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Chapter 3

Radian Measure and Dynamic Trigonometry

222

Chapter 3 Topics

• Angle Measure in Radians

• Length, Velocity and Area of a Circular sector

• Unit Circle

• Trig and Real Numbers

333

Chapter 3.1

Angle Measure in Radians

444

Overview

• Radians for Angle Measurement

• Radian Measure of Standard Angles

• Converting

• And yes, we did a bit of this before

555

The Unit Circle

• We use a “central circle”

– Circle in the x-y plane

– Center at the origin

• A central angle is an angle whose Vertex is at the center of

the circle

• An angle whose sides intersect the circle at B and C

“subtends” an arc BC

C

B

y

x

666

Radians

• A radian is the measure of an angle subtended by an arc of whose

length is the radius

• Therefore, the length of an arc, s, the radius r times the angle in

radians, 2 r = s

• Now, lets make some sense of this:

What is arc length?

– The arc whose length is the entire circle is the circumference

– The circumference is 2 r

– Therefore, the angle that encompasses the whole circle

is 2 , which in degree measure is 360

777

Example

If a circle has radius 8 cm, and the arc length is 18 cm, what is

the radian measure of an angle ?

= s/r = 18/8 = 9/4 = 2.25 rad

888

Some More Examples

Find the radian measure of the angle with arc s and

circle radius r

• s = 24 m, r = 4 m

• s = 10 ft, r = 10 ft

• s = 5.2 mm, r = 2.6 mm

• s = 11 cm, r = 20 m

999

Example (Solutions)

Find the radian measure of the angle with arc s and

circle radius r

• s = 24 m, r = 4 m; = 6

• s = 10 ft, r = 10 ft; = 1

• s = 5.2 mm, r = 2.6 mm; = 2

• s = 11 cm, r = 20 m; equivalent to 11 cm and 2000 cm:

= 11/2000

101010

Standard Angles

• Circumference C = 2 r, or an arc of length 2 r goes around

the circle one time

This is the same as saying that an arc whose length is the

circumference subtends an angle of 2

• Therefore, 360 = 2

• 90 = 2/4 = /2

• 180 = 2/2 =

• 45 = 2/8 = /4

• 60 = 2/6 = /3

• 30 = 2/12 = /6

111111

Example, Finding Angle Measure

• Find the radian measure of the following angle:

• 120=

• -225=

• 270=

121212

Example (Solutions)

• Find the radian measure of the following angle:

• 120= 2(60 ) = 2 (/3) = 2/3

• -225= - 5(45 ) = -5(/4) = -5/4 [= 2 - 5/4 = 3/4]

• 270= 3(90 ) = 3(/2)

131313

More Examples

Find the following angles in radians

• 180 =

• 30 =

• 330 =

• -225 =

141414

More Examples (Solutions)

Find the following angles in radians

• 180 = rad

• 30 =/6 rad

• 330 =11 /6 rad

• -225 = -5/4 rad

151515

Converting Degrees and Radians

360 = 2, so 180 =

• To convert from radians to degrees, multiply by 180 /

radians x deg/radians = deg

• To convert from degrees to radians, multiply by /180

deg x radians/deg = radians

161616

Example

Convert to radians:

-75 =

250 12’ =

171717

Example (Solution)

Convert to radians:

-75 = -75 ( /180 ) = 5 / 2

250 12’ = 250.2 = 259.2 ( /180 ) = 1.39

181818

Example (Solution)

Convert from radians to degrees

/ 24 =

-5 =

191919

Example

Convert from radians to degrees

/ 24 = /24 (180 / ) = 180 / 24 = 7.5

-5 = -5 (180 / ) = 900 /

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More Examples

Convert to radians:

• 230 =

• -35 =

• 200 48’ =

3.1, Even problems 38 - 64

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Example (Solutions)

Convert to radians:

•230 = 23/18

•-35 =7 /6

•200 48’ = ? For this one, you need to convert 48’ to degrees.

