Trigonometry Chapter 3 Lecture Notes Section 3.1 …faculty.nwacc.edu/tobrien/Trig Resources...

LHS Trig 8th ed Ch 3 Notes F07 OrsquoBrien

1

Trigonometry Chapter 3 Lecture Notes

Section 31 Radian Measure I Radian Measure

A Terminology

When a central angle (θ) intercepts the circumference of a circle the length of the piece subtended (cut off) is called the arc length (s)

B The radian measure of an angle (θ) is the ratio of the arc length (s) to the radius of the circle (r) ie

C If the arc length subtended by angle θ is equal to the radius (ie when s = r) then θ has a measure of 1 radian

D Since a radian is defined as a ratio of two lengths the units cancel and the measure is considered unit-less Therefore if an angle measure is written with no degree symbol it is assumed to be in radians Though it is not essential it is often customary to write radians or rads after an input measured in radians especially when doing conversions and canceling units

E In many applications of trigonometry radian measure is preferred over degree measure because it simplifies calculation and allows us to use the set of real numbers as the domain of the trig functions rather than just angles II Converting Standard Angles Between Degrees and Radians

A One full rotation measures 2π or approximately 628 radians Thus 2π radians = 360deg

π radians = 180deg o90radians 2π

= o60radians 3π

= o45radians 4π

= and o30radians 6π

=

B It is absolutely essential that you know the equivalent degree and radian measures for all standard angles This is not difficult if you think of every standard angle as a multiple of one

of our four special angles - 30deg 45deg 60deg and 90deg For example if 6

1π30 =o then

6

2π60 =o 6

3π90 =o etc If 4

1π45 =o then 4

2π90 =o 4

3π135 =o and so on

Example 1 Convert the following angles from degree measure to radian measure

a 120deg (8) b 225deg c 570deg

a 3

2π180π120 =timeso

o b 4

5π180π225 =timeso

o

c 570deg - 360deg = 210deg rarr 6

7π180π210 =timeso

o

rsθ =


2

Example 2 Convert the following inputs from radian measure to degree measure

a 4

7π rads (23) b 3

4πminus radians c

65π radians

a oo

315π

1804

7π=sdot b o

o

240π

1803

4πminus=sdotminus c o

o

150π

1806

5π=sdot

C You should memorize the following diagram

D You should memorize the following table

Degree-Radian Conversion Factors

Multiples of 30degamp 6π




30deg = 6π1

45deg = 4π1 60deg =

3π1

90deg = 2π1

60deg = 6π2

90deg = 4π2 120deg =

3π2 180deg =

2π2

90deg = 6π3

135deg = 4π3 180deg =

3π3 270deg =

2π3

120deg = 6π4

180deg = 4π4 240deg =

3π4 360deg =

2π4


3

150deg = 6π5 225deg =

4π5 300deg =

3π5

180deg = 6π6 270deg =

4π6 360deg =

3π6

210deg = 6π7 315deg =

4π7

240deg = 6π8 360deg =

4π8

270deg = 6π9

300deg =6π10

330deg =6π11

360deg =6π12

III Converting Nonstandard Angles Between Degrees and Radians

A To convert a non-standard angle from degrees to radians multiply by deg180

π and simplify

Example 3 Convert each nonstandard angle from degree measure to radians

a 174deg 50 (40) b 476925deg (45) a

b

B To convert a nonstandard angle from radians to degrees multiply by π

180deg and simplify

Example 4 Convert each radian measure to degrees In part b round to the nearest minute

a 207π (32) b 306 (50)

a b b


4

Example 5 Angle θ is an integer when measured in radians Give the radian measure of the angle (2)

Example 6 Find the exact value of each expression without using a calculator

a 3

2πcot (66) b ⎟⎠⎞

⎜⎝⎛minus

65πsin c

415πsec

a 3

2π terminates in Quadrant II and has a reference angle of 3π

33

31

OA

32πcot minus=minus==

b 6

5πminus is coterminal with

67π which terminates in Quadrant III

It has a reference angle of 6π ⎟

⎠⎞

⎜⎝⎛minus

65πsin =

21

HO

67πsin minus==

c 4

15π is coterminal with4

7π which terminates in quadrant IV

It has a reference angle of 4π

415πsec = 2

AH

47πsec ==

Section 32 Applications of Radian Measure I Arc Length

The length (s) of the arc intercepted on a circle of radius (r) by a central angle θ (measured in radians) is given by the product of the radius and the angle ie s = rmiddotθ

Caution θ must be in radians to use this formula Example 1 Find the length of the arc intercepted by a central angle o135θ = in a circle of radius r = 719 cm (10)


5

Example 2 Find the distance in kilometers between Farmersville California 36deg N and Penticton British Columbia 49deg N assuming they lie on the same north-south line The radius of the earth is 6400 km (14)

Example 3 A small gear and a large gear are meshed An 800deg rotation of the smaller gear causes the larger gear to rotate 500deg Find the radius of the larger gear if the smaller gear has a radius of 117 cm (22)

