Download Pre-Class Problems 11 for Thursday, October 23 These are the type

Pre-Class Problems 11 for Thursday, October 23
These are the type of problems that you will be working on in class. These
problems are from Lesson 8.
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 use a sketch of the graph of two cycles of
the sine and cosine functions in order to sketch two cycles of the respective
cosecant and secant functions and label the numbers on the x- and y-axes.
1.
Identify the amplitude, period, and phase shift. Sketch two cycles of the
graph of the following functions on the same side of the y-axis. Label the
numbers on the x- and y-axes as needed. You only need to label where each
cycle begins and ends.
1
x
 x
y


csc
y

6
sec


y

5
csc
2
x
a.
b.
c.
2
6 
3

4x
7
d.
y   11 sec
f.
3 

y  sec   x 

8 

h.
y  
11 
 x
2 sec  

18 
3
e.
2 

y   4 csc  x 

3 

g.
y 
3
 

csc   5 x 

7
4

i.
y 
2
8 
 6x
csc 


9
35 
 7
Additional problems available in the textbook: Page 177 … 19 - 23, 25, 26, 29, 33,
34, 35, 37, 41, 42, 45, 46, 48. Don’t use a graphing utility for Problems 41, 42, 45,
46, and 48. Examples 4 and 5 on page 174.
Solutions:
1a.
y  5 csc 2 x
Back to Problem 1.
NOTE: We will first need to sketch two cycles of the graph of the function
y  5 sin 2 x .
Amplitude of the sine function y  5 sin 2 x : 5
Amplitude of the cosecant function y  5 csc 2 x : None
2
Period of both functions :
= 
2
Phase Shift of both functions: None
y
5

x
2
x
5
NOTE: The vertical asymptotes of this cosecant function occur at the xintercepts of the sine function.
NOTE: The local maximums of the cosecant function occur at the local
minimums of the sine function. Also, the local minimums of the cosecant
function occur at the local maximums of the sine function.
1b.
y 
6 sec
x
3
Back to Problem 1.
NOTE: We will first need to sketch two cycles of the graph of the function
x
y  6 cos .
3
x
Amplitude of the cosine function y  6 cos : 6
3
Amplitude of the secant function y 
6 sec
x
: None
3
2
Period of both functions: 1 = 2   3 = 6 
3
Phase Shift of both functions: None
y
6
6
x
x

12 
x
6
NOTE: The vertical asymptotes of this secant function occur at the xintercepts of the cosine function.
NOTE: The local maximums of the secant function occur at the local
minimums of the cosine function. Also, the local minimums of the secant
function occur at the local maximums of the cosine function.
1c.
y  
1
 x
csc  

2
6 

Back to Problem 1.
NOTE: We will first need to sketch two cycles of the graph of the function
1
 x
y   sin  
.
2
6 

y  
1
1
x
 x
sin  
  sin
2
6 
2
6

NOTE: We also have that y  
1
1
x
 x
csc  
  csc
2
6 
2
6

Amplitude of the sine function y 
1
x 1
sin
:
2
2
6
Amplitude of the cosecant function y 
1
x
csc
: None
2
6
2
6
6
2


2

Period of both functions:  =
=
= 12

1
6
Phase Shift of both functions: None
y
1
2
12

24
x
1
2
NOTE: The vertical asymptotes of this cosecant function occur at the xintercepts of the sine function.
NOTE: The local maximums of the cosecant function occur at the local
minimums of the sine function. Also, the local minimums of the cosecant
function occur at the local maximums of the sine function.
1d.
y   11 sec
4x
7
Back to Problem 1.
NOTE: We will first need to sketch two cycles of the graph of the function
4x
y   11 cos
.
7
NOTE: Because of the multiplication by the number  11 , the cosine (and
the secant) cycles will be inverted.
Amplitude of the cosine function y   11 cos
4x
: 11
7
Amplitude of the secant function y   11 sec
4x
: None
7
2
7
7
7
Period of both functions: 4 = 2  
=  
=
4
2
2
7
Phase Shift of both functions: None
y
11
x
x
 11
7
2
7
x
NOTE: The vertical asymptotes of this secant function occur at the xintercepts of the cosine function.
NOTE: The local maximums of the secant function occur at the local
minimums of the cosine function. Also, the local minimums of the secant
function occur at the local maximums of the cosine function.
1e.
2 

y   4 csc  x 

3 

Back to Problem 1.
NOTE: We will first
2

y   4 sin  x 
3

need to sketch two cycles of the graph of the function

.

