Math 125 - HW3 - Solutions

HW3 Suggested Problems’ Solutions (pages 119 & 120)

11 (√32 , 12 ) means x = √32

, y = 12

sint = y = 12

cost = x = √32

tant = yx

=

12

√32

= 12.2

√3 = 1

√3 . √3√3 = √33

cott = xy

=

√3212

= √32.21

= √3

sect = 1x

= 1

√32

=1. 2

√3 = 2√3. √3√3

=¿ 2√33

csct = 1y

= 112

=¿ 1.21

= 2

33 sin45ocos45o = √22. √22

= 24

= 12

49 The point on the unit circle corresponds to θ = 2π3

= 120o is (−12 , √32 )sin2π3

= √32

cos2π3

= = - 12

tan2π3

=

√32

−12

= √32.(−21 ) = - √3

cot2π3

=

−12

√32

= - 12.2

√3 =- 1√3. √3√3 = - √3

3

sec2π3

= 1

−12

= 1. (−21 ) = - 2

csc2π3

= 1

√32

= 2√3. √3√3 =

2√33

85 (2, - 3) means x = 2, y = - 3

r = √ x2+ y2 = √4+9 = √13

sinθ = yr

= −3√13 =

−3√13

. √13√13 = -

3√1313

cosθ = xr

= 2

√13= 2

√13. √13√13 =

2√1313

tanθ = yx

= −32

cotθ = xy

= −23

secθ = rx

= √132

cscθ = ry

= −√133

HW3 Problems’ Solutions (pages 119 & 120)

16 (√22 , √22 ) means x = √22

, y = √22

sint = y = √22

cost = x = √22

tant = yx

=

√22√22

= √22.2√2

=¿1

cott = xy

=

√22√22

= √22.2√2

=¿1

sect = 1x

= 1

√22

= 1. 2

√2 = 2√2. √2√2 = √2

csct = 1y

= 1

√22

= 1. 2

√2 = 2√2. √2√2 = √2

30 sin30o – cos45o = 12

- √22

= 1−√22

54 The point on the unit circle that corresponds to θ = 11π4

= 495o is (−√22, √22 )

sin11π4

= √22

cos11π4

= −√22

tan11π4

=

√22

−√22

= √22.(−2√2 ) = -1

cot11π4

=

−√22√22

¿−22.22

= -1

sec11π4

= 1

−√22

= 1. (−2√2 ) = -2√2. √2√2 = -

2√22

= - √2

csc11π4

= 1

√22

= 1. 2

√2 = 2√2. √2√2 =

2√22

= √2

86 (-1, -2) means x = -1, y = -2

r = √ x2+ y2 = √(−1)2+(−2)2 = √1+4 = √5

sinθ = yr

= −2√5

= - 2√5. √5√5

= - 2√55

cosθ = xr

= −1√5 =

−1√5. √5√5 =

−√55

tanθ = yx

= −2−1 = 2

cotθ = xy

= −1−2 =

12

secθ = rx

= √5−1

= −√5

cscθ = ry

= √5−2

= −√52

HW3 Suggested Problems’ Solutions (page 134)

29 sinθ < 0, tanθ < 0

We know sinθ < 0 in quadrants I and IV, and tanθ < 0 in quadrants II and IV, thus they’re both negative in quadrant IV. Hence, the angle θ lies in quadrant IV.

39 sinθ = 12

, cosθ =√32

tanθ = sinθcosθ

=

12

√32

= 12.2

√3 = 1

√3 = 1√3. √3√3 = √3

3

cotθ = 1tanθ

= 1

√33

= 3

√3 = √3 .√3√3 = √3 or cotθ=¿ cosθsinθ

=

√3212

= √32.21

= √3

secθ = 1cosθ

= 1

√32

= 2

√3 = 2√3. √3√3 =

2√33

cscθ = 1sinθ

= 112

= 2

51 sinθ = 23

, tanθ < 0

Since sinθ > 0 and tanθ < 0, then we know that θ lies in quadrant II.

sin2θ + cos2θ = 1

cos2θ = 1 – sin2θ

cosθ = ±√1−sin2θ

==> cosθ = - √1−sin2θ (we choose the negative sign for cosθ because we

= - √1−(23 )2

know that θ liesin quadrant II)

= - √1−49 = - √ 9−49 = - √ 59 = - √5√9 = - √53

tanθ = sinθcosθ

=

23

−√53

= 23.(−3√5 ) = -

2

√5 = - 2√5. √5√5 = -

2√55

cotθ = 1tanθ

= 12√55

= - 5

2√5 = - √5 .√52√5 = - √5

2 (or by using cotθ =

cosθsinθ

)

secθ = 1cosθ

= 1

−√53

= - 3

√5 = - 3√5. √5√5 = -

3√55

cscθ = 1sinθ

= 123

= 32

HW3 Problems’ Solutions (page134)

34 cscθ > 0 , cosθ < 0

We know cscθ > 0 in quadrants I and II, and cosθ < 0 in quadrants II and III, thus the angle θ lies in quadrant II.

