Download Trigonometric Equations and Identities Unit Part 1 (Draft for

Trigonometric Equations and
Identities Unit, Part 1
(Level IV Pre-Calculus)
NSSAL
(Draft)
C. David Pilmer
2012
(Last Updated: March, 2013)
This resource is the intellectual property of the Adult Education Division of the Nova Scotia
Department of Labour and Advanced Education.
The following are permitted to use and reproduce this resource for classroom purposes.
 Nova Scotia instructors delivering the Nova Scotia Adult Learning Program
 Canadian public school teachers delivering public school curriculum
 Canadian non-profit tuition-free adult basic education programs
 Nova Scotia Community College instructors delivering the Academic Careers and
Connections mathematics curriculum
The following are not permitted to use or reproduce this resource without the written
authorization of the Adult Education Division of the Nova Scotia Department of Labour and
Advanced Education.
 Upgrading programs at post-secondary institutions
 Core programs at post-secondary institutions
 Public or private schools outside of Canada
 Basic adult education programs outside of Canada
Individuals, not including teachers or instructors, are permitted to use this resource for their own
learning. They are not permitted to make multiple copies of the resource for distribution. Nor
are they permitted to use this resource under the direction of a teacher or instructor at a learning
institution.
Acknowledgments
The Adult Education Division would like to thank the following university professors for
reviewing this resource to ensure all mathematical concepts were presented correctly and in a
manner that supported our learners.
Dr. Robert Dawson (Saint Mary’s University)
Dr. Genevieve Boulet (Mount Saint Vincent University)
Dr. Jeff Hooper (Acadia University)
The Adult Education Division would also like to thank the following NSCC instructors for
piloting this resource and offering suggestions during its development.
John Archibald (Truro Campus)
Elliott Churchill (Waterfront Campus)
Carissa Dulong (Truro Campus)
Barbara Gillis (Burridge Campus)
Alice Veenema (Kingstec Campus)
Table of Contents
Introduction ………………………………………………………………………………..
Negotiated Completion Date ………………………………………………………………
The Big Picture and Suggested Timelines ………………………………………………….
Pre-Calculus Relations Concept Map .……………………………………………………
Common Errors Seen in First Year University Math Classes …………………………….
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iv
v
vi
vii
Introductory Activity …………………………..………………………………………….
The Unit Circle, Part 1 ……………………………………………………………………
The Unit Circle, Part 2 ……………………………………………………………………
Special Rotations ………………………………………………………………………...…
Evaluating Trigonometric Expressions……………………………………………..……...
Introduction to Trig Equations: The Graphical Approach, Part 1……………...…………...
Introduction to Trig Equations: The Graphical Approach, Part 2 ………………………….
Trigonometric Equations, Part 1 …………………………………….……………………...
Trigonometric Equations, Part 2 …………………………………….……………………...
Trigonometric Equations, Part 3 …………………………………….……………………...
Trigonometric Equations, Part 4 …………………………………….……………………...
Introduction to Radian Measure …………………………………………..………………..
Revisiting Questions Using Radian Measure, Part 1 …………………..…………………...
Revisiting Questions Using Radian Measure, Part 2 ………………..……………………...
Trigonometric Identities; The Investigation, Part 1 ……………………..…………………
Trigonometric Identities; The Investigation, Part 2 ……………………………..…………
Trigonometric Identities and Verification Questions …………………...………………….
Trigonometric Identities and Trigonometric Equations …………………………………….
1
3
9
14
24
36
44
47
52
58
65
74
84
94
98
104
109
123
Refresher Reference Materials …………………………………………………………….
The Unit Circle and Special Rotations (In Degrees and Radians) ………………………...
List of Trigonometric Identities …………………….……………………………………..
Index ……………………………………………………………………………………….
Answers ……………………………………………………………………………………
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143
144
145
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Introduction
Level IV Pre-Calculus
Pre-Calculus is strictly designed for learners who are intending on enrolling in science or
engineering programs at the university level. The prerequisite for this course is ALP Level IV
Academic Math with a minimum mark of 85%, PSP Advanced Math 11 with a minimum mark
of 80%, or PSP Pre-Calculus Math 11 with a minimum mark of 80%. We also recommend that
the learner have a strong work ethic and the ability to work independently.
The delivery of the Pre-Calculus curriculum will likely vary from campus to campus. Many
campuses will not have the resources to offer this course on their schedule. In these cases, the
learner will have to take it as a correspondence course or as an online course. Larger campuses
may be able to offer the course, but it is likely that the learner will be placed in a class that is
predominately populated with Level IV Academic Math learners. There is a possibility that
Academic Careers and Connections (ACC) will use the ALP Pre-Calculus curriculum within
their own program. If this is the case, ACC and ALP learners may be found in the same
classroom. Ultimately, the delivery model used with Pre-Calculus will be at the discretion of the
Nova Scotia Community College and the Academic Chairs at the various campuses.
There is a strong likelihood that you may be the only Pre-Calculus learner in the class or at your
campus. For this reason, it may be a challenge to obtain assistance for extended periods of time.
This being said, we have made every effort to make the material found in this resource as
accessible as possible. One of the major differences between the Pre-Calculus resources and the
Academic Math resources can be seen in the answer key. In the Academic Math resources, the
answer keys only contained the final answers. The answer key in this resource and other PreCalculus resources is far more extensive. For many of the questions we have also provided hints,
and with the more challenging questions, we have provided complete solutions. Use this feature
judiciously; do not refer to the answer key before making a concerted effort to answer the
question on your own. Neither do we want learners to seek assistance from an instructor without
first attempting to use the answer key to resolve their problem. In the near future, we hope to
create online instructional videos to support our Pre-Calculus learners. Ultimately, learners with
a solid understanding of Academic Math concepts, a great work ethic, and the ability to work
independently, will flourish in this course. Although some of the topics can be challenging,
remember that you can do it.
Most of you will have taken ALP Level IV Academic Mathematics prior to taking this course.
In that academic course, there was a heavy emphasis on real-world application questions, and the
use of the graphing calculator. In Pre-Calculus, one will encounter application questions and
have to use graphing calculators, but to a far lesser degree. There is a greater emphasis on the
pure mathematics and the automaticity to recall relevant mathematical facts and concepts. This
parallels the expectations that one will encounter in a first year university calculus course.
Regardless of this, the material in this resource is presented in a manner to foster understanding
(opposed to encouraging memorization).
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The Trigonometric Equations and Identities Unit
Welcome to the Trigonometric Equations and Identities Unit Part 1 and 2, the longest units in
Pre-calculus. In these units, we learn about the unit circle and its associated special rotations.
We use this knowledge to solve a variety of trigonometric equations involving sines or cosines.
After completing this, we learn how to simplify trigonometric expressions using trigonometric
identities. This allows us to tackle more complex trigonometric equations, including those that
involve tangents, secants, cosecants, and cotangents. We also learn an alternative unit of
measure for rotations. Up to this point, we have only measured rotations (angles) in degrees, but
they can also be measured in radians, a concept that you learn about in this unit.
There are several concepts from Level IV Academic Math that learners may need to review as
they work through this unit. They are the following.
1. Determining the equation of a sinusoidal function; particularly application questions
2. Simplify radicals by rationalizing the denominator
3. Solving quadratic equations by factoring
4. Naming the quadrants of the Cartesian coordinate system
5. Properties associated with triangles (including isosceles and equilateral triangles).
Although refresher reference materials are provided in the appendix; do not hesitate to go back to
your academic resources if more extensive explanations are required.
The Trigonometric Equations and Identities Unit Recommended Test Schedule
There are four recommended test periods for these two units. They should follow the completion
of the following sections.
1. "Introductory Activity" through to and including "Trigonometric Equations, Part 4."
2. "Introduction to Radian Measure" through to and including "Trigonometric Identities and
Trigonometric Equations."
3. "Trigonometric Equations Involving Tangents." to "Principal Solutions, Part 2"
4. "More Trigonometric Identities, Part 1" through to and including "University Diagnostic
Tests: Trigonometry."
Please note that you are not permitted to bring the unit circle containing the coordinates of the
image points for the special rotations to any of these tests. You are, however, permitted to bring
the trigonometric identities to test 4.
The NSSAL Pre-Calculus Print Resources
If you are not affiliated with the Nova Scotia School for Adult Learning (NSSAL) and you are
choosing to use this resource, you may be wondering why some of the early topics that appear in
our Pre-Calculus resources seem to be more introductory in nature compared to topics found in
traditional Pre-Calculus programs and textbooks. The reason stems from the fact that the
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prerequisite within the ALP system for our Pre-Calculus course is Level IV Academic Math.
Traditional Pre-Calculus courses usually have an advanced level mathematics course as their
prerequisite. NSSAL does not have the resources or the number of learners needed to offer an
advanced math prerequisite. For this reason our Pre-Calculus includes introductory material that
is typically found in an advanced math prerequisite. For example in this unit, we spend a
significant amount of time learning about the unit circle, and radian measure; topics typically
covered in an advanced prerequisite. Although this is the case in this unit and other units, be
assured that we do eventually address all required Pre-Calculus concepts. To guarantee this, we
have had our curriculum and resources reviewed by multiple specialists, including university
professors in both mathematics and education.
Negotiated Completion Date
After working for a few days on this unit, sit down with your instructor and negotiate a
completion date for this unit.
Start Date:
_________________
Completion Date:
_________________
Instructor Signature: __________________________
Student Signature:
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The Big Picture and Suggested Timelines
The following flow chart shows the six required units for the 100 hour Level IV Pre-Calculus
course. These have been presented in a suggested order.
Sequences and Series (15 hours)
 Functional Form, Recursive Form, Arithmetic Sequences and
Series, Geometric Sequences and Series
Trigonometric Equations and Identities Unit, Part 1 (17 hours)
 Unit Circle, Special Rotations, Evaluating Trigonometric
Expressions, Trigonometric Equations, Radian Measure,
Trigonometric Identities
Trigonometric Equations and Identities Unit, Part 2 (14 hours)
 Trigonometric Equations Involving Tangents, Principal
Solutions, More Trigonometric Identities
Exponential Functions and Natural Logarithms Unit (7 hours)

Polynomial Functions Unit (13 hours)

Rational Functions Unit (12 hours)

Irrational Functions Unit (12 hours)

Complex Numbers Unit (10 hours)

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The Pre-Calculus Relations Concept Map
Relations
A relation shows the relationship
between two sets of numbers
Conics
Circles, ellipses, hyperbolas,
and parabolas
Function
A relation is a function when for every
member of the first set (typically x's),
there is only one corresponding member
in the second set (typically y's).
Algebraic Functions
These functions can be described
using algebraic operations (addition,
subtraction, multiplication, division,
exponentiation).
Rational Functions
The variable has an
integral constant
exponent.
e.g. y  x 1 or y 
Irrational Functions
The variable has a
non-integral constant
exponent.
1
x
Polynomial Functions
The variable has a
positive integral
constant exponent.
e.g. y  x3
Trigonometric
Functions
e.g. y  sin x , y  cot x
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Transcendental Functions
These functions cannot be described
using algebraic operations.
1
e.g. y  x 2 or y  x
Absolute Value
Functions
e.g. y  x or y  x 2
Logarithmic Functions
e.g. y  log x , y  ln x
vi
Exponential Functions
e.g. y  2 x , y  e x
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Common Errors Seen in First Year University Math Classes
Do not walk into a first year university mathematics classroom without being aware of the
following common errors. At this point in your Pre-Calculus course, you may not have
encountered all of the concepts associated with the errors listed below. However, by the end of
the course, you should be familiar with all of the concepts.

 x  y
2
is not equal to x 2  y 2 , rather it is equal to x 2  2 xy  y 2 .

 x  y
3
is not equal to x3  y 3 , rather it is equal to x3  3x2 y  3xy 2  y3 .

1
1 1
is not equal to  .
x y
x y

x 2  y 2 is not equal to x  y .
 The expression

1
y
cannot be simplified to
(i.e. We cannot cancel out the y's.)
x 1
x y
2
is not equal to 0, rather it is undefined.
0
 We know that sin 2 x  cos2 x  1 , but that does not mean that sin x  cos x  1 .
 cos  x  y  is not equal to cos x  cos y , rather it is equal to cos x cos y  sin x sin y .
 sin 2x is not equal to 2sin x , rather it equal to 2sin x cos x .
 cos1 x is not equal to
1
.
cos x
 b 0 is not equal to 0, it is equal to 1 (as long as b is not equal to 0).
 The solutions to the equations x  9 and x 2  9 are different. The solution to the
equation x  9 is 3. The solution to the equation x 2  9 is 3 or -3.
 52 and  5  are different expressions; 52  25 while  5  25 .
2
2
 log a  b  c  is not equal to log a b  log a c , rather the appropriate law of logarithms is
log a  bc   log a b  log a c .
If by the end of your Pre-Calculus course, you are still unsure why some of these common errors
are incorrect, sit down with your instructor and go over them. Understanding these errors is far
more important than merely memorizing them.
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Introductory Activity
In Level IV Academic Math, there were two units on trigonometry.
One unit focused on:
opp
adj
opp
, cos  
, tan  
hyp
adj
hyp
a
b
c
2. The Law of Sines:


sin A sin B sin C
3. The Law of Cosines: a2  b2  c2  2bc cos A
1. The Trigonometric Ratios: sin  
The other unit examined a particular type of trigonometric function called sinusoidal functions
(e.g. y  sin x and y  cos x ).
In this introductory activity, we revisit concepts from both of these units.
Part 1
Consider the right-angle triangle on the right. We are provided with the
length of one of the legs, and the length of the hypotenuse. Use your
knowledge of the trigonometric ratios to find angle  . Show your
work below. (Hint: You will need your calculator to complete the
question; make sure it is in "Degree" mode.)
2
1

Part 2
Now let's consider a similar problem that deals with sinusoidal functions. Suppose you are given
the sinusoidal function y  2sin x and asked to find x when y is equal to 1. Solve this problem
to the best of your ability without actually considering the graph of the function y  2sin x .
Show your work below. (Hint: You will need your calculator to complete the question.)
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Hopefully when you answered the questions in Part 1 and Part 2, you obtained the same answer
of 30o. Now this answer is correct for Part 1, but only partially correct for Part 2. Explain why
this is so. (Hint: Look at the graphs that have been provided below.)
Your Explanation:
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The Unit Circle, Part 1
In the previous section, we realized that we lack the skills and knowledge needed to solve
problems for which we are given a trigonometric function and asked to determine the
independent variable (typically x-values), given the dependent variable (typically y-values). In
that section, we were given the sinusoidal function y  2sin x and asked to solve for x when y
is equal to 1. We found one value for x, 30o, but realized that there were many more answers
that we missed.
This is a serious problem when one considers how many real world applications are modeled
using trigonometric functions. For example, in Level IV Academic Math, we modeled a variety
of phenomena (e.g. objects swinging, equipment vibrating, lunar cycle,…) using sinusoidal
functions, a specific type of trigonometric function. Presently we can take these functions and
solve for the dependent variable given the independent variable, but the reverse is not true.
Consider the following. The moon is always half illuminated by the sun. The portion of the
moon that we see at a particular location on the earth depends on where the moon is in its orbit
around the earth. At some time during a month, we will see a full moon. Fifteen days later at the
same location, we will see not see the moon at all (new moon).
Full
Moon
New
Moon
Full
Moon
The equation p  0.5cos 12  t  15    0.5 describes the portion, p, of the moon visible to us at
a specific location with respect to the time, t, measured in days.

If you are asked to find the portion of the moon visible on day 3, then this question is
doable given our existing skill set.
p  0.5cos 12  t  15     0.5
p  0.5cos 12  3  15     0.5
p  0.5cos  144    0.5
p  0.5  0.809   0.5
p  0.4045  0.5
p  0.096
We are given the independent variable, t, and asked to solve for the dependent variable,
p. In doing so, we discover that 9.6% of the moon is visible on day 3.

If, however, we are asked to determine the days on which 80% of the moon is visible,
then we get into trouble. For this question we know the dependent variable (p = 0.80)
and must determine the independent variable, t. Based on this situation, we understand
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that there are multiple days when 80% of the moon is visible, yet when we try to solve it,
we only get one answer.
p  0.5cos 12  t  15     0.5
0.8  0.5cos 12  t  15     0.5
0.3  0.5cos 12  t  15   
0.6  cos 12  t  15  
We will now use the cos-1 button.
4.43  t  15
t  19.43
Only one answer (Problem!)
53.1  12  t  15 
We have to learn several concepts before we can tackle these types of questions, and our starting
point is the unit circle.
The unit circle is a circle with a radius of 1 unit, centered at (0, 0). All rotations start from the
point (1, 0) located on the circumference of the circle.
(0, 0)
(1, 0)
For any rotation, there is an initial arm, a terminal arm and a vertex.
When rotating around a unit circle, the initial arm is always along the xaxis between the points (0, 0) and (1, 0), and the vertex is always at the
origin, (0, 0).
Terminal Arm
Vertex
Initial Arm
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Positive rotations are in a counterclockwise direction, and negative rotations are in a clockwise
direction.
In this case we rotated 45o from the point
(1, 0) about the point (0, 0). Since it was a
positive rotation, we moved in a
counterclockwise direction. The terminal arm
for a rotation of 45o is in the first quadrant of
the coordinate system.
In this case we rotated -120o from the point
(1, 0) about the point (0, 0). Since it was a
negative rotation, we moved in a clockwise
direction. The terminal arm for a rotation of 120o is in the third quadrant of the coordinate
system.
45o
-120o
Coterminal angles are rotations that share the same terminal arm. For example, the rotations of
150o and -210o are coterminal angles because they share the same terminal arm found in the
second quadrant.
150o
-210o
There are other rotations that are coterminal to 150o and -210o. These include 510o and -570o.
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If we list these coterminal angles as a sequence, we would have the following.
… ,-570o, -210o, 150o, 510o, …
Briefly explain the pattern we have above. Why does this pattern make sense?
Now fill in the blanks in the sequence below (i.e. identify addition coterminal angles).
… , _______, -570o, -210o, 150o, 510o, _______, …
Questions
1. For each of the following rotations,
 sketch the rotation on the coordinate system provided
 indicate in which quadrant the terminal arm is located
 state the largest negative rotation that is coterminal to the original rotation.
(a) 60o
(b) 120o
Quadrant:
Quadrant:
Coterminal Angle:
Coterminal Angle:
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(c) 330o
(d) 225o
Quadrant:
Quadrant:
Coterminal Angle:
Coterminal Angle:
2. For each of the following rotations,
 sketch the rotation on the coordinate system provided
 indicate in which quadrant the terminal arm is located
 state the smallest positive rotation that is coterminal to the original rotation.
(a) -45o
(b) -320o
Quadrant:
Quadrant:
Coterminal Angle:
Coterminal Angle:
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(c) -200o
(d) -120o
Quadrant:
Quadrant:
Coterminal Angle:
Coterminal Angle:
3. Complete each of the sequences knowing that each sequence consists of coterminal angles.
(a) … , _________, 90o, 450o, 810o, _________, …
(b) … , _________, _________, 120o, _________, 840o, _________, …
(c) … , _________, -30o, _________, _________, _________, …
(d) … , _________, _________, _________, -225o, _________, …
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The Unit Circle, Part 2
In the last section we learned about the unit
circle, terminal arms, positive rotations,
negative rotations, and coterminal angles. In
this section we learn about the coordinates of
an image point after we rotate of  degrees
about (0, 0) from the point (1, 0). For
example, if we rotated 45o on the unit circle
(as illustrated on the right), what would be the
coordinates of the point on the circumference
of the circle where the terminal arm is
located?
45o
(x, y)
The answer to this is rooted in our knowledge
of the trigonometric ratios, specifically
opp
adj
and cos  
.
sin  
hyp
hyp
In the diagram below, we rotated  degrees about (0, 0) from the point (1, 0). We constructed a
right angle triangle, where one leg represents the x-coordinate of the image point, and the other
leg represents the y-coordinate of the image point. Since we are dealing with a unit circle, the
radius of our circle (which also represents the hypotenuse of the triangle) is 1.
We now have enough information to solve
for x and y (i.e. the coordinates of the
image point) in terms of the rotation,  .
(x, y)
adj
hyp
x
cos  
1
cos   x
cos  
1
y

