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Pre-Class Problems 13 for Thursday, March 20
These are the type of problems that you will be working on in class. These
problems are from Lesson 9.
Solution to Problems on the Pre-Exam.
You can go to the solution for each problem by clicking on the problem letter.
Objective of the following problems: To find the value of the inverse sine of a
given number.
1.
Find the exact value of the following.
a.
sin
1
2
d.

3

sin  1  
 2 


g.
Arc sin
1
2
2

2 



2 

b.
3
Arc sin
2
c.
sin
e.
Arc sin 1
f.
sin  1 0
h.
sin
i.

Arc sin  

1
(  1)
1
1

2
Objective of the following problems: To find the value of the inverse tangent of a
given number.
2.
Find the exact value of the following.
1
1
3
a.
tan
d.
tan  1 ( 
3)
b.
Arc tan
e.
Arc tan 0
3
c.
Arc tan (  1 )
f.
tan  1 1
g.
 1 

Arc tan  


3


Additional problems available in the textbook: Page 186 … 5, 6, 9, 10, 12, 13, 14,
16, 17, 18. Examples 1ab and 3bc on page 181.
Solutions:
Definition The inverse sine function, denoted by sin
if and only if sin y  x , where  1  x  1 and 
Notation: Sometimes, sin
sin  1 x .
1
1
, is defined by y  sin
1
x


 y  .
2
2
x is denoted by Arc sin x .
That is, Arc sin x =
The restriction that is put on the sine function in order to make it be one-to-one
  
means that your angle answer must be in the closed interval  ,  . NOTE:
 2 2
 
The angles in the open interval  0 ,  are in the first quadrant. The angles in the
 2
  
open interval   , 0  are in the fourth quadrant. The angle 0 is on the positive x 2 
axis, the angle


is on the positive y-axis, and the angle 
is on the negative y2
2
axis.
1a.
sin  1
1
2

Answer:
6
1
is positive and not maximum positive of 1, then the angle
2
answer is in the first quadrant. That is, the angle answer is in the open
NOTE: Since
 
interval  0 ,  .
 2

1

either by Unit Circle
6
2
 1
Trigonometry or Right Triangle Trigonometry. Note that the point  ,  is
 6 2
We know that sin
on the graph of the sine function.
Back to Problem 1.
1b.
Arc sin
3
2
Answer:

3
3
is positive and not maximum positive of 1, then the angle
2
answer is in the first quadrant. That is, the angle answer is in the open
3

 

interval  0 ,  . We know that sin
either by Unit Circle
3
2
 2
Trigonometry or Right Triangle Trigonometry.
Note that the point

3
 ,

 3 2  is on the graph of the sine function.


NOTE: Since
Back to Problem 1.
1c.

2 

sin  1  


2


Answer: 

4
2
is negative and not minimum negative of  1 , then the
2
angle answer is in the fourth quadrant. That is, the angle answer is in the
  
open interval   , 0  .
 2 
NOTE: Since 

2 
1 
 , then  '


sin

If we let


2




1
    '   . That is, sin  
4

2
1
NOTE: We will show that sin
2
 sin  1
2

 . Thus,
2
4

  .

4



in Problem 1g.
4
2
2
2

 
 
We know that sin      sin
using reference angles. Note
4
2
 4
 
2 

 is on the graph of the sine function.

,

that the point  4

2


Back to Problem 1.
1d.

3

sin  1  

2 

Answer: 

3
3
is negative and not minimum negative of  1 , then the
2
angle answer is in the fourth quadrant. That is, the angle answer is in the
  
open interval   , 0  .
 2 
NOTE: Since 

3
3

1 
1



sin


'

sin

If we let
. Thus,

2  , then
2
3


3

1 
  .
sin

    '   . That is,
 2 
3
3


3

1

NOTE: We showed that sin
in Problem 1b.
2
3
3

 
sin



sin




We know that
using reference angles. Note
3
2
 3
 
3

 is on the graph of the sine function.

