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An angle is formed by two rays that have a
common endpoint. One ray is called the
initial side and the other the terminal side.
An angle is in standard position if its vertex is at
the origin of a rectangular coordinate system and
its initial side lies along the positive x-axis.
Positive angles are generated by counterclockwise rotation.
Negative angles are generated by
clockwise rotation.
An angle is called a quadrantal angle if its
terminal side lies on the x-axis or on the y-axis.
Angle  is an example of a quadrantal angle.
Angles are measured by determining the
amount of rotation from the initial side to the
terminal side. A complete rotation of the
circle is 360 degrees, or 360°.
90
0
180
An acute angle measures less than 90°.
A right angle measures 90°.
An obtuse angle measures more than 90° but
less than 180°.
A straight angle measures 180°.
360
270
Complementary Angles: The sum of any two angles that equals 90o.
Supplementary Angles:
The sum of any two angles that equals 180o.
Find the complement and supplement angles of 40o.
90o – 40o = 50o
180o – 40o = 140o
6x 1  2x 1  90
6x 1  2x 1  90
8x  2  90
means 90o
8x  88
x  11

 180
7 y  11  9 y  25  180
16 y  36  180
16 y  144
y 9
 1 
1'   
 60 

There are 60 minutes in 1 degree.
1  60'
There are 60 seconds in 1 minute.
 1   1 

1'  60" 1"     
 60   3600 
'
 3246'
8375'  8415'
 1  60'
75 min. > 60 min.,
so carry 1 degree.
8960'1837'
1837'
71 23'
We need to borrow
1 degree and
convert it to 60 min.

Convert the minutes and seconds to
fractional degrees.

8
14
21 

60 3600

Calculator help!
The whole number is the degrees.
Multiply the decimal by 60 to
determine the minutes.
34  0.267860'
3416.068'  34 16' 4"
The whole number is the minutes.
Multiply the decimal by 60 to
determine the seconds.
Round your answer according
to the directions. 21.13722222
Two angles with the same initial and terminal sides but
possibly different rotations are called coterminal angles.
Increasing or decreasing the angle measure of an angle in
standard position by an integer multiple of 360o results in a
coterminal angle. Thus, an angle of  is coterminal with
angles of    360k , where k is an integer.
Assume the following angles are in standard position.
Find a positive angle less than 360° that is coterminal
with each of the following:
a. a 400° angle
b. a –855° angle
400  360  40
 855  3603  225
CD players always spin at the same speed. Suppose a player
makes 480 revolutions per min. Through how many degrees will
a point on the edge of the CD move in 2 seconds?
Determine how many revolutions in 1 second.
480 rev 480 rev

 8 rev / sec
1 min
60 sec
8 revolutions in 1 second times 2 is 16 revolutions.
16 revolutions times 360o.
16360  5760
Symbol for parallel.
In the diagram, to the right, 2 parallel lines are
intersected by a transversal line.
Are formed by 2 intersecting lines and are the angles that open opposite of each other.
Name all the pairs of vertical angles. 1 & 4, 2 & 3, 5 & 8, and 6 & 7.
Vertical angles are always equal in measure.
Are the matching angles formed by the two intersections of the parallel lines.
Name all the pairs of corresponding angles. 1 & 5, 2 & 6, 3 & 7, and 4 & 8.
Corresponding angles are always equal in measure.
Are the angles in between the parallel lines and alternate over the transversal line.
Name all the pairs of alternate interior angles. 3 & 6 and 4 & 5.
Alternate interior angles are always equal in measure.
Are the angles outside the parallel lines and alternate over the transversal line.
Name all the pairs of alternate exterior angles. 1 & 8 and 2 & 7.
Alternate exterior angles are always equal in measure.
Are the angles in between the parallel lines and on the same side of the transversal line.
Name all the pairs of same side interior angles. 3 & 5 and 4 & 6.
Same side interior angles are always supplementary.
Sum of the angles of a triangle_______________________________.
always adds up to 180 degrees
3
1
2
All three angles are acute, less than 90o.
One angle is 90o.
One angle is greater than 90o.
All three sides are equal.
Two sides are equal.
No sides are equal.
Two triangles that are the same shape, but not
necessarily the same size.
1. Corresponding angles must be the same measure.
A
2. Corresponding sides must be proportional.
ABC ~ DEF
1. A  D
B  E
C  F
D
B
2. AB
BC CA


DE EF FD
E
C
F
Examples.
Find the measure of all angles.
180  75  40  65
2
751 65
340
404 5 75
65
1406 7
40 758 105
9 
11 12
40
10 140
75
105
Vertical angles first.
Supplementary and
Corresponding angles next.
ST  16 1  4  4
Find the measure of all angles and sides.
A
ABC ~ RST
47
R
24
12
106
27
16
B
A  R
6
3 47
106 27
S
C
4
B  S
T
C  T
A  47 106  S
180  47 106  C  T
27  C  T
AB BC
16 CA
12
4 CA