x
opp
hyp
y
sin  
1
sin   y
sin  
From this we can state that the coordinates
of the image after a rotation of  degrees
about (0, 0) from the point (1, 0) is
 cos ,sin   .
This can be written as the mapping rule R 1,0    cos  ,sin   where R stands for rotation.
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Example 1
Determine the coordinates of the image point after each of these rotations about (0, 0) from
(1, 0). State your answer using the mapping rule notation. Indicate the quadrant of the
coordinate system in which the image point is. Following this, sketch the rotation and label the
image point.
(a) 110o
(b) -38o
Answers:
To answer these questions, we need a calculator. Make sure it is in "Degree" mode.
(a)
R110 1,0    cos110,sin110
(b)
R110 1,0    0.342,0.940 
Second Quadrant
R38 1,0    cos  38  ,sin  38  
R38 1,0    0.788, 0.616 
Fourth Quadrant
(-0.342, 0.940)
110o
-38o
(0.788, -0.616)
Example 2
Without using your calculator, determine sin 90 and explain how you arrived at that answer.
Answer:
If we rotate 90o on our unit circle, we will move from the point (1, 0) to the image point
(0, 1). The coordinates of this image point would normally be found using  cos90,sin 90 .
Therefore we can conclude that cos90  0 and sin 90  1 . Since the question asked us to
find sin 90 , we can logically conclude that it is equal to 1.
Example 3
In what quadrants of the unit circle are cosines positive?
Answer:
Cosines of rotations represent the x-coordinates of the image points. The x-coordinates, or
cosines, are positive in the first and fourth quadrants.
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Questions
1. Determine the coordinates of the image point after each of these rotations about (0, 0) from
(1, 0). State your answer using the mapping rule notation. Also indicate the quadrant of the
coordinate system in which the image point is.
(a) 25o
(b) -107o
(c) 310o
(d) -234o
(e) 400o
(f) -395o
2. On the provided coordinate system, sketch the rotation and label the coordinates of the image
point of the unit circle.
(a) -280o
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(c) -65o
(d) 250o
3. Without using your calculator, determine cos180 and explain how you arrived at that
answer.
4. (a) In what quadrants of the unit circle are sines negative?
______________________
(b) In what quadrants of the unit circle are sines positive?
______________________
(c) In what quadrants of the unit circle are cosines negative?
______________________
5. (a) What is the largest possible value one can obtain for cos  ?
_____
(b) What is the smallest possible value one can obtain for cos  ?
_____
(c) What is the largest possible value one can obtain for sin  ?
_____
(d) What is the smallest possible value one can obtain for sin  ?
_____
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6. What relationship would you expect among image points of coterminal angles? Support your
answer using three coterminal angles of your choosing.
7. Refer to the diagram of a unit circle on the right. Which coordinate could represent the value
of each of the following?
(a) cos130
(b) sin 70
(C, D)
(E, F)
(c) cos 250
(d) sin 0
(e) sin 290
(f) cos  290 
(A, B)
(g) sin  230 
(h) cos  70 
(i) cos  360 
(j) sin  110 
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Special Rotations
Up to this point, if we want to find the coordinate of an image point after a rotation about (0, 0)
from (1, 0), we have to use the cosine and sine buttons on our calculator. How did people
answer such questions before calculators or tables that listed the sines and cosines of various
rotations? We can start to answer this question by looking at the special rotations; these rotations
are at increments of 30o and 45o.
30o increments: …, -120o, -90o, -60o, -30o, 0o, 30o, 60o, 90o, 120o, 150o, 180o, …
45o increments: …, -180o, -135o, -90o, -45o, 0o, 45o, 90o, 135o, 180o, 225o, 270o, …
These rotations are special because the coordinates of their image points can be found without a
calculator (or trigonometric tables). Let's see how this is accomplished by conducting the
following investigation.
Guided Investigation
Part 1: Image Point after a 45o Rotation
On the diagram below we rotate 45o about (0, 0) from (1, 0), and we want to find the coordinate
of the image point (x, y). We create a right-angle triangle that has a hypotenuse of 1, and whose
legs are x and y.
(x, y)
1
y
o
45
(0, 0)
x
(1, 0)
Part 1 Investigation Questions:
(a) What do interior angles of any triangle total?
________
(b) We already know the measure of two of the interior angles of the triangle. Based on this,
determine the measure of the third interior angle.
________
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(c) From what we learned in (b), we know that this triangle is more than simply a right-angle
triangle. Which one of these terms could also be used to describe the triangle?
(i) scalene
(ii) isosceles
(iii) equilateral
(d) Based on what we learned in (c), which one of these statements is correct?
(i) x  y  1
(ii) y  1
(iii) x  y
(e) The Pythagorean Theorem, a 2  b2  c 2 describes the relationship between the three sides of
a right-angle triangle. Using this theorem and what you learned in (d), we can now solve for
x or y. Please do so showing all your work. Do not use decimals. (Hint: At some point we
will have to rationalize the denominator.)
(f) In (e) you solved for x or y. What is the other coordinate?
________
(g) Which one of the following is the appropriate mapping rule for a rotation of 45o about a unit
circle?
 2 2
1 3
1 1
,
(i) R45 1, 0   
(iii) R45 1, 0    , 
 (ii) R45 1, 0    ,

2 2
 2 2 
2 2 
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(h) Below we draw other special rotations that are increments of 45o. We also include the rightangle triangle which is used to describe the coordinates of the image point. This triangle is
identical to the triangle that we created for the 45o rotation, but is located in a different
quadrant of the coordinate system. In each case, complete the mapping rule. No calculations
need to be done to answer these questions; you are merely using what you learned in (g).
(i)
135o
(cos135o, sin135o)
45o
(1, 0)

R135 1, 0   

(ii)
,



225o
45o
(1, 0)

R225 1, 0   

(cos225o, sin225o)
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(iii)
315o
45o,
(1, 0)
(cos315o, sin315o)

R315 1, 0   

,



(i) Remembering that coterminal angles share a terminal arm and logically an image point,
complete the following mapping rules using the exact coordinates (i.e. no decimals).
(i)

R135 1, 0   

(ii)

R45 1, 0   


(iii) R405 1, 0   


(iv) R225 1, 0   

,



,



,



,



Please Note:
You may have noticed that we did not consider all the increments of 45o. For example, we did
not consider 90o, 180o, or 270o. The reason is that the terminal arms for these rotations lie either
on the x-axis or the y-axis, making it easy to identify the coordinates of the image point.
R90 1, 0    0,1
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R180 1, 0    1, 0 
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R270 1, 0    0, 1
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Part 2: Image Point after a 30o Rotation
In Diagram A (below and to the left), we rotate 30o about (0, 0) from (1, 0), and we want to find
the coordinate of the image point (x, y). We create a right-angle triangle that has a hypotenuse is
1, and whose legs are x and y. In Diagram B (below and to the right) we redraw what we have
on the left but now we reflect the right-angle triangle in the x-axis creating the much larger
ABC .
Diagram A
Diagram B
A
(x, y)
1
30o
1
y
30o
x
B
D
C
Part 2 Investigation Questions:
(a) What do interior angles of any triangle total?
________
(b) We already know the measure of two of the interior angles of the right-angle triangle in
Diagram A. Based on this, determine the measure of the third interior angle.
________
(c) In Diagram B, we reflected on original right-angle triangle in the x-axis.
(i) Based on this, what is the measure of DBC ?
________
(ii) Based on this, what is the measure of DCB ?
________
(d) What is the measure of ABC ?
________
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(e) From what we learned in (c) and (d), which one of these terms could also be used to describe
ABC ?
(i) scalene
(ii) isosceles
(iii) equilateral
(f) Based on what we learned in (e), how long is line segment AC?
2
2
(i) 1 unit
(ii)
units
(iii) units
3
2
(g) How long is line segment AD, which happens to be equivalent to the y-coordinate of the
image point?
________
(h) Now that we have one of the coordinates of the image point, we can go back to the original
right-angle triangle in Diagram A and use the Pythagorean Theorem to solve for the other
coordinate. Please do so showing all your work. Do not use decimals.
(i) Which one of the following is the appropriate mapping rule for a rotation of 45o about a unit
circle?
 3 1
1 3
1 1
, 
(i) R30 1, 0    , 
(ii) R30 1, 0   
(iii) R30 1, 0    ,

2 2
 2 2
2 2 
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(j) Below we show other special rotations that are increments of 30o. We also include the rightangle triangle used to describe the coordinates of the image point. All the triangles are 30o60o-90o triangles with a hypotenuse of 1; these are oriented differently and, in some cases,
are located in different quadrants of our coordinate system. In each case, complete the
mapping rule. No calculations need to be done to answer these questions; you merely use
what you learned in (i).
(i)
60o
(cos60o, sin60o)
30o
60o

R60 1, 0   

(ii)
,



,



120o
(cos120o, sin120o)
30o
60o

R120 1, 0   

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(iii)
150o
(cos150o, sin150o)
60o
30o

R150 1, 0   

,



(k) Remembering that coterminal angles share a terminal arm and logically an image point,
complete the following mapping rules using the exact coordinates (i.e. no decimals).
(i)

R300 1, 0   

(ii)

R210 1, 0   


(iii) R330 1, 0   


(iv) R480 1, 0   

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,



,



,






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Questions
1. The exact coordinates of the image points for the specials rotations have not been filled in on
the unit circles found on this page and the next. Fill in these missing coordinates. No
calculations need to be completed; we merely use the information we discovered in the
previous two investigations. Once this is completed, make sure you check your answers. We
use these circles to answer many of the questions in the remaining sections of the unit.
(a) 45o Increments
 cos 90, sin 90
 ,

 cos135,



,
sin135 
 cos 45,






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


,
 cos180, sin180
 ,

sin 45 
 cos 0,
 ,
sin 0 

 cos 225,
sin 225 
 cos 315,
sin 315 












,
 cos 270, sin 270
 ,

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(a) 60o Increments
 cos120,



 cos150,



,
sin120 
 cos 90, sin 90
 ,

 cos 60,






,
sin150 



,



,



,
 cos 0,
 ,
sin 210 
 cos 330,



 cos 240, sin 240 






 cos 30, sin 30 



 cos180, sin180
 ,

 cos 210,
sin 60 
,



sin 0 

sin 330 


,




 cos 300, sin 300 

 cos 270, sin 270 
 ,

,



Important Note:
You will be expected to know the coordinates of the image points for the special rotations for
this course and for future university mathematics courses. Over the next few days it is
imperative that you learn these values.
2. Without using a calculator, evaluate each of the following. No work needs to be shown
because we merely have to use the unit circles we completed in question 1.
(a) cos135 
(b) sin150 
(c) sin 300 
(d) cos 270 
(e) sin 315 
(f) cos(30) 
(g) cos(120) 
(h) sin 405 
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Evaluating Trigonometric Expressions
In this section we learn how to evaluate trigonometric expressions and how to express them in
their most simplified form (if appropriate). All of these questions have been "cooked" as to force
you to use your knowledge of the unit circle and the special rotations to evaluate the expressions.
You are expected to develop some automaticity in recalling the sines and cosines of the special
rotations. For this reason, please minimize your use of the unit circles you completed in the
previous section (question 1); instead you are asked become so familiar with these circles that
you can work from memory.
It should also be mentioned that you should also be comfortable working with radicals, a topic
that was addressed in Level IV Academic Math. In particular, you need to know the process of
rationalizing the denominator. You might remember that by convention in mathematics, we do
not leave radicals (square roots, cube roots,…) in the denominator of an expression. The process
of getting rid of those radicals in the denominator is called rationalizing the denominator. Some
examples are provided below.
Example 1:
Rationalize the denominator in each case
(a)
(b)
2
5
(c)
6
7 3
Answer:
2
2
5


5
5
5
Answer:
6
6
3


7 3 7 3
3
13
2
Answer:
13
13

2
2
6 3
7 9

13
2

2
2
6 3
21
2 3

7

26
4

26
2

2 5
25


2 5
5

In all of the examples above there is only a single radical in the denominator. However, what do
we do when we have denominators that are sums or differences involving one or two radicals?
This is the case with the three expressions below.
4
7 2
5
2 3 1
4 2
5 2
In order to rationalize the denominators in these types of questions, we must multiply the
numerator and denominator by the conjugate of the denominator. The expressions a  b and
a  b are called conjugates. So with the first expression above we would have to multiply the
numerator and denominator by 7  2 (the conjugate of 7  2 ).
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Example 2
Rationalize the denominator for the expression
4
.
7 2
Answer:
4
4
7 2


7 2
7 2
7 2
4 7 8
49  2 7  2 7  4
4 7 8

74
4 7 8

3

Notice that we used the distributive property to
multiply the binomials in the denominator. Because
these binomials are conjugates we can see that the
2 7 and 2 7 will eventually cancel out thus
eliminating the radicals in the denominator.
Example 3
Rationalize the denominator for the expression
5
2 3 1
.
Answer:
5
5
2 3 1


2 3  1 2 3  1 2 3 1

10 3  5
4 9  2 3  2 3 1
10 3  5
12  1
10 3  5

11

Example 4
Rationalize the denominator for the expression
6 2
.
5 2
Answer:
6 2
6 2
5 2


5 2
5 2
5 2

6 10  6 4
25  10  10  4
6 10  12
3
 2 10  4

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Example 5
Rationalize the denominator for the expression
4 2 3
.
5 2 2 3
Answer:
4 2 3
4 2  3 5 2 2 3


5 2 2 3 5 2 2 3 5 2 2 3

20 4  8 6  5 6  2 9
25 4  10 6  10 6  4 9
40  13 6  6
50  12
46  13 6

38

Example 6
Rationalize the denominator for the expression
3 1
.
4 3 5
Answer:
3 1
3 1
4 3 5


4 3 5 4 3 5 4 3 5

4 9  15  4 3  5
16 9  4 15  4 15  25
12  15  4 3  5
48  5
12  15  4 3  5

43

Before learning to evaluate trigonometric expressions, let's practice rationalizing the
denominator of expressions that contain radicals.
Questions
1. Rationalize the denominator in each case. Show your work.
(a)
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(b)
5
2
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(d)
(g)
1
2 6
5
3 5
(e)
(h)
7
3 2
2 3
5 2
(f)
4 2
5 3
(i)
14 2
3 7
2. Rationalize the denominator in each case. Show your work.
1
5
(a)
(b)
7 1
3 1
(c)
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3 2
(d)
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5 3
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(e)
4
3 2 2
(f)
2 2
4 3 1
(g)
4 3
2 3 1
(h)
3 5
62 5
(i)
7
5 3
(j)
6
3 2
(k)
2 3
5 3 2
(l)
3 2
2 2 5
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(m)
2 1
3 2 1
(n)
1 2 2
3 2
(o)
3 1
5 32
(p)
1 2
42 3
(q)
2 3
2 5 3
(r)
7 2
3 7 1
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Now that we have addressed radicals, it is time that we start evaluating trigonometric
expressions.
Example 7
Find the exact numerical value of each trigonometric expression. If the answer involves radicals,
then make sure it's in simplest radical form.
(a) cos 60 sin135
(b) 5cos 60  sin150
2
(c) sin 315
(d) 7 cos2  330  sin 90 sin  225 
(e)
(g)
sin 60
cos 60
4sin  45 
3cos150
5cos 45
5cos180
(h)
6sin135  2 cos  150 
(f)
1  2sin120
Answers:
When the question asks for the "exact numerical value," then the expectation is that you will
use radicals and/or fractions (as opposed to rounding numbers off on a calculator). These
questions are formulated such that we are forced to use the sines and cosines of the special
rotations.
(a)
cos 60 sin135
(b)
 3 1
 5 
 
 2  2
 1  2 
   

 2   2 

(c)
2
4
5 3 1

2
2
5 3 1

2

(d)
sin 2 315
Note: sin 2 x means  sin x 

2
  

 2 
2
7 cos 2  330   sin 90 sin  225 
2
 3
 2
 7 
  1 

 2 
 2 
2
2
3
 7  
4 2
We have to make a common
denominator to do the subtraction.
21 2 2
 
4
4
21  2 2

4
4
4
2

4
1

2

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(e)
sin 60
cos 60
3
 2
1
2
3 1


2 2
Recall that dividing by a fraction is
equivalent to multiplying by that
fraction's reciprocal.
3 2


2 1
2 3

2
 3
(f)
3cos150
5cos 45

3
3 

2 


 2
5

 2 
3 3 5 2

2
2
3 3
2


2 5 2


3 3
5 2
Rationalize the denominator.
3 3
2


5 2
2


(g)
4sin  45 
(h)
1  2sin120

2
4 

2 


 3
1 2

 2 
5  1
 2 
3
6
  2 

 2   2 
5

3 2 3
Rationalize the denominator.
5
3 2 3


3 2 3 3 2 3
2 2
1 3
Rationalize the denominator.
2 2 1  3


1 3 1 3

2 2  2 6
1 3  3  9
15 2  5 3
9 4 3 6 3 6  9
15 2  5 3
15
3 2  3

3

2 2  2 6
2
 2 6

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3 6
10
5cos180
6sin135  2 cos  150 



