Download Chapter 5 Trigonometric Functions

Chapter 5 Trigonometric Functions
(picture here of something involving weather)
Trigonometric functions are cyclical functions. What we mean by this is that they repeat themselves
over and over again. They have a cycle. One everyday event that is cyclical is the weather,
specifically the temperature of the air. In fact you can view it as having two cycles. One is a yearly
cycle the other is a daily cycle. If you were to look at what happens, temperature wise, in St. Louis,
Missouri you would see that in the winter the average high temperature is 37.7° Fahrenheit and in the
summer the average high temperature is 89.3° Fahrenheit. With this little of information we can
construct a function that would give us the average high temperature for any month of the year for St.
Louis that would be fairly accurate. The formula we would use would be a trigonometric function


x    63.5 . Here is what the graph
2
6
called sine. The formula would look like this: y  25.8sin 
looks like with the high temperatures in January and July for three years marked by dots.
This idea along with other is what weather men and women use to help them give you the weather
forecast each day.
Chapter 5 Trigonometric Functions
2
Section 5.1 Geometry Review
Objectives

Understanding some basic terms of geometry

Understanding basics about angles

Understanding triangles
As we begin to look at our next family of functions, trigonometric functions, we need to take some
time to review geometry. We will begin with some basics.
BASIC TERMS OF GEOMETRY
Discussion 1: Basic Terms
Let’s review the definitions of a few terms: point, line and plane.
Point  A point is a dot in space that has no size. In other words, it is zero dimensional. It
has no width or height or length (we usually use capital letters to name them).
Line − A line is made up of an infinite number of points, is one dimensional and goes on
forever in both directions (we usually use the middle of the alphabet to name them).
Plane − A plane is two dimensional and is made up of an infinite number of points which
go on forever in all directions on a flat surface.
Here are illustrations of the three defined terms: point, line, plane.
Two points are called collinear if they lie on the same line. If two or more points or lines lie on the
same plane then we call them coplanar.
Question 1: Which of the following are collinear, which are coplanar and which are neither?
n
C
l
B
k
A
m
Holtfrerich & Haughn
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Section 5.1 Geometry Review
In geometry we also talk about line segments and rays. A line segment is a piece of a line that has a
starting point and an ending point. A ray is half of a line. It has a starting point but no ending point.
Here are examples of each.
Line segment
Ray
When we talk about angles in geometry we are talking about the measurement between two rays that
have a common end point (vertex). We usually use small Greek letters to represent angle names.
Here are a couple of examples of angles.
α
θ
In the first example θ, read “theta”, is the name of the angle between the two rays. In the second
example α, read “alpha”, is the name of the angle between its two rays. The point where the two rays
start is called the vertex. In trigonometry we will further define angles such that they can have
positive or negative measures of any amount. One unit of measure used for angles is degrees and the
symbol used is °. There is defined to be 360° in one full rotation around a circle. Therefore a quarter
of a circle, which creates a right angle, is 90°.
SOME BASICS ABOUT ANGLES
Let’s talk about the names we give angles in Geometry. Angles are either, acute, obtuse, right or
straight. These words have the following definitions.
Acute  An acute angle is one whose measure is less than 90 degrees (θ above).
Obtuse − An obtuse angle is one whose measure is more than 90 but less than 180 degrees
(α above).
Right − A right angle is one whose measure is exactly 90 degrees.
Straight − A straight angle is one whose measure is 180 degrees.
Discussion 2: Special relationships of Angles
If the measures of two angles add up to 90 degrees then we call the two angles complementary
angles. If the measures of two angles add up to 180 degrees then we call them supplementary angles.
Here are two examples. In example (a) α and θ are complementary. In example (b) they are
supplementary.
a)
b)
θ
α
θ
α
Chapter 5 Trigonometric Functions
4
We also have relationships that make angles interior or exterior among other things. Let’s illustrate
some of these relationships when we have parallel lines l and m.
n
1
3
5
2
l
4
6
m
7
8
Interior angles
3, 4, 5, and 6
Exterior angles
1, 2, 7, and 8
Alternate interior angles
3&6, 4&5
Corresponding angles (are equivalent)
1&5, 2&6, 3&7, 4&8
Vertical angles
1&4, 2&3, 5&8, 6&7
(are equivalent)
In the above illustration line n is called a transversal because it intersects two coplanar lines. Also we
know from geometry that alternate interior and alternate exterior angles are supplementary.
Example 1: Finding Angle Relationships
If we know that angle 4 in the above illustration is a 120° angle and that lines l and m are parallel
then what do we know about all of the other seven angles?
Solution:
From Geometry we know that corresponding
m 8 = m 4
angels will be equivalent if you have parallel
m  8 = 120°
lines. (m  4 = 120°)
Answer Q1:
All points and
lines are
coplanar with
each other
except of line k
which isn’t
coplanar with
any thing else
in this example.
Points A and B
are collinear as
are points B
and C. Also
points A and C
are collinear.
Vertical angles are always equivalent
m 1 = m 4
m 5 = m 8
m  1 = 120°
m  5 = 120°
Angles 2 and 4 are supplementary (they add
m  2 = 180° − m  4
up to 180°) so,
m  2 = 180° − 120° = 60°
Angle 2 is a corresponding angle to angle 6
m 6 = m 2
m  6 = 60°
Vertical angles are equivalent
To summarize
m 3 = m 2
m 7 = m 6
m  3 = 60°
m  7 = 60°
m  1 = m  4 = m  5 = m  8 = 120°
m  2 = m  3 = m  6 = m  7 = 60°
Holtfrerich & Haughn
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Section 5.1 Geometry Review
Question 2: If in example 1 instead of the m  4 being 120°, we knew that the m  2 = 35°, what
would be the measures of the other seven angles?
TRIANGLES
Triangles are geometric shapes that have three sides and three angles. One fact about triangles is that
the sum of the three angles of any triangle must be 180 degrees.
Example 2: Triangle Angles Equal 180°
Given the following triangle what are the measures of all of its angles?
θ + 10
2θ − 20
40°
Solution:
40 + (θ + 10) + (2θ − 20) = 180°
First we know that the sum of the three
angles is 180°
We need to solve for θ.
30 + 3θ = 180°
3θ = 150°
θ = 50°
θ + 10 = 60°
Now figure out what are the values of
(θ + 10) and (2θ − 20).
2θ − 20 = 80°
The three angles are
40°, 60°, 80°
There are three special triangles that we need to talk about. They are: right, isosceles, and equilateral.
Right  A right triangle is one that has a 90 degree angle as one of its angles. It is made up
of two legs and a hypotenuse which is the side opposite the right angle.
Isosceles − An isosceles triangle is one where two of the legs of the triangle have the same
length. This makes two of its angles equal to each other.
Equilateral − An equilateral triangle is one where all three sides are the same length. This
makes all three of its angles equal to 60 degrees.
Here are illustrations of these three types of triangles.
Leg
α
60°
Hypotenuse
β
Right
θ
θ
Isosceles
60°
60°
Equilateral
Question 3: If one of the angles in a right triangle is a 90 degree angle, then what could we say about
the other two angles?
While we are discussing triangles we need to talk about similar triangles.
Chapter 5 Trigonometric Functions
6
Question 4: What do you think it will take for two triangles to be similar?
Discussion 3: Similar Triangles
You should have answered the last question in one of two ways. Either you should have thought that
both triangles would need to have angles of equal measure or that the corresponding sides would
need to be proportional. Let’s illustrate these.
Figures a) and b) are similar
a)
b)
75°
because they have
75°
corresponding angles which
60°
60°
45°
have equal measurements.
Figures c) and d) are similar
45°
c)
because the corresponding
d)
4
2
3
6
sides are proportional.
3 1 2 1 4 1
6  2, 4  2, 8  2


4
8
Example 3: Similar Triangles
What values of x, and y will make these two triangles similar?
2
7
6
x
5
y
Solution:
Since we are looking for values which will make these two
triangles similar, we need to make sure that the
2 7 5
 
6 x y
corresponding sides have the same ratio.
This leads to the following values for x, and y.
1 7

3 x

x  21
1 5

3 y

y  15
Discussion 4: Right Triangles
Lastly, we need to point out some special right triangles and a property of right triangles.
Holtfrerich & Haughn
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Section 5.1 Geometry Review
30-60-90 right triangle
Answer Q2:
m  2=35° and,
m 2 = m 3
= m 6 =
m  7 so, they
60°
2
1
This is an important triangle to have
30°
memorized.
all are 35° .
m  2=35° and
is
supplementary
to  4 which
must then have
a measure of
145°.
m 4 = m 1
= m 5 =
m  8 so, they
all are 145° .
3
45-45-90 right triangle
45°
This is also an important triangle to have
2
1
memorized.
45°
1
If an altitude is drawn from the 90° angle to
the hypotenuse then it divides the right
Altitude
triangle into two similar triangles that are
also similar to the original one.
Now because we know that similar triangles have sides that are proportional, we can find the sides of
any 30-60-90 or 45-45-90 right triangle from knowing these special triangles.
Example 4: Finding Sides of a Right Triangle
Given the following triangles what are the lengths of all of their sides?
3
10
30°
a)
45°
b)
Solution:
a) In the first triangle since we see that it is a right
triangle with a 30° angle we can use what we know
60°
about a 30-60-90 triangle to find the other two
1
2
30°
sides.
3
We can see that the ratio of the two hypotenuses is
10
2
Each side is 5 larger.
= 5. So our triangle has sides that are 5 times
larger then the basic one. Therefore we know the
following:
b) In the second triangle since we see that it is a right
triangle with a 45° angle we can use what we know
about a 45-45-90 triangle to find the other two
10
5(1) = 5
30°
5( 3 ) = 5 3
Answer Q3:
The other two
angles must
add up to 90°.
This means that
the other two
angles are
complementary
angles.
Chapter 5 Trigonometric Functions
8
sides.
Answer Q4:
Two triangles
will be similar
if the have the
same angles or
if the sides are
proportional.
45°
2
1
45°
1
We can see that the ratio of two of the legs is
3
1
=
3 2 =3 2
3. So our triangle has sides that are 3 times larger
3
then the basic one. Therefore we know the
45°
following:
3(1) = 3
Question 5: What are the other sides for the following triangle?
1
60°
You should have noticed that the general form of the 30-60-90 and 45-45-90 triangles are as follows.
60°
45°
2 x
x
x
45°
x
2x
30°
3 x
Section Summary: Important Truths for the Study of Trigonometry
If you have two parallel lines intersected by a transversal then the follow are true:
Truth 1
Alternate interior angles have equal measure.
Truth 2
Corresponding angles have equal measure.
Truth 3
Alternate Exterior angles have equal measure.
With any intersecting lines the following is true:
Truth 4
Vertical angles have equal measure.
Two triangles will be similar when:
Truth 5
Corresponding sides are proportional.
Truth 6
They have the equivalent angles.
Angles
Truth 7
The sum of complementary angles is 90°.
Truth 8
The sum of supplementary angles is 180°.
Truth 9
The sum of three angles of a triangle is 180°.
Special Right Triangles
Truth 10
The ratio of the sides of a 30-60-90 right triangle are;
Holtfrerich & Haughn
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Section 5.1 Geometry Review
60°
2x
x
30°
3 x
Truth 11
The ratio of the sides of a 45-45-90 right triangle are;
45°
2 x
x
45°
x
Truth 12
The acute angles of a right triangle are complementary.
Chapter 5 Trigonometric Functions
10
SECTION 5.1 PRACTICE SET
(1−6) Use the figure below:
1
2
1. m 1  35 :
m 2 
2. m 1  23 :
m 2 
3. m 2  65 :
m 1 
4. m 2  72 :
m 1 
5. m1  2 x  10 and m2  3x  30 :
Answer Q5:
6. m1  5x  15 and m2  x  15 :
m 1  and m 2 
m 1 
and
m 2 
(7−12) Use the figure below:
1
1
1
3
2
60°
1
2
7. m1  68 ;
9. m2  115 ;
m2 
8. m1  59.3 ;
m1 
10. m2  123.8 ;
m2 
m1 
11. m1  3x  20 and m2  5x : m1  m2 
12. m1  2x  50 and m2  3x  20 : m1  2 
(13−20) Use the figure below:
t
7
5
4
2
13. m7  130
14. m8  70
3
8
m
line l is parallel to line m, with a transversal of t
6
l
1
Find the measure of each of the other angles.
Find the measure of each of the other angles.
15. m1  2x  100 and m7  7 x  20 Find the measure of each of the other angles.
16. m2  3x  10 and m8  5x  20 Find the measure of all the angles.
17. m5  2 x  20 and m3  5x  40 Find the measure of all the angles.
18. m4  3x  30 and m6  5x  20 Find the measure of all the angles.
Holtfrerich & Haughn
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Section 5.1 Geometry Review
(21—28) Use the figure below:
C
A
B
21. mA  50 and mB  70 ; mC 
22. mC  53 and mB  63 ; mA 
23.
mA  74 ; mB  3x  10; mC  4 x  27
mB 
m C=
24.
mB  59 ; mA  3x  11; mC  2 x  30
mA 
mC=
25.
26.
mA  3x  10; mB  2 x  20; mC  2 x  30
mA 
mB 
mC=
mA  3x  5; mB  5 x  10; mC  2 x  25
mA 
mB 
mC=
27.
mA  88 ; mB  3x  68; mC  2 x  70
mB 
mC=
28.
mB  79 ; mA  2 x  59; mC  3x  70
mA 
m C=
(29−34)
D
e
B
f
F
E
d
c
a
A
C
b
Given the two triangles are similar; answer the following questions:
29.
a  10; b  16; c  12; d  8
e f 
30.
d  5; f  2; e  4; a  6
b c
31.
d  7; f  5; e  6; b  12
a c
32.
a  8; b  13; c  10; e  6
d f 
33.
mA  42 ; mB  68
mC  mD  mE  mF 
34.
mF  33 ; mE  67
mD  mA  mB  mC 
Chapter 5 Trigonometric Functions
12
(35−42)
30o
b
c
90o
60o
a
The given right triangle is a 30−60−90 right triangle.
35. c  10
a
37. a  6
b
39. b  8 3
41.
c
a
a  2x  1
a
b
c
a
b
38. a  5 3
b
c
40. b  10
c  3x  2
b
36. c  8 3
42.
c
a
a  3x  5
a
c
c  4 x  30
b
c
(43−48)
45o
a
c
900
45o
b
The given right triangle is a 45−45−90 right triangle.
43. a  5
b
45. b  5 2
47. c  8
c
a
a
44. b  10
c
b
a
c
46. a  9 2
b
c
48. c  9 2
a
b
(49−52) Given 1 and 2 are complementary angles:
49. m1 = 53o
m 2 =
50. m2 = 48o
51. m1 = 3x+10
m2 = 2x+30
m1 =
m2 =
52. m1 = 5x−12
m2 = 3x−18
m1 =
m2 =
m1 =
(53−56) Given 1 and 2 are supplementary angles:
53. m1 = 78o
54. m2 = 103o
m2 =
55. m1 = 3x+50
m2 = 2x+30
m1 =
m2 =
56. m1 = 5x−10
m2 = 3x−50
m1 =
m 2 =
m2 =
Holtfrerich & Haughn
13
(57−58)
b
a
d
e
f
c
57. a= 4; b = 8; d = 2;
c=
58. d = 6; f = 6 3 ; b = 12;
e=
a=
c=
f=
e=
Section 5.1 Geometry Review
Chapter 5 Trigonometric Functions
14
Section 5.2 Basics of Angles
Objectives

