Download Pre-Class Problems 7

Pre-Class Problems 7 for Wednesday, February 13
Illustration of the definition of the cosine function, sine function, tangent function,
secant function, cosecant function, and cotangent function for the acute angle 
using right triangle trigonometry. Second illustration of the cosine function, sine
function, tangent function, secant function, cosecant function, and cotangent
function.
Illustration of the definition of all the six trigonometric functions for the acute
angle  using right triangle trigonometry. Second illustration of all the six
trigonometric functions.
These are the type of problems that you will be working on in class. These
problems are from Lesson 6.
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 right triangle trigonometry to find the
exact value of the six trigonometric functions for an acute angle. You will also
need to apply the Pythagorean Theorem.
1.
Find the exact value of cosine, sine, and tangent for the following angles.
a.
b.
11

7
5

6
c.

8
4
Objective of the following problems: To use right triangle trigonometry to find the
exact value of the other five trigonometric functions for an acute angle if given the
value of one of the trigonometric function of the angle.
2.
Use a right triangle to find the exact value of the other five trigonometric
functions of the angle if given the following information.
a.
cos  
b.
tan  
c.
csc  
4
and  is an acute angle
7
3
and  is an acute angle
5
8
and  is an acute angle
3
Objective of the following problems: To determine what quadrant the terminal side
of an angle is in if given the sign of two trigonometric functions of the angle.
3.
Determine what quadrant the following angles are in if given the following
information.
a.
cos   0 and sin   0
b.
sin   0 and sec   0
c.
cos   0 and csc   0
d.
sin   0 and tan   0
e.
cos   0 and cot   0
f.
tan   0 and sec   0
g.
cot   0 and csc   0
h.
csc   0 and sec   0
Additional problems available in the textbook: Page 492 … 13 – 18, 25 - 30. Page
483 … Examples1 and 2. Page 505 … 9 – 14.
Solutions:
1a.
5
7

24
NOTE: You use the Pythagorean Theorem to find the value of
cos  
adj

hyp
24
,
7
Answer: cos  
Back to Problem 1.
24
,
7
sin  
opp
5
 ,
hyp
7
sin  
5
,
7
tan  
tan  
5
24
opp

adj
24 .
5
24
11
1b.

5
6
NOTE: You use the Pythagorean Theorem to find the value of 5.
cos  
11
,
6
adj

hyp
Answer: cos  
sin  
11
,
6
opp
5
 ,
hyp
6
sin  
5
,
6
tan  
tan  
opp

adj
5
11
5
11
Back to Problem 1.
1c.

80  4 5
8
4
NOTE: You use the Pythagorean Theorem to find the value of
cos  
adj
8


hyp
4 5
2
,
5
sin  
opp
4


hyp
4 5
1
,
5
80 .
tan  
opp
4
1


adj
8
2
2
,
5
Answer: cos  
sin  
1
,
5
tan  
1
2
Back to Problem 1.
2a.
cos  
4
and  is an acute angle
7
cos  
4
7
 sec  
7
4
NOTE: We do not need to use right triangle
trigonometry to find this value.
cos  
NOTE:
4
adj

7
hyp
7
33

4
NOTE: You use the Pythagorean Theorem to find the value of
sin  
opp

hyp
33
7
csc  
hyp

opp
7
33
33 .
tan  
opp

adj
33
4
Answer: sin  
cot  
cot  
33
,
7
33
,
4
tan  
adj

opp
sec  
4
33
7
,
4
csc  
7
,
33
4
33
Back to Problem 2.
2b.
tan  
3
and  is an acute angle
5
tan  
3
5
 cot  
5
3
NOTE:
We do not need to use right
triangle trigonometry to find this value.
tan  
NOTE:
3
opp

5
adj
28
3

5
NOTE: You use the Pythagorean Theorem to find the value of
28 .
cos  
adj

hyp
5
28
sec  
hyp

adj
28
5
sin  
opp

hyp
3
csc  
hyp

opp
28
28
Answer: cos  
csc  
5
,
28
sin  
28
cot  
,
3
3
28
,
sec  
3
28
,
5
5
3
Back to Problem 2.
2c.
csc  
8
and  is an acute angle
3
csc  
8
3
 sin  
3
8
NOTE: We do not need to use right triangle
trigonometry to find this value.
sin  
NOTE:
3
opp

