 3];

86. The inner function, cos1

x, accepts values on the

interval 1,1 . Since the inner and outer functions are

inverses of each other, the domain and range are as

follows.

domain: 1,1 ; range: 1,1 87. The inner function, cos x, accepts values on the

interval , . The outer function returns values on the

interval 0, π domain: , ; range: 0, π

88. The inner function, sin x, accepts values on the

interval , . The outer function returns values

range:

π

, π

π π

2 2

on the interval ,



2 2

domain: , ; range:

π

, π

2 2



x θ

5 tan1 33

tan1 8

0.408 radians 5 5

10 tan1 33

tan1 8

0.602 radians 10 10

15 tan1 33

tan1 8

0.654 radians 15 15

20 tan1 33

tan1 8

0.645 radians 20 20

25 tan1 33

tan1 8

0.613 radians 25 25

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

interval , . The outer function returns values

on the interval π

, π

95. θ 2 tan1 21.634

1.3157 radians; 28

1.3157 180

75.4

2 2 π

domain: , ; range:

π

, π

1 21.634

2 2

96.

θ 2 tan

0.1440 radians;

90. The inner function, sin x, accepts values on the

300

0.1440 180

8.2

interval , . The outer function returns values π

on the interval 0, π domain: , ; range: 0, π

91. The functions sin1 x and cos1 x

accept values on

97.

98.

tan1

b tan1

a tan1

2 tan1

0

1.1071 square units

tan1

b tan1

a tan1 1 tan

1 (2)

the interval 1,1 . The sum of these values is always

π .

2

1.8925 square units


domain: 1,1 ; range: π 2

110. y sin1 x

1

92. The functions sin1 x and cos1 x

accept values on

y sin x 2

the interval 1,1 . The difference of these values

range from π 2

to 3π 2

domain: 1,1 ; range:

π

, 3π The graph of the second equation is the graph of the

2 2 first equation shifted up 2 units.

111. The domain of y cos1 x

is the interval

93. θ tan1 33

tan1 8

x x

[–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

max to π , and graph in a π

, π , π

by [0, 4, 1] 2 4

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.



112. y tan1

x

y 2 tan1

x The graph of the second equation is the graph of the

first equation reversed and stretched.


Sample explanation: Though this restriction works

for tangent, it is not selected simply because it is

easier to remember. Rather the restrictions are based

on which intervals will have inverses.

118. makes sense

119. does not make sense; Explanations will vary. Sample explanation:

sin1 sin 5π

sin1 2

π

113. The domain of y sin1

x is the interval 4 2 4

[–1, 1], and the range is π

, π

. Because the

120.

y 2 sin1 ( x 5)

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

y sin

1 ( x 5)

2

sin y

x 5 2

π , π

, π

by y

2 2

[–2, 3, 1] viewing rectangle.

x sin 5 2

121. 2 sin1

x π

4

sin1

x π

8

π

The graph of the second equation is the

graph of the first equation shifted left 2 units and up 1 unit.

x sin 8

122. Prove: If x > 0, tan1 x tan1 1 π

114. x 2

y tan1

x π

Since x > 0, there is an angle θ with 0 θ as 2

shown in the figure.

115.

Observations may vary.

tan θ x and tan π

θ

1 thus

2 x

tan1 x θ and tan1 1 π

θ so x 2

tan1 x tan1 1 θ

π θ

π

It seems sin1

x cos1

x π 2

for 1 x 1 . x 2 2




Sample explanation: The cosine’s inverse is not a

function over that interval.



π

123. Let α equal the acute angle in the smaller right

triangle.

tan α 8

x

so tan1 8

α x

tan(α θ ) 33

x

so tan1 33

α θ x

θ α θ α tan1 33

tan1 8 x x

124. tan A a

b

tan 22.3 a

12.1

Section 2.4

Check Point Exercises

1. We begin by finding the measure of angle B. Because

C = 90° and the sum of a triangle’s angles is 180°, we

see that A + B = 90°. Thus, B = 90° – A = 90° – 62.7°

= 27.3°.

Now we find b. Because we have a known angle, a

known opposite side, and an unknown adjacent side,

use the tangent function.

tan 62.7 8.4

b

b 8.4

4.34 tan 62.7

Finally, we need to find c. Because we have a known

angle, a known opposite side and an unknown

hypotenuse, use the sine function.

8.4

a 12.1 tan 22.3 sin 62.7

c

125.

a 4.96

cos A b

c

cos 22.3 12.1

c

c 12.1

cos 22.3

c 13.08

tan θ opposite

adjacent

tan θ 18

25

θ tan1 18

c 8.4

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.

tan 85.4 a

80 a 80 tan 85.4 994

The Eiffel tower is approximately 994 feet high.

3. Using a right triangle, we have an unknown angle, A,

a known opposite side, and a known hypotenuse.

Therefore, use the sine function.

sin A 6.7 13.8

25

θ 35.8

126. 10 cos π

x 6

A sin1 6.7

29.013.8

The wire makes an angle of approximately 29.0° with

the ground.

4. Using two right triangles, a smaller right triangle

amplitude: 10 10 corresponding to the smaller angle of elevation drawn inside a larger right triangle corresponding to the

period: 2π

2π 6

12 larger angle of elevation, we have a known angle, anπ 6 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.


Section 2.4 Applications of Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

tan 32 a

800 a 800 tan 32 499.9

tan 35 b

7. When the object is released (t 0), the ball’s

distance, d, from its rest position is 6 inches down. Because it is down, d is negative: when t = 0, d = –6. Notice the greatest distance from

rest position occurs at t = 0. Thus, we will use the

800

b 800 tan 35 560.2 equation with the cosine function, y a cos ω t, to

The height of the sculpture of Lincoln’s face is 560.2

– 499.9, or approximately 60.3 feet.

5. a. We need the acute angle between ray OD and

the north-south line through O.

The measurement of this angle is given to be 25°. The angle is measured from the south side of the north-south line and lies east of the north- south line. Thus, the bearing from O to D is S 25°E.

b. We need the acute angle between ray OC and

the north-south line through O. This angle measures 90 75 15.

This angle is measured from the south side of

the north-south line and lies west of the north-

south line. Thus the bearing from O to C is S

15° W.

6. a. Your distance from the entrance to the trail

system is represented by the hypotenuse, c, of a

right triangle. Because we know the length of

the two sides of the right triangle, we find c

using the Pythagorean Theorem.