Understanding angles

Converting between degrees and radians

Understanding linear and angular speed
ANGLES
In trigonometry, angles are made up of three parts: a vertex, an initial side, a terminal side. The initial
side is the ray where the angle begins its rotation and the terminal side is the ray where it stops. The
vertex, as stated in section 5.1, is the point where the two rays start.
θ
Terminal side
α
Initial side
When the angle rotates in a counterclockwise manner the angle will have a positive measurement
(therefore θ in the above figure has a positive measure). When it rotates in a clockwise manner it will
have a negative measurement (therefore α in the above figure has a negative measure).
Discussion 1: Measuring Angles in Degrees
There are two ways used to measure angles. The first is with degrees. In one full revolution there are
defined to be 360 degrees (360°). Here are a few examples of angles measured in degrees.
−60°
30°
−450°
405°
When we talk about angles we often will want to talk about them in what is called standard position.
Standard Position  An angle is said to be in standard position if the vertex is at the origin
of an x, y coordinate plane and the initial side is lying on the positive x-axis.
The coordinate plane is divided up into four regions called quadrants. Here is how we name each
y
quadrant.
Quadrant II
Quadrant I
x
Quadrant III
Quadrant VI
Example 1: Quadrants
In which quadrant are the terminal sides of the following angles when placed in standard position?
a) 35°
c) −190°
b) 230°
Solution:
First quadrant
35
°
a) 35°
230°
Third quadrant
b) 230°
Second quadrant
c) −190°
−190°
Question 1: With the angle, θ = 1045°, in standard position, in which quadrant does the terminal side
lie?
1045°
Angles may rotate through as many revolutions as they want. Therefore it is possible for many
different angles, an infinite number of possibilities, to have the same initial and terminal sides. When
this happens we call the angles conterminal angles.
Coterminal Angles  Two angles are called coterminal if they both have the same initial
and terminating sides. This means that when you subtract the two angle measurements you
will get a multiple of 360°. In the picture below θ − α = 360°.
θ
Terminal side
α
Example 2: Coterminal Angles
What are some coterminal angles to the following angles?
Initial side
Chapter 5 Trigonometric Functions
16
b) −135°
a) 80°
c) 340°
Solution:
−280°
(80 − (−280) = 360)
be the following since when they are
440°
(80 − (440) = −360)
subtracted from 80° you get a multiple of
800°
(80 − (800) = −720)
225°
(−135 − (225) = −360)
−495°
(−135 − (−495) = 360)
585°
(−135 − (585) = −720)
−20°
(340 − (−20) = 360)
−380°
(340 − (−380) = 720)
700°
(340 − (700) = −360)
a) Some coterminal angles to an 80° would
360.
b) Some coterminal angles to a −135°
would be the following since when they
are subtracted from −135° you get a
multiple of 360.
c) Some coterminal angles to a 340° would
be the following.
Question 2: What is a coterminal angle to θ = −210° ? (hint: there are many answers)
θ = −210°
In the degree system of measuring there are also minutes and seconds. They are defined as follows.
Minute  One minute is
1
of a degree. The symbol ' is used for minutes. Another way to
60
say this is that there are 60 minutes in one degree.
Second − One second is
1
of a minute. The symbol ″ is used for seconds. Another way to
60
say this is that there are 60 seconds in one minute.
Example 3: Converting from degrees to degrees, minutes and seconds (dms)
Convert the following from degrees to degrees minutes and seconds or vise versa.
a) 328.32°
b) −732.19°
c) 15° 48' 12″
Solution:
a) Separate the whole number from the
328°
0.32°
decimal part of the number.
0.32  60' = 19.2'
Now convert the 0.32° to minutes by
multiplying by 60 minutes.
328° 19.2'
Now do the same with the minutes that
19'
0.2'
Holtfrerich & Haughn
17
Section 5.1 Geometry Review
0.2  60″ = 12″
we did with the degrees.
The final answer is
b) Separate the whole number from the
328° 19.2' 12″
732°
0.19°
decimal part of the number.
0.19  60' = 11.4'
Now convert the 0.19° to minutes by
multiplying by 60 minutes.
−732° 11.4'
Now do the same with the minutes that
11'
0.4'
0.4  60″ = 24″
we did with the degrees.
The final answer is
c) This time we are converting a degree
minute second measurement to just
−732° 11.4' 24″
15° 48' 12″
15° +
degrees. You need to divide the minutes
by 60 and the seconds by 3600.
48
12
+
= 15 + 0.8 + 0.0033333
60
3600
15.80333333°
Let’s look at how the graphing calculator could have helped us with these last three problems.
a) 328.32°
With the TI-83/84 simply type in
With the TI-86 simply type in the
the 328.32 and then push the 2nd
328.32 and then push the 2nd X key
MATRIX key (TI-83) or 2nd APPS
(times key) for MATH then F3 for
(TI-83+,84) for ANGLE and select
ANGLE and F4 for DMS and then
number 4 for DMS and push
ENTER.
ENTER twice.
b) −732.19°
Do the same as for part a.
c) 15° 48'
With the TI-83/84 type in 15 then
With the TI-86 simply type in the
12″
get the ° symbol from the ANGLE
15 followed by the ‘ symbol found
menu. Then type in 48 and get the ‘
in the ANGLE menu. Then follow it
Answer Q1:
Fourth
quadrant
Chapter 5 Trigonometric Functions
18
symbol from the ANGLE menu.
with 48 and the ‘ symbol and then
Lastly, type in 12 and then push
12 with the ‘ symbol. Now if you
ALPHA + keys to get “. Push
have your calculator on degree
ENTER to see the answer.
mode push ENTER to see your
answer. If you are in radian mode
go change it to degree mode first
then ENTER.
Discussion 2: Measuring Angles in Radians
This second way of measuring angles is the method used most often in calculus. It will be the
primary method used in this book. Radian measure of an angle is based on the length of the
circumference of a circle (arc of the circle) created by the angle being measured in multiples of the
radius of the circle. For example, the radian measure of an angle that is one complete revolution
would be 2π rad (radians). The circumference of a whole circle is 2πr. Thus, there are 2π radii
lengths in the circumference of a circle (the circumference of a circle is slightly longer that 6 radii,
2π  6.283). So the angle that creates a whole circle would have a radian measure of 2π rad.
There are a couple of terms we need to add at this point.
Answer Q2:
Some
coterminal
angles are
150°, −570°,
510°.
Central Angle − An angle with its vertex at the center of a circle.
Arc of a Circle − An arc of a circle is a piece of the circumference of a circle that is
intersected by a central angle.
Radian − One radian is the measure of a central angle (θ) that intercepts an arc (s) that is
equal in length to the radius of the circle.
Arc Length − The arc length (s) will equal the angle θ (in radians) times the radius (r) of
the circle (s = θr ).
Here are some examples of angles measured in radian.

Half a circle

2
One quarter
of a circle
Circumference of half a circle
Circumference of a quarter of

6
One 12th
of a circle
Circumference of a twelfth of
Holtfrerich & Haughn
is
2 r
= πr. So the angle that
2
19
a circle is
Section 5.1 Geometry Review
2 r 
= r . So the
4
2
a circle is
2 r 
= r . So the
12
6
rotates through half a circle
angle that rotates through a
angle that rotates through one
has a radian measure of π.
quarter of a circle has a radian
twelfth of a circle has a radian
measure of

.
2
measure of

.
6
Example 4: Arc Length and Sector Area
What is the arc length and sector area of the figure?
θ=
4
5
r=7
Solution:
To find the arc length (s) we use the
s
formula s = θr. (This gives us part of
4
28
7
 5.6 units
5
5
the circumference of the whole circle
The outside arc, or the piece of the
2πr.)
circumference of a pizza, is 5.6π units in length.
To find the sector area (a sector is a
 4 
 5 
 72
A 
 
2
slice of pizza) the formula would be
A

2
r 2 . This comes from the fact that
the area of a circle is Ac   r 2 
A
 2  r 2 .
4
49  19.6 units squared
10
2
Another concept we need to take a look at is the concept of reference angle. A reference angle is an
acute angle (measurement between 0° and 90°) between the terminal side of an angle in standard
position and the x-axis. Let’s look at some examples.
Example 5: Reference Angles
Find the reference angle (r) for the following angles in standard position.
a) θ = 35°
b) α = 230°
c) β = −190°
d) φ = 1045°
Solution:
a) Since θ is already an acute angle from
the terminal side to the x-axis, it is its
35
°
own reference angle.
r = 35°
r
Chapter 5 Trigonometric Functions
20
b) A 230° angle from standard position
means that the terminal side has gone
230°
50° beyond the negative x-axis (180°).
r
r = 50°
c) A −190° angle has gone in a clockwise
rotation 10° past the negative x-axis.
r
−190°
r = 10°
d) A 1045° angle has made two complete
rotations and then another 325° which
means it is just 35° short of completing
1045°
its third revolution.
r
r = 35°
CONVERSION BETWEEN DEGREES AND RADIANS
Even though we will be focusing a lot of our attention on radian measure it is important that you are
comfortable with both degree and radian measurements. We also will need to be able to convert
measurements of one type into the other.
Discussion 3: Conversions from degrees to radians and vise versa
Since we know that there are 360° in a circle and 2π rad, we can create two ratios which will allow us
to convert from one form of measurement to the other. To find radian measure you will take the
degree measure times the ratio
measure times the ratio
2 rad
 rad

. To find degree measure you will take the radian
360
180
360
180
.

2 rad  rad
For example let’s convert the following two angle measurements.
a)
240°
a) We can multiply the 240° by the fraction
b) We can multiply the
5
rad
3
b)
240  rad 4


rad
1
180
3
 rad
180
5
180
rad by the fraction
3
 rad
5 rad 180

 300
3
 rad
Here is how we do these two on the calculator.
a) With the TI-83/84, type in 240 followed
a) With the TI-86, type in 240 followed by
by the ° symbol found in the ANGLE
the ° symbol found in the ANGLE menu
menu and then push ENTER if you are
and then push ENTER if you are in
in radian mode. If you are in degree
radian mode. If you are in degree mode
Holtfrerich & Haughn
21
Section 5.1 Geometry Review
mode go change it to radian mode and
go change it to radian mode and then
then push ENTER.
push ENTER.
 5 
b) With the TI-83/84, type in 
 with
 3 
 5 
b) With the TI-86, type in 
 with
 3 
parentheses followed by the r symbol
parentheses followed by the r symbol
found in the ANGLE menu and then
found in the ANGLE menu and then
push ENTER if you are in degree mode.
push ENTER if you are in degree mode.
If you are in radian mode go change it to
If you are in radian mode go change it to
degree mode and then push ENTER.
degree mode and then push ENTER.
Question 3: What is the radian measure of the angle θ that is −135° ?
LINEAR AND ANGULAR SPEED
Discussion 4: Car Speed
Visualize a car and its tires. Now consider a nail that is in one of the tires. We can talk about two
ideas in this scenario. First, how fast is the car moving forward (linear speed). Second, how fast is the
nail going around (angular speed). Let’s consider how fast the car is moving first. The tire moving
around on its circumference is what is making our car go forward. So, as the tire traces through its
circumference the car moves forward that amount (2πr is the distance traveled by the car in one
revolution of the tires). Therefore if a car has a tire with a radius of 14 inches then with one rotation
of its tires the car will have traveled 87.96 inches (2  π  14  87.96).
Let’s say that we know that a car, with 14 inch radius tires with a nail in them, is traveling such that
the tires are making 8 revolutions per second, how fast is the car moving forward?
The circumference of a tire is 2  π  14 = 87.96
inches. Multiply this by 8 since we get 8 revolutions
of the tire in one second. This will give us the
distance the car will travel in one second (in/sec).
87.96in 8rev

 703.68in/sec
1 rev
1 sec
Chapter 5 Trigonometric Functions
22
We might prefer to convert this to mph in order to
better understand how fast this is or isn’t.
703.68in 3600sec
1 mile


1 sec
1 hour 63360 in
= 39.98  40 mph
Discussion 5: Nail Speed
Now let’s take the same problem as discussion 4, but ask the question, how fast is the nail moving
around and around. Angles are a good way of measuring how we are going around and around. So, to
answer our question we need to think in terms of angles divided by time to get a rate value in terms
of radians per sec. In this example the nail is moving through 2π rad every one revolution so after 8
revolutions (one second of time) the nail will have traveled through 16π rad in one sec. Thus we
could say that the nail is traveling
16 rad
.
sec
We have the following definitions:
Linear Speed  The linear speed of an object connected to a wheel will be v =
s
, where v
t
is the velocity of the object, s is the arc length and t is the time.
Angular Speed − The angular speed of an object on a wheel will be ω =

, where ω read
t
“omega” is the angular speed, θ is the angle (in radians) and t is the time.
Question 4: What would be the linear speed (in feet/sec) of a car and the angular speed (in rad/sec) of
a nail in a tire on a car, if the tires are traveling at the rate of 10 revolutions per second with 1 foot
radii?
As you answered question 4 you should have noticed that when the radius is one the angular speed
and the linear speed have the same magnitude.
Section Summary:

Vertex  The common point where two rays begin that create an angle.

Initial side  The ray where an angle begins.

Terminal side  The ray where an angle ends.

Positive angles  Angles that rotate in a counterclockwise manner.

Negative angles  Angles that rotate in a clockwise manner.

Minute  one sixtieth of a degree.

Second  One sixtieth of a minute.

Central Angle − An angle with its vertex at the center of a circle.
Holtfrerich & Haughn

23
Section 5.1 Geometry Review
Arc of a Circle − An arc of a circle is a piece of the circumference of a circle that is
intersected by a central angle.

Radian − One radian is the measure of a central angle (θ) that intercepts an arc (s) that is
equal in length to the radius of the circle.

Arc Length − The arc length (s) will equal the angle θ (in radians) times the radius (r) of the
circle (s = θr ).

A reference angle is an acute angle (measurement between 0° and 90°) between the terminal
side of an angle in standard position and the x-axis.

Conversion  180° is equivalent to π radians. When you need to convert from one form to
another use one of the following ratios:

180

or

180
.
Linear Speed  The linear speed of an object connected to a wheel will be v =
s
, where v is
t
the velocity of the object, s is the arc length and t is the time.

Angular Speed − The angular speed of an object on a wheel will be ω =

, where ω read
t
“omega” is the angular speed, θ is the angle (in radians) and t is the time.
Answer Q3:
3

radians
4
Chapter 5 Trigonometric Functions
24
SECTION 5.2 PRACTICE SET
(1–8) For each angle in standard position, the terminal side of the angle is in which quadrant?
1. 83o
5.
3
radians
5
2. 193o
6.
8
radians
3
3. –230o
7.
5
radians
4
4. –58o
8.
32
radians
5
(9–12) For each angle give all the coterminal angles from –360o to 360o:
9. 50o
10. 190o
11. –150o
12. –210o
(13–16) For each angles give all the coterminal angles from –2 to 2:
13.
3
radians
4
14.
8
radians
5
15.
2
radians
3
16.
11
radians
6
(17–20) Convert the decimal value of a degree measurement to degrees–minutes–seconds:
17. 58.32o
18. 233.24o
19. –583.45o
20. –78.54o
(21–24) Convert from degrees–minutes–seconds to a decimal degree answer:
Answer Q4:
22. 395 4731"
23. 273 1323"
24. 52 5341"
Angular speed 21. 58 3625
would be 10*2π
since we have (25–36) For each angle give the reference angle:
10 revolutions
per sec. Thus
25. 158o
26. 321o
27. 238.5o
28. 538.4o
the answer is
20π rad/sec.
o
30. –162o
31. –323.8o
32. –58.7o
Linear speed 29. –273
would be
5
3
8
5
10*2π(1) since
33.
radians
34.
radians
35.
radians
36.
radians
we have 10
3
4
5
7
revolutions and
the radius is
(37–44) Convert the following degree measurement of an angle to a radian measurement of that
one foot. Thus
angle:
the answer is
20π feet/sec.
o
o
o
o
37. 60
38. 240
39. 165
40. 324
41. –120o
42. –300o
43. –230o
44. –55o
(44–52) Convert the following radian measurement of an angle to a degree measurement of that
angle:
45.
5
radians
3
46.
3
radians
4
47.
7
radians
9
48.
9
radians
5
49.

radians
6
50.
11
radians
3
51.
7
radians
8
52.
11
radians
7
Holtfrerich & Haughn
25
Section 5.1 Geometry Review
(53–54) For each of the following find:
a. The linear speed
b. The angular speed
53. A Ferris wheel has a radius of 30 feet and it takes 70 seconds for one revolution.
54. A bicycle wheel has a diameter of 26 inches and the wheel makes 425.5 revolutions every
minute.
(55–60) A = The area of a sector (Area bounded by the interior of the angle and the circle)
r = The radius of the circle
S = The length of the arc intersected by the central angle
 = The measurement of the central angle of a circle in radians
Use the formula A = ½ r2 and S = r
r

S
 = ½ radian
Area of the sector =
Length of arc =
56. r = 12 inches  = 5 radians
Area or the sector =
Length of arc =
55. r = 8 meters
57. r = 10 centimeters
58. r = 3 feet
=
=
2
radians
3
5
radians
4
Area =
Area of the sector =
59. r = 10 meters
 = 60o Area of the sector =
60. r = 12 inches
 = 300o
Areas of the sector =
Length of arc =
Length of arc =
Length of arc =
Length of arc =
61. A water sprinkler sprays water over a distance of 25 feet while rotating through
an angle of 120o. What is the area of the lawn is watered?
62. A person that installs sprinkler systems is asked to design a water sprinkler that will
cover a field of 50 square yards that is in the shape of a sector of a circle with radius
of 20 yards. Through what angle should the sprinkler rotate?
Chapter 5 Trigonometric Functions
26
Section 5.3 Sine, Cosine and the Unit Circle
Objectives

