Commission on Higher Education
in collaboration with the Philippine Normal University
TEACHING GUIDE FOR SENIOR HIGH SCHOOL
Precalculus
CORE SUBJECT
This Teaching Guide was collaboratively developed and reviewed by educators from public
and private schools, colleges, and universities. We encourage teachers and other education
stakeholders to email their feedback, comments, and recommendations to the Commission on
Higher Education, K to 12 Transition Program Management Unit - Senior High School
Support Team at k12@ched.gov.ph. We value your feedback and recommendations.
Published by the Commission on Higher Education, 2016
Chairperson: Patricia B. Licuanan, Ph.D.
Commission on Higher Education
K to 12 Transition Program Management Unit
Office Address: 4th Floor, Commission on Higher Education,
C.P. Garcia Ave., Diliman, Quezon City
Telefax: (02) 441-1143 / E-mail Address: k12@ched.gov.ph
DEVELOPMENT TEAM
Team Leader: Dr. Ian June L. Garces
Writers: Dr. Jerico B. Bacani, Dr. Richard B. Eden,
Mr. Glenn Rey A. Estrada, Dr. Flordeliza F. Francisco,
Mr. Mark Anthony J. Vidallo
Technical Editors: Dr. Maria Alva Q. Aberin,
Dr. Flordeliza F. Francisco, Dr. Reginaldo M. Marcelo
Copy Reader: Naomi L. Tupas
Cover Artists: Paolo Kurtis N. Tan, Renan U. Ortiz
CONSULTANTS
THIS PROJECT WAS DEVELOPED WITH THE PHILIPPINE NORMAL UNIVERSITY.
University President: Ester B. Ogena, Ph.D.
VP for Academics: Ma. Antoinette C. Montealegre, Ph.D.
VP for University Relations & Advancement: Rosemarievic V. Diaz, Ph.D.
Ma. Cynthia Rose B. Bautista, Ph.D., CHED
Bienvenido F. Nebres, S.J., Ph.D., Ateneo de Manila University
Carmela C. Oracion, Ph.D., Ateneo de Manila University
Minella C. Alarcon, Ph.D., CHED
Gareth Price, Sheffield Hallam University
Stuart Bevins, Ph.D., Sheffield Hallam University
SENIOR HIGH SCHOOL SUPPORT TEAM
CHED K TO 12 TRANSITION PROGRAM MANAGEMENT UNIT
Program Director: Karol Mark R. Yee
Lead for Senior High School Support: Gerson M. Abesamis
Lead for Policy Advocacy and Communications: Averill M. Pizarro
Course Development Officers:
Danie Son D. Gonzalvo, John Carlo P. Fernando
Teacher Training Officers:
Ma. Theresa C. Carlos, Mylene E. Dones
Monitoring and Evaluation Officer: Robert Adrian N. Daulat
Administrative Officers: Ma. Leana Paula B. Bato,
Kevin Ross D. Nera, Allison A. Danao, Ayhen Loisse B. Dalena
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Table of Contents
Introduction
1
DepEd Curriculum Guide for Precalculus
2
Unit 1:
9
Analytic Geometry (19 one-hour sessions)
Lesson 1.1: Introduction to Conic Sections and Circles . . . . . . . . 10
1.1.1: An Overview of Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.1.2: Definition and Equation of a Circle . . . . . . . . . . . . . . . . . . . . . . . 11
1.1.3: More Properties of Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.1.4: Situational Problems Involving Circles. . . . . . . . . . . . . . . . . . . . 16
Lesson 1.2: Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
1.2.1: Definition and Equation of a Parabola . . . . . . . . . . . . . . . . . . . . 24
1.2.2: More Properties of Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
1.2.3: Situational Problems Involving Parabolas . . . . . . . . . . . . . . . . 32
Lesson 1.3: Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
1.3.1: Definition and Equation of an Ellipse . . . . . . . . . . . . . . . . . . . . . 36
1.3.2: More Properties of Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
1.3.3: Situational Problems Involving Ellipses . . . . . . . . . . . . . . . . . . . 44
Lesson 1.4: Hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
1.4.1: Definition and Equation of a Hyperbola . . . . . . . . . . . . . . . . . . 50
1.4.2: More Properties of Hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
1.4.3: Situational Problems Involving Hyperbolas . . . . . . . . . . . . . . . 58
Lesson 1.5: More Problems on Conic Sections . . . . . . . . . . . . . . . . 63
1.5.1: Identifying the Conic Section by Inspection . . . . . . . . . . . . . . . 63
1.5.2: Problems Involving Di↵erent Conic Sections . . . . . . . . . . . . . . 65
iii
Lesson 1.6: Systems of Nonlinear Equations . . . . . . . . . . . . . . . . . . 70
1.6.1: Review of Techniques in Solving Systems of Linear
Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
1.6.2: Solving Systems of Equations Using Substitution . . . . . . . . . 72
1.6.3: Solving Systems of Equations Using Elimination . . . . . . . . . . 75
1.6.4: Applications of Systems of Nonlinear Equations . . . . . . . . . . 79
Unit 2:
Mathematical Induction (10 one-hour sessions)
84
Lesson 2.1: Review of Sequences and Series . . . . . . . . . . . . . . . . . . . 85
Lesson 2.2: Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
2.2.1: Writing and Evaluating Sums in Sigma Notation . . . . . . . . . 90
2.2.2: Properties of Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
Lesson 2.3: Mathematical Induction . . . . . . . . . . . . . . . . . . . . . . . . . . 99
2.3.1: Proving Summation Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
2.3.2: Proving Divisibility Statements . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
? 2.3.3: Proving Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
Lesson 2.4: The Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
2.4.1: Pascal’s Triangle and the Concept of Combination . . . . . . . . 112
2.4.2: The Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
2.4.3: Terms of a Binomial Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . 118
? 2.4.4: Approximation and Combination Identities . . . . . . . . . . . . . . . 120
Unit 3:
Trigonometry (29 one-hour sessions)
125
Lesson 3.1: Angles in a Unit Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
3.1.1: Angle Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
3.1.2: Coterminal Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
3.1.3: Arc Length and Area of a Sector . . . . . . . . . . . . . . . . . . . . . . . . . 132
Lesson 3.2: Circular Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
3.2.1: Circular Functions on Real Numbers . . . . . . . . . . . . . . . . . . . . . 139
3.2.2: Reference Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
Lesson 3.3: Graphs of Circular Functions and Situational
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
3.3.1: Graphs of y = sin x and y = cos x . . . . . . . . . . . . . . . . . . . . . . . . 148
3.3.2: Graphs of y = a sin bx and y = a cos bx . . . . . . . . . . . . . . . . . . . 150
3.3.3: Graphs of y = a sin b(x c) + d and
y = a cos b(x c) + d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
3.3.4: Graphs of Cosecant and Secant Functions . . . . . . . . . . . . . . . . 159
3.3.5: Graphs of Tangent and Cotangent Functions . . . . . . . . . . . . . 163
3.3.6: Simple Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
Lesson 3.4: Fundamental Trigonometric Identities . . . . . . . . . . . . . 180
3.4.1: Domain of an Expression or Equation . . . . . . . . . . . . . . . . . . . . 180
3.4.2: Identity and Conditional Equation . . . . . . . . . . . . . . . . . . . . . . . 182
3.4.3: The Fundamental Trigonometric Identities . . . . . . . . . . . . . . . 184
3.4.4: Proving Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . 187
Lesson 3.5: Sum and Di↵erence Identities . . . . . . . . . . . . . . . . . . . . . 193
3.5.1: The Cosine Di↵erence and Sum Identities . . . . . . . . . . . . . . . . 193
3.5.2: The Cofunction Identities and the Sine Sum and
Di↵erence Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
3.5.3: The Tangent Sum and Di↵erence Identities . . . . . . . . . . . . . . . 200
Lesson 3.6: Double-Angle and Half-Angle Identities . . . . . . . . . . . 208
3.6.1: Double-Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
3.6.2: Half-Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212
Lesson 3.7: Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . 219
3.7.1: Inverse Sine Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220
3.7.2: Inverse Cosine Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225
3.7.3: Inverse Tangent Function and the Remaining Inverse
Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
Lesson 3.8: Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 242
3.8.1: Solutions of a Trigonometric Equation . . . . . . . . . . . . . . . . . . . . 243
3.8.2: Equations with One Term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247
3.8.3: Equations with Two or More Terms . . . . . . . . . . . . . . . . . . . . . . 250
Lesson 3.9: Polar Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . 260
3.9.1: Polar Coordinates of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261
3.9.2: From Polar to Rectangular, and Vice Versa . . . . . . . . . . . . . . . 266
3.9.3: Basic Polar Graphs and Applications . . . . . . . . . . . . . . . . . . . . . 269
References
281
Biographical Notes
282
Introduction
As the Commission supports DepEd’s implementation of Senior High School (SHS), it upholds the vision
and mission of the K to 12 program, stated in Section 2 of Republic Act 10533, or the Enhanced Basic
Education Act of 2013, that “every graduate of basic education be an empowered individual, through a
program rooted on...the competence to engage in work and be productive, the ability to coexist in fruitful
harmony with local and global communities, the capability to engage in creative and critical thinking,
and the capacity and willingness to transform others and oneself.”
To accomplish this, the Commission partnered with the Philippine Normal University (PNU), the
National Center for Teacher Education, to develop Teaching Guides for Courses of SHS. Together with
PNU, this Teaching Guide was studied and reviewed by education and pedagogy experts, and was
enhanced with appropriate methodologies and strategies.
Furthermore, the Commission believes that teachers are the most important partners in attaining this
goal. Incorporated in this Teaching Guide is a framework that will guide them in creating lessons and
assessment tools, support them in facilitating activities and questions, and assist them towards deeper
content areas and competencies. Thus, the introduction of the SHS for SHS Framework.
The SHS for SHS Framework
The SHS for SHS Framework, which stands for “Saysay-Husay-Sarili for Senior High School,” is at the
core of this book. The lessons, which combine high-quality content with flexible elements to
accommodate diversity of teachers and environments, promote these three fundamental concepts:
SAYSAY: MEANING
HUSAY: MASTERY
SARILI: OWNERSHIP
Why is this important?
How will I deeply understand this?
What can I do with this?
Through this Teaching Guide,
teachers will be able to
facilitate an understanding of
the value of the lessons, for
each learner to fully engage in
the content on both the
cognitive and affective levels.
Given that developing mastery
goes beyond memorization,
teachers should also aim for deep
understanding of the subject
matter where they lead learners
to analyze and synthesize
knowledge.
When teachers empower
learners to take ownership of
their learning, they develop
independence and selfdirection, learning about both
the subject matter and
themselves.
1
2
Semester: First Semester
No. of Hours/ Semester: 80 hours/ semester
Pre-requisite (if needed):
key concepts of
conic sections and
systems of
nonlinear
equations
The learners
demonstrate an
understanding
of...
CONTENT
STANDARDS
model situations
appropriately and solve
problems accurately using
conic sections and systems
of nonlinear equations
The learners shall be able
to...
PERFORMANCE
STANDARDS
graph a circle in a rectangular coordinate system
4.
5.
6.
7.
8.
9.
10.
11.
12.
define a parabola
determine the standard form of equation of a parabola
graph a parabola in a rectangular coordinate system
define an ellipse
determine the standard form of equation of an ellipse
graph an ellipse in a rectangular coordinate system
define a hyperbola
determine the standard form of equation of a hyperbola
determine the standard form of equation of a circle
3.
2.
illustrate the different types of conic sections: parabola, ellipse,
circle, hyperbola, and degenerate cases.***
define a circle.
LEARNING COMPETENCIES
1.
The learners...
K to 12 Senior High School STEM Specialized Subject – Pre-Calculus December 2013
Analytic
Geometry
CONTENT
Page 1 of 4
STEM_PC11AG-Ia-5
STEM_PC11AG-Ib-1
STEM_PC11AG-Ib-2
STEM_PC11AG-Ic-1
STEM_PC11AG-Ic-2
STEM_PC11AG-Ic-3
STEM_PC11AG-Id-1
STEM_PC11AG-Id-2
STEM_PC11AG-Ia-4
STEM_PC11AG-Ia-3
STEM_PC11AG-Ia-2
STEM_PC11AG-Ia-1
CODE
Subject Description: At the end of the course, the students must be able to apply concepts and solve problems involving conic sections, systems of nonlinear equations,
series and mathematical induction, circular and trigonometric functions, trigonometric identities, and polar coordinate system.
Grade: 11
Core Subject Title: Pre-Calculus
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
3
key concepts of
series and
mathematical
induction and the
Binomial
Theorem.
CONTENT
STANDARDS
keenly observe and
investigate patterns, and
formulate appropriate
mathematical statements
and prove them using
mathematical induction
and/or Binomial Theorem.
PERFORMANCE
STANDARDS
recognize the equation and important characteristics of the
different types of conic sections
solves situational problems involving conic sections
illustrate systems of nonlinear equations
determine the solutions of systems of nonlinear equations using
techniques such as substitution, elimination, and graphing***
solve situational problems involving systems
of nonlinear equations
14.
15.
16.
17.
differentiate a series from a sequence
use the sigma notation to represent a series
illustrate the Principle of Mathematical Induction
apply mathematical induction in proving identities
illustrate Pascal’s Triangle in the expansion of 𝑥 + 𝑦 𝑛 for small
positive integral values of 𝑛
prove the Binomial Theorem
determine any term of 𝑥 + 𝑦 𝑛 , where 𝑛 is a positive integer,
without expanding
solve problems using mathematical induction and the Binomial
Theorem
2.
3.
4.
5.
6.
9.
7.
8.
illustrate a series
1.
18.
graph a hyperbola in a rectangular coordinate system
LEARNING COMPETENCIES
13.
K to 12 Senior High School STEM Specialized Subject – Pre-Calculus December 2013
Series and
Mathematical
Induction
CONTENT
STEM_PC11SMI-Ih-1
STEM_PC11AG-Ig-2
STEM_PC11AG-If-g-1
STEM_PC11AG-If-1
STEM_PC11AG-Ie-2
STEM_PC11AG-Ie-1
STEM_PC11AG-Id-3
CODE
Page 2 of 4
STEM_PC11SMI-Ij-2
STEM_PC11SMI-Ij-1
STEM_PC11SMI-Ii-3
STEM_PC11SMI-Ii-2
STEM_PC11SMI-Ih-2
STEM_PC11SMI-Ih-3
STEM_PC11SMI-Ih-4
STEM_PC11SMI-Ih-i-1
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
4
CONTENT
STANDARDS
key concepts of
circular functions,
trigonometric
identities, inverse
trigonometric
functions, and
the polar
coordinate
system
4. formulate and solve
accurately situational
problems involving the
polar coordinate system
3. formulate and solve
accurately situational
problems involving
appropriate trigonometric
functions
2. apply appropriate
trigonometric identities in
solving situational
problems
PERFORMANCE
STANDARDS
1. formulate and solve
accurately situational
problems involving
circular functions
illustrate the different circular functions
uses reference angles to find exact values of circular functions
determine the domain and range of the different circular functions
graph the six circular functions (a) amplitude, (b) period, and (c)
phase shift
solve problems involving circular functions
determine whether an equation is an identity or a conditional
equation
derive the fundamental trigonometric identities
derive trigonometric identities involving sum and difference of
angles
derive the double and half-angle formulas
simplify trigonometric expressions
prove other trigonometric identities
solve situational problems involving trigonometric identities
illustrate the domain and range of the inverse trigonometric
functions.
evaluate an inverse trigonometric expression.
solve trigonometric equations.
solve situational problems involving inverse trigonometric
functions and trigonometric equations
locate points in polar coordinate system
convert the coordinates of a point from rectangular to polar
systems and vice versa
solve situational problems involving polar coordinate system
3.
4.
5.
6.
7.
22.
20.
21.
17.
18.
19.
12.
13.
14.
15.
16.
10.
11.
8.
9.
illustrate angles in standard position and coterminal angles
2.
LEARNING COMPETENCIES
illustrate the unit circle and the relationship between the linear
and angular measures of a central angle in a unit circle
convert degree measure to radian measure and vice versa
1.
K to 12 Senior High School STEM Specialized Subject – Pre-Calculus December 2013
***Suggestion for ICT-enhanced lesson when available and where appropriate
Trigonometry
CONTENT
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
Page 3 of 4
STEM_PC11T-IIj-3
STEM_PC11T-IIj-2
STEM_PC11T-IIj-1
STEM_PC11T-IIi-2
STEM_PC11T-IIh-2
STEM_PC11T-IIh-i-1
STEM_PC11T-IIh-1
STEM_PC11T-IIf-1
STEM_PC11T-IIf-2
STEM_PC11T-IIf-g-1
STEM_PC11T-IIg-2
STEM_PC11T-IIe-3
STEM_PC11T-IIe-2
STEM_PC11T-IIe-1
STEM_PC11T-IId-2
STEM_PC11T-IIc-d-1
STEM_PC11T-IIb-2
STEM_PC11T-IIc-1
STEM_PC11T-IIb-1
STEM_PC11T-IIa-3
STEM_PC11T-IIa-1
STEM_PC11T-IIa-2
CODE
5
Competency
Week
Quarter
Domain/Content/
Component/ Topic
Grade Level
illustrate the different types
of conic sections: parabola,
ellipse, circle, hyperbola,
and degenerate cases
Week one
First Quarter
Analytic Geometry
Grade 11
Science, Technology,
Engineering and Mathematics
Pre-Calculus
SAMPLE
K to 12 Senior High School STEM Specialized Subject – Pre-Calculus December 2013
Arabic Number
*Put a hyphen (-) in
between letters to
indicate more than a
specific week
Lowercase
Letter/s
*Zero if no specific
quarter
Roman Numeral
Uppercase
Letter/s
First Entry
Learning Area and
Strand/ Subject or
Specialization
LEGEND
1
-
a
I
-
STEM_PC11AG
DOMAIN/ COMPONENT
Trigonometry
Series and Mathematical Induction
Analytic Geometry
Sample: STEM_PC11AG-Ia-1
Code Book Legend
K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
Page 4 of 4
T
SMI
AG
CODE
About this Teaching Guide
The Precalculus course bridges Basic Mathematics and Calculus. This course completes the
foundational knowledge on Algebra, Geometry, and Trigonometry of students who are
planning to take courses in the STEM track. It provides them with conceptual understanding
and computational skills that are crucial for Basic Calculus and future STEM courses.
Based on the Curriculum Guide for Precalculus of the Department of Education, the primary
aim of this Teaching Guide is to give Math teachers adequate stand-alone material that can be
used for each session of the Grade 11 Precalculus course.
The Guide is divided into three units: Analytic Geometry, Summation Notation and
Mathematical Induction, and Trigonometry. Each unit is composed of lessons that bring
together related learning competencies in the unit. Each lesson is further divided into sublessons that focus on one or two competencies for effective teaching and learning. Each sublesson is designed for a one-hour session, but the teachers have the option to extend the time
allotment to one-and-a-half hours for some sub-lessons.
Each sub-lesson ends with a Seatwork / Homework, which consists of exercises related to the
topic being discussed in the sub-lesson. As the title suggests, these exercises can be done in
school (if time permits) or at home. Moreover, at the end of each lesson is a set of exercises
(simply tagged as Exercises) that can be used for short quizzes and long exams. Answers,
solutions, or hints to most items in Seatwork / Homework and Exercises are provided to guide
the teachers as they solve them.
Some items in this Guide are marked with a star. A starred sub-lesson is optional and it is
suggested that these be taken only if time permits. A starred example or exercise requires the
use of a calculator.
To further guide the teachers, Teaching Notes are provided on the margins. These notes
include simple recall of basic definitions and theorems, suggested teaching methods,
alternative answers to some exercises, quick approaches and techniques in solving particular
problems, and common errors committed by students.
We hope that Precalculus teachers will find this Teaching Guide helpful and convenient to use.
We encourage the teachers to study this Guide carefully and solve the exercises themselves.
Although great effort has been put to this Guide for technical correctness and precision, any
mistake found and reported to the Team is a gain for other teachers. Thank you for your
cooperation.
6
The Parts of the Teaching Guide
This Teaching Guide is mapped and aligned to the DepEd SHS Curriculum, designed to be highly usable
for teachers. It contains classroom activities and pedagogical notes, and integrated with innovative
pedagogies. All of these elements are presented in the following parts:
1. INTRODUCTION
•
Highlight key concepts and identify the essential questions
•
Show the big picture
•
Connect and/or review prerequisite knowledge
•
Clearly communicate learning competencies and objectives
•
Motivate through applications and connections to real-life
2. MOTIVATION
•
Give local examples and applications
•
Engage in a game or movement activity
•
Provide a hands-on/laboratory activity
•
Connect to a real-life problem
3. INSTRUCTION/DELIVERY
•
Give a demonstration/lecture/simulation/hands-on activity
•
Show step-by-step solutions to sample problems
•
Give applications of the theory
•
Connect to a real-life problem if applicable
4. PRACTICE
•
Provide easy-medium-hard questions
•
Give time for hands-on unguided classroom work and discovery
•
Use formative assessment to give feedback
5. ENRICHMENT
•
Provide additional examples and applications
•
Introduce extensions or generalisations of concepts
•
Engage in reflection questions
•
Encourage analysis through higher order thinking prompts
•
Allow pair/small group discussions
•
Summarize and synthesize the learnings
6. EVALUATION
•
Supply a diverse question bank for written work and exercises
•
Provide alternative formats for student work: written homework, journal, portfolio, group/
individual projects, student-directed research project
7
On DepEd Functional Skills and CHED’s College Readiness Standards
As Higher Education Institutions (HEIs) welcome the graduates of the Senior High School program, it is
of paramount importance to align Functional Skills set by DepEd with the College Readiness Standards
stated by CHED.
The DepEd articulated a set of 21st century skills that should be embedded in the SHS curriculum across
various subjects and tracks. These skills are desired outcomes that K to 12 graduates should possess in
order to proceed to either higher education, employment, entrepreneurship, or middle-level skills
development.
On the other hand, the Commission declared the College Readiness Standards that consist of the
combination of knowledge, skills, and reflective thinking necessary to participate and succeed - without
remediation - in entry-level undergraduate courses in college.
The alignment of both standards, shown below, is also presented in this Teaching Guide - prepares
Senior High School graduates to the revised college curriculum which will initially be implemented by
AY 2018-2019.
College Readiness Standards Foundational Skills
DepEd Functional Skills
Produce all forms of texts (written, oral, visual, digital) based on:
1. Solid grounding on Philippine experience and culture;
2. An understanding of the self, community, and nation;
3. Application of critical and creative thinking and doing processes;
4. Competency in formulating ideas/arguments logically, scientifically,
and creatively; and
5. Clear appreciation of one’s responsibility as a citizen of a multicultural
Philippines and a diverse world;
Visual and information literacies
Media literacy
Critical thinking and problem solving skills
Creativity
Initiative and self-direction
Systematically apply knowledge, understanding, theory, and skills
for the development of the self, local, and global communities using
prior learning, inquiry, and experimentation
Global awareness
Scientific and economic literacy
Curiosity
Critical thinking and problem solving skills
Risk taking
Flexibility and adaptability
Initiative and self-direction
Work comfortably with relevant technologies and develop
adaptations and innovations for significant use in local and global
communities;
Global awareness
Media literacy
Technological literacy
Creativity
Flexibility and adaptability
Productivity and accountability
Communicate with local and global communities with proficiency,
orally, in writing, and through new technologies of communication;
Global awareness
Multicultural literacy
Collaboration and interpersonal skills
Social and cross-cultural skills
Leadership and responsibility
Interact meaningfully in a social setting and contribute to the
fulfilment of individual and shared goals, respecting the
fundamental humanity of all persons and the diversity of groups
and communities
Media literacy
Multicultural literacy
Global awareness
Collaboration and interpersonal skills
Social and cross-cultural skills
Leadership and responsibility
Ethical, moral, and spiritual values
8
Unit 1
Analytic Geometry
https://commons.wikimedia.org/wiki/File%3ASan Juanico Bridge 2.JPG
By Morten Nærbøe (Own work)
[CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0)
or GFDL (http://www.gnu.org/copyleft/fdl.html)],
via Wikimedia Commons
Stretching from Samar to Leyte with a total length of more than two kilometers, the San Juanico Bridge has served as one of the main thoroughfares of
economic and social development in the country since its completion in 1973.
Adding picturesque e↵ect on the whole architecture, geometric structures are
subtly built to serve other purposes. The arch-shaped support on the main span
of the bridge helps maximize its strength to withstand mechanical resonance and
aeroelastic flutter brought about by heavy vehicles and passing winds.
Lesson 1.1. Introduction to Conic Sections and Circles
Time Frame: 4 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate the di↵erent types of conic sections: parabola, ellipse, circle, hyperbola, and degenerate cases;
(2) define a circle;
(3) determine the standard form of equation of a circle;
(4) graph a circle in a rectangular coordinate system; and
(5) solve situational problems involving conic sections (circles).