We have

model the ball’s motion. Recall that a is the

maximum distance. Because the ball initially moves

down, a = –6. The value of ω can be found using the

formula for the period.

period 2π

4 ω

2π 4ω

ω 2π

π

4 2

Substitute these values into d a cos wt. The

equation for the ball’s simple harmonic motion is

d 6 cos π

t. 2

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

d 12 cos π

t, a 12 and ω π

. 4 4

a. The maximum displacement from the rest

position is the amplitude. Because a = 12, the maximum displacement is 12 centimeters.

b. The frequency, f, is π

c2 a 2 b2 (2.3)2 (3.5)2 17.54 ω

4 π 1 1

f

c 17.54 4.2 2π 2π 4 2π 8

You are approximately 4.2 miles from the

entrance to the trail system. The frequency is

1

8

cm per second.

b. To find your bearing from the entrance to the c. The time required for one cycle is the period.

trail system, consider a north-south line passing

period 2π 2π 4

π 2π 8

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.

tan θ 3.5

2.3

ω 4

π

The time required for one cycle is 8 seconds.


1. sides; angles

2. north; south



2π ωθ tan

1 3.5 56.7

2.3 Thus, 31 θ 31 56.7 87.7. Your

bearing from the entrance to the trail system is S

87.7° W.

3. simple harmonic; a ; ; ω 2π



Exercise Set 2.4

1. 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.

tan 23.5 a

10 a 10 tan 23.5 4.35

Because we have a known angle, a known adjacent side, and an unknown hypotenuse, use the cosine function.

cos 23.5 10

c

c 10

10.90 cos 23.5

In summary, B = 66.5°, a ≈ 4.35, and c ≈ 10.90.

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.

sin 54.8 a

80 a 80 sin 54.8 65.37

Because we have a known angle, a known hypotenuse,

and an unknown adjacent side, use the cosine function.

cos 54.8 b

80 b 80 cos 54.8 46.11

In summary, B 35.2, a 65.37 , and c 46.11 .

5. Find the measure of angle A. Because

C = 90°, A + B = 90°.

2. Find the measure of angle B. Because C = 90°, Thus, A 90 B 90 16.8 73.2 .

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.

tan 41.5 a

20 a 20 tan 41.5 17.69

Because we have a known angle, a known adjacent side, and an unknown hypotenuse, use the cosine function.

cos 41.5 20

c

c 20

26.70 cos 41.5

In summary, B 48.5, a 17.69 , and c 26.70 .

Because we have a known angle, a known opposite side and an unknown adjacent side, use the tangent function.

tan 16.8 30.5

a

a 30.5

101.02 tan16.8

Because we have a known angle, a known opposite

side, and an unknown hypotenuse, use the sine function.

sin 16.8 30.5

c

c 30.5

105.52 sin 16.8

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°.

3. Find the measure of angle B. Because Thus, A 90 B 90 23.8 66.2 .

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.

sin 52.6 a

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.

cos 52.6 b

54

b 54 cos 52.6 32.80

Because we have a known angle, a known opposite side, and an unknown adjacent side, use the tangent

function.

tan 23.8 40.5

a

a 40.5

91.83 tan 23.8

Because we have a known angle, a known opposite

side, and an unknown hypotenuse, use the sine

function.

sin 23.8 40.5

c

c 40.5

100.36 sin 23.8



In summary, B = 37.4°, a ≈ 42.90, and b ≈ 32.80. In summary, A 66.2, a 91.83 , and c 100.36 .



7. Find the measure of angle A. Because we have a

known hypotenuse, a known opposite side, and an

unknown angle, use the sine function.

sin A 30.4

50.2

A sin1 30.4

37.3


known opposite side, a known adjacent side, and an

unknown angle, use the tangent function.

tan A 15.3

17.6

A tan1 15.3

41.0

50.2 17.6

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

c2

(30.4)2

b2

(50.2)2

Find the measure of angle B. Because C = 90°,

A + B = 90°. Thus, B 90 A 90 41.0 49.0 .

Use the Pythagorean Theorem.

c2

a 2

b2

(15.3)2

(17.6)2

543.85

c 543.85 23.32

b2

(50.2)2

(30.4)2

1595.88 In summary, A 41.0, B 49.0 , and c 23.32 .

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.


known hypotenuse, a known adjacent side, and

unknown angle, use the cosine function.

cos A 2

7

sin A 11.2 65.8

A cos 1 2 73.4

7

A sin1 11.2 9.8

Find the measure of angle B. Because C = 90°, A + B = 90°.

65.8 Thus, B 90 A 90 73.4 16.6 .

Find the measure of angle B. Because C = 90°, A + B = 90°. Thus, B 90 A 90 9.8 80.2 .


a 2

b2

c2

(11.2)2

b2

(65.8)2


a2

b2

c2

a 2

(2)2

(7)2

a 2

(7)2

(2)2

45

a 45 6.71

b2

(65.8)2

(11.2)2

4204.2

b 4204.2 64.84

In summary, A ≈ 9.8°, B ≈ 80.2°, and b ≈ 64.84.


known opposite side, a known adjacent side, and an

unknown angle, use the tangent function.

tan A 10.8

In summary, A ≈ 73.4°, B ≈ 16.6°, and a ≈ 6.71.


known hypotenuse, a known adjacent side, and an

unknown angle, use the cosine function.

cos A 4

9

A cos1 4 63.6

24.7 9

A tan1 10.8 23.6

Find the measure of angle B. Because C = 90°,

24.7 A + B = 90°.



Thus, B 90 A 90 63.6 26.4 .Find the measure of angle B. Because C = 90°, A + B = 90°. Thus, B 90 A 90 23.6 66.4 .


c2

a 2

b2

(10.8)2

(24.7)2

726.73

c 726.73 26.96


a2

b2

c2

a 2

(4)2

(9)2

a 2

(9)2

(4)2

65

a 65 8.06

In summary, A ≈ 23.6°, B ≈ 66.4°, and

c ≈ 26.96. In summary, A 63.6, B 26.4 , and a 8.06 .



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

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

15° E. equation with the cosine function, y a cos ωt , to

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

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.

period = 2π

2 ω 2π 2ω

ω 2π

π 2

Substitute these values into d a cos ωt .

The equation for the object’s simple harmonic motion is d 8cos π t .