Understanding the definition of sine and cosine

Understanding the unit circle

Understanding the domain and range of sine and cosine

Understanding multiple inputs yielding same output
SINE AND COSINE
We are now ready to define two of the six new functions in the family called trigonometric functions.
P(x, y)
r
y
θ
x
Here is a picture of an angle θ in standard position. P is any point on the terminal side of angle θ. r is
the distance from the origin to the point P. Sine and Cosine are defined as follows.
Sine  The trigonometric function sine is defined to be sin θ =
y
.
r
Cosine − The trigonometric function cosine is defined to be cos θ =
x
.
r
Notice that these two functions are defined to be a ratio of the sides of a right triangle formed
between the terminal side of an angle and the x-axis. In essence you will always be using the
reference angle to find function values for the trigonometric functions. We now have a way of
relating x and y with an angle.
We need to remind you here about the Pythagorean Theorem.
Pythagorean Theorem  The Pythagorean Theorem states that the square of the two legs
of a right triangle added together will equal the square of the hypotenuse (a 2 + b2 = c2).
c
a
b
Example 1: Finding Function Values
Holtfrerich & Haughn
27
Section 5.1 Geometry Review
Given the following angles in standard position find the function values of sine and cosine?
8
(−3, 4)
θ
θ
4
−15
(8, −15)
−3
a)
b)
Solution:
42 + (−3)2 = r2  r2 = 25  r = 5
a) First we need to find the value of r.
Since we have a right triangle we can use
(r is positive since it represents a distance)
the Pythagorean theorem.
sin θ =
4
,
5
cos θ =
3
5
82 + (−15)2 = r2  r2 = 289  r = 17
b) First find r then sine and cosine
sin θ =
15
,
17
cos θ =
8
17
Discussion 1: Any Point Will Do
It turns out that it doesn’t matter which point you pick on the terminal side of your angle the sine and
cosine values will always be the same. Let’s look at example 1a again.
(x, y)
(−6, 8)
y
8
θ
4
−3
−6
x
You can see from the picture, that as you choose different points on the terminal side of the angle θ
you create similar triangles. And we know that similar triangles have proportional sides (Truth 5
from section 5.1). To further help you see this look at the point (−6, 8) we get sin θ =
θ=
8 4
 and cos
10 5
 6 3

. These are the same two answers that we got with the point (−3, 4). Therefore with this
10
5
angle the sin θ will always be
4
3
and the cos θ will always be
no matter which point you choose
5
5
on the terminal side of the angle.
Chapter 5 Trigonometric Functions
28
Question 1: Since it doesn’t matter which point you pick when evaluating the sine and cosine of an
angle, what happens to the definitions of sine and cosine if you choose your point on the terminal
side of your angle such that r = 1?
THE UNIT CIRCLE
The last question leads us to another way of defining sine and cosine. We can use what is called the
unit circle definition of trigonometric functions. A unit circle is a circle whose radius is one and has
this formula x2 + y2 = 1.
P (x, y)
s
1
θ
(1, 0)
If we apply the earlier definitions for sine and cosine we would get sin θ =
y
x
and cos θ = since
1
1
r = 1 in this picture. Therefore for a unit circle our original definition now becomes as follows.
When θ is in standard position as a central angle of a unit circle with point P on the
terminal side of the angle on the unit circle then,
The trigonometric function sine is defined to be
sin θ = y.
The trigonometric function cosine is defined to be cos θ = x.
Notice, that since we are working with a unit circle θ = s, the arc length of a piece of a circle
intersected by the angle in standard position equals the angle measured in radians. So we really could
view this as a way of relating x and y to the arc length of an arc from a unit circle.
We now have two ways to view what sine and cosine do for us. One is that they relate an
angle to coordinates of a point on the terminal side of an angle in standard position. The second one
is that they relate the arc length of a unit circle with the coordinates of points on that unit circle. One
relates angles to real numbers the other relates real numbers to real numbers. This distinction is
important in the study of Calculus. Let’s take a look at some of the key points on a unit circle and
how we derive them.
From section 5.1 we know the following truths.
Truth 10 − The ratio of the sides of a 30-6090 right triangle are
Truth 11 − The ratio of the sides of a 45-4590 right triangle are
Holtfrerich & Haughn
29
Section 5.1 Geometry Review
60°
2x
2 x
45°
x
x
30°
45°
x
3 x
Our radius is one so 2x = 1  x =
1
and thus
2
Our radius is one so 2 x = 1  x =
the other sides are as follows.
1
60°
30°
1
and
2
thus the other sides are as follows.
1
2
45°
1
1
2
45°
3
2
1
2
So, for a 30°, 45° and 60° central angle in a unit circle the points on the unit circle would be as
follows:
1
30°
 3 1
, 


 2 2
60°
1
=y
2
 2 2
,


 2 2 
45°
1
1 3
 ,

2 2 
1 =y
2
1
30°
45°
1
3
=x
2
=x
2
3
=y
2
60°
1
=x
2
Many of the other angles of the unit circle have these three as reference angles, so the points are just
positive or negative versions of the above points. As for 0°, 90°, 180°, and 360°, their points are
obvious since the radius of the unit circle is one.
 1 3 
 ,

 2 2 
2
 2
2
 120
,


3
2
2
3



 135
4
 3 1
5
, 

 150

6
 2 2
 1, 0 
 0,1

2
 90

3

1 3
 ,

2 2 
 2
2
60
,


2 

 2
 45
4
 3 1
, 


 30  2 2 


6
  180
0  0, 2  360
11
 330
6
7
 210
  3 1 
6
, 


2 
5
 2
 225
 2  2 
4
,


4
2 
 2
3
 1  3 
 ,

2 
 2
 240 3
 270
2
 0,  1
5
3
1, 0 
 3 1 
,
 2 2 
7


 315
4
 2  2
,


 300
2 
 2
1  3 
 ,

2 2 
Example 2: Finding Sine and Cosine Outputs
What is the sine and cosine function outputs for the following inputs?
Chapter 5 Trigonometric Functions
a) θ =

3
b) θ =
5
6
30
c) θ = 45
3
2
d) θ =
e) θ = 600
Solution:
Answer Q1:
a) To find sin
If r = 1 then
the
denominators
in the
definitions of
sine and
cosine
become 1.
Thus,
sin θ = y and
cos θ = x.

look at the y value of the
3
1 3
 ,

2 2 
point on the unit circle that corresponds
to the angle


. For the cos use the x
3
3
sin
3

=
2
3
cos
 1
=
3 2
value.
b) To find sin
5
6
look at the y value of the
 3 1
, 


 2 2
point on the unit circle that corresponds
to the angle
5
6
. For the cos
5
6
use the x
sin
5
6
=
1
2
cos
5
6
=
 3
2
value.
c) This time notice that we have a negative
angle so go clockwise around to −45°.
Now as before just look at the y and x
 2  2
,

2
 2



sin −45°=
values of the point on the unit circle.
d) As in part (c) go clockwise around
to
3
2
sin
e) Find the point at 600°.
(600° same terminal side as 240°, they
are coterminal angles)
 2
2
 0,1
. Now look at the y and x values
of the point on the unit circle.
(−45° coterminal with 315°)
3
2
cos −45°=
(
=1
 1  3
 ,
2
 2
2
2
3

coterminal with )
2
2
cos
3
2
=0



sin 600° =
 3
2
cos 600°=
1
2
We need to take some time now to talk about some patterns in the unit circle and how to use them to
help us find values for sin θ and cos θ for common angles.
Discussion 2: Ways of Finding Trigonometric Values with the Unit Circle
You should have noticed that as you go from quadrant to quadrant that many of the values for x and y
are repeated but occasionally with different signs. For example at
 3 1
5
 150 the point is
,  and at

6
 2 2
is 

6
 30 the point on the unit circle
 3 1
  3 1 
7
 210 the point is 
,  and at
,  . These


 2
6
2 
 2 2

three points have the same magnitudes for x and y because all of their angles have the same reference
angle

6
 30 . You should notice as well, that as you go around the circle the values for x goes
Holtfrerich & Haughn
through this pattern:
31
Section 5.1 Geometry Review
4
2
0
 1 1  2  3  4
3
1 1
,
,
,
,
,
1 ,
0,

 1 … and
 ,
2
2
2
2
2
2
2
2
2
2
2
then back to 1. The y values do a similar pattern:
0
2
4
2
1 1
3
3
,
,
,
,
0,
1 ,
 ,
2
2
2
2
2
2
2
2
0
 1 1  2  3  4
1 1
,
,
,
0,

 1 … and then back to 0. Also notice how, with
 ,
2
2
2
2
2
2
2
2
sin θ defined to be equal to y on the unit circle, that sin θ will always be positive in the 1st and 2nd
quadrants (above the x-axis) and negative in the 3rd and 4th (below the x-axis), likewise cos θ = x will
be positive in the 1st and 4th (right of the y-axis) but negative in the 2nd and 3rd (left of the y-axis).
Take a look at the next chart; it may help you to see these patterns more easily.
Reference angles
0°
30°
45°
60°
90°

0
0
30
45
60
90




rad
6
4
3
3
2
2
2
1
2
0
0
2
2
Q1
x value
4
1
2
Q1
y value
0
0
2
1
2
2
2
3
2
4
1
2

180
150
5
6
135
120
2
3
90
3
4

rad

2
Q2
x value
 4
 1
2
 3
2
 2
2
 1
2
0
0
2
Q2
y value
0
0
2
1
2
2
2
3
2
4
1
2

180
210
7
6
225
240
4
3
270
5
4

rad
3
2
Q3
x value
 4
 1
2
 3
2
 2
2
 1
2
0
0
2
Q3
y value
0
0
2
 1
2
 2
2
 3
2
 4
 1
2
360
2
330
11
6
315
300
5
3
270

rad
7
4
3
2
Q4
x value
4
1
2
3
2
2
2
1
2
0
0
2
Q4
y value
0
0
2
 1
2
 2
2
 3
2
 4
 1
2
Chapter 5 Trigonometric Functions
32
Ultimately, all you need to remember is the points on the unit circle in the first quadrant and in which
quadrants sine and cosine are positive and negative. With this information memorized you can find
the sin θ and the cos θ of any angle (θ) that is a multiple of the special angles: 30, 45, 60, 90.
Example 3: Finding Sine and Cosine Function Values
What is the sin θ and cos θ for the following angles?
a) −330°
b)
11
4
c) −
5
6
Solution:
a) We first need to find the angle’s
A −330° angle terminates in the first quadrant 30°
reference angle and then decide
short of the x-axis. Therefore the reference angle is
whether the function value will
30° and the function value for both sine and cosine
be positive or negative.
will be positive since the terminating side of the
angle is in the first quadrant. The answers are:
sin −330° = sin 30° =
1
2
Point on unit circle
for a 30° angle.
 3 1
, 


 2 2
3
cos −330° = cos 30° =
2
b) We first need to find the angle’s
A
reference angle and then decide
11

angle terminates in the second quadrant
4
4
whether the function value will
short of the x-axis (angle is over one revolution).
be positive or negative.
Therefore the reference angle is

and the function
4
value for sine will be positive and cosine will be
negative since the terminating side of the angle is in
the second quadrant. The answers are:
c) We first need to find the angle’s
reference angle and then decide
Point on unit circle
sin
11

= sin
=
4
4
cos
11

2
= − cos
= 
4
4
2
A−
5

angle terminates in the third quadrant
6
6
2
2
for a

angle.
4
 2
2
,

2
 2



whether the function value will
short of the x-axis. Therefore the reference angle is
be positive or negative.

and the function value for sine will be negative
6
and cosine will be negative since the terminating
side of the angle is in the third quadrant. The
answers are:
Holtfrerich & Haughn
33
sin
cos
Section 5.1 Geometry Review
5

1
= − sin
=
6
2
6
5

 3
= − cos
=
6
6
2
Point on unit circle
for a

angle.
6
 3 1
, 


 2 2
Question 2: What is the sin θ and cos θ for the following angles using the knowledge of the unit
circle you have just learned?
b) −π
a) 120°
Example 4: Finding Sine and Cosine of non-special Angles
Find the values of the following using your graphing calculator.
a) sin 50°
b) cos
5
13
c) sin (−3.56)
Solution:
a) First make sure that your
calculator is on degree mode and
then push the SIN key followed
by 50, then ENTER.
b) This time the angle is in radians,
so first make sure the calculator
is on radian mode then push the
COS key followed by 5π/13,
then ENTER
c) The last example is also in
radians and you should get the
following on your graphing
calculator.
All of this brings us to two very important ideas discussed in the chapter on functions: domain and
range. Let’s take a look at what the domain and range are for sine and cosine and their graphs.
DOMAIN AND RANGE FOR SINE AND COSINE
If you look closely at the unit circle points you will first notice that any angle could be used as an
input with the sine and cosine functions. You can have either positive or negative angles (in degrees
or radians) and you may go as many times around the circle as you wish, so the domain is all
possible angles if you are thinking in degrees or all real numbers if thinking in terms of radians. We
want to focus on the radian usage so:
Chapter 5 Trigonometric Functions
34
Domain for both sine and cosine is all real numbers (−∞, ∞).
The range is also visible from the unit circle points. If you look back at the points on the unit circle
you will see that the largest x or y ever are is 1, and the smallest they ever are is −1. Another way to
think about this is that you are talking about a unit circle (x2 + y2 = 1). On a circle of radius one, x and
y can’t be larger then 1 or smaller than −1. Therefore:
Range for both sine and cosine is [−1, 1].
From observing the points on the unit circle we can get this table for both functions.
0

6

4

3

2
3
4

5
4
3
2
7
4
2
cos θ
1
3
2
2
2
1
2
0
 2
2
−1
 2
2
0
2
2
1
sin θ
0
1
2
2
2
3
2
1
2
2
0
 2
2
−1
 2
2
0
Now, we can sketch a graph of both sine and cosine. First a partial graph of y = sin θ.
e
d

f
3
3
 135
4
y
1
2
 90 
d
f
g
g
−1
c
 45

6
c
 
6 4 3 2

4
e
  180
3
4
b
 30
0  0, 2  360
b
a
 60
a k
k

5
4
3
2
h
i
7 2
4
j
θ
7
 315
4
5
 225
4
h
3
 270
2
j
i
This isn’t the whole graph but only the graph for inputs from 0 to 2π radians (angles from the first
through the fourth quadrants). Notice that even though the angles used as inputs were from quadrants
1 through 4, the graph of the sine function (matching the inputs and outputs) is in the first and fourth
quadrants only so far.
Of course you should have realized by now that for inputs after 2π and before 0 the outputs
just repeat each other. The sine function is an example of what is called a periodic function (we go
around and around the same circle). This leads to a graph that is going to go on forever to the left and
the right of the origin.
Holtfrerich & Haughn
35
Section 5.1 Geometry Review
Answer Q2:
3
2
1
cos 120°=
2
b) sin −π = 0
cos −π = −1
a) sin 120°=
Periodic Function − A function is say to be periodic if there exists a smallest number c
such that f(x) = f(x + nc) where n is any integer.
For the sine function the value for c is 2π. Therefore the following is true for the sine function:
sin θ = sin (θ + 2nπ) where n is any integer.
Let’s look at cosine now. It is also periodic on a period of 2π. Here is its graph.
y
1



6
3
4
θ

2
3
4

5
4
7
4
3
2
2
−1
To summarize here are the graphs of both sine and cosine.
y
y
y = sin θ
1
y = cos θ
1
θ
−2π
θ
−2π
2π
2π
−1
−1
MULTIPLE INPUTS
Question 3: What are the exact values of the following and what do you think is the point of asking
this question?
a) sin
2
3
b) sin
8
3
c) sin
4
3
With trigonometric functions for each output there are many inputs (trigonometric functions are not 1
to 1). To be exact there are an infinite number of inputs that yield each possible output. So if we were
Chapter 5 Trigonometric Functions
36
to ask you, “how many angles could you find the sine of that would yield 1 as an answer?”, you
would answer many. The way we write such an answer is this way.
x=
sin x = 1 when

2
 2 n
(n any whole number)
The 2πn part of the answer is telling everybody that all revolutions around the circle that end in the
same place as the angle

will have the same answer, 1.
2
Example 5: Finding Angles That Yield Same Outputs
What angles yield the following?
a) sin θ =
2
2
b) cos θ = 0
c) cos θ =
1
2
Solution:
a) So what angles in the unit circle yield
an answer of
Answer:
2
?
2

3
and
4
4
b) So what angles in the unit circle yield
an answer of 0?
Answer:

3
and
2
2
Therefore the solutions to the question are:

4
 2 n
and
3
 2 n
4
Therefore the solutions to the question are:

2
 2 n
answer
and

2
3
 2 n or as just one
2
  n (all angles that make
cosine equal 0 are half a circle apart)
c) So what angles in the unit circle yield
an answer of
Answer:
1
?
2
2
4
and
3
3
Therefore the solutions to the question are:
2
 2 n
3
and
4
 2 n
3
Section Summary

Pythagorean Theorem states that the square of the two legs of a right triangle added
together will equal the square of the hypotenuse (a2 + b2 = c2).

