JMerrill, 2010

We know that for a function to have an
inverse function, it must be one-to-one (it
must pass the Horizontal Line Test).

From looking at a sine wave, it is obvious that
it does not pass the Horizontal Line Test.


The Unit Circle in found in section 4.7.
We will use:
◦ Radians
◦ Exact answers (mostly)
◦ Quick board review of Unit Circle, quadrants on
the wave, & converting to radian measure


In order to pass the Horizontal Line Test (so
that sin x has an inverse that is a function),
we must restrict the domain.
We restrict it
to    
 2 , 2 



  
 2 , 0 
Quadrant IV is
 
Quadrant I is
0, 2 
Answers must be in one of those two
quadrants or the answer doesn’t exist.


How do we draw inverse functions?
Switch the x’s and y’s!
Switching the x’s and y’s also
means switching the axis!


Domain/range of restricted sine wave?
D :  1,1
Domain/range of inverse?
   
R:
, 
 2 2
   
D:
, 
 2 2
R :  1,1


y = arcsin x
or
y = sin-1 x
Both mean the same thing. They mean that
you’re looking for the angle where sin y = x.


Find the exact value of:
Arcsin ½
◦ This means at what angle is the sin = ½ ?
◦ π/6
◦ (5π/6 has the same answer, but falls in QIII, so it is
not correct)



When looking for an inverse answer on the
calculator, use the 2nd key first, then hit sin,
cos, or tan.
When looking for an angle always hit the 2nd
key first.
Last example:
◦ Degree mode, 2nd, sin, .5 = 30.
◦ Radian mode: 2nd, sin, .5 = .524 (which is pi/6)


Find the value of:
Sin-1 2
◦ This means at what angle is the sin = 2 ?
◦ What does your calculator read? Why?
◦ 2 falls outside the domain of an inverse sine wave


Domain and range of restricted wave?
Domain and range of the inverse?
D :  0, 
R :  1,1
D :  1,1
R :  0, 


We must restrict the domain
Now the inverse

Quadrant I is 0, 
 2

Quadrant II is  ,  
2 


We must restrict the domain
Now the inverse
Graphing Utility: Graph the following inverse functions.
Set calculator to radian mode.
a. y = arcsin x

–1.5
1.5
–
2
b. y = arccos x
–1.5
1.5
–

c. y = arctan x
–3
3
–
Graphing Utility: Approximate the value of each expression.
Set calculator to radian mode.
a. cos–1 0.75
b. arcsin 0.19
c. arctan 1.32
d. arcsin 2.5

Previously learned notation:
◦ fog(x)
gof(x)

Find tan(arctan(-5))
◦ -5 (the tangent and its inverse cancel each other
out!)

Find arcsin(sin 53 )
  ,  
is  2 2 
◦ The domain of the sine function
. Since
5
is outside that domain, we’ll just say that the
3
5
answer is: 3 is outside the domain (unless you
remember coterminal angles and can tell me the
actual answer is  
3

Find cos(cos-)
◦ Outside the domain


Find the exact value of tan  arccos 
3
2


Draw the graph using only the info inside the
parentheses. (Easy way—completely ignore
the fact that you have inverses!)
Example:


Find the exact value of tan arccos 2 .
3
Since cos   x , x  2, r  3
y r
3
Positive so draw in Q1)
32  22  5
2
x
y
tan    5
x
2

 1  3  
Find the cos sin   5 


4
5