19. When t = 0, d = 0. Therefore, we will use the equation

south side of the north-south line and lies east of the with the sine function, y a sin ω t, to model 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

object’s motion. Recall that a is the maximum

distance. Because the object initially moves down, and

has an amplitude of 3 inches, a = –3. The value of ω can be found using the formula for the period.

period 2π

1.5 ω

2π 1.5ω

equation with the cosine function, y a cos ω t to ω 2π

4π

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.

period 2π

4 ω

1.5 3

Substitute these values into d a sin ω t. The equation

for the object’s simple harmonic motion is

d 3sin 4π

t. 3

20. When t = 0, d = 0. Therefore, we will use the equation

2π 4ω with the sine function, y a sin ω t, to model the

ω 2π

π

4 2 Substitute these values into d a cos ω t. The

equation for the object’s simple harmonic motion is

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

d 6 cos π

t. period = 2.5

ω



2 2π 2.5ω

ω 2π

4π

2.5 5 Substitute these values into d a sin ωt . The equation

for the object’s simple harmonic motion is

d 5sin 4π

t . 5



π

ω π 2

ω


d 5cos π

t, a 5 and ω π

2 2 a. The maximum displacement from the rest

position is the amplitude. Because a = 5, the maximum displacement is 5 inches.

b. The frequency, f, is

ω π

π 1 1 f

2 .

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

d 8cos π

t, a 8 and ω π

2 2


position is the amplitude.

Because a = –8, the maximum displacement is 8

inches.

2π 2π 2 2π 4 ω 2 1

The frequency is 1 4

inch per second. b. The frequency, f, is f .

2π 2π 4

c. The time required for one cycle is the period.

period 2π

2π

2π 2

4

The frequency is 1

4

inch per second.

π c. The time required for one cycle is the period.



period = 2π 2π

π

2

2 2π 4

π

d 10 cos 2π t, a 10 and ω 2π



Because a = 10, the maximum displacement is

10 inches.



d 1

sin 2t, a 1

and ω 2 2 2

b. The frequency, f, is f ω

2π

1 . a. The maximum displacement from the rest

2π 2π The frequency is 1 inch per second.


period = 2π

2π

1


Because a 1

, the maximum displacement is 2

1 inch.

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



f ω

2

1

0.32.d 6cos 2π t, a 6and ω 2π 2π 2π π



Because a = –6, the maximum displacement is 6 inches.

The frequency is approximately 0.32 cycle per second.


period 2π

2π

π 3.14

b. The frequency, f, is ω 2

f ω

2π

1. The time required for one cycle is approximately 3.14 seconds.

2π 2πThe frequency is 1 inch per second.


period 2π

2π

1


d 1

sin 2t, a 1

and ω 2 3 3



ω 2πThe time required for one cycle is

1 second.

a. The maximum displacement from the rest position is the amplitude.

Because a 1

, the maximum displacement is 3

1 inch .

3



b. The frequency, f, is f ω

2

1

0.32 . 29. x 500 tan 40 500 tan 25

2π 2π π x 653

The frequency is approximately 0.32 cycle per

second.


period = 2π

2π

π 3.14

30.

31.

x 100 tan 20 100 tan 8

x 50 x 600 tan 28 600 tan 25

ω 2 x 39The time required for one cycle is approximately 3.14 seconds.


d 5sin 2π

t, a 5 and ω 2π

3 3



Because a = –5, the maximum displacement is 5

inches.


ω 2π

2π 1 1

f 3

.

32.

33.

34.

x 400 tan 40 400 tan 28

x 123

x 300

300

tan 34 tan 64x 298

x 500

500 tan

20 tan 48

x 924

400 tan 40tan 20

2π 2π 3 2π 3 35. x tan 40 tan 20

The frequency is 1

3

cycle per second.

x 257


period = 2π

2π

2π 3

3

36.

x 100 tan 43 tan 38

tan 43 tan 38

ω 2π 2π

3

x 482



d 4 sin 3π

t, a 4 and ω 3π

2 2

a. The maximum displacement from the rest position is the amplitude. Because a = –4, the maximum displacement is 4

inches.


ω 3π

3π 1 3 f 2 .

37. d 4 cos π t

π

2

2π 2π 2 2π 4 a. 4 in.

The frequency is 3 4

cycle per second.

b. 1

2



ω

2

in. per sec

c. The time required for one cycle is the period. c. 2 sec

period = 2π

2π 2π

2 4

3π 3π 3 d.

1

The required time for one cycle is 4

3

seconds. 2



2

38. d 3cos π t π a.

1 in.

2

a. 3 in.

b. 1

in. per sec 2

c. 2 sec

2

b. 1

in. per sec 8

c. 8 sec

d. 2


unknown opposite side, a, and a known adjacent side.

Therefore, use tangent function.

tan 21.3 a 5280

a 5280 tan 21.3 2059

The height of the tower is approximately 2059 feet.

1

d.

42. 30 yd 3 ft

1 yd

90 ft

39.

2

d 2 sin π t

π

Using a right triangle, we have a known angle, an unknown opposite side, a, and a known adjacent side. Therefore, use the tangent function.

4 2

a. 2 in.

b. 1

in. per sec 8

c. 8 sec

d. −2

tan 38.7 a

90

a 90 tan 38.7 72

The height of the building is approximately 72 feet. 43. Using a right triangle, we have a known angle, a


a. Therefore, use the tangent function.

tan 23.7 305

a

a 305

695 tan 23.7

The ship is approximately 695 feet from the statue’s

base.

44. Using a right triangle, we have a known angle, a


a. Therefore, use the tangent function.

tan 22.3 200

40. d 1

sin π t

π a

4 2

a 200

488 tan 22.3

The ship is about 488 feet offshore.



45. The angle of depression from the helicopter to point P

is equal to the angle of elevation from point P to the

helicopter. Using a right triangle, we have a known

angle, a known opposite side, and an unknown

adjacent side, d. Therefore, use the tangent function.

tan 36 1000

d

d 1000

1376 tan 36

The island is approximately 1376 feet off the coast.

46. The angle of depression from the helicopter to the

stolen car is equal to the angle of elevation from the

stolen car to the helicopter. Using a right triangle, we

have a known angle, a known opposite side, and an

unknown adjacent side, d. Therefore, use the tangent

function.

tan 72 800

d

d 800

260 tan 72

The stolen car is approximately 260 feet from a point

directly below the helicopter.


a known opposite side, and a known hypotenuse.