48’ = 48/60= 0.8 degrees. Then do the multiplication:

(200.8/180)

21

2222

Convert to Degrees:

• /4 =

• /2 =

• 5 /6 =

• 6 =

22

2323

Solutions

Convert to Degrees:

•/4 = 45

•/2 =90

•5 /6 =150

•6 = 1080

23

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More Examples

Find the terminal side quadrant:

• = 3.9

• = 5.4

• = -4.3

24

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Solutions

Find the terminal side quadrant:

• = 3.9, Q 3

• = 5.4, Q 4

• = - 4.3, Q 2

25

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Summary

We can convert from radians to degrees and vice versa: The key

is that 360 degrees is 2 radians

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Chapter 3.2

Arc Length, Velocity, Area of Circular Sector

27

2828

Overview

• Use radians to compute the length of a subtended arc

• Solve problems involving angular velocity and linear velocity

• Calculate the area of a circular sector

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Length of an arc

• For , a subtended arc of a circle of radius r, the arc length is

s = r , for in radians

• For example: radius is10 cm, angle is 3.5 radians,

subtended arc is s = 10(3.5) = 35 cm

29

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Angular and Linear Velocity

Angular velocity equals the amount of rotation per unit time

• Often designated as (omega)

• = / t

For example, a Ferris wheel rotating at 10 revolutions per minute

has angular velocity of

= / t

= 10 revolutions/min

each revolution is 2 rad

= 10 (2 ) / 1 min = 20 rad/min

30

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Linear Velocity

Linear velocity is distance traveled (or change in position)

per unit time

D = r t, r = D / t

For angular motion, the distance D is the length of the arc, s

rate = v = s / t

Since s = r , v = r / t = r ( / t ) = r

Linear velocity v = r

31

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Example

A point P rotates around the circumference of a circle with radius

r = 2 ft at a constant rate. If it takes 5 sec to rotate through an

angle of 510,

a. What is the angular velocity of P

b. What is the linear velocity of P

32

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Solution

a. 510 = 510 ( / 180 )

= (510/180) = 17/6

= / t = (17/6) / 5 = (17/30) rad/sec

b. V = r = 2ft (17/30 ) = (17/15) ft/ sec

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More Examples

Point P passes through angle in time t as it travels around the

circle. Find its angular velocity in radians

• = 540 , t = 9 yr

• = 270 , t = 12 min

• = 690 , t = 5 sec

• = 300 , t = 5 hr

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Solutions

Covert to radians, = angle / time

• = 540 , t = 9 yr; 540 = 540 / 180 = 3 , = /3 rad/yr

• = 270 , t = 12 min; = /8 rad/min

• = 690 , t = 5 sec; = 23 / 30 rad/sec

• = 300 , t = 5 hr; = /3 rad/hr

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Examples

Point P travels around a circle of radius r, find its linear velocity

1. = 12 rad/min, r = 15 ft

2. = 2312 rad/sec, r = 0.01 km

3. = 282, t = 4.1 min, r = 1.2 yd

4. = 45, t = 3 hr, r = 2 mi

36

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Solutions

V = r

1. 180 ft/min

2. 72.6 mph

3. 1.44 yd/min

4. 30 mph

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Example

Using Angular Velocity to Determine Linear Velocity

The wheels on a bicycle have a radius of 13 in. How fast, mph, is

the cyclist going if the wheels turn at 300 rpm?

= 300 rev/min = 300 (2 ) / min = 600 /min

V = r = 13 in (600 /min)

1 mile = 5280 ft x 12 in /ft, 1 hour = 60 min

V = 13 in (600 /min) (60 min/hr) ( 1 mi / (5280 x 12 in)

V = 23.2 mph

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The Area of a Circular Sector

• The area of a circle is r2

• The area of one half the circle is r2/2

• The area subtended by an arbitrary angle = r2/2,

for in radians

39

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Examples

Find the area of a central angle of 3 /4 if the radius is 72 ft.

A = (3 /4 ) (72ft)2 /2 = 1944 ft2

40

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Another Example

The second hand on a wristwatch is 12 mm long. Find the area

of the watch face the second hand passes over in 20 sec

= / t = 2 / 60 rad/sec

in 20 sec, = 20 sec (2 / 60 rad/sec) = 2 / 3 rad

Area A = r2/2 = (2 / 3 rad) (12mm) 2 /2

= 48 mm 2

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Summary

• Arc Length: s = r , in radians

• Angular Velocity: = / t, in radians

• Linear Velocity: v = r

• Area of a circular sector subtended by angle : r2/2,

in radians

All assume is in radians!!!!

