Download Pre-Class Problems 6

Pre-Class Problems 6 for Monday, February 13
These are the type of problems that you will be working on in class. These
problems are from Lesson 5.
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 problem: To find the exact value of the six
trigonometric functions for an angle if given the coordinates of a point on the
terminal side of the angle. This will require the use of Radius r Trigonometry.
1.
Find the exact value of the six trigonometric functions of the angle  if the
given point is on the terminal side of  .
a.
(  3, 7 )
b.
(10 ,  6 )
c.
(  5,  8)
d.
(  9, 0)
e.
( 5 , 2)
f.
( 0,
g.
(12 , 
17 )
3)
Objective of the following problems: To find the exact value of the six
trigonometric functions for an angle whose terminal side lies on the graph of a line
and the equation of the line is given. This will require the use of Radius r
Trigonometry.
2.
Find the exact value of the six trigonometric functions for the following
angles.
a.
The terminal side of the angle  is in the fourth quadrant and lies on
3
the line y   x .
4
b.
The terminal side of the angle  is in the first quadrant and lies on the
line 5 x  2 y  0 .
c.
The terminal side of the angle  is in the third quadrant and lies on the
7
y

x.
line
3
d.
The terminal side of the angle  is in the second quadrant and lies on
the line 8 x  14 y  0 .
e.
The terminal side of the angle  is in the third quadrant and lies on the
line 4 x  y  0 .
f.
The terminal side of the angle  is in the fourth quadrant and lies on
x
the line y   .
6
Additional problems available in the textbook: Page 505 … 15 – 20, 33 - 36. Page
498 … Example 1.
Solutions:
1a.
(  3, 7 )
r 
x2  y2 
cos  
x
 
r
3
58
sin  
y

r
tan  
y
7
 
x
3
7
58
9  49 
58
58
3
sec  
r
 
x
csc  
r

y
cot  
x
3
 
y
7
58
7
NOTE: In addition to the point (  3 , 7 ) lying on the terminal side of the
2
2
angle  , it also lies on the graph of the circle x  y  58 . This is the
circle whose center is at the origin and whose radius is
58 .
Back to Problem 1.
1b.
(10 ,  6 )
r 
x2  y2 
100  36 
cos  
x
10


r
2 34
sin  
y
6
 
 
r
2 34
tan  
y
6
3
 
 
x
10
5
136 
5
34
3
34
4  34 
34  2 34
4
34
5
sec  
r

x
csc  
r
 
y
cot  
x
5
 
y
3
34
3
NOTE: In addition to the point (10 ,  6 ) lying on the terminal side of the
2
2
angle  , it also lies on the graph of the circle x  y  136 . This is the
circle whose center is at the origin and whose radius is
NOTE: Here is an easier way to simplify
r 
x2  y2 
4
34  2 34
Back to Problem 1.
100  36 
136  2 34 .
136 :
4 ( 25  9 ) 
4 ( 34 ) 
1c.
(  5,  8)
r 
x2  y2 
25  64 
89
cos  
x
 
r
5
89
sec  
r
 
x
89
5
sin  
y
 
r
8
89
csc  
r
 
y
89
8
tan  
y
8

x
5
cot  
x
5

y
8
NOTE: In addition to the point (  5 ,  8 ) lying on the terminal side of the
2
2
angle  , it also lies on the graph of the circle x  y  89 . This is the
circle whose center is at the origin and whose radius is
89 .
Back to Problem 1.
1d.
(  9, 0)
r 
x2  y2 
81  0 
81  9
cos  
x
9
   1
r
9
sec  
r
 1
x
sin  
y
0

 0
r
9
csc  
r
 undefined
y
tan  
y
0

 0
x
9
cot  
x
 undefined
y
NOTE: In addition to the point (  9 , 0 ) lying on the terminal side of the
2
2
angle  , it also lies on the graph of the circle x  y  81 . This is the
circle whose center is at the origin and whose radius is 9.
Back to Problem 1.
1e.
( 5 , 2)
r 
x2  y2 
54 
9  3
sec  
r

x
y
2

r
3
csc  
r
3

y
2
y

x
cot  
x

y
cos  
x

r
sin  
tan  
5
3
2
5
3
5
5
2
NOTE: In addition to the point ( 5 , 2 ) lying on the terminal side of the
2
2
angle  , it also lies on the graph of the circle x  y  9 . This is the
circle whose center is at the origin and whose radius is 3.
Back to Problem 1.
1f.
( 0,
r 
17 )
x2  y2 
0  17 
17
cos  
x