II Area of a Sector of a Circle

A A sector of a circle is the portion of the interior of the circle intercepted by a central angle

B The area (A) of a sector of a circle of radius r with central angle θ (measured in radians) is

given by θr21A 2=

Caution θ must be in radians to use this formula


6

Example 4 Find the area of a sector of a circle with radius 183 cm and central angle θ = 125deg Round to the nearest tenth (36)

Example 5 The Ford Model A built from 1928 to 1931 had a single windshield wiper on the driverrsquos side The total arm and blade was 10 inches long and rotated back and forth through an angle of 95deg If the wiper blade was 7 inches how many square inches of the windshield did the blade clean (42)

Section 33 The Unit Circle and Circular Functions I Introduction

In the 1600s scientists began using trigonometry to solve problems in physics and engineering Such applications necessitated extending the domains of the trigonometric functions to include all real numbers not just a set of angles This extension was accomplished by using a correspondence between an angle and the length of an arc on a unit circle (a circle with a radius of 1 centered on the origin with equation 1yx 22 =+ ) II The Unit Circle

Imagine that the real number line is wrapped around a unit circle Zero is at the point (1 0) the positive numbers wrap in a counterclockwise direction and the negative numbers wrap in a clockwise direction Each real number t corresponds to a point (x y) on the circle If central angle θ in standard position measured in radians subtends an arc length of t then according to the arc length formula (s = rθ) t = θ


7

Thus on a unit circle the measure of a central angle and the length of its arc can both be represented by the same real number t

III The Unit Circle Definitions of the Trigonometric Functions

If t is a real number and (x y) is the point on the unit circle corresponding to t then

cos t = x sin t = y tan t = 0 xxy

ne

sec t = 0 xx1

ne csc t = 0y y1

ne cot t = 0y yx

ne

Example 1 Use the diagram below to evaluate the six circular values of θ (46)

1715 θ cos minus=

1517 θ sec minus=

178 θsin =

817 θ csc =

158 θtan minus=

815 θcot minus=

IV Domains of the Circular Functions

Since x = 0 when 2π

t = and 2

3πt = sec t and tan t which both have x in the denominator are not

defined there or at any odd multiple of 2π Thus the domain of sec t and tan t is

⎭⎬⎫

⎩⎨⎧

sdot+ne2

π1)(2nt|t

Since y = 0 at 0t = πt = and 2πt = csc t and cot t which both have y in the denominator are undefined at any integer multiple of π Thus the domain of cot t and csc t is πn t|t sdotne

Sine and cosine do not have any restrictions on their domain thus their domain is ( )infininfinminus

V Finding Exact Values of Circular Functions by Using Special Points on the Unit Circle

Angle 0 or 2π 6π

4π

3π

2π π

23π

Point (1 0) ⎟⎟⎠

⎞⎜⎜⎝

⎛

21

23

⎟⎟⎠

⎞⎜⎜⎝

⎛

22

22

⎟⎟⎠

⎞⎜⎜⎝

⎛

23

21

(0 1) (-1 0) (0 -1)

The x-value of a point on the unit circle tells you cos t and the y-value tells you sin t The sign of the function values depends on where the angle terminates

Quadrant I x and y are both positive therefore all function values are positive Quadrant II x is negative and y is positive therefore only sine and cosecant are positive Quadrant III x and y are both negative therefore only tangent and cotangent are positive Quadrant IV x is positive and y is negative therefore only cosine and secant are positive


8

Example 2 Find the exact value of (a) cos θ (b) sin θ and (c) tan θ for 2πθ = (1)

The point associated with 2π is (0 1) Therefore cos θ = 0 sin θ = 1

tan θ is undefined cot θ = 0 csc θ = 1 sec θ is undefined

Example 3 Find the exact value of (a) 6

11πcsc (11) (b) 4

3πcos (22) (c) 3

10πcot

a 6

11π terminates in quadrant IV Its reference angle is 6π which is associated with

the point ⎟⎟⎠

⎞⎜⎜⎝

⎛

21

23 In quadrant IV only cosine and secant are positive

23

611πcos =

21

611πsin minus=

33

31

23

21

611πtan minus=minus=

minus=

3

326

11πsec = 26

11πcsc minus= 36

11πcot minus=

b 4

3π terminates in quadrant II It has a reference angle of 4π which is associated

with the point ⎟⎟⎠

⎞⎜⎜⎝

⎛

22

22 In quadrant II only sine and cosecant are positive

22

43πcos minus=

22

43πsin = 1

22

22

43πtan minus=

minus=

24

3πsec minus= 24

3πcsc = 14

3πcot minus=


9

c 3

10π is larger than 2π so we must first find its smallest positive coterminal angle

3

4π2π3

10π=minus which terminates in quadrant III and has a reference angle of

3π

The point associated with 3π is ⎟

⎟⎠

⎞⎜⎜⎝

⎛

23

21 and in quadrant III only tangent and

cotangent are positive

21

34πcos minus=

23

34πsin minus= 3

21

23

34πtan =

minus

minus=

23

4πsec minus= 3

323

4πcsc minus= 33

31

34πcot ==

VI Finding Approximate Values of Circular Functions

To find the approximate value of a circular function we use a calculator set on RADIAN mode For secant cosecant and cotangent we use