Therefore, use the sine function.

sin A 6

23

A sin1 6 15.1

49. Using the two right triangles, we have a known angle,

an unknown opposite side, a in the smaller triangle, b

in the larger triangle, and a known adjacent side in

each triangle. Therefore, use the tangent function.

tan 19.2 a

125 a 125 tan 19.2 43.5

tan 31.7 b

125 b 125 tan 31.7 77.2

The balloon rises approximately 77.2 – 43.5 or 33.7

feet.


corresponding to the smaller angle of elevation drawn

inside a larger right triangle corresponding to the

larger angle of elevation, we have a known angle, an

unknown opposite side, a in the smaller triangle, b in

the larger triangle, and a known adjacent side in each

triangle. Therefore, use the tangent function.

tan 53 a

330 a 330 tan 53 437.9

tan 63 b

330 b 330 tan 63 647.7

The height of the flagpole is approximately

647.7 – 437.9, or 209.8 feet (or 209.7 feet).

23

The ramp makes an angle of approximately 15.1°

with the ground.


a known opposite side, and a known adjacent side.


tan A 250

40

A tan1 250 80.9


known hypotenuse, and unknown sides. To find the

opposite side, a, use the sine function.

sin 53 a

150 a 150 sin 53 120

To find the adjacent side, b, use the cosine function.

cos 53 b

150

40 b 150 cos 53 90

The angle of elevation of the sun is approximately

80.9°.

The boat has traveled approximately 90 miles north

and 120 miles east.




known hypotenuse, and unknown sides. To find the

opposite side, a, use the sine function.

sin 64 a

40 a 40 sin 64 36

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.

tan A 6 9

A tan1 6

34

To find the adjacent side, b, use the cosine function. 9

cos 64 b

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.

tan 28 7

b

b 7

13.2 tan 28

The first ranger is 13.2 miles from the fire, to the

nearest tenth of a mile.

54. The bearing from the lighthouse to the second ship is

N 34° E. Using a right triangle, we have a known

angle, a known opposite side, and an unknown

adjacent side, b. Therefore, use the tangent function.

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.

tan θ 7

5

θ tan1 7 54.5

5

tan 34 9

b

b 9

13.3 tan 34

The first ship is about 13.3 miles from the lighthouse,

to the nearest tenth of a mile.

55. Using a right triangle, we have a known adjacent

side, a known opposite side, and an unknown angle,

A. Therefore, use the tangent function.

tan A 1.5

2

A tan 1.5

37

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.

tan θ 11

2 7

We need the acute angle between the ray that runs θ tan1 11

57.5

from your house through your location, and the 7

north-south line through your house. This anglemeasures 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.

Thus, 40 θ 40 57.5 97.5 . Because this

angle is over 90° we subtract this angle from 180° to

find the bearing from the north side of the north-

south line. The bearing of the ship

from the port is N 82.5° W.



59. The frequency, f, is

1 ω

2 2π

ω 1

2π π 2

f ω

, so 2π

71. y 6e 0.09 x cos 2π x

Because the amplitude is 6 feet, a = 6. Thus, the equation for the object’s simple harmonic motion is

d 6sin π t.

10 complete oscillations occur.

72. makes sense


1 ω

4 2π

ω 1

2π π

4 2

f ω

, so 2π


Sample explanation: When using bearings, the angle

must be less than 90 .


Sample explanation: When using bearings, north and south are listed before east and west.

Because the amplitude is 8 feet, a = 8. Thus, the equation for the object’s simple harmonic motion is

d 8sin π

t . 2


Sample explanation: Frequency and Period are

inverses of each other. If the period is 10 seconds

then the frequency is 1

0.1 oscillations per

61. The frequency, f, is f ω

, so 2π

10 second.

264 ω

2π ω 264 2π 528π

Thus, the equation for the tuning fork’s simple

harmonic motion is d sin 528π t.

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.

sin 76.6 r


98,100, 000 ω

2π

f ω

, so 2π

r 112

sin 76.6(r 112) r r

sin 76.6 112 sin 76.6 r

r r sin 76.6 112 sin 76.6

ω 98,100, 000 2π 196, 200, 000π

Thus, the equation for the radio waves’ simple

harmonic motion is d sin 196, 200, 000π t .


r 1 sin 76.6 112 sin 76.6

r 112 sin 76.6

4002 1 sin 76.6

The Earth's radius is approximately 4002 miles.

70. y 4e0.1x

cos 2 x



3

com

plet

e

osci

llati

ons

occ

ur.

Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

x coordinates

0 (0, 0)

π

8

π

8 , 3

π

4

π

4 , 0

3π

8

3π

8 , 3

π

2

(2π , 0)

77. Let d be the adjacent side to the 40° angle. Using the

right triangles, we have a known angle and unknown Chapter 2 Review Exercises

sides in both triangles. Use the tangent function. 1. The equation y 3sin 4 x is of the form

tan 20 h

75 d

y A sin Bx

A 3 3.

with A = 3 and B = 4. The amplitude is

h (75 d ) tan 202π 2π π

Also, tan 40 h

d The period is

B π

π π

. The quarter-period is 4 2

h d tan 40 2 1 . The cycle begins at x = 0. Add

Using the transitive property we have (75 d ) tan 20 d tan 40

75 tan 20 d tan 20 d tan 40

d tan 40 d tan 20 75 tan 20

d (tan 40 tan 20) 75 tan 20

d 75 tan 20

tan 40 tan 20

4 2 4 8 quarter-periods to generate x-values for the key points.

x 0

x 0 π

π 8 8

x π

π

π

8 8 4

Thus, h d tan 40x

π π

3π

75 tan 20 tan 40 48

tan 40 tan 20The height of the building is approximately 48 feet.


4 8 8

x 3π

π

π


79. sec x cot x 1

cos x

1

or csc x

cos x sin x sin x

80. tan x csc x cos x sin x

1

cos x

1

cos x sin x 1

81. sec x tan x 1

sin x

1 sin x

cos x cos x cos x


and graph one complete cycle of the given function.



x

coordinates

0

(0, 2)

π

(π , 0)

2π

(2π , 2)

3π

(3π , 0)

4π

(4π , 2)

x coordinates

0 (0, –2)

π

4

π

4 , 0

π

2

π

2 , 2

3π

4

3π

4 , 0

π (π , 2)

B

4

2. The equation y 2 cos 2 x is of the form 3. The equation y 2cos

1 x

is of the form

y A cos Bx with A = –2 and B = 2. The amplitude 2

is A 2 2. The period is 2π

2π

π. The y A cos Bx with A = 2 and B

1 . The amplitude

B 2 2


. The cycle begins at x = 0. Add

is A 2 2. The period is 2π

2π 2π 2 4π .

