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TRIGONOMETRY
CHAPTER 3: RADIAN MEASURE AND THE CIRCULAR FUNCTIONS
3.1 RADIAN MEASURE
 RADIAN:
o The figure below shows an angle  in standard position along
with a circle of radius r. The vertex of  is at the center of
the circle. Angle  intercepts an arc on the circle equal in
length to the radius of the circle. Therefore, angle  is said to
have a measure of 1 radian.
r
r

r
 CONVERTING BETWEEN DEGREES AND RADIANS
o The circumference of a circle (the distance around a circle) is
given by the formula C  2 r , where r is the radius of the
circle. This tells us that there are 2 radians around a circle.
Consider an angle of 360°. This is a complete circle, so 360°
intercepts an arc on the circle equal in length to 2 times the
radius of the circle. Therefore, we have 360  2 radians. This
gives us the following relationship:
180 
1
1
360    2  radians   radians

2
2
1
So we have,
180   radians
1 

180
radian
1 radian 

180

Example: Convert 45° to radians
 

45  45 
radian 
 180



4
radian
5
radian to degrees
6
5
5  180 
radian 


6
6   
 150

Example: Convert

Example: Find cot
cot
3
4
3
3

 cot  180 
4
4

 cot 135

 1
2
 EQUIVALENT ANGLE MEASURES IN DEGREES AND RADIANS

0° or 0 radians

30° or
radians
6

45° or
radians
4

60° or
radians
3

90° or
radians
2
2
120° or
radians
3
3
135° or
radians
4
5
150° or
radians
6
180° or  radians
7
210° or
radians
6
5
225°or
radians
4
4
240° or
radians
3
3
270° or
radians
2
5
300° or
radians
3
7
315° or
radians
4
11
330° or
radians
6
360° or 2 radians
sin 
0
cos 
1
tan 
0
1
2
3
2
3
3
cot 
Undefined
3
2
2
2
2
1
1
3
2
1
2
1
0
Undefined
3
2
 12
 3

2
2
1
2

3
2
-1

3
3
csc 
Undefined
2
2 3
3
2
2
3
3
2
2 3
3
0
Undefined
1
-2
2 3
3
3
2
2
sec 
1

3
3
-1
 2
2
 3
 2 33
2
0
 12
-1
 23
0
3
3
Undefined
3
-1
 2 33
Undefined
-2

2
2

1
1
 2
 2

3
2
3
3
-2
 2 33
0
Undefined
-1
2
 2 33
2
 2
-1
2
2
 12
3
0
Undefined

3
2
1
2
 3

2
2
2
2
-1
 12
3
2
0
1

3
3
0

3
3
-1
 3
2 3
3
-2
Undefined
1
Undefined
3
3.2
APPLICATIONS OF RADIAN MEASURE
 ARC LENGTH OF A CIRCLE: The length of an arc is proportional to
the measure of its central angle.
T
Q
s
r

R
radians
1 radian
O
r
P
r
s 

r 1
o Arc Length: The length s of the arc intercepted on a circle of
radius r by a central angle of measure  radians is given by the
product of the radius and the radian measure of the angle, or
s  r ,  in radians.

Example: Find the radius of a circle in which a central angle
of 2 radians intercepts an arc of length 3 feet.
s
s  r  r 

3
feet
2
o Area of a Sector: The area of a sector of a circle of radius r and
central angle  is given by
r
A
1 2
r  ,  in radians.
2
4

Example: Find the area of a sector of a circle having a
radius of 18.3 m and a central angle of 125°.
First you have to change the angle to radian measure. So we
   25
radians . Now we can solve the problem
have 125 

 180  36
using the area for a sector of a circle.
1 2
r
2
1
 25 
 18.32  

2
 36 
365 m 2
A
3.3
CIRCULAR FUNCTIONS OF REAL NUMBERS
 THE CIRCULAR FUNCTIONS
o The radius (r) is always 1. Therefore, from our earlier equation
(derived from the distance formula) we have 1  x 2  y 2 .
o Trigonometric functions for the circular functions
 Recall that that the radian measure of  is related to the
arc length s. Here, r = 1, so s = r  gives us s =  .
y
y
x
tan s  , x  0
sin s   y
cos s   x
x
1
1

1
x
1
csc s  , y  0
cot s  , y  0
sec s  , x  0
y
y
x
(0, 1)
(-1, 0)
(1, 0)
(0, -1)
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
Since sin s = y and cos s = x, we can replace x and y in the
equation x 2  y 2  1 and obtain the identity cos 2 s  sin 2 s  1 .
o Domain and Range of the Circular Functions
 Since the ordered pair (x, y) represents a point on the unit
circle,
1  x  1
and
 1  y  1,



so  1  cos s  1
and
 1  cos s  1.
The domain of cos s and sin s is the set of all real
numbers, since cos s and sin s exist for any value of s
The domains of tan s and sec s are restricted since x

has a value of 0 at
plus any multiple of  , positive
2
or negative. So we write the domains for tan s and



sec s as follows:  s | s   n , n   .
2


The domains of cot s and csc s are restricted since y
has a value of 0 at any multiple of  , positive or
negative. So we write the domains for cot s and csc s
as follows: s | s  n , n   .
 The Unit Circle: The figure below shows the relationship between
degree and radian measure. This is something you need to memorize!
It will make the rest of the course much easier.
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3.4
LINEAR AND ANGULAR VELOCITY
 Linear Velocity
o In many situations we need to know how fast a point on a circle
is moving or how fast the central angle is changing. Suppose
that point P moves at a constant speed along a circle of radius r
and center O. The measure of how fast the position of P is
changing is called linear velocity. If v represents velocity, then
we have
distance
time
s
v
t
velocity 
where s is the length of the arc traced by point P at time t.
This formula is just d  rt with s as the distance, v as the rate,
and t is still time.
P
s
O
r
B
 Angular Velocity
o Consider the figure above. As point P moves along the circle,
OP rotates around the origin. Since OP is the terminal side of
POB , the measure of the angle changes as P moves along the
7
circle. The measure of how fast POB is changing is called
angular velocity. Angular velocity is denoted by  is given as

  ,  in radians,
t
where  is the measure of POB at time t.  is expressed as
radians per unit of time.
 Recall from section 3.2 that the arc length of a circle
was measured by s  r ,  in radians . Now that we know
the meaning of  , we have an alternate formula for
linear velocity.
v
s r


 r   r .
t
t
t
This formula relates linear and angular velocities.

Note that a radian is a “pure number”, with no units
associated with it. This is why you get some length
per unit of time with no extra unit like degrees.
o Example: Radius of a gear. A gear is driven by a chain that
travels 1.46 m/s. Find the radius of the gear if it
makes 46 rev/minute.
Solution: v = 1.46 m/s. One complete revolution is 2  radians,
46  2  23
46

and 46 seconds is
minute, so  
60
60
15
radians per second. Now we have only one unknown in
the formula for linear velocity v  r . Substituting
for v and  we have
1.46
 23 
v  r  1.46  r 
  r  23  0.303 meters.
 15 
15
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