Lesson Outline
(1) Introduction of the four conic sections, along with the degenerate conics
(2) Definition of a circle
(3) Derivation of the standard equation of a circle
(4) Graphing circles
(5) Solving situational problems involving circles
Introduction
We introduce the conic sections, a particular class of curves which sometimes
appear in nature and which have applications in other fields. In this lesson, we
discuss the first of their kind, circles. The other conic sections will be covered in
the next lessons.
1.1.1. An Overview of Conic Sections
We introduce the conic sections (or conics), a particular class of curves which
oftentimes appear in nature and which have applications in other fields. One
of the first shapes we learned, a circle, is a conic. When you throw a ball, the
trajectory it takes is a parabola. The orbit taken by each planet around the sun
is an ellipse. Properties of hyperbolas have been used in the design of certain
telescopes and navigation systems. We will discuss circles in this lesson, leaving
parabolas, ellipses, and hyperbolas for subsequent lessons.
• Circle (Figure 1.1) - when the plane is horizontal
• Ellipse (Figure 1.1) - when the (tilted) plane intersects only one cone to form
a bounded curve
10
• Parabola (Figure 1.2) - when the plane intersects only one cone to form an
unbounded curve
• Hyperbola (Figure 1.3) - when the plane (not necessarily vertical) intersects
both cones to form two unbounded curves (each called a branch of the hyperbola)
Figure 1.1
Figure 1.2
Figure 1.3
We can draw these conic sections (also called conics) on a rectangular coordinate plane and find their equations. To be able to do this, we will present
equivalent definitions of these conic sections in subsequent sections, and use these
to find the equations.
There are other ways for a plane and the cones to intersect, to form what are
referred to as degenerate conics: a point, one line, and two lines. See Figures 1.4,
1.5 and 1.6.
Figure 1.4
Figure 1.5
Figure 1.6
1.1.2. Definition and Equation of a Circle
A circle may also be considered a special kind of ellipse (for the special case when
the tilted plane is horizontal). For our purposes, we will distinguish between
these two conics.
11
See Figure 1.7, with the point C(3, 1) shown. From the figure, the distance
of A( 2, 1) from p
C is AC = 5. By the distance formula, the distance of B(6, 5)
from C is BC = (6 3)2 + (5 1)2 = 5. There are other points P such that
P C = 5. The collection of all such points which are 5 units away from C, forms
a circle.
Figure 1.7
Figure 1.8
Let C be a given point. The set of all points P having the same
distance from C is called a circle. The point C is called the center of
the circle, and the common distance its radius.
The term radius is both used to refer to a segment from the center C to a
point P on the circle, and the length of this segment.
See Figure 1.8, where a circle is drawn. It has center C(h, k) and radius r > 0.
A point P (x, y) is on the circle if and only if P C = r. For any such point then,
its coordinates should satisfy the following.
p
(x
(x
PC = r
h)2 + (y
h)2 + (y
k)2 = r
k)2 = r2
This is the standard equation of the circle with center C(h, k) and radius r. If
the center is the origin, then h = 0 and k = 0. The standard equation is then
x2 + y 2 = r 2 .
Example 1.1.1. In each item, give the standard equation of the circle satisfying
the given conditions.
(1) center at the origin, radius 4
p
(2) center ( 4, 3), radius 7
12
(3) circle in Figure 1.7
(4) circle A in Figure 1.9
(5) circle B in Figure 1.9
(6) center (5, 6), tangent to the y-axis
(7) center (5, 6), tangent to the x-axis
(8) has a diameter with endpoints A( 1, 4) and B(4, 2)
Figure 1.9
Solution. (1) x2 + y 2 = 16
(2) (x + 4)2 + (y
3)2 = 7
(3) The center is (3, 1) and the radius is 5, so the equation is (x
25.
3)2 + (y
1)2 =
(4) By inspection, the center is ( 2, 1) and the radius is 4. The equation is
(x + 2)2 + (y + 1)2 = 16.
(5) Similarly by inspection, we have (x
3)2 + (y
2)2 = 9.
(6) The center is 5 units away from the y-axis, so the radius is r = 5 (you can
make a sketch to see why). The equation is (x 5)2 + (y + 6)2 = 25.
(7) Similarly, since the center is 6 units away from the x-axis, the equation is
(x 5)2 + (y + 6)2 = 36.
13
(8) The center C is the midpoint of A and B: C = 1+4
, 4+2 = 32 , 3 . The
2 q 2
q
2
29
radius is then r = AC =
1 32 + (4 3)2 =
. The circle has
4
equation x
3 2
2
+ (y
3)2 =
29
.
4
2
Seatwork/Homework 1.1.2
Find the standard equation of the circle being described in each item.
p
(1) With center at the origin, radius 11
Answer: x2 + y 2 = 11
(2) With center ( 6, 7), tangent to the y-axis Answer: (x + 6)2 + (y
7)2 = 36
(3) Has a diameter with endpoints A( 3, 2) and B(7, 4)
Answer: (x
2)2 + (y
3)2 = 26
1.1.3. More Properties of Circles
After expanding, the standard equation
✓
◆2
3
x
+ (y
2
3)2 =
29
4
can be rewritten as
x2 + y 2
3x
6y
5 = 0,
an equation of the circle in general form.
If the equation of a circle is given in the general form
Ax2 + Ay 2 + Cx + Dy + E = 0,
A 6= 0,
or
x2 + y 2 + Cx + Dy + E = 0,
we can determine the standard form by completing the square in both variables.
Completing the square in an expression like x2 + 14x means determining Teaching Notes
the
the term to be added that will produce a perfect polynomial square. Since the Recall
technique of
2
coefficient of x is already 1, we take half the coefficient of x and square it, and completing the
This was
we get 49. Indeed, x2 + 14x + 49 = (x + 7)2 is a perfect square. To complete square.
introduced in
the square in, say, 3x2 + 18x, we factor the coefficient of x2 from the expression: Grade 9.
3(x2 + 6x), then add 9 inside. When completing a square in an equation, any
extra term introduced on one side should also be added to the other side.
Example 1.1.2. Identify the center and radius of the circle with the given equation in each item. Sketch its graph, and indicate the center.
(1) x2 + y 2
6x = 7
14
(2) x2 + y 2
14x + 2y =
(3) 16x2 + 16y 2 + 96x
14
40y = 315
Solution. The first step is to rewrite each equation in standard form by completing the square in x and in y. From the standard equation, we can determine the
center and radius.
(1)
x2 6x + y 2 = 7
x2 6x + 9 + y 2 = 7 + 9
(x 3)2 + y 2 = 16
Center (3, 0), r = 4, Figure 1.10
(2)
x2
x2 14x + y 2 + 2y = 14
14x + 49 + y 2 + 2y + 1 = 14 + 49 + 1
(x 7)2 + (y + 1)2 = 36
Center (7, 1), r = 6, Figure 1.11
(3)
Teaching Notes
A common mistake
committed by
students is to add 9
and 25
only. They
16
16x2 + 96x + 16y 2 40y = 315
✓
◆
5
2
2
16(x + 6x) + 16 y
y = 315
2
✓
◆
✓ ◆
5
25
25
2
2
16(x + 6x + 9) + 16 y
y+
= 315 + 16(9) + 16
2
16
16
✓
◆2
5
16(x + 3)2 + 16 y
= 484
4
✓
◆2
✓ ◆2
5
484
121
11
2
(x + 3) + y
=
=
=
4
16
4
2
Center
3, 54 , r = 5.5, Figure 1.12.
2
often forget the
multiplier outside
the parenthesis.
Figure 1.10
Figure 1.11
15
Figure 1.12
In the standard equation (x h)2 + (y k)2 = r2 , both the two squared
terms on the left side have coefficient 1. This is the reason why in the preceding
example, we divided by 16 at the last equation.
Seatwork/Homework 1.1.3
Identify the center and radius of the circle with the given equation in each item.
Sketch its graph, and indicate the center.
(1) x2 + y 2
5x + 4y = 46
Answer: center
(2) 4x2 + 4y 2 + 40x
5
,
2
2 , radius
15
2
= 7.5, Figure 1.13
32y = 5
Answer: center ( 5, 4), radius
13
2
= 6.5, Figure 1.14
Figure 1.14
Figure 1.13
1.1.4. Situational Problems Involving Circles
We now consider some situational problems involving circles.
? Example 1.1.3. A street with two lanes, each 10 ft wide, goes through a
semicircular tunnel with radius 12 ft. How high is the tunnel at the edge of each
lane? Round o↵ to 2 decimal places.
16
Solution. We draw a coordinate system with origin at the middle of the highway,
as shown. Because of the given radius, the tunnel’s boundary is on the circle
x2 + y 2 = 122 . Point P is the point on the arc just above the edge of a lane, so
its x-coordinate is 10. Wepneed its y-coordinate. We then solve 102 + y 2 = 122
for y > 0, giving us y = 2 11 ⇡ 6.63 ft.
2
Example 1.1.4. A piece of a broken plate was dug up in an archaeological site.
It was put on top of a grid, as shown in Figure 1.15, with the arc of the plate
passing through A( 7, 0), B(1, 4) and C(7, 2). Find its center, and the standard
Teaching Notes equation of the circle describing the boundary of the plate.
A perpendicular
bisector of a
segment is the line
that passes
through the
midpoint of the
segment and is
perpendicular to
the segment.
Figure 1.15
Figure 1.16
Solution. We first determine the center. It is the intersection of the perpendicular
17
bisectors of AB and BC (see Figure 1.16). Recall that, in a circle, the perpendicular bisector of any chord passes through the center. Since the midpoint M
of AB is 7+1
, 0+4
= ( 3, 2), and mAB = 41+70 = 12 , the perpendicular bisector
2
2
of AB has equation y 2 = 2(x + 3), or equivalently, y = 2x 4.
Since the midpoint N of BC is 1+7
, 4+2
= (4, 3), and mBC = 27 41 = 13 ,
2
2
the perpendicular bisector of BC has equation y 3 = 3(x 4), or equivalently,
y = 3x 9.
The intersection of the two lines y = 2x 4 and y = 3x 9 is (1, 6) (by
solving a system of linear equations). We can take the radius as the distance of
this point from any of A, B or C (it’s most convenient to use B in this case). We
then get r = 10. The standard equation is thus (x 1)2 + (y + 6)2 = 100.
2
Seatwork/Homework 1.1.4
? 1. A single-lane street 10 ft wide goes through a semicircular tunnel with radius
9 ft. How high is the tunnel at the edge of each lane? Round o↵ to 2 decimal
places.
Answer: 7.48 ft
2. An archeologist found the remains of an ancient wheel, which she then placed
on a grid. If an arc of the wheel passes through A( 7, 0), B( 3, 4) and C(7, 0),
locate the center of the wheel, and the standard equation of the circle defining
its boundary.
Answer: (0, 3), x2 + (y + 3)2 = 58
Exercises 1.1
1. Identify the center and radius of the circle with the given equation in each
item. Sketch its graph, and indicate the center.
(a) x2 + y 2 = 49
Answer: center (0, 0), r = 7
(b) 4x2 + 4y 2 = 25
Answer: center (0, 0), r =
(c) x
7 2
4
2
+ y+
(d) x2 + y
12x
(e) x2 + y 2 + 8x
3 2
4
=
169
16
10y =
Answer: center
12
3
4
,r=
13
4
Answer: center (6, 5), r = 7
9y = 6
(f) x2 + y 2 + 10x + 12y =
7
,
4
5
2
Answer: center ( 4, 4.5), r =
13
2
12
Answer: center ( 5, 6), r = 7
(g) 2x2 + 2y 2
14x + 18y = 7
(h) 4x2 + 4y 2
20x + 40y =
Answer: center (3.5, 4.5), r = 6
p
Answer: center (2.5, 5), r = 30
p
7
14
Answer: center
,
,
r
=
2
5
3
3
p
5 1
Answer: center
, , r = 10
2 2
5
(i) 9x2 + 9y 2 + 42x + 84y + 65 = 0
(j) 2x2 + 2y 2 + 10x = 2y + 7
18
(a)
(b)
(c)
(d)
(e)
(f)
19
(g)
(h)
(i)
(j)
2. Find the standard equation of the circle which satisfies the given conditions. Teaching Notes
To determine the
p
equation of a
2
2
(a) center at the origin, radius 2 2
Answer: x + y = 8 circle, we just need
(b) center at (15, 20), radius 9
(c) center at (5, 6), through (9, 4)
Answer: (x
Answer: (x
to determine the
15)2 + (y + 20)2 = 81 center and the
5)2 + (y
6)2 = 20
Solution. The radius is the distance from the center to (9, 4):
p
p
(5 9)2 + (6 4)2 = 20.
(d) center at ( 2, 3), tangent to the x-axis
Answer: (x + 2)2 + (y
3)2 = 9
(e) center at ( 2, 3), tangent to the y-axis
Answer: (x + 2)2 + (y
3)2 = 4
20
radius.
(f) center at ( 2, 3), tangent to the line y = 8
Answer: (x + 2)2 + (y
3)2 = 25
Solution. We need to determine the radius. This is best done by
sketching the center and line, to see that the center ( 2, 3) is 5 units
away from the nearest point on the line, ( 2, 8) (which is the point of
tangency).
(g) center at ( 2, 3), tangent to the line x = 10
Answer: (x + 2)2 + (y
3)2 = 64
(h) center in the third quadrant, tangent to both the x-axis and y-axis,
radius 7
Answer: (x + 7)2 + (y + 7)2 = 49
(i) a diameter with endpoints ( 9, 2) and (15, 12)
Answer: (x
(j) concentric with x2 + y 2 + 2x
3)2 + (y
4y = 5, radius is 7
Answer: (x + 1)2 + (y
7)2 = 169
2)2 = 49
Solution. Two circles are said to be concentric if they have the same
center.
The standard equation of the given circle is
2
(x + 1) + (y 2)2 = 10. Thus, the circle we’re looking for has center
( 1, 2) and radius 7.
(k) concentric with x2 + y 2
8x
10y = 16 and 4 times the area
Answer: (x 4)2 + (y 5)2 = 100
Solution. The given circle has standard equation
(x
4)2 + (y
5)2 = 52 .
Its radius is 5, so its area is 25⇡ sq. units. The circle we are looking
for should have area 100⇡ sq. units, so its radius is 10.
(l) concentric with x2 + y 2
x2 + y 2 14x + 6y = 33
10x
6y =
2, same radius as
Answer: (x 5)2 + (y 3)2 = 25
(m) center at C(3, 4), tangent to the line y = 13 x 13
Answer: (x
Teaching Notes
The radius drawn
to a point on the
circle is
perpendicular to
the line tangent to
the circle at that
point.
3)2 + (y
4)2 = 10
Solution. (A sketch will greatly help in understanding the argument.)
If P is the point of tangency, then line CP is perpendicular to the given
tangent line. Since the tangent line has slope 13 , line CP has slope 3.
Because it passes through C, line CP has equation y 4 = 3(x 3),
or y = 3x + 13. Solving the system {y = 13 x 13 , y = 3x + 13}
yields p
x = 4 and y = 1, the p
coordinates of P . The radius is then
2
2
CP = (3 4) + (4 1) = 10.
21
(n) center at ( 4, 3), tangent to the line y = 4x 30
Answer: (x + 4)2 + (y
3)2 = 17
Solution. (Similar to the previous problem) Let P be the point of
tangency, so line CP is perpendicular to the tangent line. The tangent
line has slope 4, so line CP has slope 14 . Line CP passes through C,
so it has equation y 3 = 14 (x + 4), or y = 14 x + 4. Solving the system
{y = 4x 30, y = 14 x + 4} yields x = 8 and y = 2, the coordinates
p
p
of P . The radius is then CP = ( 4 + 8)2 + (3 2)2 = 17.
? 3. A seismological station is located at (0, 3), 3 km away from a straight
shoreline where the x-axis runs through. The epicenter of an earthquake
was determined to be 6 km away from the station.
(a) Find the equation of the curve that contains the possible location of
the epicenter.
Answer: x2 + (y + 3)2 = 62
(b) If furthermore, the epicenter was determined to be 2 km away from
the shore, find its possible coordinates (rounded o↵ to two decimal
places).
Answer: (±3.32, 2)
Solution. Since the epicenter is 6 units away from (0, 3), it could be any
of the points of a circle with center (0, 3) and radius 6. The equation is
then x2 + (y + 3)2 = 62 . Next, we solve thispequation for x if y = 2, and we
get x2 = 62 (2 + 3)2 = 11, and so x = ± 11 ⇡ ±3.32.
4. A ferris wheel is elevated 1 m above ground. When a car reaches the highest
point on the ferris wheel, its altitude from ground level is 31 m. How far
away from the center, horizontally, is the car when it is at an altitude of
25 m?
Answer: 12 m
Solution. The ferris wheel, as shown
below, is drawn 1 unit above the xaxis (ground level), center on the yaxis, and highest point at y = 31.
The diameter is thus 30, and the radius 15. We locate the center at
(0, 16), and write the equation of the
circle as x2 + (y 16)2 = 152 .
If y = 25, we have x2 + (25 16)2 =
152 , so x2 = 152 92 = 144, and
x = ±12. (Clearly, there are two
points on the ferris wheel at an altitude of 25 m.) Thus, the car is 12 m
away horizontally from the center.
22
? 5. A window is to be constructed as shown, with its upper boundary the arc
of a circle having radius 4 ft and center at the midpoint of base AD. If the
vertical side is to be 34 as long as the base, find the dimensions (vertical side
and base) of this window. Round o↵ your final answer to 1 decimal place.
Answer: base 4.44 ft, side 3.33 ft
Solution. We put two lines corresponding to the x-axis and y-axis, as shown,
with the origin coinciding with the midpoint of the window’s base. This
origin is the center of the circle containing the arc. The equation of the circle
is then x2 + y 2 = 16. Let n be length of the base AD, so the side AD has
2
2
length 34 n. Point B then has coordinates n2 , 3n
. Therefore, n2 + 3n
=
4
4
16
p
16. Solving this for n > 0 yields n = 13 . The base is then n ⇡ 4.44 ft and
the side 34 n ⇡ 3.33 ft.
4
Lesson 1.2. Parabolas
Time Frame: 3 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) define a parabola;
(2) determine the standard form of equation of a parabola;
(3) graph a parabola in a rectangular coordinate system; and
(4) solve situational problems involving conic sections (parabolas).
23
Lesson Outline
(1) Definition of a parabola
(2) Derivation of the standard equation of a parabola
(3) Graphing parabolas
(4) Solving situational problems involving parabolas
Introduction
A parabola is one of the conic sections. We have already seen parabolas which
open upward or downward, as graphs of quadratic functions. Here, we will see
parabolas opening to the left or right. Applications of parabolas are presented
at the end.
1.2.1. Definition and Equation of a Parabola
Consider the point F (0, 2) and the line ` having equation y = 2, as shown in
Figure 1.17. What are the distances of A(4, 2) from F and from `? (The latter
is taken as the distance of A from A` , the point on ` closest to A). How about
the distances of B( 8, 8) from F and from ` (from B` )?
AF = 4
BF =
p
( 8
0)2 + (8
and
AA` = 4
2)2 = 10
and
BB` = 10
There are other points P such that P F = P P` (where P` is the closest point on
line `). The collection of all such points forms a shape called a parabola.
Figure 1.17
Figure 1.18
Let F be a given point, and ` a given line not containing F . The set of
all points P such that its distances from F and from ` are the same, is
called a parabola. The point F is its focus and the line ` its directrix.
24
Consider a parabola with focus F (0, c) and directrix ` having equation x = c.
See Figure 1.18. The focus and directrix are c units above and below, respectively,
the origin. Let P (x, y) be a point on the parabola so P F = P P` , where P` is the
point on ` closest to P . The point P has to be on the same side of the directrix
as the focus (if P was below, it would be closer to ` than it is from F ).
PF
p
x2 + (y c)2
x2 + y 2 2cy + c2
x2
= P P`
= y ( c) = y + c
= y 2 + 2cy + c2
= 4cy
The vertex V is the point midway between the focus and the directrix. This
equation, x2 = 4cy, is then the standard equation of a parabola opening upward
with vertex V (0, 0).
Suppose the focus is F (0, c) and the directrix is y = c. In this case, a
point P on the resulting parabola would be below the directrix (just like the
focus).
Instead of opening upward, it will open downward. Consequently, P F =
p
x2 + (y + c)2 and P P` = c y (you may draw a version of Figure 1.18 for
this case). Computations similar to the one done above will lead to the equation
x2 = 4cy.
We collect here the features of the graph of a parabola with standard equation
x2 = 4cy or x2 = 4cy, where c > 0.
(1) vertex : origin V (0, 0)
• If the parabola opens upward, the vertex is the lowest point. If the
parabola opens downward, the vertex is the highest point.
(2) directrix : the line y =
c or y = c
• The directrix is c units below or above the vertex.
(3) focus: F (0, c) or F (0, c)
• The focus is c units above or below the vertex.
25
• Any point on the parabola has the same distance from the focus as it
has from the directrix.
(4) axis of symmetry: x = 0 (the y-axis)
• This line divides the parabola into two parts which are mirror images
of each other.
Example 1.2.1. Determine the focus and directrix of the parabola with the
given equation. Sketch the graph, and indicate the focus, directrix, vertex, and
axis of symmetry.
(1) x2 = 12y
(2) x2 = 6y
Solution. (1) The vertex is V (0, 0) and the parabola opens upward. From 4c =
12, c = 3. The focus, c = 3 units above the vertex, is F (0, 3). The directrix,
3 units below the vertex, is y = 3. The axis of symmetry is x = 0.
(2) The vertex is V (0, 0) and the parabola opens downward. From 4c = 6, c = 32 .
The focus, c = 32 units below the vertex, is F 0, 32 . The directrix, 32 units
above the vertex, is y = 32 . The axis of symmetry is x = 0.
26
Example 1.2.2. What is the standard equation of the parabola in Figure 1.17?
Solution. From the figure, we deduce that c = 2. The equation is thus x2 =
8y.
2
Seatwork/Homework 1.2.1
1. Give the focus and directrix of the parabola with equation x2 = 10y. Sketch
the graph, and indicate the focus, directrix, vertex, and axis of symmetry.
Answer: focus 0, 52 , directrix y = 52
2. Find the standard equation of the parabola with focus F (0, 3.5) and directrix
y = 3.5.
Answer: x2 = 14y
1.2.2. More Properties of Parabolas
The parabolas we considered so far are “vertical” and have their vertices at the
origin. Some parabolas open instead horizontally (to the left or right), and some
have vertices not at the origin. Their standard equations and properties are given
in the box. The corresponding computations are more involved, but are similar
to the one above, and so are not shown anymore.
In all four cases below, we assume that c > 0. The vertex is V (h, k), and it
lies between the focus F and the directrix `. The focus F is c units away from
the vertex V , and the directrix is c units away from the vertex. Recall that, for
any point on the parabola, its distance from the focus is the same as its distance
from the directrix.
27
(x
(x
h)2 = 4c(y
h)2 =
4c(y
k)
(y
k)
(y
k)2 = 4c(x
k)2 =
4c(x
h)
h)
directrix `: horizontal
directrix `: vertical
axis of symmetry: x=h, vertical
axis of symmetry: y=k, horizontal
The following observations are worth noting.
• The equations are in terms of x h and y k: the vertex coordinates are
subtracted from the corresponding variable. Thus, replacing both h and k
with 0 would yield the case where the vertex is the origin. For instance, this
replacement applied to (x h)2 = 4c(y k) (parabola opening upward) would
yield x2 = 4cy, the first standard equation we encountered (parabola opening
upward, vertex at the origin).
• If the x-part is squared, the parabola is “vertical”; if the y-part is squared,
the parabola is “horizontal.” In a horizontal parabola, the focus is on the left
or right of the vertex, and the directrix is vertical.
• If the coefficient of the linear (non-squared) part is positive, the parabola
opens upward or to the right; if negative, downward or to the left.
28
Example 1.2.3. The figure shows the graph of parabola, with only its focus and
Teaching Notes vertex indicated. Find its standard equation. What is its directrix and its axis
In finding the
of symmetry?
equation of a
parabola, we just
need to determine
the vertex and the
value of c.
Solution. The vertex is V (5, 4) and the focus is F (3, 4). From these, we
deduce the following: h = 5, k = 4, c = 2 (the distance of the focus from the
vertex). Since the parabola opens to the left, we use the template (y k)2 =
4c(x h). Our equation is
(y + 4)2 =
8(x
5).
Its directrix is c = 2 units to the right of V , which is x = 7. Its axis is the
horizontal line through V : y = 4.
The standard equation (y + 4)2 = 8(x 5) from the preceding example can
be rewritten as y 2 + 8x + 8y 24 = 0, an equation of the parabola in general
form.
If the equation is given in the general form Ax2 + Cx + Dy + E = 0 (A and C
are nonzero) or By 2 + Cx + Dy + E = 0 (B and C are nonzero), we can determine
the standard form by completing the square in both variables.
Example 1.2.4. Determine the vertex, focus, directrix, and axis of symmetry
of the parabola with the given equation. Sketch the parabola, and include these
points and lines.