r
0
 0
17
sec  
r
 undefined
x
sin  
y

r
17
csc  
r
1
y
tan  
y

x
17
 undefined
0
cot  
x

y
17
1
0
 0
17
NOTE: In addition to the point ( 0 ,
17 ) lying on the terminal side of the
2
2
angle  , it also lies on the graph of the circle x  y  17 . This is the
circle whose center is at the origin and whose radius is 17 .
Back to Problem 1.
1g.
(12 , 
r 
3)
x2  y2 
144  3 
147 
3  49 
49
3  7 3
cos  
12 3
4 3
x
12



r
73
7
7 3
sec  
7 3
r

x
12
sin  
3
y
1
 
 
r
7
7 3
csc  
r
 7
y
tan  
3
y
 
x
12
cot  
12 3
x
12
 
 
 4 3
y
3
3
NOTE: In addition to the point (12 , 
3 ) lying on the terminal side of the
2
2
angle  , it also lies on the graph of the circle x  y  147 . This is the
circle whose center is at the origin and whose radius is 147  7 3 .
Back to Problem 1.
2a.
The terminal side of the angle  is in the fourth quadrant and lies on the line
3
y   x.
4
3
x in the
4
fourth quadrant. Since x-coordinates are positive in the fourth quadrant, then
we will need to pick a positive number for x. We will find the y-coordinate
3
y


x . Picking the number 4 for x, we
of the point using the equation
4
have the following.
NOTE: We will need to find a point that is on the line y  
x
y
4
3
y
3
3
( 4 )   (1)   3
4
1
NOTE: The point ( 4 ,  3 ) is in the fourth quadrant and lies on the line
3
y   x . Since the terminal side of the angle  is in the fourth quadrant
4
and lies on this line, then the point ( 4 ,  3 ) also lies on the terminal side of
.
r 
x2  y2 
cos  
x
4

r
5
16  9 
25  5
sec  
r
5

x
4
sin  
y
3
 
r
5
csc  
r
5
 
y
3
tan  
y
3
 
x
4
cot  
x
4
 
y
3
NOTE: In addition to the point ( 4 ,  3 ) lying on the terminal side of the
2
2
angle  , it also lies on the graph of the circle x  y  25 . This is the
circle whose center is at the origin and whose radius is 5.
Back to Problem 2.
2b.
The terminal side of the angle  is in the first quadrant and lies on the line
5x  2 y  0.
NOTE: We will need to find a point that is on the line 5 x  2 y  0 in the
first quadrant. Since x-coordinates are positive in the first quadrant, then we
will need to pick a positive number for x. We will find the y-coordinate of
5
the point using the equation y  x . This is the equation that we obtain
2
when we solve the given equation 5 x  2 y  0 for y.
5x  2 y  0  5x  2 y  y 
5
x
2
Picking the number 2 for x, we have the following.
x
y
2
5
y
5
5
( 2 )  (1)  5
2
1
NOTE: The point ( 2 , 5 ) is in the first quadrant and lies on the line
5 x  2 y  0 . Since the terminal side of the angle  is in the first quadrant
and lies on this line, then the point ( 2 , 5 ) also lies on the terminal side of  .
r 
x2  y2 
4  25 
29
cos  
x

r
2
29
sec  
r

x
29
2
sin  
y

r
5
29
csc  
r

y
29
5
tan  
y
5

x
2
cot  
x
2

y
5
NOTE: In addition to the point ( 2 , 5 ) lying on the terminal side of the angle
 , it also lies on the graph of the circle x 2  y 2  29 . This is the circle
whose center is at the origin and whose radius is
29 .
Back to Problem 2.
2c.
The terminal side of the angle  is in the third quadrant and lies on the line
7
y
x.
3
7
x in the third
3
quadrant. Since x-coordinates are negative in the third quadrant, then we will
need to pick a negative number for x. We will find the y-coordinate of the
7
y

x . Picking the number  3 for x, we have
point using the equation
3
the following.
NOTE: We will need to find a point that is on the line y 
x
3
y

