Ch 3 Notes F07 O’Brien
LHS Trig 8th ed
Trigonometry
Chapter 3 Lecture Notes
Section 3.1
I.
Radian Measure
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.
C.
II.
The radian measure of an angle (θ) is the ratio
of the arc length (s) to the radius of the circle (r),
i.e. θ  s
r
If the arc length subtended by angle θ is equal to
the radius (i.e., 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.
Converting Standard Angles Between Degrees and Radians
A.
One full rotation measures 2π or approximately 6.28 radians. Thus 2π radians = 360°;
π
π
π
π
π radians = 180°; radians  90 ; radians  60 ; radians  45 ; and radians  30 .
2
3
4
6
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
1π
of our four special angles - 30°, 45°, 60°, and 90°. For example, if 30 
, then
6
2π
3π
1π
2π
3π
60 
, 90 
, etc. If 45 
, then 90 
; 135 
; and so on.
6
6
4
4
4
Example 1:
Convert the following angles from degree measure to radian measure.
a. 120° (#8)
c. 570°
π
5π

4
180
π
7π
c. 570° - 360° = 210°  210 

6
180
a. 120  
π
2π

3
180 
b. 225°
b. 225 
1
Ch 3 Notes F07 O’Brien
LHS Trig 8th ed
Example 2:
C.
D.
Convert the following inputs from radian measure to degree measure.
a.
7π
rads (#23)
4
b. 
4π
radians
3
c.
5π
radians
6
a.
7π 180 

 315 
4
π
b. 
4π 180 

 240 
3
π
c.
5π 180 

 150 
6
π
You should memorize the following diagram.
You should memorize the following table.
Degree-Radian Conversion Factors
Multiples of 30°&
30° =
60° =
90° =
120° =
1π
6
2π
6
3π
6
4π
6
π
6
Multiples of 45°&
1π
4
2π
90° =
4
3π
135° =
4
4π
180° =
4
45° =
π
4
Multiples of 60°&
60° =
1π
3
2π
120° =
3
3π
180° =
3
4π
240° =
3
π
3
Multiples of 90°&
90° =
π
2
1π
2
2π
180° =
2
3π
270° =
2
4π
360° =
2
2
Ch 3 Notes F07 O’Brien
LHS Trig 8th ed
5π
6
6π
180° =
6
7π
210° =
6
8π
240° =
6
9π
270° =
6
10 π
300° =
6
11π
330° =
6
12 π
360° =
6
5π
4
6π
270° =
4
7π
315° =
4
8π
360° =
4
150° =
III.
300° =
Converting Nonstandard Angles Between Degrees and Radians
A.
To convert a non-standard angle from degrees to radians, multiply by
Example 3
a.
5π
3
6π
360° =
3
225° =
π
and simplify.
180
Convert each nonstandard angle from degree measure to radians.
174° 50' (#40)
b. 47.6925° (#45)
a.
b.
B.
To convert a nonstandard angle from radians to degrees, multiply by
Example 4
a.
a.
180
and simplify.
π
Convert each radian measure to degrees. In part b, round to the nearest minute.
7π
(#32)
20
b. 3.06 (#50)
b.
b.
3
Ch 3 Notes F07 O’Brien
LHS Trig 8th ed
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.
cot
c.
 5π 
b. sin   
 6 
c. sec
15π
4
2π
π
terminates in Quadrant II and has a reference angle of .
3
3
2π A
1
3
cot
 

