Transcript Section 3.2 The Inverse Trigonometric Functions [Continued]

Section 3.1 – The Inverse
Sine, Cosine and Tangent
Functions
Continued
5
7.85
0
7.85
5
Horizontal line test – tan x is not one-to-one
If we limit the
domain of the
tangent function to
be (–π/2, π/2) we
have a function that
is one-to-one…it
will have an

inverse.

5

2
1.57
0
2
5
1.57
The inverse tangent of x
1
y  tan x
where 

2
y
tan tan x   x
1

1

means
tan tan x  x

2
x  tan y
and    x  
where
where


 x
2
2
  x  
1
Characteristics of y  tan x
1
Domain of y  tan x is the Range of y  tan x:
 x
1
Range of y  tan x is the Domain of y  tan x:


 y
2
2
x
y
y


x
2
1
y  tan x
2
2

2
2
y  tan x

0
2
Find the exact value of tan
1
  3.
Find the value in (–π/2, π/2) whose tangent is
  tan
1
  3
tan    3

where

where


3

2

2
 
 

2

2
 3
Section 3.2 The Inverse
Trigonometric Functions
[Continued]
Remaining Inverse Trig Functions
1
y  sec x means x  sec y
where x  1 ( x  1 and x  1), 0  y   , y 

2
1
y  csc x means x  csc y
where x  1 ( x  1 and x  1), 1

2
 y

2
y  cot x means x  cot y
where    x  , and 0  y  
,y0

1 1
Find the exact value of csc cos  .

4
1 1
  cos
4
where
0 
1
cos 
4

1
0 
since cos   0
2
4
y  1  4  16
2
y
2
2
y  15
4

y
r
csc 
y
4
4 15
csc 

15
15
x
1
4 15

1 1
csc cos   csc 

4
15

1

Find the exact valu e of csc tan 1
  tan 1, 
1
tan   1, 


4

2

2
 
 

2

2
So, we need to find
 
csc 
4
We know
tan   1
r  a  b  1 1
2
2
2
r 2
2
( a, b)
y
2
2


4
1
1
x
2
  r
csc tan 1  csc   
 2
1
4 b

1

1
Use a calculator to approximat e sec 3.

1
We have   sec 3, for 0     ,  
sec   3, 0     ,  

2
2
But we (most likely) do not have a sec button
on our calculator, so
1
1
cos  

sec  3
1
  cos  
3
1
  1.23