6 3
10 2
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Questions (Continued)
3. Find the exact numerical value of each trigonometric expression. If the answer involves
radicals, then make sure it's in simplest radical form. We strongly recommend that you not
look at the unit circle to find the sines and cosines of the special rotations. Over the next few
weeks you are expected to know these sines and cosines from memory.
(a) sin 60 cos300
(b) 3sin 45 cos180
(c) 9sin120 cos135
(d) cos2 150
(e) 4cos2 225 sin  120 
(f) 7sin 2 240 sin 270
(g) sin  240  cos  45 
(h) 3cos30  sin  210 
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(i) sin135  sin  270 
(j) cos 45 cos330  sin 30
(k) 3cos  30 sin 240  cos315
(l) 5cos2 225  7sin 300
(m)
cos180
sin 270
(n)
4 cos 60
6sin 90
(o)
3cos 0
2sin 45
(p)
sin 30
cos 30
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(q)
(s)
(u)
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5sin135
7 cos  330 
sin  270   2cos 45
4sin  240 
5
2 cos 45  1
(r)
3sin150
5cos 225
(t)
4 cos 330  3cos180
6 cos  45 
(v)
7
4sin120  sin 90
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(w)
5
sin  225   2 cos  30 
(x)
(y)
6cos 210
4sin120  cos180
(z)
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sin  90 
6cos  60   4cos 315
2cos 45  sin 270
5  6sin135
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Introduction to Trig Equations: The Graphical Approach, Part 1
First Equation: 3 = 2sinx° +1
Suppose we are asked to solve the trigonometric equation 3  2sin x  1. One approach,
although not the most accurate method, is to graph the sinusoidal function y  2sin x  1 and see
whick x-values correspond to the y-value of 3. Below we have the graphs of the functions
y  2sin x  1 and y = 3. Note that these two functions intersect at multiple points; four of
which we can see on the displayed coordinate system.
So how do we describe all the points of intersection, whose x-coordinates actually respresent the
solutions to the original trigonometric equation, 3  2sin x  1. Let's describe these points in
three different ways.
1. Using Words
Our first point of intersection occurs when x equals 90o. To find the x-coordinates of the
other points of intersection, we move forwards and backwards from 90o in increments of
360o. For example if we go 360o to the right of 90o, then we obtain 450o, the
x-coordinate of another point of intersection. Similarly, if we go 360o to the left of 90o,
then we obtain -270o, the x-coordinate of another point of intersection.
2. Using a Sequence
… , -630o, -270o, 90o, 450o, …
Notice that the values in the sequence are changing by increments of 360o. Based on this,
we can extend the sequence so that it reads as the following.
… , -990o, -630o, -270o, 90o, 450o, 810o, …
3. Using an Algebraic Expression
x  90  360k , k I
At first most people don't understand what this expression means so let's start explaining
some keys points. At the end of the expression, we have k I . This means that the
variable k is a member of the set of integers. Integers are positive and negative whole
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numbers (…, -2, -1, 0, 1, 2, …). So the values of k in the expression 90  360k are
restricted to integers. Let's substitute in integer values for k into the expression and see
what we get.
When k = 0,
When k = 1,
When k = 2,
90  360k
90  360k
90  360k
90  360  0 
90  360 1
90  360  2 
90  0
90
90  360
450
90  720
810
When k = -1,
When k = -2,
When k = -3,
90  360k
90  360k
90  360k
90  360  1
90  360  2 
90  360  3
90   360 
90   720 
90   1080 
270
630
990
Notice that the numbers we generated using the expression 90  360k , k I are actually
identical to those found in the sequence … , -990o, -630o, -270o, 90o, 450o, 810o, … that
we previously used to describe the solution to the original trigonometric equation. The
algebraic expression is just a more concise way to describe the sequence.
Second Equation: 1 = 2sinx° +1
Suppose we alter the equation slightly so that it now reads 1  2sin x  1 . If we want to solve
this trigonometric equation graphically, we start by graphing the functions y  2sin x  1 and
y = 1. That is what we have done below.
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As before, the x-coordinates of the points of intersection between these two functions represent
the solutions to the trigonometric equation, 1  2sin x  1 . Now we can see seven points of
intersection on the coordinate system, but we realize that there are more than those. How can we
describe those points? Let's use the techniques we used with the first equation.
1. Using Words
The first point of intersection occurs when x equals 0o. To find the x-coordinates of the
other points of intersection, we move forwards and backwards from 0o at increments of
180o.
2. Using a Sequence
…, -720o, -640o, -360o, -180o, 0o, 180o, 360o, 640o, 720o, …
3. Using an Algebraic Expression
x  0  180k , k I
If you are uncertain whether or not this algebraic expression is correct, then you can
evaluate it using different integer values (…, -2, -1, 0, 1, 2, …) for k.
When k = 0,
When k = 1,
When k = 2,
When k = 3,
0  180k
0  180k
0  180k
0  180k
0  180  0 
0  180 1
0  180  2 
0  180  3
0  0
0
0  180
180
0  360
360
0  640
640
When k = -1,
When k = -2,
When k = -3,
When k = -4
0  180k
0  180k
0  180k
0  180k
0  180  1
0  180  2 
0  180  3
0  180  4 
0   180 
0   360 
0   640 
0   720 
180
360
640
720
Notice that the numbers we generated using the expression 0  180k , k I are actually
identical to those found in the sequence …, -720o, -640o, -360o, -180o, 0o, 180o, 360o,
640o, 720o, … The algebraic expression must thus be correct.
Third Equation: -1 = 2sinx° +1
Suppose we alter the equation slightly so that it now reads 1  2sin x  1 . If we want to solve
this trigonometric equation graphically, we start by graphing the functions y  2sin x  1 and
y = -1. That is what we have done on the next page.
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Let's describe the x-coordinates for the points of intersection (i.e. the solutions to the equation)
using the three techniques.
1. Using Words
The first point of intersection occurs when x equals 270o. To find the x-coordinates of the
other points of intersection, we move forwards and backwards from 270o at increments of
360o.
2. Using a Sequence
…, -810o, -450o, -90o, 270o, 630o, 990o, …
3. Using an Algebraic Expression
x  270  360k , k I
Could the solution to this equation be written in the forms listed below?
x  630  360k , k I or x  90  360k , k I or x  450  360k , k I
Yes, because they all generate the sequence …, -810o, -450o, -90o, 270o, 630o, 990o, …
Please Note:
Do not assume, based on the three trigonometric equations that have presented in this section,
that the algebraic expression will always have 360k or 180k in it. These values will
change if the original sinusoidal function has been subjected to a horizontal stretch (HS affects
the period.).
e.g.
Sequence: …, -150o, -60o, 30o, 120o, 210o, 300o, …
Corresponding Algebraic Expression: x  30  90k , k I
e.g.
Sequence: …, -14o, -4o, 6o, 16o, 26o, 36o, …
Corresponding Algebraic Expression: x  6  10k , k I
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Questions
1. A graph of the function y  2cos x  3 is provided below. Use this graph to solve each of
the trigonometric equations below. Write your answers in the three forms: using words,
using a sequence, and using an algebraic expression.
(a) 3  2cos x  3
Words:
Sequence:
Algebraic Expression:
(b) 5  2cos x  3
Words:
Sequence:
Algebraic Expression:
(c) 1  2cos x  3
Words:
Sequence:
Algebraic Expression:
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2. A graph of the function y  4sin 2 x  1 is provided below. Use this graph to solve each of
the trigonometric equations below. Write your answers in the three forms: using words,
using a sequence, and using an algebraic expression.
(a) 5  4sin 2 x 1
Words:
Sequence:
Algebraic Expression:
(b) 3  4sin 2 x 1
Words:
Sequence:
Algebraic Expression:
(c) 1  4sin 2 x 1
Words:
Sequence:
Algebraic Expression:
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3. A graph of the function
y  cos3x  4 is
provided. Use this graph
to solve each of the
trigonometric equations
below. Write your
answers in the two
forms: using a sequence,
and using an algebraic
expression.
(a) 3  cos3x  4
Sequence:
Algebraic Expression:
(b) 5  cos3x  4
Sequence:
Algebraic Expression:
(c) 4  cos3x  4
Sequence:
Algebraic Expression:
4. A graph of the function
y  2cos 18  x  4    is
provided. Use this graph
to solve each of the
trigonometric equations
below. Write your
answers in the two
forms: using a sequence,
and using an algebraic
expression.
(a) 2  2cos 18  x  4   
Sequence:
Algebraic Expression:
(b) 0  2cos 18  x  4   
Sequence:
Algebraic Expression:
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5. Match the appropriate algebraic expressions and sequences. Please note that I stands for
integers (…, -2, -1, 0, 1, 2, …), and W stands for whole numbers (0, 1, 2, 3, …).
(a)
10  60k , k I
30o, 70o, 110o, 150o, 190o, …
(b)
20  30k , k I
…, -340o, -160o, 20o, 200o, 380o, …
(c)
30  40k , kW
15o, 40o, 65o, 90o, 115o, …
(d)
40  60k , kW
…, -40o, -10o, 20o, 50o, 80o, …
(e)
70  120k , k I
15o, 30o, 45o, 60o, 75o, …
(f)
20  180k , k I
40o, 100o, 160o, 220o, 280o, …
(g)
5  30k , kW
…, -170o, -50o, 70o, 190o, 310o, …
(h)
15  25k , kW
5o, 35o, 65o, 95o, 125o, …
(i)
10  35k , k I
…, -110o, -50o, 10o, 70o, 130o, …
(j)
0  15k , kW
…, -60o, -25o, 10o, 45o, 80o, …
6. The graph of the sinusoidal function y  3sin x  5 is provided below. If you are asked to
solve the trigonometric equation 1  3sin x  5 , what do you conclude? Explain your
reasoning.
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Introduction to Trig Equations: The Graphical Approach, Part 2
You may have noticed that in the examples and questions you encountered in the last section, the
points of intersection were always found on the maximums, minimums, or on the sinusoidal axis.
This was deliberate. The questions are a little more challenging to solve graphically when the
points of intersection occur at other locations. The reason is that the x-coordinates of the points
of intersection have to be described using two different sequences and therefore two different
algebraic expressions. Let's try one of these questions.
Example 1
Using the graph below, solve the trigonometric equation 1  2cos x .
Answer:
Let's start by listing the x-coordinates of the points of intersection that we can see on the
coordinate system.
-420o, -300o, -60o, 60o, 300o, 420o
These x-values are not changing by the same increment. For example the difference between
-420oand -300o is 120o, while the difference between the next two values, -300oand -60o, is
240o.
What we do is split the values into two separate sequences, using every second value.
Sequence #1: -300o, 60o, 420o
Sequence #2: -420o, -60o, 300o
Notice that in both of these sequences, the values are changing by increments of 360o. We
can use this to expand the sequences and then create the algebraic expression for each.
…, -660o, -300o, 60o, 420o, 780o, …
…, -780o, -420o, -60o, 300o ,660o, …
x  60  360k , k I
x  300  360k , k I
Therefore our final answer is:
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 60  360k , k I
x
300  360k , k I
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Questions
1. Using the graph of the sinusoidal function y  4sin x  1 , solve the following trigonometric
equations. Show your work.
(a) 1  4sin x  1
(b) 3  4sin x  1
2. Using the graph of the sinusoidal function y  2sin  6 x  3 , solve the following
trigonometric equations. Show your work.
(a) 2  2sin  6 x  3
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(b) 4  2sin  6 x  3
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3. Using the graph of the sinusoidal function y  6cos  3  x  40    , solve the following
trigonometric equations. Show your work.
(a) 6  6cos  3  x  40   
(b) 3  6cos  3  x  40   
(c) 0  6cos  3  x  40   
(d) 3  6cos  3  x  40   
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Trigonometric Equations, Part 1
In the next three sections we will solve trigonometric equations using our knowledge of the unit
circle, specifically the coordinates of the image points for rotations at 30o and 45o intervals. As
mentioned previously, you are expected to know the coordinates of these image points for these
special rotations. What should you be able to do at this stage?


If someone asks us what is the cosine of 135o, then it is expected that we can quickly
2
respond 
.
2
3
Similarly, if someone asks us what rotations between 0o and 360o have a sine of 
,
2
then it is expected that we can quickly respond 240o and 300o.
We stress that this level of familiarity with these special rotations is required in the remaining
sections of this unit.

We are also expected to understand coterminal angles and that they occur every 360o.
3
Suppose we are asked to determine what rotations have a sine of 
. It is expected
2
that we know that 240o and 300o are two possible answers. To capture all the other
possible answers (i.e. the coterminal angles) we add and subtract increments of 360o from
240o and 300o. That means our answers can be written as 240  360k , k I and
300  360k , k I . Please note that when we write k I , we are saying that the variable k
is a member of integers (…, -2, -1, 0, 1, 2,…).
Example 1
Solve 2sin x  3  0 .
Answer:
2sin x  3  0
2sin x  3
Add
3
2
 60  360k , k I
x
120  360k , k I
Divide both sides of the equation by 2.
sin x 
3 to both sides of the equation.
Based on our knowledge of the unit circle, we know that
3
the rotations of 60o and 120 o have a sine of
. The
2
360k , k I is added to both to capture all the coterminal
angles.
Example 2
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Solve
2  2cos x  0 .
Answer:
2  2 cos x  0
2 cos x   2
2
2
135  360k , k I
x
225  360k , k I
cos x  
Subtract
2 from both sides of the equation.
Divide both sides of the equation by 2.
Based on our knowledge of the unit circle, we know that
2
the rotations of 135o and 225 o have a cosine of 
.
2
The 360k , k I is added to both to capture all the
coterminal angles.
Example 3
Solve 2 sin x  1  0 .
Answer:
2 sin x  1  0
2 sin x  1
1
sin x  
2
sin x  
1
2

2
2
2
2
 225  360k , k I
x
315  360k , k I
Subtract 1 from both sides of the equation.
Divide both sides of the equation by 5.
Rationalize the denominator by multiplying the
numerator and denominator by 2 .
sin x  
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We know that the rotations of 225o and 315o (plus their
2
coterminal angles) have a sine of 
.
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Example 4
Solve 5cos x  4  9 , where -180o  x  720o.
Answer:
5cos x  4  9
This question is similar to the previous
examples however it is not asking for all
possible values of x, only those between
and/or equal to -180o and 720o.
5cos x  9  4
5cos x  5
5
cos x  
5
cos x  1
x  180  360k , k I
x  ..., -540, -180, 180, 540, 900,...
We tackle the question in the same
manner but we add a few more steps. We
find all the possible values of x (as we did
in the previous questions). We then
describe those values as a sequence.
Finally we identify those values that are
restricted between -180o and 720o.
Based on the restriction in the question:
x  -180, 180, and 540
Example 5
Given the sinusoidal function y  2sin x , find the values of x between -720o and 360o for which
y is equal to 1.
Answer:
y  2sin x
Start by taking the equation of the
sinusoidal function and substituting 1 for
y. After that is done, divide both sides of
the equation by 2.
1  2sin x
1
sin x 
2
 30  360k , k I
x
150  360k , k I
As with the previous example, we find all
the possible values of x, describe those
values as two sequences, and finally
identify those values between -720o and
360o.
..., -1050, -690, -330, 30, 390,...
x
 ..., -930, -570, -210, 150, 510,...
Based on the restriction in the question:
x  -690, -570, -330, -210, 30, and 150
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Questions
1. Solve the following trigonometric equations. We strongly recommend that you do not look
at the unit circle to solve these; you are expected to know the sines and cosines
corresponding to the special rotations at this point in the course.
(a) sin x  1  0
(b) 2cos x  3  0
(c) 7cos x  6  1
(d) 5  4sin x  5
(e) 2sin x  2  0
(f)
(g) 2 3 sin x  1  0 , -720o  x  0o
(h) 5  8cos x  1 , -360o  x  360o
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2 cos x  8  9
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2. (a) Given the sinusoidal function y  6sin x , find the values of x for which y is equal to -3.
(b) Which of the three graphs represents the mathematical situation described in (a)? Please
note that for each of the graphs below, each increment on the vertical axis represents 1,
and each increment on the horizontal axis represents 90o.
(i)
(ii)
(iii)
3. (a) Given the sinusoidal function y  3cos x  2 , find the values of x between -360o and
180o for which y is equal to 2.
(b) Which of the three graphs represents the mathematical situation described in (a)? Please
note that for each of the graphs below, each increment on the vertical axis represents 1,
and each increment on the horizontal axis represents 90o.
(i)
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(ii)
(iii)
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Trigonometric Equations, Part 2
When one sees sin 2 x in an equation it actually means  sin x  . Similarly when we read
2
cos2 x , it means  cos x  . So what do we do when we have a trigonometric equation in which
the cosine or sine is squared? The answer is that we use our knowledge of quadratic equations to
solve these trigonometric equations. For example, knowing how to solve a quadratic equation
such as y 2  4 y  5  0 , allows us to solve a trigonometric equation such as
cos2 x  4cos x  5  0 .
2
Before we consider these types of trigonometric equations, let's take a few minutes to refresh
ourselves with quadratic equations. The first thing that we must remember is that the two
techniques for solving quadratic equations (i.e. factoring and the quadratic formula) require that
we first set the equation equal to zero. In this unit, we only deal with equations that can be
factored. The factoring techniques that we learned in Level IV Academic Math were common
factoring, factoring difference of squares, factoring perfect square trinomials, inspection, and
decomposition. If you have forgotten how to factor, go to the Factoring Refresher Sheet and the
Factoring Flow Chart Sheet found in the appendix of this unit.
Now let's solve a trigonometric equation that requires knowledge of quadratic equations.
.
Example 1
Solve sin 2 x  4sin x  3  0 .
Answer:
sin 2 x  4sin x  3  0
Let y represent the sin x. Many people find
the equation easier to factor when it is in
this form.
Let y  sin x
y2  4 y  3  0
Factor by inspection (i.e. find two numbers
that multiply to give +3 and add to give
+4.) If the product of the two factors,
(y + 1) and (y + 3) is 0, then at least one of
the factors must be equal to 0. Solve for y.
 y  1 y  3  0
y 1  0
y  1
or
or
y3 0
y  3
sin x  1
x  270  360k , k I
or
sin x  3
no solution
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Substitute sin x back into the equation for y.
The first part can be solved using the unit
circle. There is no solution to the second
part because the sine of rotations (as well as
the cosines of rotations) are always between
or equal to 1 and -1; no rotation has a sine
of -3.
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C. D. Pilmer
Example 2
Solve 2cos2 x  3 cos x  0 .
Answer:
2 cos 2 x  3 cos x  0
Let y  cos x
Start by letting y represent cos x.
2 y2  3 y  0


y 2y  3  0
y0
or
We common factor out a y. If the product
of the two factors, y and 2 y  3 is 0, then
at least one of the factors must be equal to
0. Solve for y.
2y  3  0
2y  3
y
cos x  0
or
x  90  180k , k I
3
2
3
2
 30  360k , k I
x
330  360k , k I
cos x 
Substitute cos x back into the equation for
y. Now we solve for x using the unit circle.
Example 3
Solve cos2 x  1 .
Answer:
cos 2 x  1
Start by letting y represent cos x.
Let y  cos x
y2  1
y2 1  0
Set the quadratic equation equal to zero.
Factor the difference of squares. Once
factored, solve for y.
 y  1 y  1  0
y 1  0
y 1
or
cos x  1
or
x  0  360k , k I
y 1  0
y  1
cos x  1
x  180  360k , k I
Substitute cos x back into the equation for
y. Now we solve for x using the unit circle.
Can be simplified to:
x  0  180k , k I
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Example 4
Solve 2sin 2 x  3sin x  7  6 , where -360o  x  360o.
Answer:
2sin 2 x  3sin x  7  6
Start by letting y represent sin x.
Let y  sin x
2 y2  3y  7  6
2 y2  3y 1  0
Set the quadratic equation equal to zero.
-1  -2 = 2
-1 + -2 = -3
Since the leading numerical coefficient is a
number other than one, we factor using
decomposition. We start by finding two
numbers that multiply to give +2 (product
of 1 and 2), and add to give -3 (the
coefficient of the middle term), break up
the middle term using these numbers, group
the terms, and factor.
2 y 2  1y  2 y  1  0
2y
2

 1y   2 y  1  0
y  2 y  1  1 2 y  1  0
 2 y  1 y  1  0
2 y 1  0
2y 1
1
y
2
or
or
1
or
2
 30  360k , k I
x
150  360k , k I
sin x 
y 1  0
y 1
Solve for y.
sin x  1
Substitute sin x back into the equation for y.
Solve for all possible values of x.
x  90  180k , k I
x  330, -270, -210, 30, 90, and 150
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The question set restrictions on x (-360o  x
 360o). Consider this when stating the
final answer.
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Questions:
Solve each of the following trigonometric equations. Show all your work. Do not use a
calculator.
1.
cos2 x  4cos x  5  0
2.
sin 2 x  1  0
3.
2sin 2 x  sin x  0
4.
2cos2 x  7cos x  3  1
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5.
2sin 2 x   2 sin x
6.
4cos2 x  1  0
7.
5cos2 x  5cos x  0
8.
sin 2 x  8sin x  9  2
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9.
2cos2 x  5cos x  3  0 , 0o  x  720o
11. 4sin 2 x  1  0 , -720o  x  0o
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10.
2 sin 2 x  sin x  0 , -360o  x  360o
12. 2 3 cos2 x  cos x  0 , -360o  x  180o
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Trigonometric Equations, Part 3
Up to this point we have dealt with equations such as sin x  1 or 2cos2 x  cos x  0 where
the arguments are simply x. However, what do we do when we have an equation such as
sin  2  x  10     1 where the argument is 2  x  10  ? The problem is still very manageable,
there are just a few more steps.
Example 1
Solve sin  2  x  10     1 .
Answer:
sin  2  x  10     1
2  x  10    90  360k , k I
Do not try to manipulate the equation algebraically
in the beginning. Rather find the rotations that have
a sine of 1. By doing so, you are solving for
2  x  10  .
x  10  45  180k , k I
Now we divide both sides of the equation by 2.
x  55  180k , k I
Our last step is to add 10 to both sides of the
equation so that we can solve for x.
Example 2
Solve cos  3  x  5    
3
.
2
Answer:
3
2
150


360
k , k  I

3  x  5   
210  360k , k I
Do not try to manipulate the equation algebraically
in the beginning. Rather find the rotations that have
3
a cosine of 
. By doing so, you are solving for
2
3  x  5 .
50  120k , k I
x  5  
70  120k , k I
Now we divide both sides of the equation by 3.
45  120k , k I
x
65  120k , k I
Our last step is to subtract 5 from both sides of the
equation so that we can solve for x.
cos  3  x  5    
Why do we care about arguments that are not simply x? Remember that when we find the
equation of a sinusoidal function, it is of the form y  k cosax  c   d . Notice that the
argument is a  x  c  . If we have the equation of a sinusoidal function of this form and are asked
to find x given y, then the skills we learned in the first two examples come into play.
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Example 3
For y  6sin 10  x  7     2 , find x given y is equal 1.
Answer:
y  6sin 10  x  7     2
1  6sin 10  x  7     2
3  6sin 10  x  7   
1
 sin 10  x  7   
2
 30  360k , k I
10  x  7    
150  360k , k I
 3  36k , k I
x  7  
15  36k , k I
10  36k , k I
x
22  36k , k I
Start by substituting 1 in for y.
Add 2 to both sides of the equation.
Divide both sides of the equation by 6, and simplify.
Find the rotations that have a sine of
1
.
2
Divide both sides of the equation by 10.
Add 7 to both sides of the equation.
Example 4
For f  x   8cos  5  x  2    , find x if f  x   4 2 and 80  x  80 .
Answer:
f  x   8cos  5  x  2   
4 2  8cos  5  x  2   
4 2
 cos  5  x  2   
8
2
 cos  5  x  2   
2
 45  360k , k I
5  x  2   
315  360k , k I
 9  72k , k I
x  2  
63  72k , k I
 7  72k , k I
x
61  72k , k I
 ..., -137, -65, 7, 79,151,...
x
..., -83, -11, 61, 133, 205,...
x  -65, -11, 7, 61, and 79
NSSAL
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Start by substituting 4 2 in for f  x  .
Divide both sides of the equation by 8.
Simplify the fraction.
Find the rotations that have a cosine of
2
.
2
Divide both sides of the equation by 5.
Subtract 2 from both sides of the equation.
Now consider the restrictions on x. ( 80  x  80 )
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At the beginning of this unit, we stated that we lacked the skills to solve sinusoidal application
problems for which we have to find the independent variable given the dependent variable. We
can now start considering some of these types of problems.
Example 5
An acrobat is swinging back and forth on a trapeze. The distance between the acrobat and a
vertical support beam can be modeled using a sinusoidal function. At t = 1 seconds, the acrobat
is 10 metres from the beam, the maximum distance from the beam. At t = 3 seconds, the acrobat
is 2 metres from the beam, the minimum distance. When is the acrobat 8 metres from the
vertical support beam?
Answer:
We cannot answer this question without finding the equation of the sinusoidal function that
models this situation. Before we do that, let's draw a graph using the information provided
and our understanding of sinusoidal functions.
Now we determine the equation of the sinusoidal function which will be of the form
y  k cosax  c   d .
max + min
Equation of Sinusoidal Axis: h 
2
10  2
h
2
(Vertical Translation)
h

6

Amplitude = max – sinusoidal axis
= 10 - 6
= 4 (Vertical Stretch)
period
4
1
Period = 4
Horizontal Stretch = 


360
360 90
Coordinates of a Maximum: (1, 10)
Horizontal Translation = 1
Equation of the Sinusoidal Function: d  4 cos90t  1  6
Now we can substitute 8 in for d, and solve the resulting trigonometric equation.
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d  4 cos  90  t  1    6
8  4 cos  90  t  1    6
2  4 cos  90  t  1  
2
 cos  90  t  1  
4
1
 cos  90  t  1  
2
 60  360k , k I
90  t  1  
300  360k , k I
Please note that the degree symbol has been
dropped because we are solving for time,
which in this case is measured in seconds.
2
 3  4k , k I
Divide both sides of the equation by 90. We
t 1  
simplified the fractions in the same step.
10  4k , k I
 3
5
 3  4k , k I
3
Add 1 or to both sides of the equation.
t
3
13  4k , k I
 3
We have to remember that time, t, is on the horizontal axis, and that time must be positive.
We have to alter our answer slightly so that it reads as follows. Please note that W represents
whole numbers (0, 1, 2, 3, …).
5
 3  4k , kW
t
13  4k , kW
 3
Questions
1. Solve each of the following trigonometric equations.
(a) cos  4  x  9     1
(b) sin  5  x  3    1
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(c) sin  6  x  12     
(e) 
3
2
1
 cos 12  x  5  
2
(d) cos 15  x  1    
(f)
2
2
3
1

 sin   x  15  
2
2

2. For y  10sin  24  x  1   , find x given y is equal 5 3 .
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3. For f  x   7sin  3  x  20     9 , find x if f  x   9 and 150  x  150 .
1