θcos1θ sec =

θsin 1θ csc = and

θtan 1θcot =

Example 4 Find the approximate value of sec (-83429) (32)

sec (-83429) = ( )83429 cos1

minus asymp -21291

VII Determining a Number with a Given Circular Function Value

A To determine without a calculator what input generates a specified output we must be able

to recognize our special point values in all their alternate forms for example 3

323

2=

22

21

= and 33

31

= And we must know which functions are positive in each quadrant

(A S T C)

Example 5 Find the exact value of t if 21sin t minus= in the interval ⎥⎦

⎤⎢⎣⎡

23ππ

Do not use a calculator ( 58)

A sine value of 21 indicates the reference angle tprime is

3π and since the given interval

is quadrant III t must be 3

4π3ππ =+

B To determine with a calculator what input generates a specified output we use an inverse trig function and the fact that if ( ) ytfxn = then ( ) tyfxn 1 =minus


10

Example 6 Find the values of t in ⎥⎦⎤

⎢⎣⎡

2π0 such that (a) cos t = 7826 (50) and

(b) csc t = 10219 (54) a Since the given interval is in radians we set the mode to RADIANS and type ( )7826cos 1minus asymp 6720

b Again the mode should be radians but to find t = ( )02191csc 1minus we must first say

csc t = 10219 implies 10219

1sin t = and 36341021911sin 1 asymp⎟

⎠⎞

⎜⎝⎛= minust

VIII Applying Circular Functions

Example 7 The temperature in Fairbanks Alaska is modeled by ( ) ( ) 25101x3652π37sinxT +⎥⎦

⎤⎢⎣⎡ minus=

where T(x) is the temperature in degrees Fahrenheit on day x with x = 1 corresponding to January 1 and x = 365 corresponding to December 31 Use the model to estimate the temperature on March 1 (74)

Since March 1 corresponds to day 60 ( ) ( ) F125101603652π37sin60T oasymp+⎥⎦

⎤⎢⎣⎡ minus=

Section 34 Linear and Angular Speed

I Arc Length θrs sdot= Alternate form tωrs sdotsdot=

For a circle with radius r and central angle θ measured in radians if θ intercepts an arc length of s then θrs sdot=

Hint For two pulleys connected by a belt or for two intermeshed gears 1s must equal 2s so 2211 θrθr sdot=sdot

II Linear Speed tsv = Alternate forms

tθrv sdot

= and ωrv sdot=

Consider a fly sitting on the tire of a bicycle If we measure the linear distance (s) the fly travels per unit of time (t) we are describing its linear speed (v) Linear speed is measures in units such as miles per hour and feet per second

III Angular Speed tθω = Alternate form

trsωsdot

=

If we measure the angle (θ) the fly travels through per unit of time (t) we are describing its angular speed (ω) Angular speed is measured in units such as radians per second and degrees per minute To convert revolutions per minute to an angular speed we multiply by 2π since there are 2π radians in one revolution Note The symbol for angular speed ω is the Greek letter omega


11

IV The Relationship between Linear Speed and Angular Speed ωrv sdot=

The linear speed (v) and the angular speed (ω) of a point moving in a circular path with radius r are related by the formula ωrv sdot=

Hint For two pulleys connected by a belt or for two intermeshed gears 1v must equal 2v so 2211 ωrωr sdot=sdot Example 1 Suppose that point P is on a circle with radius r and ray OP is rotating with angular

speed ω If r = 30 cm 10πω = radians per sec and t = 4 sec find the following (4)

a the angle generated by P in time t

tωθ tθω sdot=rarr= rarr

52π4

10πθ =sdot= radians

b the distance traveled by P along the circle in time t

tωrs sdotsdot= rarr cm 12π410π30s =sdotsdot=

c the linear speed of P

ωrv sdot= rarr 3π10π30v =sdot= cm per sec

Example 2 Given 8

3πθ = radians and 24πω = radians per min find t (10)

tθω = rarr

ω1θ

ωθt sdot== rarr 9

π24

83πt =sdot= min

Example 3 Given r = 8 cm and 5

9πω = radians per sec find v (14)

ωrv sdot= rarr 5

72π5

9π8v =sdot= cm per sec

Example 4 Given m5

12πs = m23r = and

52πω = find t (22)

tωrs sdotsdot= rarr ωr

stsdot

= rarr 43π5

512π

52π

23

512π

t =sdot=sdot

= sec

Example 5 Find ω for a line from the center to the edge of a CD revolving at 300 times per min (26)