3 O
3
3
a.
b.
2π
(#66)
3
5π
7π
is coterminal with
which terminates in Quadrant III.
6
6
π
7π O
1
 5π 
 
It has a reference angle of . sin    = sin
6
6
H
2
 6 

15π
7π
is coterminal with
which terminates in quadrant IV.
4
4
π
15π
7π H
  2
It has a reference angle of . sec
= sec
4
4
4
A
************************************************************************************
Section 3.2
I.
Applications of Radian Measure
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, i.e., s = r∙θ.
Caution: θ must be in radians to use this formula.
Example 1
Find the length of the arc intercepted by a central angle θ  135  in a circle of radius
r = 71.9 cm. (#10)
4
Ch 3 Notes F07 O’Brien
LHS Trig 8th ed
Example 2
Find the distance in kilometers between Farmersville, California, 36° N, and
Penticton, British Columbia, 49° 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 80.0° rotation of the smaller gear
causes the larger gear to rotate 50.0°. Find the radius of the larger gear if the smaller
gear has a radius of 11.7 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
1
given by A  r 2 θ .
2
Caution: θ must be in radians to use this formula.
5
Ch 3 Notes F07 O’Brien
LHS Trig 8th ed
Example 4
Find the area of a sector of a circle with radius 18.3 cm and central angle θ = 125°.
Round to the nearest tenth. (#36)
Example 5
The Ford Model A, built from 1928 to 1931, had a single windshield wiper on the
driver’s side. The total arm and blade was 10 inches long and rotated back and
forth through an angle of 95°. If the wiper blade was 7 inches, how many square
inches of the windshield did the blade clean? (#42)
************************************************************************************
Section 3.3
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 x 2  y 2  1 ).
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 = r·θ), t = θ.
6
Ch 3 Notes F07 O’Brien
LHS Trig 8th ed
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
y
cos t = x
sin t = y
tan t = , x  0
x
1
1
x
sec t = , x  0
csc t = , y  0
cot t = , y  0
x
y
y
Use the diagram below to evaluate the six circular values of θ. (#46)
Example 1:
cos θ  
17
15
17
csc θ 
8
15
cot θ  
8
15
17
sec θ  
8
17
8
tan θ  
15
sin θ 
IV. Domains of the Circular Functions
Since x = 0 when
t
π
2
and
t
3π
2
, sec t and tan t, which both have x in the denominator, are not
defined there or at any odd multiple of
π
2
π
. Thus the domain of sec t and tan t is t | t  (2n  1)   .

2
Since y = 0 at t  0 , t  π , and t  2π , 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 t | t  n  π.
Sine and cosine do not have any restrictions on their domain, thus their domain is  ,   .
V.
Finding Exact Values of Circular Functions by Using Special Points on the Unit Circle
Angle
0 or 2π
π
6
π
4
π
3
π
2
π
3π
2
Point
(1, 0)
 3 1

, 
 2 2


 2
2 

,
 2
2 

1 3
 ,

2 2 


(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:
Quadrant II:
Quadrant III:
Quadrant IV:
x and y are both positive, therefore all function values are positive.
x is negative and y is positive, therefore only sine and cosecant are positive.
x and y are both negative, therefore only tangent and cotangent are positive.
x is positive and y is negative, therefore only cosine and secant are positive.
7
Ch 3 Notes F07 O’Brien
LHS Trig 8th ed
Example 2:
π
2
Find the exact value of (a) cos θ, (b) sin θ, and (c) tan θ for θ  . (#1)
The point associated with
π
is (0, 1). Therefore, cos θ = 0; sin θ = 1;
2
tan θ is undefined; cot θ = 0; csc θ = 1; sec θ is undefined.
Example 3:
3π
10π
(#22); (c) cot
.
3
4
π
11π
a.
terminates in quadrant IV. Its reference angle is which is associated with
6
6
 3 1
the point  ,  . In quadrant IV, only cosine and secant are positive.
 2 2
1
11π
1
11π  2
1
3
11π
3
sin
 ;
tan



;
;
cos

3
6
2
6
3
6
2
3
2
11π
11π
11π 2 3
 2 ;
cot
 3
; csc
sec

6
6
6
3
Find the exact value of (a) csc
b.
11π
6
(#11); (b) cos
3π
π
terminates in quadrant II. It has a reference angle of which is associated
4
4
with the point
3π
2
;
cos

4
2
sec
3π
 2;
4
 2
2  .

,
 2
2 

In quadrant II, only sine and cosecant are positive.
3π
2
;
sin

4
2
csc
3π
 2;
4
2
3π
tan
 2
4
cot
2

2
 1
3π
 1
4
8
Ch 3 Notes F07 O’Brien
LHS Trig 8th ed
c.
10π
is larger
3
10π
4π
 2π 
3
3
than 2π, so we must first find its smallest positive coterminal angle.
which terminates in quadrant III and has a reference angle of
The point associated with
π
3
π
3
.
1 3
is  ,  and in quadrant III only tangent and
2
2 
cotangent are positive.
4π
1
cos

3
2
sec
4π
 2 ;
3
;
4π
3
;
sin

3
2
csc
4π
2 3
;

3
3
3
4π  2
tan

 3
1
3

2
cot
4π
1
3


3
3
3
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 sec θ 
Example 4:
1
1
1
, csc θ 
, and cot θ 
.
cos θ
sin θ
tan θ
Find the approximate value of sec (-8.3429). (#32)
sec (-8.3429) =
1
≈ -2.1291
cos  8.3429 
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,
1
2