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Chapter 3.3

The Unit Circle

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Overview

• Locate points on a unit circle

• Use special triangles to find points on a unit circle

• Define the 6 trig functions in in terms of points on the

unit circle

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Unit Circle

• Circle of radius 1

• Center at the origin

45

x

y

(0,0)

(1,0)

(0,1)

(-1,0)

(0,1)

4646

Exercise

• Find a point of the unit circle if y = 1/2

46

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Solution

x2 + y2 = 1

x = ± sqrt (1 – ¼) = ± sqrt(3)/2

47

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Symmetry

• Find the quadrant containing (-3/5, -4/5) and verify it is on the

unit circle

48

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Solution

• Quadrant : x and y < 0, so is in quadrant 3

• Check: x2 + y2 = 1

(3/5) 2 + (4/5) 2 = (9 + 16)/25 = 1

49

50

More on Symmetry

• If the point (a,b) is on the unit circle, so are

(-a, b)

(a, -b)

(-a, -b)

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Special Triangles

Find points on the unit circle associated with /4: /4: /2 triangle

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Solution

• In quadrant 1, the triangle has x and y values sqrt(2)/2

• In quadrant 2 it is (- sqrt(2)/2, sqrt(2)/2)

• In quadrant 3, both signs are –

• In quadrant 4, x is positive, y is negative

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More Examples

• Do the same for a /6: /3: /4 triangle

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Solutions

(± sqrt(3)/2, ± 1/2)

(± 1/2, ± sqrt(3)/2)

54

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Points Associated with Rotation

Find the points on the unit circle associated with

a. 5/6

b. 4 /3

c. 7 /4

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Solution

a. (-sqrt(3)/2, 1/2)

b. (-1/2, - sqrt(3)/2)

c. (sqrt(2)/2, -sqrt(2)/2)

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Trig Functions and Rotations

Find the six trig functions for = 5 /4

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Solution

Quadrant 3

• cos = -1/sqrt(2)

• sin = -1/sqrt(2)

• tan = 1

• sec = -sqrt(2)

• csc = -sqrt(2)

• cot = 1

58

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Summary

• Can find points on the unit circle for angles in radians

• Can calculate trig functions for points on the unit circle

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Chapter 3.4

Trigonometry of Real Numbers

6161

Overview

• Define trig functions in terms of a real number t

• Find the number associated with special values of the

trig functions

• Find the real number t associated with any trig value

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Real Numbers

• Integers … -2, -1, 0, 1, 2, …

• Rationals: any number that can be expressed as a fraction

Rationals include integers

• Reals: the non-imaginary numbers. Includes 2 and all

rational numbers

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Trigonometry of Real Numbers

• Work with the unit circle: r=1.

if r = 1, then arc length s = r =

• That way, any function of is a function of arc length, s

• We can also treat reference arc as reference angle

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Examples

Give the six trig functions of:

a. 11 / 6

b. 3 / 2

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Solution

a. Quadrant 4:

cos = sqrt(3)/2

sin = -1/2

tan = -1/sqrt(3)

b. Along y axis

cos = 0

sin = -1

tan = undefined (-)

Given these, can find csc, sec, cot

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More Examples

• csc (/6) =

• csc (5/6) =

• csc (11/6) =

• csc (- /6) =

• csc (-17/6) =

6767

Solutions

• csc (/6) = 2

• csc (5/6) = 2 ; /6 away from 180

• csc (11/6) = - 2 ; /6 away from 360

• csc (- /6) = - 2 ; /6 away from 0

• csc (-17/6) = -2 ; /6 away from -18 /6 = -3 = -180

6868

Examples

• cot =

• cot 0 =

• cot /2 =

• cot 3 /2

6969

Solutions

• cot = undefined

• cot 0 =undefined

• cot /2 = 0

• cot 3 /2 = 0

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Special Angles

Find t such that

a. cos t = -1

2

in quadrant 2

b. tan t = 3

in quadrant 3

7171

Solutions

Find t such that

a. cos t = -1/sqrt(2); Q2 (like a 45 deg angle)

t = 3 /4

b. tan t = sqrt (3); Q3(like a 60 deg angle)

t = 4 /3

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Summary

• Know how to find special angles on the unit circle

• Can calculate the trig functions of these angles

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73

Chapter 3 Review

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7474

• Give the reference angle for:

7 /6

24 /3

75

Solution

• Give the reference angle for:

7 /6 : / 6

24 /3: 8 or 0

75

76

Find

• Sin /3 =

• Cos 2 /3 =

• Tan /3 =

• Cos 7 /4 =

• Tan 7/4 =

76

77

Solution

77

78

• Verify that (1/3, - 2 sqrt(2) / 3) is a point on the unit circle.

• Find the value of all the trig functions associated with

this point

78

79

Solution

79

80

• A camera crew rids a cart on a circular arc. The radius of the

arc is 75 ft and can sweep an angle of 172.5º in 20 sec.

– Find the length of the track in feet

– Find the angular velocity of the cart

– Find the linear velocity of the cart.

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81

Solution

• A camera crew rids a cart on a circular arc. The radius of the

arc is 75 ft and can sweep an angle of 172.5º in 20 sec.

– Find the length of the track in feet

• S=r = 75 ft (172.5)(/180) = 225.8 ft

– Find the angular velocity of the cart

• ω = /time = 172.5(/180) /20 = 0.15 rad/sec

– Find the linear velocity of the cart

• V = r ω = (0.15 rad/sec) (75ft) = 11.29 ft/sec

81

82

• Find t, between 0 and 2 if

Sin t = -1/2 in Q3

Sec t = 2 sqrt(3)/3 in Q4

Tan t = -1 in Q2

82

83

Solution

• Find t, between 0 and 2 if

Sin t = -1/2 in Q3: t= 7 /6

Sec t = 2 sqrt(3)/3 in Q4: t = 11 /6

Tan t = -1 in Q2: t = 3 /4

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84

Example

• If (20/29, 21/29) is a point on the central unit circle, use

symmetry to find 3 other points on the circle.

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85

Solution

• If (20/29, 21/29) is a point on the central unit circle, use

symmetry to find 3 other points on the circle.

(-20/29, 21/29)

(-20/29, -21/29)

(20/29, -21/29)

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86

• Convert to radians:

300

-7239’

86

87

• Convert to radians:

300: 5 /3

-7239’= -1.26; First change the minutes to seconds

= 0.0548

87

88

• Convert to degrees:

9.29

-3 /2

45

88

89

Solution

• Convert to degrees:

9.29 = 532.3

-3 /2 = -270

45 = 2578.3

Remember, to convert to degrees, multiply by 180 /

89

90

• Assume Memphis is in directly north of New Orleans, at

90 º W longitude. Find the distance between cities, in km, if

the radius of the earth is 6000 km, Memphis is at 35º north,

and New Orleans is at 29.6º.

90

91

Solution

• Assume Memphis is in directly north of New Orleans, at

90 º W longitude. Find the distance between cities, in km, if

the radius of the earth is 6000 km, Memphis is at 35º north,

and New Orleans is at 29.6º.

Angle is 35º - 29.6 º = 5.4º

5.4º (/180) = 0.094 rad

Radius is 6000 km,

so arclength is 6000 ( 0.094) = 565 km

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Chapter 3 Summary Test

With additional Chapter 3 problems

93

1. What angles in [0, 2) make this true?

Sin t = - 3 / 2

94

Solution

1. What angles in [0, 2) make this true?

Sin t = - 3 / 2

t=7/6, 11/6

95

2. Given (3/4, - sqrt(7)/4) is a point on the unit circle, find all 6 trig

functions.

96

Solution

2. Given (3/4, - sqrt(7)/4) is a point on the unit circle, find all 6 trig

functions.

x =3/4, y = -sqrt(7)/4, r = 1

cos = 3/4 sec = 4/3

sin = -sqrt(7)/4 csc = -4/sqrt(7)

tan = -sqrt(7)/3 cot = -3/sqrt(7)

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3. Find the complement of 67 22’ 39”

98

Solution

3. Find the complement of 67 22’ 39”;

Need 90 – the angle.