=sdot=rev 1rad 2π

min 1rev 300ω 600π radians per min


12

Example 6 Find v for a point on the tread of a tire of radius 18 cm rotating 35 times per min (32)

rotation1radians 2π

min 1rotations 3518cmv sdotsdot= =1260π cm per min

Example 7 The earth revolves on its axis once every 24 hours Assuming that Earthrsquos radius is 6400 km find the following (38)

a angular speed of Earth in radians per day and radians per hour

2πrotation 1radians 2π

day 1 rotation 1ω =sdot= radians per day

12π

hours 24day 1

day 1radians 2πω =sdot= radians per hour

b linear speed at the North Pole or South Pole

Since r = 0 at the North or South Pole v = 0 at the North or South Pole c linear speed at Quito Ecuador a city on the equator

At the equator r = 6400 km so 12800π2π6400v =sdot= km per day

or 533π12π6400v asympsdot= km per hour

d linear speed at Salem Oregon (halfway from the equator to the North Pole)

Since Salem is halfway between the equator and the North Pole if we draw a radius from the center of the earth to Salem it forms a 45deg angle with the equator Thus

6400

r45sin =o rarr o45sin 6400r sdot=

km 4525483423200226400r asymp=sdot=

Therefore ==sdot= 26400π2π23200v 284344508 asymp 28000 km per day

or ==sdot=3

2800π12π23200v

1184768784 asymp 1200 km per hr


2

Example 2 Convert the following inputs from radian measure to degree measure

a 4

7π rads (23) b 3

4πminus radians c

65π radians

a oo

315π

1804

7π=sdot b o

o

240π

1803

4πminus=sdotminus c o

o

150π

1806

5π=sdot

C You should memorize the following diagram

D You should memorize the following table

Degree-Radian Conversion Factors





30deg = 6π1

45deg = 4π1 60deg =

3π1

90deg = 2π1

60deg = 6π2

90deg = 4π2 120deg =

3π2 180deg =

2π2

90deg = 6π3

135deg = 4π3 180deg =

3π3 270deg =

2π3

120deg = 6π4

180deg = 4π4 240deg =

3π4 360deg =

2π4


3

150deg = 6π5 225deg =

4π5 300deg =

3π5

180deg = 6π6 270deg =

4π6 360deg =

3π6

210deg = 6π7 315deg =

4π7

240deg = 6π8 360deg =

4π8

270deg = 6π9

300deg =6π10

330deg =6π11

360deg =6π12



π and simplify


a 174deg 50 (40) b 476925deg (45) a

b


180deg and simplify


a 207π (32) b 306 (50)

a b b


4



a 3

2πcot (66) b ⎟⎠⎞

⎜⎝⎛minus

65πsin c

415πsec

a 3


33

31

OA


b 6




⎠⎞

⎜⎝⎛minus

65πsin =

21

HO

67πsin minus==

c 4




415πsec = 2

AH

47πsec ==





5






given by θr21A 2=



6







7





ne

sec t = 0 xx1

ne csc t = 0y y1

ne cot t = 0y yx

ne


1715 θ cos minus=

1517 θ sec minus=

178 θsin =

817 θ csc =

158 θtan minus=

815 θcot minus=



t = and 2



⎭⎬⎫

⎩⎨⎧

sdot+ne2

π1)(2nt|t




Angle 0 or 2π 6π

4π

3π

2π π

23π

Point (1 0) ⎟⎟⎠

⎞⎜⎜⎝

⎛

21

23

⎟⎟⎠

⎞⎜⎜⎝

⎛

22

22

⎟⎟⎠

⎞⎜⎜⎝

⎛

23

21

(0 1) (-1 0) (0 -1)




8





11πcsc (11) (b) 4

3πcos (22) (c) 3

10πcot

a 6


the point ⎟⎟⎠

⎞⎜⎜⎝

⎛

21


23

611πcos =

21

611πsin minus=

33

31

23

21


minus=

3

326

11πsec = 26

11πcsc minus= 36

11πcot minus=

b 4



⎞⎜⎜⎝

⎛

22


22

43πcos minus=

22

43πsin = 1

22

22

43πtan minus=

minus=

24

3πsec minus= 24

3πcsc = 14

3πcot minus=


9

c 3


3

4π2π3


3π


⎟⎠

⎞⎜⎜⎝

⎛

23



21

34πcos minus=

23

34πsin minus= 3

21

23

34πtan =

minus

minus=

23

4πsec minus= 3

323

4πcsc minus= 33

31

34πcot ==



θcos1θ sec =

θsin 1θ csc = and

θtan 1θcot =


sec (-83429) = ( )83429 cos1

minus asymp -21291




323

2=

22

21

= and 33

31


(A S T C)