4. For g  x   4cos   x  50     1 , find x if g  x   1 and 1080  x  1080 .
3

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5. Nita is riding on a Ferris wheel with her daughter. Based on her work in her adult education
math class, she realizes that her height above the ground with respect to time can be modeled
using a sinusoidal function. At t = 0 seconds, she is at her minimum height of 1 metre. At
t = 9 seconds, she reaches her maximum height of 13 metres. Nine seconds later she returns
to the minimum height of 1 metre. At what times will she be at a height of 10 metres?
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Trigonometric Equations, Part 4
Up to this point, all the trigonometric equations that we have encountered have resulted in the
1
2
3
sine or cosine of a rotation being equal to 0, 1 ,  , 
, or 
. This was done
2
2
2
deliberately by the author of this resource so that learners would be forced to remember the sines
and cosines associated with the special rotations. Obviously real-life problems do not always
work out to be these values. What happens when we have questions such as the ones below?
sin x  0.789
cos x  
sin  3  x  6     0.12
2
7
For these questions, we reach for our calculators and use either the sin 1 or cos 1 button. Now
there is a problem when we do this; the calculator only produces one solution (the principal
solution), not the infinite number of solutions we typically associate with trigonometric
equations. Consider the first equation sin x  0.789 . When we use the calculator, we obtain
the following. (Make sure the calculator is in "degree" mode.)
We arrive at an answer of approximately 52.1o. That rotation is in the first
quadrant of the coordinate system. We know that there are an infinite
number of coterminal angles to 52.1o and these can be captured by
including 360k , k I . But we still have a problem; we know that there is
another set of rotations that also have a sine of 0.789. Since 0.789 is a
positive value, we know that the terminal arm for these rotations must be in the second quadrant,
but how do we determine them? We don't know how just yet. Therefore, we only have half of
the answer.
52.1  360k , k I
x
 ?  360k , k I
Looking for Patterns: Sine
We are going to start by considering sines we are familiar with. Using your knowledge of the
unit circle, solve the following trigonometric equations where 0  x  360 . There are two
rotations/answers for each of the equations. Write your answers in the space provided.
1st Rotation
2nd Rotation
sin x  0
Look for a pattern between the
first and second rotation.
Specifically, how can you use
the first rotation to find the
second rotation? (This was the
problem in the question
sin x  0.789 above.)
1
sin x 
2
2
2
3
sin x 
2
sin x 
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Given the first rotation, which we call  (theta), what expression below allows one to find the
second rotation with the same sine? Only one of these six choices is correct.
90  
(i)
(iv) 180  
(ii) 90  
(iii) 180  
(v) 360  
(vi) 360  
Looking for Patterns: Cosine
We are going to start by considering cosines we are familiar with. Using your knowledge of the
unit circle, solve the following trigonometric equations where 0  x  360 . There are two
rotations/answers for each of the equations. Write your answers in the space provided.
1st Rotation
2nd Rotation
3
2
2
cos x 
2
1
cos x 
2
cos x 
Look for a pattern between the
first and second rotation.
Specifically, how can you use
the first rotation to find the
second rotation?
cos x  0
Given the first rotation, which we call  (theta), what expression below allows one to find the
second rotation with the same cosine? Only one of these six choices is correct.
90  
(i)
(iv) 180  
(ii) 90  
(iii) 180  
(v) 360  
(vi) 360  
Example 1
Solve sin x  0.789 .
Answer:




1
2
3
, 
, or 
, then we know that we are not
2
2
2
dealing with one of the special rotations. We have to use our calculator. Make sure it is
in "degree" mode.
The inverse sine ( sin 1 ) of 0.789 is approximately 52.1o. This rotation is found in the
first quadrant.
There is a rotation in the second quadrant with the same sine; it can be found using the
formula 180   . We find it to be approximately 127.9o.
We still have to tack on 360k , k I to capture all the coterminal angles. Our final
answer is:
 52.1  360k , k I
x
127.9  360k , k I
Since the sine is not equal to 0, 1 , 
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Example 2
Solve cos x  0.362 .
Answer:




1
2
3
,
, or 
, then we know that we have
2
2
2
to use our calculator. Make sure it is in "degree" mode.
The inverse cosine ( cos 1 ) of -0.362 is approximately 111.2o. This rotation is found in
the second quadrant.
There is a rotation in the third quadrant with the same cosine; it can be found using the
formula 360   . We find it to be approximately 248.8o.
We still have to tack on 360k , k I to capture all the coterminal angles. Our final
answer is:
111.2  360k , k I
x
248.8  360k , k I
Since the cosine is not equal to 0, 1 , 
For the trigonometric equations that are quadratic in nature (such as those in examples 3 and 4),
often one can arrive at a stage when there are two similar but different courses of action.
Specifically one may use their knowledge of the unit circle and the special rotations to complete
one part of the question, and then use the inverse sine or cosine features on your calculator to
complete another part of the question. Your instructor and future professors want learners to
know when it is appropriate to use these techniques. Learners who blindly grab the calculator
for every question will ultimately be doing a disservice to themselves.
Example 3
Find the zeroes of g  x   3sin 2 x  sin x .
Answer:
g  x   3sin 2 x  sin x
0  3sin 2 x  sin x
 Zeroes (or x-intercepts) occur when y (or g(x)) equals 0.
Let y  sin x
2
3y  y  0
 Common Factoring
y  3 y  1  0
y0
or
3y 1  0
y
sin x  0
(Unit Circle)
x  0  180k , k I
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1
3
(Calculator and the " 180   " Rule)
19.5  360k , k I
x
199.5  360k , k I
sin x  
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Example 4
If g  x   10cos2 x and h  x   cos x  3 , find all values of x for which g  x   h  x  .
Answer:
g  x  h  x
10cos2 x  cos x  3
10cos2 x  cos x  3  0
Let y  cos x
2
10 y  y  3  0
(+5)  (-6) = -30
(+5) + (-6) = -1
2
10 y  5 y  6 y  3  0
10 y
2
 Set the equation equal to 0 by subtracting 2 from both sides.
 We will have to factor using decomposition.

 5 y   6 y  3  0
5 y  2 y  1  3  2 y  1  0
 2 y  1 5 y  3  0
2 y 1  0
1
y
2
or
1
2
(Unit Circle)
120  360k , k I
x
240  360k , k I
cos x  
5y  3  0
3
y
5
3
cos x 
5
(Calculator and the " 360   " Rule)
 53.1  360k , k I
x
306.9  360k , k I
Example 5
For f  x   sin  3  x  5   , find x if f  x   0.4
Answer:
f  x   sin  3  x  5  
0.4  sin  3  x  5  
 23.6  360k , k I
3  x  5   
156.4  360k , k I
 7.9  120k , k I
x  5  
52.1  120k , k I
 Calculator and the " 180   " Rule
12.9  120k , k I
x
57.1  120k , k I
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Example 6
A small portion of a water wheel is submerged below the surface of a river. A nail is located on
the circumference of the wheel. As the river flows, the current causes the water wheel to rotate.
The height of the nail relative to the surface of the water changes as the wheel rotates. The
height of the nail with respect to time can be modeled using a sinusoidal function. At t = 1
second, the nail is at the maximum height of 3.8 metres. At t = 6 seconds, the nail is at the
minimum height of -1.2 metres. When will the nail be at a height of 0.3 metres relative to the
surface of the water?
Answer:
Draw the graph and then determine the
equation of the sinusoidal function.
Equation of Sinusoidal Axis:
max + min
h
2
h
3.8   1.2 
2
h  1.3 (Vertical Translation)

Amplitude = max – sinusoidal axis
= 3.8 – 1.3
= 2.5 (Vertical Stretch)
Period = 10
Horizontal Stretch = 
Coordinates of a Maximum: (1, 3.8)
period 10
1


360
360 36
Horizontal Translation = 1
Equation of Function: h  2.5cos  36  t  1    1.3
Now substitute 0.3 in for h, and solve for t.
0.3  2.5cos  36  t  1    1.3
1  2.5cos  36  t  1  
0.4  cos  36  t  1  
 Subtracted 1.3 from both sides of equation.
 Divided both sides of equation by 2.5
113.6  360k , k I
36  t  1  
 Calculator and the " 360   " Rule
246.4  360k , k I
3.16  10k , k I
t 1  
6.84  10k , k I
4.16  10k , k I
t
 This corresponds to what we see on the graph.
7.84  10k , k I
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Questions:
Note: Make sure your calculator is in "degree" mode.
1. Solve the following trigonometric equations.
(a) sin x  0.63
(b) cos x  0.38
(c) cos x  0.17
(d) sin x  0.74
(e) 4cos2 x  cos x  0
(f) 4sin 2 x  7sin x  3  0
(g) 3cos2 x  4cos x  3  1
(h) cos 10  x  2     0.87
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(i) sin  2  x  20     0.13
(j) cos  9  x  1    0.93
2. If f  x   8sin 2 x and g  x   5sin x , find all values of x for which f  x   g  x  .
3. Find the zeroes of h  x   25sin 2 x  1.
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4. For f  x   7 cos  5  x  3    1 , find x if f  x   2 .
5. A skyscraper is swaying in high wind conditions. At t = 3 seconds, the top floor of the
building swayed to the extreme right 21 cm (+21 cm). At t = 12 seconds, the top floor
swayed to the extreme left 21 cm (-21 cm). This back and forth swaying motion can be
modeled using a sinusoidal function. When will the building be 7 cm to the right of its
resting position (i.e. +7 cm)?
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6. As the shaft on an electric motor rotates, the shaft itself vibrates slightly. Even accurately
constructed shafts will vibrate at high speeds. Stress is exerted on the shaft as it vibrates and
this stress is measured in megaPascals (MPa). The relationship between the stress exerted on
the shaft and time can be modeled using a sinusoidal function. For a particular shaft, rotating
a specific speed, the following information was collected. At t = 0 seconds, a minimum
stress of 4 MPa was exerted on the shaft. At t = 0.03 seconds, a maximum stress of 10 MPa
was exerted on the shaft. When does the shaft experience 9 MPa of stress?
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Introduction to Radian Measure
Up to this point in our mathematics education, we have only measured rotations or angles in
degrees. There is, however, another unit of angle measure that is extensively used in higher level
mathematics, physics, and engineering courses. This unit of measure is the radian.
A radian is the measure of the angle subtended at the centre of the circle by an arc equal in
length to the radius of the circle. This definition has probably left you scratching your head as to
what it means. Consider the three rotations that are shown below. All three rotations,  , are the
same size but they occur about circles with different radii. The length of the arc (or arc length)
that subtends angle  also differs in each case. However, the arc length is equal to the radius in
all three cases. When the arc length equals the radius, we know that we have a radian measure of
1 (Calculations are shown below.).
Circle A:
arc length
radius
1

1
1
number of radians =
arc length = 1

radius = 1
Circle B
arc length
radius
2

2
1
arc length = 2
number of radians =

radius = 2
Circle C
arc length = 3
arc length
radius
3

3
1
number of radians =

radius = 3
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Well this might be fine but what is the relationship between the degree, a unit of measure we are
very familiar with, and the radian, this new unit of angle measure? To understand this
relationship, let's start by considering a rotation of 180o on the unit circle (See the diagram
below.).
180o
(-1,0)
(0,0)
(1,0)
If we want to determine the number of radians, we need to know the arc length and the radius.
Since we are dealing with a unit circle, we know that the radius is 1. However, we need to
determine the arc length. This arc length happens to represent half the circumference of the unit
circle. If we calculate the circumference of the circle and take half of this, then we will know the
arc length (Shown below on the left.). After that, we can determine the radian measure (Shown
below on the right.).
Circumference  2 r
number of radians =
 2 1
 2

arc length
radius

1

The arc length is half of the circumference.
We learn that 180o is equivalent to  radians.
Notice that we expressed our answer in terms
of  , rather than rounding our answer to 3.14.
Such an approximate value would not be
equivalent to 180o.
Arc Length  Half of the Circumference
1
  2
2

Now let's consider a rotation of 90o on the unit circle (See the diagram below.).
(0,1)
90o
(0,0)
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We now determine the number of radians. We know the radius (1) but we need to find the arc
length. Since this arc length represents one-quarter of the circumference of the unit circle, it is
fairly easy to determine.
Circumference  2 r
number of radians =
 2 1

 2
The arc length is one-quarter of the circumference.
Arc Length  Quarter of the Circumference
1
  2
4
2

4

arc length
radius
 2
1


2
Notice that we expressed our answer as

radians, rather than rounding our
2
answer to 1.57.

2
Up to this point we have learned that:
 180o is equivalent to  radians

 90o is equivalent to
radians
2
We now use this information to convert other degree measures to radians. Five examples are
provided below. In each case, we take an intuitive approach to completing the conversion.
 360o is 2 times larger than 180o (  radians), therefore 360o must be equivalent to 2
radians.

 45o is one-quarter of 180o (  radians), therefore 45o must be equivalent to
radians.
4

 60o is one-third of 180o (  radians), therefore 60o must be equivalent to
radians.
3
 120o can be viewed as either two-thirds of 180o (  radians) or as 2 times larger than 60o

( radians; determined in previous example.). Regardless of the approach, we realize
3
2
that 120o is equivalent to
radians.
3

1
 270o is 180o (  radians) plus 90o ( radians), therefore 270o must be equivalent to 1 
2
2
3
radians, which is typically written as
radians
2
When we consider the above conversions, we notice that in each case we are dealing with special
rotations. Upon completion of this section, we are expected to know all of the special rotations
in radian measure.
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What do we do when we are not dealing with a special rotation? We treat it as a ratio and
proportion question using the fact that 180o is equivalent to  radians.
Convert 50o to radian measure.
radians
degrees


x
50
180
50
x
180
5
x
radians
18
Example 1
Convert the following degree measures to radians.
(a) 135o
(b) -240o
(c) 72o
(d) -126o
Answers:
When we are dealing with special rotations, as is the case in (a) and (b), we use the intuitive
approach to complete the conversion (rather than relying on the ratio and proportion method).
(a) 135o can be viewed as:
 three-quarters of 180o (  radians), or

 3 times 45o ( radians; previously determined.).
4
Regardless of how we approach it, we realize that 135o is equivalent to
3
radians.
4
(b) We initially ignore the negative, and then re-incorporate it in the last step.
240o can be viewed as:

 4 times 60o ( radians; previously determined.). , or as
3
2
 2 times 120o (
radians; previously determined.).
3
4
Regardless of how we approach it, we realize that 240o is equivalent to
radians.
3
4
Therefore -240o is equivalent to 
radians.
3
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radians
degrees
(c)
radians
degrees
(d)


x
72
180
72
x
180
2
x
radians
5


(72 and 180 are divisible by 36.)
x
126
180
126
(126 and 180 are divisible by 18.)
x
180
7
x
radians
20
Example 2
Convert the following radian measures to degrees.
(a) 3
(c)
(b)
5
8
7
6
(d) -0.9
Answers:
As in example 3, when we are dealing with special rotations, such as those in (a) and (b), we
use the intuitive approach to complete the conversion (rather than relying on the ratio and
proportion method).
We know that we are dealing with a special rotation if the radian measure is expressed in
4 5 
9 7
terms of  , and the denominator is either 1, 2, 3, 4, or 6 (e.g.
).
,
, , 
,
1
2 3
4
6
We provide the thought process that one might engage in to solve these first two questions;
learners are expected to do this in their heads, as opposed to writing everything out. Please
note that radian measures are not always expressed in terms of  , as is the case in
question (d).
(a) If  radians is equivalent to 180o, then 3 radians must be equivalent to 540o  3 180  .
7
7
radians as   radians. Converting this to degrees means that it
6
6
7
can be viewed as 180 . If one-sixth of 180o is 30o, then seven-sixths of 180o are 210o.
6
(b) One can think of
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(c)
radians
degrees
5

 8
180
x
 x  180
900
5
900
x 5
5
8
x 

x
(d)
radians
degrees
900 
or 112.5
8

0.9
180
x
 x  180  0.9 

 x  162
x
162 

or  51.6
Example 3
Determine the radian measure of the central angle that intercepts an arc of 3 cm in a circle of
radius 5 cm.
Answer:
The most difficult part of this question is determining what is known and what needs to be
found. In this case, we know the arc length (3 cm) and the radius (5 cm), and we need to find
the number of radians.
arc length
number of radians =
radius
3
 radians or 0.6 radians
5
In prior questions, we often expressed the radian measure in terms of  . This is not always
the case, as illustrated in this question.
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Example 4
Find the length of the intercepted arc for a central angle of 
3
radians in a circle with a radius
4
of 6 cm.
Answer:
As in example 3, the most difficult part of this question is extracting the given information
and determining what needs to be found. In this case, we are given the number of
 3 
radians  
 and the radius (6 cm), and are asked to find the arc length.
 4 
arc length
number of radians =
radius
3 x


4 6
 3 
6 
x
 4 
18
x
4
9
We can have a negative arc length. It just indicates that the arc is
x
cm
formed by rotating in a clockwise direction.
2
Example 5
We know that the coordinates of an image point after a rotation of  degrees about (0, 0) from
(1, 0) are  cos  ,sin   . This can be expressed in a mapping rule. Mapping rules in degree
measure are provided below. Rewrite these mapping rules in radian measure.
(a) R135 1,0    cos135,sin135

2 2
  
,

2
2 

(b) R300 1,0    cos300,sin 300
1
3
  , 

2 
2
Answers:
3
3 

(a) R3 1, 0    cos ,sin

4
4 

4
5
5 

(b) R5 1, 0    cos
,sin

3
3 

3

2 2
  
,

 2 2 
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  , 

2 
2
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Questions
1. Convert the following degree measures to radians.
(a) 720o
(b) -360o
(c) 30o
(d) 210o
(e) 225o
(f) -300o
(g) 130o
(h) -208o
2. Convert the following radian measures to degrees.
11
(a) 5
(b)
6
(c)
4
3
(e) 
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(g) 
4
9
(h) 1.3
3. Determine the radian measure of the central angle that intercepts an arc of 4 cm in a circle
of radius 6 cm.
4. Find the length of the intercepted arc for a central angle of
5
radians in a circle with a
3
radius of 9 cm.
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5. In each case, we are provided with a sequence of special rotations in degree measure. Our
mission is to convert the sequence to radian measure. This has been partially completed for
us. Fill in the blanks.
(a) … , 0o, 30o, 60o, 90o, 120o, 150o, 180o, …
   2
   3

 0
  1
…, 
or  , 
or 0  , 
or  , 
or  ,
6  6
3  6
2
 6
  6
(b) … , 0o, 45o, 90o, 135o, 180o, 225o, 270o, …
   2

 0
  1
…, 
or  , 
or 0  , 
or  ,
4  4
2
 4
  4
,
,
,
,
,
6. The questions below refer to the sequences we were dealing with in the previous question.
(a) For the first sequence, in degree measure, the values are changing by increments of 30o.
For the corresponding sequence, in radian measure, by what increment are the values
changing?
_______
(b) for the second sequence, in degree measure, the values are changing by increments of
45o. For the corresponding sequence, in radian measure, by what increment are the
values changing?
_______
7. Mapping rules in degree measure are provided below. Rewrite these mapping rules in radian
measure.
(a) R315 1,0    cos315,sin 315
(b) R210 1,0    cos 210,sin 210
 2
2
 
,

2 
 2
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
3 1
  
,  
2
2

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Revisiting Questions Using Radian Measure, Part 1
In this section we revisit a number of questions that we previously answered in degree measure.
Now we answer them in radian measure. The most common mistake made by learners, in this
section and the ones that follow, is that they are unwilling to move from thinking in degrees to
thinking in radians. These individuals first determine their answers in degrees and in their last
step covert the answers to radians. Please do not do this. Get your mind attuned to working
solely in radians; it is initially challenging but learners ultimately benefit from this practice.
Please note that for the remainder of this unit we only work in radian measure.
We start by revisiting the unit circle and the special rotations. These rotations and their
corresponding image points are found on this page and the next. Make sure you learn these
values. (You will not be able to bring them in when writing a test.)