Sine is defined as the y value of the point on a unit circle on the terminal side of angle θ
(sinθ = y).

Cosine is defined as the x value of the point on a unit circle on the terminal side of angle θ
(cosθ = x).
Holtfrerich & Haughn
37
Section 5.1 Geometry Review
 0,1

2
 90

3

1 3
 ,

2 2 
 2
2
60
,


2
2



 45
4
 3 1
, 


 30  2 2 


6

0  0
1, 0 
The first quadrant of the unit circle
sin θ > 0
cos θ < 0
sin θ > 0
cos θ > 0
sin θ < 0 sin θ < 0
cos θ < 0 cos θ > 0

The sign of the outputs of sine and cosine are:

The domain for both sine and cosine is (−∞, ∞)

The range for both sine and cosine is [−1, 1]

A function is periodic if for some smallest c, f(x) = f(x + cn) where n is any integer

Sine and cosine have a period of 2π.
Answer Q3:
2
3
sin
=
3
2
sin
8
3
=
3
2
4
3
=
3
2
The point is that
these are all
coterminal angles
and thus they all
have the same
answer.
sin
Chapter 5 Trigonometric Functions
38
SECTION 5.3 PRACTICE SET
(1–8) Use the Pythagorean theorem to find the following values:
a
b
c
1. a = 9
b = 16
c=
2. a = 10 b = 24 c =
3. a = 6 c = 10 b =
4. a = 15 c = 39 b =
5. b = 5 c = 9 a =
6. b = 7 c = 11 a =
7. a = 4 b = 8 c =
8. a = 6 c = 11 b =
(9–18) Find the value of sin  and cos  for the following: (Don’t use the calculator)
sin
9.
cos

3

3

10.

5

6
5
cos

6
sin
12.
15.
18.
sin(150 ) 
cos(150 ) 
sin(30 ) 
cos(30 ) 
11

6
11
cos

6
2

3
2
cos

3
13.
7

6
7
cos

6
5

6
5
cos

6
14.
sin
16.
19.
22.
sin
24.
cos 60 
11.
sin
sin
21.
sin 60 
sin120 
cos120 
sin 330 
cos330 
5

3
5
cos

3
cos150 
sin 210 
cos 210 


6

cos

6
sin
17.
2

3
2
cos

3
sin
20.
23.
sin
25.
sin150 
26.
sin(120 ) 
cos(120 ) 
sin(330 ) 
cos(330 ) 
Holtfrerich & Haughn
27.
30.
39
sin
sin 45 
28.
cos 45 
cos
sin 225 
31.
cos 225 
7

4
7
cos

4
sin
33.
34.
17

6
17
cos

6
39.
42.
45.
4

4

29.

sin(225 ) 
cos(225 ) 
sin(315 ) 
cos(315 ) 
cos180 
sin 270 
cos 270 
5

4
5
cos

4
sin
32.
35.
37.
40.
sin  
cos  
41.
43.
sin(2 ) 
cos(2 ) 
44.
sin
sin180 
5

4
5
cos

4
sin
13

3
13
cos

3
sin
36.

Section 5.1 Geometry Review
38.
sin 510 
cos510 
sin(780 ) 
cos(780 ) 
3

2
3
cos

2
sin
sin(360 ) 
cos(360 ) 


2

cos

2
sin
sin(90 ) 
46.
cos(90 ) 
(47–54) For each angle in standard position find the value of sin  and cos .
r

(x, y)

47.
x3 y4
sin   cos  
48.
x  5 y  12
sin   cos  
50.
x  10 y  24
sin   cos  
51.
x  4
sin  
49.
x  6
sin  
y  8
cos  
r 5
Terminal side in the 2nd quadrant
cos  
Chapter 5 Trigonometric Functions
52.
y  5
sin  
54.
x  6 y  3
sin   cos  
40
r  13
Terminal side is in the 4th quadrant
cos  
53.
x  8
sin  
y4
cos  
(55–64) Give the angle values for  between –2 and 2 that yield the following answer.
(Give the answers radians)
55. sin  
3
2
56. cos  
1
2
59. sin   1
60. cos   1
62. sin   0
63. sin  
58. sin   
61. cos   0
64. cos  
3
2
57. cos   
1
2
 2
2
 2
2
(65–74) Give the angle values for  between –360o and 360o that yield the following answer.
(Give the answers degrees)
65. sin  
 3
2
66. cos  
68. sin  
1
2
69. sin   1
70. cos   1
72. sin   0
73. sin  
71. cos   0
74. cos  
 3
2
67. cos  
1
2
2
2
2
2
75. What is the domain for sin ?
76. What is the domain for cos ?
77. What is the range for cos ?
78. What is the range for sin ?
79. What is the period for cos  and what does the period tell us?
80. What is the period for sin  and what does the period tell us?
(81–92) Use the calculator to approximate the value of sin  and cos  for the given value of .
(Approximate to 4 decimal places)
Holtfrerich & Haughn
81.
84.
41
sin 53 
82.
cos 53 
sin 324 
85.
cos 324 
3

7
87.
3
cos

7
sin
sin
90.
cos

9

9
sin153 
cos153 
sin(157 ) 
cos(157 ) 
(93–94)
86.
sin 239 
cos 239 
sin(138 ) 
cos(138 ) 
9

7
89.
9
cos

7
3

11
91.
3
cos

11
11

5
92.
11
cos

5
sin

83.
13

8
88.
13
cos

8
sin

Section 5.1 Geometry Review
sin
sin
(a,b)
r 
y

b
1
x
a
93. a = 3 and b =4; a.) x = b.) y = c.) r =
d.) Find the value of sin  and cos  by using both the rectangular and unit circle definition.
e.) How do the values compare?
94. a = 5 and b =12; a.) x = b.) y = c.) r =
d.) Find the value of sin  and cos  by using both the rectangular and unit circle definition.
e.) How do the values compare?
Chapter 5 Trigonometric Functions
42
Section 5.4 Tangent, Cotangent, Secant, Cosecant
Objectives

Understanding the definition and graphs of tangent and cotangent

Understanding the definition and graphs of secant and cosecant
TANGENT AND COTANGENT
We are now ready to define the last four trigonometric functions using the unit circle.
Given a central angle θ in standard position whose terminal side intersects the unit circle at
the point (x, y) the following is defined.
Tangent − Tangent is defined to be tan θ =
y sin 
=
.
x cos 
P (x, y)
1
1
x
Cotangent − Cotangent is defined to be cot θ = =
.
y tan 
Secant − Secant is defined to be sec θ =
θ
x
1
1

.
x cos 
Cosecant − Cosecant is defined to be csc θ =
y
(1, 0)
1
1
.

y sin 
Let’s look at a table of values for all six of the trigonometric functions using just the first quadrant.
0

6

4

3

2
y = sin θ
0
1
2
2
2
3
2
1
x = cos θ
1
3
2
2
2
1
2
0
sin 
= tan θ
cos 
0
=0
1
1
2 = 1
3
3
2
2
2 =1
2
2
3
2 = 3
1
2
1
= undefined
0
1
= cot θ
tan 
1
= undefined
0
1
= 3
1
3
1
=1
1
1
3
1 0
 =0
1 1
0
1
= sec θ
cos 
1
=1
1
1
2
=
3
3
2
1
2
= 2

2
2
2
1
=2
1
2
1
= undefined
0
1
= csc θ
sin 
1
= undefined
0
1
=2
1
2
1
2
= 2

2
2
2
1
2
=
3
3
2
1
=1
1
Holtfrerich & Haughn
43
Section 5.1 Geometry Review
As you look at the above table notice how a function’s output value for an angle is equal to its cofunction’s output of the angle’s complement. A couple of examples are that sin
tan 0 = cot


= cos or that the
6
3

. To state this more mathematically:
2
Cofunction Property − For any angle θ the following is true.
sin θ = cos (

− θ),
2
tan θ = cot (

− θ),
2
sec θ = csc (

− θ)
2
Discussion 1: The Graph of Tangent
Let’s look at the graphs of these new functions. First, let’s look at the graph of the tangent function.
If you look back at the table you see that as the inputs go from 0 to
increasing in value 0,
=

3
 1.047 and θ =

2
1
, 1,
3
3 , undefined. The question is what is happening between θ
 1.57 ? Let’s turn to our graphing calculator and see.
It looks as though the function values, as we get close to
approaches

, the outputs, tan θ, are
2

, are getting very large. In fact as θ
2

the tan θ approaches ∞.
2
Let’s go backwards now from θ = 0 to θ =

.
2
0

6

4

3

2
y = sin θ
0
1
2
 2
2
 3
2
−1
x = cos θ
1
3
2
2
2
1
2
0
1
2 = 1
3
3
2
 2
 3
2 = −1
2 =− 3
1
2
2
2


Once again the question is what is happening between θ =
and θ =
?
3
2
sin 
= tan θ
cos 
0
=0
1
1
= undefined
0
Question 1: After looking at the first quadrant what is your thoughts on what will happen here in the
fourth quadrant?
Chapter 5 Trigonometric Functions
44
If we continued to follow this process and looked in both the positive and negative directions we’d
quickly discover that the tangent function is periodic with a period of π. Here is the graph of tangent
near the origin.
3
2

2
Because the tangent function is undefined at


2
3
2


, and every multiple of π added to , we get many
2
2

vertical asymptotes   n , n any integer  . From the graph of tangent we see that the domain for
2

tangent will be (−∞, ∞) except for


and any added π multiple to
2
2
 3         3 
,
,
,
, ,
…}. The range is (−∞, ∞). You can also see from the graph of the
2   2 2   2 2 
 2
{… 
tangent function that it does not have a period of 2π, like sine and cosine, but has a period of π.
Discussion 2: The Graph of Cotangent
Since the cotangent function is just the reciprocal of the tangent function all we need to do to graph it
is to look at the reciprocals of the output values of tangent.
tan θ

3
4

2

4
0

4

2
3
4

0
1
undefined
−1
0
1
undefined
−1
0
1
0
−1
undefined
1
0
−1
undefined
cot θ undefined
Remember that undefined happened when we had
1
0
so its reciprocal would be .
0
1
From this we can determine the graph of cotangent.
Holtfrerich & Haughn
45
3
2

2
Section 5.1 Geometry Review

2
3
2
From the graph of cotangent we see that the domain is (−∞, ∞) except for 0 and every multiple of π
added to 0  0  n , n any integer  , {…(−2π, −π), (−π, 0), (0, π), (π, 2π)…}. Also, you can see from the
graph of the cotangent function that it has a period of π just like the tangent function. That should be
no surprise, cotangent was just the reciprocal of tangent. We’d expect them to have a lot of
similarities.
Question 2: What is the range of cotangent?
SECANT AND COSECANT
Discussion 3: The Graph of Secant
Secant is the reciprocal of cosine, so we just need to look at the cosine function and use the
reciprocals of its output values.

cos θ −1
sec θ −1
3
4

2
 2
 0.71
2
0

4

4
0
2
 0.71
2
1
2
 0.71
2

2
3
4

0
 2
 0.71
2
−1
2
2
2
2
  2  1.4 undefined
  2  1.4 1
  2  1.4 undefined
  2  1.4 −1
2
2
2
2
As with the other functions, when you approach an input where the output is undefined your outputs
are approaching either ∞ or −∞. The graph of secant is,
1
3
2

−1
2

2
3
2
Answer Q1:
The same but
going to
negative
infinite.
Chapter 5 Trigonometric Functions
46
Once again like the tangent function (which has cosine in the denominator
has vertical asymptotes at
sin 
 tan  ) this graph
cos 


and every multiple of π added to . We get many vertical asymptotes
2
2




 n , n any integer  . The domain is (−∞, ∞) except for and every multiple of π added to
,
2
2
2

at 
 3         3 
,
,
,
,
,
…}. The range is (−∞, −1)  (1, ∞). But unlike the tangent
2   2 2   2 2 
 2
{… 
function you can see from the graph that it has a period of 2π, just like the cosine function.
Discussion 4: The Graph of Cosecant
Cosecant is the reciprocal of sine, so we just need to look at the sine function and use the reciprocals
of its output values.

sin θ
0
3
4

2
 2
 0.71 −1
2
sec θ undefined  2  1.4
−1

4
0
 2
 0.71
2
0
 2  1.4 undefined

4

2
2
 0.71
2
1
2  1.4
1
3
4
2
 0.71
2

0
2  1.4 undefined
As with the other functions, when you approach an input where the output is undefined your outputs
are approaching either ∞ or −∞. The graph of cosecant is,

2
3
2
3
2
1
−1

2
This time, like with the cotangent function (which has sine in the denominator
cos 
 cot  ), the
sin 
graph has vertical asymptotes at 0 and every multiple of π added to 0  0  n , n any integer  . The
domain is (−∞, ∞) except for 0 and every multiple of π added to 0 {…(−2π, −π), (−π, 0), (0, π), (π,
2π)…}. The range is (−∞, −1)  (1, ∞). But unlike the cotangent function you can see from the
graph that it has a period of 2π, just like the sine function.
Example 1: Finding Trigonometric Function Values
What are the following values?
Holtfrerich & Haughn
a) sec

3
47
b) csc
5
6
c) cot
2
3
Section 5.1 Geometry Review
d) tan
15
4
Solution:
a) To find the sec

we first must realize that secant is the
3
sec
reciprocal of cosine (cos θ = x). Second we must notice
that the reference angle is

. The unit circle point
3

1
=
3 cos 
3

1
3
 . Lastly we
2 2 
1
1
2
=2
associated with that reference angle is  ,

realize that the original angle is in the first quadrant where
every trigonometric function has a positive answer.
b) To find the csc
5
we first must realize that cosecant is
6
csc
the reciprocal of sine (sin θ = y). Second we notice that the
reference angle is

. The unit circle point associated with
6
sin
 3 1
. Third, we know that sine
,
 2 2 


=

in third quadrant 
is negative in the third quadrant. 
 6

c) To find the cot
2
we first must realize that cotangent is
3
the reciprocal of tangent (tan θ =
that the reference angle is
cot
y
). Second we notice
x
1
3
 . Third, we
2 2 
1
1
 
2
Answer Q2:
15
we first must notice that the reference
4
tan
(−∞, ∞)
1
 3 


 2 
 1 
 2 


1
=  2 =
3
2
know that tangent is negative in the second quadrant.
quadrant where tangent is negative.
6
2
1
1
=
=
2

3
tan
 tan
3
3
=

. The unit circle point
3

angle is . The unit circle point associated with that
4
 2 2
reference angle is 
,
 . Second, we calculate that
 2 2 
15
is just short of two revolutions, so it is in the fourth
4

= −2
associated with that reference angle is  ,
d) To find the tan
1
=
that reference angle is 
 5
5
1
=
5
6
sin
6
15

= − tan
4
4
= −1
1
3
Chapter 5 Trigonometric Functions
48
Here is a little idea that may help you remember which functions are positive and negative in which
quadrants. The acronym to memorize is ASTC (all students take calculus). All, stands for all trig
functions positive in the first quadrant. Students, stands for only sine and its reciprocal (cosecant) are
positive in the second quadrant. Take, stands for only tangent and its reciprocal (cotangent) are
positive in the third quadrant. Calculus, stands for only cosine and its reciprocal (secant) are positive
in the fourth quadrant.
Students
All
Take
Calculus
Let’s do one more example to remind you of the first definition once again.
Example 2: Finding Trigonometric Values
What are the values of the following given the picture of θ? (Hint: you will want to apply the first
definition and not the unit circle definition.)
a) tan θ
b) sin θ
c) sec θ
d) cos θ
e) cot θ
P (5, 6)
θ
Solution:
a) Tangent is defined to be
b) Sine is defined to be
y
.
x
y
. We need to find r
r
(x2 + y2 = r2).
tan θ =
6
5
52  62  r 2
r 2  61
r  61
c) Secant is the reciprocal of cosine so secant
sin θ =
6
61
sec θ =
61
5
cos θ =
5
61
r
is defined to be .
x
d) Cosine is defined to be
x
.
r
e) Cotangent is the reciprocal of tangent so it
is defined to be
x
.
y
cot θ =
5
6
Holtfrerich & Haughn
49
Section 5.1 Geometry Review
Example 3: Using the Calculator
What are the output values of the following trig functions?
a) tan 17°
b) cot 126°
c) sec (78°)
d) csc 341°
Solution:
A word of caution is needed here. When using the graphing calculator with trigonometric
functions it is very important as to whether the calculator is set in radian or degree mode.
The best rule of thumb is: if the angles are given in radians then the calculator should be in
radian mode and if the angles are given in degrees then the calculator should be in degree
mode.
a) To find the tan 17° simply push the TAN
key followed by 17, then ENTER. Be
sure that your calculator is in degree
mode.
b) To find the cot 126° use the fact that
cotangent is the reciprocal of tangent.
Push 1 then / then TAN followed by
126, then ENTER.
c) To find the sec (78°) use the fact that
secant is the reciprocal of cosine. Push 1
then / then COS followed by 78, then
ENTER.
d) To find the csc 341° use the fact that
cosecant is the reciprocal of sine. Push 1
then / then SIN followed by 341, then
ENTER.
The last thing we should mention here is that sine is an odd function and cosine is an even function.
You might remember from chapter 3 that an odd function is a function whose graph is symmetric to
the origin and an even function is a function whose graph is symmetric to the y-axis. It is true that for
even functions f(−x) = f(x) and for odd functions f(−x) = −f(x). The other four trigonometric functions
can be found from these two, thus the following is true for the six trigonometric functions.
Odd Properties − For any angle θ the following is true.
sin (−θ) = −sin (θ), csc (−θ) = −csc (θ),
tan (−θ) = −tan (θ),
Even Properties − For any angle θ the following is true.
cos (−θ) = cos (θ), sec (−θ) = sec (θ)
cot (−θ) = −cot (θ)
Chapter 5 Trigonometric Functions
50
Section Summary:

The graphs of the six trig functions are:

Notice that in order to graph cosecant, secant, and cotangent with the graphing calculator
you need to use the fact that they are the reciprocals of sine, cosine and tangent.

