Download Pre-Class Problems 21 for Thursday, April 18 Problems which are

Pre-Class Problems 21 for Thursday, April 18
Problems which are due at the beginning of class:
3
   2  , then find the exact value of the six
2

trigonometric functions of the angle .
2
1a.
If sin   
3
8
1b.
If cot  
13
6
and
and      
trigonometric functions of the angle

, then find the exact value of the six
2

.
2
7
.
12
2.
Find the exact value of sin
3.
One website that you used for these pre-class problems other than mine.
These are the type of problems that you will be working on in class.
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.
1a.
3
   2  , then find the exact value of the six
2

trigonometric functions of the angle .
2
If cos  
5
8
and
1b.
If sin   
2
6
and      

, then find the exact value of the six
2

trigonometric functions of the angle .
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 267 … 53 - 60, 63 - 72.
Example 6 on page 262.
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
= 
13
4
1  cos 
2
= 
1 
2
5
8
= 
5
8  8
= 
2
8
1 
8  5
16
cos

 
2
13

 sec  
4
2
1  cos 
2
1 
sin


2
sin
3


4

 csc 
2
4
2
3
=
4
13
5
8
3


4
2
 
tan 
=

2
13
cos

2
4
sin
Using the half-angle formula tan
tan

 
2
= 
3
13
1  cos 
1  cos 
=
2
Using Basic Identities to find tan
=

5
8  8
=
2
8
1 
8  5
=
16
3
4

:
2
3
13

 
2
5
8
5
1 
8
1  cos 

to find tan :
1  cos 
2
1 
=

5
8  8
5 8 = 
1 
8
1 
8  5
8  5
tan
3
13


 
 cot  
2
2
13
3
Answer:
1b.
sin   
13

sec
 
,
2
4
cos

 
2
sin


2
tan

 
2
3

csc

,
2
4
3
13
, cot
4
13
4
3

 
2
13
3
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
34
6
cos    cos  '  
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  cos 
2
=

34 

1  

6 

=
2
34
6
1 
2
=
1 
34
6  6
=
6
2
18  3
=
6  34 3

=
12
3
18  3 34
36
34
6  34
12


2
6 
12

6  34 6 
6(6 
sin

=
6
cos

6  34
12
34 ) =

 
2
1 
 sec
34
6  6
= 
2
6
18  3 34
36
=
36  6
1  cos 
2
= 
12
6  34
12 ( 6  34 )
=
36  34
34
34


2
=
12 ( 6  34 )
=
2
34

34 

1  


6


= 
2
= 
6  34
12
18  3
6
34
= 
34
6
1
2
6  34 3

=
12
3
=
sin
6  34
12

 
2

= 
6 
12

6  34
6 
6(6 
 csc
34
34
34 ) = 

 
2
36  6
= 

3( 6 
34 )
3( 6 
34 )
(6 
(6 
6 
34
6 
34
34 ) 2
34 ) ( 6 
70  12 34
2
= 
= 
34 )
12 ( 6  34 )
2
34

:
2
18  3 34


sin

6
2
tan

 =
2
18  3 34
cos
2
6

=
12 ( 6  34 )
= 
36  34
= 
Using Basic Identities to find tan
12
6  34
 
18  3
34
18  3
34
= 
= 
70  12 34
2

tan
 
Using the half-angle formula
2
6 
34
6 
34
=

6 
34
6 
34
36  12 34  34
36  34
= 
=
35  6 34
1  cos 

to find tan :
1  cos 
2
tan


1  



1  



1  cos 

 
=
2
1  cos 
1 
1 
34
6  6

6 =
34
6
6 
34
6 
34





34  =

6 
34
6
= 
34
6
34
6
1 
1 
35  6 34
=
(from our
work above)
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:
6 
34
6 
34
cos


2
sin

 
2
tan

 
2
, you will obtain 
18  3 34
, sec
6
18  3
6
35  6
35  6 34 .


2
36  6
34
, csc

 
2
36  6
34
34 , cot

 
2
35  6
34
34
1c.
tan   
NOTE:
17

   
and
8
2
Back to Problem 1.

      is in the second 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 sec   tan   1 to find sec  which
will give us cos  :
sec   tan   1 and tan   
2
sec 2  
2
17
17
1 
 sec 2  
64
8
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
=
1
=
18
1

18

1  

2
8

9
1 2

=
18 2
1

 sec

2
18
1  cos 
2
=
1 
=
2
=
36
8
9
2
8
9  9
=
2
9
1 
=
2
6
18
 8
1   
 9
=
2
1 
2
8
9
8
9  9
=
2
9
1 
=
9  8
=
18
sin


