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Point P(x, y) is the point on the terminal arm of angle  ,an angle in
standard position, that intersects a circle.
y
sin  
r
x
cos  
r
y
tan  
x
P(x, y)
r2 = x2 + y2
r=
√x2
+
r
y2
y

x
csc 
sec 
cot 
r
y
r
x
x
y
The three reciprocal ratios are defined as follows:
cosecant 
1
sine
secant 
1
cosine
Math 30-1
cotangent 
1
tangent
1
Finding the Trig Ratios of an Angle in Standard Position
The point P(-2, 3) is on the terminal arm of  in standard position.
Does point P(-2, 3) lie on the unit circle?
Determine the exact value of the
six trigonometric ratios for angle .
P(-2, 3)
13
3
-2
r2 = x2 + y2
r2 = (-2)2 + (3)2
r2 = 4 + 9
r2 = 13
r = √ 13
No, the radius of a unit circle is 1.

3
sin  
13
13
csc  
3
2
cos   
13
13
sec   
2
3
tan   
2
Math 30-1
2
cot   
3
2
P( )   x, y 
is a point on the terminal arm of angle θ that intersects the unit circle
x
cos  
cos  x
r
y
sin   y
sin  
r
P( )   cos ,sin  
y
tan  
x
Math 30-1
3
 5 12 
,  lies at the intersection of the unit circle
 13 13 
The point A  
and the terminal arm of an angle θ in standard position.
Diagram
sin   y
 5 12 
A  , 
 13 13 
12
sin  
13
13
csc 
12
13
5
sec  
cos  x cos   
13
5

r=1
tan  
Math 30-1
5
y
12
tan   
cot  
x
5
12
4
The Unit Circle
 1 3
 2, 2 


 1 1 
,


2 2

 3 1
 2 , 2 


(0, 1)
120   2 90
3


2
135  3
4

 1 1 
 ,

2
2


60   
3
45  
30   
150   5
6
6
0  2  360 
210  7 
6
3
2
4

240  4
 1
3
  2 , 2 


3
6
315  7 
4
300   5
(0, -1)
(1, 0)
330   11
270 
225   5
 1

, 1 

2
2

 3 1
 2 ,2


4
(-1, 0) 180   
 3 1
  2 , 2 


1 3
2, 2 


3
1
3
 2 , 2 


Math 30-1
 3 1
 2 , 2 


Exact Values for
the trigonometric
ratios can be
determined using
multiples of
special angles for
points P(θ) on the
unit circle.
P( )   x, y 
P( )   cos ,sin  
 1

, 1 

2
 2
5
Exact Values For Trigonometric Ratios
1. sin
3300
RA = 300
quadrant IV
4
3. tan

3

RA =
quadrant III 3
= 1
2
3
3
5
2. cos
 
6
2

RA =
quadrant II 6
  1
4. cos   =
 3 2

RA =
quadrant IV 3
Math 30-1
6
Using the unit circle
Determine the exact value of:
5
3
sin

3
2
3
cos 210  
2
0
4
sec
3
 2
cot 135  1
0
1
 2 
cos  
  2
 3 
3
tan 690  
3
0
csc3  undefined
 
tan      3
 3
Math 30-1
7
Approximate Values for Trig Ratios (four decimal places)
The mode of calculator must match the domain of the angle,
degrees or radians.
1. sin 250 = 0.4226
2. cos 3750 = 0.9659
 
3. tan     0.5773
 6
4. csc 1.57 =
1
   1.1547
5. sec    
 6
6.
cot 2700=
0
Math 30-1
8
 5 12 
P , 
 13 13 
2
 5   12 
   
 13   13 
Does point P lie on the unit circle? Yes
2

25 144

169 169
1
If Point P is the point on the terminal arm of angle  that intersects
the unit circle, in which quadrant does P lie? III
Determine the exact values of the 6 trigonometric ratios.
sin   
12
13
5
cos   
13
csc   
13
12
sec   
12
tan  
13
13
5
cot  
Math 30-1
13
12
9
Determine the exact value of the trig ratios given
5 3
sin    ,
   2
7
2

sin   
Must be in Quad IV
cos  
-5
x 2  72  52
x  72  52
2 6
7
tan   
x

5
7
5
2 6
5 6

12
7
csc   
5
7
sec  
x  24
x2 6
7
2 6
7 6

12
2 6
cot   
5
Math 30-1
10
Using Technology
sin 30º=
trig
function
angle
1
2 or 0.5
trig ratio
Given sin 30º
Given sin θº = ½
Asking for the sine ratio value
when angle θ is 30°
Asking for the value of angle θ
when the sine ratio is ½
Enter sin 30°, the answer is a ratio
Use the inverse of the
sine ratio which gives the angle.
Enter sin-1 (½), the answer is an angle.
When a positive ratio is
used, the calculator will
display the reference angle.
The mode of calculator must
match the domain of the angle,
degrees or radians.
Math 30-1
11
Deteriming the Measure of an Angle Given the Ratio
Determine the measure of θ.
0 ≤ θ < 3600
cos θ = -0.6691
trig
angle trig ratio
function
The trig ratio is negative, indicating that the x-coordinate is negative,
therefore, the angle θ would be found in Quadrants II or III.
The trig ratio is not an exact value:
use inverse cosine.
Domain is in degrees.
cos1 (0.6691)  48
The reference angle is 480.
Alternate Method
nSolve(cos(x)=0.6691,x)
θ = 1320 or 2280
Math 30-1
12
 5 12 
,   lies at the intersection of the unit circle
 13 13 
The point A 
and the terminal arm of an angle θ in standard position.
Determine the measure of angle θ for each domain.
0    360
0    2
Degrees
Point A is in quadrant IV
Point A is in quadrant IV
5
x
so
13
5
cos 1   
 13 
Radians
x
67.4
Reference
5
so
13
5
cos 1   
 13 
  360  67.4
  2 1.18
  292.6
  5.10
Math 30-1
1.18
Reference
13
Determine Angle  , Given an Exact Trigonometric Ratio
Determine the value of angle .
0 ≤  < 2
1
1. sin 
2
RA =

4
y is positive in
Q1 or Q2

1
2. tan  
3

RA =
6
 3
4
,
4
y/x is negative in
Q2 or Q4
1 x is negative in
3. cos  
2 Q2 or Q3
2 4 


,
RA =
3
3
3
5 11

,
6
6
Math 30-1
14
Find the Measure of an Angle  , Given an Exact Trigonometric Ratio
0 ≤  < 2
1. sec   2
1
cos   
2
RA =

4
3. csc  2
sin  
1
2
RA =

6
x is neg in
Q2 & Q3
3 5 

,
4
4
y is pos in
Q1 & Q2
2. cot  3
y/x is pos in
Q1 & Q3
1
tan  
3

RA =

6
4. cos  0

  , 5
6
6
x value is 0
Math 30-1
7
 ,
6 6
Quadrantal

 3
2
,
2
15
Jackie stated that
sin 30  cos 60
Is she correct?
What is the relationship between the angles?
Complete each equality.
sin 60  cos 30
sin 30  sin 150
cos 45  sin 45
cos 60  cos 300
Math 30-1
16
Page 201
1, 2, 3, 5a,b, 6, 7, 8, 10, 11a,c, 12, 13, 14, 16, 19
Math 30-1
17