Domain and range of the six trig functions and their period.
Domain
Range
Periodic
Sine Function
(−∞, ∞)
[−1, 1]
2π
Cosine Function
(−∞, ∞)
[−1, 1]
2π

 …}

(−∞, ∞)
π
{…(−2π, −π), (−π, 0), (0, π), (π, 2π)…}
(−∞, ∞)
π

 …}

(−∞, −1)  (1, ∞)
2π
{…(−2π, −π), (−π, 0), (0, π), (π, 2π)…}
(−∞, −1)  (1, ∞)
2π
Tangent Function
Cotangent Function
Secant Function
Cosecant Function
 3         3
,
,
,
, ,
2   2 2   2 2
 2
{… 
 3         3
,
,
,
, ,
2   2 2   2 2
 2
{… 
1
1
1
, sec θ =
, csc θ =
.
tan 
cos 
sin 

Reciprocal relationships cot θ =

The first definitions of the six trigonometric functions sometimes refered to as the coordinate
definitions are: sin θ =

y
x
y
r
r
x
, cos θ = , tan θ = , csc θ = , sec θ = , cot θ = .
r
r
x
x
y
y
The second definitions which are the unit circle definitions are:
sin θ = y, cos θ = x, tan θ =
y
1
1
x
, csc θ = , sec θ = , cot θ = .
x
x
y
y

The even functions are cosine and secant [ f(−x) = f(x)]

The odd functions are sine, cosecant, tangent, and cotangent [ f(−x) = −f(x)]
Holtfrerich & Haughn
51
Section 5.1 Geometry Review
SECTION 5.4 PRACTICE SET
(116) For the given value of  find the exact value for tan , cot , sec  and csc .
1.   120
2.   210
3.   350
4.   30
5.   225
6.   135
7.  
3
radians
4
10.  
11
radians
6
13.   180
8.  
7
radians
4
11.  
4
radians
3
14.   270
9.  
7
radians
6
12.  

radians
3
15.  
3
radians
2
16.    radians
(17−26) For the given value of  approximate to 4 decimal places the value of tan , cot , sec 
and csc .
17.   138
18.   233
19.   351
20.   78
21.   331
22.   153
23.  
5
radians
7
26.  
11
radians
5
24.  
7
radians
5
25.  
9
radians
8
(27−32) Give all the other angles in degrees from –360o to 360o that will give the same value as each
of the following:
27. tan 105o
28. cot 225o
29. sin 321o
30. cos 78o
31. sec 70o
32. csc 330o
(33−38) Give all the other angles in radians from −2 to 2 that will give the same value as each
of the following:
33. cot
3
5
34. tan

5
35. sin
7
5
36. cos
8
5
37. sec
9
5
38. csc
3
8
Chapter 5 Trigonometric Functions
52
(39−46) The angle  is in standard position and (x, y) is a point on the terminal side of the angle.
For the given value of (x, y) find the value for tan , cot , sec  and csc :
39. (3, 4)
40. (5, 12)
41. (5, 3)
42. (4, 2)
43. (0, 3)
44. (0, 4)
45. (2, 0)
46. (5, 0)
(4752)
4
then csc  =?
7
4
48. If cos  
then sin  = ?
5
47. If sin  
49. If tan  
9
then cot  = ?
5
50. If cot  
11
then tan  = ?
3
51. If csc  
7
then sin  = ?
3
52. If sec  
9
then cos  = ?
4
(5360) Use the calculator to test the following relationships:
53. tan 38o 
54. cot 123o 
55. sin 238o 
56. cos 332o 
57. tan
3

5
58. cot 59o 
sin 38
cos 38
cos123
sin123


1

csc 238
1
sec 332

1

3
cot
5
1
tan 59

Holtfrerich & Haughn
53
59. sec
7

8
1

7
cos
8
60. csc
3

11
1

3
sin
11
Section 5.1 Geometry Review
(6164)
(a,b)
r 
b

1 y
x
a
Use the value of (a, b) that is given to find r and (x, y). Use (a, b) and r to find the value of tan ,
cot , sec  and csc  using the coordinate system definition. Use (x, y) and the unit circle
definition to find the value of the four functions. (Hint: to find the value of (x, y) use similar
triangles)
61. (a, b) = (3, 4)
(x, y) =?
tan  = ?
cot  = ?
sec  = ?
csc  = ?
62. (a, b) = (5, 12)
(x, y) =?
tan  = ?
cot  = ?
sec  = ?
csc  = ?
63. (a, b) = (3, 6)
(x, y) =?
tan  = ?
cot  = ?
sec  = ?
csc  = ?
64. (a, b) = (10, 10)
(x, y) =?
tan  = ?
cot  = ?
sec  = ?
csc  = ?
65. What did you find out about the coordinate definition answers and the unit circle answers from
the problems 61 to 64?
66. What is the domain of tan ?
67. What is the range of tan ?
68. What is the range of cot ?
69. What is the domain of cot ?
70. What is the domain of sec ?
71. What is the range of sec ?
72. What is the range of csc ?
73. What is the domain of csc ?
Chapter 5 Trigonometric Functions
54
Section 5.5 Inverse Trigonometric Functions
Objectives

Understanding the inverse of sine and cosine

Understanding all the other inverse trigonometric functions

Understanding the composition of trigonometric functions
As we begin our look at the inverses of trigonometric functions we need to review what was
discussed in chapter 1. Here is a list of facts from chapter 1.

Functions are simply special relations where for each input there is only one output.

Notation is y = f(x) where x is the input, the independent variable, and f(x) is the output, the
dependent variable.

Domain is the set of all possible inputs, and the range is the set of all resulting outputs.

You can add, subtract, multiply, divide and compose functions.

An inverse “undoes” something by doing the opposite operations in the opposite order.

A relation is one-to-one if for each output there is only one input.

For a relation to have an inverse function it must be a function and it must be one-to-one.

There are two ways to find an inverse function: one is to solve for the independent variable
and then switch the two variables; the second, is to switch the two variables and then solve
for the dependent variable. In short, either solve for x then switch x and y or switch x and y
and then solve for y.

If the function g is the inverse of function f , then the domain of f(x) (Df) is the same as the
range of g(x) (Rg) and the range of f(x) (Rf) is the same as the domain of g(x) (Dg) .

Inverse function graphs are mirror images of each other around the line y = x .
INVERSE SINE AND INVERSE COSINE
Discussion 1: Finding Inverse Sine
Let’s take a look at the sine graph once again.
y
y = sin θ
1
θ
−2π
2π
−1
At first glance you should notice that the graph passes the vertical line test so we can visually see that
it is a function (no vertical lines will cross our graph more than once). But next, you should also
Holtfrerich & Haughn
55
Section 5.1 Geometry Review
notice that it doesn’t pass the horizontal line test for determining if something is one-to-one (no
horizontal lines cross the graph more than once). Thus, to quote a famous statement “Houston we
have a problem”! We will get around this problem by adding a restriction to the domain of the sine
function. This is similar to what you saw us do with the function x2. By restricting how much of the
graph we are going to look at we can get something that will pass the horizontal line test.
Question 1: What part of the graph, and thus what domain, should we restrict ourselves too?
There are many possible answers you could have given to question 1 that would be correct. The one
   
,  . Notice that
 2 2
most often used in the science/mathematics world is to restrict the domain to 
this keeps the most amount of the graph that one can and still maintain a one-to-one function and it is
near the origin. Thus as we begin talking about the inverse of sine we will use this graph for sine.
y
1

2

4

4

2
θ
−1
In order to find an inverse we can switch the two variables and solve.
Original function
y = sin x
Switch the two variables
x = sin y
Now solve for y
Can’t solve, don’t know how. This happened
to us when we discussed the logarithm
function. There we made a definition, so
here we will do the same.
   
Inverse Sine − Given, y = sin x with domain  ,  and range [−1, 1], inverse sine is
 2 2
defined to be:
   
,  and x = sin y.
 2 2
y = sin -1 x, where the domain is [−1, 1] and the range is 
(Notice how the independent and dependent variables switch positions in inverse functions.)
Chapter 5 Trigonometric Functions
56
The notation sin -1 is sometimes written arcsin. Thus, the two ways you will see the function inverse
sine written are: y = sin 1 x, or y = arcsin x.
We can make a graph of sin -1 x by using the fact that the graph of an inverse is the mirror image
of the original graph about the line y = x. Here is a sketch of sin -1 x, a TI calculator screen shot of
inverse sine, and a calculator screen shot of the sine and the inverse sine functions and the line y = x.
y

2

4
x
−1

4
1

2
Notice the mirror image we have in the third picture.
Example 1: Finding Inverse Sine Values
What are the missing values in the following table for the function y = sin-1 x?
x
−1
 2
2
1
2
0
1
2
3
2
1
y
Solution:
First remember that the inputs of the inverse
x
y
x
y
are the outputs of the original function, so a

2
−1

6
1
2

3
 3
2

4
2
2

4

3

2
3
2

6
 2
2
1
2
0
0
table of values for y = sin x should be
helpful to us.
1
Holtfrerich & Haughn
57
Section 5.1 Geometry Review
All we need to do is take the inputs from y = sin x and use those for the outputs of
y = sin -1 x.
x
−1
 2
2
1
2
0
1
2
3
2
1
y



4

6
0

6

3

2
Example 2: Finding Inverse Sine Values from the Calculator
What are the values of the following?
a) sin -1 (0.2)
b) sin -1 (−0.92)
c) sin -1 (1.2)
Solution:
a) To find the inverse sine of a number that
doesn’t yield a standard angle we will use
the graphing calculator. One word of
caution, make sure the calculator is on
radian mode or degree mode depending
on the type of angle answer you are
wanting.
In this example the angle answers are in
radian measure. Here is what we see on
our calculators. SIN −1 is above the SIN
key.
b) Type this into the calculator to see the
answer.
c) Do the same as for the first two
examples.
Notice that we get an error message about
the domain. Our domain for sin -1 x is [−1,1]
and so 1.2 can’t be used as an input value.
Question 2: With the function y = sin x, we are either substituting in real numbers, that are angle
measurements in radian measure, or angles in degree measure for the input values and we get real
number outputs. What are the kinds of inputs and outputs that correspond to the function y = sin -1 x?
Answer Q1:
We would look
at the first
part of the
graph near the
origin that
goes uphill.
The domain
would be
   
 2 , 2 .


Chapter 5 Trigonometric Functions
58
In essence, we have one function with angle inputs yielding number answers and the inverse function
where we have number inputs yielding angle answers. Keeping this in mind could really save you
some mistakes down the road in trigonometry.
Please keep in mind also that we defined an angle in radian measure to be the arc length of a
piece of a unit circle. So the angle is a real number which we need it to be for many areas of
mathematics but yet these same real numbers are representing an angle.
Discussion 2: Finding Inverse Cosine
Here is a graph of the cosine function.
y
y = cos θ
1
θ
−2π
2π
−1
Notice that cosine isn’t one-to-one just like the sine function isn’t one-to-one, unless we restrict its
domain. With this function the standard definition for inverse cosine is:
Inverse Cosine − Given y = cos x with domain 0,   and range [−1, 1], inverse cosine is
defined to be:
y = cos -1 x, where the domain is [−1, 1] and the range is 0,   and x = cos y.
The notation cos -1 is sometimes written arccos. Thus, the two ways you will see the function inverse
cosine written are: y = cos 1 x, or y = arccos x.
We can make a graph of cos -1 x by using the fact that the graph of an inverse is the mirror image of
the original graph about the line y = x. Here is a sketch of cos -1 x, a TI calculator screen shot of
inverse cosine, and a calculator screen of cosine, inverse cosine and the line y = x.
y
π
3
4

2

4
−1
x
1
Holtfrerich & Haughn
59
Section 5.1 Geometry Review
Notice the mirror image we have in the third picture.
Example 3: Finding Inverse Cosine Values
What are the missing values in the following table for the function y = cos-1 x?
x
1
2
 2
2
−1
0
1
2
3
2
1
y
Solution:
First remember that the inputs of the inverse
are the outputs of the original function, so a
x
y
x
y
0
1
2
3
1
2

6
3
2
3
4
 2
2

4
2
2
5
6
 3
2

3
1
2
π
−1

2
0
table of values for y = cos x should be
helpful to us.
All we need to do is take the inputs from y = cos x and use those for the outputs of
y = cos -1 x.
x
y
−1
 2
2
1
2
0
1
2
3
2
1
π
3
4
2
3

2

3

6
0
Example 4: Finding Outputs for sin -1 x and cos -1 x
What are the values of the following?
a) cos -1
Solution:
2
2
b) cos -1
 3
2
c) sin -1
2
2
d) cos -1 (1.5)
Answer Q2:
The inputs are
now real
numbers and
the outputs
are angles
either, in real
number form
as radians or
angle form in
degrees.
Chapter 5 Trigonometric Functions
2
we need to think
2
a) To find the cos -1
2
= cos x. Inputs in the
2
find x such that
60
So, what angle makes cos x =
Answer:
2
?
2


rad (the angle
does also
4
4
inverse function are outputs in the
along with many others but they aren’t in the
original function.
range of inverse cosine [0, π])
b) To find the cos -1
find x such that
c) To find the sin -1
find x such that
 3
we need to think
2
 3
= cos x.
2
2
we need to think
2
2
= sin x.
2
Thus, cos -1

2
=
4
2
The angle is
5
7
. (the angle
does also
6
6
along with many others but they aren’t in the
range of inverse cosine [0, π])
Thus, cos -1
5
 3
=
6
2
The angle is

3
. (the angle
does also
4
4
along with many others but they aren’t in the
   
range of inverse sine  ,  )
 2 2
Thus, sin -1
d) The value of cos -1 (1.5) can’t be
determined.

2
=
4
2
1.5 is not in the domain of the inverse cosine
function. Another way to look at this is that
there isn’t any angle where cos x = 1.5.
Here is a good rule of thumb.
If you are asked to find an inverse trigonometric function value, try thinking in terms of
what angle would yield the input for the inverse function. This is very much like finding
the log 5 125. There we talked about thinking in terms of “What exponent should 5 be
raised to in order to get 125 as an answer”. With trigonometry if you are asked to find
sin -1 (0.5) you should be thinking what angle does one input into the sine function in order
to get 0.5 as an answer. In this example the angle would be a

angle.
6
THE OTHER FOUR INVERSES
Discussion 3: Finding More Inverses
We now must look at the other four trigonometric functions. Here are their graphs.
Holtfrerich & Haughn
61
Section 5.1 Geometry Review
y = tan x
y = cot x
y = sec x
y = csc x
Notice that like sine and cosine none of these are one-to-one functions. We will need to restrict the
domain on each of these functions in order to find their inverse functions.
For y = tan x let’s take the branch of the tangent function near the origin.
   
Therefore, the domain we will consider is  ,  . This means that the inverse will have a range of
 2 2
   
-1
 2 , 2  and a domain of (−∞, ∞). Here is the graph of tan x or if you will arctan x.


For y = cot x let’s take the branch of the cotangent function to the right of the y-axis.
Therefore, the domain we will consider is (0, π). This means that the inverse will have a range of (0,
π) and a domain of (−∞, ∞). Here is the graph of cot -1 x or if you will arccot x.
Chapter 5 Trigonometric Functions
62
There isn’t a standard way of restricting the secant function (y = sec x).
 


