Download Pre-Class Problems 24

Pre-Class Problems 24 for Friday, April 21
These are the type of problems that you will be working on in class.
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: If given the value of a trigonometric function
of an angle and information pertaining the location of the angle, then to use the
half-angle formulas to find the values of the six trigonometric functions of the half
angle.
3
   2  , then find the exact value of the six
2

trigonometric functions of the angle .
2
5
8
1a.
If cos  
1b.
If sin   
and
2
6
and      
trigonometric functions of the angle

, then find the exact value of the six
2

.
2

    , then find the exact value of the six
2

trigonometric functions of the angle .
2
1c.
If tan   
1d.
If csc  
17
8
and
3
3
and  2     
, then find the exact value of the six
2
5

trigonometric functions of the angle .
2
Objective of the following problems: To use the half-angle formulas to find the
value of a trigonometric function of a certain angle.
2.
Find the exact value of the following.
a.
sin

12
b.
cos 75 
e.
sec
7
8
f.
 3 
tan  

 8 
c.
cot
5
12
d.
csc (  165  )
Additional problems available in the textbook: Page 598 … 39 – 44, 47 – 50.
Example 4 – 6 on page 595.
Solutions:
COMMENT: You will be given the following Half-Angle Formulas:
cos

 
2
1  cos 
2
sin

 
2
1  cos 
2
tan

 
2
1  cos 
1  cos 
These formulas are derived at the end of this set of pre-class problems.
1a.
cos  
5
3
   2
and
8
2
Back to Problem 1.
NOTE: The value of cos  for the half-angle formulas has been given to us.
In order to use the half-angle formulas, we also need to know what quadrant

is in:
2
3
3


   2 

  
is in the second quadrant
2
4
2
2
NOTE: Cosine is negative, sine is positive, and tangent is negative in the
second quadrant.
cos

 
2
= 
cos
1  cos 
2
5
8
1 
= 
2
= 
5
8  8
= 
2
8
1 
8  5
16
13
4

 
2
13

 sec  
4
2
1  cos 
2
1 
sin


2
sin
3


4

 csc 
2
4
2
3
=
4
13
5
8
2
Using Basic Identities to find tan
5
8  8
=
2
8
1 
=

:
2
8  5
=
16
3
4
3


4
2
 
tan 
=

2
13
cos

2
4
sin
Using the half-angle formula tan
tan

 
2
= 
tan
1  cos 
1  cos 
=

3
13

 
2
5
8
5
1 
8
1  cos 

to find tan :
1  cos 
2
1 
=

3
13
3
13


 
 cot  
2
2
13
3
Answer:

13
sec
 
,
2
4
cos

 
2
sin


2
tan

 
2
3

csc

,
2
4
3
13
, cot
4
13
4
3

 
2
13
3
5
8  8
5 8 = 
1 
8
1 
8  5
8  5
1b.
sin   
2

and      
2
6
NOTE:      
Back to Problem 1.

  is in the third quadrant
2
In order to use the half-angle formulas, we will need to find cos  . You can
use the Pythagorean Identities or Right Triangle Trigonometry to find cos  .
2
2
Using the Pythagorean Identity cos   sin   1 to find cos  :
cos 2   sin 2   1 and sin   
cos 2  
34
 cos   
36
2
2
1 
 cos 2  
36
6
34
 cos   
6
34
since  is the III
6
quadrant. NOTE: Cosine is negative in the third quadrant.
Using Right Triangle Trigonometry to find cos  :
sin   
2
 sin  ' 
6
2
:
6
6
2
NOTE: sin  ' 
opp
hyp
'
34
cos    cos  '  
34
6
NOTE: cos  ' 
adj
and cosine is negative in the third quadrant.
hyp
In order to use the half-angle formulas, we also need to know what quadrant

is in:
2
    





 

 

is in the fourth quadrant
2
2
2
4
2
NOTE: Cosine is positive, sine is negative, and tangent is negative in the
fourth quadrant.
cos