(a) y 2
5x + 12y =
16
(b) 5x2 + 30x + 24y = 51
29
Solution. (1) We complete the square on y, and move x to the other side.
y 2 + 12y = 5x 16
y 2 + 12y + 36 = 5x 16 + 36 = 5x + 20
(y + 6)2 = 5(x + 4)
The parabola opens to the right. It has vertex V ( 4, 6). From 4c = 5, we
get c = 54 = 1.25. The focus is c = 1.25 units to the right of V : F ( 2.75, 6).
The (vertical) directrix is c = 1.25 units to the left of V : x = 5.25. The
(horizontal) axis is through V : y = 6.
(2) We complete the square on x, and move y to the other side.
5x2 + 30x =
5(x2 + 6x + 9) =
5(x + 3)2 =
(x + 3)2 =
24y + 51
24y + 51 + 5(9)
24y + 96 = 24(y
24
(y 4)
5
4)
In the last line, we divided by 5 for the squared part not to have any coefficient. The parabola opens downward. It has vertex V ( 3, 4).
From 4c = 24
, we get c = 65 = 1.2. The focus is c = 1.2 units below V :
5
F ( 3, 2.8). The (horizontal) directrix is c = 1.2 units above V : y = 5.2. The
(vertical) axis is through V : x = 3.
30
Example 1.2.5. A parabola has focus F (7, 9) and directrix y = 3. Find its
standard equation.
Solution. The directrix is horizontal, and the focus is above it. The parabola
then opens upward and its standard equation has the form (x h)2 = 4c(y k).
Since the distance from the focus to the directrix is 2c = 9 3 = 6, then c = 3.
Thus, the vertex is V (7, 6), the point 3 units below F . The standard equation is
then (x 7)2 = 12(y 6).
2
Seatwork/Homework 1.2.2
1. Determine the vertex, focus, directrix, and axis of symmetry of the parabola
with equation x2 6x + 5y = 34. Sketch the graph, and include these points
and lines.
Answer: vertex (3, 5), focus (3, 6.25), directrix y =
31
3.75, axis x = 3
2. A parabola has focus F ( 2, 5) and directrix x = 6. Find the standard
equation of the parabola.
Answer: (y + 5)2 = 16(x 2)
1.2.3. Situational Problems Involving Parabolas
We now solve some situational problems involving parabolas.
Example 1.2.6. A satellite dish has a shape called a paraboloid, where each
cross-section is a parabola. Since radio signals (parallel to the axis) will bounce
o↵ the surface of the dish to the focus, the receiver should be placed at the focus.
How far should the receiver be from the vertex, if the dish is 12 ft across, and 4.5
ft deep at the vertex?
Solution. The second figure above shows a cross-section of the satellite dish drawn
on a rectangular coordinate system, with the vertex at the origin. From the
problem, we deduce that (6, 4.5) is a point on the parabola. We need the distance
of the focus from the vertex, i.e., the value of c in x2 = 4cy.
x2 = 4cy
62 = 4c(4.5)
62
c=
=2
4 · 4.5
Thus, the receiver should be 2 ft away from the vertex.
2
Example 1.2.7. The cable of a suspension bridge hangs in the shape of a
parabola. The towers supporting the cable are 400 ft apart and 150 ft high.
If the cable, at its lowest, is 30 ft above the bridge at its midpoint, how high is
the cable 50 ft away (horizontally) from either tower?
32
Solution. Refer to the figure above, where the parabolic cable is drawn with
its vertex on the y-axis 30 ft above the origin. We may write its equation as
(x 0)2 = a(y 30); since we don’t need the focal distance, we use the simpler
variable a in place of 4c. Since the towers are 150 ft high and 400 ft apart, we
deduce from the figure that (200, 150) is a point on the parabola.
x2 = a(y 30)
2002 = a(150 30)
2002
1000
a=
=
120
3
The parabola has equation x2 = 1000
(y
30), or equivalently,
3
2
y = 0.003x + 30. For the two points on the parabola 50 ft away from the
towers, x = 150 or x = 150. If x = 150, then
y = 0.003(1502 ) + 30 = 97.5.
Thus, the cable is 97.5 ft high 50 ft away from either tower. (As expected, we
get the same answer from x = 150.)
2
Seatwork/Homework 1.2.3
? 1. A satellite dish in the shape of a paraboloid is 10 ft across, and 4 ft deep at
its vertex. How far is the receiver from the vertex, if it is placed at the focus?
Round o↵ your answer to 2 decimal places. (Refer to Example 1.2.6.)
Answer: 1.56 ft
Exercises 1.2
1. Determine the vertex, focus, directrix, and axis of symmetry of the parabola
with the given equation. Sketch the graph, and include these points and lines.
33
(a) x2 =
(d) x2 + 6x + 8y = 7
4y
(b) 3y 2 = 24x
(c) y +
5 2
2
(e) y 2
=
5 x
9
2
12x + 8y =
(f) 16x2 + 72x
40
112y =
221
Answer:
Item
Vertex
Focus
Directrix
Axis of Symmetry
(a)
(0, 0)
(0, 1)
y=1
x=0
(b)
(0, 0)
(2, 0)
(c)
(4.5, 2.5)
(3.25, 2.5)
x = 5.75
(d)
( 3, 2)
( 3, 0)
y=4
(e)
(2, 4)
(5, 4)
(f)
( 2.25, 1.25)
( 2.25, 3)
x=
x=
y=
2
1
0.5
y=0
y=
2.5
x=
3
y=
4
x=
(a)
(b)
(c)
(d)
(e)
(f)
2.25
2. Find the standard equation of the parabola which satisfies the given conditions. Teaching Notes
It is helpful to
Answer: (y + 9)2 =
(a) vertex (1, 9), focus ( 3, 9)
34
16(x
a diagram for
1) draw
each item.
(b) vertex ( 8, 3), directrix x =
10.5
Answer: (y
3)2 = 10(x + 8)
(c) vertex ( 4, 2), focus ( 4, 1)
Answer: (x + 4)2 =
(d) focus (7, 11), directrix x = 1
(e) focus (7, 11), directrix y = 4
12(y
2)
Answer: (y
11)2 = 12(x
4)
Answer: (x
7)2 = 14(y
7.5)
(f) vertex ( 5, 7), vertical axis of symmetry, through the point P (7, 11)
Answer: (x + 5)2 = 8(y + 7)
Solution. Since the axis is vertical and P is above the vertex, then the
parabola opens upward and has equation of the form (x + 5)2 = 4c(y + 7).
We plug the coordinates of P : (7 + 5)2 = 4c(11 + 7). We then get c = 2.
Thus, we have (x + 5)2 = 8(y + 7).
(g) vertex ( 5, 7), horizontal axis of symmetry, through the point P (7, 11)
Answer: (y + 7)2 = 27(x + 5)
Solution. Since the axis is horizontal and P is to the right of the vertex,
then the parabola opens to the right and has equation of the form (y +
7)2 = 4c(x + 5). We plug the coordinates of P : (11 + 7)2 = 4c(7 + 5).
We then get c = 6.75. Thus, we have (y + 7)2 = 27(x + 5).
3. A satellite dish shaped like a paraboloid, has diameter 2.4 ft and depth 0.9 ft.
If the receiver is placed at the focus, how far should the receiver be from the
vertex?
Answer: 0.4 ft
4. If the diameter of the satellite dish from the previous problem is doubled, with
the depth kept the same, how far should the receiver be from the vertex?
Answer: 1.6 ft
? 5. A satellite dish is shaped like a paraboloid, with the receiver placed at the
focus. It is to have a depth of 0.44 m at the vertex, with the receiver placed
0.11 m away from the vertex. What should the diameter of the satellite dish
be?
Answer: 0.88 m
? 6. A flashlight is shaped like a paraboloid, so that if its light bulb is placed at
the focus, the light rays from the bulb will then bounce o↵ the surface in a
focused direction that is parallel to the axis. If the paraboloid has a depth of
1.8 in and the diameter on its surface is 6 in, how far should the light source
be placed from the vertex?
Answer: 1.25 in
7. The towers supporting the cable of a suspension bridge are 1200 m apart and
170 m above the bridge it supports. Suppose the cable hangs, following the
shape of a parabola, with its lowest point 20 m above the bridge. How high is
the cable 120 m away from a tower?
Answer: 116 m
4
35
Lesson 1.3. Ellipses
Time Frame: 3 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) define an ellipse;
(2) determine the standard form of equation of an ellipse;
(3) graph an ellipse in a rectangular coordinate system; and
(4) solve situational problems involving conic sections (ellipses).
Lesson Outline
(1) Definition of an ellipse
(2) Derivation of the standard equation of an ellipse
(3) Graphing ellipses
(4) Solving situational problems involving ellipses
Introduction
An ellipse is one of the conic sections that most students have not encountered
formally before, unlike circles and parabolas. Its shape is a bounded curve which
looks like a flattened circle. The orbits of the planets in our solar system around
the sun happen to be elliptical in shape. Also, just like parabolas, ellipses have
reflective properties that have been used in the construction of certain structures
(shown in some of the practice problems). We will see some properties and
applications of ellipses in this section.
1.3.1. Definition and Equation of an Ellipse
Consider the points F1 ( 3, 0) and F2 (3, 0), as shown in Figure 1.19. What is the
sum of the distances of A(4, 2.4) from F1 and from F2 ? How about the sum of
the distances of B (and C(0, 4)) from F1 and from F2 ?
Teaching Notes
AF1 + AF2 = 7.4 + 2.6 = 10
BF1 + BF2 = 3.8 + 6.2 = 10
CF1 + CF2 = 5 + 5 = 10
There are other points P such that P F1 + P F2 = 10. The collection of all such
points forms a shape called an ellipse.
36
You may review
the distance
formula.
Figure 1.19
Figure 1.20
Let F1 and F2 be two distinct points. The set of all points P , whose
distances from F1 and from F2 add up to a certain constant, is called
an ellipse. The points F1 and F2 are called the foci of the ellipse.
Given are two points on the x-axis, F1 ( c, 0) and F2 (c, 0), the foci, both c
units away from their center (0, 0). See Figure 1.20. Let P (x, y) be a point on
the ellipse. Let the common sum of the distances be 2a (the coefficient 2 will
make computations simpler). Thus, we have P F1 + P F2 = 2a.
p
P F1 = 2a
(x + c)2 + y 2 = 2a
P F2
p
(x c)2 + y 2
p
4a (x c)2 + y 2 + x2
x2 + 2cx + c2 + y 2 = 4a2
p
a (x c)2 + y 2 = a2 cx
⇥
⇤
a2 x2 2cx + c2 + y 2 = a4 2a2 cx + c2 x2
(a2 c2 )x2 + a2 y 2 = a4 a2 c2 = a2 (a2
b 2 x 2 + a2 y 2 = a2 b 2
x2 y 2
+ 2 =1
a2
b
c2 )
by letting b =
p
a2
2cx + c2 + y 2
c2 , so a > b
p
When we let b = a2 c2 , we assumed a > c. To see why this is true, look at
4P F1 F2 in Figure 1.20. By the Triangle Inequality, P F1 + P F2 > F1 F2 , which
implies 2a > 2c, so a > c.
We collect here the features of the graph of an ellipse with standard equation
p
x2 y 2
+ 2 = 1, where a > b. Let c = a2 b2 .
2
a
b
37
(1) center : origin (0, 0)
(2) foci : F1 ( c, 0) and F2 (c, 0)
• Each focus is c units away from the center.
• For any point on the ellipse, the sum of its distances from the foci is 2a.
(3) vertices: V1 ( a, 0) and V2 (a, 0)
• The vertices are points on the ellipse, collinear with the center and foci.
• If y = 0, then x = ±a. Each vertex is a units away from the center.
• The segment V1 V2 is called the major axis. Its length is 2a. It divides
the ellipse into two congruent parts.
(4) covertices: W1 (0, b) and W2 (0, b)
• The segment through the center, perpendicular to the major axis, is the
minor axis. It meets the ellipse at the covertices. It divides the ellipse
into two congruent parts.
• If x = 0, then y = ±b. Each covertex is b units away from the center.
• The minor axis W1 W2 is 2b units long. Since a > b, the major axis is
longer than the minor axis.
Example 1.3.1. Give the coordinates of the foci, vertices, and covertices of the
ellipse with equation
x2 y 2
+
= 1.
25
9
Sketch the graph, and include these points.
Solution. With a2 = 25 and b2 = 9, we have a = 5, b = 3, and c =
foci: F1 ( 4, 0), F2 (4, 0)
p
vertices: V1 ( 5, 0), V2 (5, 0)
covertices: W1 (0, 3), W2 (0, 3)
38
a2
b2 = 4.
Example 1.3.2. Find the (standard) equation of the ellipse whose foci are
F1 ( 3, 0) and F2 (3, 0), such that for any point on it, the sum of its distances
from the foci is 10. See Figure 1.19.
Solution. We have 2a = 10 and c = 3, so a = 5 and b =
equation is
x2 y 2
+
= 1.
25 16
p
a2
c2 = 4. The
2
Seatwork/Homework 1.3.1
1. Give the coordinates of the foci, vertices, and covertices of the ellipse with
x2
y2
equation
+
= 1. Sketch the graph, and include these points.
169 25
Answer: foci: F1 ( 12, 0) and F2 (12, 0), vertices: V1 ( 13, 0) and V2 (13, 0),
covertices: W1 (0, 5) and W2 (0, 5)
2. Find the equation in standard form of the ellipse whose foci are F1 ( 8, 0) and
F2 (8, 0), such that for any point on it, the sum of its distances from the foci
x2
y2
is 20.
Answer:
+
=1
100 36
39
1.3.2. More Properties of Ellipses
Some ellipses have their foci aligned vertically, and some have centers not at
the origin. Their standard equations and properties are given in the box. The
derivations are more involved, but are similar to the one above, and so are not
shown anymore.
Center
Corresponding Graphs
(0, 0)
x2 y 2
+ 2 =1
a2
b
x2 y 2
+ 2 =1
b2
a
(h, k)
(x
h)2
a2
+
k)2
(y
b2
=1
h)2
(x
b2
+
(y
k)2
a2
=1
major axis: horizontal
major axis: vertical
minor axis: vertical
minor axis: horizontal
40
p
In all four cases above, a > b and c = a2 b2 . The foci F1 and F2 are c
units away from the center. The vertices V1 and V2 are a units away from the
center, the major axis has length 2a, the covertices W1 and W2 are b units away
from the center, and the minor axis has length 2b. Recall that, for any point on
the ellipse, the sum of its distances from the foci is 2a.
In the standard equation, if the x-part has the bigger denominator, the ellipse
is horizontal. If the y-part has the bigger denominator, the ellipse is vertical.
Example 1.3.3. Give the coordinates of the center, foci, vertices, and covertices
of the ellipse with the given equation. Sketch the graph, and include these points.
(x + 3)2 (y 5)2
+
=1
24
49
(2) 9x2 + 16y 2 126x + 64y = 71
(1)
p
Solution. p(1) From a2 = 49 and b2 = 24, we have a = 7, b = 2 6 ⇡ 4.9, and
c = a2 b2 = 5. The ellipse is vertical.
center:
foci:
vertices:
covertices:
( 3, 5)
F1 ( 3, 0), F2 ( 3, 10)
V1 ( 3, 2), V2 ( 3, 12)
p
W1 ( 3 2 6, 5) ⇡ ( 7.9, 5)
p
W2 ( 3 + 2 6, 5) ⇡ (1.9, 5)
41
(2) We first change the given equation to standard form.
9(x2 14x) + 16(y 2 + 4y) = 71
9(x2 14x + 49) + 16(y 2 + 4y + 4) = 71 + 9(49) + 16(4)
9(x 7)2 + 16(y + 2)2 = 576
(x 7)2 (y + 2)2
+
=1
64
36
p
p
We have a = 8 and b = 6. Thus, c = a2 b2 = 2 7 ⇡ 5.3. The ellipse is
horizontal.
center:
foci:
vertices:
covertices:
(7, 2)
p
2 7, 2) ⇡ (1.7, 2)
p
F2 (7 + 2 7, 2) ⇡ (12.3, 2)
F1 (7
V1 ( 1, 2), V2 (15, 2)
W1 (7, 8), W2 (7, 4)
Example 1.3.4. The foci of an ellipse are ( 3, 6) and ( 3, 2). For any point
on the ellipse, the sum of its distances from the foci is 14. Find the standard
equation of the ellipse.
42
Solution. The midpoint ( 3, 2) of the foci is the center of the ellipse. The
ellipse is vertical (because the foci are vertically
p aligned)pand c = 4. From the
given sum, 2a = 14 so a = 7. Also, b = a2 c2 = 33. The equation is
(x + 3)2 (y + 2)2
+
= 1.
2
33
49
p
p
Example 1.3.5. An ellipse has vertices (2
61, 5) and (2 + 61, 5), and
its minor axis is 12 units long. Find its standard equation and its foci.
Solution. The midpoint (2, 5)pof the vertices is the center of the ellipse, which is
horizontal. Each vertex is a = 61 units away from the center. From the length of
(x 2)2 (y + 5)2
the minor axis, 2b = 12 so b = 6. The standard equation is
+
=
61
36
p
1. Each focus is c = a2 b2 = 5 units away from (2, 5), so their coordinates
are ( 3, 5) and (7, 5).
2
Seatwork/Homework 1.3.2
1. Give the coordinates of the center, foci, vertices, and covertices of the ellipse
with equation 41x2 + 16y 2 + 246x 192y + 289 = 0. Sketch the graph, and
include these points.
Answer:
center C( 3,
p
p6), foci F1 ( 3, 1) and F2 ( 3, 11), vertices V1 ( 3, 6
41) and V2 ( 3, 6 + 41), covertices W1 ( 7, 6) and W2 (1, 6)
43
2. An ellipse has vertices ( 10, 4) and (6, 4), and covertices ( 2, 9) and
( 2, 1). Find its standard equation and its foci.
p
p
(x + 2)2 (y + 4)2
Answer:
+
= 1, foci ( 2
39, 4) and ( 2 + 39, 4)
64
25
1.3.3. Situational Problems Involving Ellipses
We now apply the concept of ellipse to some situational problems.
? Example 1.3.6. A tunnel has the shape of a semiellipse that is 15 ft high at
the center, and 36 ft across at the base. At most how high should a passing truck
be, if it is 12 ft wide, for it to be able to fit through the tunnel? Round o↵ your
answer to two decimal places.
Solution. Refer to the figure above. If we draw the semiellipse on a rectangular
coordinate system, with its center at the origin, an equation of the ellipse which
contains it, is
x2
y2
+
= 1.
182 152
To maximize its height, the corners of the truck, as shown in the figure, would
have to just touch the ellipse. Since the truck is 12 ft wide, let the point (6, n)
be the corner of the truck in the first quadrant, where n > 0, is the (maximum)
height of the truck. Since this point is on the ellipse, it should fit the equation.
Thus, we have
62
n2
+
=1
182 152
2
n = 15
2
✓
1
62
182
◆
p
n = 10 2 ⇡ 14.14 ft
44
2
Example 1.3.7. The orbit of a planet has the shape of an ellipse, and on one
of the foci is the star around which it revolves. The planet is closest to the star
when it is at one vertex. It is farthest from the star when it is at the other vertex.
Suppose the closest and farthest distances of the planet from this star, are 420
million kilometers and 580 million kilometers, respectively. Find the equation of
the ellipse, in standard form, with center at the origin and the star at the x-axis.
Assume all units are in millions of kilometers.
Solution. In the figure above, the orbit is drawn as a horizontal ellipse with
center at the origin. From the planet’s distances from the star, at its closest
and farthest points, it follows that the major axis is 2a = 420 + 580 = 1000
(million kilometers), so a = 500. If we place the star at the positive x-axis,
then it is c = 500 420 = 80 units away from the center. Therefore, we get
b2 = a2 c2 = 5002 802 = 243600. The equation then is
x2
y2
+
= 1.
250000 243600
The star could have been placed on the negative x-axis, and the answer would
still be the same.
2
Seatwork/Homework 1.3.3
? 1. The arch of a bridge is in the shape of a semiellipse, with its major axis at the
water level. Suppose the arch is 20 ft high in the middle, and 120 ft across its
major axis. How high above the water level is the arch, at a point 20 ft from
the center (horizontally)? Round o↵ to 2 decimal places. Refer to Example
1.3.6.
Answer: 18.86 ft
45
Exercises 1.3
1. Give the coordinates of the center, vertices, covertices, and foci of the ellipse
with the given equation. Sketch the graph, and include these points.
x2
y2
(a)
+
=1
169 25
x2
y2
(b)
+
=1
144 169
(c) 4x2 + 13y 2 = 52
(x + 7)2 (y 4)2
+
=1
16
25
(e) 9x2 + 16y 2 + 72x 96y + 144 = 0
(d)
(f) 36x2 + 20y 2
Answer:
144x + 120y
396 = 0
Item
Center
Vertices
Covertices
Foci
(a)
(0, 0)
(±13, 0)
(0, ±5)
(±12, 0)
(b)
(0, 0)
(±12, 0)
(0, ±5)
(c)
(0, 0)
(0, ±13)
p
(± 13, 0)
(0, ±2)
(±3, 0)
(d)
( 7, 4)
( 7, 1)
( 11, 4)
( 7, 1)
( 7, 9)
( 3, 4)
( 8, 3)
( 4, 0)
( 7, 7)
p
( 4 ± 7, 3)
(0, 3)
( 4, 6)
p
(2 ± 2 5, 3)
(e)
(f)
( 4, 3)
(2, 3)
(2, 9)
(2, 3)
(2, 3)
(2, 7)
(2, 1)
(a)
(b)
46
(c)
(d)
(e)
(f)
2. Find the standard equation of the ellipse which satisfies the given conditions.
(a) foci ( 7, 6) and ( 1, 6), the sum of the distances of any point from the
(x + 4)2 (y 6)2
foci is 14
Answer:
+
=1
49
40
(b) center (5, 3), horizontal major axis of length 20, minor axis of length 16
(x 5)2 (y 3)2
Answer:
+
=1
100
64
(c) major axis of length 22, foci 9 units above and below the center (2, 4)
(x 2)2 (y 4)2
Answer:
+
=1
40
121
(d) covertices ( 4, 8) and (10, 8), a focus at (3, 12)
(x 3)2 (y 8)2
Answer:
+
=1
49
65
Solution. The midpoint of the covertices is the center, (3, 8). From this
point, the given focus is c = 4 units away. Since b = 7 (the distance from
the center to a covertex), then a2 = b2 + c2 = 65. The ellipse then has
(x 3)2 (y 8)2
equation
+
= 1.
49
65
(e) focus ( 6, 2), covertex ( 1, 5), horizontal major axis
(x + 1)2 (y + 2)2
Answer:
+
=1
74
49
Solution. Make a rough sketch of the points to see that the center is to
the right of the given focus, and below the given covertex. The center is
thus ( 1, 2). It follows that c = 5, b = 7, so a2 = b2 + c2 = 74. The
(x + 1)2 (y + 2)2
ellipse then has equation
+
= 1.
74
49
3. A semielliptical tunnel has height 9 ft and a width of 30 ft. A truck that is
about to pass through is 12 ft wide and 8.3 ft high. Will this truck be able to
pass through the tunnel?
Answer: No
47
4. A truck that is about to pass through the tunnel from the previous item is 10
ft wide and 8.3 ft high. Will this truck be able to pass through the tunnel?
Answer: Yes
5. An orbit of a satellite around a planet is an ellipse, with the planet at one
focus of this ellipse. The distance of the satellite from this star varies from
300, 000 km to 500, 000 km, attained when the satellite is at each of the two
vertices. Find the equation of this ellipse, if its center is at the origin, and the
vertices are on the x-axis. Assume all units are in 100, 000 km.
Answer:
x2
16
6. The orbit of a planet around a star is described by the equation
y2
+
y2
15
=1
x2
+
640,000
= 1, where the star is at one focus, and all units are in millions of
630,000
kilometers. The planet is closest and farthest from the star, when it is at the
vertices. How far is the planet when it is closest to the sun? How far is the
planet when it is farthest from the sun?
Answer: 700 million km, 900 million km
Solution. The ellipse has center at the origin, and major axis on the x-axis.
Since a2 = 640, 000, then a = 800, so p
the verticespare V1 ( 800, 0) and
2
V2 (800, 00). Since b = 630, 000, then c = a2 b2 = 10, 000 = 100. Suppose the star is at the focus at the right of the origin (this choice is arbitrary,
since we could have chosen instead the focus on the left). Its location is then
F (100, 0). The closest distance is then V2 F = 700 (million kilometers) and
the farthest distance is V1 F = 900 (million kilometers).
7. A big room is constructed so that the ceiling is a dome that is semielliptical
in shape. If a person stands at one focus and speaks, the sound that is made
bounces o↵ the ceiling and gets reflected to the other focus. Thus, if two
people stand at the foci (ignoring their heights), they will be able to hear each
other. If the room is 34 m long and 8 m high, how far from the center should
each of two people stand if they would like to whisper back and forth and hear
each other?
Answer: 15 m
48
Solution. We could put a coordinate system with the floor of the room on
the x-axis, and the center of the room at the origin, as shown in the figures.