2
17
=
18
17 2

=
18 2
34

 csc

6
2


2
tan

 =
2
cos
2
34
6 
2
6


2
1  cos 
1  cos 
9  8
=
9  8
tan


2
=
17
=
1
17
17  cot


2

:
2
34
2
Using the half-angle formula tan
tan
34
6
6 34
3 34
6


34
17
34
Using Basic Identities to find tan
sin
34
=
36


2

1  


1  

1
17

17
1  cos 

tan
to find
:
1  cos 
2
8

9
8 =

9
8
9
8
1 
9
1 
8
9  9
8 9 =
1 
9
1 
=
Answer:
1d.
csc  
cos


2
2


, sec
2
6
sin


2
34
3 34


, csc
6
2
17
tan


2
17 , cot


2
18
1
17
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  :
sin  
5
 sin  ' 
3
5
:
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
= 
5
= 
6
5
6
= 
= 
30
6
1
2
2
3
= 
2
3  3
= 
2
3
1 
3  2
6
cos

 
2
5
6
 sec

 
2
sin

 
2
1  cos 
2

1
6
1


=
=
6
6
6
sin

1

 
 csc  
2
2
6
= 
6
5
1 
=
1
5
1  cos 
1  cos 
30



2
2
3
2
1 
3
1 
=
3  2
6

:
2
6
Using the half-angle formula tan


2
= 
2
3  3
= 
2
3
1 
6
sin
tan
2
3
2
Using Basic Identities to find tan
6



6 
2
tan 
=

2
30
cos

2
6
30
5
 
1
5
1  cos 

to find tan :
1  cos 
2
2
3  3
2 3 =
1 
3
1 
=
3  2
=
3  2
1
5
tan

1


 cot 
2
2
5
Answer:
2a.
sin
5
cos

 
2
30

 
, sec
6
2
sin
6


 
, csc  
2
2
6
tan

1


cot

,
2
2
5
30
5
6
5

12
NOTE:
Back to Problem 2.




is one-half of
or is twice .
12
12
6
6
We will use the half-angle formula sin






with
, which means that
.
2
12
6
NOTE: The angle

 
2
1  cos 
2
to find sin

is in the first quadrant. Sine is positive in the first
12
quadrant.
sin


12
1  cos
2

12

6
3
2
1
=
2
1 
=
3
2  2
=
2
2
2 
3
4
2 
=
2
NOTE: cos
Answer:
2b.
3


6
3
2
2 
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
3
2
3
3
2
1
2
=
2 
Answer:
2c.
cot
3
2
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
 

formula tan
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
1  cos
6
5
1  cos
6
3
2  2
2 =
3
2

3

1  

2 


3 =

1  


2


=
2 
3
2 
3
1
1
3
2
3
2
=
NOTE: cos
tan
5

12
3
5

  cos  
6
6
2
2 
3
2 
3
(2 
=
(2 
7  4
3 )2
3 )(2 
3)
=
5

12
2 
3
2 
3

4  4 3  3
=
4  3
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  ,
sin


sin
(

165

)
formula
to find
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  cos (  330  )
= 
2
3
2
1 
2
= 
3
2
1 
2
=
1 

3
2  2
= 
2
2
2 
3
4
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 
3
2
3
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 
= 
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

7

1  cos 
7

formula cos  
to find sec
with
, which means
2
8
8
2
2
7
that  
.
4
7
NOTE: The angle
is in the second quadrant. Cosine is negative in the
8
second quadrant.
cos
7
 
8
2 

4
cos
2 
7
 
8
2 
2
2
4  2
2
2
= 
2
2
2  2
=
2
2
2
2
2 
2
2
2
2
 sec
2 
= 
1 
2
2

2 
2 
7

 cos

4
4
2


= 
2
2
1 
= 
2
2
NOTE: cos
7
4
1  cos
7
 
8
= 
2 
2
2
2
2 
2 
2
(2 
= 
=
2
2
2 )(2 
2
2 
2
2)
2

=
2
2
=

2
2(2 
2
Answer: 
2f.
2 )
4  2
= 
4  2
2
2
 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
3 

4 

=
3 

4 
2 
2
2 
2

2

1  
 2 




2 =

1  
 2 


= 
2 
2
2 
2

1
1
2 
2
2 
2
2
2
2
2
=


(2 
2 )2
(2 
2 )(2 
3  2
2
2)
4  4 2  2
= 
4  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
7

3
cot



   2  . Put
if
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 2   tan 2   1 and tan   
sec 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  ' 
7
4
adj
and cosine is positive in the fourth quadrant.
hyp
NOTE: cos  ' 
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  


cos
 
, we obtain that
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  


tan
 
, we obtain that
2
2
Back to the half-angles formulas above.
1  cos 2 
2
1  cos 
1  cos 
1  cos 2 
1  cos 2 
