Download Sullivan Algebra and Trigonometry: Section 7.5

Sullivan Algebra and
Trigonometry: Section 6.5
Unit Circle Approach; Properties
of the Trig Functions
Objectives of this Section
• Find the Exact Value of the Trigonometric Functions Using
the Unit Circle
• Determine the Domain and Range of the Trigonometric
Functions
• Determine the Period of the Trigonometric Functions
• Use Even-Odd Properties to Find the Exact Value of the
Trigonometric Functions
The unit circle is a circle whose radius
is 1 and whose center is at the origin.
Since r = 1:
s  r
becomes
s
(0, 1) y
s 

x
(1, 0)
(-1, 0)
(0, -1)
(0, 1) y
 t
P = (a, b)

(-1, 0)
(1, 0)
(0, -1)
x
Let t be a real number and let P = (a, b) be
the point on the unit circle that corresponds
to t.
The sine function associates with t the
y-coordinate of P and is denoted by
sint  b
The cosine function associates with t the
x-coordinate of P and is denoted by
cost  a
If a  0 , the tangent function is defined as
b
tan t 
a
If b  0 , the cosecant function is defined as
1
csc t 
b
If a  0 , the secant function is defined as
1
sect 
a
If b  0 , the cotangent function is defined as
a
cot t 
b
1
15 
 be
Let t be a real number and let P   ,

4
4


the point on the unit circle that corresponds to t.
Find the exact value of the six trigon ometric
functions.

a , b    1 4 ,

15
sint  b  
4
15

4 
1
cost  a 
4
a , b 
  1 , 
 4
15

b
4 
tan t  
1
a
4
1
1
csct  

b  15
4
1 1
sect  
4
a 1
4
1
a
4 
cot t  
b  15
4
15

4 
15
4
4 15

15
15
1
15

15
15
(0, 1) y
 t
P = (a, b)

(-1, 0)
(1, 0)
(0, -1)
x
If   t radians, the six trigonometric
functions of the angle  are
defined as
sin   sin t
cos  cos t
tan   tan t
csc  csc t
sec  sec t
cot   cot t
y

a
b
r
x
For an angle  in standard position, let P  (a, b)
be any point on the terminal side of  . Let r
equal the distance from the origin to P. Then
b
sin  
r
a
cos  
r
b
tan   , a  0
a
r
r
a
csc   , b  0 sec   , a  0 cot   , b  0
b
a
b
5
Given that sec =
and sin > 0, find the
2
exact value of the remaining five trigonometric
functions.
x 2  y 2  25
P=(a,b)

(5, 0)
5
r
sec  
 , so r  5, a  2
2 a
b
a  b  r with b > 0 since sin  > 0
r
2
2
2
 2  b  5
2
2
2
4  b  25
2
b  21
2
b
21
a  2, b  21, r  5
a 2
cos  
r
5
b
21
21
tan   

a 2
2
b
21
sin   
r
5
r
5
5 21
csc  

b
21
21
a 2
2 21
cot   

b
21
21
(0, 1) y
 t
P = (a, b)

(-1, 0)
(1, 0)
(0, -1)
x
The domain of the sine function is the set of
all real numbers.
The domain of the cosine function is the set of
all real numbers.
The domain of the tangent function is the set
of all real numbers except odd multiples of
 
 2 90 .
The domain of the secant function is the set
of all real numbers except odd multiples of
 2 90 .
 
The domain of the cotangent function is the
set of all real numbers except integral
multiples of  180 .
The domain of the cosecant function is the
set of all real numbers except integral
multiples of  180 .
Range of the Trigonometric Functions
Let P = (a, b) be the point on the unit
circle that corresponds to the angle  .
Then, -1 < a < 1 and -1 < b < 1.
sin   b
cos  a
 1  sin   1
 1  cos  1
sin   1
cos  1
1
1
csc  

1
sin 
b
csc    1 or csc   1
1
1
sec  

1
cos 
a
sec    1 or sec   1
   tan   
   cot   
A function f is called periodic if there is a
positive number p such that whenever  is
in the domain of f , so is   p, and
f   p  f  p
If there is a smallest such number p, this
smallest value is called the (fundamental)
period of f.
Periodic Properties
sin  2   sin 
csc  2   csc
cos  2   cos
sec  2   sec
tan     tan 
cot     cot 
Find the exact valu e of
(a) sec 390
7
(b) cot
4


(a) sec 390  sec 30  360




2 3
 sec 30 
3

3
 7 
 3

( b ) cot 
    cot
 1
  cot 
4
 4 
 4

Even-Odd Properties
sin     sin 
csc     csc
cos    cos
sec    sec
tan     tan 
cot     cot 
Find the exact valu e of

(a) sin  30

( a ) sin  30




 
(b) cot   
 4
 sin 30

1
 
2

  
(b) cot  
 1
   cot
4
 4 