The major axis has length 34, and the height of the room is half of the minor
y2
x2
axis. The ellipse that contains the ceiling then has equation 17
2 + 82 = 1. The
p
p
distance of a focus from the center is c = a2 b2 = 172 82 = 15. Thus,
the two people should stand 15 m away from the center.
8. A whispering gallery has a semielliptical ceiling that is 9 m high and 30 m
long. How high is the ceiling above the two foci?
Answer: 5.4 m
Solution. As in the previous problem, put a coordinate system with the floor
of the room on the x-axis, and the center of the room at the origin. The major
axis has length 30, and half the minor axis is 9. The ellipse that contains the
y2
x2
ceiling then has equation 15
2 + 92 = 1. The distance of a focus from the center
p
p
is c = a2 b2 = 152 92 = 12. If we put x = 12 in the equation of the
2
2
ellipse, we get 12
+ y92 = 1. Solving for y > 0 yields y = 27
= 5.4. The height
152
5
of the ceiling above each focus is 5.4 m.
4
Lesson 1.4. Hyperbolas
Time Frame: 3 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) define a hyperbola;
(2) determine the standard form of equation of a hyperbola;
(3) graph a hyperbola in a rectangular coordinate system; and
(4) solve situational problems involving conic sections (hyperbolas).
Lesson Outline
(1) Definition of a hyperbola
(2) Derivation of the standard equation of a hyperbola
(3) Graphing hyperbolas
(4) Solving situational problems involving hyperbolas
49
Introduction
A hyperbola is one of the conic sections that most students have not encountered formally before, unlike circles and parabolas. Its graph consists of two
unbounded branches which extend in opposite directions. It is a misconception
that each branch is a parabola. This is not true, as parabolas and hyperbolas
have very di↵erent features. An application of hyperbolas in basic location and
navigation schemes are presented in an example and some exercises.
1.4.1. Definition and Equation of a Hyperbola
Consider the points F1 ( 5, 0) and F2 (5, 0) as shown in Figure 1.21. What is the
absolute value of the di↵erence of the distances of A(3.75, 3) from F1 and from
F2 ? How about the absolute value of the di↵erence of the distances of B 5, 16
3
from F1 and from F2 ?
|AF1
|BF1
AF2 | = |9.25 3.25| = 6
16 34
BF2 | =
=6
3
3
There are other points P such that |P F1 P F2 | = 6. The collection of all such
points forms a shape called a hyperbola, which consists of two disjoint branches.
For points P on the left branch, P F2 P F1 = 6; for those on the right branch,
P F1 P F2 = 6.
Figure 1.21
Figure 1.22
Let F1 and F2 be two distinct points. The set of all points P , whose
distances from F1 and from F2 di↵er by a certain constant, is called a
hyperbola. The points F1 and F2 are called the foci of the hyperbola.
In Figure 1.22, given are two points on the x-axis, F1 ( c, 0) and F2 (c, 0), the
foci, both c units away from their midpoint (0, 0). This midpoint is the center
50
Figure 1.23
Figure 1.24
of the hyperbola. Let P (x, y) be a point on the hyperbola, and let the absolute
value of the di↵erence of the distances of P from F1 and F2 , be 2a (the coefficient
2 will make computations simpler). Thus, |P F1 P F2 | = 2a, and so
p
(x + c)2 + y 2
p
(x
c)2 + y 2 = 2a.
Algebraic manipulations allow us to rewrite this into the much simpler
x2
a2
y2
= 1,
b2
where b =
p
c2
a2 .
p
When we let b = c2 a2 , we assumed c > a. To see why this is true, suppose
that P is closer to F2 , so P F1 P F2 = 2a. Refer to Figure 1.22. Suppose also
that P is not on the x-axis, so 4P F1 F2 is formed. From the triangle inequality,
F1 F2 + P F2 > P F1 . Thus, 2c > P F1 P F2 = 2a, so c > a.
Now we present a derivation. For now, assume P is closer to F2 so P F1 > P F2 ,
P F2 = 2a.
Teaching Notes and P F1
If it is assumed
that P is closer to
F1 , then the same
equation will be
obtained because
of symmetry.
P F1 = 2a + P F2
p
p
(x + c)2 + y 2 = 2a + (x c)2 + y 2
⇣p
⌘2 ⇣
⌘2
p
2
2
2
2
(x + c) + y
= 2a + (x c) + y
p
cx a2 = a (x c)2 + y 2
⇣ p
⌘2
(cx a2 )2 = a (x c)2 + y 2
(c2
a2 )x2
a2 y 2 = a2 (c2
a2 )
b 2 x 2 a2 y 2 = a2 b 2
x2 y 2
=1
a2
b2
by letting b =
51
p
c2
a2 > 0
We collect here the features of the graph of a hyperbola with standard equation
x2 y 2
= 1.
a2
b2
p
Let c = a2 + b2 .
(1) center : origin (0, 0)
(2) foci : F1 ( c, 0) and F2 (c, 0)
• Each focus is c units away from the center.
• For any point on the hyperbola, the absolute value of the di↵erence of
its distances from the foci is 2a.
(3) vertices: V1 ( a, 0) and V2 (a, 0)
• The vertices are points on the hyperbola, collinear with the center and
foci.
• If y = 0, then x = ±a. Each vertex is a units away from the center.
• The segment V1 V2 is called the transverse axis. Its length is 2a.
(4) asymptotes: y = ab x and y =
b
x,
a
the lines `1 and `2 in Figure 1.24
• The asymptotes of the hyperbola are two lines passing through the center which serve as a guide in graphing the hyperbola: each branch of
the hyperbola gets closer and closer to the asymptotes, in the direction
towards which the branch extends. (We need the concept of limits from
calculus to explain this.)
• An aid in determining the equations of the asymptotes: in the standard
2
2
equation, replace 1 by 0, and in the resulting equation xa2 yb2 = 0, solve
for y.
• To help us sketch the asymptotes, we point out that the asymptotes
`1 and `2 are the extended diagonals of the auxiliary rectangle drawn
in Figure 1.24. This rectangle has sides 2a and 2b with its diagonals
intersecting at the center C. Two sides are congruent and parallel to
the transverse axis V1 V2 . The other two sides are congruent and parallel
to the conjugate axis, the segment shown which is perpendicular to the
transverse axis at the center, and has length 2b.
Example 1.4.1. Determine the foci, vertices, and asymptotes of the hyperbola
with equation
x2 y 2
= 1.
9
7
Sketch the graph, and include these points and lines, the transverse and conjugate
axes, and the auxiliary rectangle.
52
2
2
Solution. With
p a = 9 and
p b = 7, we have
a = 3, b = 7, and c = a2 + b2 = 4.
foci: F1 ( 4, 0) and F2 (4, 0)
vertices: V1 ( 3, 0) and V2 (3, 0)
p
p
asymptotes: y = 37 x and y = 37 x
The graph is shown at the right. The conjup
gate axis drawn has its endpoints b = 7 ⇡
2.7 units above and below the center.
2
Example 1.4.2. Find the (standard) equation of the hyperbola whose foci are
F1 ( 5, 0) and F2 (5, 0), such that for any point on it, the absolute value of the
di↵erence of its distances from the foci is 6. See Figure 1.21.
Solution. We have 2a = 6 and c = 5, so a = 3 and b =
x2 y 2
hyperbola then has equation
= 1.
9
16
p
c2
a2 = 4. The
2
Seatwork/Homework 1.4.1
1. Determine foci, vertices, and asymptotes of the hyperbola with equation
x2
y2
= 1.
16 20
Sketch the graph, and include these points and lines, along with the auxiliary
rectangle.
Answer: foci
F1 ( 6, 0) and
F2 (6, 0), vertices V1 ( 4, 0) and V2 (4, 0), asympp
p
5
5
totes y = 2 x and y = 2 x
p
2. Find the equation
in
standard
form
of
the
hyperbola
whose
foci
are
F
(
4
2, 0)
1
p
and F2 (4 2, 0), such that for any point on it, the absolute value of the
x2
y2
di↵erence of its distances from the foci is 8.
Answer:
=1
16 16
53
1.4.2. More Properties of Hyperbolas
The hyperbolas we considered so far are “horizontal” and have the origin as their
centers. Some hyperbolas have their foci aligned vertically, and some have centers
not at the origin. Their standard equations and properties are given in the box.
The derivations are more involved, but are similar to the one above, and so are
not shown anymore.
Center
Corresponding Hyperbola
(0, 0)
x2
a2
y2
=1
b2
y2
a2
x2
=1
b2
(h, k)
(x
h)2
a2
k)2
(y
b2
=1
(y
k)2
a2
h)2
(x
b2
=1
transverse axis: horizontal
transverse axis: vertical
conjugate axis: vertical
conjugate axis: horizontal
54
p
In all four cases above, we let c = a2 + b2 . The foci F1 and F2 are c units
away from the center C. The vertices V1 and V2 are a units away from the center.
The transverse axis V1 V2 has length 2a. The conjugate axis has length 2b and is
perpendicular to the transverse axis. The transverse and conjugate axes bisect
each other at their intersection point, C. Each branch of a hyperbola gets closer
and closer to the asymptotes, in the direction towards which the branch extends.
The equations of the asymptotes can be determined by replacing 1 in the standard
equation by 0. The asymptotes can be drawn as the extended diagonals of the
auxiliary rectangle determined by the transverse and conjugate axes. Recall that,
for any point on the hyperbola, the absolute value of the di↵erence of its distances
from the foci is 2a.
In the standard equation, aside from being positive, there are no other restrictions on a and b. In fact, a and b can even be equal. The orientation of the
hyperbola is determined by the variable appearing in the first term (the positive
term): the corresponding axis is where the two branches will open. For example,
if the variable in the first term is x, the hyperbola is “horizontal”: the transverse
axis is horizontal, and the branches open to the left and right in the direction of
the x-axis.
Example 1.4.3. Give the coordinates of the center, foci, vertices, and asymptotes of the hyperbola with the given equation. Sketch the graph, and include
these points and lines, the transverse and conjugate axes, and the auxiliary rectangle.
(y + 2)2 (x 7)2
=1
25
9
(2) 4x2 5y 2 + 32x + 30y = 1
(1)
2
2
Solution.
(1) From
p a = 25 and b = 9, we have a = 5, b = 3, and c =
p
a2 + b2 = 34 ⇡ 5.8. The hyperbola is vertical. To determine the asymp2
(x 7)2
totes, we write (y+2)
= 0, which is equivalent to y + 2 = ± 53 (x 7).
25
9
We can then solve this for y.
center: C(7, 2)
foci: F1 (7, 2
p
34) ⇡ (7, 7.8) and F2 (7, 2 +
vertices: V1 (7, 7) and V2 (7, 3)
asymptotes: y = 53 x
41
3
and y =
5
x
3
+
p
34) ⇡ (7, 3.8)
29
3
The conjugate axis drawn has its endpoints b = 3 units to the left and right
of the center.
55
(2) We first change the given equation to standard form.
4(x2 + 8x) 5(y 2 6y) = 1
4(x2 + 8x + 16) 5(y 2 6y + 9) = 1 + 4(16) 5(9)
4(x + 4)2 5(y 3)2 = 20
(x + 4)2 (y 3)2
=1
5
4
p
p
We have a = 5 ⇡ 2.2 and b = 2. Thus, c = a2 + b2 = 3. The hyperbola
2
(y 3)2
is horizontal. To determine the asymptotes, we write (x+4)
= 0
5
4
2
p
which is equivalent to y 3 = ± 5 (x + 4), and solve for y.
center: C( 4, 3)
foci: F1 ( 7, 3) and F2 ( 1, 3)
p
p
vertices: V1 ( 4
5, 3) ⇡ ( 6.2, 3) and V2 ( 4 + 5, 3) ⇡ ( 1.8, 3)
asymptotes: y =
p2 x
5
+
p8
5
+ 3 and y =
p2 x
5
p8
5
+3
The conjugate axis drawn has its endpoints b = 2 units above and below
the center.
56
Example 1.4.4. The foci of a hyperbola are ( 5, 3) and (9, 3). For any point
on the hyperbola, the absolute value of the di↵erence of its of its distances from
the foci is 10. Find the standard equation of the hyperbola.
Solution. The midpoint (2, 3) of the foci is the center of the hyperbola. Each
focus is c = 7 units away from the center. From the given di↵erence, 2a = 10 so
a = 5. Also, b2 = c2 a2 = 24. The hyperbola is horizontal (because the foci are
horizontally aligned), so the equation is
(x
2)2
25
(y + 3)2
= 1.
24
2
Example 1.4.5.
p A hyperbola has vertices ( 4, 5) and ( 4, 9), and one of its
foci is ( 4, 2
65). Find its standard equation.
Solution. The midpoint (
which is vertical (because
a = 7 units away from the
the center. Thus, b2 = c2
4, 2) of the vertices is the center of the hyperbola,
the vertices are vertically aligned).
Each vertex is
p
center. The given focus is c = 65 units away from
a2 = 16, and the standard equation is
(y
2)2
49
(x + 4)2
= 1.
16
57
2
Seatwork/Homework 1.4.2
1. Give the coordinates of the center, foci, vertices, and asymptotes of the hyperbola with equation 9x2 4y 2 90x 32y = 305. Sketch the graph, and
include these points and lines, along with the auxiliary rectangle.
p
p
Answer: center C(5, 4), foci F1 (5, 4 2 13) and F2 (5, 4+2 13), vertices
V1 (5, 10) and V2 (5, 2), asymptotes y = 32 x + 72 and y = 32 x 23
2
2. A hyperbola has vertices (1, 9) and (13, 9), and one of its foci is ( 2, 9). Find
(x 7)2 (y 9)2
its standard equation.
Answer:
=1
36
45
1.4.3. Situational Problems Involving Hyperbolas
We now give an example on an application of hyperbolas.
Example 1.4.6. An explosion is heard by two stations 1200 m apart, located at
F1 ( 600, 0) and F2 (600, 0). If the explosion was heard in F1 two seconds before
it was heard in F2 , identify the possible locations of the explosion. Use 340 m/s
as the speed of sound.
Solution. Using the given speed of sound, we deduce that the sound traveled
340(2) = 680 m farther in reaching F2 than in reaching F1 . This is then the
di↵erence of the distances of the explosion from the two stations. Thus, the
explosion is on a hyperbola with foci are F1 and F2 , on the branch closer to F1 .
58
We have c = 600 and 2a = 680, so a = 340 and b2 = c2 a2 = 244400.
The explosion could therefore be anywhere on the left branch of the hyperbola
y2
x2
= 1.
2
115600
244400
Seatwork/Homework 1.4.3
? 1. Two stations, located at M ( 1.5, 0) and N (1.5, 0) (units are in km), simultaneously send sound signals to a ship, with the signal traveling at the speed of
0.33 km/s. If the signal from N was received by the ship four seconds before
the signal it received from M , find the equation of the curve containing the
y2
x2
possible location of the ship.
Answer: 0.4356
= 1 (right branch)
1.8144
Exercises 1.4
1. Give the coordinates of the center, foci, vertices, and the asymptotes of the
hyperbola with the given equation. Sketch the graph, and include these points
and lines.
x2
36
y2
(b)
25
(c) (x
(a)
y2
=1
64
x2
=1
16
1)2 y 2 = 4
(y + 2)2 (x + 3)2
=1
15
10
(e) 3x2 2y 2 42x 16y =
(d)
(f) 25x2
67
39y 2 + 150x + 390y =
225
59
Answer:
Item
Center
Vertices
Foci
(a)
(0, 0)
(±6, 0)
(b)
(0, 0)
(0, ±5)
(c)
(1, 0)
(d)
( 3, 2)
( 1, 0), (3, 0)
p
( 3, 2 ± 15)
(±10, 0)
p
(0, ± 41)
p
(1 ± 2 2, 0)
(e)
(7, 4)
(3, 4), (11, 4)
(f)
( 3, 5)
( 3, 0), ( 3, 10)
( 3, 7), ( 3, 3)
p
(7 ± 2 10, 4)
( 3, 3), ( 3, 13)
Item
Asymptotes
(a)
y = ± 43 x
y = ± 54 x
(b)
(c)
(d)
(e)
(f)
y=x
x+1
q
y = ± 32 x ± 3 32 2
q
q
3
y = ± 2 x ⌥ 7 32 4
q
1, y =
y = ± p539 x ±
(a)
p15
39
+5
(b)
60
(c)
(d)
(e)
(f)
2. Find the standard equation of the hyperbola which satisfies the given conditions.
(a) foci ( 4, 3) and ( 4, 13), the absolute value of the di↵erence of the
distances of any point from the foci is 14
(y 5)2 (x + 4)2
Answer:
=1
49
15
(b) vertices ( 2, 8) and (8, 8), a focus (12, 8)
(x 3)2 (y 8)2
Answer:
=1
25
56
(c) center ( 6, 9), a vertex ( 6, 15), conjugate axis of length 12
(y 9)2 (x + 6)2
Answer:
=1
25
36
(d) asymptotes y = 43 x + 13 and y = 43 x + 41
, a vertex ( 1, 7)
3
(x 5)2 (y 7)2
Answer:
=1
36
64
Solution. The asymptotes intersect at (5, 7). This is the center. The
distance of the given vertex from the center is a = 6. This vertex and
center are aligned horizontally, so the hyperbola has equation of the form
(x h)2
(y k)2
= 1. The asymptotes consequently have the form y k =
a2
b2
61
± ab (x h), and thus, have slopes ± ab . From the given asymptotes,
Since a = 6, then b = 8. The standard equation is then
(x
(e) asymptotes y = 13 x +
5
3
5)2
36
(y
b
a
= 43 .
7)2
= 1.
64
+ 73 , a focus (1, 12)
(y 2)2 (x 1)2
Answer:
=1
10
90
Solution. The asymptotes intersect at (1, 2). This is the center. The
distance of the given focus from the center is c = 10. This focus and
center are aligned vertically, so the hyperbola has equation of the form
(y k)2
(x h)2
= 1. The asymptotes consequently have the form y k =
a2
b2
a
± b (x h), and thus, have slopes ± ab . From the given asymptotes, ab = 13 ,
so b = 3a.
c2 = 100 = a2 + b2 = a2 + (3a)2 = 10a2
and y =
1
x
3
Thus, a2 = 10, and b2 = 9a2 = 90. The standard equation is
(y
2)2
10
(x
1)2
= 1.
90
3. Two control towers are located at points Q( 500, 0) and R(500, 0), on a
straight shore where the x-axis runs through (all distances are in meters).
At the same moment, both towers sent a radio signal to a ship out at sea, each
traveling at 300 m/µs. The ship received the signal from Q 3 µs (microseconds)
before the message from R.
(a) Find the equation of the curve containing the possible location of the
x2
y2
ship.
Answer:
= 1 (left branch)
202500 47500
(b) Find the coordinates (rounded o↵ to two decimal places) of the ship if it
is 200 m from the shore (y = 200).
Answer: ( 610.76, 200)
Solution. Since the time delay between the two signals is 3 µs, then the di↵erence between the distances traveled by the two signals is 300 · 3 = 900 m. The
ship is then on a hyperbola, consisting of points whose distances from Q and R
(the foci) di↵er by 2a = 900. With a = 450 and c = 500 (the distance of each
focus from the center, the origin), we have b2 = c2 a2 = 5002 4502 = 47500.
y2
x2
Since a2 = 202500, the hyperbola then has equation 202500
= 1. Since
47500
the signal from Q was received first, the ship is closer to Q than R, so the
ship is on the left branch of this hyperbola. Using y = 200, we then solve
x2
2002
= 1 for x < 0 (left branch), and we get x ⇡ 610.76.
202500
47500
4
62
Lesson 1.5. More Problems on Conic Sections
Time Frame: 2 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) recognize the equation and important characteristics of the di↵erent types of
conic sections; and
(2) solve situational problems involving conic sections.
Lesson Outline
(1) Conic sections with associated equations in general form
(2) Problems involving characteristics of various conic sections
(3) Solving situational problems involving conic sections
Introduction
Inspecting the equation can lead us to the right conic section for its graph,
and set us on the right step towards analyzing it. We will also look at problems
that use the properties of the di↵erent conic sections, allowing us to synthesize
what has been covered so far.
1.5.1. Identifying the Conic Section by Inspection
The equation of a circle may be written in standard form
Ax2 + Ay 2 + Cx + Dy + E = 0,
that is, the coefficients of x2 and y 2 are the same. However, it does not follow
that if the coefficients of x2 and y 2 are the same, the graph is a circle.
(A)
(B)
General Equation
2x2 + 2y 2 2x + 6y + 5 = 0
x2 + y 2 6x 8y + 50 = 0
Standard Equation
2
2
x 12 + y + 32 = 0
(x 3)2 + (y 4)2 = 25
graph
point
empty set
For a circle with equation (x h)2 + (y k)2 = r2 , we have r2 > 0. This is
not the case for the standard equations of (A) and (B).
In (A), because the sum of two squares can only be 0 if and only if each square
is 0, it follows that x 12 = 0 and y + 32 = 0. The graph is thus the single point
1
, 32 .
2
In (B), no real values of x and y can make the nonnegative left side equal to
the negative right side. The graph is then the empty set.
63
Let us recall the general form of the equations of the other conic sections. We
may write the equations of conic sections we discussed in the general form
Ax2 + By 2 + Cx + Dy + E = 0.
Some terms may vanish, depending on the kind of conic section.
(1) Circle: both x2 and y 2 appear, and their coefficients are the same
Ax2 + Ay 2 + Cx + Dy + E = 0
Example: 18x2 + 18y 2
24x + 48y
5=0
Degenerate cases: a point, and the empty set
(2) Parabola: exactly one of x2 or y 2 appears
Ax2 + Cx + Dy + E = 0 (D 6= 0, opens upward or downward)
By 2 + Cx + Dy + E = 0 (C 6= 0, opens to the right or left)
Examples: 3x2 12x + 2y + 26 = 0 (opens downward)
2y 2 + 3x + 12y 15 = 0 (opens to the right)
(3) Ellipse: both x2 and y 2 appear, and their coefficients A and B have the same
sign and are unequal
Examples: 2x2 + 5y 2 + 8x
4x2 + y 2 16x
10y 7 = 0 (horizontal major axis)
6y + 21 = 0 (vertical major axis)
If A = B, we will classify the conic as a circle, instead of an ellipse.
Degenerate cases: a point, and the empty set
(4) Hyperbola: both x2 and y 2 appear, and their coefficients A and B have different signs
Examples: 5x2 3y 2 20x 18y
4x2 + y 2 + 24x + 4y
22 = 0 (horizontal transverse axis)
36 = 0 (vertical transverse axis)
Degenerate case: two intersecting lines
The following examples will show the possible degenerate conic (a point, two
intersecting lines, or the empty set) as the graph of an equation following a similar
pattern as the non-degenerate cases.
(1) 4x2 + 9y 2
16x + 18y + 25 = 0
=)
=)
(2) 4x2 + 9y 2
16x + 18y + 61 = 0
=)
=)
64
(x
2)2
(y + 1)2
=0
32
22
one point: (2, 1)
(x
2)2
+
+
32
empty set
(y + 1)2
=
22
1
(3) 4x2
9y 2
16x
18y + 7 = 0
=)
=)
(y + 1)2
=0
32
22
2
two lines: y + 1 = ± (x
3
(x
2)2
2)
A Note on Identifying a Conic Section
by Its General Equation
It is only after transforming a given general equation to standard
form that we can identify its graph either as one of the degenerate
conic sections (a point, two intersecting lines, or the empty set) or as
one of the non-degenerate conic sections (circle, parabola, ellipse, or
hyperbola).
Seatwork/Homework 1.5.1
The graphs of the following equations are (nondegenerate) conic sections. Identify
the conic section.
(1) 5x2
3y 2 + 10x
(2) 2y 2
5x
12y = 22
Answer: hyperbola
12y = 17
(3) 3x2 + 3y 2 + 42x
Answer: parabola
12y =
154
Answer: circle
2
(4) 3x + 6x + 4y = 18
2
(5) 7x + 3y
(6)
2
Answer: parabola
14x + 12y =
4x2 + 3y 2 + 24x
14
Answer: ellipse
12y = 36
Answer: hyperbola
1.5.2. Problems Involving Di↵erent Conic Sections
The following examples require us to use the properties of di↵erent conic sections
at the same time.
Example 1.5.1. A circle has center at the focus of the parabola y 2 + 16x + 4y =
44, and is tangent to the directrix of this parabola. Find its standard equation.
Solution. The standard equation of the parabola is (y + 2)2 = 16(x 3). Its
vertex is V (3, 2). Since 4c = 16 or c = 4, its focus is F ( 1, 2) and its directrix
is x = 7. The circle has center at ( 1, 2) and radius 8, which is the distance
from F to the directrix. Thus, the equation of the circle is
(x + 1)2 + (y + 2)2 = 64.
2
Example 1.5.2. The vertices and foci of 5x2 4y 2 + 50x + 16y + 29 = 0 are,
respectively, the foci and vertices of an ellipse. Find the standard equation of
this ellipse.
