Chapter 7

The Trigonometric Functions

1

Chapter 7 Topics

• Radian Measure

• Trig Functions of angles

• Evaluating Trig Functions

• Algebra and Trig Functions

• Right Triangle Trig

2

Chapter 7.1

Real Numbers

3

444

Overview

• Radians for Angle Measurement

• Radian Measure of Standard Angles

• Converting

Why Radians?

• Degrees are a bit artificial, radians refer to an actual measure

• Definition: If we put the vertex of an angle at the center of a

circle of radius r and let s be the arc length intercepted, then

the angle, is defined as

= s/r

• Because of the definition, angle is now related to a

measureable quantity, that is the ratio of two lengths, s and r

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Radians

• Now, lets make some sense of this:

– 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

• Bottom line, the angle of a whole circle is 2 ;

from that you can determine all other angle measures

Therefore:

• Find the angle measure in radians of the following:

7

Degree Radian

360 2

180

90

270

60

30

45

Angle Measures

8

Degree Radian

360 2

180

90 /2

270 3/2

60 /3

30 /6

45 /4

Converting:

• Let d be an angle in degrees and be its measure in radians

• 180 degrees is

• Need to have 180/d be the same as / , or

d/180 = /

• This is the same as /180 = /d

•

9

Example

• Find 36 deg in radians:

36/180 = /

36/180 = 1/5 = 0.2

So = 0.2

• We will not ask you to do much converting, but you will need

to learn the common angles in radians

10

Angles in Degrees and Radians

Angle 0 30 45 60 90 120 135 150 180

Radian 0 /6 /4 /3 /2 2/3 3/4 5/6

11

Angle 180 210 225 240 270 300 315 330 360

Radian 7/6 5/4 4/3 3/2 5/3 7/4 11/6 2

121212

Example, Finding Angle Measure

• Find the radian measure of the following angle:

• 120=

• -225=

• 270=

131313

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)

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More Examples

Find the following angles in radians

• 180 =

• 30 =

• 330 =

• -210 =

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Solutions

Find the following angles in radians

• 180 = rad

• 30 =/6 rad

• 330 =11 /6 rad

• -210 = -180-30 = -7 /6 rad

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Example

Convert to radians:

-75 =

250 12’ =

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Example (Solution)

Convert to radians:

-75 = -45 - 30 = - /4 - /6 = - 5/12

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

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Example (Solution)

Convert from radians to degrees

/ 24 =

-5 =

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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’ =

2121

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)

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Convert to Degrees:

• /4 =

• /2 =

• 5 /6 =

• 6 =

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Solutions

Convert to Degrees:

•/4 = 45

•/2 =90

•5 /6 =150

•6 = 1080

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More Examples

Find the terminal side quadrant:

• = 3.9

• = 5.4

• = -4.3

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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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Arc Length 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

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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

282828

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

Sector Area

• The area of a sector is the portion of a circle with angle and

radius r

– A whole circle has area r2

– The area of a sector of angle is r2/2

• Think about it. This formula gives the following for a circle of

radius 1:

– Angle , half circle, area = r2/2

– Angle 2 , whole circle, area = r2

– Angle /4, 1/8 of a circle, area = r2/8

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Summary

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

key is that 360 degrees is 2 radians

• Arc length is r , where is in radians

• Sector area is r2/2

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Angular Velocity

Angular velocity equals the amount of rotation per unit time

• Often designated as (omega)

• = / t, the angle per time

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

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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

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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

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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

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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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Examples

• A belt goes around two wheels

• The radius of the smaller is 5cm and the larger 15 cm

• The angular speed of the larger wheel is 1800 RPM

• Find the angular speed of the larger in radians, the linear

speed of a point on the edge of the larger, the angular speed

of the smaller, and the angular speed of the smaller in rpm

40

Solution

• The radius of the smaller is 5cm and the larger 15 cm

• The angular speed of the larger wheel is 1800 RPM

• Find the angular speed of the larger in radians, the linear

speed of a point on the edge of the larger, the angular speed

of the smaller, and the angular speed of the smaller in rpm

• 3000 pi rad/min

• 54000 pi cm/min

• 10800 pi cm/min; the linear velocity has to be the same; v=r.