2
1 
1  cos 
2
=
34
6  6
=
2
6
18  3
=

34 

1  


6


=
2
6  34
12
=
34
6
1 
2
6  34 3

=
12
3
=
18  3 34
36
34
6
cos


2
6  34
12
6 
12

6  34 6 
 sec
34
34
=


2
12
6  34
12 ( 6  34 )
=
36  34
=
12 ( 6  34 )
=
2
6(6 
sin

 
2
1 

1  cos 
2
18  3 34
36
sin

 
2
= 
= 
6  34
12
6 
12

6  34
6 
6(6 
34

34 

1  

6 

= 
2
= 
34
6  6
= 
2
6


36  6
34 ) =
6  34
12
18  3
= 
34
6
1
2
=
6  34 3

=
12
3
34
6
 csc
34
34
34 ) = 

 
2
=
12 ( 6  34 )
= 
36  34
= 
36  6
Using Basic Identities to find tan
12
6  34

:
2
34
12 ( 6  34 )
2
18  3 34


sin

6
2
tan

 =
2
18  3 34
cos
2
6

= 

3( 6 
34 )
3( 6 
34 )
(6 
(6 
6 
34
6 
34
34 ) 2
34 ) ( 6 
70  12 34
2
= 
= 
34 )


1  cos 

 
=
2
1  cos 
1 
1 
34
6  6

6 =
34
6
work above)
= 
70  12 34
2
Using the half-angle formula tan
tan
 
18  3
34
18  3
34
= 
6 
34
6 
34
=

6 
34
6 
34
36  12 34  34
=
36  34
= 
35  6 34

1  cos 

 
to find tan :
2
1  cos 
2

1  



1  


6 
34
6 
34





34  =

6 
34
6
= 
1 
1 
35  6 34
34
6
34
6
=
(from our
tan

 
2
6 
34
6 
34
 cot

 
2
6 
34
6 
34
NOTE: I’ll let you verify that if you rationalize the denominator in the
answer 
Answer:
1c.
tan   
6 
34
6 
34
cos


2
sin

 
2
tan

 
2
, you will obtain 
18  3 34
, sec
6
18  3


2
36  6
34
, csc

 
2
36  6
34
34 , cot

 
2
35  6
34
34
6
35  6
35  6 34 .
17

   
and
8
2
Back to Problem 1.

      is in the second quadrant
NOTE:
2
In order to use the half-angle formulas, we will need to find cos  . You can
use the Pythagorean Identities or Right Triangle Trigonometry to find cos  .
2
2
Using the Pythagorean Identity sec   tan   1 to find sec  which
will give us cos  :
sec 2   tan 2   1 and tan   
17
17
1 
 sec 2  
64
8
sec 2  
81
9
9
 sec   
 sec    since  is the II
64
8
8
quadrant. NOTE: Secant is negative in the second quadrant.
sec   
9
8
 cos   
8
9
Using Right Triangle Trigonometry to find cos  :
tan   
17
 tan  ' 
8
17
:
8
17
9
NOTE: tan  ' 
opp
adj
'
8
cos    cos  '  
NOTE: cos  ' 
8
9
adj
and cosine is negative in the second quadrant.
hyp
In order to use the half-angle formulas, we also need to know what quadrant

is in:
2





    



is in the first quadrant
2
4
2
2
2
NOTE: Cosine is positive, sine is positive, and tangent is positive in the first
quadrant.
cos