65
Solution. We first write the equation of the hyperbola in standard form:
(x + 5)2
16
(y
2)2
= 1.
20
For this hyperbola, using the notations ah , bh , and ch to refer to a, b, and p
c of
the standard
equation of the hyperbola, respectively, we have ah = 4, bh = 2 5,
p
2
2
ch = ah + bh = 6, so we have the following points:
center: ( 5, 2)
vertices: ( 9, 2) and ( 1, 2)
foci: ( 11, 2) and (1, 2).
It means that, for the ellipse, we have these points:
center: ( 5, 2)
vertices: ( 11, 2) and (1, 2)
foci: ( 9, 2) and ( 1, 2).
In this case, ce = 4 and ae = 6, so that be =
equation of the ellipse is
p
a2e
c2e =
p
20. The standard
(x + 5)2 (y 2)2
+
= 1.
36
20
2
Seatwork/Homework 1.5.2
1. Find the standard equation of all circles having center at a focus of 21x2
4y 2 + 84x 24y = 36 and passing through the farther vertex.
Answer: (x + 7)2 + (y + 3)2 = 49, (x
3)2 + (y + 3)2 = 49
2. Find the standard equation of the hyperbola one branch of which has focus and
vertex that are the same as those of x2 6x + 8y = 23, and whose conjugate
axis is on the directrix of the same parabola.
(y 6)2 (x 3)2
Answer:
=1
4
12
Exercises 1.5
1. The graphs of the following equations are non-degenerate conic sections. Identify the conic section.
(a) 5x2 + 7y 2
40x
2
30y =
(b) 5y + 2x
(c) 3x2
3y 2 + 12x
28y =
73
Answer: ellipse
49
Answer: parabola
12y = 5
Answer: hyperbola
(d) 3x2 + 3y 2 + 12x + 12y = 4
Answer: circle
66
(e) 2x2 + 24x
5y =
57
Answer: parabola
2. The graphs of the following equations are degenerate conic sections. What are
the specific graphs?
(a) x2 + 3y 2
(b) 9x2
4x + 24y =
4y 2 + 18x
(c) 3x2 + 5y 2
6x
52
Answer: point: (2, 4)
16y = 7
Answer: lines: y + 2 = ± 32 (x + 1)
20y =
25
Answer: empty set
3. An ellipse has equation 25x2 + 16y 2 + 150x 32y = 159. Find the standard
equations of all parabolas whose vertex is a focus of this ellipse and whose
focus is a vertex of this ellipse.
Answer: (x + 3)2 = 8(y + 2), (x + 3)2 = 32(y + 2), (x + 3)2 =
and (x + 3)2 = 8(y 4)
32(y
4),
Solution. The standard equation of the ellipse is
(x + 3)2 (y 1)2
+
= 1.
16
25
Its center is ( 3, 1). Since a = 5 and b = 4, we get c = 3, so the vertices are
P ( 3, 4) and S( 3, 6), while its foci are Q( 3, 2) and R( 3, 4). We then
get four parabolas satisfying the conditions of the problem. The focal distance
indicated below is the distance from the vertex to the focus.
vertex
focus
focal distance
standard equation
Q( 3, 2)
P ( 3, 4)
2
(x + 3)2 =
Q( 3, 2)
S( 3, 6)
8
(x + 3)2 = 32(y + 2)
R( 3, 4)
P ( 3, 4)
8
R( 3, 4)
S( 3, 6)
2
(x + 3)2 =
8(y + 2)
32(y
(x + 3)2 = 8(y
4)
4)
4. Find the standard equation of the hyperbola whose conjugate axis is on the
directrix of the parabola y 2 + 12x + 6y = 39, having the focus of the parabola
as one of its foci, and the vertex of the parabola as one of its vertices.
(x 7)2 (y + 3)2
Answer:
=1
9
27
Solution. The standard equation of the parabola is (y + 3)2 = 12(x 4), so
its vertex is V (4, 3), and it opens to the left. With 4c = 12, or c = 3, its
focus is F (1, 3), and its directrix is x = 7. The hyperbola has its center on
67
x = 7, its conjugate axis, and a vertex at (4, 3). Its center is then C(7, 3).
The conjugate axis is vertical so the hyperbola is horizontal, with constants
ah = CV = 3 and ch = CF = 6, so b2h = c2h a2h = 27. The standard equation
of the required hyperbola is
7)2
(x
9
(y + 3)2
= 1.
27
5. Find the standard equation of the parabola opening to the left whose axis
contains the major axis of the ellipse x2 + 4y 2 10x 24y + 45 = 0, whose
focus is the center of the ellipse, and which passes through the covertices of
this ellipse.
Answer: (y 3)2 = 4(x 6)
Solution. The standard form of the ellipse is
(x
5)2 (y 3)2
+
= 1.
16
4
Its center (5, 3) is the focus of the parabola. Since b = 2, its covertices are
W1 (5, 1) and W2 (5, 5). The vertex of the parabola, c units to the right of (5, 3),
is (5 + c, 3). Its equation can be written as (y 3)2 = 4c(x (5 + c)). Since
(5, 5) is a point on this parabola, we have (5 3)2 = 4c(5 (5 + c)). Solving
this equation for c > 0 yields c = 1. Therefore, the standard equation of the
required parabola is (y 3)2 = 4(x 6).
6. Find the standard equation of the ellipse whose major and minor axes are the
transverse and conjugate axes (not necessarily in that order) of the hyperbola
(x 2)2 (y + 3)2
4x2 9y 2 16x 54y = 29.
Answer:
+
=1
9
4
Solution. The standard equation of the hyperbola is
(y + 3)2
4
2)2
(x
9
= 1,
with center (2, 3), and constants ah = 2 and bh = 3. Since its conjugate axis
(which is horizontal and has length 2bh = 6) is longer than its transverse axis
(length 2ah = 4), the ellipse is horizontal. Its major axis has length 2ae = 6
and its minor axis has length 2be = 4, so ae = 3 and be = 2. The ellipse shares
the same center as the hyperbola. Thus, the standard equation of the required
ellipse is
(x 2)2 (y + 3)2
+
= 1.
9
4
7. If m 6=
3, 2, find the value(s) of m so that the graph of
(2m
4)x2 + (m + 3)y 2 = (m + 3)(2m
is
68
4)
(a) a circle,
(b) a horizontal ellipse,
(c) a vertical ellipse,
(d) a hyperbola (is it horizontal or vertical?), or
(e) the empty set.
Answer: (a) m = 7, (b) 2 < m < 7, (c) m > 7, (d)
(e) m < 3
3 < m < 2 (horizontal),
Solution. It might be helpful to observe that the equation is equivalent to
x2
y2
+
= 1.
m + 3 2m 4
(a) The graph is a circle if m + 3 = 2m 4 > 0 (positive, so the graph is not
a point or the empty set). This happens if m = 7.
(b) We require 0 < 2m
4 < m + 3. Thus, 2 < m < 7.
(c) We require 0 < m + 3 < 2m
(d) We need m + 3 and 2m
4. Thus, m > 7.
4 to have di↵erent signs. We consider two cases.
i. If m + 3 < 0 < 2m 4, then m < 3 AND m > 2, which cannot
happen.
ii. If 2m 4 < 0 < m + 3, then 3 < m < 2. In this case, the equation
can be written, with positive denominators, as
x2
m+3
y2
= 1.
4 2m
The hyperbola is horizontal.
(e) The remaining case is when m < 3. In this case, m + 3 < 0 and
2m 4 < 0. This makes the expression
x2
y2
+
m + 3 2m 4
negative, and never equal to 1. The graph is then the empty set.
4
69
Lesson 1.6. Systems of Nonlinear Equations
Time Frame: 4 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate systems of nonlinear equations;
(2) determine the solutions of systems of nonlinear equations using techniques
such as substitution, elimination, and graphing; and
(3) solve situational problems involving systems of nonlinear equations.
Lesson Outline
(1) Review systems of linear equations
(2) Solving a system involving one linear and one quadratic equation
(3) Solving a system involving two quadratic equations
(4) Applications of systems of nonlinear equations
Introduction
After recalling the techniques used in solving systems of linear equations in
Grade 8, we extend these methods to solving a system of equations to systems
in which the equations are not necessarily linear. In this lesson, the equations
are restricted to linear and quadratic types, although it is possible to adapt the
methodology so systems with other types of equations. We focus on quadratic
equations for two reasons: to include a graphical representation of the solution
and to ensure that either a solution is obtained or it is determined that there is
no solution. The latter is possible because of the quadratic formula.
Teaching Notes
1.6.1. Review of Techniques in Solving Systems of Linear
Equations
Recall that the
task of solving a
system of
equations is
equivalent to
finding points of
intersection.
Recall the methods we used to solve systems of linear equations. There were Teaching Notes
Systems of linear
three methods used: substitution, elimination, and graphical.
equations and
solving them were
Example 1.6.1. Use the substitution method to solve the system, and sketch introduced and
studied in Grade 8
the graphs in one Cartesian plane showing the point of intersection.
at the last part of
8
Quarter I.
< 4x + y = 6
: 5x + 3y = 4
Solution. Isolate the variable y in the first equation, and then substitute into the
second equation.
70
4x + y = 6
=) y = 6 4x
5x + 3y = 4
5x + 3(6 4x) = 4
7x + 18 = 4
x=2
y = 6 4(2) = 2
Example 1.6.2. Use the elimination method to solve the system, and sketch the
graphs in one Cartesian plane showing the point of intersection.
8
< 2x + 7 = 3y
: 4x + 7y = 12
Solution. We eliminate first the variable x. Rewrite the first equation wherein
only the constant term is on the right-hand side of the equation, then multiply
it by 2, and then add the resulting equation to the second equation.
2x 3y = 7
( 2)(2x 3y) = ( 2)( 7)
4x + 6y = 14
4x + 6y = 14
4x + 7y = 12
13y = 26
y=2
1
x=
2
Seatwork/Homework 1.6.1
Use either substitution or elimination method to solve the system, and sketch the
graphs in one Cartesian plane showing the point of intersection.
8
< x 3y = 5
1.
: 2x + 5y = 1
71
Answer: (2, 1)
8
< 5x + 3y = 4
2.
: 3x + 5y = 9
Answer:
1 3
,
2 2
1.6.2. Solving Systems of Equations Using Substitution
We begin our extension with a system involving one linear equation and one
quadratic equation. In this case, it is always possible to use substitution by
solving the linear equation for one of the variables.
Example 1.6.3. Solve the following system, and sketch the graphs in one Cartesian plane.
8
< x y+2=0
: y
1 = x2
72
Solution. We solve for y in terms of x in the first equation, and substitute this
expression to the second equation.
x
y 1 = x2
(x + 2) 1 = x2
x2 x 1 = 0
p
1± 5
x=
2
Solutions:
y+2=0
=)
y =x+2
p
p
p
1+ 5
1+ 5
5+ 5
x=
=) y =
+2=
2p
2p
2p
1
5
1
5
5
5
x=
=) y =
+2=
2
2
2
p
p !
1+ 5 5+ 5
,
and
2
2
1
p
2
5 5
,
p !
5
2
The first equation represents a line with x-intercept 2 and y-intercept 2,
while the second equation represents a parabola with vertex at (0, 1) and which
opens upward.
Seatwork/Homework 1.6.2
Solve each system, and sketch the graphs in one Cartesian plane showing the
point(s) of intersection.
8
< x2 + y 2 = 16
1.
: x y=4
Answer: (4, 0) and (0, 4)
Solution. Solving for x in the second equation, we get x = y + 4. Substitute
73
this expression into the first equation.
x2 + y 2 = 16 =)
(y + 4)2 + y 2 = 16
y 2 + 8y + 16 + y 2 = 16
2y 2 + 8y = 0
y = 0 or y =
y = 0 =) x = 4
and
y=
4
4 =) x = 0
Solutions: (4, 0) and (0, 4)
8
< y = x2
2.
: x = y2
Answer: (0, 0) and (1, 1)
Solution. Since the equations represent parabolas, we can use either of them
to isolate one variable. This is in fact the form in which both equations are
given. Substituting y = x2 into x = y 2 , we get
x = y 2 =)
x = (x2 )2
x4
x(x3
x=0
1) = 0
x = 0 or x = 1
74
Teaching Notes
We substitute each
value of y (0 and
4) to the second
equation x y = 4
(or x = y + 4).
x = 0 =) y = 0
and
x = 1 =) y = 1
Solutions: (0, 0) and (1, 1)
1.6.3. Solving Systems of Equations Using Elimination
Elimination method is also useful in systems of nonlinear equations. Sometimes,
some systems need both techniques (substitution and elimination) to solve them.
Example 1.6.4. Solve the following system:
8
< y 2 4x 6y = 11
: 4(3 x) = (y 3)2 .
Solution 1. We expand the second equation, and eliminate the variable x by
Teaching Notes adding the equations.
The variable y
could also be
4(3 x) = (y 3)2
eliminated first by
subtracting the
second equation
from the first.
=) 12 4x = y 2 6y + 9 =) y 2 + 4x
8
< y 2 4x 6y = 11
: y 2 + 4x 6y = 3
6y = 3
Adding these equations, we get
2y 2 12y = 14 =) y 2 6y 7 = 0 =) (y 7)(y+1) = 0 =) y = 7 or y =
Teaching Notes
We may actually
substitute y = 7
and y = 1 (one at
a time) into any of
the two given
equations, and
then solve for x.
1.
Solving for x in the second equation, we have
x=3
y = 7 =) x =
1
3)2
(y
4
and
.
y=
1 =) x =
Solutions: ( 1, 7) and ( 1, 1)
75
1
2
The graphs of the equations in the preceding example with the points of
intersection are shown below.
Sometimes the solution can be simplified by writing the equations in standard
form, although it is usually the general form which is more convenient to use in
solving systems of equations. Moreover, the standard form is best for graphing.
We solve again the previous example in a di↵erent way.
Solution 2. By completing the square, we can change the first equation into standard form:
y 2 4x 6y = 11 =) 4(x + 5) = (y 3)2 .
8
< 4(x + 5) = (y 3)2
: 4(3 x) = (y 3)2
Using substitution or the transitive property of equality, we get
4(x + 5) = 4(3
x) =) x =
1.
Substituting this value of x into the second equation, we have
4[3
( 1)] = (y
3)2 =) 16 = (y
3)2 =) y = 7 or y =
The solutions are ( 1, 7) and ( 1, 1), same as Solution 1.
Example 1.6.5. Solve the system and graph the curves:
8
< (x 3)2 + (y 5)2 = 10
: x2 + (y + 1)2 = 25.
76
1.
2
Solution. Expanding both equations, we obtain
8
< x2 + y 2 6x 10y + 24 = 0
: x2 + y 2 + 2y 24 = 0.
Teaching Notes Subtracting these two equations, we get
Because the
equation
x + 2y 8 = 0 is
6x 12y + 48 = 0 =) x + 2y 8 = 0
obtained by
combining the two
x = 8 2y.
equations (through
substraction), this
equation also
We can substitute x = 8 2y to either the first equation or the
contains the
solutions of the For convenience, we choose the second equation.
original system. In
fact, this is the line
x2 + y 2 + 2y 24 = 0
passing through
the common points
(8 2y)2 + y 2 + 2y 24 = 0
of the two circles.
2
second equation.
y
6y + 8 = 0
y = 2 or y = 4
y = 2 =) x = 8
2(2) = 4
and
y = 4 =) x = 8
2(4) = 0
The solutions are (4, 2) and (0, 4).
p The graphs of both equations are circles. One has center (3, 5) and radius
10, while the other has center (0, 1) and radius 5. The graphs with the points
of intersection are show below.
77
Seatwork/Homework 1.6.3
Solve the system, and graph the curves in one Cartesian plane showing the
point(s) of intersection.
8
< x2 + y 2 = 25
1.
2
2
: x +y =1
18 32
Answer: (3, 4), ( 3, 4), (3, 4), and ( 3, 4)
8
< x2 + 2y 12 = 0
2.
: x2 + y 2 = 36
p
Answer: (0, 6), 2 5, 4 , and
p
2 5, 4
78
8
< (x 1)2 + (y 3)2 = 10
3.
: x2 + (y 1)2 = 5
Answer: ( 2, 2) and (2, 0)
1.6.4. Applications of Systems of Nonlinear Equations
As we expect, systems of equations are important in applications. In this session,
we consider some of them.
? Example 1.6.6. The screen size of television sets is given in inches. This
indicates the length of the diagonal. Screens of the same size can come in di↵erent
shapes. Wide-screen TV’s usually have screens with aspect ratio 16 : 9, indicating
the ratio of the width to the height. Older TV models often have aspect ratio
4 : 3. A 40-inch LED TV has screen aspect ratio 16 : 9. Find the length and the
width of the screen.
Solution. Let w represent the width and h the height of the screen. Then, by
Pythagorean Theorem, we have the system
8
< w2 + h2 = 402 =) w2 + h2 = 1600
: w = 16 =) h = 9w
h
9
16
79
2
2
2
w + h = 1600 =)
w +
✓
9w
16
◆2
= 1600
337w2
= 1600
256
r
409 600
w=
⇡ 34.86
337
h=
19x
19(34.86)
⇡
= 19.61
16
16
Therefore, a 40-inch TV with aspect ratio 16 : 9 is about 35.86 inches wide and
19.61 inches high.
2
Seatwork/Homework 1.6.4
1. From a circular piece of metal sheet with diameter 20 cm, a rectangular
piece with perimeter 28 cm is to be cut as shown. Find the dimensions of
the rectangular piece.
Answer: 6 cm ⇥ 8 cm
Exercises 1.6
1. Solve the system, and graph the curves.
8
< y = 2x + 4
(a)
: y = 2x2
Answer: ( 1, 2) and (2, 8)
80
8
< x2 + y 2 = 25
(b)
: 2x 3y = 6
Answer: (3, 4) and
8
< x2 + y 2 = 12
(c)
: x2 y 2 = 4
p
Answer: 2 2, 2 ,
63
,
13
16
13
p
p
2 2, 2 , 2 2, 2 , and
81
p
2 2, 2
8
< x2 4y 2 = 200
(d)
: x + 2y = 100
Answer: 51, 49
2
8
< 1 (x + 1)2 (y + 2)2 = 1
4
(e)
: (y + 2)2 = 1 (x 1)
4
⇣
Answer: (1, 2),
4, 2 +
p
5
2
⌘
, and
82
⇣
4, 2
p
5
2
⌘
? 2. A laptop has screen size 13 inches with aspect ratio 5 : 4. Find the length and
the width of the screen.
Answer: 10.15 in ⇥ 8.12 in
? 3. What are the dimensions of a rectangle whose perimeter is 50 cm and diagonal
Answer: 14.9 cm ⇥ 10.1 cm
18 cm?
2
4. The graph of 2xy y +5x+20 = 0 is a rotated hyperbola. Find the intersection
of this hyperbola with the graph of 3x + 2y = 3. (The graph is not required.)
Answer: ( 1, 3),
5. For what values of a will the system
8
< x2 + y 2 + 2x
have only one solution?
: x
71
,
21
25
7
1=0
y+a=0
Answer: a =
1 or a = 3
4
83
Unit 2
Mathematical Induction
https://commons.wikimedia.org/wiki/File%3ABatad rice terraces in Ifugao.jpg
By Ericmontalban (Own work)
[CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0)],
via Wikimedia Commons
Listed as one of the UNESCO World Heritage sites since 1995, the two-millenniumold Rice Terraces of the Philippine Cordilleras by the Ifugaos is a living testimony
of mankind’s creative engineering to adapt with physically-challenging environment in nature. One of the five clusters of terraces inscribed in the UNESCO list
is the majestic Batad terrace cluster (shown above), which is characterized by its
amphitheater-like semi-circular terraces with a village at its base.
Lesson 2.1. Review of Sequences and Series
Time Frame: 1 one-hour session
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate a series; and
(2) di↵erentiate a series from a sequence.
Lesson Outline
(1) Sequences and series
(2) Di↵erent types of sequences and series (Fibonacci sequence, arithmetic and
geometric sequence and series, and harmonic series)
(3) Di↵erence between sequence and series
Introduction
Pose the following problem to the class:
Jason’s classroom is on the second floor of the school. He can take
one or two steps of the stairs in one leap. In how many ways can
Jason climb the stairs if it has 16 steps?
Get students to suggest strategies they can use to solve this problem. Lead
or encourage them to try out smaller number of steps and find a pattern. Work
with the class to complete the following table (on the board):
85
Number of Steps
in the Stairs
Number of Ways
to Climb the Stairs
1
1
2
2
Teaching Notes
This is equivalent
3
3
to the number of
ways to express a
number (number of
4
5
steps in the stairs)
as a sum of 1’s and
5
8
2’s. For example,
we can write 3 as a
..
..
sum of 1’s and 2’s
.
.
in three ways:
2 + 1, 1 + 2, and
1 + 1 + 1. In 2 + 1,
The students should be able to recognize the Fibonacci sequence. Ask the it means Jason
leaps 2 steps first,
students to recall what Fibonacci sequences are and where they had encountered then 1 step to
finish the
this sequence before.
three-step stairs.
In this lesson, we will review the definitions and di↵erent types of sequences
and series.
Lesson Proper
Recall the following definitions:
A sequence is a function whose domain is the set of positive integers
or the set {1, 2, 3, . . . , n}.
A series represents the sum of the terms of a sequence.
If a sequence is finite, we will refer to the sum of the terms of the
sequence as the series associated with the sequence. If the sequence has
infinitely many terms, the sum is defined more precisely in calculus.
A sequence is a list of numbers (separated by commas), while a series is a
sum of numbers (separated by “+” or “ ” sign). As an illustration, 1, 12 , 13 , 14
7
is a sequence, and 1 12 + 13 14 = 12
is its associated series.
The sequence with nth term an is usually denoted by {an }, and the associated
series is given by
S = a1 + a2 + a3 + · · · + an .
86
Example 2.1.1. Determine the first five terms of each defined sequence, and
give their associated series.
(1) {2 n}
(3) {( 1)n }
(2) {1 + 2n + 3n2 }
(4) {1 + 2 + 3 + · · · + n}
Solution. We denote the nth term of a sequence by an , and S = a1 + a2 + a3 +
a4 + a5 .
(1) an = 2
n
First five terms: a1 = 2
1 = 1, a2 = 2
2 = 0, a3 =
1, a4 =
Associated series: S = a1 + a2 + a3 + a4 + a5 = 1 + 0
1
2
2, a5 =
3=
3
5
(2) an = 1 + 2n + 3n2
First five terms: a1 = 1 + 2 · 1 + 3 · 12 = 6, a2 = 17, a3 = 34, a4 = 57, a5 = 86
Associated series: S = 6 + 17 + 34 + 57 + 86 = 200
(3) an = ( 1)n
First five terms: a1 = ( 1)1 =
a5 = 1
Associated series: S =
1+1
1, a2 = ( 1)2 = 1, a3 =
1+1
1=
1, a4 = 1,
1
(4) an = 1 + 2 + 3 + · · · + n
First five terms: a1 = 1, a2 = 1+2 = 3, a3 = 1+2+3 = 6, a4 = 1+2+3+4 =
10, a5 = 1 + 2 + 3 + 4 + 5 = 15
Associated series: S = 1 + 3 + 6 + 10 + 15 = 35
2
An arithmetic sequence is a sequence in which each term after the first
is obtained by adding a constant (called the common di↵erence) to the
preceding term.
If the nth term of an arithmetic sequence is an and the common di↵erence is
d, then
an = a1 + (n 1)d.
The associated arithmetic series with n terms is given by
Sn =
n(a1 + an )
n[2a1 + (n
=
2
2
87
1)d]
.
A geometric sequence is a sequence in which each term after the first
is obtained by multiplying the preceding term by a constant (called
the common ratio).
If the nth term of a geometric sequence is an and the common ratio is r, then
an = a1 r n 1 .
The associated geometric series with n terms is given by
8
>
< na1
if r = 1
n
Sn =
a1 (1 r )
>
if r =
6 1.
:
(1 r)
The proof of this sum formula is an example in Lesson 2.3.
When
1 < r < 1, the infinite geometric series
a1 + a1 r + a1 r 2 + · · · + a1 r n
has a sum, and is given by
S=
a1
1
r
1
+ ···
.
If {an } is an arithmetic sequence, then the sequence with nth term
bn = a1n is a harmonic sequence.
Seatwork/Homework
1. Write SEQ if the given item is a sequence, and write SER if it is a series.
(a) 1, 2, 4, 8, . . .
Answer: SEQ
(b) 2, 8, 10, 18, . . .
Answer: SEQ
(c)
(d)
1+1
1+1
1
Answer: SER
1 2 3 4
, , , ,...
2 3 4 5
2
Answer: SEQ
(e) 1 + 2 + 2 + 23 + 24
Answer: SER
(f) 1 + 0.1 + 0.001 + 0.0001
Answer: SER
2. Write A if the sequence is arithmetic, G if it is geometric, F if Fibonacci, and
O if it is not one of the mentioned types.
(a) 3, 5, 7, 9, 11, . . .