42 sinθ = 2√23

, cosθ = - 13

tanθ = sinθcosθ

=

2√23

−13

= 2√23.(−31 ) = - 2√2

cotθ = 1tanθ

= 1

−2√2 = - 12√2

. √2√2 = - √2

4 (or by using cotθ =

cosθsinθ

)

secθ = 1cosθ

= 1

−13

= - 3

cscθ = 1sinθ

= 12√23

= 32√2 =

32√2

. √2√2 =

3√24

52 cosθ = - 14

, tanθ > 0

Since cosθ < 0 and tanθ > 0, thus θ must lie in quadrant III. Hence, sinθ < 0

sin2θ + cos2θ = 1sin2θ = 1 – cos2θsinθ = ± √1−cos2θ ==> sinθ = - √1−cos2θ (since sinθ < 0)

= - √1−(−14 )2

= - √1− 116

= - √ 16−116 = - √ 1516 = -

√15√16

= -

√154

tanθ = sinθcosθ

=

−√154

−14

= √154

. 41

= √15

cotθ = 1tanθ

= 1

√15 =

1√15

. √15√15

= √1515

(or by using cotθ = cosθsinθ

)

secθ = 1cosθ

= 1

−14

= - 4

cscθ = 1sinθ

= 1

−√154

= - 4

√15 = -

4√15

. √15√15

= - 4 √1515

Suggested Problems’ Solutions (page 148)

51 y = 2 sin( 12 x) Step 1: The amplitude = | A| = |2| ¿>¿ [ -|2|, |2|] ¿>¿ [- 2, 2]

Step 2: For this function, ω = 12

The period = 2πω

= 2π12

= 4π

Divide [0, 2πω

] = [0, 4π] into 4 equal lengths

|____________|_____________|_____________|____________|

0 π 2π 3π 4π

0+4 π2

= 4 π2

= 2π

0+2π2

= 2π2

= π

2π+4 π2

= 6π2

= 3π

Step 3: Make the table to find five key points

x 0 π 2π 3π 4 π

12x

0

π2

π

3π2

2π

sin12x

0 1 0 -1 0

y = 2sin12x

0 2 0 -2 0

Step 4: Connect the points: (0, 0), (π , 2), (2π , 0), (3π , -2), (4π , 0)

y = 2sin(

12

x)

57 y = 5cos(πx) – 3

Step 1: The amplitude = |A| = |5| ¿>¿ [-|5|, |5|] ¿>¿ [- 5, 5]

Step 2: For this function, ω = π

The period = 2πω

= 2ππ

= 2

Divide [0, 2πω

] = [0, 2] into 4 equal lengths

|__________|___________|___________|____________|

0 12

1 32

2

0+22

= 1

0+12

= 12

1+22

= 32


x 0

12

1 32

2

πx 0

π2

π 3π2

2π

cos(πx) 1 0 -1 0 1

5cos(πx) 5 0 -5 0 5

y =5cos(πx) - 3 2 -3 -8 -3 2

Step 4: Connect the points: (0, 2), (12

, -3), (1, -8), (32

, -3), (2, 2)

y = 5cos(

π

x) - 3

61 y = 5 – 3sin(2x)

Step 1: The amplitude = |A| = |-3| ¿>¿ [-|-3|, |-3|] ¿>¿ [- 3, 3]


The period = 2πω

= 2π2

= π

Divide [0, 2πω

] = [0, π] into 4 equal lengths

|___________|___________|___________|___________|

0 π4

π2

3π4

π

0+π2

= π2

0+ π22

= π4

π2+π

2 = 3 π22

= 3π4


x 0

π4

π2

3π4

π

2x 0

π2

π

3π2

2π

sin(2x) 0 1 0 -1 0

-3sin(2x) 0 -3 0 3 0

y = 5 – 3sin(2x) 5 2 5 8 5

Step 4: Connect the points: (0, 5), (π4

, 2), (π2

, 5), (3π4

, 8), (π , 5)

y = 5 – 3sin(2x)