 

3 
The two common choices for restricting the domain are 0,    ,   or 0,    ,  . The first
 2 2 
 2  2 
makes an identity work, the second allows for both positive and negative sloped tangent lines to the
 
3 

curve. In this book we are going to choose the later, 0,    ,  . This means that the inverse
 2  2 
 

3 
will have a range of 0,    ,  and a domain of (−∞, −1]  [1, ∞). Here is the graph of sec -1 x
 2  2 
or if you will arcsec x.
There isn’t a standard way of restricting the cosecant function (y = csc x).
 






3 
The two common choices for restricting the domain are  ,0    0,  or  0,     ,  . The
2 
 2   2
 2 
first makes an identity work, the second allows for both positive and negative sloped tangent lines to



3 
the curve. In this book we are going to choose the later,  0,     ,  . This means that the inverse
2 
 2 



3 
will have a range of  0,     ,  and a domain of (−∞, −1]  [1, ∞). Here is the graph of csc -1 x
2 
 2 
or if you will arccsc x.
Holtfrerich & Haughn
63
Section 5.1 Geometry Review
Here is a summary of the six inverses and their domains, ranges and graphs. It will be very important
to have this information memorized!
Function
Domain
Range
y = sin -1 x
[−1, 1]
   
 2 , 2


y = cos -1 x
[−1, 1]
[0, π]
y = tan -1 x
(−∞, ∞)
   
 2 ,2


y = cot -1 x
(−∞, ∞)
 0,  
y = sec -1 x
(−∞, −1]  [1, ∞)
    3 
0, 2    , 2 

 

y = csc -1 x
(−∞, −1]  [1, ∞)
    3 
 0, 2     , 2 

 

Graph
Chapter 5 Trigonometric Functions
64
Example 5: Evaluating Inverse Trigonometric Functions
What are the values of the following:
a) sin -1 (0.5)
 2
 2
b) cos -1 





c) tan -1  3

d) sec -1 (−2)
e) csc -1 (1)
Solution:
a) To find the sin -1 (0.5) it might be
The unit circle points with y coordinate of 0.5
easiest to think solve 0.5 = sin x.
are associated with the angles of
(Answers must come from 1st or 4th quadrants)
   
Inverse sine range is  ,  .
 2 2
Therefore,
b) As with part (a), solving for x in the
equation

5
and
.
6
6
2
= cos x might be best.
2
sin -1 (0.5) =

.
6
The unit circle points with x coordinate of
are associated with the angles of


and
.
4
4
(Answers must come from 1st and 2nd quadrants)
Inverse cosine range is [0, π].
Therefore,
c) Like the previous two examples, solve
 2 
= .
 2  4


cos -1 
The unit circle points, where
− 3 = tan x.

   
, .
 2 2

2
and
.
3
3
(Answers must come from 1st and 4th quadrants)
Inverse tangent range is 
Therefore,
d) To find the answer solve −2 = sec x. It


tan -1  3 =

.
3
The unit circle points, where x is
1
, are the
2
points associated with the angles
2
4
and
.
3
3
might be easier if we think,
1
cos x

1
= cos x.
2
    3 
Inverse secant range is 0,    ,  .
 2  2 
e) To find the answer solve 1 = csc x. It
might be easier if we think,
1=
1
sin x

(Answers must come from 1st and 3rd quadrants)
Therefore,
sec -1 (−2) =
4
.
3
The unit circle point, where y is 1, is the point
associated with the angle
1 = sin x.


y
is  3 , are the
x
points associated with the angles
−2=
2
2

.
2
There is only one choice and it is in our range.


3 
Inverse cosecant range is  0,     ,  .
2 
 2 
Therefore,
csc -1 (1) =

.
2
Holtfrerich & Haughn
65
Section 5.1 Geometry Review
Discussion 4: Evaluating Inverse Trigonometric Functions with the Calculator
Let’s see how we could evaluate the following using the graphing calculator.
a) cos1 0.34
b) tan 11.23
c) cot 1  13
d) sec1  2.7 
e) csc11.1
a) Here we just simply push 2nd and the COS
key and then our number followed by
ENTER (answer in degrees since the
calculator was set to degree mode).
b) Here we just simply push 2 nd and the
TAN key and then our number followed
by ENTER (answer in degrees since the
calculator was set to degree mode).
c) To graph, or evaluate cot 1 x for some
value of x, you must know that
1

 tan 1/ x 
cot 1 x  
1

 tan 1/ x   
x0
x0
.
Thus for our problem we will use the
second half of the formula since our input
Make sure that the calculator is in radian
value is negative.
mode if you use π, or degree mode if you use
180° in the formula.
d) To graph, or evaluate sec1 x for some
value of x, you must know that
1

 cos 1/ x 
sec1 x  
1

 cos 1/ x   2
x 1
x  1
.
Thus for our problem we will use the
second half of the formula since our input
Make sure that the calculator is in radian
value is negative.
mode if you use 2π, or degree mode if you
use 360° in the formula.
e) To graph, or evaluate csc1 x for some
value of x, you must know that
1

 sin 1/ x 
csc1 x  
1

 sin 1/ x   
x 1
x  1
.
Thus for our problem we will use the first
half of the formula since our input value is
Make sure that the calculator is in radian
Chapter 5 Trigonometric Functions
66
mode if you use π, or degree mode if you use
positive.
180° in the formula.
COMPOSITION OF TRIGONOMETRIC FUNCTIONS
Discussion 5: Composing Trigonometric Functions
To serve as a refresher, remember that if we had f(x) = x2 − 4 and g(x) = x + 2 then (f g)(3) = f (g(3))
= f (5) (since g(3) = 3 + 2 = 5) = 21 (since 5 2 − 4 = 21). Composition stated simply, is just
substitution. We will need to be able to do the same with trigonometric functions. Let’s look at and
evaluate the following problems.
a) sin (cos -1 0.5)
b) tan -1 (sin π)
a) We start with the inner most

   
d) sin -1  sin 

 6 

c) sec (sec -1 (−1))
cos -1 0.5 =
-1
parenthesis and find the cos 0.5

3
(since cos

= 0.5 and the angle
3

is in the range of inverse
3
cosine.)
Now find the sin of our answer.
sin

3
=
.
3
2
Therefore,
b) Once again begin by finding sin π.
sin π = 0
sin (cos -1 0.5) =
3
.
2
(since π puts the terminal side on
the negative x-axis which has the
unit circle point of (−1, 0))
-1
Now evaluate tan 0.
-1
tan 0 = 0
Therefore,
c) Start with finding sec -1 (−1)
tan -1 (sin π) = 0.
sec -1 (−1) = π (since negative input values yield

outputs in the interval  ,

Now evaluate sec π.
sec π = −1
3
2

)

(you should not be surprised by this
answer. Remember that inverses
cancel each other out and you get
just the input as the answer. See
discussion following question 3.)
Therefore,
  

 6 
d) We start with sin 
sec (sec -1 (−1)) = −1

   1
sin 
 = 2 (the reference angle is 6 in the
6


fourth quadrant)
Holtfrerich & Haughn
67
Now evaluate sin -1
1
.
2
sin -1
Section 5.1 Geometry Review
1

=
2
6


    
 = 6
 6 
sin -1  sin 
Therefore,

5 
Question 3: What is the value of cos -1  cos  ?
4 

As you answered the last question you may have thought the answer was
5
. That shows that you
4
are understanding the concept of the composition of inverses, but it isn’t correct in this case because
the inverse cosine function only exists with a restricted domain on cosine. Therefore,
5
can’t be
4
the correct answer because when we defined inverse cosine we had used the part of cosine from 0 to

5
5 
π. In order to evaluate cos -1  cos  we need to first change the angle
to an equivalent one in
4
4 

the proper domain. For the cosine function that would be
reference angle

cos -1  cos

3
5
3
. Both
and
have the same
4
4
4

5 

and the cosine of both is negative. Therefore, cos -1  cos  can be changed to
4
4 

3
3 
which then can be evaluated and we get the answer
. Our statement in discussion
4
4 
5c, “…inverses cancel each other out and you just get the input as the answer”, is only true when you
are working with 1 to 1 functions or, the restricted domains of non 1 to 1 functions. If you are using
input values that are not in the restricted domains of non 1 to 1 functions, this rule will not hold true.
For problems similar to these, but where the values are not the standard ones, we sometimes
need to go back to our original definitions with the right triangles. Let’s do one last example.
Example 6: Finding Compositions Using Triangles
What are the values of the following?
a) cos (sin -1
3
)
5
b) cot (csc -1 (−3))
c) tan (cos -1
7
)
9
d) sec (sec -1 x)
Solution:
a) First we could use the graphing
calculator, but we’ll get to that
later. What we can do in order to
r=5
θ
y=3
x=?
get the answer by hand is sketch a
picture of an angle where the sine
3
of it would yield the answer .
5
By the Pythagorean Theorem we can find that x=4.
Chapter 5 Trigonometric Functions
68
Remember from section 5.3 sine
was defined to equal
y
.
r
We now have a picture of the answer to sin -1
cos θ =
3
=θ.
5
x 4
= .
r 5
Therefore,
cos (sin -1
3
4
)=
5
5
b) We will use the same approach
here as we did for part (a). We
need to draw a picture of an angle
where csc θ = −3
3
1
r
1
= csc θ =
=
= (notice that we put
y
1
sin 
y
r
the negative sign in the denominator since r is in
the numerator and it can’t be a negative value) so,
we get a picture that looks like this.
Notice that we drew the picture of
the angle in the third quadrant
x=?
since that is the defined restricted
y = −1
domain for cosecant with negative
outputs.
θ
r=3
x2 = 32 − (−1)2 = 9 − 1 = 8, so x =  8  2.83 (x in
x
y
Definition section 5.3, cot θ =
our picture is in the negative direction)
cot θ =
x  8
 8
=
1
y
Therefore,
(reciprocal of tangent)
cot (csc -1 (−3)) =
8
c) We need to use the same approach
r=9
as we have and start off by drawing
a picture of the angle cos -1
We know that cos θ =
definition cos θ =
7
.
9
7
and the
9
x
.
r
x=7
y2 = 92 − 72 = 81 − 49 = 32, so y =
tan θ =
sec -1 x. We know that sec θ =
and the definition
x
1
tan (cos -1
7
4 2
)=
9
7
We will draw our picture with r = x and the x
coordinate equal to 1.
r=x
θ
sec θ =
1
r
1
= = .
x
cos 
x
r
32  5.66
y
32 4 2

=
x
7
7
Therefore,
d) We draw a picture of the angle
y=?
θ
x=1
y=?
Holtfrerich & Haughn
69
sec θ =
The answer should not be a
surprise. Inverses undo each other,
Section 5.1 Geometry Review
r
x
= =x
x 1
Therefore,
sec (sec -1 x) = x
as long as we are working within
the restricted domains for
Answer Q3:
5
is not in the
4
trigonometric functions.
Question 4: What is the value of cos (sin -1 a)?
Section Summary:

Here are the domains, ranges and graphs of the six trigonometric inverses.
Function
Domain
Range
y = sin -1 x
[−1, 1]
   
 2 , 2


y = cos -1 x
[−1, 1]
[0, π]
y = tan -1 x
(−∞, ∞)
   
 2 ,2


y = cot -1 x
(−∞, ∞)
 0,  
y = sec -1 x
(−∞, −1]  [1, ∞)
    3 
0, 2    , 2 

 

Graph
domain for
cosine inverse.
Thus we change
to an angle
which has the
same value,
3
which is
.
4
So now we
evaluate
3 

cos -1  cos 

=
3
.
4
4 
Chapter 5 Trigonometric Functions
y = csc -1 x

(−∞, −1]  [1, ∞)
70
    3 
 0, 2     , 2 

 

Remember, when working with a problem like trig -1 (trig θ), the trigonometric inverse of a
trigonometric function, that we must stay within the restricted defined domains and ranges.
Holtfrerich & Haughn
71
Section 5.1 Geometry Review
SECTION 5.5 PRACTICE SET
(1–24) Evaluate each expression without using the calculator. (Give your answer in radians)
 1 
1. sin 1  
 2
1
2. arcsin  
2
 1 
3. arccos 

 2
 1 
4. cos 1 

 2
 3
5. arcsin 
 2 


 3
6. sin 1 
 2 


 1 
7. cos1  
 2
1
8. arccos  
2
 1 
9. sin 1 

 2
 1 
10. arcsin 

 2
 3
11. arccos 
 2 


 3
12. cos 1 
 2 




14. tan 1


 1 
17. tan 1 

 3
 1 
18. arctan 

 3
 1 
19. arc cot 

 3
 1 
20. cot 1 

 3
21. arctan 1
22. tan 1  1
23. cot 1  1
24. arccot 1
13. arctan  3
16. arc cot  3
 3
15. cot 1
 3
Answer Q4:
We make a
triangle first
which is a
picture of the
angle
sin -1 a = θ.
r=1
θ
(25–48) Evaluate each expression without using the calculator. (Give your answer in degrees)
25. arcsin(1)
26. sin 1 (1)
27. cos1 (1)
28. arccos(1)
29. sin 1 (0)
30. arccos(0)
31. arctan(0)
32. cot 1 (0)
33. sec1 ( 2)
34. arcsec( 2)
35. arccsc( 2)
36. csc1 ( 2)
37. csc1 (2)
38. arc csc(2)
39. arcsec(2)
40. sec1 (2)
 2 
41. sec1 

 3
 2 
42. arc sec 

 3
 2 
43. arc csc 

 3
 2 
44. csc 1 

 3
45. arcsec(1)
46. sec1 (1)
47. csc1 (1)
48. arccsc(1)
(49–80) Evaluate each expression without using a calculator.
x=
y=a
1  a2
The cos θ =
x  1  a2
Chapter 5 Trigonometric Functions
72
49. arcsin(sin 45o )
   
50. sin 1  sin   
  3 
    
51. sin 1  sin 

  6 
52. arcsin(sin(45o ))
53. arccos(cos 60o )

  
54. cos 1  cos   
 4 


 3
55. cos 1  cos 
 4

56. arccos(cos150o )

 1 
57. sin  arcsin   
 2 


 3 
59. cos  arccos 
 2  





 1 
60. cos  cos 1   
 2 





 2 
58. sin  sin 1 
 2  





61. tan arctan
 3 

62. cot cot 1
 3 
63. arctan(tan 45o )
   
64. cot 1  cot   
  4 
   
65. arc cot  cot   
  6 
66. tan 1 tan 30

    
68. arc cot  cot 

  3 

 3 
69. sin  arcsin   
 5 


 3 
70. cos  arccos   
 5 


 4 
71. cos  cos 1   
 5 


 4 
72. sin  sin 1   
 5 

73. sin  arcsin  x  
74. cos  arccos  x  

 1 
75. cos  cos 1   
 x 


 1 
76. sin  sin 1   
 x 

77. tan  arctan  x  
78. cot  arc cot  x  

 1 
79. cot  cot 1   
 x 


 1 
80. tan  tan 1   
 x 

67. tan 1 tan(60



(81–96) Evaluate each expression without using a calculator.

 3 
81. sin  tan 1   
 4 


 3 
82. cos  arctan   
 4 


 4 
83. tan  arcsin   
 5 


 4 
84. cot  sin 1   
 5 


 5 
85. tan  cot 1   
 12  


 5 
86. cot  arctan   
 12  


 5 
87. sin  arccos   
 13  


 5 
88. cos  sin 1   
 13  


 12  
89. sec  tan 1   
 5 

Holtfrerich & Haughn
73
Section 5.1 Geometry Review

 12  
90. csc  arc cot   
 5 


 12  
91. sec  arcsin   
 13  


 12  
92. csc  cos 1   
 13  


 1 
93. sin  cos 1   
 3 


 1 
94. cos  arcsin   
 3 


 5 
95. tan  arccos 
 5  





 5 
96. cot  sin 1 
 5  




(97–102) Give the domain and range of each function.
97. sin–1 
98. arccos 
99. arctan 
100. cot–1 
101. sec–1 
102. arccsc 
(103–126) Use the calculator to evaluate each expression. (If possible)
3
103. tan 1  
2
7
104. arctan  
3
5
105. arc cot  
7
5
106. cot 1  
3
 3 
107. tan 1  
 8 
 7 
108. arctan  
 6 
 11 
109. arc cot 

 7 
 5 
110. cot 1  
 8 
4
111. sin 1  
9
3
112. arcsin  
7
3
113. arccos  
7
5
114. cos1  
 11 
 4 
115. arcsin  
 9 
 3 
116. sin 1  
 7 
 5 
117. cos1  
 11 
 3 
118. arccos  
 7 
7
119. arcsec  
5
 15 
120. sec1  
 11 
 11 
121. csc1  
5
8
122. arc csc  
7
5
123. arcsin  
3
 7 
124. cos1  
 5 
8
125. cos 1  
 3
 8 
126. arcsin  
 5 
1
127. arcsec  
5
 2 
128. csc1  
 3 
(129–140) Arcsin , Arccos , Arctan  and Arccot  represent the inverse relations of each
corresponding trigonometric function. For each of the following give an expression
that will give all the values for the inverse relations.
Chapter 5 Trigonometric Functions
74
1
129. Arcsin  
2
 3
130. Arccos 
 2 


 1 
131. Arccos  
 2
 3
132. Arc sin 
 2 


133. Arctan 1
134. Arccot 1
135. Arccot  1
136. Arctan  1
137. Arctan
138. Arccot
 3

139. Arccot  3

 3

140. Arctan  3

141. For problems 123 thru 128 why did the calculator give you an error reading?
Holtfrerich & Haughn
75
Section 5.1 Geometry Review
Section 5.6 More On Graphing Trigonometric Functions
Objectives

Understanding sine and cosine translations
In chapter 3 we looked at vertical and horizontal shifts in a graph and what causes them along with
rotations about the x-axis (flipping), rotations about the y-axis (rotating), stretching and compressing.
Let’s look at these same ideas but with trigonometric functions.
TRANSLATIONS OF THE SINE AND COSINE FUNCTIONS
Let’s see a summary of the chapter 3 material that relates to what we wish to learn in this section.