7
y
7
7
(  3)  
(1)  
3
1
7
NOTE: The point (  3 , 
7 ) is in the third quadrant and lies on the line
7
x . Since the terminal side of the angle  is in the third quadrant and
3
lies on this line, then the point (  3 ,  7 ) also lies on the terminal side of
.
y
r 
x2  y2 
97 
16  4
cos  
x
3
 
r
4
sec  
r
4
 
x
3
sin  
7
y
 
r
4
csc  
r
4
 
y
7
tan  
y

x
cot  
x

y
7
3
3
7
NOTE: In addition to the point (  3 , 
7 ) lying on the terminal side of the
2
2
angle  , it also lies on the graph of the circle x  y  16 . This is the
circle whose center is at the origin and whose radius is 4.
Back to Problem 2.
2d.
The terminal side of the angle  is in the second quadrant and lies on the line
8 x  14 y  0 .
NOTE: We will need to find a point that is on the line 8 x  14 y  0 in the
second quadrant. Since x-coordinates are negative in the second quadrant,
then we will need to pick a negative number for x. We will find the y4
coordinate of the point using the equation y   x . This is the equation
7
that we obtain when we solve the given equation 8 x  14 y  0 for y.
8 x  14 y  0  14 y   8 x  y  
8
4
x   x
14
7
Picking the number  7 for x, we have the following.
x
y
7
4
y
4
4
(  7 )  (1)  4
7
1
NOTE: The point (  7 , 4 ) is in the second quadrant and lies on the line
8 x  14 y  0 . Since the terminal side of the angle  is in the second
quadrant and lies on this line, then the point (  7 , 4 ) also lies on the terminal
side of  .
r 
x2  y2 
cos  
x
 
r
7
65
sin  
y

r
tan  
y
4
 
x
7
4
65
49  16 
65
65
7
sec  
r
 
x
csc  
r

y
cot  
x
7
 
y
4
65
4
NOTE: In addition to the point (  7 , 4 ) lying on the terminal side of the
2
2
angle  , it also lies on the graph of the circle x  y  65 . This is the
circle whose center is at the origin and whose radius is
Back to Problem 2.
65 .
2e.
The terminal side of the angle  is in the third quadrant and lies on the line
4 x  y  0.
NOTE: We will need to find a point that is on the line 4 x  y  0 in the
third quadrant. Since x-coordinates are negative in the third quadrant, then
we will need to pick a negative number for x. We will find the y-coordinate
of the point using the equation y  4 x . This is the equation that we obtain
when we solve the given equation 4 x  y  0 for y.
4x  y  0  y  4x
Picking the number  1 for x, we have the following.
x
y
1
4
y  4 (  1)   4
NOTE: The point (  1,  4 ) is in the third quadrant and lies on the line
4 x  y  0 . Since the terminal side of the angle  is in the third quadrant
and lies on this line, then the point (  1,  4 ) also lies on the terminal side of
.
r 
x2  y2 
1  16 
17
cos  
x
 
r
1
17
sec  
r
 
x
17
sin  
y
 
r
4
17
csc  
r
 
y
17
4
tan  
y
 4
x
cot  
x
1

y
4
NOTE: In addition to the point (  1,  4 ) lying on the terminal side of the
2
2
angle  , it also lies on the graph of the circle x  y  17 . This is the
circle whose center is at the origin and whose radius is
17 .
Back to Problem 2.
2f.
The terminal side of the angle  is in the fourth quadrant and lies on the line
x
y .
6
x
in the fourth
6
quadrant. Since x-coordinates are positive in the fourth quadrant, then we
will need to pick a positive number for x. We will find the y-coordinate of
x
the point using the equation y   . Picking the number 6 for x, we have
6
the following.
NOTE: We will need to find a point that is on the line y  
x
y
6
1
y
6
 1
6
NOTE: The point ( 6 ,  1 ) is in the fourth quadrant and lies on the line
x
y   . Since the terminal side of the angle  is in the fourth quadrant and
6
lies on this line, then the point ( 6 ,  1 ) also lies on the terminal side of  .
r 
x2  y2 
cos  
x

r
6
37
36  1 
37
sec  
r

x
37
6
sin  
y
 
r
1
37
tan  
y
1
 
x
6
csc  
r
 
y
cot  
x
 6
y
37
NOTE: In addition to the point ( 6 ,  1 ) lying on the terminal side of the
2
2
angle  , it also lies on the graph of the circle x  y  37 . This is the
circle whose center is at the origin and whose radius is
37 .
Back to Problem 2.
Solution to Problems on the Pre-Exam:
9.
If the terminal side of  passes through the point ( 4 ,  7 ) , then find the
exact value of sec  and tan  .
tan  
r 
10.
Back to Page 1.
y
7
 
x
4
x2  y2 
sec  
r

x
sec  
65
4
16  49 
65
65
4
tan   
7
4
If the terminal side of  is in the III quadrant and lies on the line
15 x  6 y  0 , then find the exact value of sin  and cot  .
15 x  6 y  0  15 x  6 y  y 
x
y
2
5
cot  
r 
y
5
5
(  2 )   (1)   5
2
1
x
2

y
5
x2  y2 
sin  
15
5
x  x
6
2
y
 
r
sin   
5
29
4  25 
29
5
29
cot  
2
5