2
1  cos 
2
9  8
=
18
cos
sin


2


2
sin


2
8

9
1 2

=
18 2
17
=
18
=

1  

2
8

9
17 2

=
18 2
34

 csc

6
2
8
9
1 
=
2
=
36
1

 sec

2
18
1  cos 
2
9  8
=
18
=
1
=
18
1

18

1  

2
=
2
2
6
18
1 
=
34
=
36
8
9
2

:
2
8
9  9
=
2
9
1 
=
34
6
6 34
3 34
6


34
17
34
Using Basic Identities to find tan
8
9  9
=
2
9
1 


2
tan

 =
2
cos
2
sin
34
6 
2
6
34
2
Using the half-angle formula tan
tan


2
1  cos 
1  cos 
9  8
=
9  8
tan


2
Answer:
=
17
=
1
17
17  cot


2



2

1  


1  

17
1  cos 

to find tan :
1  cos 
2
8

9
8 =

9
1
17
cos


2
2


, sec
2
6
sin


2
34
3 34


, csc
6
2
17
tan


2
17 , cot


2
18
1
17
8
9
8
1 
9
1 
8
9  9
8 9 =
1 
9
1 
=
1d.
csc  
3
3
and  2     
2
5
NOTE:  2     
Back to Problem 1.
3
  is in the first quadrant
2
In order to use the half-angle formulas, we will need to find cos  . You can
use the Pythagorean Identities or Right Triangle Trigonometry to find cos  .
2
2
Using the Pythagorean Identity cos   sin   1 to find cos  :
csc  
3
 sin  
5
5
3
cos 2   sin 2   1 and sin  
cos 2  
5
5
1 
 cos 2  
9
3
4
2
2
 cos   
 cos  
since  is the I
9
3
3
quadrant. NOTE: Cosine is positive in the first quadrant.
Using Right Triangle Trigonometry to find cos  :
5
5
sin  
 sin  ' 
:
3
3
3
5
NOTE: sin  ' 
opp
hyp
'
2
cos   cos  ' 
2
3
NOTE: cos  ' 
adj
and cosine is positive in the first quadrant.
hyp
In order to use the half-angle formulas, we also need to know what quadrant

is in:
2
 2    
3

3

  
 

is in the third quadrant
2
2
4
2
NOTE: Cosine is negative, sine is negative, and tangent is positive in the
third quadrant.
cos

1  cos 
 
2
2
= 
cos
sin
5
= 
6

 
2

 
2
5
6
5
6
= 
= 
 sec
1  cos 
2
1
2
3
2
= 
2
3  3
= 
2
3
3  2
6
2
3  3
= 
2
3
3  2
6
1 
30
6

 
2
= 
6
5
1 
2
 
2
3
30
5
= 
1 

6
1
1


=
=
6
6
6
sin

1

 
 csc  
2
2
6
6
Using Basic Identities to find tan
6



6 
2
tan 
 =
2
30
cos

2
6
sin
6
30
Using the half-angle formula tan
=
1
5
tan

1


 cot 
2
2
5

 
2


2
2
3
2
1 
3


2
cos

1 
tan
Answer:
1  cos 
1  cos 

:
2
=
1
5
1  cos 

tan
to find
:
1  cos 
2
2
3  3
2 3 =
1 
3
1 
=
5
30

 
, sec
6
2
30
5
3  2
=
3  2
1
5
2a.
sin
sin
6


 
, csc  
2
2
6
tan

1



, cot
2
2
5
6
5

12
Back to Problem 2.




is one-half of
or is twice .
12
12
6
6
NOTE:
We will use the half-angle formula sin
with




, which means that   .
2
12
6

 
2
1  cos 
2
to find sin

12

NOTE: The angle
is in the first quadrant. Sine is positive in the first
12
quadrant.
sin


12
2 
=
1  cos

6
2
3
2
NOTE: cos


6
3
2
3
2
1
=
2
1 
=
3
2  2
=
2
2
2 
3
4
2 
Answer:
2b.
3
2
cos 75 
Back to Problem 2.
NOTE: 75  is one-half of 150  or 150  is twice 75  .
We will use the half-angle formula cos

 
2

 75  , which means that   150  .
with
2
1  cos 
to find cos 75 
2
NOTE: The angle 75  is in the first quadrant. Cosine is positive in the first
quadrant.
cos 75  
1 