⎤⎢⎣⎡

23ππ





4π3ππ =+



10


⎢⎣⎡






⎠⎞

⎜⎝⎛= minust



⎤⎢⎣⎡ minus=



⎤⎢⎣⎡ minus=






tθrv sdot

= and ωrv sdot=



trsωsdot

=



11







52π4






Example 2 Given 8


tθω = rarr

ω1θ

ωθt sdot== rarr 9

π24

83πt =sdot= min



ωrv sdot= rarr 5

72π5


Example 4 Given m5

12πs = m23r = and

52πω = find t (22)


stsdot

= rarr 43π5

512π

52π

23

512π

t =sdot=sdot

= sec


=sdot=rev 1rad 2π



12








12π

hours 24day 1








6400


km 4525483423200226400r asymp=sdot=


or ==sdot=3

2800π12π23200v

1184768784 asymp 1200 km per hr


3

150deg = 6π5 225deg =

4π5 300deg =

3π5

180deg = 6π6 270deg =

4π6 360deg =

3π6

210deg = 6π7 315deg =

4π7

240deg = 6π8 360deg =

4π8

270deg = 6π9

300deg =6π10

330deg =6π11

360deg =6π12



π and simplify


a 174deg 50 (40) b 476925deg (45) a

b


180deg and simplify


a 207π (32) b 306 (50)

a b b


4



a 3

2πcot (66) b ⎟⎠⎞

⎜⎝⎛minus

65πsin c

415πsec

a 3


33

31

OA


b 6




⎠⎞

⎜⎝⎛minus

65πsin =

21

HO

67πsin minus==

c 4




415πsec = 2

AH

47πsec ==





5






given by θr21A 2=



6







7





ne

sec t = 0 xx1

ne csc t = 0y y1

ne cot t = 0y yx

ne


1715 θ cos minus=

1517 θ sec minus=

178 θsin =

817 θ csc =

158 θtan minus=

815 θcot minus=



t = and 2



⎭⎬⎫

⎩⎨⎧

sdot+ne2

π1)(2nt|t




Angle 0 or 2π 6π

4π

3π

2π π

23π

Point (1 0) ⎟⎟⎠

⎞⎜⎜⎝

⎛

21

23

⎟⎟⎠

⎞⎜⎜⎝

⎛

22

22

⎟⎟⎠

⎞⎜⎜⎝

⎛

23

21

(0 1) (-1 0) (0 -1)




8





11πcsc (11) (b) 4

3πcos (22) (c) 3

10πcot

a 6


the point ⎟⎟⎠

⎞⎜⎜⎝

⎛

21


23

611πcos =

21

611πsin minus=

33

31

23

21


minus=

3

326

11πsec = 26

11πcsc minus= 36

11πcot minus=

b 4



⎞⎜⎜⎝

⎛

22


22

43πcos minus=

22

43πsin = 1

22

22

43πtan minus=

minus=

24

3πsec minus= 24

3πcsc = 14

3πcot minus=


9

c 3


3

4π2π3


3π


⎟⎠

⎞⎜⎜⎝

⎛

23



21

34πcos minus=

23

34πsin minus= 3

21

23

34πtan =

minus

minus=

23

4πsec minus= 3

323

4πcsc minus= 33

31

34πcot ==



θcos1θ sec =

θsin 1θ csc = and

θtan 1θcot =


sec (-83429) = ( )83429 cos1

minus asymp -21291




323

2=

22

21

= and 33

31


(A S T C)


⎤⎢⎣⎡

23ππ





4π3ππ =+



10


⎢⎣⎡






⎠⎞

⎜⎝⎛= minust



⎤⎢⎣⎡ minus=



⎤⎢⎣⎡ minus=






tθrv sdot

= and ωrv sdot=



trsωsdot

=



11







52π4






Example 2 Given 8


tθω = rarr

ω1θ

ωθt sdot== rarr 9

π24

83πt =sdot= min



ωrv sdot= rarr 5

72π5


Example 4 Given m5

12πs = m23r = and

52πω = find t (22)


stsdot

= rarr 43π5

512π

52π

23

512π

t =sdot=sdot

= sec


=sdot=rev 1rad 2π



12








12π

hours 24day 1








6400


km 4525483423200226400r asymp=sdot=


or ==sdot=3

2800π12π23200v

1184768784 asymp 1200 km per hr


4



a 3

2πcot (66) b ⎟⎠⎞

⎜⎝⎛minus

65πsin c

415πsec

a 3


33

31

OA


b 6




⎠⎞

⎜⎝⎛minus

65πsin =

21

HO

67πsin minus==

c 4




415πsec = 2

AH

47πsec ==





5






given by θr21A 2=



6







7





ne

sec t = 0 xx1

ne csc t = 0y y1

ne cot t = 0y yx

ne


1715 θ cos minus=

1517 θ sec minus=

178 θsin =

817 θ csc =

158 θtan minus=

815 θcot minus=



t = and 2



⎭⎬⎫

⎩⎨⎧

sdot+ne2

π1)(2nt|t




Angle 0 or 2π 6π

4π

3π

2π π

23π

Point (1 0) ⎟⎟⎠

⎞⎜⎜⎝

⎛

21

23

⎟⎟⎠

⎞⎜⎜⎝

⎛

22

22

⎟⎟⎠

⎞⎜⎜⎝

⎛

23

21

(0 1) (-1 0) (0 -1)




