Section 4.7 Notes
Inverse Trig Functions
Name two points on y = 2x + 5.
(0, 5) and (1, 7)
Find the inverse of y = 2x + 5.
Switch the x and y and solve for y.
x  2y  5
2y  x 5
1
5
y  x
2
2
Name two points on the inverse.
(5, 0) and (7, 1)
Inverse Sine Function
Graph two cycles of y = sin x. Label the key
points.
State the domain and range of this function.
D: (-∞, ∞)
(-3π/2, 1)

(π/2, 1)
R: [-1, 1]
(-π, 0) (0, 0)
(-2π, 0)




(-π/2, -1)





(2π, 0)
(π, 0)



(3π/2, -1)

Write the inverse of the key points on the previous
graph.
Graph these new points on the graph. Think about
how you must label the x- and y-axes.
Graph the inverse of the previous graph.
Is it a function?

(0, 2π)


(-1, 3π/2)

Not a function
(0, π)


(1, π/2)
(0, 0)

(-1, -π/2)




(0 -π)



(1, -3π/2)
(0, -2π)
To make the inverse on the last slide a function it
must pass the vertical line test.
This is done by restricting the range to
  
  2 , 2 
Regraph inverse sine using this range.






(1, π/2)
(0, 0)

(-1, -π/2)







The range for inverse sine is in what quadrants
using the unit circle?
1st and 4th Quadrants
/2
-/2
Inverse Cosine Function
Graph two cycles of y = cos x. Label the key
points.
State the domain and range of this function.
(0, 1)
(2π, 1)

(-2π, 1)

(π/2, 0)
(-π/2, 0)
(-3π/2, 0)



(-π, -1)





(3π/2, 0)

(π, -1)



D: (-∞, ∞)
R: [-1, 1]
Write the inverse of the key points on the previous
graph.
Graph these new points on the graph. Think about
how you must label the x- and y-axes.
Graph the inverse of the previous graph.
Is it a function?
Not a function



(-1, π)
(1, 2π)
(0, 3π/2)



(0, π/2)
(1, 0)




(-1, -π)
(0, -π/2)




(0, -3π/2)
(1, -2π)
To make the inverse on the last slide a function it
must pass the vertical line test.
This is done by restricting the range to
0, 



(-1, π)



(0, π/2)
(1, 0)








The range for inverse cosine is in what quadrants
using the unit circle?
1st and 2nd Quadrants

0
Inverse Tangent Function
Graph two cycles of y = tan x. Label the
asymptotes and zeros.






(π, 0)








(0, 0)





Write the inverse of the zeros on the previous
graph.
Graph these new points on the graph. Think about
how you must label the x- and y-axes.
Graph the inverse of the previous graph.
Is it a function?


Not a function

(0, π)


(0, 0)














To make the inverse on the last slide a function it
must pass the vertical line test.
This is done by restricting the range to
  
 , 
 2 2



















The range for inverse tangent is in what quadrants
using the unit circle?
1st and 4th Quadrants
/2
-/2
Write the following inverses using three different
notations.
y = sin x
x = sin y
y = sin-1 x
y = arcsin x
y = cos x
x = cos y
y = cos-1 x
y = arccos x
y = tan x
x = tan y
y = tan-1 x
y = arctan x
Example 1
Evaluate the following.

a.
arcsin  1  
2
Rewrite as sin y  1.
  
R :  , 
 2 2
b.
arctan(0) = 0
tan y = 0
  
R: , 
 2 2
c.
 1  2
arccos    
 2 3
1
cos y  
2
R :  0, 

3

d. sin  

3
 2 
1
3
sin y  
2
  
R :  , 
 2 2
e.
 3 
tan 

 3  6
1
3
tan y 
3
  
R: , 
 2 2
Example 2
Evaluate the following.
a.
sin–1(sin (–/2))
sin–1(-1) = –/2
  
R :  , 
 2 2
b.

  5  
sin sin     
3
  3 
1
 3
sin  

 2 
1
  
R :  , 
 2 2
Example 3
2

Find the exact value of tan  arccos  .
3

2
adj 2
Let  = arccos , then cos  
 .
3
hyp 3
y
3
32  22  5
θ
2
x
2
opp

5
tan  arccos   tan  

3
adj

2
2nd Day
Example 4
a.
Let y = arccos x. Find sin y
x
cos y = x 
1
sin y = 1 x
1
2
1 x
y
x
2
b.
x
Let y  arccos   . Find tan y.
3
x
cos y 
3
9  x2
tan y 
x
3
9 x
y
x
2
c.
 x 
Let y  arctan 
 . Find csc y.
 2
x
tan y 
2
csc y 
x2  2
x
x2  2
x
y
2
Example 5
A security car with its spotlight on is parked 20
meters from a warehouse. Consider θ and x as
shown in the figure.
a.
Write θ as a function of x.
x
tan  
20
x
  arctan
20
b.
Find θ when x = 5 meters and x = 12
meters.
5
  arctan
20
 14.036
12
  arctan
20
 30.964