Increments
4



 cos , sin 
2
2

 0,1
3
3 

 cos , sin

4
4 


2 2
,
 

 2 2 
 cos  ,
 1, 0 
sin  
5
5 

, sin
 cos

4
4 


2
2
,
 

2 
 2
NSSAL
©2012



 cos , sin 
4
4

 2 2
,


 2 2 
 cos 0,
1, 0 
3
3 

 cos , sin

2
2 

 0, 1
84
sin 0 
7
7 

, sin
 cos

4
4 

 2
2
,


2 
 2
Draft
C. D. Pilmer

Increments
6
2
2 

, sin
 cos

3
3 

 1 3
  ,

 2 2 



 cos , sin 
2
2

 0,1
5
5 

, sin
 cos

6
6 


3 1
, 
 
2
2

 cos  ,
 1, 0 



 cos , sin 
3
3

1 3
 ,

2 2 



 cos , sin 
6
6

 3 1
, 

2
2

sin  
 cos 0,
1, 0 
sin 0 
11
11 

, sin
 cos

6
6 

 3 1
,  

2
 2
7
7 

, sin
 cos

6
6 


3 1
,  
 
2
 2
4
4 

, sin
 cos

3
3 

 1
3
3
3 

  , 

 cos , sin

2
2
2
2 



 0, 1
5
5 

, sin
 cos

3
3 

1
3
 , 

2 
2
We can now use these revised unit circles to answer a variety of questions.
Example 1
Find the exact numerical value of each trigonometric expression. If the answer involves radicals,
then make sure it's in simplest radical form.
4
11
sin
10sin
3



3
6
(a) sin 2   
(b)

5
4
  6 cos
1  4sin
3
3
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Answers:
(a)
 3
sin 2 
 4

 

sin
4
3


 6 cos 
3
3
2

2
2
 
2 
1
6 
2
2
 
4

(b)
3
2
3

1 
3
   
 3 
2  2

1 
3 1
   
 
2  2 3 
1 
3
   

2  6 
3 
3
   

6  6 

11
6
5
1  4sin
3
 1
10   
 2


3
1 4 

 2 
10sin

5
1 2 3

5
1 2 3

1 2 3 1 2 3

5  10 3
1 4 9
5  10 3
1  12
5  10 3
5  10 3

or
11
11

3 3
6
Example 2
For each of the following, we are supplied with a sequence of coterminal angles. Fill in the
blanks.
 7 13
(a) … , _____, ,
,
, _____, _____, …
3 3
3
3
(b) …, _____, _____,  , _____, …
4
Answers:
Previously we learned that coterminal angles are angles or rotations that share the same
terminal arm. When we worked in degrees, we considered sequences of coterminal angles
such as … , -405o, -45o, 315o, 675o, … In this example, and all other sequences of coterminal
angles, we recognized that the values changed by increments of 360o. This should make
sense in that we are making complete revolutions (360o) to get back to the same terminal
arm. Using this logic, a sequence of coterminal angles in radian measure should be changing
by increments of 2 ( 2 is equivalent to 360o).
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C. D. Pilmer
6
.
3
5  7 13 19 25
…,  , ,
,
,
,
, …
3 3 3
3
3
3
(a) Changing by increments of 2 or
(b) Changing by increments of 2 or
…, 
19
11
3 5
…
, 
, 
,
4
4
4
4
Example 3
Solve each of the following equations.
(a) cos x  1
 
 
1
(c) cos  3  x     
6 
2
 
8
.
4
(b) 2sin x  2  0,  3  x  3
(d) 2sin 2 x  3 sin x  0
Answers:
Notice that these equations do not possess degree symbols; this indicates to us that we must
answer in radian measure. For these questions, such as the ones we previously encountered
earlier in this unit, we know that there are multiple solutions. In the past we used the
expression 360k , k I to capture all the coterminal angles. However, at that time we were
working in degrees. Now that we are working in radians, we use the expression 2 k , k I
to capture the coterminal angles.
(a) cos x  1
Use our revised unit circles on the previous two pages to identify all the rotations that
have a cosine of -1.
x    2 k , k I
(b) 2sin x  2  0,  3  x  3
2sin x   2
2
sin x  
2
 5
 4  2 k , k I
x
We cannot leave it here because there was a restriction.
 7  2 k , k I
 4
19
11
3 5 13 21

... ,  4 ,  4 ,  4 , 4 , 4 , 4 , ...
x
 ... ,  17 ,  9 ,   , 7 , 15 , 23 , ...

4
4
4 4
4
4
11
9
3
 5
7
x
, 
, 
,  ,
, and
4
4
4
4 4
4
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C. D. Pilmer
 
 
1
(c) cos  3  x     
6 
2
 
 2
 2 k , k I
   3

3 x    
6   4

 2 k , k I
 3
 2 2
  k , k I
  9 3
x 
6  4 2
  k , k I
 9 3
 2   2
 9  6   3  k , k I


x
 4     2  k , k I
 9 6  3


Start by solving for 3  x   .
6

Divide both sides of the equation by 3.
Add
 4 3  2
 18  18   3  k , k I


x
 8  3   2  k , k I
 18 18  3
 7 2
 18  3  k , k I
x
11  2  k , k I
 18 3

to both sides of the equation.
6
We need a common denominator.
(d) 2sin 2 x  3 sin x  0
Let y  sin x
2 y2  3 y  0


y 2y  3  0
y0
We common factored.
or
2y  3  0
2y  3
y
sin x  0
x  0   k , k I
NSSAL
©2012
or
3
2
sin x 
3
2

 3  2 k , k I
x
 2  2 k , k I
 3
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C. D. Pilmer
Questions
1. For each of the following, we are supplied with a sequence of coterminal angles. Fill in the
blanks.
(a) … ,  ,  , 3 , 5 , _____, _____, _____, …
(b) … , _____,
5 17 29
,
,
, _____, _____, …
6
6
6
(c) …, _____, _____, 
(d) …, _____, _____,

4
, _____, …
5
, _____, …
3
2. State the smallest positive rotation that is coterminal to each rotation below.
(a) 
(c) 

2
7
4
(b) 
(d) 
4
3

6
3. State the largest negative rotation that is coterminal to each rotation below.
(a)

3
(c) 
3
2
(b)
7
6
(d)
5
4
4. Find the exact numerical value of each trigonometric expression. If the answer involves
radicals, then make sure it's in simplest radical form.
7

 2    
(a) 5sin
(b) 7sin 2 
cos
 sin   
4
6
 3   2
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(c) cos
5
 2
cos  
6
 3
 11 
cos  

 6 
(e)

sin
4
NSSAL
©2012
7

  sin
4

 3
(d) 7 cos 2 
 4
7 sin
(f)
90

 4 
  5sin  


 3 
7
6
 
5cos   
 6
Draft
C. D. Pilmer
sin

 
6
(g) 3cos 2    
 4  cos 
6
(i)
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5
 
3  2 cos   
 6
(h)
6
4 cos

4
1
 3 
sin  

 2 
(j)
7
4
4 cos
 6sin
4
3
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5. Solve the following equations.
3
(a) sin x  
2
(b) 2cos x  2  0
(c) sin x  1,  3  x  3
(d) 1  2sin x  0,  2  x  2
 
 
3
(e) sin  2  x    
4  2
 
 
 
2
(f) cos  3  x    
6  2
 
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(g) 2sin 2 x  2 sin x  0,    x  
(h) 2cos2 x  cos x  1
6. Given that f  x   cos2 x and g  x   3cos x  4 , solve for all values of x so that
f  x  g  x .
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Revisiting Questions Using Radian Measure, Part 2
In the previous section, we answered a variety of questions that were primarily concerned with
trigonometric expressions and equations. These questions were "cooked" such that you were
forced to work only with the special rotations (now in radian measure). Obviously we have to go
beyond questions that solely involve the special rotations.
Example 1
Solve sin x  0.8 .
Answer:
1
2
3
, 
, or 
, then we know that we are not
2
2
2
dealing with one of the special rotations. We will have to use our calculator. Since we are
now working in radians, we have to make sure the calculator is in "radian" mode. When we
evaluate sin 1 0.8 , we obtain 0.927 radians. To capture all the coterminal angles, we include
2 k , k I . However, we still have a problem; we know that there is another set of rotations
that also have a sine of 0.8. Since 0.8 is a positive value, we know that the terminal arm for
these rotations must be in the second quadrant. So up to this point, we have the following.
0.927  2 k , k I
x
 ?  2 k , k I
This problem is easily solved. Previously we learned that for questions involving sines, the
other set of rotations in degrees can be found using the " 180   " rule. Now that we are
working in radians, it becomes the "    " rule. Therefore the final answer is the following.
0.927  2 k , k I
x
(Note:   0.927  3.049 )
3.049  2 k , k I
Since the sine is not equal to 0, 1 , 
Example 2
Solve cos x  0.587 .
Answer:
1
2
3
, 
, or 
, then we know that we are not
2
2
2
dealing with one of the special rotations. Using our calculator in "radian" mode, we
determine that cos1  0.587   2.198 . This ultimately gives us only one set of answers.
Since the sine is not equal to 0, 1 , 
2.198  2 k , k I
x
 ?  2 k , k I
This problem is easily solved. Previously we learned that for questions involving cosines,
the other set of rotations in degrees can be found using the " 360   " rule. Now that we are
working in radians, it becomes the "2    " rule. Therefore the final answer is the following.
2.198  2 k , k I
x
(Note: 2  2.198  4.085 )
4.085  2 k , k I
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Example 3
 
2
Solve sin  5  x 
9
 
3

   .
7

Answer:
 
2  
3
sin  5  x 
  
9 
7
 
2  0.443  2 k , k I

5 x 

9   3.585  2 k , k I

2

0.089   k , k I

2 
5
x

2
9 
0.717   k , k I

5
2

0.787  5  k , k I
x
 0.019  2  k , k I

5
 3
Find sin 1    and use the "    " rule.
 7
Divide both sides by 5.
Subtract
2
(approx. 0.698) from both sides.
9
Example 4
Solve 6cos2 x  cos x  2  0 .
Answer:
6cos2 x  cos x  2  0
Let y  cos x
0  6 y2  y  2
(+4)  (-3) = -12
(+4) + (-3) = 1
0  6 y2  4 y  3 y  2
0  (6 y 2  4 y)  (3 y  2)
0  2 y(3 y  2)  1(3 y  2)
0  (3 y  2)(2 y 1)
3y  2  0
or
2
y
or
3
2
cos x  
or
3
 2.301  2 k , k I
x
3.883  2 k , k I
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Factor by Decomposition
2 y 1  0
1
y
2
1
cos x 
2

 3  2 k , k I
x
 5  2 k , k I
 3
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Questions
1. Solve each of the following trigonometric equations. Use exact values when appropriate.
(a) cos x  0.347
(b) 7sin x  3  0
(c) sin x  0.834,  2  x  2
 
2
(d) cos  4  x 
7
 
(e) 5sin 2 x  7sin x  2
(f) 81cos2 x  25  0
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3

  
5

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 
 
2. For g  x   sin  3  x    ,
  12  
(a) find x when g  x   0.729 .
(b) find g  3.5 .
3. Find the zeroes of f  x   7 cos2 3x  5cos3x .
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Trigonometric Identities; The Investigation, Part 1
Up to this point, we only worked with the trigonometric functions y  sin x , y  cos x and
transformations of these two functions. These functions were first classified as periodic
functions because they repeat at regular intervals. We went further to say that they were a
specific type of periodic function called sinusoidal functions because the graphs looked like
symmetrical waves.
y  cos x
y  sin x
There are other trigonometric functions that we need to learn about: y  tan x , y  csc x ,
y  sec x and y  cot x .
tan stands for tangent
 csc stands for cosecant
 sec stands for secant
 cot stands for cotangent
The graphs of these four trigonometric functions are provided below.
y  tan x
y  csc x
y  sec x
y  cot x
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The following questions involve the four new trigonometric functions we graphed on the
previous page.
(a) Are these periodic and sinusoidal functions, or are they strictly periodic functions?
x
(b) In Level IV Academic Math we learned that exponential functions of the form y  ab c have
an asymptote (specifically a horizontal asymptote) at y = 0. An asymptote is a line that a
function approaches but never touches. In the case of exponential functions, the curve
approachedsthe x-axis (y = 0) but never touches it. All four of the new trigonometric
functions possess asymptotes. However, in these cases we are dealing with vertical
asymptotes. Let's consider the function y  tan x . The asymptotes that we can see have the
3


3
following equations: x  
, x   , x  , and x 
(but we know that there are
2
2
2
2
more). We can use the following algebraic expression to describe all the vertical asymptotes
for the trigonometric function y  tan x :
x

2
 k , k  I
(In Degrees: x  90  180k , k I )
For the remaining trigonometric functions, determine the algebraic expression that describes
the vertical asymptotes.
(i)
y  csc x
(ii) y  sec x
(iii) y  cot x
On the next page you are provided with a table containing the special rotations and their
corresponding sines, cosines, tangents, cosecants, secants and cotangents. In answering the
questions that follow, we find the relationship(s) among sin x , cos x , tan x , csc x , sec x , and
cot x . Please note that any number, other than zero, divided by zero is undefined.
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x
sin x
cos x
tan x
csc x
sec x
cot x
0
0
1
0
undefined
1
undefined

6

4

3

2
2
3
3
4
5
6
1
2
3
3
2
2 3
3
3
2
2
3
2
3
2
2
2
1
2
1
2
2
1
3
2 3
3
2
3
3
1
0
undefined
1
undefined
0
 3
2 3
3
-2

0
3
2
2
2
1
2

1
2
2
2
3

2

-1
-1

3
3
0
2
2
undefined
 2


3
3
-1
2 3
3
 3
-1
undefined
We provide hints for the questions that follow.
 Think about reciprocals and quotients.
 Remember that in mathematics, we do not leave radicals in the denominator. When this
occurs, we rationalize the denominator.
(c) Look at the columns for sin x and csc x . What is the relationship between these values (i.e.
how can you use the sines of these rotations to determine the cosecants of those same
rotations)? Use some of the rotations to illustrate the relationship you discovered.
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(d) Look at the columns for cos x and sec x . What is the relationship between these values (i.e.
how can you use the cosines of these rotations to determine the secants of those same
rotations)? Use some of the rotations to illustrate the relationship you discovered.
(e) Look at the columns for tan x and cot x . What is the relationship between these values (i.e.
how can you use the tangents of these rotations to determine the cotangents of those same
rotations)? Use some of the rotations to illustrate the relationship you discovered.
(f) Look at the columns for sin x , cos x and tan x . What is the relationship amongst these
values (i.e. how can you use the sines and cosines of these rotations to determine the tangents
of those same rotations)? Use some of the rotations to illustrate the relationship you
discovered.
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(g) Look at the columns for sin x , cos x and cot x . What is the relationship amongst these
values (i.e. how can you use the sines and cosines of these rotations to determine the
cotangents of those same rotations)? Use some of the rotations to illustrate the relationship
you discovered.
(h) Circle the correct relationships. Of the sixteen presented, eight are correct.
sec x 
1
sin x
tan x 
cos x
sin x
sin x 
1
csc x
sec x 
1
tan x
cot x 
cos x
sin x
csc x 
1
sec x
cot x 
sin x
cos x
csc x 
1
sin x
sec x 
1
cos x
tan x 
csc x
sin x
tan x 
1
cot x
cos x 
1
sec x
cot x 
1
csc x
cot x 
1
tan x
tan x 
sin x
cos x
sin x 
1
cos x
(i) Complete the following table. Use exact values (i.e. no decimal values).
x
7
6
5
4
4
3
3
2
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sin x
1

2

2
2
cos x
3

2
2

2
tan x
102
csc x
sec x
cot x
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(j) The rotations, that have a positive value for their sine, have their terminal arm located in the
first or second quadrants. The rotations, that have a positive value for their cosine, have their
terminal arm located in the first or fourth quadrants. In what quadrants can the terminal arms
for rotations that have a positive tangent be found?
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Trigonometric Identities; The Investigation, Part 2
In the first part of this investigation, you discovered five relationships which are actually referred
to as trigonometric identities (i.e. rules that are true for all rotations). The identities can be
broken down into two categories: reciprocal identities and quotient identities.
Reciprocal Identities
1
csc x 
sin x
sec x 
1
cos x
Quotient Identities
sin x
tan x 
cos x
cot x 
cos x
sin x
cot x 
1
tan x
There are other identities that we are will discover in the second part of this investigation.
Step 1
Let's start by considering the unit circle. Remember that
when we rotate  degrees from (1, 0) about (0, 0), the
coordinates of the image point are  cos  ,sin   . Notice
that a right-angle triangle is formed on the diagram.
Explain how the identity sin 2   cos2   1 can be
obtained from this diagram.
 cos ,sin  
 0,
0, 00 

1, 0 
Step 2
To obtain the next identity, take the identity sin 2   cos2   1 , divide everything in the equation
by cos2  , and simplify the equation using the reciprocal and quotient identities.
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Step 3
To obtain the next identity, take the identity sin 2   cos2   1 , divide everything in the equation
by sin 2  , and simplify the equation using the reciprocal and quotient identities.
Example 1
1

If sin x  and 0  x  , then find:
3
2
(a) csc x
(b) cos x
Answer:
The restriction 0  x 

2
(c) tan x
is supplied so that we only have to consider rotations between and

. This makes the question far easier to answer because we don't have to
2
consider negative values for sines, cosines, tangents, secants, cosecants, and cotangents.
1
(a) csc x 
Translated: cosecant and sine are reciprocals of one another.
sin x
1
If sin x  , then we can conclude that csc x = 3.
3
equal to 0 and
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(b) We discovered that the identity sin 2   cos2   1 shows the relationship between the
sine of a rotation and the cosine of that same rotation. We can use this to solve for cos x .
sin 2 x  cos 2 x  1
2
1
2
   cos x  1
3
1
 cos 2 x  1
9
8
cos 2 x 
9
8
cos x  
9
8
cos x 
9
cos x 
In the second last step we omitted the  because the
question restricted us to rotations between and equal

to 0 and
. Rotations in the first quadrant have
2
cosines that are positive.
2 2
3
(c) From part 1 of the investigation, we discovered that tan x 
sin x
.
cos x
sin x
cos x
1
tan x  3
2 2
3
1 2 2
tan x  
3
3
1
3
tan x  
3 2 2
1
tan x 
2 2
tan x 
tan x 
tan x 
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1
2 2

2
2
2
4
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Question:
3

and 0  x  , then find:
4
2
(a) sec x
(b) sin x
(c) cot x

3 5
and 0  x  , then find:
2
5
(a) sin x
(b) cos x
(c) tan x
1. If cos x 
2. If csc x 
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1

and 0  x  , then find:
2
2
(a) cot x
(b) sec x
3. If tan x 
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(c) cos x
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Trigonometric Identities and Verification Questions
In the last two parts of the investigation, we discovered the following three sets of trigonometric
identities.
Reciprocal Identities
1
csc x 
sin x
sec x 
1
cos x
Quotient Identities
sin x
tan x 
cos x
cot x 
cos x
sin x
Pythagorean Identities
sin 2 x  cos2 x  1
tan 2 x  1  sec2 x
cot x 
1
tan x
1  cot 2 x  csc2 x
The third set of identities is referred to as the Pythagorean Identities because they are derived
from the Pythagorean Theorem. There are other identities that we learn about in this unit, but we
need to become familiar with these ones before we start working with additional identities.
Equations Versus Identities
We have been talking about identities in the last two sections. Although we have used the term
identity repeatedly, some learners may want a formal definition. An identity is an equation that
is true for all values of the variable.
 The equation 2 x  6  8 is not an identity because it is only true when x = 1.
 The equation 2 x  6  2  x  3 is an identity because it is true for all values of x.