8





11πcsc (11) (b) 4

3πcos (22) (c) 3

10πcot

a 6


the point ⎟⎟⎠

⎞⎜⎜⎝

⎛

21


23

611πcos =

21

611πsin minus=

33

31

23

21


minus=

3

326

11πsec = 26

11πcsc minus= 36

11πcot minus=

b 4



⎞⎜⎜⎝

⎛

22


22

43πcos minus=

22

43πsin = 1

22

22

43πtan minus=

minus=

24

3πsec minus= 24

3πcsc = 14

3πcot minus=


9

c 3


3

4π2π3


3π


⎟⎠

⎞⎜⎜⎝

⎛

23



21

34πcos minus=

23

34πsin minus= 3

21

23

34πtan =

minus

minus=

23

4πsec minus= 3

323

4πcsc minus= 33

31

34πcot ==



θcos1θ sec =

θsin 1θ csc = and

θtan 1θcot =


sec (-83429) = ( )83429 cos1

minus asymp -21291




323

2=

22

21

= and 33

31


(A S T C)


⎤⎢⎣⎡

23ππ





4π3ππ =+



10


⎢⎣⎡






⎠⎞

⎜⎝⎛= minust



⎤⎢⎣⎡ minus=



⎤⎢⎣⎡ minus=






tθrv sdot

= and ωrv sdot=



trsωsdot

=



11







52π4






Example 2 Given 8


tθω = rarr

ω1θ

ωθt sdot== rarr 9

π24

83πt =sdot= min



ωrv sdot= rarr 5

72π5


Example 4 Given m5

12πs = m23r = and

52πω = find t (22)


stsdot

= rarr 43π5

512π

52π

23

512π

t =sdot=sdot

= sec


=sdot=rev 1rad 2π



12








12π

hours 24day 1








6400


km 4525483423200226400r asymp=sdot=


or ==sdot=3

2800π12π23200v

1184768784 asymp 1200 km per hr


5






given by θr21A 2=



6







7





ne

sec t = 0 xx1

ne csc t = 0y y1

ne cot t = 0y yx

ne


1715 θ cos minus=

1517 θ sec minus=

178 θsin =

817 θ csc =

158 θtan minus=

815 θcot minus=



t = and 2



⎭⎬⎫

⎩⎨⎧

sdot+ne2

π1)(2nt|t




Angle 0 or 2π 6π

4π

3π

2π π

23π

Point (1 0) ⎟⎟⎠

⎞⎜⎜⎝

⎛

21

23

⎟⎟⎠

⎞⎜⎜⎝

⎛

22

22

⎟⎟⎠

⎞⎜⎜⎝

⎛

23

21

(0 1) (-1 0) (0 -1)




8





11πcsc (11) (b) 4

3πcos (22) (c) 3

10πcot

a 6


the point ⎟⎟⎠

⎞⎜⎜⎝

⎛

21


23

611πcos =

21

611πsin minus=

33

31

23

21


minus=

3

326

11πsec = 26

11πcsc minus= 36

11πcot minus=

b 4



⎞⎜⎜⎝

⎛

22


22

43πcos minus=

22

43πsin = 1

22

22

43πtan minus=

minus=

24

3πsec minus= 24

3πcsc = 14

3πcot minus=


9

c 3


3

4π2π3


3π


⎟⎠

⎞⎜⎜⎝

⎛

23



21

34πcos minus=

23

34πsin minus= 3

21

23

34πtan =

minus

minus=

23

4πsec minus= 3

323

4πcsc minus= 33

31

34πcot ==



θcos1θ sec =

θsin 1θ csc = and

θtan 1θcot =


sec (-83429) = ( )83429 cos1

minus asymp -21291




323

2=

22

21

= and 33

31


(A S T C)


⎤⎢⎣⎡

23ππ





4π3ππ =+



10


⎢⎣⎡






⎠⎞

⎜⎝⎛= minust



⎤⎢⎣⎡ minus=



⎤⎢⎣⎡ minus=






tθrv sdot

= and ωrv sdot=



trsωsdot

=



11







52π4






Example 2 Given 8


tθω = rarr

ω1θ

ωθt sdot== rarr 9

π24

83πt =sdot= min



ωrv sdot= rarr 5

72π5


Example 4 Given m5

12πs = m23r = and

52πω = find t (22)


stsdot

= rarr 43π5

512π

52π

23

512π

t =sdot=sdot

= sec


=sdot=rev 1rad 2π



12








12π

hours 24day 1








6400


km 4525483423200226400r asymp=sdot=


or ==sdot=3

2800π12π23200v

1184768784 asymp 1200 km per hr


6







7





ne

sec t = 0 xx1

ne csc t = 0y y1

ne cot t = 0y yx

ne


1715 θ cos minus=

1517 θ sec minus=

178 θsin =

817 θ csc =

158 θtan minus=

815 θcot minus=



t = and 2



⎭⎬⎫

⎩⎨⎧

sdot+ne2

π1)(2nt|t




Angle 0 or 2π 6π

4π

3π

2π π

23π

Point (1 0) ⎟⎟⎠

⎞⎜⎜⎝

⎛

21

23

⎟⎟⎠

⎞⎜⎜⎝

⎛

22

22

⎟⎟⎠

⎞⎜⎜⎝

⎛

23

21

(0 1) (-1 0) (0 -1)