1 2

quarter-periods to generate x-values for the key 4πpoints. x 0

The quarter-period is π. The cycle begins at x 4

x 0 π

π 4 4

x π

π

π

4 4 2

x π

π

3π

2 4 4

x 3π

π

π 4 4


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

x 0 π π

x π π 2π

x 2π π 3π

x 3π π 4π






x coordinates

0 (0, 0)

1

2

1

1 (1, 0)

3

2

3

2 (2, 0)

x

coordinates

0

(0, 0)

3

2

3 1 2 2

3

(3, 0)

9

2

9 1 ,

6

(6, 0)

π

4. The equation y 1

sin π


5. The equation

y A sin Bx

y sin π x is of the form

with A = –1 and B = π . The amplitude

y A sin Bx with A = 1 2

and B = π

. The amplitude 3

is A 1 1. The period is 2π B

2π

π

2. The

is A 1 1

. The period is quarter-period is 2

1

. The cycle begins at x = 0.

2 2

2π

2π 2π

3 6. The quarter-period is

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

B 3

π x 0

6 3 . The cycle begins at x = 0. Add quarter-

4 2 periods to generate x-values for the key points. x 0

x 0 3

3

2 2

x 3

3

3 2 2

x 3 3

9 2 2

x 9

3

6 2 2


x 0 1

1

2 2

x 1

1

1 2 2

x 1 1

3

2 2

x 3

1

2 2 2


, 1 2

, , 1

2 2







x coordinates

π (π , 0)

3π

2

3π

, 2

2π (2π , 0)

5π

2

5π

2 , 2

3π (3π , 0)

x

coordinates

0

(0, 3)

3π

2

3π

2 , 0

3π

(3π , 3)

9π

2

9π , 0

6π

(6π , 3)

B

3

6. The equation y 3cos x

is of the form 7. The equation y 2 sin( x π ) is of the form

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

y A cos Bx with A = 3 and B = 1

. The amplitude amplitude is A 2 2. The period is

3 2π 2π C πis A 3 3. 2π . The phase shift is

B 1 π . The

B 1

The period is 2π

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

2π π

.

1 4 2

The cycle begins at x π . Add quarter-periods toperiod is

6π 3π

. The cycle begins at x = 0. Add 4 2

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

x 0 3π

3π

2 2

x 3π

3π

3π 2 2

x 3π 3π

9π

2 2

x 9π

3π

6π 2 2


generate x-values for the key points. x π

x π π

3π 2 2

x 3π

π

2π 2 2

x 2π π

5π 2 2

x 5π

π

3π 2 2


2

2 Connect the five key points with a smooth curve





x coordinates

π (π , 3)

π 2

π

, 0

0 (0, 3)

π

2

π

2 , 0

π (π , 3)

x

coordinates

π 8

π

, 3

8 2

π

8

π , 0

3π

8

3π 3

8 ,

2

5π

8

5π

8 , 0

7π

8

7π 3

8 ,

2

4

x

π

8. y 3cos( x π ) 3cos( x (π )) 3 π 3 π

The equation y 3cos( x (π )) is of the form 9. y cos 2 x

2 4 cos 2 x

2 4

y A cos( Bx C ) with A = –3, B = 1, and C = π.

3

π

The amplitude is A 3 3. The equation y cos 2 x is of 2 4

The period is 2π

2π

2π . The phase shift is 3B 1 the form y A cos( Bx C ) with A ,

2

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

B 1 4 2 B = 2, and C = . The amplitude is 4

cycle begins at x π . Add quarter-periods to 3 3

generate x-values for the key points. x π

x π π

π 2 2

A . 2 2

The period is 2π B

π

2π

2

π. The phase shift is

π π x 0

C π 1 π π . The quarter-period is .

2 2 B 2 4 2 8 4

x 0 π

π 2 2

x π

π

π 2 2





π

8

π π π x

8 4 8

x π

π

3π

8 4 8

2

x 3π

π

5π

8 4 8

x 5π

π

7π



8



x

coordinates

π 4

( π

, 0) 4

0

0, 5

2

π

4

π

4 , 0

π

2

π 5

2 ,

2

3π

4

3π

, 0

2

B π

B



10.

y 5

sin 2 x

π 5

sin 2 x

π

2 2 2 2

The equation

y 5

sin 2 x

π

is of 4

2 2 Connect the five key points with a smooth curve and

the form y A sin( Bx C ) with A 5

, 2


B = 2, and C = π

. The amplitude is 2

A 5

5

. 2 2

The period is 2π

2π

π. The phase shift is B 2

π

C π 1 π π . The quarter-period is .

B 2 2 2 4 4


. Add quarter-periods to4 11. The equation y 3sin

π x 3π

is of

generate x-values for the key points. 3

π x

the form y A sin( Bx C ) with A = –3,

4

π π x 0

B = π

, and C = 3π. The amplitude is 3

A 3 3.

4 4 The period is

2π 2π

2π 3

6. The phase shift

x 0 π

π

4 4

π 3

C 3π 3

x π

π

π

4 4 2

is π 3

3π 9. The quarter-period is π

x π

π

3π


6 3 . The cycle begins at x = 9. Add quarter-

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


x

coordinates

9

(9, 0)

21

2

21

2 , 3

12

(12, 0)

27

2

27 , 3

15

(15, 0)

x coordinates

0 (0, 1)

π

4

π

4 , 2

π

2

π

, 1

3π

4

3π

4 , 0

π (π , 1)

x 9 12. The graph of y sin 2 x 1 is the graph of

x 9 3

21 y sin 2 x shifted one unit upward. The period for

2 2

x 21

3

12

both functions is 2π

π. The quarter-period is π

. 2 4

2 2

x 12 3

27

2 2

x 27

3

15 2 2


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

x 0 π

π 4 4

x π

π

π

4 4 2

x π

π

3π

2 4 4

3π πx π

4 4


2

Connect the five key points with a smooth curve and 2





3

13. The graph of y 2 cos 1

x 2 is the graph of 3

y 2 cos 1

x shifted two units downward. The period for both functions is 3

2π 2π 3 6π . The quarter-period is

6π 3π

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

1 4 2

for the key points. x 0

x 0 3π

3π

2 2

x 3π

3π

3π 2 2

x 3π 3π

9π

2 2

x 9π

3π

6π 2 2


x

coordinates

0

(0, 0)

3π

2

3π

2 , 2

3π

(3π , 4 )

9π

2

9π

2 , 2

6π

(6π , 0)