3

1  

2 

=
2
1  cos 150 
=
2
3
2  2
=
2
2
2 
2 
3
4
=
NOTE: cos 150    cos 30   
2 
Answer:
2
3
2
3
2
3
3
2
1
2
=
2c.
cot
5
12
NOTE:
Back to Problem 2.
5
5
5
5
is one-half of
or
is twice
.
6
6
12
12
Since cotangent is the reciprocal of tangent, then we will use the half-angle

1  cos 

5
5
tan



formula
to find tan
with
, which
2
1  cos 
2
12
12
5


means that
.
6
NOTE: The angle
5
is in the first quadrant. Tangent is positive in the first
12
quadrant.
tan
5

12
1 
1 
5
6
5
1  cos
6
1  cos
3
2  2
2 =
3
2
NOTE: cos

3


1  

2 


3 =


1  

2 

=
2 
3
2 
3
3
5

  cos  
6
6
2
1
1
3
2
3
2
=
tan
5

12
2 
3
2 
3
(2 
=
(2 
7  4
5

12
3)
2 
3
2 
3

4  4 3  3
=
4  3
3 )2
3 )(2 
=
2 
3
2 
3
7  4
1

2 
3
2 
3
3
=
3
Answer:
2d.
 cot
7  4
3
csc (  165  )
Back to Problem 2.
NOTE:  165  is one-half of  330  or  330  is twice  165  .
Since cosecant is the reciprocal of sine, then we will use the half-angle


1  cos 
  165  ,
 
formula sin
to find sin (  165  ) with
2
2
2
which means that    330  .
NOTE: The angle  165  is in the third quadrant. Sine is negative in the
third quadrant.
sin (  165  )  

1 
1  cos (  330  )
= 
2
3
2  2
= 
2
2
2 
3
4
= 
3
2
1 
2
2 
2
3
= 
3
2
1 
2
=
NOTE: cos (  330  )  cos 30  
2 
sin (  165  )  
2



2 
2
3
2 
3
4  3
Answer:  2
2e.
sec
3
3
2 
3
= 
2 
2
= 
2 
1
3
2
(2 
= 2
2 
2 
3
=
3
3 )(2 
2 
3)
=
3
3
7
8
NOTE:
2
 csc (  165  )  
2
2 
3
2
Back to Problem 2.
7
7
7
7
is one-half of
or
is twice
.
8
4
4
8
Since secant is the reciprocal of cosine, then we will use the half-angle

1  cos 

7
7

formula cos  
to find sec
with
, which means
2
8
8
2
2
7
that  
.
4
NOTE: The angle
7
is in the second quadrant. Cosine is negative in the
8
second quadrant.
cos
7
 
8
2 

4
cos
2 

2

2
2
2 
2
4  2
2(2 
2
2 
2
= 
2 )
4  2
2
2
 sec
2 
2
Answer: 
2
2

= 
2
2
2  2
=
2
2
2
2 
7
 
8
1 
2
7

 cos

4
4
2

2 
= 
2
2
1 
= 
2
2
NOTE: cos
7
4
1  cos
2
7
 
8
= 
2 
2
2
= 
2
4  2
2
2 
2 
2
(2 
= 
2
=
2
2
2 )(2 
2
2 
2
2)
2

=
2
2
=
2f.
 3 
tan  

 8 
NOTE: 
Back to Problem 2.
3
3
3
3
is one-half of 
or 
is twice 
.
8
4
4
8

1  cos 
 3 
 

to find tan  
2
1  cos 
 8 

3
3
 
with
, which means that   
.
4
2
8
We will use the half-angle formula tan
NOTE: The angle 
3
is in the fourth quadrant. Tangent is negative in the
8
fourth quadrant.
 3 
tan  
  