8





11πcsc (11) (b) 4

3πcos (22) (c) 3

10πcot

a 6


the point ⎟⎟⎠

⎞⎜⎜⎝

⎛

21


23

611πcos =

21

611πsin minus=

33

31

23

21


minus=

3

326

11πsec = 26

11πcsc minus= 36

11πcot minus=

b 4



⎞⎜⎜⎝

⎛

22


22

43πcos minus=

22

43πsin = 1

22

22

43πtan minus=

minus=

24

3πsec minus= 24

3πcsc = 14

3πcot minus=


9

c 3


3

4π2π3


3π


⎟⎠

⎞⎜⎜⎝

⎛

23



21

34πcos minus=

23

34πsin minus= 3

21

23

34πtan =

minus

minus=

23

4πsec minus= 3

323

4πcsc minus= 33

31

34πcot ==



θcos1θ sec =

θsin 1θ csc = and

θtan 1θcot =


sec (-83429) = ( )83429 cos1

minus asymp -21291




323

2=

22

21

= and 33

31


(A S T C)


⎤⎢⎣⎡

23ππ





4π3ππ =+



10


⎢⎣⎡






⎠⎞

⎜⎝⎛= minust



⎤⎢⎣⎡ minus=



⎤⎢⎣⎡ minus=






tθrv sdot

= and ωrv sdot=



trsωsdot

=



11







52π4






Example 2 Given 8


tθω = rarr

ω1θ

ωθt sdot== rarr 9

π24

83πt =sdot= min



ωrv sdot= rarr 5

72π5


Example 4 Given m5

12πs = m23r = and

52πω = find t (22)


stsdot

= rarr 43π5

512π

52π

23

512π

t =sdot=sdot

= sec


=sdot=rev 1rad 2π



12








12π

hours 24day 1








6400


km 4525483423200226400r asymp=sdot=


or ==sdot=3

2800π12π23200v

1184768784 asymp 1200 km per hr


7





ne

sec t = 0 xx1

ne csc t = 0y y1

ne cot t = 0y yx

ne


1715 θ cos minus=

1517 θ sec minus=

178 θsin =

817 θ csc =

158 θtan minus=

815 θcot minus=



t = and 2



⎭⎬⎫

⎩⎨⎧

sdot+ne2

π1)(2nt|t




Angle 0 or 2π 6π

4π

3π

2π π

23π

Point (1 0) ⎟⎟⎠

⎞⎜⎜⎝

⎛

21

23

⎟⎟⎠

⎞⎜⎜⎝

⎛

22

22

⎟⎟⎠

⎞⎜⎜⎝

⎛

23

21

(0 1) (-1 0) (0 -1)




8





11πcsc (11) (b) 4

3πcos (22) (c) 3

10πcot

a 6


the point ⎟⎟⎠

⎞⎜⎜⎝

⎛

21


23

611πcos =

21

611πsin minus=

33

31

23

21


minus=

3

326

11πsec = 26

11πcsc minus= 36

11πcot minus=

b 4



⎞⎜⎜⎝

⎛

22


22

43πcos minus=

22

43πsin = 1

22

22

43πtan minus=

minus=

24

3πsec minus= 24

3πcsc = 14

3πcot minus=


9

c 3


3

4π2π3


3π


⎟⎠

⎞⎜⎜⎝

⎛

23



21

34πcos minus=

23

34πsin minus= 3

21

23

34πtan =

minus

minus=

23

4πsec minus= 3

323

4πcsc minus= 33

31

34πcot ==



θcos1θ sec =

θsin 1θ csc = and

θtan 1θcot =


sec (-83429) = ( )83429 cos1

minus asymp -21291




323

2=

22

21

= and 33

31


(A S T C)


⎤⎢⎣⎡

23ππ





4π3ππ =+



10


⎢⎣⎡






⎠⎞

⎜⎝⎛= minust



⎤⎢⎣⎡ minus=



⎤⎢⎣⎡ minus=






tθrv sdot

= and ωrv sdot=



trsωsdot

=



11







52π4






Example 2 Given 8


tθω = rarr

ω1θ

ωθt sdot== rarr 9

π24

83πt =sdot= min



ωrv sdot= rarr 5

72π5


Example 4 Given m5

12πs = m23r = and

52πω = find t (22)