π


x

0 π

4

π

2

3π

4

π 5π

4

3π

2

7π

4

2π

y1 3sin x 0 2.1 3 2.1 0 2.1 3 2.1 0

y2 cos x 1 0.7 0 0.7 1 0.7 0 0.7 1

y 3sin x cos x 1 2.8 3 1.4 1 2.8 3 1.4 1


x

0 π

4

π

2

3π

4

π 5π

4

3π

2

7π

4

2π

y1 sin x 0 0.7 1 0.7 0 0.7 1 0.7 0

y cos 1

x 2

2

1

0.9

0.7

0.4

0

0.4

0.7

0.9

1

y sin x cos 1

x 2

1

1.6

1.7

1.1

0

1.1

1.7

1.6

1

16. a. At midnight x = 0. Thus, y 98.6 0.3sin π

0 11π

12 12

98.6 0.3sin 11π

12

98.6 0.3(0.2588) 98.52

The body temperature is about 98.52°F.

b. period: 2π

2π

2π 12

24 hours

B 12

π


2

c. Solve the equation

π x

11π π

12 12 2

π x

π 11π

6π

11π

17π

12 2 12 12 12 12

x 17π

12

17 12 π

The body temperature is highest for

x = 17.

y 98.6 0.3sin π

17 11π

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 x

11π 3π

x 11

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.

17. Blue:

This is a sine wave with a period of 480.

12 12 2

π x

3π 11π

18π

11π

29π

12 2 12 12 12 12

Since the amplitude is 1,

B 2π

2π

π

period 480 240

A 1.

x 29π

12

29

The equation is y sin π

x.

12 π 240

The body temperature is lowest for x = 29.

y 98.6 0.3sin π

29 11π

12 12

Red:

This is a sine wave with a period of 640.

Since the amplitude is 1, A 1.

98.6 0.3sin 3π 2π 2π π

B period 640 320

98.6 0.3(1) 98.3

29 hours after midnight or 5 hours after

midnight, at 5 A.M., the body temperature is

The equation is y sin

π x.

320

98.3°F. 18. Solve the equations

π π

e. The graph of

y 98.6 0.3sin π

x 11π

is 2 x

2 and 2 x

2

12 12 π π

of the form

y D A sin( Bx C ) with A = 0.3,

x 2 x

2

2 2

B π

, C 11π

, and D = 98.6. The π π12 12 x x

4 4

amplitude is A 0.3 0.3. The period Thus, two consecutive asymptotes occur at

from part (b) is 24. The quarter-period is π x

and x

π .


24 6. The phase shift is

4

C 11π

11π 12 12 11. The cycle begins at x

π

4 4

π π 0

x-intercept 4 4 0 2 2

B 12 12 π An x-intercept is 0 and the graph passes through (0,

= 11. Add quarter-periods to generate x-values

for the key points.

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.


π π

Use the two consecutive asymptotes. x π

and 20. Solve the equations

4 x π π

and x π π

x π

, to graph one full period of 2 2

y 4 tan 2 x from

4 x x π

π

π

to π

. 2 2

4 4 x 3π

x π



2 2


x 3π π

and x . 2 2

3π π 2π

x-intercept 2 2 π 2 2


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

points on the graph midway between an x- intercept

and the asymptotes have y-coordinates of –1

and 1. Use the two consecutive asymptotes,19. Solve the equations 3π π

x and x , to graph one full period ofπ π x

and π x

π 2 2

4 2 4 2 3π π

y tan( x π ) from to .x

π 4 x

π 4 2 2

2 π

x 2

2 π

x 2



x = –2 and x = 2.

x-intercept 2 2

0

0 2 2

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



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. x

π π

and x

π π

4 2 4 2

x π π

x π π

2 4 2 4

x π

x 3π


4 4 Thus, two consecutive asymptotes occur at

x π 3π

and x . 4 4

π 3π π π

x-intercept 4 4 2 2 2 4




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


4




x π 3π

and x , to graph one full period of 4 4

y tan x

π from

π to 3π

. Continue the

4 4 4

pattern and extend the graph another full period to the

right.


π x 0 and π

x π

2 2

x 0

x π 2

π


3x 0 and 3x π

x 2


x = 0 and x = 2.

x-intercept 0 2

2

1 2 2

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

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

x 0 x π 3

2 intercept and the asymptotes have y-coordinates of


x = 0 and x π

. 3

1 and

2


2

π π π

x = 0 and x = 2, to graph one full period of0

x-intercept 3

3

1 π2 2 6 y cot x from 0 to 2. Continue the pattern and

An x-intercept is

π , 0 .

π and the graph passes through

6

2 2 extend the graph another full period to the right.

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

two consecutive asymptotes, x = 0 and x π

, to 3

graph one full period of y 2 cot 3x from 0 to π

. 3



x

π

24. Solve the equations Use these key points to graph y 3cos 2π x from 0

x π

0 and x π

π to 1. Extend the graph one cycle to the right. Use the

2 2 graph to obtain the graph of the reciprocal function.

x 0 π

x π π Draw vertical asymptotes through the x-intercepts,

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

x π

x π

2 2


π π and x .

2 2

π π 0

x-intercept 2 2 0 2 2

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

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




y 2 sin π x .

Use the two consecutive asymptotes, x π 2

and The equation is of the form

A = –2 and B π .

y A sin Bx with

x π

, to graph one full period of

y 2 cot x

π amplitude: A 2 2

2 2 period: 2π

2π

2

from π

B π to . Continue the pattern and extend the 2 1

2 2 Use quarter-periods, , to find

graph another full period to the right. 4 2 x-values for the five key points. Starting with

x = 0, the x-values are 0, 1

, 2

1, 3

, 2 . Evaluating the 2

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

1 , 2

, 1, 0,

3 , 2

, (2, 0) . Use these key points

2 2

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


y 3cos 2π x . guides to graph y 2 csc π x.


and B 2π .

y A cos Bx with A = 3

amplitude: A 3 3

period: 2π

2π

1 B 2π




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

, 4

1 ,

3 , 1 . Evaluating the function at each value of x,

2 4 the key points are (0, 3),

1 , 0 ,

1 , 3

,

3 , 0 , (1, 3) .

4 2

4


27. Graph the reciprocal cosine function, function at each value of x, the key points

y 3cos( x π ) . The equation is of the form 3π 5 5π 5

y A cos( Bx C ) with A = 3, B = 1, and C π. are (π , 0), , , (2π , 0) ,

2 2 , ,

2 2 (3π , 0).

amplitude: A 3 3

5

period: 2π

2π

2π Use these key points to graph y sin( x π ) from

B

phase shift:

1

C π π

2 π to 3π . Extend the graph one cycle to the right.