The equation sin x  1 is not an identity because it is only true when
x  270  360k , k I
The equation sin 2 x  cos2 x  1 is an identity because it is true for all values of x.
Verification Questions (or Proofs)
In this section, we focus on verification questions. For these questions we prove that one side of
trigonometric equation is equal to the other side of the equation. This is done using the
trigonometric identities listed above. Since these are formal proofs, we must show all our work
such that any learner can follow our logical progression of thought. Many learners find this a
challenging topic because there is not always an obvious way to start answering each question;
you can start down a certain path and even though everything is mathematically correct, it
doesn't take you in the desired direction. It takes time for learners to recognize some of the
subtle patterns that may indicate the appropriate course of action; this only comes with practice,
hard work, and some heavy frontal lobe work. Be patient and remember that you can do this.
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Five Strategies
In this section, we focus on five strategies for completing the verification questions.
1. Express everything in terms of sine and cosine (most overused strategy)
2. Expand
3. Factor
4. Make a common denominator to add or subtract
5. Multiply by the conjugate
With each of these proofs, we ask you to start with one side of the equation (typically the more
complicated side) and apply a number of identities and operations to convert it to the other side
of the equation. We insist that you manipulate only one side of the equation. Why is this so?
Consider the following proofs submitted by learners.
Learner A
Learner A is asked whether the equation 3  x  7   2  x  1  x  23 is an identity.
The learner submits the following.
3  x  7   2  x  1  x  23
3x  21  2 x  2  x  23
x  23  x  23
This learner manipulates only one side of the equation and converts it to the other side of the
equation. This procedure correctly demonstrates that we are dealing with an identity (or that
the original equation is true for all values of x).
Learner B
Learner B is asked whether the equation x2  1  x2  1 is an identity.
The learner submits the following.
x2 1  x2  1
 Learner subtracted x 2 from both sides of the equation.
1  1
2
2
 Learner squared both sides of the equation.
 1  1
11
This learner manipulates both sides of the equation. Since the last step shows the two sides
of the equation being equal to each other, then this learner believes he/she is dealing with an
identity. This is wrong. Try any value for x; you see that the original equation does not
work. If you choose an x-value of 3, the left side of the original equation is equal to 8, and
the right side of the original equation is equal to 10. Eight is not equal to ten; we are not
dealing with an identity. The problem with this proof occurs when the learner squared both
sides of the equation.
Learner B's incorrect submission clearly demonstrates why mathematicians insist that we
manipulate only one side of the equation when completing verification questions.
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Example 1
Verify csc x tan x cos x  1 .
Answer:
In this case, we express everything in terms of sines and cosines. Learners tend to like this
strategy and hence overuse it, making many questions even more challenging.
csc x tan x cos x  1
 1  sin x 


 cos x  1
 sin x  cos x 
sin x cos x
1
sin x cos x
11
We used a reciprocal identity and a quotient identity.
Example 2
Prove 1  sin x 1  sin x   cos2 x .
Answer:
In this case, we multiply the two binomials (i.e. expanding) using the distributive property.
1  sin x 1  sin x   cos2 x
1  sin x  sin x  sin 2 x  cos2 x
1  sin 2 x  cos2 x
cos2 x  cos2 x
If sin 2 x  cos2 x  1 , then cos2 x  1  sin 2 x .
Example 3
Transform one side of the equation sin x  cos2 x sin x  sin3 x to prove that it is equal to the
other side.
Answer:
In this case we factor the left hand side of the equation. Specifically we will common factor
out sin x from both terms.
sin x  cos2 x sin x  sin3 x
sin x 1  cos 2 x   sin 3 x

If sin 2 x  cos2 x  1 , then sin 2 x  1  cos2 x .

sin x sin 2 x  sin 3 x
sin x  sin x
3
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Example 4
Verify the statement
1
 tan x  sec2 x cot x .
tan x
Answer:
We make a common denominator so that we can add the two terms. This is accomplished by
multiplying the numerator and denominator of the second term by tan x .
1
 tan x  sec2 x cot x
tan x
1
 tan x tan x 
2


  sec x cot x
tan x  1
tan x 
1
tan 2 x

 sec2 x cot x
tan x tan x
1  tan 2 x
 sec2 x cot x
tan x
sec2 x
 sec2 x cot x
tan x
 1 
2
sec2 x 
  sec x cot x
 tan x 
2
sec x cot x  sec2 x cot x
Example 5
sin x
Prove
 csc x  cot x .
1  cos x
Answer:
In this case, we multiply the numerator and denominator by the conjugate of 1  cos x .
sin x
 csc x  cot x
1  cos x
sin x
1  cos x

 csc x  cot x
1  cos x 1  cos x
sin x 1  cos x 
 csc x  cot x
1  cos 2 x
sin x 1  cos x 
 csc x  cot x
sin 2 x
1  cos x
 csc x  cot x
sin x
1
cos x

 csc x  cot x
sin x sin x
csc x  cot x  csc x  cot x
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Conjugates are generally used when we are
attempting to create one of the versions of the
Pythagorean identities.
sin 2 x  1  cos 2 x
cos 2 x  1  sin 2 x
tan 2 x  sec 2 x  1
cot 2 x  csc 2 x  1
Notice that in the third line of this
verification, we were able to create the
desired version of the Pythagorean identity.
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For many verification questions, more than one strategy may need to be used and more than one
pathway can be used to complete the proof. Such is the case in Example 6.
Example 6
Verify that sec2 x  cot 2 x  cos2 x   cot 2 x .
Answer A:
sec2 x  cot 2 x  cos 2 x   cot 2 x
sec2 x cot 2 x  sec2 x cos2 x  cot 2 x
2
 1   cos x   1  2
2

 2   
 cos x  cot x
2
2
 cos x   sin x   cos x 
1
 1  cot 2 x
sin 2 x
csc2 x  1  cot 2 x
cot 2 x  cot 2 x
Strategy #2: Expand
Strategy #1: Change to Sines and Cosines
If 1  cot 2 x  csc2 x , then cot 2 x  csc2 x  1 .
Answer B:
sec2 x  cot 2 x  cos 2 x   cot 2 x
sec2 x cot 2 x  sec2 x cos2 x  cot 2 x
2
 1   cos x   1  2
2

 2   
 cos x  cot x
2
2
 cos x   sin x   cos x 
1
 1  cot 2 x
sin 2 x
1
sin 2 x

 cot 2 x
sin 2 x sin 2 x
1  sin 2 x
 cot 2 x
2
sin x
cos 2 x
 cot 2 x
2
sin x
cot 2 x  cot 2 x
Strategy #2: Expand
Strategy #1: Change to Sines and Cosines
Strategy #4: Make a Common Denominator
We are now ready to try answering some of these questions on our own. Complete solutions to
these questions can be found in the answer section of this resource. Please refrain from using
these solutions until the last possible moment. It usually takes time for your brain to become
attuned to answering these types of questions. If you run to the answer key too quickly, you'll
have difficulty mastering the skills needed for these types of problems. The struggle is actually
part of the learning process. Initially we group questions that use the same strategy together so
that you may start to see certain patterns.
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Questions
1. Tylena successfully completed the following proof using trigonometric identities. Her
friend, Anne, has provided written explanations for each step of Tylena's proof. Your
mission is to match each explanation with the appropriate step in the proof.
Tylena has to prove that
tan x
 csc x  cot x
sec x  1
tan x
sec x  1

 csc x  cot x
sec x  1 sec x  1
tan x  sec x  1
 csc x  cot x
sec2 x  1
tan x  sec x  1
 csc x  cot x
tan 2 x
sec x  1
 csc x  cot x
tan x
sec x
1

 csc x  cot x
tan x tan x
1
 1 
sec x 
 csc x  cot x

 tan x  tan x
sec x cot x  cot x  csc x  cot x
 1  cos x 


  cot x  csc x  cot x
 cos x  sin x 
1
 cot x  csc x  cot x
sin x
csc x  cot x  csc x  cot x
Anne's Explanations:
A Break the expression into two
separate expressions with the same
denominator.
C Express part of the equation in terms
of sines and cosines.
E Multiply the numerator and
denominator by the conjugate
G Use a reciprocal identity to change
1
to cot x .
tan x
I Use a reciprocal identity to change
1
to csc x .
sin x
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B
Complete the multiplication using the
distributive property.
D
Both the numerator and denominator
are divisible by tan x .
The cos x in the numerator and
denominator give a quotient of 1.
Use one of the Pythagorean identities
to convert sec2 x  1 to tan 2 x .
F
H
J
114
Dividing by tan x is the same as
1
multiplying by
.
tan x
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C. D. Pilmer
2. Verify each of the following statements. The same strategy is used throughout.
(a) sin x sec x  tan x
(b) sec x cot x sin x  1
(c) sin x cos x cot x  cos2 x
(d) cos2 x tan 2 x  sin 2 x
(e) 5tan 2 x csc x cot x  5sec x
(f) Of the five strategies that we presented, which strategy was used in answering all five
preceding questions?
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3. Prove.
(a) csc x  cos x  tan x   cot x  sec x
(b) 1  cos x 1  cos x   sin 2 x
(c)  tan x  1  sec2 x  2 tan x
(d) sec2 x cot 2 x  cos2 x  csc2 x  1

2

(e)  sin x  cos x    sin x  cos x   2
2
2
(f) Of the five strategies that we presented, which strategy was used in answering all five
preceding questions?
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4. Show that each equation is an identity.
(a) tan x sin 2 x  tan x cos2 x  tan x
(b) sec x  sec x tan 2 x  sec3 x
(c) 4csc x  4csc x cos2 x  4sin x
(d) 3sin x csc2 x  3sin x  3csc x cos2 x
(e) cot 2 x sin 2 x  cot 2 x cos2 x  cos2 x csc2 x
(f) Of the five strategies that we presented, which strategy was used in answering all five
preceding questions?
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5. Verify each identity.
1
(a)
 cos x  tan x sin x
cos x
(c)
(b) cot x 
1
 csc2 x tan x
cot x
1
1

 csc2 x sec2 x
2
2
sin x cos x
(d) Of the five strategies that we presented, which strategy was used in answering all three
preceding questions?
6. Prove each of the following statements.
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(a)
cos x
 sec x  tan x
1  sin x
(b)
1
 csc2 x  cot x csc x
1  cos x
(c)
tan x
 csc x  cot x
sec x  1
(d)
1
 tan x sec x  tan 2 x
csc x  1
(e) Of the five strategies that we presented, which strategy was used in answering all four
preceding questions?
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7. Verify each of the following. Please note that we have not grouped questions that use the
same strategy together. Learners must be able to recognize which strategy/strategies is/are
appropriate for answering each question.
(a) tan 2 x csc2 x  tan x  cos x csc x
(b) csc x  sin x  cos x   1  cot x
(c) csc x tan x sec x  sec2 x
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(d)
120
1
 sin x  cos 2 x csc x
sin x
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(e)
cot x
 csc x tan x  tan x
csc x  1
(g) 4sec x 
4
 4 tan 2 x cos x
sec x
(i) 5cot 2 x sin x tan x  5cos x
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(f) sin x  csc x  sin x   cos2 x
(h) 3cos x  3tan 2 x cos x  3sec x
(j) 1  cot x 1  cot x   csc2 x  2cot x
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2
 2cot 2 x sec x  2cot 2 x
sec x  1
(l)
sec x sin x

 cot x
sin x cos x
(m) 4cot 3 x csc2 x  4cot 3 x  4cot 5 x
(n)
tan x
 csc2 x tan x  csc x
1  cos x
(k)
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Trigonometric Identities and Trigonometric Equations
Trigonometric identities are particularly important when it comes to solving complex
trigonometric equations. The identities often allow us to simplify the equations into more
manageable forms. As stated earlier in this unit, we are only going to work in radian measure.
When it is appropriate (i.e. when dealing with special rotations), supply answers in their exact
values.
Example 1
Solve 2cot x sin x  1 .
Answer:
2cot x sin x  1
 cos x 
2
 sin x  1
 sin x 
2 cos x  1
1
cos x  
2
 2
 3  2 k , k I
x
 4  2 k , k I
 3
We used one of the quotient identities.
Use your knowledge of the unit circle and the
special rotations to find the exact values.
Example 2
Solve csc x  3  0 , where 2  x  3 .
Answer:
csc x  3  0
csc x  3
1
3
sin x
1
sin x 
3
0.340  2 k , k I
x
2.802  2 k , k I
We used one of the reciprocal identities.
We took the reciprocal of both sides of the equation.
Use the sin-1 button on your calculator to find the
principle solution (i.e. 0.340). Make sure your
calculator is in "Radian" mode. The other value,
2.802, is found using the "    " rule.
...,  12.226,  5.943, 0.340, 6.623, 12.906,...
x
 ...,  9.764,  3.481, 2.802, 9.085, 15.368,...
x  5.943,  3.481, 0.340, 2.802, 6.623, and 9.085
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Example 3
 
 
Solve sec  4  x     2
3 
 
Answer:
 
 
sec  4  x     2
3 
 
1
 2
 
 
cos  4  x   
3 
 
 
 
cos  4  x    
3 
 
 
 
cos  4  x    
3 
 
We used one of the reciprocal identities.
1
2
Take the reciprocal of both sides of the equation.
1
2

2
2
Rationalize the denominator.
 
 
2
cos  4  x    
3  2
 

 2 k , k I
   4

4 x    
3   7

 2 k , k I
 4
 1
  k , k I
  16 2
x 
3  7 1
  k , k I
 16 2
    1
  16  3   2  k , k I


x
 7     1  k , k I
 16 3  2
  3 16  1
  48  48   2  k , k I


x
 21  16   1  k , k I

 48
48  2


We have solved for 4  x   at this stage.
3

Divide both sides of the equation by 4.
Add

to both sides of the equation.
3
Create a common denominator to do the addition.
 19 1
 48  2  k , k I
x
 37  1  k , k I
 48 2
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Example 4
 
 
8
If g  x   csc  3  x    , then find x when g  x    .
6 
5
 
Answer:
 
 
g  x   csc  3  x   
6 
 
 
8
 
  csc  3  x   
5
6 
 
 
 
8
csc  3  x     
6 
5
 
1
8

5
 
 
sin  3  x   
6 
 
 
 
5
sin  3  x     
6 
8
 
  0.675  2 k , k I

3 x    
6   3.817  2 k , k I

2

0.225   k , k I

 
3
x 
2
6 
1.272   k , k I

3

 2
 0.225  6   3  k , k I


x
 1.272     2  k , k I

 
6 3
We used one of the reciprocal identities.
Take the reciprocal of both sides of the equation.


We used our calculator to solve for 3  x   .
6

Divide both sides of the equation by 3.
Subtract

(or 0.524) from both sides of the equation.
6
2


0.749

 k , k I

3
x
 0.748  2  k , k I

3
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Example 5
Find the zeroes of f  x   4sin 2 x  12cos x tan x  5 .
Answer:
f  x   4sin 2 x  12cos x tan x  5
0  4sin 2 x 12cos x tan x  5
 sin x 
0  4sin 2 x  12cos x 
5
 cos x 
0  4sin 2 x 12sin x  5
0  4 y  12 y  5
(-10)  (-2) = 20
(-10) + (-2) = -12
0  4 y 2  10 y  2 y  5
0  (4 y 2  10 y )  (2 y  5)
2
Zeroes (or x-intercepts) occur when y (or f(x)) equal 0.
We used one of the reciprocal identities.
Let y  sin x
We factored by decomposition.
0  2 y  2 y  5   1 2 y  5 
0   2 y  5  2 y  1
2y 5  0
or
2y  5
5
y
2
5
2
no solution
sin x 
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or
or
or
2 y 1  0
2y 1
1
y
2
1
sin x 
2

 6  2 k , k I
x
 5  2 k , k I
 6
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Example 6
Solve sec2 x  2sec x  8  0 .
Answer:
sec2 x  2sec x  8  0
Let y  sec x
y2  2 y  8  0
 y  2 y  4  0
y20
y  2
sec x  2
1
 2
cos x
1
cos x  
2
 2
 3  2 k , k I
x
 4  2 k , k I
 3
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We factored by inspection.
or
or
or
or
or
y40
y4
sec x  4
1
4
cos x
1
cos x 
4
1.318  2 k , k I
x
4.965  2 k , k I
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Example 7
If f ( x)  cos 4 x  1 and g  x   sin 2 4 x , find all values of x for which f  x   g  x  .
Answer:
f  x  g  x
cos 4 x  1  sin 2 4 x
 sin 2 4 x  cos 4 x  1  0
 1  cos2 4 x  cos 4 x  1  0


We used one of the Pythagorean identities.
1  cos 2 4 x  cos 4 x  1  0
cos 2 4 x  cos 4 x  0
Let y  cos 4 x
y2  y  0
We common factored.
y  y  1  0
y0
cos 4 x  0
4x 
x
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

2
  k , k I
1
  k , k I
8 4
or
or
or
y 1 0
y  1
cos 4 x  1
4x    2 k , k I
x

1
  k , k I
4 2
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Questions
1. Solve each of the following trigonometric equations. Use exact values where appropriate.
(Note: Most of these questions have been "doctored" so that you must use exact values.)
(a) tan x cos x  
2
2
(c) 2cot x sin x  3  0
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(b) csc x  2  0
(d) 3sec x  4  0 , where 2  x  2
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sec x  2  0 , where 2  x  
(e) 2cos x  sin x csc x  0
(f)
(g) 3csc x  2 3  0
(h) cot x sec x  7
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(i)
sec5x  2
 
 
(k) sec  2  x     1
4 
 
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(j)


csc  x    2
3

(l)
 
 
csc  3  x      2
6 
 
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 
5   9
(m) csc  6  x 
 
8  4
 
1
 
(n) sec   x     5
7 
4
(o) 6cos2 x  3cot x sin x  0
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(p) csc2 x  5csc x  6
(q) 3sin 2 x  10 tan x cos x  6  2
(r)
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2 cot x sec x  csc2 x
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(s) tan 2 x  sec x  5  0
2. Find the zeroes of g  x   csc3x  1 .
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2 3


3. For h  x   sec  x   , then find x when h  x   
.
3
6

4. If f  x   sec2 3x and h  x   10  3sec3x , find x when f  x   h  x  .
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Appendix
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Refresher Reference Materials
1. Determining the Equation of a Sinusoidal Function
Although all sinusoidal functions can be described as transformations of y  sin x or
y  cos x , most people find it easier to describe them just as a transformation of y  cos x .
When determining the equation we have to find four key attributes (equation of the
sinusoidal axis, amplitude, period, coordinates of one maximum) and connect these to four
different transformations (vertical translation, vertical stretch, horizontal stretch, horizontal
translation).
 The equation of the sinusoidal axis is equivalent to the vertical translation.
 The amplitude is equivalent to the vertical stretch.
 The period can be used to determine the horizontal stretch. This is accomplished
using the following formula.
period
HS =
360
 If we are transforming y  cos x , then the x-coordinate of one maximum is
equivalent to the horizontal translation.
Remember that for equations of the form y  k cosax  c   d :
1
VS = k, VT = d, HS = , and HT = c
a
Example
Determine the equation of the
sinusoidal function based its graph.
5
4
3
2
y 1
0
-1 0
5
10
15
20
25
30
35
40
-2
-3
x
Answer:
You need to examine the graph and extract the four key attributes.
VT = 1 (Sinusoidal Axis)
VS = 3 (Amplitude)
period 20
1
HS =


360
360 18
HT = 5 (x-coordinate of a maximum)
Therefore: y  3 cos18x  5  1
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2. Simplify Radicals by Rationalizing the Denominator
There are times when we encounter radical expressions like
2
where the radical is in the
5
denominator of the expression. In mathematics, if a radical expression has a radical in its
denominator it is not considered to be in its most simplified form. We use a process called
rationalizing the denominator to transform radical expressions with one or more radicals in
the denominator to an equivalent expression without radicals in the denominator.
Example 1
2
Simplify
.
5
Answer:
2
5

2

2 5

2 5
5
5

5
5
25
We must rationalize the denominator because we have the 5
in the denominator of the rational expression. We will
5
multiply the expression by
, which is equivalent to 1. This
5
will change the appearance of our expression but not its value
(Multiplying by 1 does not change the value.). In the
denominator when we multiply 5 by 5 , we obtain 25 .
The square root of 25 is 5. Now we have eliminated the radical
in the denominator.
Example 2
Simplify
6
7 3
.
Answer:
6
7 3


6
7 3
6 3
7 9
6 3
21
2 3

7

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
3
3
To rationalize the denominator we will multiply the expression
3
by
, which is equivalent to 1. This will change the
3
appearance of our expression but not its value. By doing this
we obtain 7 9 in the denominator which is equivalent to 21
(7  3).
Notice that the radical expression can be reduced because the
number 3 divides evenly into both the numerator and
denominator.
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3. Solving Quadratic Equations by Factoring
We need to use the zero-factor property (see below) to solve quadratic functions. It is only
applied once the quadratic equation is set equal to zero and put in its factored form.
Zero-Factor Property
If a and b are real numbers and a  b  0 , then a = 0 or b = 0. Translated this means
that if the product of two numbers is 0, then at least one of the numbers must be 0. One
number must be zero, but it is possible that both numbers are 0.
Example 1
Given x 2  3x  30  2 , find x.
Answer:
x 2  3x  30  2
x 2  3x  28  0
x  7x  4  0
x  7  0 or x  4  0
x  7
x4
Set the equation equal to zero by adding 2 to both sides.
We have to do this because the zero-factor property only
applies when the product is 0.
Factor the quadratic equation completely. In this case we
factored by inspection.
We now use the zero-factor property. If the product of
two binomials is 0, then one of the binomials must be
equal to 0. We can now solve the two resulting equations.
We end up with two real roots (i.e. two answers for x).
Example 2
Solve each of the following.
(a) 6a 2  7  15a  7
(c) 12d  4d 2  5
(b) 2  9 x2  30 x  27
(d) 25t 2  49
Answers:
(a) 6a 2  7  15a  7
6a 2  15a  0
3a2a  5  0
Therefore:
3a  0
o
r
3a 0

3 3
a0
(b)
0  9 x 2  30 x  25
0  9 x 2  30 x  25
2a  5  0
2a  5
5
a
2
We used common factoring with
this question.
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2  9 x 2  30 x  27
139
0   3x  5
Therefore:
3x  5  0
3x  5
5
x
3
2
We factored a perfect square
trinomial in this question.
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C. D. Pilmer
(c)
12d  4d 2  5
(d)
25t 2  49  0
0  4d 2  12d  5
-10  (-2) = 20
-10 + (-2) = -12
0  4d 2  10d  2d  5