54000/5 = 10800

• Rpm = /2 pi, so10800 pi/2 pi = 5400 rpm

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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 7.2

Trig Functions of Angles Using Radians

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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)

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Exercise

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

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Solution

x2 + y2 = 1

x = ± sqrt (1 – ¼) = ± sqrt(3)/2

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Symmetry

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

unit circle

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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

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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)

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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

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

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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

6060

Chapter 7.3

Evaluating Trig Fucntions

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

• In this section we cover reference angles

6262

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 sqrt (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

Reference Angle

• A reference angle is an acute angle associated with an angle

formed by the x axis and the terminal ray of the angle. It is,

necessarily, /2

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Give the Reference Angles

• 70 deg

• 115 deg

• 200 deg

• 400 deg

• - 80 deg

• - 325 deg

65

Solution

• 70 deg; 70

• 115 deg; 65

• 200 deg; 20

• 400 deg; 40

• - 80 deg; 80

• - 325 deg; 55

66

Give the reference angle

• 4/3

• 5

• 5/6

• 7/8

67

Solution

• 4/3; /3

• 5 ;

• 5/6 ; /6

• 7/8 ; /8

• 21/10 ; /10

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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) =

7272

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

7373

Examples

• cot =

• cot 0 =

• cot /2 =

• cot 3 /2

7474

Solutions

• cot = undefined

• cot 0 =undefined

• cot /2 = 0

• cot 3 /2 = 0

7575

Special Angles

Find t such that

a. cos t = -1/sqrt(2)

in quadrant 2

b. tan t = sqrt (3)

in quadrant 3

7676

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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What you need to know

• Determine any reference angle

• Know how to find special angles on the unit circle

• Calculate the trig functions of these special angles

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78

Chapter 7.4

Algebra and the Trig Functions

78

Notation

• sin of the quantity a is written sin a or sin (a); parentheses are

not required

• When the argument of sin is an expression parentheses are

required: sin of the quantity (a + b) is written sin(a+b)

• The expression sin2a means (sin a)(sin a)

• The expression sin2 (a+b) means (sin(a+b))(sin(a+b))

• The expression sin (a+b) 2 means sin of the quantity (a+b)2 ,

the function is not squared, only its argument

79

Manipulating Trig Functions

• Add sin a and 1/cos a

• Simplify (Sec a + 1)/ (sin + tan)

80

Solution

• Add sin a and 1/cos a

find a common denominator: (sin a cos a + 1)/cos a

• Simplify (Sec a + 1)/ (sin a+ tan a)

Replace sec and tan by sin and cos

(1/cos a + 1) / (sin a + sin a / cos a)

Multiply top and bottom by cos a

(1 + cos a)/(sin a cos a + sin a)

This becomes 1+cos 𝑎

(sin 𝑎)(1+cos 𝑎)= 1/sin a or csc a

81

Using identities to calculate functions

• Sin a = 2/3 in the second 2nd quad, find cos a and tan a

82

Solution

• Sin a = 2/3 in the second 2nd quad, find cos a and tan a

• 1 – sin2 a = cos2 a = 1-4/9 = 5/9

• Cos2 a = 5/9 or cos a = 5 /3

83

Example

• Simplify sin a/cos a + cos a/sin a

84

Solution

• Simplify sin a/cos a + cos a/sin a

• Get a common denominator, sin a cos a

• (sin2 a + cos2 a )/(sin a cos a) = 1/(sin a cos a)

85

Example

• cos a/(1-sin a) = (1+sin a)/cos a

86

Solution

• cos a/(1-sin a) = (1+sin a)/cos a

• Multiply both sides by cos a (1 – sin a) to remove the

denominators

• cos2 a = 1 – sin2 a which is the Pythagorean identity

• Note, in this case we manipulated both sides of the equation.