B 1 function. Draw vertical asymptotes through the x-


π

, to find 4 2

x-values for the five key points. Starting with

intercepts, and use them as guides to graph

y 5

csc( x π ). 2

x π , the x-values are π , π

, 2

0, π

, π . 2

Evaluating the function at each value of x, the key

points are π , 3, π

, 0 , 0, 3 , 2

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

2

y 3cos( x π ) from π to π . Extend the graph

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

graph of the reciprocal function. Draw vertical

asymptotes through the x-intercepts, and use them as

29. Let θ sin1 1 , then sin θ 1 .

guides to graph y 3sec( x π ). π π

The only angle in the interval , that satisfies

2 2

sin θ 1 is π

. Thus θ π

, or sin1 1 π

. 2 2 2

30. Let θ cos1 1 , then cosθ 1 .

The only angle in the interval 0, π that satisfies

cosθ 1 is 0 . Thus θ 0 , or cos1 1 0 .


y 5

sin( x π ) . 31. Let θ tan

1 1 , then tan θ 1 .

π π


2 y A sin( Bx C ) with

The only angle in the interval 2

, 2

that satisfies

π π πA

5 ,

B = 1, and C π . tan θ 1 is . Thus θ , or tan

1 1 .

2

amplitude:

A 5

5

4 4 4

3 3

2 2 32. Let θ sin1

, then sin θ .


2

2

period: 2π

2π

2π

B 1 The only angle in the interval

π

, π

that satisfies

phase shift: C π 2 2

π

B 1 3 π π sin θ is . Thus θ , or


π

, to find 4 2

2 3 3

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

sin1

3

π

.

2

3

the x-values are π , 3π

, 2π , 5π

, 3π . Evaluating the 2 2


3

2

33. Let θ cos1

1

, then cosθ 1

1 3 3

2 2 38. Let θ cos

2 , then cosθ

2 . The only

.

The only angle in the interval 0, π that satisfies

angle in the interval 0, π that satisfies

cosθ 1 is

2π . Thus θ

2π , or

cosθ 3

is 5π

.

2 3 3 2 6

cos1 1

2π

. 3

5π 3

2 3 Thus, tan cos1

tan .

34. Let θ tan1 3

, then tan θ .

2 6 3

3

3

39. Let θ tan

1 3 , then tan θ

3 .


, π

that satisfies

3 3

π π

2 2 The only angle in the interval 2

, 2

that satisfies

3

tan θ 3

is π

. 6

tan θ 3

3

is π

. 6

Thus θ π

, or tan1 3

π

.

1 3 π

Thus csc tan

csc

2 .6 3 6

3

6


2 . The only angle


3

2 2 4 4

in the interval π

, π


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

2 2 2 π

. 4

Thus,

cos sin1 2 cos

π 2

.

2

4 2

2 2 2

36. Let θ cos1 0 , then cosθ 0 . The only angle in r x y

2 2 2

the interval 0, π that satisfies cosθ 0 is π

. r 4 3

Thus,

2

sin cos1

0 sin π

1 .

r2

25

r 5

cos tan1 3

cosθ x

4

37. Let θ sin1 1

, then sin θ 1 . The only

4 r 5

2 2


angle in the interval π

, π

that satisfies

2 2

sin θ 1 2

is π

. 6

Thus, tan sin1

1

tan π

3

.

2

6 3


.

,

2 2 2

41. Let θ cos1 3

, then cosθ 3

. x y r

5 5 (4)2 y 2 52

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

y 2

25 16 9

y 9 3


tan cos1

4

tan θ 3

5

4

x

2 y

2 r

2

44. Let θ tan1 1

,

32

y 2

52 3

y 2

25 9 16

y 16 4

Because tan θ is negative, θ is in quadrant IV and x 3 and y 1 .

r2

x2

y 2

sin cos1 3 sin θ

y

4 r2 32 (1)2

5 r 5

42. Let θ sin1 3

, then sin θ 3

r2

10

r 10

5 5 1 4

y 1 10sin tan

sin θ

Because sin θ is negative, θ is in

quadrant IV. 5 r 10 10

45.

x

π , x is in

π , π

, so sin1 sin π

π

3 2 2 3 3

46.

x 2π

, x is not in π

, π

. x is in the domain of

3 2 2

sin x , so

x2

(3)2

52

sin 1 sin 2π sin 1 3

π

x2

y 2

r 2

x2 25 9 16

3 2 3

1 2π 1 1

x 16 4 47. sin cos 3

sin 2

tan sin1 3

tan θ y

3

1 1 1

5

x 4 Let θ sin

, then sin θ . The only

2

π π

2 that satisfies

43. Let θ cos1 4

, then cosθ 4

. angle in the interval

2 2


5 5 1 π π

Because cosθ is negative, θ is in

quadrant II. sin θ is . Thus, θ , or

2 6 6

sin1 cos 2π

sin1 1

π

. 3 2 6


48. Let θ tan1 x , then tan θ

x . 50. Find the measure of angle B. Because C 90 ,

2 2

r2

x2

22

r2

x2

y 2

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.

sin 22.3 a

10 a 10 sin 22.3 3.79

We have a known angle, a known hypotenuse, and an unknown adjacent side. Use the cosine function.

cos 22.3 b

10

r x2 4


b 10 cos 22.3 9.25 In summary, B 67.7, a 3.79 , and b 9.25 .

51. Find the measure of angle A. Because C 90 ,

A B 90 . Thus,

cos tan

1 x cosθ

2 2

2 x2 4

x2

4

x2

4

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.

tan 37.4 6

49. Let θ sin1 1

, then sin θ 1

. a

x x

a 6

7.85 tan 37.4

Use the Pythagorean theorem to find the third side, b.

12

b2

x2

b2 x2 1

We have a known angle, a known opposite side, and

an unknown hypotenuse. Use the sine function.

sin 37.4 6

c

c 6

9.88 sin 37.4

b x2

1 In summary, A 52.6, a 7.85 , and c 9.88 .


expression.

1 1 x x x2 1

sec sin

secθ x

x2 1 x2

1


52. Find the measure of angle A. We have a known

hypotenuse, a known opposite side, and an unknown

angle. Use the sine function.

sin A 2

7

A sin1 2

16.6


unknown opposite side, h, and a known adjacent side.


tan 40 h

60 h 60 tan 40 50 yd

7 The second building is 50 yds taller than the first.

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.

a2

b2

c2

22

b2

72

b2

72

22

45

Total height = 40 50 90 yd .