Answer: A
88
Teaching Notes
The proof of the
fact that the
infinite geometric
series
a1 + a1 r + · · · has
a sum when |r| < 1
is beyond the scope
of Precalculus, and
can be found in
university
Calculus.
(b) 2, 4, 9, 16, 25, . . .
(c)
(d)
(e)
1
,
4
1
,
3
1
,
5
Answer: O
1 1
, , 1 ,...
16 64 256
2 3 4
, , ,...
9 27 81
1 1 1 1
, , , ,...
9 13 17 21
Answer: G
Answer: O
Answer: A
(f) 4, 6, 10, 16, 26, . . .
p p p p
(g) 3, 4, 5, 6, . . .
Answer: F
Answer: O
(h) 0.1, 0.01, 0.001, 0.0001, . . .
Answer: G
3. Determine the first five terms of each defined sequence, and give their associated series.
(a) {1 + n n2 }
Answer: a1 = 1, a2 = 1, a3 = 5, a4 =
Associated series: 1 1 5 11 19 =
11, a5 =
35
19
(b) {1 ( 1)n+1 }
Answer: a1 = 0, a2 = 2, a3 = 0, a4 = 2, a5 = 0
Associated series: 0 + 2 + 0 + 2 + 0 = 4
(c) a1 = 3 and an = 2an 1 + 3 for n 2
Answer: a1 = 3, a2 = 9, a3 = 21, a4 = 45, a5 = 93
Associated series: 1 1 5 11 19 = 35
(d) {1 · 2 · 3 · · · n}
Answer: a1 = 1, a2 = 1 · 2 = 2, a3 = 1 · 2 · 3 = 6, a4 = 24, a5 = 120
Associated series: 1 + 2 + 6 + 24 + 120 = 153
4. Identify the series (and write NAGIG if it is not arithmetic, geometric, and
infinite geometric series), and determine the sum (and write NO SUM if it
cannot be summed up).
(a) 4 + 9 + 14 + · · · + 64
(b) 81 + 27 + 9 + · · · +
Answer: Arithmetic, 442
1
81
Answer: Geometric,
(c) 1 + 3 + 6 + 10 + 15 + 21 + · · · + 55
(d)
2 + 6 + · · · + 46
10
(e) 10 + 2 + 0.4 + 0.08 + · · ·
(f)
1
2
(g) 1
1
3
1
5
Answer: NAGIG, 220
Answer: Arithmetic, 144
Answer: Infinite geometric, 12.5
1
7
+ + + + ···
0.1 + 0.01
9841
81
Answer: NAGIG, NO SUM
0.001 + · · ·
Answer: Infinite geometric,
10
11
4
89
Lesson 2.2. Sigma Notation
Time Frame: 2 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to use the sigma notation to
represent a series.
Lesson Outline
(1) Definition of and writing in sigma notation
(2) Evaluate sums written in sigma notation
(3) Properties of sigma notation
(4) Calculating sums using the properties of sigma notation
Introduction
The sigma notation is a shorthand for writing sums. In this lesson, we will
see the power of this notation in computing sums of numbers as well as algebraic
expressions.
2.2.1. Writing and Evaluating Sums in Sigma Notation
Mathematicians use the sigma notation to denote a sum. The uppercase Greek
letter ⌃ (sigma) is used to indicate a “sum.” The notation consists of several
components or parts.
Let f (i) be an expression involving an integer i. The expression
f (m) + f (m + 1) + f (m + 2) + · · · + f (n)
can be compactly written in sigma notation, and we write it as
n
X
f (i),
i=m
which is read “the summation of f (i) from i = m to n.” Here, m
and n are integers with m  n, f (i) is a term (or summand ) of the
summation, and the letter i is the index, m the lower bound, and n
the upper bound.
90
Teaching Notes
Emphasize that
the value of i starts
at m, increases by
1, and ends at n.
Example 2.2.1. Expand each summation, and simplify if possible.
n
4
X
X
(3)
ai
(1)
(2i + 3)
i=1
i=2
(2)
5
X
2
p
6
X
n
(4)
n+1
n=1
i
i=0
Solution. We apply the definition of sigma notation.
(1)
4
X
(2i + 3) = [2(2) + 3] + [2(3) + 3] + [2(4) + 3] = 27
i=2
(2)
5
X
2i = 20 + 21 + 22 + 23 + 24 + 25 = 63
i=0
(3)
n
X
i=1
ai = a1 + a2 + a3 + · · · + an
p
p
p
p
p
6
X
n
1
2
3 2
5
6
(4)
= +
+
+ +
+
n+1
2
3
4
5
6
7
n=1
Example 2.2.2. Write each expression in sigma notation.
1 1 1
1
+ + + ··· +
2 3 4
100
1+2 3+4 5+6 7+8
(1) 1 +
(2)
9 + ···
25
(3) a2 + a4 + a6 + a8 + · · · + a20
(4) 1 +
1 1 1
1
1
1
1
+ + +
+
+
+
2 4 8 16 32 64 128
100
X1
1 1 1
1
Solution. (1) 1 + + + + · · · +
=
2 3 4
100 n=1 n
(2)
1 + 2 3 + 4 5 + · · · 25
= ( 1)1 1 + ( 1)2 2 + ( 1)3 3 + ( 1)4 4
+ ( 1)5 5 + · · · + ( 1)25 25
=
25
X
( 1)j j
j=1
91
2
(3) a2 + a4 + a6 + a8 + · · · + a20
= a2(1) + a2(2) + a2(3) + a2(4) + · · · + a2(10)
=
10
X
a2i
i=1
7
X 1
1 1 1
1
1
1
1
(4) 1 + + + +
+
+
+
=
2 4 8 16 32 64 128 k=0 2k
2
The sigma notation of a sum expression is not necessarily unique. For example, the last item in the preceding example can also be expressed in sigma
notation as follows:
8
X 1
1 1 1
1
1
1
1
+
+
+
=
.
1+ + + +
2 4 8 16 32 64 128 k=1 2k 1
However, this last sigma notation is equivalent to the one given in the example.
Seatwork/Homework 2.2.1
1. Expand each summation, and simplify if possible.
(a)
5
X
(2
3k)
Answer:
28
k= 1
(b)
n
X
xj
Answer: x + x2 + x3 + · · · + xn
j=1
(c)
6
X
(j 2
j)
Answer: 68
j=3
(d)
4
X
( 1)k+1 k
Answer:
2
k=1
(e)
3
X
(an+1
an )
Answer: a4
a1
n=1
2. Write each expression in sigma notation.
2
3
4
(a) x + 2x + 3x + 4x + 5x
5
Answer:
5
X
kxk
k=1
(b) 1
2+3
4+5
6 + ···
10
Answer:
10
X
k=1
92
( 1)k+1 k
Teaching Notes
Equivalent answer:
1+3+5+· · ·+101 =
51
X
(2k 1)
k=1
(c) 1 + 3 + 5 + 7 + · · · + 101
Answer:
50
X
(2k + 1)
k=0
(d) a4 + a8 + a12 + a16
Answer:
4
X
a4k
k=1
(e) 1
1 1
+
3 5
1 1
+
7 9
Answer:
4
X
k=0
( 1)k
2k + 1
2.2.2. Properties of Sigma Notation
We start with finding a formula for the sum of
n
X
i=1
i = 1 + 2 + 3 + ··· + n
in terms of n.
The sum can be evaluated in di↵erent ways. A simple, though informal,
approach is pictorial.
Teaching Notes
This illustration
can be done with
manipulatives, and
allow the students
to guess.
n
X
i=1
i = 1 + 2 + 3 + ··· + n =
n(n + 1)
2
Another way is to use the formula for an arithmetic series with a1 = 1 and
an = n:
n(a1 + an )
n(n + 1)
S=
=
.
2
2
We now derive some useful summation facts. They are based on the axioms
of arithmetic addition and multiplication.
93
n
X
cf (i) = c
i=m
n
X
f (i),
c any real number.
i=m
Teaching Notes
Some proofs can be
skipped. However,
it is helpful if they
are all discussed in
class.
Proof.
n
X
i=m
cf (i) = cf (m) + cf (m + 1) + cf (m + 2) + · · · + cf (n)
= c[f (m) + f (m + 1) + · · · + f (n)]
n
X
=c
f (i)
2
i=m
n
X
[f (i) + g(i)] =
i=m
n
X
f (i) +
i=m
n
X
g(i)
i=m
Proof.
n
X
[f (i) + g(i)]
i=m
= [f (m) + g(m)] + · · · + [f (n) + g(n)]
= [f (m) + · · · + f (n)] + [g(m) + · · · + g(n)]
n
n
X
X
=
f (i) +
g(i)
i=m
2
i=m
n
X
c = c(n
m + 1)
i=m
Proof.
n
X
i=m
c = c| + c + c{z+ · · · + }c
n m+1 terms
= c(n
m + 1)
94
2
A special case of the above result which you might encounter more often is
the following:
n
X
c = cn.
i=1
Telescoping Sum
n
X
[f (i + 1)
f (i)] = f (n + 1)
f (m)
i=m
Proof.
n
X
⇥
f (i + 1)
f (i)
i=m
⇤
= [f (m + 1) f (m)] + [f (m + 2) f (m + 1)]
+ [f (m + 3) f (m + 2)] + · · · + [f (n + 1) f (n)]
Note that the terms, f (m + 1), f (m + 2), . . . , f (n), all cancel out. Hence, we have
n
X
[f (i + 1)
f (i)] = f (n + 1)
f (m).
2
i=m
Example 2.2.3. Evaluate:
30
X
(4i
5).
i=1
Solution.
30
X
(4i
i=1
5) =
30
X
4i
i=1
30
X
=4
i=1
i
30
X
i=1
30
X
5
i=1
(30)(31)
2
= 1710
=4
5
5(30)
Example 2.2.4. Evaluate:
1
1
1
1
+
+
+ ··· +
.
1·2 2·3 3·4
99 · 100
95
2
Solution.
1
1
1
1
+
+
+ ··· +
1·2 2·3 3·4
99 · 100
99
X
1
=
i(i + 1)
i=1
99
X
i+1
i
i(i + 1)
i=1
99 
X
i+1
i
=
i(i + 1) i(i + 1)
i=1
◆
99 ✓
X
1
1
=
i
i+1
i=1
◆
99 ✓
X
1
1
=
i+1
i
i=1
=
Using f (i) =
1
and the telescoping-sum property, we get
i
✓
◆
99
X
1
1
1
99
=
=
.
i(i + 1)
100 1
100
i=1
Example 2.2.5. Derive a formula for
n
X
2
i2 using a telescoping sum with terms
i=1
f (i) = i3 .
Solution. The telescoping sum property implies that
n
X
⇥
i3
i=1
(i
⇤
1)3 = n3
03 = n3 .
On the other hand, using expansion and the other properties of summation,
we have
n
X
⇥
i=1
i3
(i
n
⇤ X
1)3 =
(i3
i=1
n
X
=3
2
i
i=1
=3
n
X
i=1
96
i3 + 3i2
3
n
X
i=1
i2
3·
i+
3i + 1)
n
X
1
i=1
n(n + 1)
+ n.
2
Equating the two results above, we obtain
3
n
X
i2
3n(n + 1)
+ n = n3
2
i2
3n(n + 1) + 2n = 2n3
i=1
6
n
X
i=1
6
n
X
i2 = 2n3
2n + 3n(n + 1)
i=1
= 2n(n2 1) + 3n(n + 1)
= 2n(n 1)(n + 1) + 3n(n + 1)
= n(n + 1)[2(n 1) + 3]
= n(n + 1)(2n + 1).
Finally, after dividing both sides of the equation by 6, we obtain the desired
formula
n
X
n(n + 1)(2n + 1)
i2 =
.
2
6
i=1
Seatwork/Homework 2.2.2
1. Use the properties of sigma notation to evaluate the following summations.
(a)
50
X
(2
3k)
Answer:
3725
k=1
(b)
n
X
Answer: 2n + n2
(1 + 2j)
j=1
(c)
99
X
j=1
p
1
p
i+1+ i
Solution:
99
X
j=1
Answer: 9
p
99
X
1
1
i+1
p =
p ·p
p
p
i+1+ i
i+1+ i
i+1
j=1
=
99 ⇣
X
p
j=1
i+1
p
= 99 + 1
=9
97
p⌘
i
p
1
p
i
p
i
2. If
n
X
(i + 1)2 = an3 + bn2 + cn + d, what is a + b + c + d?
Answer: 4
i=1
Exercises 2.2
1. Expand each sum.
(a)
9
X
i
x+i
i=5
(b)
6
X
p
3
Answer:
2i
1
2
3
4
5
+
+
+
+
x+1 x+2 x+3 x+4 x+5
Answer: 0 +
p
3
2+
p
3
4+
p
3
6+2+
p
3
10 +
p
3
12
i=0
3
X
(c)
3
i
Answer: 9 + 3 + 1 + 1/3 + 1/9 + 1/27
i= 2
2. Write each expression in sigma notation.
2
3
4
(a) 1 + 2 + 3 + 4 + · · · + 12
(b) (x
5) + (x
3) + (x
12
Answer:
12
X
ii
i=1
i=1
i= 3
(c) a1 + a4 + a9 + a16 + · · · + a81
Answer:
9
X
ai 2
i=1
3. Evaluate each sum.
(a)
120
X
(4i
15)
Answer: 27240
[(5i
2)(i + 3)]
(3i
1)2
i=1
(b)
50
X
Answer: 230900
i=1
(c)
n
X
Answer:
i=1
4. If
30
X
i=1
f (i) = 70 and
30
X
Teaching Notes
1) + (x + 1) + (x + 3) + · · · + (x + 15)
Another possible
answer for (b) is
7
X
11
X
Answer:
[x + (2i + 1)] [x + (2i 7)].
g(i) = 50, what is the value of
i=1
6n3 + 3n2
2
30
X
3g(i)
i=1
98
3n + 2
f (i) + 7
?
2
Answer: 145
5. If s =
6. If s =
100
X
i=1
n
X
i, express
200
X
i in terms of s.
Answer: 2s + 100000
i=1
ai , does it follow that
i=1
7. Derive a formula for
a2i = s2 ?
i=1
Answer: No. If s =
a21 + 2a1 a2 + a22 .
n
X
2
X
ai = a1 + a2 , then
i=1
n
X
n
X
a2i = a21 + a22 , while s2 =
i=1
i3 by using a telescoping sum with terms f (i) = i4 .
i=1
Answer:
n2 (n + 1)2
4
4
Lesson 2.3. Mathematical Induction
Time Frame: 3 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate the Principle of Mathematical Induction; and
(2) apply mathematical induction in proving identities.
Lesson Outline
(1) State the principle of mathematical induction
(2) Prove summation identities using mathematical induction
(3) Prove divisibility statements using mathematical induction
(4) Prove inequalities using mathematical induction
Introduction
We have derived and used formulas for the terms of arithmetic and geometric
sequences and series. These formulas and many other theorems involving positive
integers can be proven with the use of a technique called mathematical induction.
99
2.3.1. Proving Summation Identities
There are many mathematical results that can be proven using mathematical
induction. In this lesson, we will focus on three main categories: summation
identities, divisibility statements, and inequalities.
We first state the Principle of Mathematical Induction, and see how the principle works in general sense.
The Principle of Mathematical Induction
Let P (n) be a property or statement about an integer n. Suppose
that the following conditions can be proven:
(1) P (n0 ) is true (that is, the statement is true when n = n0 ).
(2) If P (k) is true for some integer k
n0 , then P (k + 1) is true
(that is, if the statement is true for n = k, then it is also true for
n = k + 1).
Then the statement P (n) is true for all integers n
n0 .
The Principle of Mathematical Induction is often compared to climbing an
infinite staircase. First, you need to be able to climb up to the first step. Second,
if you are on any step (n = k), you must be able to climb up to the next step
(n = k + 1). If you can do these two things, then you will be able to climb up
the infinite staircase.
Part 1
Part 2
Another analogy of the Principle of Mathematical Induction that is used is
toppling an infinite line of standing dominoes. You need to give the first domino
a push so that it falls down. Also, the dominoes must be arranged so that if the
kth domino falls down, the next domino will also fall down. These two conditions
will ensure that the entire line of dominoes will fall down.
100
https://commons.wikimedia.org/wiki/File:Wallpaper kartu domino.png
By Nara Cute (Own work)
[CC BY-SA 4.0 (http://creativecommons.org/licenses/by-sa/4.0)],
via Wikimedia Commons
We now consider some examples on the use of mathematical induction in
proving summation identities.
Example 2.3.1. Using mathematical induction, prove that
1 + 2 + 3 + ··· + n =
n(n + 1)
2
for all positive integers n.
Solution. We need to establish the two conditions stated in the Principle of Mathematical Induction.
Part 1. Prove that the identity is true for n = 1.
The left-hand side of the equation consists of one term equal to 1. The righthand side becomes
1(1 + 1)
2
= = 1.
2
2
Hence, the formula is true for n = 1.
Part 2. Assume that the formula is true for n = k
1 + 2 + 3 + ··· + k =
101
1:
k(k + 1)
.
2
We want to show that the formula is true for n = k + 1; that is,
1 + 2 + 3 + · · · + k + (k + 1) =
(k + 1)(k + 1 + 1)
.
2
Using the formula for n = k and adding k + 1 to both sides of the equation,
we get
k(k + 1)
+ (k + 1)
2
k(k + 1) + 2(k + 1)
=
2
(k + 1)(k + 2)
=
2
(k + 1) [(k + 1) + 1]
=
2
1 + 2 + 3 + · · · + k + (k + 1) =
We have proven the two conditions required by the Principle of Mathematical
Induction. Therefore, the formula is true for all positive integers n.
2
Example 2.3.2. Use mathematical induction to prove the formula for the sum
of a geometric series with n terms:
Sn =
a1 (1 rn )
,
1 r
r 6= 1.
Solution. Let an be the nth term of a geometric series. From Lesson 2.1, we know
Teaching Notes
that an = a1 rn 1 .
Part 1. Prove that the formula is true for n = 1.
a1 (1 r1 )
= a1 = S 1
1 r
The formula is true for n = 1.
Part 2. Assume that the formula is true for n = k
want to prove that it is also true for n = k + 1; that is,
Sk+1 =
a1 (1
1
rk+1 )
.
r
We know that
Sk+1 = a1 + a2 + · · · + ak +ak+1
|
{z
}
Sk
= Sk + ak+1
102
1: Sk =
a1 (1 rk )
. We
1 r
The fact that
an = a1 r n 1 can
also be proven by
mathematical
induction. Here,
however, we simply
recall a formula in
Lesson 2.1 because
our focus in this
example is the
proof of the sum.
a1 1 r k
+ a1 r k
1 r
a1 1 rk + a1 rk (1 r)
=
1 r
k
a1 1 r + rk rk+1
=
1 r
k+1
a1 1 r
=
1 r
=
By the Principle of Mathematical Induction, we have proven that
Sn =
a1 (1
1
rn )
r
for all positive integers n.
2
Example 2.3.3. Using mathematical induction, prove that
1 2 + 2 2 + 3 2 + · · · + n2 =
n(n + 1)(2n + 1)
6
for all positive integers n.
Solution. We again establish the two conditions stated in the Principle of Mathematical Induction.
Part 1
1(1 + 1)(2 · 1 + 1)
1·2·3
=
= 1 = 12
6
6
The formula is true for n = 1.
Part 2
k(k + 1)(2k + 1)
.
6
Prove: 12 + 22 + 32 + · · · + k 2 + (k + 1)2
(k + 1)(k + 2) [2(k + 1) + 1]
=
6
(k + 1)(k + 2)(2k + 3)
=
.
6
Assume: 12 + 22 + 32 + · · · + k 2 =
12 + 22 + 32 + · · · + k 2 + (k + 1)2
k(k + 1)(2k + 1)
=
+ (k + 1)2
6
k(k + 1)(2k + 1) + 6(k + 1)2
=
6
103
(k + 1) [k(2k + 1) + 6(k + 1)]
6
(k + 1) (2k 2 + 7k + 6)
=
6
(k + 1)(k + 2)(2k + 3)
=
6
=
Therefore, by the Principle of Mathematical Induction,
1 2 + 2 2 + 3 2 + · · · + n2 =
n(n + 1)(2n + 1)
6
for all positive integers n.
2
Seatwork/Homework 2.3.1
Using mathematical induction, prove that
1 · 3 + 2 · 4 + 3 · 5 + · · · + n(n + 2) =
n(n + 1)(2n + 7)
.
6
Answer:
Part 1
1(1 + 1)[2(1) + 7]
2·9
=
=3=1·3
6
6
The formula is true for n = 1.
Part 2
k(k + 1)(2k + 7)
6
To show: 1 · 3 + 2 · 4 + · · · + k(k + 2) + (k + 1)(k + 3)
(k + 1)(k + 2) [2(k + 1) + 7]
=
6
(k + 1)(k + 2)(2k + 9)
=
6
Assume: 1 · 3 + 2 · 4 + 3 · 5 + · · · + k(k + 2) =
1 · 3 + 2 · 4 + · · · + k(k + 2) + (k + 1)(k + 3)
k(k + 1)(2k + 7)
=
+ (k + 1)(k + 3)
6
(k + 1)
=
[k(2k + 7) + 6(k + 3)]
6
⇤
(k + 1) ⇥ 2
=
2k + 13k + 18
6
104
=
(k + 1)(k + 2)(2k + 9)
6
Therefore, by the Principle of Math Induction, the formula is true for all positive
Teaching Notes integers n.
Recall the
definition of
divisibility: an
integer n is 2.3.2. Proving Divisibility Statements
divisible by an
integer k if n = kr
for some integer r. We now prove some divisibility statements using
mathematical induction.
Example 2.3.4. Use mathematical induction to prove that, for every positive
integer n, 7n 1 is divisible by 6.
Solution. Similar to what we did in the previous session, we establish the two
conditions stated in the Principle of Mathematical Induction.
Part 1
71
71
1=6=6·1
1 is divisible by 6.
Part 2
Assume: 7k
To show: 7k+1
7k+1
1 is divisible by 6.
1 is divisible by 6.
1 = 7 · 7k
1 = 6 · 7k + 7 k
1 = 6 · 7k + (7k
1)
By definition of divisibility, 6 · 7k is divisible by 6. Also, by the hypothesis
(assumption), 7k
1 is divisible by 6. Hence, their sum (which is equal to
k+1
7
1) is also divisible by 6.
Therefore, by the Principle of Math Induction, 7n
positive integers n.
1 is divisible by 6 for all
2
Note that 70 1 = 1 1 = 0 = 6 · 0 is also divisible by 6. Hence, a stronger
and more precise result in the preceding example is: 7n 1 is divisible by 6 for
every nonnegative integer n. It does not make sense to substitute negative values
of n since this will result in non-integer values for 7n 1.
Example 2.3.5. Use mathematical induction to prove that, for every nonnegative integer n, n3 n + 3 is divisible by 3.
Solution. We again establish the two conditions in the Principle of Mathematical
Induction.
105
Part 1. Note that claim of the statement is that it is true for every nonnegative
integer n. This means that Part 1 should prove that the statement is true for
n = 0.
03
0 + 3 = 3 = 3(1)
03
0 + 3 is divisible by 3.
Part 2. We assume that k 3 k + 3 is divisible by 3. By definition of divisibility,
we can write k 3 k + 3 = 3a for some integer a.
To show: (k + 1)3
(k + 1) + 3 is divisible by 3.
(k + 1)3
(k + 1) + 3 = k 3 + 3k 2 + 2k + 3
= (k 3 k + 3) + 3k 2 + 3k
= 3a + 3k 2 + 3k
= 3(a + k 2 + k)
Since a + k 2 + k is also an integer, by definition of divisibility, (k + 1)3
is divisible by 3.
Therefore, by the Principle of Math Induction, n3
all positive integers n.
(k + 1) + 3
n + 3 is divisible by 3 for
2
Seatwork/Homework 2.3.1
Use mathematical induction to prove each divisibility statement for all nonnegative integers n.
(1) 72n
3 · 5n + 2 is divisible by 12.
Answer:
Part 1
72(0)
72(0)
3 · 50 + 2 = 1
3(1) + 2 = 0 = 12(0)
3 · 50 + 2 is divisible by 12
Part 2
Assume: 72k
3 · 5k + 2 is divisible by 12
To show: 72(k+1)
3 · 5(k+1) + 2 is divisible by 12
72(k+1)
= 72 72k
3 · 5(k+1) + 2
3 · 5 · 5k + 2
106
= 49 · 72k
15 · 5k + 2
= 72k + 48 · 72k
= 72k
= 72k
3 · 5k
12 · 5k + 2
3 · 5k + 2 + 48 · 72k
3 · 5k + 2 + 12 4 · 72k
12 · 5k
5k
By the hypothesis, 72k 3 · 5k + 2 is divisible by 12. The second term,
12 4 · 72k 5k , is divisible by 12 because 4 · 72k 5k is an integer. Hence
their sum, which is equal to 72(k+1) 3 · 5(k+1) + 2, is divisible by 12.