8
hyp
8
3

55
NOTE: You use the Pythagorean Theorem to find the value of
cos  
adj

hyp
55
8
sec  
hyp

adj
8
55
tan  
opp

adj
3
55
cot  
adj

opp
55
3
Answer: cos  
55
,
8
sin  
sec  
8
,
55
cot  
3
,
8
tan  
55 .
3
,
55
55
3
Back to Problem 2.
3a.
cos   0 and sin   0
Using Method 1, which is Unit Circle Trigonometry, from Lesson 6:
Recall Unit Circle Trigonometry: P (  )  ( cos  , sin  )
cos   0  x  0
sin   0  y  0
Answer: IV
NOTE: The quadrant where x-coordinates are positive and y-coordinates are
negative is the fourth quadrant.
Using Method 2 from Lesson 6:
sin   0
y
cos   0
y
X
x
X
x
X
X
Answer: IV
Back to Problem 3.
3b.
sin   0 and sec   0
Using Method 1, which is Unit Circle Trigonometry, from Lesson 6:
Recall Unit Circle Trigonometry: P (  )  ( cos  , sin  )
sin   0  y  0
sec   0  cos   0  x  0
Answer: II
NOTE: The quadrant where x-coordinates are negative and y-coordinates are
positive is the second quadrant.
NOTE: Cosine and secant are reciprocals of each other. The sign of
reciprocals is the same.
Using Method 2 from Lesson 6:
sec   0 or cos   0
y
sin   0
y
X
X
X
x
x
X
Answer: II
Back to Problem 3.
3c.
cos   0 and csc   0
Using Method 1, which is Unit Circle Trigonometry, from Lesson 6:
Recall Unit Circle Trigonometry: P (  )  ( cos  , sin  )
cos   0  x  0
csc   0  sin   0  y  0
Answer: III
NOTE: The quadrant where x-coordinates are negative and y-coordinates are
negative is the third quadrant.
NOTE: Sine and cosecant are reciprocals of each other. The sign of
reciprocals is the same.
Using Method 2 from Lesson 6:
csc   0 or sin   0
y
cos   0
y
X
x
X
x
X
X
Answer: III
Back to Problem 3.
3d.
sin   0 and tan   0
Using Method 1, which is Unit Circle Trigonometry, from Lesson 6:
Recall Unit Circle Trigonometry: P (  )  ( cos  , sin  )
sin   0  y  0
tan   0 :
(  )  tan  
y
()

 x  0
x
?
Answer: IV
NOTE: The quadrant where x-coordinates are positive and y-coordinates are
negative is the fourth quadrant.
NOTE: A negative divided by a positive will produce a negative result.
Division with unlike signs will produce a negative.
Using Method 2 from Lesson 6:
tan   0
y
sin   0
y
X
x
X
x
X
X
Answer: IV
Back to Problem 3.
3e.
cos   0 and cot   0
Using Method 1, which is Unit Circle Trigonometry, from Lesson 6:
Recall Unit Circle Trigonometry: P (  )  ( cos  , sin  )
cos   0  x  0
cot   0  tan   0 :
(  )  tan  
y
?

 y  0
x
()
Answer: III
NOTE: The quadrant where x-coordinates are negative and y-coordinates are
negative is the third quadrant.
NOTE: Tangent and cotangent are reciprocals of each other. The sign of
reciprocals is the same.
NOTE: A negative divided by a negative will produce a positive result.
Division with like signs will produce a positive.
Using Method 2 from Lesson 6:
cot   0 or tan   0
y
cos   0
y
X
X
x
x
X
X
Answer: III
Back to Problem 3.
3f.
tan   0 and sec   0
Using Method 1, which is Unit Circle Trigonometry, from Lesson 6:
Recall Unit Circle Trigonometry: P (  )  ( cos  , sin  )
sec   0  cos   0  x  0
tan   0 :
(  )  tan  
y
?

 y  0
x
()
Answer: II
NOTE: The quadrant where x-coordinates are negative and y-coordinates are
positive is the second quadrant.
NOTE: Cosine and secant are reciprocals of each other. The sign of
reciprocals is the same.
NOTE: A positive divided by a negative will produce a negative result.
Division with unlike signs will produce a negative.
Using Method 2 from Lesson 6:
sec   0 or cos   0
y
tan   0
y
X
X
x
x
X
X
Answer: II
Back to Problem 3.
3g.
cot   0 and csc   0
Using Method 1, which is Unit Circle Trigonometry, from Lesson 6:
Recall Unit Circle Trigonometry: P (  )  ( cos  , sin  )
csc   0  sin   0  y  0
cot   0  tan   0 :
(  )  tan  
y
()

 x  0
x
?
Answer: II
NOTE: The quadrant where x-coordinates are negative and y-coordinates are
positive is the second quadrant.
NOTE: Sine and cosecant are reciprocals of each other. The sign of
reciprocals is the same.
NOTE: Tangent and cotangent are reciprocals of each other. The sign of
reciprocals is the same.
NOTE: A positive divided by a negative will produce a negative result.
Division with unlike signs will produce a negative.
Using Method 2 from Lesson 6:
cot   0 or tan   0
y
X
csc   0 or sin   0
y
X
x
X
x
X
Answer: II
Back to Problem 3.
3h.
csc   0 and sec   0
Using Method 1, which is Unit Circle Trigonometry, from Lesson 6:
Recall Unit Circle Trigonometry: P (  )  ( cos  , sin  )
csc   0  sin   0  y  0
sec   0  cos   0  x  0
Answer: IV
NOTE: The quadrant where x-coordinates are positive and y-coordinates are
negative is the fourth quadrant.
NOTE: Sine and cosecant are reciprocals of each other. The sign of
reciprocals is the same.
NOTE: Cosine and secant are reciprocals of each other. The sign of
reciprocals is the same.
Using Method 2 from Lesson 6:
sec   0 or cos   0
y
csc   0 or sin   0
y
X
x
X
x
X
X
Answer: IV
Back to Problem 3.
Solution to Problems on the Pre-Exam:
11.
7
Given:
Back to Page 1.
Find the exact value of
a. cos  
3
3
4
4

b. csc  
4
7
8.
If tan   0 and sec   0 , then  lies in which quadrant?
Using Method 1, which is Unit Circle Trigonometry, from Lesson 6:
Recall Unit Circle Trigonometry: P (  )  ( cos  , sin  )
sec   0  cos   0  x  0
(  )  tan  
tan   0 :
y
?

 y  0
x
()
Answer: II
NOTE: The quadrant where x-coordinates are negative and y-coordinates are
positive is the second quadrant.
NOTE: Cosine and secant are reciprocals of each other. The sign of
reciprocals is the same.
NOTE: A positive divided by a negative will produce a negative result.
Division with unlike signs will produce a negative.
Using Method 2 from Lesson 6:
sec   0 or cos   0
y
tan   0
y
X
X
x
X
Answer: II
x
X