,

that the point  3
2 

Back to Problem 1.
1e.
Arc sin 1
Answer:

2
NOTE: Since 1 is the maximum positive of 1, then the angle answer is not in
the first quadrant. So, the angle answer is on one of the coordinate axes,
namely the positive y-axis. The angle which is in the closed interval

  
 2 , 2  and lies on the positive y-axis is the angle 2 . We know that
sin

 1 by Unit Circle Trigonometry. Note that the point
2
 
 , 1 is on the
2 
graph of the sine function.
Back to Problem 1.
1f.
sin  1 0
Answer: 0
NOTE: Since 0 is not positive, then the angle answer is not in the first
quadrant. Since 0 is not negative, then the angle answer is not in the fourth
quadrant. So, the angle answer is on one of the coordinate axes, namely the
  
positive x-axis. The angle which is in the closed interval  ,  and lies
 2 2
on the positive x-axis is the angle 0. We know that sin 0  0 by Unit Circle
Trigonometry. Note that the point 0 , 0 is on the graph of the sine function.
Back to Problem 1.
1g.
Arc sin
2
2
Answer:

4
2
is positive and not maximum positive of 1, then the angle
2
answer is in the first quadrant. That is, the angle answer is in the open
2

 

interval  0 ,  . We know that sin
either by Unit Circle
4
2
 2
Trigonometry or Right Triangle Trigonometry. Note that the point

2 
 ,

 4 2  is on the graph of the sine function.


NOTE: Since
Back to Problem 1.
1h.
sin
1
(  1)
Answer: 

2
NOTE: Since  1 is the minimum negative of  1 , then the angle answer is
not in the fourth quadrant. So, the angle answer is on one of the coordinate
axes, namely the negative y-axis. The angle which is in the closed interval

  

,

and
lies
on
the
negative
y-axis
is
the
angle
. We know that
 2 2 
2
 
sin      1 by Unit Circle Trigonometry. Note that the point
 2
 

  ,  1 is on the graph of the sine function.
 2

Back to Problem 1.
1i.
 1
Arc sin   
 2
Answer: 

6
1
is negative and not minimum negative of  1 , then the
2
angle answer is in the fourth quadrant. That is, the angle answer is in the
  
open interval   , 0  .
 2 
NOTE: Since 
1

 1
 . Thus,
If we let   Arc sin    , then  '  Arc sin
2
6
 2


 1
    '   . That is, Arc sin      .
6
6
 2
1


NOTE: We showed that Arc sin
in Problem 1a.
2
6

1
 
sin



sin




We know that
using reference angles. Note that
6
2
 6
1
 
the point   ,   is on the graph of the sine function.
2
 6
Back to Problem 1.
1
Definition The inverse tangent function, denoted by tan , is defined by
y  tan  1 x if and only if tan y  x , where x is any real number and



 y  .
2
2
1
Notation: Sometimes, tan x is denoted by Arc tan x .
tan  1 x .
That is, Arc tan x =
The restriction that is put on the tangent function in order to make it be one-to-one
  
means that your angle answer must be in the open interval   ,  . NOTE:
 2 2
 
The angles in the open interval  0 ,  are in the first quadrant. The angles in the
 2
  
open interval   , 0  are in the fourth quadrant. The angle 0 is on the positive x 2 
axis.
From Lesson 2, you had the following diagram to help you find the value of the
tangent of the three Special Angles in the I quadrant.

6

 30  

 45  
4
3
 60  
|
|
|

1


3
1
3
Tangent
Now, reverse the arrows to have a diagram to help you find the value of the inverse
1
tangent of these three numbers of 1, 3 , and
:
3

6
 30  

 45  
4

3
 60  



|
|
|
1
3
1
Inverse Tangent
3
2a.
tan
1
1
3
Answer:

6
1
NOTE: Since
is positive, then the angle answer is in the first quadrant.
3
 
That is, the angle answer is in the open interval  0 ,  . We know that
 2

1
tan

either by memorization or Right Triangle Trigonometry. Note
6
3
 1 

 is on the graph of the tangent function.
that the point  6 ,
3 

Back to Problem 2.
2b.
Arc tan
3
Answer:

3
3 is positive, then the angle answer is in the first quadrant.
 