stsdot

= rarr 43π5

512π

52π

23

512π

t =sdot=sdot

= sec


=sdot=rev 1rad 2π



12








12π

hours 24day 1








6400


km 4525483423200226400r asymp=sdot=


or ==sdot=3

2800π12π23200v

1184768784 asymp 1200 km per hr


8





11πcsc (11) (b) 4

3πcos (22) (c) 3

10πcot

a 6


the point ⎟⎟⎠

⎞⎜⎜⎝

⎛

21


23

611πcos =

21

611πsin minus=

33

31

23

21


minus=

3

326

11πsec = 26

11πcsc minus= 36

11πcot minus=

b 4



⎞⎜⎜⎝

⎛

22


22

43πcos minus=

22

43πsin = 1

22

22

43πtan minus=

minus=

24

3πsec minus= 24

3πcsc = 14

3πcot minus=


9

c 3


3

4π2π3


3π


⎟⎠

⎞⎜⎜⎝

⎛

23



21

34πcos minus=

23

34πsin minus= 3

21

23

34πtan =

minus

minus=

23

4πsec minus= 3

323

4πcsc minus= 33

31

34πcot ==



θcos1θ sec =

θsin 1θ csc = and

θtan 1θcot =


sec (-83429) = ( )83429 cos1

minus asymp -21291




323

2=

22

21

= and 33

31


(A S T C)


⎤⎢⎣⎡

23ππ





4π3ππ =+



10


⎢⎣⎡






⎠⎞

⎜⎝⎛= minust



⎤⎢⎣⎡ minus=



⎤⎢⎣⎡ minus=






tθrv sdot

= and ωrv sdot=



trsωsdot

=



11







52π4






Example 2 Given 8


tθω = rarr

ω1θ

ωθt sdot== rarr 9

π24

83πt =sdot= min



ωrv sdot= rarr 5

72π5


Example 4 Given m5

12πs = m23r = and

52πω = find t (22)


stsdot

= rarr 43π5

512π

52π

23

512π

t =sdot=sdot

= sec


=sdot=rev 1rad 2π



12








12π

hours 24day 1








6400


km 4525483423200226400r asymp=sdot=


or ==sdot=3

2800π12π23200v

1184768784 asymp 1200 km per hr


9

c 3


3

4π2π3


3π


⎟⎠

⎞⎜⎜⎝

⎛

23



21

34πcos minus=

23

34πsin minus= 3

21

23

34πtan =

minus

minus=

23

4πsec minus= 3

323

4πcsc minus= 33

31

34πcot ==



θcos1θ sec =

θsin 1θ csc = and

θtan 1θcot =


sec (-83429) = ( )83429 cos1

minus asymp -21291




323

2=

22

21

= and 33

31


(A S T C)


⎤⎢⎣⎡

23ππ





4π3ππ =+



10


⎢⎣⎡






⎠⎞

⎜⎝⎛= minust



⎤⎢⎣⎡ minus=



⎤⎢⎣⎡ minus=






tθrv sdot

= and ωrv sdot=



trsωsdot

=



11







52π4






Example 2 Given 8


tθω = rarr

ω1θ

ωθt sdot== rarr 9

π24

83πt =sdot= min



ωrv sdot= rarr 5

72π5


Example 4 Given m5

12πs = m23r = and

52πω = find t (22)


stsdot

= rarr 43π5

512π

52π

23

512π

t =sdot=sdot

= sec


=sdot=rev 1rad 2π



12








12π

hours 24day 1








6400


km 4525483423200226400r asymp=sdot=


or ==sdot=3

2800π12π23200v

1184768784 asymp 1200 km per hr


10


⎢⎣⎡






⎠⎞

⎜⎝⎛= minust



⎤⎢⎣⎡ minus=



⎤⎢⎣⎡ minus=






tθrv sdot

= and ωrv sdot=



trsωsdot

=



11







52π4






Example 2 Given 8


tθω = rarr

ω1θ

ωθt sdot== rarr 9

π24

83πt =sdot= min



ωrv sdot= rarr 5

72π5


Example 4 Given m5

12πs = m23r = and

52πω = find t (22)


stsdot

= rarr 43π5

512π

52π

23

512π

t =sdot=sdot

= sec


=sdot=rev 1rad 2π



12








12π

hours 24day 1








6400


km 4525483423200226400r asymp=sdot=


or ==sdot=3

2800π12π23200v

1184768784 asymp 1200 km per hr


11







52π4






Example 2 Given 8


tθω = rarr

ω1θ

ωθt sdot== rarr 9

π24

83πt =sdot= min



ωrv sdot= rarr 5

72π5


Example 4 Given m5

12πs = m23r = and

52πω = find t (22)


stsdot

= rarr 43π5

512π

52π

23

512π

t =sdot=sdot

= sec


=sdot=rev 1rad 2π



12








12π

hours 24day 1








6400


km 4525483423200226400r asymp=sdot=


or ==sdot=3

2800π12π23200v

1184768784 asymp 1200 km per hr


12








12π

hours 24day 1








6400


km 4525483423200226400r asymp=sdot=


or ==sdot=3

2800π12π23200v

1184768784 asymp 1200 km per hr

Trigonometry Chapter 3 Lecture Notes Section 3.1 …faculty.nwacc.edu/tobrien/Trig Resources...

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Transcript of Trigonometry Chapter 3 Lecture Notes Section 3.1 …faculty.nwacc.edu/tobrien/Trig Resources...

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