We could have worked with just one.

87

Examples

• 1/sin a – sin a = cot a cos a

88

Solution

• 1/sin a – sin a = cot a cos a

• Get rid of cot and get the left a common denominator

• (1 – sin2 a)/sin a = cos2 a / sin a

• Holds by Pythagorean identity

89

Example

• (cos a – sin a)2 + 2sin a cos a = 1

90

Solution

• (cos a – sin a)2 + 2sin a cos a = 1

• Multiply out parentheses

• Cos2 a + sin2 a – 2 sin a cos a + 2 sin a cos a = 1

• By Pythagoras becomes:

• 1 + 0 = 1

91

Example

• Tan a tan b = (tan a + tan b)/(cot a + cot b)

92

Solution

• Tan a tan b = (tan a + tan b)/(cot a + cot b)

• Eliminate the denominator

tan a tan b(cot a + cot b) = tan a + tan b

• Multiply left side out, remembering tan a cot a = 1

the left becomes tan b + tan a, which is the right

93

How to prove isn’t identity

• Show cos A + cos B = cos (A+B) is not an identity

• To prove something isn’t an identity, only need to show that it

doesn’t work for one value!

94

Solution

• Show cos A + cos B = cos (A+B) is not an identity

• Try A = B = 60 deg, cos 60 = ½

• Does ½ + ½ =1 = cos 120? Cos 120 is – cos 60 = - ½ NO

95

Which is an identity?

• (Csc a – cot a) = (1 + cos a)(1-cos a)

• Cot a + sin a/(1+cos a) = csc a

96

Solution

• (Csc a – cot a) = (1 + cos a)(1-cos a)

The right side becomes 1 – cos2a = 1,

the left is (1- cos a)/sin a; there are many values for

which that is not = 1; Not an identity

• Cot a + sin a/(1+cos a) = csc a

Simplify by multiplying by sin a

cos a + sin2 a/(1 + cos a) = 1

• Replace sin2 a by (1-cos2 a)

• We have cos a + 1 – cos a on the left which = 1

97

What you need to know

• Notation: sin2 a vs sin a2

• Need to be comfortable in manipulating trig functions to prove

identities

• Need to be able to disapprove identities by substituting a

value

98

Chapter 7.5

Right Angle Trigonometry

99

• We have already completed most of this section

• The section uses a given trig value to find the remainder,

e.g., if sin = 3/5, find cos and tan

100

Cofunctions

• Sin (90 – A) = cos A; If two acute angles, A and B, are

complementary, then sin A = cos B

• We have the three angles of the triangle as A, B, 90

• Clearly 90 – A = B if they are complementary

• From the picture, sin A = a/c = cos B, etc.

• Additionally, cos (90 – A ) = sin B

101

A

B

b

ac

Using Cofunctions

• Given cos (75) = a find sin (15)

102

Solution

• Given cos (75) = a find sin (15)

• We know that cos (90 – x) = sin x, so sin (15) = a

103

Trig values in other quadrants

• We have done these

104

More Examples

105

106106

• Give the reference angle for:

7 /6

24 /3

107

Solution

• Give the reference angle for:

7 /6 : / 6

24 /3: 8 or 0

107

108

Find

• Sin /3 =

• Cos 2 /3 =

• Tan /3 =

• Cos 7 /4 =

• Tan 7/4 =

108

109

Solution

109

110

• 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

110

111

Solution

111

112

• 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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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

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114

Example

• 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

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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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116

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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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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118

• Convert to radians:

300

-7239’

118

119

• Convert to radians:

300: 5 /3

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

= 0.0548

119

120

• Convert to degrees:

9.29

-3 /2

45

120

121

Solution

• Convert to degrees:

9.29 = 532.3

-3 /2 = -270

45 = 2578.3

Remember, to convert to degrees, multiply by 180 /

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122

Example

• 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.

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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