Vertical translation: f(x) + k, means the graph shifts up (k > 0) or down (k < 0) k units. Notice
that by adding something to a function you change the y values (vertical shift).

Horizontal translation: f(x + h), means the graph shifts left (h > 0) or right (h < 0) h units.
Notice that by adding something to the x variable in a function you change the x values
(horizontal shift).

Rotations about the x-axis (flipping): f(x), means the graph flips upside down. Notice that by
multiplying by a negative a function changes its y value signs (Flipped upside down).

Rotations about the y-axis (rotating): f(x), means the graph rotates around the y-axis. Notice
that by multiplying the x variable by a negative a function changes its x value signs (rotating
around the y-axis).

Stretching or compressing: af(x), means the graph is narrower if a is larger than 1 and wider
if a is less than 1. Notice that multiplying a function by a number changes its y value
magnitudes (stretching or compressing the graph).
These same properties will hold true with any function so let’s see how it looks in the world of
trigonometry.
Question 1: How do you think the graph of y = sin (x) + 4 will differ from y = sin (x)?
Discussion 1: Vertical and Horizontal Translations
Let’s look at how the following functions differ from their parent functions.
a) y = cos (x) − 2
b) y = sin (x −

)
4
a) The parent function would be
y = cos (x). Since we see that this
function is subtracting 2 from
c) y = sin (x +

)+3
3
As you can see from the graphing calculator the
graph of our function is 2 units lower.
Chapter 5 Trigonometric Functions
76
every output in the parent function
the graph should move down 2
units.
b) The parent function would be
y = sin (x). Since we see that this

function tells us to subtract
4
As you can see from the graphing calculator the
graph of our function is

4
 0.7854 units to the
right.
from every input in the parent
function the graph should move
right

units.
4
c) The parent function would be
y = sin (x). Since we see that this

function is adding
to every
3
As you can see from the graphing calculator the
graph of our function is

3
 1.0472 units to the left
and 3 units up.
input and adding 3 to every output
as compared to the parent function
the graph should move left

3
units and up 3 units.
We have now seen a few examples that fit right in with what was learned in chapter 3. We see here
as before that by adding to the dependent variable (y) the graph moves up or down and that by adding
to the independent variable (x) the graph moves left or right. Let’s try a couple of more examples.
Example 1: Vertical Stretching and Compressing (Amplitude)
How do the graphs of the following different from their parent graphs?
a) y = 3sin (x)
b) y =
1
cos (x)
2
c) y = −5sin (x)
Solution:
a) The parent function would be
y = sin (x). Since we see that this
As you can see from the graphing calculator the
graph of our function is 3 times taller.
function is multiplying every
output in the parent function by 3
the graph should stretch by a
factor of 3.
b) The parent function would be
y = cos (x). Since we see that this
As you can see from the graphing calculator the
Holtfrerich & Haughn
77
function is multiplying every
Section 5.1 Geometry Review
graph of our function is
output in the parent function by
1
2
1
times shorter.
2
the graph should compress by a
factor of
1
.
2
c) The parent function would be
As you can see from the graphing calculator the
y = sin (x). Since we see that this
graph of our function is 5 times taller but flipped
function is multiplying every
upside down because of the negative sign.
output in the parent function by
−5 the graph should stretch by a
factor of 5 and be flipped upside
down.
The magnitude of the coefficient in front of a trigonometric function is called the amplitude. It effects
how much the graph is stretched or compressed vertically. The sign of the coefficient determines if
the graph is rotated about the x-axis (flipped upside down or not) by whether the coefficient is
positive or negative.
It is now time to address horizontal stretching and compressing. This is a bit more complicated and
affects two concepts: period and phase shift. Let’s do a few examples.
Discussion 2: Horizontal Stretching and Compressing (Period)
Here we are going to look at examples of the form y = sin (bx). Let’s see how the following functions
differ from their parent functions.
a) y = sin (2x)
x
b) y = cos  
2
a) The parent function would be
c) y = cos (3x)
d) y = sin (−x)
2π  6.28,   3.14
Answer Q1:
Probably it
will be four
units higher
(up four).
y = sin (x). Let’s see how the two compare.
Notice that the second graph is more
compact. The parent graph must go from 0
to 2π before covering a whole period (sine
is 2π periodic). But, the function
y = sin (2x), as we can see, has a period of
only π. The second function has been
compressed horizontally by a factor of two.
b) The parent function would be
y = cos (x). Let’s see how the two compare.
4π  12.57
Chapter 5 Trigonometric Functions
78
Notice that the second graph is more
spread out. The parent graph must go from
0 to 2π before covering a whole period
(cosine is 2π periodic). But, the function
x
y = cos   , as we can see, has a period of
2
 
4π. The second function has been stretched
by a factor of two.
c) The parent function would be
2  6.28
y = cos (x). Let’s see how the two compare.
Notice that the second graph is more
compact. The parent graph must go from 0
to 2π before covering a whole period. But,
the function y = cos (3x), as we can see,
has a period of
2
 2.0944
3
2
. The second function
3
has been compressed by a factor of three.
d) The parent function would be
2π  6.28
y = sin (x). Let’s see how the two compare.
Your first thought might be that the second
graph is flipped upside down from the first.
But, that can’t be correct because that only
happens when you are multiplying the
whole function by a negative. If you
remember from chapter 3, when
multiplying x by a negative you rotate the
graph around the y-axis. That is what has
happened in this case.
Let’s see if you understand the pattern yet after just a few examples.
Question 2: How do we figure out whether ae function has been compressed or stretched horizontally
and by what factor? (In essence this question is asking how do we find the period of a sine or cosine
function.)
Holtfrerich & Haughn
79
Section 5.1 Geometry Review
The period of a sine or cosine function isn’t difficult to evaluate but you do need to take some care or
you will quickly find out how easy it is to make a mistake.
If given a function y = asin (bx − c) or y = acos (bx − c), where b is a positive number, then
the period is calculated as follows.
Period =
2
.
b
The last new concept is phase shift. Let’s once again look at a few examples.
Discussion 3: A More Complete Look at Horizontal Shifts (Phase shift)
Let’s find the phase shift for each of the following functions.
a) y = sin (x − π)
b) y = sin (2x +

)
2
c) y = cos (3x −
a) This is a problem just like the one we looked at

)
2
We now know that the graph of the
in discussion 1(b). This time though let’s
function y = sin (x − π) will be shifted
approach the problem not from a chapter 3
π units to the right from its parent
approach but from the idea of figuring out how
function y = sin (x).
far we have moved from the origin (this method
could be used in chapter 3 as well). What we
will do is set the input quantity equal to zero to
see what x-value will yield a zero input value.
y = sin (x − π)
given
(x − π) = 0
set equal to zero
x=π
solved for x
b) As in part (a) above let’s set the input quantity
equal to zero and then solve for x. This will
We now know that the graph of the
function y = sin (2x +
help us to find the phase shift for this function.
shifted
y = sin (2x +
(2x +

)
2

)=0
2
given

) will be
2

units to the left from its
4
parent function y = sin (x).
set equal to zero
2x =

2
subtracted
x=

4
multiplied by

2
1
2
Chapter 5 Trigonometric Functions
80
You should also notice that this graph will have
a period of π since
2 2

 .
b
2
c) As in part (a) above let’s set the input quantity
equal to zero and then solve for x. This will
We now know that the graph of the
function y = cos (3x −
help us to find the phase shift for this function.
shifted
y = cos (3x −
(3x −

)
2

)=0
2

units to the right from its
6
parent function y = cos (x).
set equal to zero

2
added

6
multiplied by
3x =
x=
given

) will be
2

2
1
3
You should also notice that this graph will have
a period of
2
2 2

since
.
3
b
3

6
 0.5236
The General Form
If given a function y = asin (bx − c) + d or y = acos (bx − c) + d then the following are
true.

a is the amplitude of the function (how the graph has been stretched or
compressed vertically).


2
= period of the function (how long it takes for the function to repeat itself).
b
bx − c = 0  x =
c
is the phase shift (how much the graph has been shifted
b
horizontally).
Answer Q2:
It looks as
though you
just divide 2π
by the number
in front of x.

d is the amount the graph has been shifted vertically.

If a  0 , then the graph is rotated about the x-axis (flipped vertically).

If b  0 , then the graph is rotated horizontally.

Domain (∞, ∞)

Range 1 a  d ,1 a  d  , a stretches the range and d shifts it up or down.
Holtfrerich & Haughn
81
Section 5.1 Geometry Review
Example 2: Sketching Graphs of Sine and Cosine
What are the graphs of the following functions?
a) y = 2cos (6x + 5π) − 3
b) y = −sin (2x − 3π) + 1
Solution:
a) From the function we see that a = 2, b = 6,
As we sketch the graph we center it on
c = −5π, and d = −3. Therefore, we know that
the line y = −3 (shifted down 3), make
the graph is shifted down 3, that it will have
it go up and down 2 units from the
an amplitude of 2 ( a  2 ) and a period of
center line (amplitude = 2), repeat itself
2 2 

 . We find the phase shift by
b
6
3
solving (6x + 5π) = 0 for x.
every

5
, and we start the graph
to
3
6
the left of the origin.
(6x + 5π) = 0
6x = − 5π
x=
5
6
5
 2.618
6
b) From the function we see that a = −1, b = 2,
As we sketch the graph we center it on
c = 3π, and d = 1. Therefore, we know that
the line y = 1 (shifted up 1), make it go
the graph is flipped vertically (turned upside
up and down 1 units from the center
down) since a is negative, shifted up 1, that it
line (amplitude = 1), repeat itself every
will have an amplitude of 1 ( a  1  1 ) and
π, and we start the graph
a period of
2 2

  . We find the phase
b
2
shift by solving (2x − 3π) = 0 for x.
3
to the
2
right of the origin. The negative a
means the graph is flipped.
(2x − 3π) = 0
2x = 3π
x=
3
2
3
 4.71
2
Let’s take a look at how we could use this information to help us model a real life event.
Discussion 4: Weather Model
Chapter 5 Trigonometric Functions
82
As you are aware the weather from day to day doesn’t change much. Or the weather from year to
year follows a pattern, hot in the summer and cooler in the winter. Since weather tends to be cyclical,
repeats itself, the sine or cosine functions can be used to model temperature fairly well. In section 1.1
discussion 8 we looked at how temperature was a function of time. Here is the table we used.
Month
Jan.
0
Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov.
1
2
3
4
5
6
7
8
9
10
Dec.
11
Average
54
58
62 70 79
88
94
92
86
75
62
54
Temp.
Let’s use the information we have learned about translating sine and cosine to derive a function we
could use to model this yearly temperature change. Let’s be loose about the exactness of our model.
We need to find values for a, b, c, and d for the formula y = asin (bx − c) + d.
Question 3: What do you think the d in the formula might represent in a real life example?
First, we should find a number for d. d gives
the function its vertical translation. It would
represent the average between the high and
Given that the high is 94 and the low is 54
 94  54 
 = 74. Since
 2 
the average would be 
the low. The a, amplitude, would tell us how
d = 74, a would be 20, the amount up and
far the high and low are from the average.
down to the high and low.
d = 74, a = 20
Next, we need to find a value for b.
The period in this example is 12, it takes 12
Remember that b is involved with the
months before you begin to repeat.
period.
Therefore,
2

 12  2  12b  b 
b
6
Lastly, we need to find a value for c. c is
The x then is either month 3 or 4. They are
involved with the phase shift. We need to set
near the average value and the months
bx  c  0 to find phase shift. In this example
following them have higher temps. Thus we
we have found b, which is

and x would
6
need to be the month when it is the average
temperature and increasing because this is
the starting point for sine. The point at the
origin of the parent sine function.
need to solve this for c.

6
 3  c  0


2
c
Our model then is,


x   + 74
6
2

y = 20sin 
Here is a graph of the model along with the data points.
Holtfrerich & Haughn
83
Section 5.1 Geometry Review
As you can see our model does fit the data fairly well.
Section Summary:
If given a function y = asin (bx − c) + d or y = acos (bx − c) + d then the following are true.

a is the amplitude of the function (how the graph has been stretched or compressed
vertically). It tells you how far it is to the high and low points on the graph from the average
height (d).

2
= period of the function (how long it takes for the function to repeat itself).
b
c
is the phase shift (how much the graph has been shifted horizontally).
b

bx − c = 0  x =

d is the amount the graph has been shifted vertically (average height).

If a  0 , then the graph is rotated about the x-axis (flipped vertically).

If b  0 , then the graph is rotated horizontally.
Chapter 5 Trigonometric Functions
84
SECTION 5.6 PRACTICE SET
(1–46)
For each function give:
a.) The amplitude
b.) The period
c.) The phase shift
d.) The vertical change
e.) Flipped vertically or horizontally
f.) The domain of the function
g.) The range of the function
h.) Graph sketched
1. f ( )  sin   2
2. f ( )  sin   4
4. f ( )  sin   3
5. f ( )  sin   50


7. f ( )  sin   
6



8. f ( )  sin   
4


Answer Q3:
It is the
1
amount the
10. f ( )  sin 
graph gets
2
shifted up so
this would be
the average
13. f    sin  2 
temperature
over the entire
year. It’s the
middle
16. f    sin  3 
amount.
11. f ( ) 
3. f ( )  sin   1


6. f ( )  sin    50

9. f ( )  3sin 
1
sin 
2
12. f ( )  4sin 
 3 
15. f    sin   
 4 
2 
14. f    sin   
3 
3

17. f    2sin  3 
4

1 2


  1 18. f    sin      2
2 3
6

1

2 

 1
19. f    3sin      1 20. f   
sin  2    2 21. f    sin 2  120  3
3
4
3
3
 2


1

22. f    3sin    15   1 23. f    cos  4
4

24. f    cos  1
25. f    cos  3
27. f    cos   30

28. f    cos   150

26. f    cos  5


29. f    cos   60




30. f    cos   120

31. f    2cos
32. f    3cos
33. f    3cos
34. f    2cos
35. f    cos 3 
1 
36. f    cos   
2 
 1 
37. f    cos   
 3 
38. f    cos  2 


39. f    cos    
3



40. f    cos    
6

5 

41. f    cos   

6 

7 

42. f    cos  

6 

Holtfrerich & Haughn
85
Section 5.1 Geometry Review
3 
2 


43. f    2cos  3 
  1 44. f    3cos  2 
  4 45. f    2cos   45  3
4 
3 






46. f    3cos 3  180  2
(47–48)
47. The following is a table for average temperatures by month for a certain area of the country.
Month
Jan.
0
Feb.
1
March
2
April
3
May
4
June
5
July
6
Aug.
7
Sept.
8
Oct.
9
Nov.
10
Dec.
11
Average
Temp.
49
51
55
62
69
87
93
91
87
80
68
58
a. Find an equation of y = asin(bx – c) + d that fits this data.
b. Check the equation and the data by graphing the data points and the equation on the same graph
using the graphing calculator.
48. The following is a table for average temperatures by month for a certain area of the country.
Month
Jan.
0
Feb.
1
March
2
April
3
May
4
June
5
July
6
Aug.
7
Sept.
8
Oct.
9
Nov.
10
Dec.
11
Average
Temp.
54
57
63
74
83
92
96
94
90
81
67
59
a. Find an equation of y = asin(bx – c) + d that fits this data.
b. Check the equation and the data by graphing the data points and the equation on the same graph
using the graphing calculator.
Chapter 5 Trigonometric Functions
86
Chapter 5 Review
Topic
Intersecting
Section
5.1
lines
Key Points
If you have two parallel lines intersected by a transversal then the follow are
true:
Truth 1
Alternate interior angles have equal measure.
Truth 2
Corresponding angles have equal measure.
Truth 3
Alternate Exterior angles have equal measure
With any intersecting lines the following is true:
Truth 4
Triangles
Angles
Special
5.1
Vertical angles have equal measure
Two triangles will be similar when:
5.1
5.1
Right
Truth 5
Corresponding sides are proportional.
Truth 6
They have the equivalent angles.
Truth 7
The sum of complementary angles is 90°.
Truth 8
The sum of supplementary angles is 180°.
Truth 9
The sum of three angles of a triangle is 180°.
Truth 10
The ratio of the sides of a 30-60-90 right triangle are;
Truth 11
The ratio of the sides of a 45-45-90 right triangle are;
Triangles
60°
x
2x
45°
2 x
x
30°
3 x
Angles
5.1
Basics of
5.2
Truth 12

Trig.
45°
x
The acute angles of a right triangle are complementary.
Vertex  The common point where two rays begin that create an
angle.