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


d, in the smaller triangle. Therefore, use the tangent

function.

tan 68 25

b 45 6.71 dIn summary, A 16.6, B 73.4 , and b 6.71 .

d 125

50.5

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.

tan A 1.4

3.6

tan 68We now have a known angle, a known adjacent side, and an unknown opposite side, h, in the larger

triangle. Again, use the tangent function.

tan 71 h

A tan1 1.4 21.3

50.5

h 50.5 tan 71 146.7

3.6 The height of the antenna is 146.7 125 , or 21.7 ft,

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.

c2

a 2

b2

(1.4)2

(3.6)2

14.92

c 14.92 3.86

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 north-

south line. Thus the bearing from O to A is N35°E.

58. We need the acute angle between ray OA and the

In summary, A 21.3, B 68.7 , and c 3.86 . north-south line through O. This angle measures 90 55 35 . This angle measured from the south


unknown opposite side, h, and a known adjacent side.


tan 25.6 h

80 h 80 tan 25.6

38.3

The building is about 38 feet high.

side of the north-south line and lies west of the north-

south line. Thus the bearing from O to A is S35°W.


known adjacent side, and an unknown opposite side,

d. Therefore, use the tangent function.

tan 64 d

12 d 12 tan 64 24.6

The ship is about 24.6 miles from the lighthouse.


ω

π

60.

62. d 1 2

sin 4t

a 1

and ω 4 2

a. maximum displacement:

a 1 1

cm

a. Using the figure,

2 2

b. f ω

4

2

0.64

B 58 32 90

Thus, use the Pythagorean Theorem to find the

distance from city A to city C.

8502

9602

b2

2π 2π π frequency: 0.64 cm per second

c. period: 2π

2π

π

1.57

b2

722, 500 921, 600 ω 4 2

b2

1, 644,100

b 1, 644,100 1282.2

The distance from city A to city B is about

1282.2 miles.

The time required for one cycle is about 1.57 seconds.

63. Because the distance of the object from the rest

position at t 0 is a maximum, use the form

2π

b. Using the figure,

tan A opposite

960

1.1294 adjacent 850

A tan1 (1.1294) 48

180 58 48 74

d a cos ωt . The period is

2 2π

ω

ω 2π

π 2

so, ω

The bearing from city A to city C is S74°E. Because the amplitude is 30 inches, a 30 .

61.

d 20 cos π

t 4

a 20 and ω π

4 a. maximum displacement:

because the object starts below its rest position

a 30 . the equation for the object’s simple

harmonic motion is d 30 cos π t . 64. Because the distance of the object from the rest

position at t 0 is 0, use the form d a sin ωt . The

a 20 20 cm period is

2π so

ω π

π ω

b. f 4 1

1

2π

2π 2π 4 2π 8 5 ω

frequency: 1

8

cm per second ω

2π 5

c. period: 2π

2π

2π 4

8 1 1

π Because the amplitude is 4 4

inch, a . a is 4



negative since the object begins pulled down. The

equation for the object’s simple harmonic motion is

d 1

sin 2π

t. 4 5


x coordinates

0 (0, 0)

π

4

π

4 , 3

π

2

π , 0

3π

4

3π , 3

π (π , 0)

x

coordinates

π

2

π

2 , 2

π

π , 0

3π

2

3π , 2

2π

2π , 0

5π

2

5π

2 , 2

Chapter 2 Test 2. The equation y 2 cos

x

π is of the form

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

with A = 3 and B = 2. The amplitude is A 3 3. y A cos( Bx C ) with A = –2, B = 1, and

The period is 2π

2π

π. The quarter-period is π

. C =

π . The amplitude is

2

A 2 2.

B 2 4The cycle begins at x = 0. Add quarter-periods to The period is

2π

2π 2π . The phase shift is


x 0 B 1

C π

π

2π π

x 0 π

π

2

B 1 2 . The quarter-period is .

4 2

4 4

x π

π

π

4 4 2

x π

π

3π

2 4 4

x 3π

π

π 4 4


The cycle begins at x = π



x π

2

x π

π

π 2 2

x π π

3π 2 2

x 3π

π

2π 2 2

x 2π π

5π 2 2


2

4



2




Chapter 2 Test Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions

2


x π and

x π

2 2 2 2

π x x

π 2

2 2

x π x π

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

x-intercept π π

0

0 2 2

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

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

x π and x π , to graph one

full period of y 2 tan x 2

from π

to π .


sin π x. The equation is of the form 2

y A sin Bx with A = 1 2

and B π.

amplitude: A 1

1

2 2

period: 2π

2π

2 B π


1

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

x-values are 0, 1

, 1, 3

, 2. Evaluating the function at each value of x, the key points are 2 2

(0, 3), 1

, 1

, 1, 0, 3

, 1

, (2, 0) . 2 2 2 2

Use these key points to graph y 1

sin π x from 0 to 2. Use the graph to obtain the graph of the reciprocal function. 2

Draw vertical asymptotes through the x-intercepts, and use them as guides to graph y 1

csc π x. 2


.


x

0 π

4

π

2

3π

4

π 5π

4

3π

2

7π

4

2π

y 1

sin x 1

2

0

0.4

0.5

0.4

0

0.4

0.5

0.4

0

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

y 1

sin x 2 cos x 2

2

1.8

0.5

1.1

2

1.8

0.5

1.1

2

6. Let θ cos1 1

, then cosθ 1

2 2

Because cosθ is negative, θ is in quadrant II.

x2

y2

r 2

(1)2

y2

22

y2

4 1 3

y 3

tan cos1

1

tan θ y

3

3 2

x 1

7. Let θ cos1 x , then cosθ

x .

3 3

Because cosθ is positive, θ is in quadrant I.

x2

y 2

r 2

x2

y 2

32

y 2

9 x2

y 9 x2

1 x y 9 x 2

sin cos sin θ

3

r 3


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.

sin 21 a

13 a 13sin 21 4.7

We have a known angle, a known hypotenuse, and an unknown adjacent side. Use the cosine function.

cos 21 b

13 b 13 cos 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. maximum displacement:

a 6

6 in.

b. f ω

π

1

2π 2π 2

frequency: 1 2

in. per second

c. period = 2π

2π

2 ω π


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Chapter 2 Graphs of the Trigonometric Functions; …...Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric FunctionsSec tion 2.1 Graphs of Sine and Cosine Functions