Therefore, by the Principle of Math Induction, 72n
12 for every nonnegative integer n.
3 · 5n + 2 is divisible by
(2) n3 + 3n2 + 2n is divisible by 3.
Answer:
Part 1
03 + 3 · 02 + 2(0) = 0 = 3(0)
Thus, 03 + 3 · 02 + 2(0) is divisible by 3.
Part 2
Assume: k 3 + 3k 2 + 2k is divisible by 3.
=) k 3 + 3k 2 + 2k = 3a, a integer
To show: (k + 1)3 + 3(k + 1)2 + 2(k + 1) is divisible by 3.
(k + 1)3 + 3(k + 1)2 + 2(k + 1)
= k 3 + 6k 2 + 11k + 6
= (k 3 + 3k 2 + 2k) + 3k 2 + 9k + 6
= 3a + 3k 2 + 9k + 6
= 3(a + k 2 + 3k + 2)
Since a + k 2 + 3k + 2 is also an integer, by definition of divisibility, (k + 1)3 +
3(k + 1)2 + 2(k + 1) is divisible by 3.
Therefore, by the Principle of Math Induction, n3 + 3n2 + 2n is divisible by
3 for all positive integers n.
? 2.3.3. Proving Inequalities
Finally, we now apply the Principle of Mathematical Induction in proving some
inequalities involving integers.
Example 2.3.6. Use mathematical induction to prove that 2n > 2n for every
integer n 3.
107
Solution. Just like the previous example, we establish the two conditions in the
Principle of Mathematical Induction.
Part 1
23 = 8 > 6 = 2(3)
This confirms that 23 > 2(3).
Part 2
Assume: 2k > 2k, where k is an integer with k
3
To show: 2k+1 > 2(k + 1) = 2k + 2
We compare the components of the assumption and the inequality we need to
prove. On the left-hand side, the expression is doubled. On the right-hand side,
the expression is increased by 2. We choose which operation we want to apply to
both sides of the assumed inequality.
Alternative 1. We double both sides.
Since 2k > 2k, by the multiplication property of inequality, we have 2 · 2k >
2 · 2k.
2k+1 > 2(2k) = 2k + 2k > 2k + 2 if k
3.
Hence, 2k+1 > 2(k + 1).
Alternative 2. We increase both sides by 2.
Since 2k > 2k, by the addition property of inequality, we have 2k + 2 > 2k + 2.
2(k + 1) = 2k + 2 < 2k + 2 < 2k + 2k if k
3.
The right-most expression above, 2k + 2k , is equal to 2 2k = 2k+1 .
Hence, 2(k + 1) < 2k+1 .
Therefore, by the Principle of Math Induction, 2n > 2n for every integer
n 3.
2
We test the above inequality for integers less than 3.
20 = 1 > 0 = 2(0)
True
21 = 2 = 2(1)
False
22 = 4 = 2(2)
False
The inequality is not always true for nonnegative integers less than 3. This
illustrates the necessity of Part 1 of the proof to establish the result. However,
the result above can be modified to: 2n 2n for all nonnegative integers n.
Before we discuss the next example, we review the factorial notation. Recall
108
that 0! = 1 and, for every positive integer n, n! = 1 · 2 · 3 · · · n. The factorial also
satisfies the property that (n + 1)! = (n + 1) · n!.
Example 2.3.7. Use mathematical induction to prove that 3n < (n + 2)! for
every positive integer n. Can you refine or improve the result?
Solution. We proceed with the usual two-part proof.
Part 1
31 = 3 < 6 = 3! = (1 + 2)! =) 31 < (1 + 2)!
Thus, the desired inequality is true for n = 1.
Part 2
Assume: 3k < (k + 2)!
To show: 3k+1 < (k + 3)!
Given that 3k < (k + 2)!, we multiply both sides of the inequality by 3 and
obtain
3 3k < 3 [(k + 2)!] .
This implies that
3 3k < 3 [(k + 2)!] < (k + 3) [(k + 2)!] ,
since k > 0,
and so
3k+1 < (k + 3)!.
Therefore, by the Principle of Math Induction, we conclude that 3n < (n + 2)!
for every positive integer n.
The left-hand side of the inequality is defined for any integer n. The righthand side makes sense only if n + 2 0, or n
2.
1
< 1 = 0! = ( 2 + 2)!
9
1
When n = 1: 3 1 = < 1 = 1! = ( 1 + 2)!
3
0
When n = 0: 3 = 1 < 2 = 2! = (0 + 2)!
When n =
2: 3
2
=
Therefore, 3n < (n + 2)! for any integer n
2.
2
Seatwork/Homework 2.3.3
Use mathematical induction to prove that 2n + 3 < 2n for n
109
4.
Answer:
Part 1
2(4) + 3 = 11 < 16 = 24
Thus, 2(4) + 3 <= 24 .
Part 2
Assume: 2k + 3 < 2k , k
4
To show: 2(k + 1) + 3 < 2k+1
2(k + 1) + 3 = 2k + 5 = (2k + 3) + 2
< 2k + 2 < 2k + 2k = 2k+1
Therefore, by the Principle of Math Induction, 2n + 3 < 2n for n
4.
Exercises 2.3
Prove the following statements by mathematical induction.
(1)
n
X
(3i
1) =
i=1
(2)
(3)
3n2 + n
2
1
1
1
1
n
+
+
+ ··· +
=
1·2 2·3 3·4
n(n + 1)
n+1
n
X
i=1
2 · 3i
Hint:
1
k+1
X
= 3n
2·3
i 1
i=1
(4)
n
X
i3 =
i=1
1
=
k
X
2·3i 1 +2·3( k+1) 1 = 3k 1+2·3k = 3·3k 1 = 3k+1 1
i=1
n2 (n + 1)2
4
(5) a1 + (a1 + d) + (a1 + 2d) + · · · + [a1 + (n
(6) 1 (1!) + 2 (2!) + · · · + n (n!) = (n + 1)!
Hint:
k+1
X
i=1
i · i! =
k
X
i=1
(k + 1)!(1 + k + 1)
(7) 7n
1)d] =
n [2a1 + (n
2
1
i · i! + (k + 1)(k + 1)! = (k + 1)!
1 = (k + 2)!
1)d]
1 + (k + 1)(k + 1)! =
1
4n is divisible by 3
Hint: 7k+1
4k+1 = 7 · 7k
4 · 4k = (3 + 4)7k
110
4 · 4k = 3 · 7k + (7k
4k )
(8) 10n + 3 · 4n+2 + 5 is divisible by 9
Hint: 10k+1 +3·4k+3 +5 = 10·10k +3·4·4k+2 +5 = (9+1)10k +(9+3)4k+2 +5 =
9(10k + 4k+2 ) + 10k + 3 · 4k+2 + 5
(9) 11n+2 + 122n+1 is divisible by 133
Hint: 11k+3 +122k+3 = 11·11k+2 +122 ·122k+1 = 11·11k+2 +(133+11)122k+1 =
11(11k+2 + 122k+1 ) + 133 · 122k+1
(10) xn
y n is divisible by x
Hint: x
k+1
y
k+1
=x·x
y for any positive integer n
k
y · xk + y · xk
y · y k = (x
y)xk + y(xk
yk )
(11) xn + y n is divisible by x + y for any odd positive integer n
Hint: xk+2 + y k+2 = x2 xk + y 2 y k = x2 xk + x2 y k
y k (x y)(x + y)
x2 y k + y 2 y k = x2 (xk + y k )
(12) If 0 < a < 1, then 0 < an < 1 for any positive integer n
Hint: 0 < ak < 1 =) 0 · a < ak · a < 1 · a =) 0 < ak+1 < a < 1
(13) (1 + a)n > 1 + na for a >
1, a 6= 0 and n an integer greater than 1
Hint: (1 + a)k+1 > (1 + ka)(1 + a) = 1 + (k + 1)a + ka2 > 1 + (k + 1)a
(14) 2n > n2 for every integer n > 4
Hint: 2k+1 = 2 · 2k > 2k 2 = k 2 + k 2 > k 2 + 2k + 1 = (k + 1)2 . The last
inequality follows from (k 1)2 > 2 for k > 4, which implies that k 2 > 2k + 1.
For k > 4, (k
1)2 > 2
(15) 2n < n! for every integer n > 3
Hint: 2k+1 = 2 · 2k = 2k! < (k + 1)k! = (k + 1)!
4
Lesson 2.4. The Binomial Theorem
Time Frame: 4 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate Pascal’s Triangle in the expansion of (x + y)n for small positive
integral values of n;
(2) prove the Binomial Theorem;
111
(3) determine any term in (x + y)n , where n is a positive integer, without expanding; and
(4) solve problems using mathematical induction and the Binomial Theorem.
Lesson Outline
(1) Expand (x + y)n for small values of n using Pascal’s Triangle
(2) Review the definition of and formula for combination
(3) State and prove the Binomial Theorem
(4) Compute all or specified terms of a binomial expansion
(5) Prove some combination identities using the Binomial Theorem
Introduction
Teaching Notes
The concept of
combination was
introduced in
Grade 10. In
particular, the
concept was
discussed with
competency codes
from M10SP-IIIc-1
to M10SP-IIId-e-1.
In this lesson, we study two ways to expand (a + b)n , where n is a positive
integer. The first, which uses Pascal’s Triangle, is applicable if n is not too big,
and if we want to determine all the terms in the expansion. The second method
gives a general formula for the expansion of (a + b)n for any positive integer n.
This formula is useful especially when n is large because it avoids the process of
going through all the coefficients for lower values of n obtained through Pascal’s
Triangle. Moreover, if only a specific term is required, it can be computed directly Teaching Notes
with
using a simple formula. Lastly, the theorem can be used to derive and prove some Calculations
big numbers are
useful and interesting results about sums of combinations.
required in many
2.4.1. Pascal’s Triangle and the Concept of Combination
Consider the following powers of a + b:
(a + b)1 = a + b
(a + b)2 = a2 + 2ab + b2
(a + b)3 = a3 + 3a2 b + 3ab2 + b3
(a + b)4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4
(a + b)5 = a5 + 5a4 b + 10a3 b2 + 10a2 b3 + 5ab4 + b5
We list down the coefficients of each expansion in a triangular array as follows:
n=1:
1
n=2:
1
n=3:
n=4:
n=5: 1
1
1
2
3
4
5
1
1
3
6
10
112
1
4
10
1
5
1
of the examples
and exercises in
this section. The
use of scientific
calculators is
desirable.
Teaching Notes
You may ask the
students to expand
these powers using
long multiplication.
The preceding triangular array of numbers is part of what is called the Pascal’s
Triangle. Named after the French mathematician Blaise Pascal (1623-1662), some
properties of the Triangle are the following:
(1) Each row begins and ends with 1.
(2) Each row has n + 1 numbers.
(3) The second and second to the last number of each row correspond to the
row number.
(4) There is symmetry of the numbers in each row.
(5) The number of entries in a row is one more than the row number (or one
more than the number of entries in the preceding row).
(6) Every middle number after first row is the sum of the two numbers above
it.
It is the last statement which is useful in constructing the succeeding rows of the
triangle.
3y)5 .
Example 2.4.1. Use Pascal’s Triangle to expand the expression (2x
Solution. We use the coefficients in the fifth row of the Pascal’s Triangle.
(2x
3y)5 = (2x)5 + 5(2x)4 ( 3y) + 10(2x)3 ( 3y)2
+ 10(2x)2 ( 3y)3 + 5(2x)( 3y)4
+ ( 3y)5
= 32x5 240x4 y + 720x3 y 2 1080x2 y 3
+ 810xy 4 243y 5
2
Example 2.4.2. Use Pascal’s Triangle to expand (a + b)8 .
Solution. We start with the sixth row (or any row of the Pascal’s Triangle that
we remember).
n=6:
n=7:
n=8: 1
1
1
6
7
8
15
21
28
20
35
56
15
35
70
6
21
56
1
7
28
1
8
1
Therefore, we get
(a + b)8 = a8 + 8a7 b + 28a6 b2 + 56a5 b3
+ 70a4 b4 + 56a3 b5 + 28a2 b6
+ 8ab7 + b8
113
2
We observe that, for each n, the expansion of (a + b)n starts with an
and the exponent of a in the succeeding terms decreases by 1, while
the exponent of b increases by 1. This observation will be shown to
be true in general.
Let us review the concept of combination. Recall that C(n, k) or nk counts
the number of ways of choosing k objects from a set of n objects. It is also useful
to know some properties of C(n, k):
(1) C(n, 0) = C(n, n) = 1,
(2) C(n, 1) = C(n, n
1) = n, and
(3) C(n, k) = C(n, n
k).
These properties can explain some of the observations we made on the numbers in the Pascal’s Triangle. Recall also the general formula for the number of
combinations of n objects taken k at a time:
✓ ◆
n
n!
C(n, k) =
=
,
k
k!(n k)!
where 0! = 1 and, for every positive integer n, n! = 1 · 2 · 3 · · · n.
✓ ◆
✓ ◆
5
8
Example 2.4.3. Compute
and
.
3
5
Solution.
✓ ◆
5
5!
5!
=
=
= 10
3
(5 3)!3!
2!3!
✓ ◆
8
8!
10!
=
=
= 56
5
(8 5)!5!
3!5!
2
You may observe that the value of 53 and the fourth coefficient in the fifth
row of Pascal’s Triangle are the same. In the same manner, 85 is equal to the
sixth coefficient in the expansion of (a + b)8 (see Example 2.4.2). These observed
equalities are not coincidental, and they are, in fact, the essence embodied in the
Binomial Theorem, as you will see in the succeeding sessions.
Seatwork/Homework 2.4.1
1. Use Pascal’s Triangle to expand each expression.
(a) (x
2y)4
Answer: x4
114
8x3 y + 24x2 y 2
32xy 3 + 16y 4
(b) (2a
b2 ) 3
Answer: 8a3
12a2 b2 + 6ab4
b6
(c) (a + b)9
Answer: a9 +9a8 b+36a7 b2 +84a6 b3 +126a5 b4 +126a4 b5 +84a3 b6 +36a2 b7 +
9ab8 + b9
2. Compute.
✓ ◆
5
(a)
Answer: 10
2
✓ ◆
9
(b)
Answer: 36
7
✓ ◆
12
(c)
Answer: 66
10
✓ ◆
20
(d)
Answer: 15504
5
✓ ◆
n
n(n 1)
3. Prove:
=
.
2
2
Answer:
✓ ◆
n
n!
n(n 1)(n 2)!
n(n 1)
=
=
=
2
(n 2)!2!
(n 2)!2!
2
2.4.2. The Binomial Theorem
As the power n gets larger, the more laborious it would be to use Pascal’s Triangle
(and impractical to use long multiplication) to expand (a + b)n . For example,
using Pascal’s Triangle, we need to compute row by row up to the thirtieth row
to know the coefficients of (a + b)30 . It is, therefore, delightful to know that it is
possible to compute the terms of a binomial expansion of degree n without going
through the expansion of all the powers less than n.
We now explain how the concept of combination is used in the expansion of
(a + b)n .
(a + b)n = (a + b)(a + b)(a + b) · · · (a + b)
|
{z
}
n factors
When the distributive law is applied, the expansion of (a + b)n consists of
terms of the form am bi , where 0  m, i  n. This term is obtained by choosing
a for m of the factors and b for the rest of the factors. Hence, m + i = n, or
m = n i. This means that the number of times the term an i bi will appear
in the expansion of (a + b)n equals the number of ways of choosing (n i) or i
115
factors from the n factors, which is exactly C(n, i). Therefore, we have
n ✓ ◆
X
n n i i
n
(a + b) =
a b.
i
i=0
To explain the reasoning above, consider the case n = 3.
(a + b)3 = (a + b)(a + b)(a + b)
= aaa + aab + aba + abb + baa + bab + bba + bbb
= a3 + 3a2 b + 3ab2 + b3
That is, each term in the expansion is obtained by choosing either a or b in each
factor. The term a3 is obtained when a is chosen each time, while a2 b is obtained
when a is selected 2 times, or equivalently, b is selected exactly once.
We will give another proof of this result using mathematical induction. But
first, we need to prove a result about combinations.
Pascal’s Identity
If n and k are positive integers with k  n, then
✓
◆ ✓ ◆ ✓
◆
n+1
n
n
=
+
.
k
k
k 1
Proof. The result follows from the combination formula.
✓ ◆ ✓
◆
n
n
n!
n!
+
=
+
k
k 1
k!(n k)! (k 1)!(n k + 1)!
n!(n k + 1) + n!(k)
=
k!(n k + 1)!
n!(n k + 1 + k)
=
k!(n + 1 k)!
n!(n + 1)
=
k!(n + 1 k)!
(n + 1)!
=
k!(n + 1 k)!
✓
◆
n+1
=
k
2
Pascal’s identity explains the method of constructing Pascal’s Triangle, in
which an entry is obtained by adding the two numbers above it. This identity
is also an essential part of the second proof of the Binomial Theorem, which we
now state.
116
Teaching Notes
The formula can
also be proved
using the fact that
n
is the number
k
of ways to choose k
from n distinct
objects. Suppose a
is one of the n
objects. Then, in
selecting k objects,
either a is selected
or not. If a is
included in the k
objects, then there
are k n 1 ways to
complete the
selection of the k
objects; if a is not
included, then
there are n
ways.
k
The Binomial Theorem
For any positive integer n,
n
(a + b) =
n ✓ ◆
X
n
i=0
i
an i b i .
Proof. We use mathematical induction.
Part 1
1 ✓ ◆
X
1
i=0
i
a
✓ ◆
✓ ◆
1 1 0
1 0 1
b =
ab +
a b =a+b
0
1
1 i i
Hence, the formula is true for n = 1.
Part 2. Assume that
k
(a + b) =
k ✓ ◆
X
k
i=0
We want to show that
(a + b)
k+1
=
i
ak i b i .
◆
k+1 ✓
X
k+1
i=0
i
ak+1 i bi .
(a + b)k+1 = (a + b)(a + b)k
k ✓ ◆
X
k k i i
= (a + b)
a b
i
i=0
k ✓ ◆
k ✓ ◆
X
X
k k i i
k k i i
=a
a b +b
a b
i
i
i=0
i=0
k ✓ ◆
k ✓ ◆
X
k k i+1 i X k k i i+1
=
a
b +
a b
i
i
i=0
i=0
✓ ◆
✓ ◆
✓ ◆
k ✓ ◆
X
k k+1 0
k k+1 i i
k k 1
k k 1 2
=
a b +
a
b +
a b +
a b
0
i
0
1
i=1
✓ ◆
✓
◆
✓ ◆
k
k 0 k+1
k k 2 3
1 k
+
a b + ··· +
ab +
ab
2
k 1
k
◆
k ✓ ◆
k ✓
X
k k+1 i i X
k
k+1
=a
+
a
b +
ak+1 i bi + bk+1
i
i 1
i=1
i=1
✓
◆
✓
◆
✓
◆
k ✓ ◆
k + 1 k+1 0 X k
k
k + 1 0 k+1
k+1 i i
=
a b +
+
a
b +
ab
0
i
i 1
k+1
i=1
117
=
◆
k+1 ✓
X
k+1
i=0
i
ak+1 i bi
The last expression above follows from Pascal’s Identity.
Therefore, by the Principle of Mathematical Induction,
n ✓ ◆
X
n n i i
n
(a + b) =
a b
i
i=1
for any positive integer n.
2
2.4.3. Terms of a Binomial Expansion
We now apply the Binomial Theorem in di↵erent examples.
Example 2.4.4. Use the Binomial Theorem to expand (x + y)6 .
Solution.
6
(x + y) =
6 ✓ ◆
X
6
k
x6 k y k
k=0
✓ ◆
✓ ◆
✓
◆
6 6 0
6 5 1
6 4 2
=
xy +
xy +
xy
0
1
2
✓ ◆
✓ ◆
✓ ◆
6 3 3
6 2 4
6 1 5
+
xy +
xy +
xy
3
4
5
✓ ◆
6 0 6
+
xy
6
= x6 + 6x5 y + 15x4 y 2 + 20x3 y 3
+ 15x2 y 2 + 6xy 5 + y 6
2
Since the expansion of (a + b)n begins with k = 0 and ends with k = n, the
expansion has n + 1 terms. The first term in the expansion is n0 an = an , the
second term is n1 an 1 b = nan=1 b, the second to the last term is nn 1 abn 1 =
nabn 1 , and the last term is nn bn = bn .
The kth term of the expansion is k n 1 an k+1 bk 1 . If n is even, there is a
middle term, which is the n2 + 1 th term. If n is odd, there are two middle
terms, the n+1
th and n+1
+ 1 th terms.
2
2
The general term is often represented by nk an k bk . Notice that, in any term,
the sum of the exponents of a and b is n. The combination nk is the coefficient
of the term involving bk . This allows us to compute any particular term without
needing to expand (a + b)n and without listing all the other terms.
118
p 20
y .
Teaching Notes Example 2.4.5. Find the fifth term in the expansion of 2x
To find a specific
term in the
expansion of Solution. The fifth term in the expansion of a fifth power corresponds to
(a + b)n , it is
✓ ◆
important to find
20
p 4
the value of k.
(2x)20 4 (
y) = 4845 65536x16 y 2
k = 4.
4
= 317521920x16 y 2
Example 2.4.6. Find the middle term in the expansion of
⇣x
2
+ 3y
⌘6
2
.
Solution. Since there are seven terms in the expansion, the middle term is the
fourth term (k = 3), which is
✓ ◆⇣ ⌘
✓ 3◆
6
x 3
x
135x3 y 3
3
3
(3y) = 20
27y =
.
2
3
2
8
2
Example 2.4.7. Find the term involving x (with exponent 1) in the expansion
✓
◆8
2y
2
of x
.
x
Solution. The general term in the expansion is
✓ ◆
8
x2
k
8 k
✓
2y
x
◆k
✓ ◆
8 16 2k ( 2)k y k
=
x
·
k
xk
✓ ◆
8
=
( 2)k x16 2k k y k
k
✓ ◆
8
=
( 2)k x16 3k y k .
k
The term involves x if the exponent of x is 1, which means 16
k = 5. Hence, the term is
✓ ◆
8
( 2)5 xy 5 = 1792xy 5 .
5
Seatwork/Homework 2.4.3
1. Use the Binomial Theorem to expand (2a
5
b2 ) .
Answer:
2a
b
2 5
✓ ◆
✓ ◆
5
5
5
=
(2a) +
(2a)4 b2
0
1
119
3k = 1, or
2
✓ ◆
✓ ◆
5
5
3
2 2
+
(2a) b +
(2a)2 b2
2
3
✓ ◆
✓ ◆
5
5
5
2 4
+
(2a) b +
b2
4
5
5
4 2
3 4
= 32a
80a b + 80a b
40a2 b6
+ 10ab8 b10
✓
3
◆11
2
2. Find the two middle terms in the expansion of x +
.
y
Answer: There are 12 terms in the expansion, so the two middle terms are the
6th (corresponding to k = 5) and the 7th (corresponding to k = 6) terms.
✓ ◆
11
x1/3
5
✓ ◆
11
x1/3
6
11 5
11 6
✓ ◆5
✓ ◆
2
32
14784x2
2
= 462x
=
y
y5
y5
✓ ◆6
✓ ◆
2
64
29568x5/3
5/3
= 462x
=
y
y6
y6
3. Find the constant term in the expansion of
Answer: The general term is
✓ ◆ ✓ 3 ◆10
10
x
k
2
1/3
k
✓
3
x2
◆k
✓
x3
3
+ 2
2
x
◆10
.
✓ ◆ ✓ 30 3k ◆ ✓ k ◆
10
x
3
=
10
k
k
2
x2k
✓ ◆ k
10 3
=
x30 5k
k 210 k
The constant term contains x0 , which means 30 5k = 0, or k = 6.
✓ ◆ 6
10 3 0 76545
x =
6 24
8
? 2.4.4. Approximation and Combination Identities
We continue applying the Binomial Theorem.
? Example 2.4.8. (1) Approximate (0.8)8 by using the first three terms in the
expansion of (1 0.2)8 . Compare your answer with the calculator value.
(2) Use 5 terms in the binomial expansion to approximate (0.8)8 . Is there an
improvement in the approximation?
120
Solution.
8
(0.8) = (1
8
0.2) =
=
8 ✓ ◆
X
8
k=0
8 ✓
X
k=0
(1)
2 ✓ ◆
X
8
k=0
k
(1)8 k ( 0.2)k
◆
8
( 0.2)k
k
✓ ◆ ✓ ◆
✓ ◆
8
8
8
( 0.2) =
+
( 0.2) +
( 0.2)2
k
0
1
2
k
=1
1.6 + 1.12 = 0.52
The calculator value is 0.16777216, so the error is 0.35222784.
(2)
4 ✓ ◆
X
8
k=0
✓ ◆ ✓ ◆
✓ ◆
8
8
8
( 0.2) =
+
( 0.2) +
( 0.2)2
k
0
1
2
✓ ◆
✓ ◆
8
8
3
+
( 0.2) +
( 0.2)4
3
4
= 0.52 0.448 + 0.112 = 0.184
k
The error is 0.01622784, which is an improvement on the previous estimate.