2
2
; and
1
3

2
3

2 3
;
3
3
. And we must know which functions are positive in each quadrant
3
(A S T C).
Example 5
Find the exact value of t if sin t  
1
3π
in the interval π,  .
2
 2
Do not use a calculator (# 58).
1
π
indicates the reference angle, t  , is and since the given interval
2
3
π 4π
is quadrant III, t must be π   .
3 3
A sine value of
B.
To determine, with a calculator, what input generates a specified output, we use an inverse
trig function and the fact that if fxn t   y ; then fxn 1y  t .
9
Ch 3 Notes F07 O’Brien
LHS Trig 8th ed
π
Find the values of t in 0,  such that (a) cos t = .7826. (#50) and
Example 6

2
(b) csc t = 1.0219 (#54).
a. Since the given interval is in radians, we set the mode to RADIANS and type
cos1 .7826  ≈ .6720.
b. Again, the mode should be radians, but to find t = csc11.0219  we must first say
csc t = 1.0219 implies sin t 
1
1 
and t  sin 1
  1.3634 .
1.0219
 1.0219 
VIII. Applying Circular Functions
The temperature in Fairbanks, Alaska is modeled by Tx   37sin 
2π
x  101  25
 365

Example 7
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, T60   37sin 
2π
60  101   25  1 F .
365


************************************************************************************
Section 3.4
Linear and Angular Speed
I.
Arc Length
s  r θ
Alternate form: s  r  ω  t
For a circle with radius r and central angle θ measured in radians, if θ intercepts an arc length of s,
then s  r  θ .
Hint:
II.
For two pulleys connected by a belt or for two intermeshed gears, s1 must equal s 2 , so
r1  θ1  r2  θ 2 .
Linear Speed
v
s
t
Alternate forms: v 
r θ
and v  r  ω
t
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
ω
θ
t
Alternate form: ω 
s
rt
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.
10
Ch 3 Notes F07 O’Brien
LHS Trig 8th ed
IV. The Relationship between Linear Speed and Angular Speed
v  rω
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 v  r  ω .
Hint:
For two pulleys connected by a belt or for two intermeshed gears, v1 must equal v 2 ,
so r1  ω1  r2  ω2 .
Example 1
a.
Suppose that point P is on a circle with radius r, and ray OP is rotating with angular
π
speed ω. If r = 30 cm, ω 
radians per sec, and t = 4 sec, find the following. (#4)
10
the angle generated by P in time t
ω
b.
θ
 θ  ωt
t
 s  30 
π
 3π cm per sec
10
3π
π
radians and ω 
radians per min, find t. (#10)
8
24
θ
1
3π 24

 9 min
 t   θ
 t
ω
ω
8 π
Given θ 
ω
Example 3
π
 4  12π cm
10
the linear speed of P
v  r  ω  v  30 
Example 2
π
2π
4 
radians
10
5
the distance traveled by P along the circle in time t
s  r ωt
c.
 θ
θ
t
9π
radians per sec, find v. (#14)
5
9π 72π

cm per sec
 v  8
5
5
Given r = 8 cm and ω 
v  rω
12π
3
2π
m , r  m , and ω 
, find t. (#22)
5
2
5
12π
s
5  12π  5  4 sec
s  r ωt  t 
 t
3 2π
rω
5 3π

2 5
Example 4
Given s 
Example 5
Find ω for a line from the center to the edge of a CD revolving at 300
times per min. (#26)
ω
300 rev 2π rad

 600π radians per min
1 min
1 rev
11
Ch 3 Notes F07 O’Brien
LHS Trig 8th ed
Example 6
Find v for a point on the tread of a tire of radius 18 cm, rotating 35
times per min. (#32)
v  18cm 
Example 7
a.
35 rotations 2π radians

=1260π cm per min
1 min
1 rotation
The earth revolves on its axis once every 24 hours. Assuming that Earth’s radius
is 6400 km, find the following. (#38)
angular speed of Earth in radians per day and radians per hour
1 rotation 2π radians

 2π radians per day
1 day
1 rotation
2π radians
1 day
π
radians per hour
ω


1 day
24 hours 12
ω
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 v  6400  2π  12800π km per day
π
or v  6400   533π km per hour
12
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 45° angle with the
equator. Thus
r
sin 45 
 r  6400  sin 45
6400
2
r  6400 
 3200 2  4525.4834 km
2
Therefore, v  3200 2  2π  6400π 2 
28434.4508 ≈ 28000 km per day
π 800π 2

or v  3200 2  
12
3
1184.768784 ≈ 1200 km per hr
12