NOTE: Because of the multiplication by the number  4 , the sine (and the
cosecant) cycles will be inverted.
2 

: 4
Amplitude of the sine function y   4 sin  x 
3 

2

y


4
csc
x


Amplitude of the cosecant function
3


 : None

Period of both functions: 2 
2
Phase Shift of both functions:
units to the left
3
NOTE: The cycle will start at 
2
because of the shift.
3
NOTE: Starting Point + Period = 
Thus, the cycle will end at
2
2
6
4
 2 = 

=
.
3
3
3
3
4
.
3
2
4
and ends at
, the cycle will have to
3
3
cross the y-axis. With the up and down oscillation of the sine cycle, I do not
want to cross the y-axis. Remember, where one cycle ends, another cycle
4
begins. So, we will sketch the cycle that starts at
as our first cycle and
3
the cycle that comes after this one as our second cycle.
NOTE: Since the cycle starts at 
4
4
6
 2 =

3
3
3
10 
Thus, our first sketched cycle will end at
.
3
NOTE: Starting Point 1 + Period =
=
10 
.
3
10 
10 
6
16 
 2 =

=
. Thus, the
3
3
3
3
16 
second sketched cycle will end at
.
3
y
Starting Point 2 + Period =
4

2
3
4
3
10 
3
16 
3
x
4
NOTE: The vertical asymptotes of this cosecant function occur at the xintercepts of the sine function.
NOTE: The local maximums of the cosecant function occur at the local
minimums of the sine function. Also, the local minimums of the cosecant
function occur at the local maximums of the sine function.
1f.
3 

y  sec   x 

8 

Back to Problem 1.
NOTE: We will first need to sketch two cycles of the graph of the function
3 

y  cos   x 
.
8 

 
3 
3 

y  cos   x 
  cos   x  
8 
8 

 
 
3 
Amplitude of the cosine function y  cos   x   : 1
8 
 
 
3 
Amplitude of the secant function y  sec   x   : None
8 
 
Period of both functions:
2
 = 2
Phase Shift of both functions:
3
units to the right
8
NOTE: The first cycle will start at
3
because of the shift.
8
NOTE: Starting Point 1 + Period =
first cycle will end at
19
.
8
3
3
16
19
 2 =

=
. Thus, the
8
8
8
8
Starting Point 2 + Period =
19
19
16
35
 2 =

=
. Thus, the second
8
8
8
8
35
cycle will end at
.
8
y
1
x
1
3
8
19
8
35
8
x
NOTE: The vertical asymptotes of this secant function occur at the xintercepts of the cosine function.
NOTE: The local maximums of the secant function occur at the local
minimums of the cosine function. Also, the local minimums of the secant
function occur at the local maximums of the cosine function.
1g.
y 
3


csc   5 x  
7
4

Back to Problem 1.
NOTE: We will first need to sketch two cycles of the graph of the function
3


y  sin   5 x   .
7
4

y 

 
3
 3
 
3
 


sin   5 x    sin  5  x 
   sin 5  x 

7
4
7
20 
7
20 



 


 1


5

NOTE: The
was obtained by
=
=
.
20
4
20
4 5
3
NOTE: Because of the multiplication by the number  , the sine (and the
7
cosecant) cycles will be inverted.
Amplitude of the sine function y  
 
3
  3
sin 5  x 
 :
7
20

 7

Amplitude of the cosecant function y  
Period of both functions:
 
3
 
csc 5  x 
 : None
7
20 
 
2
5

Phase Shift of both functions:
units to the right
20
NOTE: The first cycle will start at

because of the shift.
20
NOTE: Starting Point 1 + Period =
Thus, the first cycle will end at
Starting Point 2 + Period =
second cycle will end at
17 
.
20
9
.
20