 5t  7  5t  7   0
Therefore:
5t  7  0
5t  7
7
t
5

0  4d  10d   2d  5
0  2d 2d  5  12d  5
0  2d  52d  1
Therefore:
or
2d  5  0
2d  1  0
2d  5
2d  1
5
1
d
d
2
2
2
25t 2  49
or
5t  7  0
5t  7
7
t
5
We factored a difference of
squares in this question.
We factored by decomposition in
this question.
4. Quadrants of the Cartesian Coordinate System
The Cartesian coordinate system is divided into four quadrants by our two axes. The vertical
axis was typically labeled the y-axis and the horizontal axis was typically labeled the x-axis.
In the first quadrant, both the x-coordinates and y-coordinates are positive numbers. In the
second quadrant, the x-coordinates are negative numbers while the y-coordinates are positive
numbers.
y-axis
5
4
3
2nd quadrant 2 1st quadrant
1
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
x-axis
-1
-2
3rd quadrant -3
4th quadrant
-4
-5
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5. Properties Associated with Triangles
(a) Interior Angles of a Triangle
The interior angles of any triangle add up to 180o.
In the diagram:
ABC  BAC  ACB
 48  25  107
 180
A
B
(b) Equilateral Triangles
All sides are the same length, and all interior angles are
60o.
In the diagram:
AC  AB  BC
ABC  BAC  ACB  60
(c) Isosceles Triangles
Two sides are of equal length, and two interior angles
are equal.
In the diagram:
AC  AB
ABC  ACB
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C
B
A
C
A
B
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The Unit Circle and Special Rotations (In Degrees and Radians)
You are expected to know all the information on this unit circle. You will not be permitted to
bring it in when you are tested on material from this unit.
Downloaded from http://en.wikipedia.org/wiki/File:Unit_circle_angles_color.svg on July 16,
2012. File from Wikipedia Commons with permission from author, Jim.belk, to share.
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List of Trigonometric Identities
Reciprocal Identities
csc x 
1
sin x
sec x 
1
cos x
cot x 
cos x
sin x
cot x 
1
tan x
Quotient Identities
tan x 
sin x
cos x
Pythagorean Identities
sin 2 x  cos2 x  1
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tan 2 x  1  sec2 x
143
1  cot 2 x  csc2 x
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Index
" 180   " Rule, 66
" 360   " Rule, 67
"    " Rule, 94
" 2   " Rule, 94
Conjugate, 24
Cosecant, 98
Cotangent, 98
Coterminal Angles, 5
Pythagorean Identities, 109
Quotient Identities, 104, 109
Radian Measure, 74
Reciprocal Identities, 104, 109
Secant, 98
Special Rotations, 14, 84
Tangent, 98
Terminal Arm, 4
Unit Circle, 4
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Answers
Introductory Activity (pages 1 to 2)
Part 1
opp
hyp
1
sin  
2
  30
sin  
Part 2
y  2sin x
1  2sin x
1
 sin x
2
x  30
Now use the sin-1 button on your calculator.
Now use the sin-1 button on your calculator.
Explanation:
You were provided with the graphs of the functions y  2sin x and y  1 . Notice that these
two functions intersect at multiple points. That means that for the function y  2sin x , there are
multiple values of x that correspond to a y-value of 1. The value of 30o is just one of these
multiple values.
The Unit Circle, Part 1 (pages 3 to 8)
Pattern: Increasing and decreasing by increments of 360o
This pattern should make sense because every time you rotate 360o, you end up
pointing in the same direction.
Fill in the Blanks:
… , -930o, -570o, -210o, 150o, 510o, 870o, …
1. (a) Quadrant: First
Coterminal: -300o
(b) Quadrant: Second
Coterminal: -240o
(c) Quadrant: Fourth
Coterminal: -30o
(d) Quadrant: Third
Coterminal: -135o
2. (a) Quadrant: Fourth
Coterminal: 315o
(c) Quadrant: Second
Coterminal: 160o
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(b) Quadrant: First
Coterminal: 40o
(d) Quadrant: Third
Coterminal: 240o
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3
(a)
(b)
(c)
(d)
… , -270o, 90o, 450o, 810o, 1170o, …
… , -600o, -240o, 120o, 480o, 840o, 1200o, …
… , -390o, -30o, 330o, 690o, 1050o, …
… , -1305o, -945o, -585o, -225o, 135o, …
The Unit Circle, Part 2 (pages 9 to 13)
1. (a) R25 1,0    0.906, 0.423
First Quadrant
(b) R107 1,0    0.292,  0.956 
Third Quadrant
(c) R310 1,0    0.643,  0.766 
Fourth Quadrant
(d) R234 1,0    0.588, 0.809 
Second Quadrant
(e) R400 1,0    0.766,0.643
First Quadrant
(f) R395 1,0    0.819, 0.574 
Fourth Quadrant
2. (a) (0.174, 0.985)
(c) (0.423, -0.906)
(b) (-0.643, 0.766)
(d) (-0.342, -0.940)
3. If we rotate 180o on our unit circle, we will move from the point (1, 0) to the image point
(-1, 0). The coordinates of this image point would normally be found using
 cos180,sin180 . Therefore we can conclude that cos180  1 and sin180  0 . Since
the question asked us to find cos180 , we can logically conclude that it is equal to -1.
4. (a) Third and Fourth
(b) First and Second
(c) Second and Third
5. (a) 1
(c) 1
(b) -1
(d) -1
6. We know that coterminal angles share the same terminal arm, therefore it is logical that they
also share the same image point. I have decided to confirm this using the coterminal angles
-20o, 340o, and 700o.
R20 1, 0    cos  20  ,sin  20  
R340 1, 0    cos 340,sin 340 
R20 1, 0    0.940, 0.342 
R340 1, 0    0.940, 0.342 
R700 1, 0    cos 700,sin 700 
R700 1, 0    0.940, 0.342 
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7. (a)
(c)
(e)
(g)
(i)
E
G
J
F
A
(b)
(d)
(f)
(h)
(j)
D
B
C
I
H
Special Rotations (pages 14 to 23)
Part 1 Investigation Questions
(a)
(b)
(c)
(d)
360o
45o
(ii) isosceles
(iii) x  y
(e) I've solved for y here, but you just as easily could have solved for x (same procedure).
a 2  b2  c 2
In (d) we concluded that x  y
x 2  y 2  12
y 2  y 2  12
2 y2  1
1
y2 
2
1
2
y
y
1
2
Ignore the negative answer; our rotation is in the first quadrant
1
2
1
y
2
y
1
2

2
2
2
y
2
y
Rationalize the denominator.
(f) Since I solved for y in (e), I'm now solving for x in (f).
2
x
2
 2 2
,
(g) (i) R45 1, 0   

 2 2 
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
2 2
,
(h) (i) R135 1, 0    

 2 2 

2
2
,
(ii) R225 1, 0    

2 
 2
 2
2
,
(iii) R315 1, 0   

2 
 2

2
2
,
(i) (i) R135 1, 0    

2 
 2
 2
2
,
(ii) R45 1, 0   

2 
 2
 2 2
,
(iii) R405 1, 0   

2
2 


2 2
,
(iv) R225 1, 0    

2
2 

(Reason: Second Quadrant: x's negative, y's positive)
(Reason: Third Quadrant: x's negative, y's negative)
(Reason: Fourth Quadrant: x's positive, y's negative)
(Reason: Coterminal with 225o)
(Reason: Coterminal with 315o)
(Reason: Coterminal with 45o)
(Reason: Coterminal with 135o)
Part 2 Investigation Questions
(a) 360o
(b) 60o
(c) (i) 30o
(ii) 60o
(d) 60o
(e) (iii)equilateral
(f) (i) 1 unit
1
(g)
2
(h) a 2  b2  c 2
x 2  y 2  12
2
1
x2     1
2
1
x2   1
4
3
x2 
4
3
x
4
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In (g), we discovered that y equals
148
1
.
2
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x
x
x
3
4
3
Ignore the negative answer; our rotation is in the first quadrant
4
3
2
 3 1
, 
(i) (ii) R30 1, 0   
 2 2
1 3
(j) (i) R60 1, 0    ,

2 2 
 1 3
(ii) R120 1, 0     ,

 2 2 

3 1
, 
(iii) R150 1, 0    
 2 2
1 3
(k) (i) R300 1, 0    ,

2 2 

3 1
, 
(ii) R210 1, 0    
 2 2
 3 1
, 
(iii) R330 1, 0   
 2 2
 1 3
(iv) R480 1, 0     ,

 2 2 
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1. (a) 45o Increments
 cos 90,
 0,1
 cos135,
sin 90 
sin135 
 cos 45,

2 2
,
 

 2 2 
 cos180,
 1, 0 
 cos 0,
1, 0 
sin 225 

2
2
,
 

2 
 2
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 2 2
,


 2 2 
sin180 
 cos 225,
sin 45 
 cos 315,
 cos 270,
 0, 1
150
sin 270 
sin 0 
sin 315 
 2
2
,


2 
 2
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(b) 60o Increments
 cos120,
sin120 
 cos 90,
 0,1
 1 3
  ,

 2 2 
 cos150, sin150 
sin 90 
 cos 60,
1 3
 ,

2 2 
 cos 30,

3 1
, 
 
 2 2
 cos180,
 1, 0 
sin 30 
 3 1
, 

 2 2
sin180 
 cos 210,
 cos 0,
1, 0 
sin 210 
 cos 330,

3 1
,  
 
2
 2
 cos 240, sin 240 
 1
3
  , 

2 
 2
2
2
3
(c) sin 300  
2
2
(e) sin 315  
2
1
(g) cos(120)  
2
2. (a) cos135  
NSSAL
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sin 60 
sin 0 
sin 330 
 3 1
,  

2
 2
 cos 300, sin 300 
1
3
,


 cos 270, sin 270  2 2 
 0, 1
(b) sin150 
1
2
(d) cos 270  0
3
2
2
(h) sin 405 
2
(f) cos(30) 
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Evaluating Trigonometric Expressions (pages 24 to 35)
3
3
6
(d)
12
5
(g)
3
1. (a)
7 1
6
27  9 2
(c)
7
6 2 4
(e)
7
24  4 3
(g)
11
7 5 7 3
(i)
2
30  2 6
(k)
73
74 2
(m)
17
17  7 3
(o)
71
2 10  6  2 15  3
(q)
17
2. (a)
5 2
2
7 2
(e)
6
6
(h)
5
15
5
4 6
(f)
15
2 14
(i)
3
(b)
(c)
5 3 5
2
15 2  10
(d)
14
8 6 2 2
(f)
47
9 5  15
(h)
8
(b)
(j) 6 3  6 2
(l) 4  10
1  5 2
7
2 3 2 2  6
(p)
2
21  7  2 14  2
(r)
62
(n)
3
4
3
(c)
4
(b) 
(e) 1
(f)
3. (a)
(g)
(i)
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3 2
2
22
2
(d)
(h)
(j)
152
3 2
2
9 6

4
21

4
3 3 1
2
6 2
4
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9  2 2
4
(k)
(m) 1
3 2
2
5 6
(q)
21
3 6
(s)
6
(o)
(u) 5 2  5
20 2  5 3
29
18  3 3
(y)
11
(w)
(l)
(n)
(p)
(r)
(t)
(v)
57 3
2
1
3
3
3
3 2

10
2 6 3 2
6
14 3  7
11
(x) 3  2 2
(z)
2 2 1
7
Introduction to Trig Equations: The Graphical Approach, Part 1 (pages 36 to 43)
1. (a) Words: Our first point of intersection occurs when x equals 90o. To find the xcoordinates of the other points of intersection, we move forwards and
backwards from 90o at increments of 180o.
Sequence: …, -450o, -270o, -90o, 90o, 270o, 450o, …
Algebraic Expression: x  90  180k , k I
(b) Words: Our first point of intersection occurs when x equals 180o. To find the xcoordinates of the other points of intersection, we move forwards and
backwards from 180o at increments of 360o.
Sequence: …, -900o, -540o, -180o, 180o, 540o, 900o, …
Algebraic Expression: x  180  360k , k I
(c) Words: Our first point of intersection occurs when x equals 0o. To find the xcoordinates of the other points of intersection, we move forwards and
backwards from 0o at increments of 360o.
Sequence: …, -1080o, -720o, -360o, 0o, 360o, 720o, 1080o, …
Algebraic Expression: x  0  360k , k I
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2. (a) Words: Our first point of intersection occurs when x equals 135o. To find the xcoordinates of the other points of intersection, we move forwards and
backwards from 135o at increments of 180o.
Sequence: …, -405o, -225o, -45o, 135o, 315o, 495o, …
Algebraic Expression: x  135  180k , k I
(b) Words: Our first point of intersection occurs when x equals 45o. To find the xcoordinates of the other points of intersection, we move forwards and
backwards from 45o at increments of 180o.
Sequence: …, -495o, -315o, -135o, 45o, 225o, 405o, …
Algebraic Expression: x  45  180k , k I
(c) Words: Our first point of intersection occurs when x equals 0o. To find the xcoordinates of the other points of intersection, we move forwards and
backwards from 0o at increments of 90o.
Sequence: …, -270o, -180o, -90o, 90o, 180o, 270o,…
Algebraic Expression: x  0  90k , k I
3. (a) Sequence: …, -300o, -180o, -60o, 60o, 180o, 300o, …
Algebraic Expression: x  60  120k , k I
(b) Sequence: …, -360o, -240o, -120o, 0o, 120o, 240o, 360o, …
Algebraic Expression: x  0  120k , k I
(c) Sequence: …, -150o, -90o, -30o, 30o, 90o, 150o, …
Algebraic Expression: x  30  60k , k I
4. (a) Sequence: …, -36o, -16o, 4o, 24o, 44o, …
Algebraic Expression: x  4  20k , k I
(b) Sequence: …, -21o, -11o, -1o, 9o, 19o, 29o, …
Algebraic Expression: x  9  10k , k I
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5. Match the appropriate algebraic expressions and sequences. Please note that I stands for
integers (…, -2, -1, 0, 1, 2, …), and W stands for whole numbers (0, 1, 2, 3, …).
(a)
10  60k , k I
(c)
30o, 70o, 110o, 150o, 190o, …
(b)
20  30k , k I
(f)
…, -340o, -160o, 20o, 200o, 380o, …
(c)
30  40k , kW
(h)
15o, 40o, 65o, 90o, 115o, …
(d)
40  60k , kW
(b)
…, -40o, -10o, 20o, 50o, 80o, …
(e)
70  120k , k I
(j)
15o, 30o, 45o, 60o, 75o, …
(f)
20  180k , k I
(d)
40o, 100o, 160o, 220o, 280o, …
(g)
5  30k , kW
(e)
…, -170o, -50o, 70o, 190o, 310o, …
(h)
15  25k , kW
(g)
5o, 35o, 65o, 95o, 125o, …
(i)
10  35k , k I
(a)
…, -110o, -50o, 10o, 70o, 130o, …
(j)
0  15k , kW
(i)
…, -60o, -25o, 10o, 45o, 80o, …
6. No solution
Introduction to Trig Equations: The Graphical Approach, Part 2 (pages 44 to 46)
210  360k , k I
1. (a) x  
330  360k , k I
 30  360k , k I
(b) x  
150  360k , k I
 5  60k , k I
2. (a) x  
25  60k , k I
35  60k , k I
(b) x  
55  60k , k I
3. (a) x  100  120k , k I
(c) x  10  60k , k I
20  120k , k I
(b) x  
60  120k , k I
 0  120k , k I
(d) x  
80  120k , k I
Trigonometric Equations, Part 1 (pages 47 to 51)
1. (a) x  270  360k , k I
(c) x  0  360k , k I
NSSAL
©2012
 30  360k , k I
(b) x  
330  360k , k I
(d) x  0  180k , k I
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 45  360k , k I
(e) x  
135  360k , k I
(g) x  -480, -420, -120, and -60
 45  360k , k I
(f) x  
315  360k , k I
(h) x  -240, -120, 120, and 240
210  360k , k I
2. (a) x  
330  360k , k I
(b) Graph (ii)
3. (a) x  -270, -90, and 90
(b) Graph (iii)
Trigonometric Equations, Part 2 (pages 52 to 57)
1.
2.
3.
Hints:
Inspection
cos x  5 or cos x  1
x  180  360k , k I
Difference of Squares
sin x  1 or sin x  1
Answer can be simplified.
x  90  180k , k I
Common
x  0  180k , k I
sin x  0 or sin x  
4.
Final Answer:
1
2
210  360k , k I
x
330  360k , k I
Set equation equal to zero.
Decomposition
120  360k , k I
x
240  360k , k I
1
cos x  4 or cos x  
2
5.
6.
Difference of Squares
Answer can be simplified.
7.
Common
8.
9.
Inspection
Decomposition
x  0  360k , k I
Common
Rationalize the denominator.
10.
x  0  180k , k I
Set equation equal to zero.
Common Factor
NSSAL
©2012
225  360k , k I
x
315  360k , k I
 60  180k , k I
x
120  180k , k I
x  90  180k , k I
x  0  360k , k I
x  270  360k , k I
x  0, 360, and 720
x  -360, -315, -225, -180, 0, 45,
135, 180, and 360
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x  -690, -570, -510, -390, -330,
-210, -150, and -30
11.
Difference of Squares
12.
Common
Rationalize the denominator.
x  90  180k , k I
150  360k , k I
x
210  360k , k I
x  -270°, -210, -150, -90 and 90
Trigonometric Equations, Part 3 (pages 58 to 64)
1. (a) x  36  90k , k I
28  60k , k I
(c) x  
38  60k , k I
15  30k , k I
(e) x  
25  30k , k I
(b) x  57  72k , k I
10  24k , k I
(d) x  
16  24k , k I
105  720k , k I
(f) x  
225  720k , k I
 9  15k , k I
2. x  
11.5  15k , k I
3. x  -100, -40, 20, 80, and 140
4. x  -770, -410, 310, and 670
5. Hints:
h  6cos  20  t  9     7
Answer:
12  18k , kW
t
 6  18k , kW
NSSAL
©2012
12  18k , k I
t
24  18k , k I
Note that the second set of answers had to be changed slightly
(24 changed to 6) to ensure that the first solution within the
whole numbers is captured.
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C. D. Pilmer
Trigonometric Equations, Part 4 (pages 65 to 73)
Make sure your calculator is in "degree" mode.
 39.1  360k , k I
1. (a) x  
140.9  360k , k I
 99.8  360k , k I
(c) x  
260.2  360k , k I
 75.5  360k , k I
(e) x  
284.5  360k , k I
x  90  180k , k I
 48.2  360k , k I
(g) x  
311.8  360k , k I
16.2  180k , k I
(i) x  
 66.3  180k , k I
 67.7  360k , k I
(b) x  
292.3  360k , k I
47.7  360k , k I
(d) x  
 227.7  360k , k I
 48.6  360k , k I
(f) x  
131.4  360k , k I
x  270  360k , k I
17.1  36k , k I
(h) x  
23.0  36k , k I
 1.4  40k , k I
(j) x  
36.6  40k , k I
 38.7  360k , k I
2. x  
141.3  360k , k I
x  0  180k , k I
 11.5  180k , k I
3. x  
11.5  180k , k I
16.6  72k , k I
4. x  
49.4  72k , k I
5. Hints:
d  21cos  20  t  3  
 6.5  18k , kW
Answer: t  
17.5  18k , kW
6. Hints:
s  3cos  6000  t  0.03    7
0.038  0.06k , kW
Answer: t  
0.022  0.06k , kW
NSSAL
©2012
 6.5  18k , k I
t
17.5  18k , k I
0.038  0.06k , k I
t
0.082  0.06k , k I
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C. D. Pilmer
Introduction to Radian Measure (pages 74 to 83)
1. (a) 4

(c)
6
5
(e)
4
13
(g)
18
2. (a)
(c)
(e)
(g)
3.
900o
240o
-450o
-80o
(b) 2
7
(d)
6
5
(f) 
3
52
(h) 
45
(b)
(d)
(f)
(h)
330o
-315o
420o
74.5o
2
3
4. 15 cm
5. (a)
2 5
,
,
3
6
(b)
3
5 3
, ,
,
4
4
2
6. (a)

6
(b)

4
7
7 

7. (a) R7 1, 0    cos
,sin

4
4 

4
 2
2
 
,

2 
 2
7
7 

(b) R7 1, 0    cos
,sin

6
6 

6

3 1
  
,  
2
 2
Revisiting Questions Using Radian Measure, Part 1 (pages 84 to 93)
1. (a) … ,  ,  , 3 , 5 , 7 , 9 , 11 , …
7 5 17 29 41 53
(b) … , 
,
,
,
,
,
,…
6
6
6
6
6
6
17
9
 7
(c) …, 
, 
,  ,
, …
4
4
4 4
7  5 11
(d) …, 
, , ,
, …
3
3 3 3
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2
(c)
(b)

4
3. (a) 
(c)
2
3
11
(d)
6
3
2
(a)

2
7
3
5
6
3
(d) 
4
(b) 
Hint(s):
4.
Answer
(a)
 1  3 
5    

 2   2 
(b)
 3
7 
  1
2



3  1  
2
 
      

 2  2   2 
Make a common denominator.
(c)
(d)
(e)
(f)
(g)
NSSAL
©2012

2

5 3
4
21
1
or  5
4
4
32 2
4
2

 3
2
7  
  5 

 2 
 2 
3
2
2
2
Rationalize the denominator.
 1
7 
 2
 3
5

 2 
Rationalize the denominator.
1
2
 2
2
3 
 
2
3


2
Rationalize the denominator.
Make a common denominator.
75 3
2
6
2

7 3
15
92 3
6
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C. D. Pilmer
(h)
(i)
(j)
5.
Hint(s):
6
 2
4
 1
 2 
Rationalize the denominator using the conjugate.
5
 3
3  2

 2 
Rationalize the denominator using the conjugate.
1
 2 
3
4
  6 

 2   2 
Rationalize the denominator using the conjugate.
Answer
12 2  6
7
15  5 3
6
2 2  3 3
19
(a)
Hint(s):
None provided.
Answer
 4
 3  2 k , k I
x
 5  2 k , k I
 3
(b)
None provided.
(c)
x

 4  2 k , k I
x
 7  2 k , k I
 4
5

3
x   ,  , and
2
2
2
(d)
(e)
(f)
NSSAL
©2012
3
 2 k , k I
2

 6  2 k , k I
x
 5  2 k , k I
 6
x

 2 k , k I
   3

2 x    
4   2

 2 k , k I
 3
 2
 2 k , k I
   3

3 x    
6   4

 2 k , k I
 3
11
7 
5
, 
, , and
6
6 6
6
 5
 12   k , k I
x
 7   k , k I
 12
  2
 12  3  k , k I
x
 5  2  k , k I
 12 3
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C. D. Pilmer
(g)
(h)
Hint(s):
Common Factor
x  0   k , k I
Answer
 5
 4  2 k , k I
x
 7  2 k , k I
 4
Set equation equal to zero.
Factor by decomposition
1
cos x  1 or cos x  
2
x   , 
3