 8 
=



1 
1 

1  cos  


1  cos  

2
2  2

2 =
2
2
(2 
2 )2
(2 
2 )(2 
3  2
2
2)
3 

4 

=
3 

4 
2 
2
2 
2
= 

2

1  
 2 




2 =

1  
 2 


= 
2 
2
2 
2
4  4 2  2
= 
4  2

2
2
2
2
1
1
2 
2
2 
2
6  4
2
=
2
=
2

 3 
cos



cos




NOTE:
4
2
 4 
Answer: 
3  2
2
Solution to Problems on the Pre-Exam:
23.
Find the exact value of cos
Back to Page 1.
7

3
   2  . Put
if cot   
and
2
3
2
a box around your answer. (4 pts.)
From the Formula Sheet, we have that cos

 
2
1  cos 
.
2

3
3

   2 

  . Thus, the angle
is in the II quadrant.
2
4
2
2
Thus, cos

 
2
1  cos 
2
since cosine is negative in the second
quadrant.
We will need to find cos  . You can use the Pythagorean Identities or Right
Triangle Trigonometry to find cos  .
cot   
7
3
 tan   
3
7
3
   2    is the IV quadrant
2
2
2
Using the Pythagorean Identity sec   tan   1 to find sec  which
will give us cos  :
sec   tan   1 and tan   
2
sec 2  
2
3
9
 sec 2  
1 
7
7
4
16
 sec  
 sec   
7
7
4
since  is the IV
7
quadrant. NOTE: Secant is positive in the fourth quadrant.
sec  
4
 cos  
7
7
4
Using Right Triangle Trigonometry to find cos  :
tan   
3
 tan  ' 
7
3
:
7
4
NOTE: tan  ' 
opp
adj
3
'
7
cos   cos  ' 
NOTE: cos  ' 
7
4
adj
and cosine is positive in the fourth quadrant.
hyp
cos

 
2



24.
4 
1  cos 
2
7
= 
8
8  2
= 
4 
7
8
7
4
1 

2
= 
2
8  2 7
16
7
4  4
=
2
4
=
7
4
4 
7
8
or

Find the exact value of tan
NOTE:
8  2
7
4
5
. (5 pts.) Put a box around your answer.
8
5
5
5
5
is one-half of
or
is twice
.
8
4
4
8
We will use the half-angle formula tan
with
= 
2
1 

1  cos 
5
 
to find tan
2
1  cos 
8

5
5

, which means that  
.
2
8
4
NOTE: The angle
second quadrant.
5
is in the second quadrant. Tangent is negative in the
8
tan
=


5
1  cos
4
5
1  cos
4
5
 
8

2
2  2

2 =
2
2
1 
1 
(2 
(2 
2 )(2 
3  2
2
NOTE: cos

2 )2
2 
2
2 
2
2)
=


2

1  
 2 




2 =

1  
 2 


2 
2
2 
2
= 
= 
1
1
2 
2
2 
2

4  4 2  2
= 
4  2
2
2
2
2
2 
2
2 
2
6  4
2
=
2
=
2
5

  cos  
4
4
2
or

3  2
2
Deriving the Half-Angle Formulas
2
Using the double angle formula cos 2   2 cos   1 and solving for cos  , we
obtain the following:
2
cos 2   2 cos 2   1  1  cos 2  2 cos 2   cos  
1  cos 2 
2

cos   
1  cos 2 
.
2
Setting  


, we obtain that cos  
2
2
1  cos 
2
2
Using the double angle formula cos 2   1  2 sin  and solving for sin  , we
obtain the following:
2
cos 2  1  2 sin 2   2 sin 2   1  cos 2   sin  
sin   
1  cos 2 
.
2
Setting  


 
, we obtain that sin
2
2
1  cos 
2
1  cos 2 
sin 
1  cos 2 
2
tan 2  


 tan   
1  cos 2 
cos 2 
1  cos 2 
2
2
Setting  


, we obtain that tan  
2
2
Back to the half-angles formulas above.
1  cos 2 
2
1  cos 
1  cos 
1  cos 2 
1  cos 2 