Also, 67 = 66 + 59’ + 60” – need this to subtract

22 37’ 21”

99

4. Given cot (/8) = sqrt(2) + 1, find tan2 (3 /8)

100

Solution

4. Given cot (/8) = sqrt(2) + 1, find tan2 (3 /8)

Note that /8 + 3 /8 = /2. These are cofunctions!

All that remains is to square [sqrt(2)+1]

we get 2 sqrt(2)+3;

since (a+b)2 = a2 + b2 + 2ab

101

5. Given A = 9x, B = (6x+4) , and C = 7x , find the measures

of the angles A, B, C in triangle ABC

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Solution

5. Given A = 9x, B = (6x+4) , and C = 7x , find the measures

of the angles A, B, C in triangle ABC

9x + 6x + 4 + 7x = 180; the sum of the angles is 180

22 x = 180 – 4

x = 176/22 = 8

A = 72 , B = 52 , c = 56 .

Check 99+52+56 = 180

103

6. You stand 457 m from a tower. The angle of elevation to the

top of the tower is 30 deg. How tall is the tower?

104

Solution

6. You stand 457 m from a tower. The angle of elevation to the

top of the tower is 30 deg. How tall is the tower?

tan 30 = H/457, where H is height.

tan 30 = 1/sqrt(3), so H = 457 sqrt(3)

105

7. One angle in a right triangle is 3115’18”, find the other two in

radians

106

Solution

7. One angle in a right triangle is 3115’18”, find the other two

One angle is a right angle, /2. The other two must sum to 90

Again, need to convert to decimal degrees, then calculate

(90 – angle) ( / 180) = 1.025

107

8. Find the value of the trig functions if the point (-5, -8) is on the

terminal side of the angle

108

Solution

8. Find the value of the trig functions if the point (-5, -8) is on the

terminal side of the angle.

Pythagorean thm: 25 + 64 = 89, hyp = sqrt (89)

We are in quadrant 3, so only tan and cot are positive

Cos = -5/sqrt(89)

Sin = -8/sqrt(89)

Tan = 8/5

109

10. t = / 6; find the trig functions

110

Solution

10. t = / 6; find the trig functions

/ 6 corresponds to a 30 deg angle.

Cos = sqrt(3)/2

Sin = 1/2

Tan = 1/(sqrt(3))

From these you can get the other three

111

13. A conveyor belt moves on rollers 2 in radius, turning at

252 rpm. Find the angular velocity of the rollers.

How fast are your groceries moving on the belt?

112

Solution

13. A conveyor belt moves on rollers 2 in radius, turning at

252 rpm. Find the angular velocity of the rollers.

How fast are your groceries moving on the belt?

Angular velocity = angle/time; 252 rpm is 252 (2)/min

linear velocity is just r (angular velocity) = 2(252)(2)in/min

113

15. Verify (sin x cos x + cos x) / (sin x + sin2 x) = cot x

114

Solution

15. Verify (sin x cos x + cos x) / (sin x + sin2 x) = cot x

Take a cos x out of the left numerator, and sin x out of the

denominator:

cos x (sin x + 1)/ [sin x (sin x + 1)] = cot x

remove (sin x + 1) from top and bottom, leaving

cos x / sin x = cot x

115

16. If t = 5.37, in what quadrant does it terminate?

116

Solution

16. If t = 5.37, in what quadrant does it terminate?

2 is 6.28.. t is close to that, and certainly greater than 3 /2, so

is in the 4th quadrant

117

18. Find the values of sin, cos, tan, if = 225

118

Solution

18. Find the values of sin, cos, tan, if = 225

Reference angle is 225 – 180 = 45 , Q 3.

Cos = -1/sqrt(2)

Sin = -1/sqrt(2)

Tan = 1

119

23. Given that (sin A)(tanA) >0 and (cos2A)(sin A) < 0, in which

quadrant is the terminal side of A?

120

Solution

23. Given that (sin A)(tanA) >0 and (cos2A)(sin A) < 0, in which

quadrant is the terminal side of A?

If ab > 0, the a and b are either both > 0 or both < 0. Therfore,

Need sin and tan both > or both < 0, but (cos2A)> 0, which

means that sin <0, so tan < 0.

Sin and Tan are < 0 in Q4.