HW3 Problems’ Solutions (page 148)

52 y = 2cos(14

x)



The period = 2πω

= 2π14

= 2π . 41

= 8π

Divide [0, 2πω

] = [0, 8π] into 4 equal lengths

|___________|____________|____________|____________|

0 2π 4π 6π 8π

0+8 π2

= 8π2

=¿4π

0+4 π2

= 4 π2

=¿2π

4 π+8 π2

= 12π2

= 6π


x 0 2π 4π 6π 8π

14

x 0

π2

π

3π2

2π

cos(14

x) 1 0 -1 0 1

y = 2cos(14

x) 2 0 -2 0 2

Step 4: Connect the points: (0, 2), (2π , 0), (4π , -2), (6π ,0), (8π , 2)

y = 2cos(

14

x)

58 y = 4sin(π2

x) – 2


Step 2: For this function, ω = π2

The period = 2πω

= 2ππ2

= 2π . 2π

= 4

Divide [0, 2πω


|___________|___________|___________|____________|

0 1 2 3 4

0+42

= 2

0+22

= 1

2+42

= 3


x 0 1 2 3 4

π2

x 0

π2

π

3π2

2π

sin(π2

x) 0 1 0 -1 0

4sin(π2

x) 0 4 0 -4 0

y = 4sin(π2

x) - 2 -2 2 -2 -6 -2

Step 4: Connect the points: (0, -2), (1, 2), (2, -2), (3, -6), (4, -2)

y = 4sin(

π2

x) – 2

60 y = -3cos(π4

x) + 2

Step 1: The amplitude = |A| = |-3| ¿>¿ [-|-3|, |-3|] ¿>¿ [-3, 3]

Step 2: For this function, ω = π4

The period = 2πω

= 2ππ4

= 2π . 4π

= 8

Divide [0, 2πω


|__________|____________|____________|____________|

0 2 4 6 8

0+82

= 4

0+42

= 2

4+82

= 6

Step 3: Make the table to fine five key points

x 0 2 4 6 8

π4

x 0

π2

π 3π2

2π

cos(π4

x) 1 0 -1 0 1

-3cos(π4

x) -3 0 3 0 -3

y = -3cos(π4

x) +

2

-1 2 5 2 -1

Step 4: Connect the points: (0, -1), (2, 2), (4, 5), (6, 2), (8, -1)

y = -3cos(

π4

x) + 2

HW Problems’ Solutions (page 157)

34 y = 2cotx – 1

Step 1: The period = πω

= π1

= π

Step 2: Adjacent vertical asymptotes: ωx = 0 and ωx = π

1x = 0 and 1x = π

Step 3: Divide (0, π) into 4 equal lengths

|__________|___________|___________|___________|

0 π4

π2

3π4

π

0+π2

= π2

0+ π22

= π4

π2+π

2 = 3 π22

= 3π4

Step 4: Make the table for first-quarter pointπ4

, midpoint π2

, third-quarter point 3π4

Step 5:

Connect the points: (π2

, -1), (π4

, 1), (3π4

, -3)

x

π2

π4

3π4

cotx 0 1 -1

2cotx 0 2 -2

y = 2cotx - 1 -1 1 -3

y = 2cotx – 1

37 y = 12

tan(14

x) – 2

Step 1: The period = πω

= π14

= 4π

Step 2: Adjacent vertical asymptotes: ωx = −π2

and ωx = π2

14

x = −π2

and 14

x = π2

¿>¿ x = - 2π and x = 2π

Step 3: Divide ( -2π , 2π) into 4 equal lengths

|__________|___________|___________|____________|

-2π -π 0 π 2π

−2π+2 π

2 = 0

−2π+02

= - π

0+2π2

= π

Step 4: Make the table for the first-quarter point - π , midpoint 0, third-quarter point π

x -π 0 π

14

x -π4

0

π4

tan(14

x) -1 0 1

12

tan(14

x) -12

0

12

y = 12

tan(14

x) - 2 -52

-2 -

32

Step 5: Connect the points: (-π , -52

), (0, - 2), (π ,−32

)

y =

12

tan(

14

x) – 2

Math 125 - HW3 - Solutions

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