2

20
5
=

8
9

=
.
20
20
20
9
2
9
8
17 


=
=
. Thus, the
20
5
20
20
20
y
3
7

20

9
20
17 
20
x
3
7
NOTE: The vertical asymptotes of this cosecant function occur at the xintercepts of the sine function.
NOTE: The local maximums of the cosecant function occur at the local
minimums of the sine function. Also, the local minimums of the cosecant
function occur at the local maximums of the sine function.
1h.
y  
11 
 x
2 sec  

18 
3
Back to Problem 1.
NOTE: We will first need to sketch two cycles of the graph of the function
11 
 x
y   2 cos  
.
18 
3
y  
11 
x
2 cos  
  
18 
3
1 
11 
2 cos   x 

6 
3 
11
1
11
11
11


3
1 =
NOTE: The
was obtained by
=
=
18
3
18
6
6
11
.
6
NOTE: Because of the multiplication by the number 
the secant) cycles will be inverted.
2 , the cosine (and
Amplitude of the cosine function y  
1 
11 
2 cos   x 
 :
6 
3 
Amplitude of the secant function y  
1 
11 
2 sec   x 
 : None
6 
3 
2
2
Period of both functions: 1 = 2   3 = 6 
3
Phase Shift of both functions:
11
units to the left
6
NOTE: The cycle will start at 
11
because of the shift.
6
NOTE: Starting Point + Period = 
11
11
36 
 6 = 

=
6
6
6
25 
25 
. Thus, the cycle will end at
.
6
6
11
25 
and ends at
, the cycle will have
6
6
to cross the y-axis. With the up and down oscillation of the cosine cycle, I do
not want to cross the y-axis. Remember, where one cycle ends, another cycle
25 
begins. So, we will sketch the cycle that starts at
as our first cycle and
6
the cycle that comes after this one as our second cycle.
NOTE: Since the cycle starts at 
25 
25 
36 
61
 6 =

=
.
6
6
6
6
61
Thus, our first sketched cycle will end at
.
6
NOTE: Starting Point 1 + Period =
61
61
36 
97 
 6 =

=
. Thus,
6
6
6
6
97 
the second sketched cycle will end at
.
6
Starting Point 2 + Period =
y
5
|
x
11

6
5
|
25 
6
|
61 
6
|
97 
6
x
1i.
y 
2
8 
 6x
csc 


9
35 
 7
Back to Problem 1.
NOTE: We will first need to sketch two cycles of the graph of the function
2
8 
 6x
y  sin 

.
9
35 
 7
y 
6 
2
8 
2
4
 6x
sin 

  sin   x 
9
35 
9
15
 7
7 
NOTE: The



8
6
8 7
4 1
4
4



was obtained by
=
=
=
.
5
3
35
7
35 6
15
15
Amplitude of the sine function y 
6 
2
4   2
sin   x 
 :
9
15  9
7 
Amplitude of the cosecant function y 
6 
2
4  
csc   x 
 : None
9
15 
7 
2
7
7
7
Period of both functions: 6 = 2  
=  
=
6
3
3
7
Phase Shift of both functions:
4
units to the right
15
NOTE: The first cycle will start at
4
because of the shift.
15
NOTE: Starting Point 1 + Period =
Thus, the first cycle will end at
39 
.
15
4
7
4
35 
39 


=
=
.
15
3
15
15
15
39 
7
39 
35 
74 


=
=
. Thus,
15
15
15
15
3
74 
the second cycle will end at
.
15
Starting Point 2 + Period =
y
2
9
4
15
x

39 
15
74 
15
x
2
9
Reducing the fraction
39 
, we obtain our required sketch.
15
y
2
9
4
15
x

2
9
13
5
74 
15
x
Solution to Problems on the Pre-Exam:
18.
Back to Page 1.
Sketch the graph of two cycles of the following function on the same side of
the y-axis. Label the numbers on the x- and y-axes as needed. Label where
the cycles begin and end. (6 pts.)
y  
4
x
sec
9
6
Amplitude: None
2
6
6
Period =  = 2  
= 2
= 12
1

6
Period = 12
Phase Shift: None
NOTE: Because of the multiplication by the number 
4
, the cosine (and the
9
secant) cycles will be inverted.
y
4
9
12
x

4
9
24
x