### Slide 1

```1.3 Definition II: Right Triangle Trigonometry
If  is an acute angle of a right triangle, then
opp
opp
sin  
, cos  
, tan  
,
hyp
hyp
hyp
hyp
csc  
, sec  
, cot  
.
opp
opp
Trigonometric Functions
• Let (x, y) be a point other the origin on the terminal side of an angle
 in standard position. The distance from the point to the origin is
The six trigonometric functions of  are defined as follows.
r  x2  y 2 .
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
Example: Finding Function
Values
• The terminal side of angle  in standard
position passes through the point (12, 16).
Find the values of the six trigonometric
functions of angle .
r  x 2  y 2  122  162
(12, 16)
16
 144  256  400  20

12
Example: Finding Function
Values continued
• x = 12
y 16 4
sin   

r 20 5
x 12 3
cos  

r 20 5
y 16 4
tan   

x 12 3
y = 16
r = 20
r 20 5
csc  

y 16 4
r 20 5
sec  

x 12 3
x 12 3
cot   

y 16 4
Example: Finding Function
Values
• Find the six
trigonometric function
values of the angle 
in standard position, if
the terminal side of 
is defined by
x + 2y = 0, x  0.
• We can use any point
on the terminal side of
 to find the
trigonometric function
values.
Example: Finding Function
Values continued
• Choose x = 2
x  2y  0
2  2y  0
2 y  2
y  1
• The point (2, 1) lies
on the terminal side,
 22  (1) 2  5.
andr the
corresponding value
of r is
• Use the definitions:
y 1 1 5
5
sin   



r
5
5
5 5
x
2
2
5 2 5




r
5
5
5 5
y
1
r
tan    
csc    5
x
2
y
cos 
sec 
r
5

x
2
cot  
x
 2
y
Example: Function Values
• Find the values of the six trigonometric functions for
an angle of 270.
• First, we select any point on the terminal side of a
270 angle. We choose (0, 1). Here x = 0, y = 1
and r = 1. 1
0
sin 270 
 1
1
1
tan 270 
undefined
0
1
sec 270  undefined
0
cos 270 
0
1
1
csc 270 
 1
1
0
cot 270   0
1
Commonly Used Function
Values

sin  cos 
tan 
cot 
sec 
1
0
0
1
0
undefined
90
1
0
undefine
d
0
180
0
1
0
undefined
270
1
0
0
360
0
1
undefine
d
0
undefined
csc 
undefine
d
undefined
1
1
undefine
d
undefined
1
1
undefine
d
Reciprocal Identities
•
1
sin  
csc
1
cos 
sec
1
tan  
cot 
1
csc 
sin 
1
sec 
cos
1
cot  
tan 
Example: Find each function
value.
2
• cos  if sec  =
3
• Since cos  is the
reciprocal of sec 
1
1 3
cos 
 2
sec 3 2
15
• sin  if csc   
3
1
3
sin  

15
15

3
3 • 15
3 • 15


15
15 • 15

15
5
angle  that satisfies tan  > 0, cot  > 0.
• tan  > 0 in quadrants I and III
• cot  > 0 in quadrants I and III
Ranges of Trigonometric
Functions
• For any angle  for which the indicated
functions exist:
• 1. 1  sin   1 and 1  cos   1;
• 2. tan  and cot  can equal any real
number;
• 3. sec   1 or sec   1 and
csc   1 or csc   1.
(Notice that sec  and csc  are never
between 1 and 1.)
Identities
• Pythagorean
sin 2   cos 2   1,
tan 2   1  sec 2  ,
1  cot 2   csc 2 
• Quotient
sin 
 tan 
cos 
cos 
 cot 
sin 
Example: Other Function Values
• Find sin and cos if tan  = 4/3 and  is in
• Since is in quadrant III, sin and cos will both
be negative.
• sin and cos must be in the interval [1, 1].
Example: Other Function Values
continued
• We use the identity
tan 2   1  sec 2 
2
4
2

1

s
ec

 
3
16
 1  sec 2 
9
25
 sec 2 
9
5
  sec
3
3
  cos 
5
tan 2   1  sec 2 
Since sin 2   1  cos 2  ,
 3
sin 2   1    
 5
9
sin 2   1 
25
16
sin 2  
25
4
sin   
5
2
Example: If
and  is in quadrant II,
find each function value.
5
tan   
3
tan 2   1  sec 2 
• a) sec 
2
Look for an identity that
relates tangent and
secant.
tan   1  sec 
2
2
 5
2


1

sec

 3 
25
 1  sec 2 
9
34
 sec 2 
9
sec   
34
9
34
sec   
3
Example: If
and  is in quadrant II,
find each function value continued
5
tan   
3
• b) sin 
sin 
tan  
cos
cos tan   sin 
 1

 sec

 tan   sin 

 3 34   5 

     sin 
 34   3 
5 34
 si n 
34
• c) cot ()
1
cot(  ) 
tan(  )
1
cot(  ) 
 tan 
1
3
cot( ) 

   53  5
Example: Express One Function
in Terms of Another
1 cot 2 x  csc 2 x
1
1

2
1 cot x csc 2 x
1
2

sin
x
2
1 cot x
• Express cot x in
terms of sin x.

1
2

sin
x
2
1 cot x
1
sin x 
1 cot 2 x
 1 cot 2 x
sin x 
1 cot 2 x
Example: Rewriting an Expression in Terms
of Sine and Cosine
• Rewrite cot   tan  in terms of sin  and cos  .
•
cos sin 
cot   tan  

sin  cos 
cos 2 
sin 2 


sin  cos sin  cos
cos 2   sin 2 

sin  cos
(cos  sin  )(cos  sin  )

sin  cos
Example: Working with One Side
• Prove the identity
(tan 2 x  1)(cos2 x  1)   tan 2 x
(tan 2 x  1)(cos 2 x  1)   tan 2 x
 sin 2 x 
2
2

1
(cos
x

1)


tan
x
 cos 2 x 
sin 2 x
2
2
sin x 

cos
x

1


tan
x
2
cos x
2
sin
x
2
2
2
sin x  cos x 

1


tan
x
2
cos x
2
sin 2 x
2
1

1


tan
x
2
cos x
sin 2 x
2



tan
x
2
cos x
 tan 2 x   tan 2 x
Example: Working with One Side
• Prove the identity
1
 csc x  sin x
sec x tan x
right side
1
 csc x  sin x
sec x tan x
1

 sin x
sin x
1
sin 2 x


sin x sin x
• continued
1
1  sin 2 x

sec x tan x
sin x
cos 2 x

sin x
cos x cos x

sin x 1
 cot x cos x
1
1

tan x sec x
1
1

sec x tan x sec x tan x
```