That is, the angle answer is in the open interval  0 ,  . We know that
 2

tan
 3 either by memorization or Right Triangle Trigonometry. Note
3


that the point  , 3  is on the graph of the tangent function.
3

NOTE: Since
Back to Problem 2.
2c.
Arc tan (  1 )
Answer: 

4
NOTE: Since  1 is negative, then the angle answer is in the fourth quadrant.
  
That is, the angle answer is in the open interval   , 0  .
 2 
If we let   Arc tan (  1 ) , then  '  Arc tan 1 

. Thus,
4


. That is, Arc tan (  1 )   .
4
4

NOTE: We will show that Arc tan 1 
in Problem 2f.
4
  '  

 
  1 using reference angles. Note that
We know that tan      tan
4
4


 

the point   ,  1 is on the graph of the tangent function.
 4

Back to Problem 2.
2d.
tan  1 ( 
Answer: 
3)
NOTE: Since 

3
3 is negative, then the angle answer is in the fourth
  
quadrant. That is, the angle answer is in the open interval   , 0  .
 2 
1
If we let   tan ( 
  '  
1
3 
3 ) , then  '  tan

1
. That is, tan ( 
3
1
3 
NOTE: We showed that tan
3)  

.
3

. Thus,
3

in Problem 2b.
3
 
We know that tan   
 3
 
that the point   ,  3
 3
  tan

 
3
3 using reference angles. Note

 is on the graph of the tangent function.

Back to Problem 2.
2e.
Arc tan 0
Answer: 0
NOTE: Since 0 is not positive, then the angle answer is not in the first
quadrant. Since 0 is not negative, then the angle answer is not in the fourth
quadrant. So, the angle answer is on one of the coordinate axes, namely the
  
positive x-axis. The angle which is in the open interval   ,  and lies on
 2 2
the positive x-axis is the angle 0. We know that tan 0  0 by Unit Circle
Trigonometry. Note that the point 0 , 0 is on the graph of the tangent
function.
Back to Problem 2.
2f.
tan  1 1
Answer:

4
NOTE: Since 1 is positive, then the angle answer is in the first quadrant.
 
That is, the angle answer is in the open interval  0 ,  . We know that
 2

 1 either by memorization or Right Triangle Trigonometry. Note
4
 
that the point  , 1 is on the graph of the tangent function.
4 
tan
Back to Problem 2.
2g.
 1 

Arc tan  

3 

Answer: 

6
1
is negative, then the angle answer is in the fourth
3
  
quadrant. That is, the angle answer is in the open interval   , 0  .
 2 
NOTE: Since 
 1 

 , then  '  Arc tan 1   . Thus,


Arc
tan
If we let

6
3 
3

 1 

  .
    '   . That is, Arc tan  
6
3 
6

1


NOTE: We showed that Arc tan
in Problem 2a.
6
3

1
 
tan



tan




We know that
using reference angles. Note
6
3
 6
 
1 


,

that the point  6
 is on the graph of the tangent function.
3


Back to Problem 2.
Solution to Problems on the Pre-Exam:
16.
Back to Page 1.
Find the exact value of each of the following. (3 pts. each)
a.
sin  1 1
Answer:

2
NOTE: Since 1 is the maximum positive of 1, then the angle answer is
not in the first quadrant. So, the angle answer is on one of the
coordinate axes, namely the positive y-axis. The angle which is in the

  
closed interval  ,  and lies on the positive y-axis is the angle .
2
 2 2
We know that sin

 1 by Unit Circle Trigonometry. Note that the
2
 
point  , 1 is on the graph of the sine function.
2 
c.

3

Arc sin  


2


Answer: 

3
3
is negative and not minimum negative of  1 ,
2
then the angle answer is in the fourth quadrant. That is, the angle
  
answer is in the open interval   , 0  .
 2 
NOTE: Since 

3
3





Arc
sin


'

Arc
sin

If we let
,
then
. Thus,
 2 
2
3



3


  .
Arc
sin

    '   . That is,

2 
3
3

3

 
sin



sin




We know that
using reference angles.
3
2
 3
 
3

 is on the graph of the sine function.

,

Note that the point  3
2 

d.
Answer: 
tan  1 (  1)

4
NOTE: Since  1 is negative, then the angle answer is in the fourth
  
quadrant. That is, the angle answer is in the open interval   , 0  .
 2 
1
1
If we let   tan (  1) , then  '  tan 1 
  '  


1
. That is, tan (  1)   .
4
4

. Thus,
4

 
  1 using reference angles.
We know that tan      tan
4
 4
 

Note that the point   ,  1 is on the graph of the tangent function.
 4