,  , 0, and 
4
4
x  0  2 k , k I
 2
 3  2 k , k I
x
 4  2 k , k I
 3
6. Hints: Set equal to zero, factor by inspection, cos x  1 or cos x  4
Answer: x    2 k , k I
Revisiting Questions Using Radian Measure, Part 2 (pages 94 to 97)
1.216  2 k , k I
1. (a) x  
5.067  2 k , k I
0.443  2 k , k I
(b) x  
 3.585  2 k , k I
0.986  2 k , k I
(c) Hints: x  
2.156  2 k , k I
Due to the restriction, we want values between or approximately equal to -6.28 and
6.28.
Answer: x = -5.297, -4.127, 0.986, and 2.156
2  2.214  2 k , k I

(d) Hint: 4  x 

7  4.069  2 k , k I

1

1.451  2  k , k I
Answer: x  
1.915  1  k , k I

2
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©2012
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(e) Hints: Factor by Decomposition, sin x 

2
or sin x  1
5
 2 k , k I
2
0.412  2 k , k I
x
2.730  2 k , k I
Answer: x 
(f) Hints: Factoring Difference of Squares, cos x 
0.982   k , k I
Answer: x  
 5.301   k , k I
2.160   k , k I
x
 4.123   k , k I
Answer can be simplified to:
5
5
or cos x  
9
9
0.982   k , k I
x
2.160   k , k I
  0.817  2 k , k I

2. (a) Hint: 3  x    
 12  2.325  2 k , k I
2

0.534  3  k , k I
Answer: x  
1.037  2  k , k I

3
(b) -0.286
3. Hints: Common Factoring, cos3x  0 or cos 3x 

5
, Solve for 3x before solving for x.
7
1
  k , k I
6 3
2

0.258  3  k , k I
x
1.836  2  k , k I

3
Answer: x 
Trigonometric Identities; The Investigation, Part 1 (pages 98 to 103)
(a) periodic
(b) (i)
NSSAL
©2012
x  0   k , k I
(ii) x 

2
  k , k I
163
(iii) x  0   k , k I
Draft
C. D. Pilmer
(c) csc x 
1
sin x
e.g. csc
(d) sec x 

3

1
sin

3

1
3
2
2
2
3 2 3
 1
 1




2
3
3
3
3
3
3
2
1
cos x
e.g. sec

3
1
1
2
2
2
2 2
 2 


 1   



 2
  1  

4 cos 3
2 
2
2
2
2
2



4
2
1
tan x
2
1
1
1
3
3
e.g. cot





2  3
3
3
3
3
tan
3
(e) cot x 
(f) tan x 
sin x
cos x
3
2
3
4  2  1
e.g. tan

3
4 cos 
2

4
2
sin
(g) cot x 
cos x
sin x

3

6  2  3 1  32  2 3  3
e.g. cot 
1
6 sin 
2 2
2 1
2
6
2
cos
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164
Draft
C. D. Pilmer
(h) sec x 
1

sin x
tan x 
cos x

sin x
sin x 
1

csc x
sec x 
1

tan x
cot x 
cos x

sin x
csc x 
1

sec x
cot x 
sin x

cos x
csc x 
1

sin x
sec x 
1

cos x
tan x 
csc x

sin x
tan x 
1

cot x
cos x 
1

sec x
cot x 
1

csc x
cot x 
1

tan x
tan x 
sin x

cos x
sin x 
1

cos x
(i)
x
7
6
5
4
4
3
3
2
tan x
csc x
3
3
-2
2
2
3

2
cos x
3

2
2

2
1

2
1
 2
-1
0
undefined
sin x
1

2

3

sec x
2 3

3
cot x
 2
1
2 3
3
-2
3
3
-1
undefined
0
3
(j) first and third quadrants
Trigonometric Identities; The Investigation, Part 2 (pages 104 to 108)
Step 1
Since we are dealing with a unit circle, we know that the radius is equal to one. This radius also
represents the hypotenuse of our right triangle. So at this point we know that the hypotenuse of
our right triangle is 1 and the legs are cos  and sin  . We also know that the Pythagorean
Theorem ( a 2  b2  c 2 ) describes the relationship between the three sides of a right triangle.
a 2  b2  c 2
 sin     cos 
2
2
sin 2   cos2   1
NSSAL
©2012
 12
(The Desired Identity)
165
Draft
C. D. Pilmer
Step 2
sin 2   cos2   1
sin 2  cos 2 
1


2
2
cos  cos  cos 2 
sin x
sin 2 x
Note: If tan x 
, then tan 2 x 
cos x
cos 2 x
1
12
1
If sec x 
, then sec2 x 
or
2
cos x
cos 2 x
cos x
(New Identity)
tan 2   1  sec2 
Step 3
sin 2   cos2   1
sin 2  cos 2 
1


2
2
sin  sin  sin 2 
cos x
cos 2 x
Note: If cot x 
, then cot 2 x 
sin 2 x
sin x
1
12
1
If csc x 
, then csc2 x 
or
2
sin x
sin 2 x
sin x
(New Identity)
1  cot 2   csc2 
1. (a) sec x 
2. (a) sin x 
4
3
5
3
3. (a) cot x  2
NSSAL
©2012
(b) sin x 
7
4
(c) cot x 
3 7
7
(b) cos x 
2
3
(c) tan x 
5
2
(c) cos x 
2 5
5
(b) sec x 
5
2
166
Draft
C. D. Pilmer
Trigonometric Identities and Verification Questions (pages 109 to 122)
1.
tan x
sec x  1

 csc x  cot x
sec x  1 sec x  1
tan x  sec x  1
 csc x  cot x
sec2 x  1
tan x  sec x  1
 csc x  cot x
tan 2 x
sec x  1
 csc x  cot x
tan x
sec x
1

 csc x  cot x
tan x tan x
1
 1 
sec x 
 csc x  cot x

 tan x  tan x
E
B
H
D
A
J
sec x cot x  cot x  csc x  cot x
G
 1  cos x 


  cot x  csc x  cot x
 cos x  sin x 
1
 cot x  csc x  cot x
sin x
C
F
csc x  cot x  csc x  cot x
I
With the remaining questions in this section we have only presented one of many acceptable
proofs.
2.
(a)
sin x sec x  tan x
 1 
sin x 
  tan x
 cos x 
sin x
 tan x
cos x
tan x  tan x
(b)
sec x cot x sin x  1
 1   cos x 


 sin x  1
 cos x   sin x 
cos x sin x
1
cos x sin x
11
(c)
sin x cos x cot x  cos2 x
 cos x 
2
sin x cos x 
  cos x
 sin x 
sin x cos 2 x
 cos 2 x
sin x
cos 2 x  cos 2 x
(d)
cos2 x tan 2 x  sin 2 x
 sin 2 x 
2
cos 2 x 
  sin x
2
 cos x 
NSSAL
©2012
sin 2 x  sin 2 x
167
Draft
C. D. Pilmer
(e)
5 tan 2 x csc x cot x  5sec x
(f)
Express everything in terms of sines
and cosines.
(b)
1  cos x 1  cos x   sin 2 x
 sin 2 x   1   cos x 
5


  5sec x
2
 cos x   sin x   sin x 
 sin 2 x cos x 
5
  5sec x
2
2
cos
x
sin
x


 1 
5
  5sec x
 cos x 
5sec x  5sec x
3.
(a)
(c)
csc x  cos x  tan x   cot x  sec x
csc x cos x  csc x tan s  "
1  cos x  cos x  cos 2 x  "
 1 
 1  sin x 

 cos x  

 "
 sin x 
 sin x  cos x 
cos x
1

"
sin x cos x
cot x  sec x  cot x  sec x
1  cos 2 x  "
sin 2 x  sin 2 x
 tan x  1  sec2 x  2 tan x
 tan x  1 tan x  1  "
(d)
2

sec 2 x cot 2 x  sec 2 x cos 2 x  "
2
 1   cos x   1 



 cos x  "
2
2
2
 cos x   sin x   cos x 
1
1  "
sin 2 x
csc 2 x  1  csc 2 x  1
tan 2 x  tan x  tan x  1  "
tan 2 x  1  2 tan x  "
sec 2 x  2 tan x  sec 2 x  2 tan x
(e)

sec 2 x cot 2 x  cos 2 x  csc 2 x  1
 sin x  cos x    sin x  cos x   2
 sin x  cos x  sin x  cos x    sin x  cos x  sin x  cos x   2
2
2
sin 2 x  sin x cos x  sin x cos x  cos 2 x  sin 2 x  sin x cos x  sin x cos x  cos 2 x  2
sin 2 x  cos 2 x  sin 2 x  cos 2 x  2
11  2
22
(f)
NSSAL
©2012
Expand (i.e. Multiply through using the distributive property)
168
Draft
C. D. Pilmer
4.
(a)
tan x sin 2 x  tan x cos 2 x  tan x
(b)
tan x 1  tan x


sec x  sec x   sec
tan x  tan x
sec3 x  sec3 x


sec x 1  tan 2 x  sec3 x
tan x sin 2 x  cos 2 x  tan x
(c)
2
4 csc x  4 csc x cos 2 x  4sin x
(d)
3
x
3sin x csc 2 x  3sin x  3csc x cos 2 x


4 csc x  sin x   4sin x
3sin x csc 2 x  1  3csc x cos 2 x
 1  2
4
 sin x  4sin x
 sin x 
4sin x  4sin x
 cos 2 x 
2
3sin x 
  3csc x cos x
2
 sin x 
3cos 2 x
 3csc x cos 2 x
sin x
 1  2
2
3
 cos x  3csc x cos x
 sin x 
3csc x cos 2 x  3csc x cos 2 x

4 csc x 1  cos 2 x  4sin x

3sin x cot 2 x  3csc x cos 2 x
2
(e)
sec x  sec x tan 2 x  sec3 x
cot 2 x sin 2 x  cot 2 x cos 2 x  cos 2 x csc 2 x


cot 2 x sin 2 x  cos 2 x  cos 2 x csc 2 x
cot 2 x 1  cos 2 x csc 2 x
cot 2 x  cos 2 x csc 2 x
 cos 2 x 
2
2

  cos x csc x
2
 sin x 
 1 
cos 2 x  2   cos 2 x csc 2 x
 sin x 
cos 2 x csc 2 x  cos 2 x csc 2 x
(f)
NSSAL
©2012
Factor, specifically common factor
169
Draft
C. D. Pilmer
5.
(a)
1
 cos x  tan x sin x
cos x
1
cos x cos x


 tan x sin x
cos x
1
cos x
1
cos 2 x

 tan x sin x
cos x cos x
1  cos 2 x
 tan x sin x
cos x
sin 2 x
 tan x sin x
cos x
 sin x 

 sin x  tan x sin x
 cos x 
tan x sin x  tan x sin x
(b)
cot x 
(c)
1
1

 csc 2 x sec 2 x
2
2
sin x cos x
1
cos 2 x
1
sin 2 x



"
sin 2 x cos 2 x cos 2 x sin 2 x
cos 2 x
sin 2 x

"
sin 2 x cos 2 x sin 2 x cos 2 x
cos 2 x  sin 2 x
"
sin 2 x cos 2 x
1
"
2
sin x cos 2 x
 1  1 
 2 
 "
2
 sin x   cos x 
csc 2 x sec 2 x  csc 2 x sec 2 x
(d)
Create a common denominator to
complete the addition or subtraction.
NSSAL
©2012
170
1
 csc 2 x tan x
cot x
cot x cot x
1


 csc 2 x tan x
1
cot x cot x
2
cot x
1

 csc 2 x tan x
cot x cot x
cot 2 x  1
 csc 2 x tan x
cot x
csc 2 x
 csc 2 x tan x
cot x
 1 
2
csc 2 x 
  csc x tan x
cot
x


2
csc x tan x  csc 2 x tan x
Draft
C. D. Pilmer
6.
(a)
cos x
 sec x  tan x
1  sin x
cos x 1  sin x

 sec x  tan x
1  sin x 1  sin x
cos x 1  sin x 
 sec x  tan x
1  sin 2 x
cos x 1  sin x 
 sec x  tan x
cos 2 x
1  sin x
 sec x  tan x
cos x
1
sin x

 sec x  tan x
cos x cos x
sec x  tan x  sec x  tan x
(c)
tan x
 cot x  csc x
sec x  1
tan x
sec x  1

 cot x  csc x
sec x  1 sec x  1
tan x  sec x  1
 cot x  csc x
sec 2 x  1
tan x  sec x  1
 cot x  csc x
tan 2 x
sec x  1
 csc x  cot x
tan x
sec x
1

 csc x  cot x
tan x tan x
 1 

 sec x  cot x  csc x  cot x
 tan x 
cot x sec x  cot x  csc x  cot x
(b)
1
 tan x sec x  tan 2 x
csc x  1
1
csc x  1

"
csc x  1 csc x  1
csc x  1
"
csc 2 x  1
csc x  1
"
cot 2 x
csc x
1

"
2
cot x cot 2 x
 1 
2
 2  csc x  tan x  "
 cot x 
tan 2 x csc x  tan 2 x  "
 cos x   1 


  cot x  csc x  cot x
 sin x   cos x 
1
 cot x  csc x  cot x
sin x
csc x  cot x  csc x  cot x
(e)
NSSAL
©2012
1
 csc 2 x  cot x csc x
1  cos x
1
1  cos x

"
1  cos x 1  cos x
1  cos x
"
1  cos 2 x
1  cos x
"
sin 2 x
1
cos x
 2 "
2
sin x sin x
 cos x   1 
csc 2 x  

 "
 sin x   sin x 
csc 2 x  cot x csc x  csc 2 x  cot x csc x
 sin 2 x   1 
2


  tan x  "
2
 cos x   sin x 
sin x
 tan 2 x  "
2
cos x
 sin x   1 
2


  tan x  "
 cos x   cos x 
tan x sec x  tan 2 x  tan x sec x  tan 2 x
Multiply the numerator and denominator by the conjugate
171
Draft
C. D. Pilmer
7.
(a)
tan 2 x csc 2 x  tan x  cos x csc x
(b)


tan x  cot x   cos x csc x
csc x sin x  csc x cos x  "
tan x csc 2 x  1  cos x csc x
 1 
 1 

 sin x  
 cos x  "
 sin x 
 sin x 
cos x
1
"
sin x
1  cot x  1  cot x
2
2
 sin x   cos x 

  cos x csc x


2
 cos x   sin x 
cos x
 cos x csc x
sin x
 1 
cos x 
  cos x csc x
 sin x 
cos x csc x  cos x csc x
(c)
csc x tan x sec x  sec 2 x
(d)
 1  sin x  1 
2



  sec x
sin
x
cos
x
cos
x




1
 sec 2 x
2
cos x
sec 2 x  sec 2 x
NSSAL
©2012
csc x  sin x  cos x   1  cot x
172
1
 sin x  cos 2 x csc x
sin x
1
sin x sin x


 cos 2 x csc x
sin x
1
sin x
2
1
sin x

 cos 2 x csc x
sin x sin x
1  sin 2 x
 cos 2 x csc x
sin x
cos 2 x
 cos 2 x csc x
sin x
 1 
2
cos 2 x 
  cos x csc x
 sin x 
2
cos x csc x  cos 2 x csc x
Draft
C. D. Pilmer
(e)
(f)
cot x
 csc x tan x  tan x
csc x  1
cot x
csc x  1

"
csc x  1 csc x  1
cot x  csc x  1
"
csc 2 x  1
cot x  csc x  1
"
cot 2 x
csc x  1
"
cot x
1 
 csc x  1 
 "
 cot x 
 csc x  1 tan x  "
sin x  csc x  sin x   cos 2 x
sin x csc x  sin 2 x  cos 2 x
 1 
2
2
sin x 
  sin x  cos x
 sin x 
1  sin 2 x  cos 2 x
cos 2 x  cos 2 x
csc x tan x  tan x  csc x tan x  tan x
(g)
4
 4 tan 2 x cos x
sec x
4sec x sec x
4


 4 tan 2 x cos x
1
sec x sec x
4sec 2 x
4

 4 tan 2 x cos x
sec x
sec x
2
4sec x  4
 4 tan 2 x cos x
sec x
4 sec 2 x  1
 4 tan 2 x cos x
sec x
4 tan 2 x
 4 tan 2 x cos x
sec x
 1 
2
4 tan 2 x 
  4 tan x cos x
 sec x 
2
4 tan x cos x  4 tan 2 x cos x
(h)
5cot 2 x sin x tan x  5cos x
(j)
4sec x 

(i)


3cos x  sec x   3sec x
3cos x 1  tan 2 x  3sec x
2
 1 
3cos x 
  3sec x
2
 cos x 
 1 
3
  3sec x
 cos x 
3sec x  3sec x

 cos 2 x 
 sin x 
5
 sin x 
  5cos x
2
 cos x 
 sin x 
5cos x  5cos x
3cos x  3 tan 2 x cos x  3sec x
1  cot x 1  cot x   csc2 x  2 cot x
1  cot x  cot x  cot 2 x  "
1  2 cot x  cot 2 x  "
1  cot 2 x  2 cot x  "
csc2 x  2 cot x  csc 2 x  2 cot x
NSSAL
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C. D. Pilmer
(k)
2
 2 cot 2 x sec x  2 cot 2 x
sec x  1
2
sec x  1

"
sec x  1 sec x  1
2  sec x  1
"
sec 2 x  1
2sec x  2
"
tan 2 x
1
 2sec x  2   2   "
 tan x 
 2sec x  2  cot 2 x  "
(l)
sec x sin x

 cot x
sin x cos x
sec x cos x sin x sin x



 cot x
sin x cos x cos x sin x
sec x cos x
sin 2 x

 cot x
sin x cos x sin x cos x
sec x cos x  sin 2 x
 cot x
sin x cos x
 1 
2

 cos x  sin x
 cos x 
 cot x
sin x cos x
1  sin 2 x
 cot x
sin x cos x
cos 2 x
 cot x
sin x cos x
cos x
 cot x
sin x
cot x  cot x
(n)
tan x
 csc 2 x tan x  csc x
1  cos x
tan x 1  cos x

"
1  cos x 1  cos x
tan x 1  cos x 
"
1  cos 2 x
tan x  tan x cos x
"
sin 2 x
 sin x 
tan x  
 cos x
 cos x 
"
sin 2 x
tan x  sin x
"
sin 2 x
tan x sin x

"
sin 2 x sin 2 x
1
 1 
"
 2  tan x 
sin x
 sin x 
2 cot 2 x sec x  2 cot 2 x  "
(m)
4 cot 3 x csc 2 x  4 cot 3 x  4 cot 5 x


x  cot x   4 cot
4 cot 3 x csc 2 x  1  4 cot 5 x
4 cot 3
2
5
x
4 cot 5 x  4 cot 5 x
csc 2 x tan x  csc x  csc 2 x tan x  csc x
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C. D. Pilmer
Trigonometric Identities and Trigonometric Equations (pages 123 to 135)
 5
 4  2 k , k I
1. (a) x  
 7  2 k , k I
 4
 7
 6  2 k , k I
(b) x  
11  2 k , k I
 6
 5
 6  2 k , k I
x

(c)

 7  2 k , k I
 6
(d) x  3.864,  2.419, 2.419, and 3.864
 2
 3  2 k , k I
(e) x  
 5  2 k , k I
 3
(f)

 3  2 k , k I
(g) x  
 2  2 k , k I
 3
0.143  2 k , k I
(h) x  
2.999  2 k , k I
(i)
 2 2
 15  5  k , k I
x
 4  2  k , k I
 15 5
(k) x 
3
  k , k I
4
1

2.040  3  k , k I
(m) x  
2.410  1  k , k I

3
NSSAL
©2012
x
5
3
3
, 
, and
4
4
4
 5
 6  2 k , k I
(j) x  
 3  2 k , k I
 2
(l)
 2
 4  3  k , k I
x
 5  2  k , k I
 12 3
 6.639  8 k , k I
(n) x  
17.595  8 k , k I
175
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C. D. Pilmer
(o) Hints: Reciprocal Identity, Common Factor, cos x  0 or 

1
2
  k , k I
2
 2
Answer:
 3  2 k , k I
x
 4  2 k , k I
 3
x
(p) Hints: Factor by Inspection, sin x  
1
1
or 
2
3
 7
 6  2 k , k I
x
11  2 k , k I
Answer:
 6
0.340  2 k , k I
x
 3.481  2 k , k I
(q) Hints: Reciprocal Identity, Factor by Decomposition, sin x 
0.730  2 k , k I
Answer: x  
2.412  2 k , k I
2
or  4
3
(r) Hints: Common Factor, Rationalize Denominator, sin x  0 or
x  0   k , k I
2
2

 4  2 k , k I
Answer:
x
 3  2 k , k I
 4
(s) Hints: Pythagorean Identity, Factor by Inspection, cos x  
1
1
or
2
3
 2
 3  2 k , k I
x
 4  2 k , k I
Answer:
 3
1.231  2 k , k I
x
5.052  2 k , k I
NSSAL
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C. D. Pilmer
2. x 

2
  k , k I
6 3
   2 k , k I

3. x   4
 3  2 k , k I
4. Hints: Factor by Inspection, cos 3x 
 2
 9  3  k , k I
x
 5  2  k , k I
 9 3
Answer:
2

0.591  3  k , k I
x
1.504  2  k , k I

3
NSSAL
©2012
1
1
or 
2
5
177
Draft
C. D. Pilmer