RS
ST
TR
3
6
1
6
CA  4  6 1  24
4 16

1 ST
Definitions of Trigonometric Functions of Any Angle.
Let  be any angle in standard position and let P = (x, y) be a
point on the terminal side of  . If r  x 2  y 2 is the distance
from (0, 0) to (x, y), the six trigonometric functions of  are
Pythagorean Theorem
defined by the following ratios:
y
sin  
r
r
csc  , y  0
y
x
cos 
r
y
tan   , x  0
x
r
sec  , x  0
x
x
cot   , y  0
y
What do you notice about the fractions?
Reciprocals!
r 2  x2  y 2
Let P = (8, 15) be a point on the terminal side of  . Find each of
the six trigonometric functions of  . P = (8, 15) is a point on the
terminal side of  . x = 8 and y = 15.
r  x 2  y 2  82  152  64  225  289  17
sin   
y
15

r
17
x


cos  
8

r
17
y
15
tan   

x
8
17
r

csc  
y 15
r


sec  
x

17
8
8
x

cot   
15
y
P  8,15
r  17
x 8
y  15
Let P = (1, –3) be a point on the terminal side of  . Find each of
the six trigonometric functions of  . P = (1, –3) is a point on the
terminal side of  . x = 1 and y = –3.
r  x 2  y 2  (1)2  (3)2  1  9  10
x 1
y
3
10
 3 10


sin   


r
10 10
10
r  10
Rationalize the denominator!
P  1,3
10
x
1
10

cos   

10
r
10 10
y
3

 3
x
1
x
1
1
cot    

y 3
3
tan   
r


sec  
y  3
10
 10
1
x
r
10
10
csc   

y
3
3

Find the six trigonometric function values of the angle  in standard
position, if the terminal side of  is defined by 3x + 4y = 0, x < 0.
x < 0 means that the terminal side is in
quadrant 2 or 3!
Solve the linear equation for y so it is in slope
intercept form, y = mx + b.
3x  4 y  0
4 y  3x  0
 4,3
3
y
x0
4

Start at (0, 0) and
slope of -3/4.
Since we are in quadrant 2 or 3, we will
have to reverse the slope…up 3, left 4.
sin   
y 3

r 5
x 4


cos   
r
5
y
3
tan    
x 4
r 5
csc   
y 3
r
5


sec   
x 4
x 4
cot    
y
3
x  4, y  3
r
 42  32
 16  9  25  5
Section 1.4 Reciprocal Identities.
y
1
sin   

r
csc  
Flip the fractions
1
y
csc  rr 1

csc
csc
yy   csc  r
1
sin
Use this same concept for the other trig. functions we get the rest of the identities
x 1
cos  
r  
sec
tan   
y 1
x  
cot
r


sec  
1
xcos 
x 1
cot   
ytan  
We know that r is always positive and
x < r & y < r.


Sin & csc
cos & sec
Students tan & cot
S
y
r
r
y
 rx
 rx


y  sin   
Take
A
All
sin & csc
C
Cos & sec
tan & cot Calculus
Blue = Positive and Red = Negative
y
x
x
y
y  cos  
sin & csc
cos & sec
Tan & cot
T
sin & csc
cos & sec
tan & cot
y
r
x
r
Since x and y are < r, the
fraction < 1. If x and y are
negative, the fraction > -1.
R : 1  y  1
R :  1,1
y  tan   
y  cot  
y
x
x
y
Since x and y can be all
real numbers, dividing
real numbers will still
be real numbers.
R :  , 
y  sec   rx
y  csc  
r
y
Since x and y are < r, the
fraction > 1. If x and y are
negative, the fraction < -1.
R : y 1
R :  ,1 1, 
Positive
Negative
Use “All Students Take Calculus”
Quadrant 2 has the
sign representation
for both conditions,
therefore, the
terminal side is in
quadrant 2.
S
+– 2
3
A

+
 5
T
2 y
sin    
3 r
Use the Pythagorean
Theorem.
– C
x2  y 2  r 2
x 2  2 2  32
x2  4  9
x has to be negative
because we are in
quadrant 2.
x2  5
x 5
 5

r
3
x


cos  
tan   
2
y
2 5


 5
x
5
Rationalize the denominator!
x  5
cot    
y
2
r


sec  
x

3
 5
3
r

csc  
2
y

3 5
5
Divide by r2
r2
r2
r2
x2 y2
 2 1
2
r
r
2
2
 x  y
    1
r r
cos  2  sin  2  1
cos 2   sin 2   1
Divide by x2
x2
x2
Divide by y2
x2
y2
y2 r 2
1 2  2
x
x
2
2
 y r
1     
 x  x
y2
y2
x2
r2
1  2
2
y
y
2
x
r
   1   
 y
 y
2
1  tan    sec  
cot  2  1  csc  2
1  tan 2   sec 2 
cot 2   1  csc 2 
2
2
Since we flipped the sine and cosine, the
reciprocal of tangent is cotangent.
y
r
x
r
y
  tan  
x
cot  
Negative
+
–
S
A
+ sin    y 
4
– T
 3
C
x  3
cos   
r
4
 3   y
2
13
4
r
13
y
tan    
x  3
 42
3  y 2  16
2
y 2  13
y  13
Positive, Quad. 2
Positive
Using identities may be quicker.
Pythagorean and Quotient.
sin 2    cos 2    1
2
2
3
sin     4  1
 
sin 2    163  1
13 3
Sine is
13
sin 2    16


Positive, so
3
3
sin    413 no + sign.
39
13



sin

39
3
tan   
 43 
cos   4
3
Positive
Negative
Negative
Quadrant 3 is (-x, -y).
Building the x-y axis's
and using “All Students
Take Calculus” helps
with signs!
r x y
2
2
r  32  42
r  9  16  25  5
y 4
sin    
r
5
x 3
cos   
r
5
S
A
y
4
tan   

x
3
– 33
T
–44
r
C
Using Pythagorean & Reciprocal Identities.
tan 2    1  sec 2  
43 2  1  sec2  
16
9
 1  sec  
2
25
 

sec
9
 53  sec 
3
 
5  cos 
2
Cosine was given
as Negative.
sin 2    1  cos 2  
sin    1   53 
2
sin     54
Sine was given
as Negative.
 54