Initial side  The ray where an angle begins.

Terminal side  The ray where an angle ends.

Positive angles  Angles that rotate in a counterclockwise manner.

Negative angles  Angles that rotate in a clockwise manner.

Minute  one sixtieth of a degree.

Second  One sixtieth of a minute.

Central Angle − An angle with its vertex at the center of a circle.

Arc of a Circle − An arc of a circle is a piece of the circumference of
a circle that is intersected by a central angle.

Radian − One radian is the measure of a central angle (θ) that
intercepts an arc (s) that is equal in length to the radius of the circle.
Holtfrerich & Haughn
5.2
87

Section 5.1 Geometry Review
A reference angle is an acute angle (measurement between 0° and
90°) between the terminal side of an angle in standard position and
the x-axis.

Conversion  180° is equivalent to π radians. When you need to
convert from one form to another use one of the following ratios:
180
or



180
.
Linear Speed  The linear speed of an object connected to a wheel
s
, where v is the velocity of the object, s is the arc length
t
will be v =
and t is the time.

Angular Speed − The angular speed of an object on a wheel will be ω
=

, where ω read “omega” is the angular speed, θ is the angle (in
t
radians) and t is the time.
Sine Cosine
5.3

Pythagorean Theorem states that the square of the two legs of a right
and the
triangle added
Unit Circle
together will equal the square of the hypotenuse (a2 + b2 = c2).

Sine is defined as the y value of the point on a unit circle on the
terminal side of angle θ (sinθ = y).

Cosine is defined as the x value of the point on a unit circle on the
terminal side of angle θ (cosθ = x).

The first quadrant of the unit circle and the sign of the outputs of sine
and cosine are:
 0,1

2
 90

3

1 3
 ,

2 2 
 2
2
60
,


2
2



 45
4
 3 1
, 


 30  2 2 


6
0  0
1, 0 
sin θ > 0
cos θ < 0
sin θ > 0
cos θ > 0
sin θ < 0 sin θ < 0
cos θ < 0 cos θ > 0

The domain for both sine and cosine is (−∞, ∞)

The range for both sine and cosine is [−1, 1]

A function is periodic if for some smallest c, f(x) = f(x + cn) where n
is any integer

Sine and cosine have a period of 2π.
Chapter 5 Trigonometric Functions
All Trig.
5.4
88
The graphs of the six trig functions are:
Functions
Domain and range of the six trig functions and their period.
Function
Domain
Range
Periodic
Sine
(−∞, ∞)
[−1, 1]
2π
Cosine
(−∞, ∞)
[−1, 1]
2π
{… 
   
,  …}
 2 2
(−∞, ∞)
π
{… (−π, 0) …}
(−∞, ∞)
π
{… 
   
,  …}
 2 2
(−∞, −1)  (1, ∞)
2π
{… (−π, 0) …}
(−∞, −1)  (1, ∞)
2π
Tangent
Cotangent
Secant
Cosecant
Reciprocal relationships cot θ =
1
1
1
, sec θ =
, csc θ =
.
tan 
cos 
sin 
The first definitions of the six trigonometric functions sometimes refered to as
the coordinate definitions are: sin θ =
sec θ =
y
x
y
r
, cos θ = , tan θ = , csc θ = ,
r
r
x
y
r
x
, cot θ = .
x
y
The second definitions which are the unit circle definitions are:
sin θ = y, cos θ = x, tan θ =
y
1
1
x
, csc θ = , sec θ = , cot θ = .
x
x
y
y
The even functions are cosine and secant [ f(−x) = f(x)]
The odd functions are sine, cosecant, tangent, and cotangent [ f(−x) = −f(x)]
Holtfrerich & Haughn
Inverse
5.5
89
Section 5.1 Geometry Review
Here are the domains, ranges and graphs of the six trigonometric inverses.
Trig.
Function
Domain
Range
Graph
y = sin -1 x
[−1, 1]
   
 2 , 2


y = cos -1 x
[−1, 1]
[0, π]
y = tan -1 x
(−∞, ∞)
   
 2 ,2


y = cot -1 x
(−∞, ∞)
 0,  
Functions
 

3 


3 
y = sec -1 x (−∞, −1]  [1, ∞) 0,    , 
 2  2 

y = csc -1 x (−∞, −1]  [1, ∞)  0,     , 
2 
 2 
Shifting
5.6
Remember, when working with a problem like trig -1 (trig θ), the trigonometric
inverse of a trigonometric function, that we must stay within the restricted
defined domains and ranges.
If given a function y = asin (bx − c) + d or y = acos (bx − c) + d then:

a is the amplitude of the function (how the graph has been stretched
or compressed vertically). It tells you how far it is to the high and low
points on the graph from the average height (d).

2
= period of the function (how long it takes for the function to
b
repeat itself).

bx − c = 0  x =
c
is the phase shift (how much the graph has been
b
shifted horizontally).

d is the amount the graph has been shifted vertically (average height).

If a  0 , then the graph is rotated about the x-axis (flipped vertically).
If b  0 , then the graph is rotated horizontally.
CHAPTER 5 REVIEW HOMEWORK
Section 5.1
1
2
(1−2)
1. m 1 = 53o; m 2 =
2. m 1 = 3x + 20; m 2 = 5x −10; m 1=
m 2 =
1
2
(3−4)
3. m 2 = 33o; m 1 =
4. m 1 = 3x − 10; m 2 = 7x −20; m 1=
m2 =
t
7
5
4
2
8
m
line l is parallel to line m, with a transversal of t
6
3
l
1
(5−6)
5. m1 = 1100 find the measure of each of the other angles.
6. m5 = 2x + 10 and m1 = 3x + 20 find the measure of all of the angles.
C
A
B
(7−8)
7. mA = 33o; mC = 62o mB =
D
e
8. mB = 50o; mA = 2x + 10 mC =
B
f
F
E
d
c
a
A
C
b
(9−10) Given triangle DEF is similar to triangle ABC answer the following questions:
9. e = 6; d = 10; f = 4; c = 12 a = b =
10. mE = 38o; mF =62o; mD =
mA =
mB =
mC =
Section 5.1 Geometry Review
30o
b
c
90o
60o
a
(11−13) Given the right triangle is a 30o−60o−90o right triangle answer the following questions:
11. a = 10 b = c =
12. b= 7 3 a = c =
13. c = 12 a = b =
45o
a
c
900
45o
b
(14−16) Given the right triangle is a 45o−45o−90o right triangle answer the following questions:
14. a = 9 b = c =
15. b = 15 a = c =
16. c = 5 2 a =
b=
(1718) Given 1 and 2 are complementary:
17. m1 = 50o m2 =
18. m1 = 3x  20; m2 = 5x  20
m1 =
m2 =
(1920) Given 1 and 2 are supplementary:
19. m1 = 133o m2 =
20. m1 = 5x + 10; m2 = 3x + 50
m1 =
m2 =
Section 5.2
(2124) For each angle in standard position the terminal side is in which quadrant?
7

radians
21. 78o
22.
23. 153o
24.  radians
8
3
(2528) Convert the following degree measurement of an angle to a radian measurement of that
angle.
25. 30o
26. 120o
27. 60o
28. 150o
(2933) Convert the following radian measurement of an angle to a degree measurement of that
angle.
7
3

4
radians
radians
radians
29.
30.
31.  radians
32. 
3
5
5
3
33. A bicycle wheel has a diameter of 24 inches and the wheel makes 440.5 revolutions every
minute.
a. What is the linear speed of the bicycle wheel?
b. What is the angular speed of the bicycle wheel?
Chapter 5 Trigonometric Functions
r

(3435)
S
34. r = 8 inches
 = 45o
a. area of the sector =
b. length of the arc =
35. r = 8 feet
=

3
a. area of the sector =
b. length of the arc =
radians
36. A water sprinkler sprays water over a distance of 20 feet while rotating through an angle of 90 o.
What is the area of the lawn watered?
Section 5.3
(3740)
a
b
Use the Pythagorean theorem to find the following values:
c
37. a = 12; c = 16
b=
38. a = 25; b = 65
39. c = 10; b = 12
a=
40. a = 8; c = 5
c=
b=
(4148) Without using the calculator find the value of sin  and cos  for each of the following.

2
sin 
sin

3
6
41.
42.
2

cos

cos 
3
6
43.
45.
47.
sin 210o 
cos 210o 
sin  
cos  
sin  45o 
cos  45o 
44.
sin 315o 
cos 315o 
sin
46.
cos
48.

2

2


sin  240o 
cos  240o 
Section 5.1 Geometry Review
(4952) For each angle in standard position find the value of sin  and cos .
r

49.
51.
52.
(x,y)

x  10
sin  
y  24
cos  
x  8
r  10 terminal side in the 3rd quadrant.
sin  
cos  
y  10
r  26 terminal side in the 4th quadrant.
sin  
50.
x  9
sin  
y  12
cos  
cos  
(5356) Give the angle values of  in radians between 2 and 2 that yield the following
answers.
53. sin   
1
2
54. sin  
2
2
55. cos  
2
2
56. cos   
3
2
(5760) Give the angle values of  in degrees between 360o and 360o that yield the following
answers.
57. sin  
2
2
58. sin   
3
2
59. cos   
1
2
60. cos  
3
2
(6164) Use the calculator to approximate to 3 decimal places the value of sin  and cos  for the
given value of .
11
11
61.  = 67o
62.  = 163o
63.  =
64.  = 
9
5
Section 5.4
(6570) Without using the calculator find the exact value of tan  , cot , sec  and csc 
for each of the following values of .
4

65.  = 60o
66.  = 150o
67.  =  radians
68.  =
radians
3
4
69.  = 180o
70.  =

2
(7176) Use the calculator to approximated to 3 decimal places the value of tan , cot , sec 
and csc  for each of the following values of .
7
11
radians
radians
71.  = 135o
72.  = 59o
73.  =
74.  = 
9
7
75.  = 583o
76.  = 733o
Chapter 5 Trigonometric Functions
(7780) Give the angle values of  in degrees between 360o and 360o that yield the following
answers.
77. tan 233o
78. cot 83o
79. sec 157o
80. csc 321o
(8184) Give the angle values of  in radians between 2 and 2 that yield the following
answers.
3
7
5
7
81. cot
82. tan
83. csc
84. sec
7
5
7
4
(8588) For each angle in standard position find the value of tan , cot , sec  and csc .
r

(x,y)

85. x = 6 and y = 8
86. x = 10 and y = 26
87. x = 5 and y = 4
88. x = 3 and y = 6
89. If sin  =
3
then csc  = ?
5
90. If tan  =
5
then cot  = ?
7
91. If sec  =
3
then cos  = ?
8
92. If cot  =
3
then tan  = ?
5
Section 5.5
(93104) Evaluate each expression exactly without using the calculator.
(Give your answers in degrees and radians)


 1 
3
3
93. sin 1 
94. cos 1  
95. arctan  


 2 
 2 
3





 1 
96. arccot 

 3
 3
97. sec1 

 2 
 1
99. arcsin   
 2
1
100. arccos  
2
102. cot 1  0 
103. arcsec  2 

3
98. csc 1  

 2 
101. tan 1  0 
104. arccsc  2
Section 5.1 Geometry Review
(105114) Evaluate each of the following without using a calculator.

3
3



105. arccos  cos 
106. ccos  arccos 
107. sin  sin 1 
5
5
4




3 

109. tan 1  tan

7 

5

111. cot  arccot 
9

112. arccot cot 58o

108. sin 1 sin48o

7

110. tan  tan 1 
3


113. arcsin sin x 
114. sin  arcsin x 
(115118) Evaluate each expression without using a calculator.
5
12 
1



115. sin  arccot
116. cos  arctan
117. tan  sin 1 


12
5
4






3
118. cot  cos 1


3 

(119128) Use the calculator to evaluate each expression if possible.
 2
3
 3
119. sin 1  
120. cos 1   
121. arccos   
 3
4
 4
3
122. arcsin  
7
9
123. tan 1  
5
9
124. sin 1  
5
 9
125. arccot   
 5
 17 
126. arcsec  
 12 
 8
127. csc1   
 5
3
128. sec1  
5
Section 5.6
(129146) For each function give:
a. The amplitude
b. The period
c. The phase shift
d. The vertical change
e. Flipped vertically or horizontally
f. The domain of the function
g. The range of the function
h. Sketch a graph
129. f ( )  cos  3
130. f ( )  sin   5
5 

132. f ( )  cos  

6 

133. f ( )  cos   60o
135. f ( )  sin  3 
136. f ( )  cos  2 



131. f ( )  sin    
3



134. f ( )  sin   120o
137. f ( )  cos  2 

Chapter 5 Trigonometric Functions
138. f ( )  sin  3 
1
139. f ( )   sin  
2
140. f ( )  2cos  
141. f ( )  3cos  
1
142. f ( )  sin  
3


143. f ( )  3sin  2    1
6

3

144. f ( )  2cos  3 
4



2


146. f ( )  2sin 3  210o  3


1
145. f ( )   cos 2  45o  1
2
Section 5.1 Geometry Review
CHAPTER 5 EXAM
(1−2) Give the reference angle for each angle.
1. 232o
2.
3
radians
7
3. Convert 305 o to radians.
4. Convert
5
radians to degrees.
9
5. A Ferris wheel has a radius of 40 feet and takes 90 seconds for one revolution.
a. What is the linear speed of the Ferris wheel?
b. What is the angular speed of the Ferris wheel?
6. A water sprinkler sprays water a distance of 20 feet while rotating through an angle of 150 o.
What area of the lawn is watered?
(7−10) Find the value of sin θ, cos θ, tan θ, cot θ, sec θ and csc θ for each of the following angles.
8.  
7.   150o
5
radians
3
9.   120o

10.    radians
4
(11−14) For each angle in standard position find the value of sin θ, cos θ, tan θ, cot θ, csc θ and
csc θ.
(x,y)
r


11. x = 6
y = −8
13. x = 9
r = 15, terminal side in the first quadrant
14. y = 24
12. x = −15
y = −36
r = 26, terminal side in the second quadrant
(15−20) Give the angle values in radians for θ between 2 and 2 that yield the following
answer.
1
2
15. cos  
3
2
16. sin  
18. cot   
1
3
19. sec  2
21. For sin θ
a. What is the domain?
b. What is the range?
17. tan  3
20. csc  2
Chapter 5 Trigonometric Functions
22. For cos θ
a. What is the domain?
b. What is the range?
23. For tan θ
a. What is the domain?
b. What is the range?
24. For cot θ
a. What is the domain?
b. What is the range?
25. For sec θ
a. What is the domain?
b. What is the range?
26. For csc θ
a. What is the domain?
b. What is the range?
(27−32) Evaluate without a calculator

27. arcsin  


30. arccot  

 3
1 

2
28. cos 1  
2
29. tan 1
1 

3
31. arcsec  2 
32. sec 1  
1
 


2 

3
(33−38) Evaluate to 3 decimal places with a calculator

33. sin 133o



 3 
35. tan  
 7 


 11 
38. csc 

 5 
34. cos 35o
 5 
36. cot   
 11 
37. sec 239o
(39−44) Evaluate in radians to 3 decimal places with a calculator
39. sin 1  0.39
40. arccos  0.932 
41. arctan  3.214 
42. cot 1  2.532 
43. arcsec 1.932 
(45−46) For each function give:
a. The amplitude
b. The period
c. The phase shift
d. The vertical change
e. Flipped vertically or horizontally
f. The domain of the function
g. The range of the function
h. Sketch a graph


45. f    3sin  2    4

3
44. csc1  2.359 


46. f    2cos  3    3
2