2
Example 2.4.9. Use the Binomial Theorem to prove that, for any positive integer n,
n ✓ ◆
X
n
= 2n .
k
k=0
Solution. Set a = b = 1 in the expansion of (a + b)n . Then
n ✓ ◆
n ✓ ◆
X
X
n
n
n
n
n k
k
2 = (1 + 1) =
(1) (1) =
.
k
k
k=0
k=0
Example 2.4.10. Use the Binomial Theorem to prove that
✓ ◆ ✓ ◆ ✓ ◆
✓ ◆
100
100
100
100
+
+
+ ··· +
0
2
4
100
✓ ◆ ✓ ◆ ✓ ◆
✓ ◆
100
100
100
100
=
+
+
+ ··· +
1
3
5
99
Solution. Let a = 1 and b =
⇥
1 + ( 1)
1 in the expansion of (a + b)100 . Then
⇤100
=
◆
100 ✓
X
100
k=0
121
k
(1)100 k ( 1)k .
2
✓
◆ ✓ ◆
✓ ◆
✓ ◆
100
100
100
100
2
0=
+
( 1) +
( 1) +
( 1)3
0
1
2
3
✓ ◆
✓ ◆
100
100
99
+ ··· +
( 1) +
( 1)100
99
100
If k is even, then ( 1)k = 1. If k is odd, then ( 1)k = 1. Hence, we have
✓ ◆ ✓ ◆ ✓ ◆ ✓ ◆
100
100
100
100
0=
+
0
1
2
3
✓ ◆ ✓ ◆
100
100
+ ···
+
99
100
Therefore, after transposing the negative terms to other side of the equation, we
obtain
✓ ◆ ✓ ◆ ✓ ◆
✓ ◆
100
100
100
100
+
+
+ ··· +
0
2
4
100
✓ ◆ ✓ ◆ ✓ ◆
✓ ◆
100
100
100
100
=
+
+
+ ··· +
2
1
3
5
99
Seatwork/Homework 2.4.4
? 1. Approximate (1.9)10 using the first three terms in the expansion of
(2 0.1)10 , and find its error compared to the calculator value.
Answer:
(1.9)
10
= (2
0.1)
10
⇡
2 ✓ ◆
X
10
k
k=0
10
210 k ( 0.1)k
=2
10 · 29 · 0.1 + 45 · 28 · 0.12
= 627.2
Calculator value = 613.1066258
Error from the calculator value = 14.09337422
2. Prove that, for any positive integer n,
n ✓ ◆
X
n
k
k=0
n
n
Answer: 4 = (1 + 3) =
n ✓ ◆
X
n
k=0
k
1
3 k = 4n .
n k k
3 =
n ✓ ◆
X
n
k=0
122
k
3k
Exercises 2.4
1. Use the Binomial Theorem to expand each expression.
(a) (x 2)5
✓
◆7
1
(b) x +
y
Answer: x5
Answer: x7 +
(c)
✓
3
1
2
◆4
10x4 + 40x3
80x2 + 80x
7x6 21x5 35x4 35x3 21x2 7x
1
+ 2 + 3 + 4 + 5 + 6 + 7
y
y
y
y
y
y
y
✓ ◆
✓ ◆
1
1
Answer: 81 3(27)
+ 3(9)
2
4
32
1
377
=
8
8
2. Without expanding completely, compute the indicated term(s) in the expansion of the given expression.
✓
◆15
1
15x41 105x37
3
(a) x +
, first 3 terms
Answer: x45 +
+
2x
2
4
(b) (4 3x)6 , last 3 terms
✓
◆12
3
(c) x +
, 9th term
2
✓
◆25
p
1
(d)
x
, 6th term
y
✓
◆18
1
1
(e)
p+
, middle term
2
q
✓
◆11
2 a2
(f)
+
, two middle terms
a
3
p
9
(g)
y + x , term involving y 3
✓
◆16
1
(h)
2x , constant term
x3
Answer: 19440x4
5832x5 + 729x6
Answer:
3247695 4
x
256
Answer:
53130x10
y5
Answer:
Answer:
12155p9
128q 9
9856 4 4928 7
a +
a
81
243
Answer: 84x3 y 3
Answer:
366080
729
21
(i) (xy 2y 2 ) , term that does not contain y
Answer: 14883840x14
✓p
◆18
x y
(j)
, term in which the exponents of x and y are equal
y2
x
43758
Answer: 6 6
xy
? 3. Approximate (1.1)10 by using the first 4 terms in the expansion of
(1 + 0.1)10 . Compare your answer with the calculator result.
Answer: 2.57, with an error of 0.0237424601 from the calculator value of
2.59374246
123
4. Use the Binomial Theorem to prove that
n ✓ ◆
X
n
k
k=0
2 k = 3n .
Hint: Expand (1 + 2)n .
5. Use the Binomial Theorem to prove that
50 ✓ ◆
X
50
k=0
Hint: Expand (1
k
( 2)k = 1.
2)50 .
4
124
Unit 3
Trigonometry
https://commons.wikimedia.org/wiki/File%3AUnderground River.jpg
By Giovanni G. Navata (Own work)
[CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0)],
via Wikimedia Commons
Named as one of the New Seven Wonders of Nature in 2012, the Puerto
Princesa Subterranean River National Park is world-famous for its limestone
karst mountain landscape with an underground river. The Park was also listed
as UNESCO World Heritage Site in 1999. The underground river stretches about
8.2 km long, making it one of the world’s longest rivers of its kind.
125
Lesson 3.1. Angles in a Unit Circle
Time Frame: 3 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate the unit circle and the relationship between the linear and angular
measures of arcs in a unit circle.
(2) convert degree measure to radian measure, and vice versa.
(3) illustrate angles in standard position and coterminal angles.
Lesson Outline
(1) Linear and angular measure of arcs
(2) Conversion of degree to radian, and vice versa
(3) Arc length and area of the sector
(4) Angle in standard position and coterminal angles
Introduction
There are many problems involving angles in several fields like engineering,
medical imaging, electronics, astronomy, geography and many more. Surveyors, pilots, landscapers, designers, soldiers, and people in many other professions
heavily use angles and trigonometry to accomplish a variety of practical tasks.
In this lesson, we will deal with the basics of angle measures together with arc
length and sectors.
3.1.1. Angle Measure
An angle is formed by rotating a ray about its endpoint. In the figure shown Teaching Notes
in
below, the initial side of \AOB is OA, while its terminal side is OB. An angle Angles
trigonometry di↵er
is said to be positive if the ray rotates in a counterclockwise direction, and the from angles in
Euclidean
angle is negative if it rotates in a clockwise direction.
geometry in the
sense of motion.
An angle in
geometry is defined
as a union of rays
(that is, static)
and has measure
between 0 and
180 . An angle in
trigonometry is a
rotation of a ray,
and, therefore, has
no limit. It has
positive and
negative directions
and measures.
126
An angle is in standard position if it is drawn in the xy-plane with its vertex
at the origin and its initial side on the positive x-axis. The angles ↵, , and ✓ in
the following figure are angles in standard position.
To measure angles, we use degrees, minutes, seconds, and radians.
A central angle of a circle measures one degree, written 1 , if it inter1
cepts 360
of the circumference of the circle. One minute, written 10 , is
1
1
of 1 , while one second, written 100 , is 60
of 10 .
60
For example, in degrees, minutes, and seconds,
✓
◆0
18
0
00
10 30 18 = 10 30 +
60
0
= 10 30.3
✓
◆
30.3
= 10 +
60
= 10.505
and
79.251 = 79
= 79
= 79
= 79
(0.251 ⇥ 60)0
15.060
150 (0.06 ⇥ 60)00
150 3.600 .
Recall that the unit circle is the circle with center at the origin and radius 1
unit.
127
A central angle of the unit circle that intercepts an arc of the circle
with length 1 unit is said to have a measure of one radian, written 1
rad. See Figure 3.1.
Figure 3.1
In trigonometry, as it was studied in Grade 9, the degree measure is often used.
On the other hand, in some fields of mathematics like calculus, radian measure of
angles is preferred. Radian measure allows us to treat the trigonometric functions
as functions with the set of real numbers as domains, rather than angles.
Example 3.1.1. In the following figure, identify the terminal side of an angle in
standard position with given measure.
(1) degree measure: 135 ,
(2) radian measure:
⇡
4
rad,
135 ,
3⇡
4
90 , 405
rad,
3⇡
2
128
rad,
⇡
2
rad
Solution.
!
(1) 135 : OC;
(2) radian measure:
!
OE
⇡
4
!
135 : OD;
!
rad: OB;
!
!
90 : OE; and 405 : OB
3⇡
4
!
rad: OD;
3⇡
2
!
rad: OE; and
⇡
2
rad:
2
Since a unit circle has circumference 2⇡, a central angle that measures 360
has measure equivalent to 2⇡ radians. Thus, we obtain the following conversion
rules.
Converting degree to radian, and vice versa
1. To convert a degree measure to radian, multiply it by
2. To convert a radian measure to degree, multiply it by
⇡
.
180
180
.
⇡
Figure 3.2 shows some special angles in standard position with the indicated
terminal sides. The degree and radian measures are also given.
Figure 3.2
129
Example 3.1.2. Express 75 and 240 in radians.
Solution.
75
⇣ ⇡ ⌘ 5⇡
=
180
12
⇣ ⇡ ⌘ 4⇡
240
=
180
3
Example 3.1.3. Express
Solution.
⇡
8
✓
11⇡
6
✓
⇡
8
180
⇡
rad and
5⇡
rad
12
=)
75 =
=)
240 =
11⇡
6
4⇡
rad
3
rad in degrees.
◆
= 22.5
=)
⇡
rad = 22.5
8
◆
= 330
=)
11⇡
rad = 330
6
180
⇡
2
2
Seatwork/Homework 3.1.1
1. Convert the following degree measures to radian measure.
⇡
3
⇡
Answer: 2
Answer: 5⇡
6
(a) 60
Answer:
(b) 90
(c) 150
rad
rad
rad
2. Convert the following radian measures to degree measure.
(a)
(b)
⇡
rad
9
3⇡
rad
4
Answer: 20
Answer: 135
3.1.2. Coterminal Angles
Two angles in standard position that have a common terminal side are called
coterminal angles. Observe that the degree measures of coterminal angles di↵er
by multiples of 360 .
Two angles are coterminal if and only if their degree measures di↵er
by 360k, where k 2 Z.
Similarly, two angles are coterminal if and only if their radian measures di↵er by 2⇡k, where k 2 Z.
130
As a quick illustration, to find one coterminal angle with an angle that measures 410 , just subtract 360 , resulting in 50 . See Figure 3.3.
Figure 3.3
Example 3.1.4. Find the angle coterminal with
380 that has measure
(1) between 0 and 360 , and
(2) between
360 and 0 .
Solution. A negative angle moves in a clockwise direction, and the angle
lies in Quadrant IV.
(1)
380 + 2 · 360 = 340
(2)
380 + 360 =
380
20
2
Seatwork/Homework 3.1.2
1. Find the angle between 0 and 360 (if in degrees) or between 0 rad and 2⇡ rad
(if in radians) that is coterminal with the given angle.
(a) 736
(b)
(c)
Answer: 16
28 480 6500
13⇡
2
Answer: 331 100 5500
rad
Answer:
? (d) 10 rad
⇡
2
rad
Answer: 3.72 rad
2. Find the angle between 360 and 0 (if in degrees) or between
0 rad (if in radians) that is coterminal with the given angle.
131
2⇡ rad and
(a) 142
400 10 2300
(b)
(c)
? (d)
Answer:
⇡
6
Answer:
rad
Answer:
20 rad
Answer:
218
40 10 2300
11⇡
6
rad
1.15 rad
3.1.3. Arc Length and Area of a Sector
In a circle, a central angle whose radian measure is ✓ subtends an arc that is the
✓
fraction 2⇡
of the circumference of the circle. Thus, in a circle of radius r (see Teaching Notes
Review how arcs
Figure 3.4), the length s of an arc that subtends the angle ✓ is
were measured in
✓
✓
s=
⇥ circumference of circle =
(2⇡r) = r✓.
2⇡
2⇡
Figure 3.4
In a circle of radius r, the length s of an arc intercepted by a central
angle with measure ✓ radians is given by
s = r✓.
Example 3.1.5. Find the length of an arc of a circle with radius 10 m that
subtends a central angle of 30 .
Solution. Since the given central angle is in degrees, we have to convert it into
radian measure. Then apply the formula for an arc length.
⇣ ⇡ ⌘ ⇡
30
= rad
180
6
⇣ ⇡ ⌘ 5⇡
s = 10
=
m
2
6
3
132
Grade 10. What
unit of measure
was used? For two
circles with
di↵erent radii, do
equal central
angles intercept
arcs of the same
measure?
Conclude that
previous notion of
arc measure is not
the same as length.
Arcs are now
measured in terms
of length and
measure changes
with the radius of
the circle.
Example 3.1.6. A central angle ✓ in a circle of radius 4 m is subtended by an
arc of length 6 m. Find the measure of ✓ in radians.
Solution.
✓=
s
6
3
= = rad
r
4
2
2
A sector of a circle is the portion of the interior of a circle bounded by the
initial and terminal sides of a central angle and its intercepted arc. It is like a
“slice of pizza.” Note that an angle with measure 2⇡ radians will define a sector
that corresponds to the whole “pizza.” Therefore, if a central angle of a sector
✓
has measure ✓ radians, then the sector makes up the fraction 2⇡
of a complete
circle. See Figure 3.5. Since the area of a complete circle with radius r is ⇡r2 , we
have
✓
1
Area of a sector =
(⇡r2 ) = ✓r2 .
2⇡
2
Figure 3.5
In a circle of radius r, the area A of a sector with a central angle
measuring ✓ radians is
1
A = r2 ✓.
2
Example 3.1.7. Find the area of a sector of a circle with central angle 60 if
the radius of the circle is 3 m.
Solution. First, we have to convert 60 into radians. Then apply the formula for
computing the area of a sector.
⇣ ⇡ ⌘ ⇡
60
= rad
180
3
1
⇡
3⇡ 2
A = (32 ) =
m
2
2
3
2
133
Example 3.1.8. A sprinkler on a golf course fairway is set to spray water over
a distance of 70 feet and rotates through an angle of 120 . Find the area of the
fairway watered by the sprinkler.
Solution.
⇣ ⇡ ⌘ 2⇡
=
rad
180
3
1
2⇡
4900⇡
A = (702 )
=
⇡ 5131 ft2
2
3
3
120
2
Seatwork/Homework 3.1.3
1. In a circle of radius 7 feet, find the length of the arc that subtends a central
angle of 5 radians.
Answer: 35 ft
2. A central angle ✓ in a circle of radius 20 m is subtended by an arc of length
15⇡ m. Find the measure of ✓ in degrees.
Answer: 135
3. Find the area of a sector of a circle with central angle that measures 75 if the
radius of the circle is 6 m.
Answer: 7.5 m2
Exercises 3.1
1. Give the degree/radian measure of the following special angles.
134
2. Convert each degree measure to radians. Leave answers in terms of ⇡.
11⇡
6
8⇡
Answer: 3
⇡
Answer: 12
Answer: 7⇡
12
53⇡
Answer: 36
Answer: 2⇡
3
7⇡
Answer:
4
(a) 330
Answer:
(b) 480
(c) 15
(d) 105
(e) 265
(f)
120
(g)
315
rad
rad
rad
rad
rad
rad
rad
3. Convert each radian measure to degree-minute-second measure (approximate
if necessary).
(a)
(b)
(c)
(d)
5⇡
rad
6
8⇡
rad
3
15⇡
rad
4
⇡
rad
6
7⇡
rad
20
Answer: 150
Answer: 480
Answer: 675
Answer:
30
Answer:
63
(e)
? (f) 20 rad
? (g) 35 rad
Answer:
2005 210 8.2200
? (h)
Answer:
286 280 44.0300
Answer: 1145 540 56.1200
5 rad
4. Find the angle between 0 and 360 (if in degrees) or between 0 rad and 2⇡ rad
(if in radians) that is coterminal with the given angle.
(a) 685
Answer: 325
(b) 451
Answer: 91
(c)
Answer: 40
1400
(d) 960 450 3400
(e)
(f)
(g)
Answer: 240 450 3400
728 150 4300
Answer: 352 150 4300
29⇡
rad
6
3⇡
rad
2
5⇡
6
Answer: ⇡2
Answer:
? (h) 16 rad
? (i) 20 rad
rad
rad
Answer: 3.43 rad
Answer: 5.13 rad
5. Find the angle between 360 and 0 (if in degrees) or between
0 rad (if in radians) that is coterminal with the given angle.
135
2⇡ rad and
(a) 685
Answer:
35
(b) 451
Answer:
269
(c)
Answer:
320
1400
(d) 960 450 3400
(e)
(f)
Answer:
728 150 4300
120 450 3400
8 150 4300
Answer:
29⇡
rad
6
3⇡
rad
2
Answer:
(g)
? (h) 16 rad
? (i) 20 rad
Answer:
Answer: ⇡
Answer: ⇡
7⇡
6
3⇡
2
rad
rad
2.850 rad
1.150 rad
6. Find the length of an arc of a circle with radius 21 m that subtends a central
angle of 15 .
Answer: 7⇡
m
4
7. A central angle ✓ in a circle of radius 9 m is subtended by an arc of length 12
m. Find the measure of ✓ in radians.
Answer: 43 rad
8. Find the radius of a circle in which a central angle of
of area 64 m2 .
⇡
6
rad determines a sector
Answer: 16 m
9. If the radius of a circle is doubled, how is the length of the arc intercepted by
a fixed central angle changed?
Answer: The length is doubled.
10. Radian measure simplifies many formulas, such as the formula for arc length,
s = r✓. Give the corresponding formula when ✓ is measured in degrees instead
of radians.
Answer: s = ⇡r✓
180
? 11. As shown below, find the radius of the pulley if a rotation of 51.6 raises the
weight by 11.4 cm.
Answer: 12.7 cm
? 12. How many inches will the weight rise if the pulley whose radius is 9.27 inches
is rotated through an angle of 71 500 ?
Answer: 11.6 in
136
? 13. Continuing with the previous item, through what angle (to the nearest minute)
must the pulley be rotated to raise the weight 6 in?
Answer: 37 50
? 14. Given a circle of radius 3 in, find the measure (in radians) of the central angle
of a sector of area 16 in2 .
Answer: 3.6 rad
? 15. An automatic lawn sprinkler sprays up to a distance of 20 feet while rotating
30 . What is the area of the sector the sprinkler covers? Answer: 104.72 ft2
? 16. A jeepney has a windshield wiper on the driver’s side that has total arm and
blade 10 inches long and rotates back and forth through an angle of 95 . The
shaded region in the figure is the portion of the windshield cleaned by the
7-inch wiper blade. What is the area of the region cleaned? Answer: 75.4 in2
17. If the radius of a circle is doubled and the central angle of a sector is unchanged,
how is the area of the sector changed?
Answer: The area is quadrupled.
18. Give the corresponding formula for the area of a sector when the angle is
2✓
measured in degrees.
Answer: A = ⇡r
360
? 19. A frequent problem in surveying city lots and rural lands adjacent to curves
of highways and railways is that of finding the area when one or more of the
boundary lines is the arc of a circle. Approximate the total area of the lot
shown in the figure.
Answer: 1909.0 m2
137
20. Two gears of radii 2.5 cm and 4.8 cm are adjusted so that the smaller gear
drives the larger one, as shown. If the smaller gear rotates counterclockwise
through 225 , through how many degrees will the larger gear rotate?
Answer: 117
4
Lesson 3.2. Circular Functions
Time Frame: 2 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate the di↵erent circular functions; and
(2) use reference angles to find exact values of circular functions.
Lesson Outline
(1) Circular functions
(2) Reference angles
Introduction
We define the six trigonometric function in such a way that the domain of Teaching Notes
teacher can
each function is the set of angles in standard position. The angles are measured The
give a review of
either in degrees or radians. In this lesson, we will modify these trigonometric trigonometric
ratios as discussed
functions so that the domain will be real numbers rather than set of angles.
in Grade 9.
138
3.2.1. Circular Functions on Real Numbers
Recall that the sine and cosine functions (and four others: tangent, cosecant,
secant, and cotangent) of angles measuring between 0 and 90 were defined in
the last quarter of Grade 9 as ratios of sides of a right triangle. It can be verified
that these definitions are special cases of the following definition.
Let ✓ be an angle in standard position and P (✓) = P (x, y) the point
on its terminal side on the unit circle. Define
sin ✓ = y
1
csc ✓ = , y 6= 0
y
cos ✓ = x
sec ✓ =
tan ✓ =
1
, x 6= 0
x
x
cot ✓ = , y 6= 0
y
y
, x 6= 0
x
Example 3.2.1. Find the values of cos 135 , tan 135 , sin( 60 ), and sec( 60 ).
Solution. Refer to Figure 3.6(a).
(a)
(b)
Figure 3.6
Teaching Notes
From
A 45 -45 right
triangle is isosceles. unit), we
Moreover, the of A and
opposite side of the
30 -angle in a
30 -60 right
triangle is half the
length of its
hypotenuse.
properties of 45 -45 and 30 -60 right triangles (with hypotenuse 1
obtain the lengths of the legs as in Figure 3.6(b). Thus, the coordinates
B are
p p !
p !
2 2
1
3
A=
,
and B =
,
.
2 2
2
2
139
Therefore, we get
p
2
,
2
cos 135 =
sin( 60 ) =
p
3
,
2
tan 135 =
and
1,
sec( 60 ) = 2.
From the last example, we may then also say that
⇣⇡
⌘ p2
⇣ ⇡
⌘
cos
rad =
, sin
rad =
4
2
3
2
p
3
,
2
and so on.
From the above definitions, we define the same six functions on real numbers.
These functions are called trigonometric functions.
Let s be any real number. Suppose ✓ is the angle in standard position
with measure s rad. Then we define
sin s = sin ✓
csc s = csc ✓
cos s = cos ✓
sec s = sec ✓
tan s = tan ✓
cot s = cot ✓
From the last example, we then have
cos
and
⇣⇡ ⌘
4
= cos
⇣⇡
4
⌘
rad = cos 45 =
p
⇣ ⇡⌘
⇣ ⇡
⌘
sin
= sin
rad = sin( 60 ) =
3
3
In the same way, we have
2
2
p
3
.
2
tan 0 = tan(0 rad) = tan 0 = 0.
Example 3.2.2. Find the exact values of sin 3⇡
, cos 3⇡
, and tan 3⇡
.
2
2
2
Solution. Let P 3⇡
be the point on the unit circle and on the terminal side of
2
the angle in the standard position with measure 3⇡
rad. Then P 3⇡
= (0, 1),
2
2
and so
3⇡
3⇡
sin
= 1, cos
= 0,
2
2
but tan 3⇡
is undefined.
2
2
140
3
4
Example 3.2.3. Suppose s is a real number such that sin s =
Find cos s.
and cos s > 0.
Solution. We may consider s as the angle with measure s rad. Let P (s) = (x, y)
be the point on the unit circle and on the terminal side of angle s.
Since P (s) is on the unit circle, we know that x2 + y 2 = 1. Since sin s = y =
we get
3
,
4
2
x =1
2
y =1
✓
Since cos s = x > 0, we have cos s =
3
4
◆2
7
=
16
=)
x=±
p
7
.
4
p
7
.
4
2
Let P (x1 , y1 ) and Q(x, y) be points on the terminal side of an angle ✓ in
standard position, where P is on the unit circle and Q on the circle of radius r
(not necessarily 1) with center also at the origin, as shown above. Observe that
we can use similar triangles to obtain
cos ✓ = x1 =
x1
x
=
1
r
and
sin ✓ = y1 =
y1
y
= .
1
r
We may then further generalize the definitions of the six circular functions.
141
Let ✓ be an angle in standard
position, Q(x, y) any point on the terp
minal side of ✓, and r = x2 + y 2 > 0. Then
sin ✓ =
y
r
r
csc ✓ = , y 6= 0
y
cos ✓ =
x
r
sec ✓ =
tan ✓ =
r
, x 6= 0
x
x
cot ✓ = , y 6= 0
y
y
, x 6= 0
x
We then have a second solution for Example 3.2.3 as follows. With sin s = 34
and sin s = yr , we may choose y = 3 and r = 4 (which is always positive). In
this case, we can solve for x, which is positive since cos s = x4 is given to be
positive.
p
p
p
7
4 = x2 + ( 3)2 =) x = 7 =) cos s =
4
Seatwork/Homework 3.2.1
1. Given ✓, find the exact values of the six circular functions.
(a) ✓ = 30
Answer:
sin 30 = 12 , cos 30 =
p
p
2 3
, cot 30 = 3
3
(b) ✓ = 3⇡
4
p
3
,
2
tan 30 =
p
3
,
3
csc 30 = 2, sec 30 =
p
p
p
2
2
3⇡
Answer: sin 3⇡
=
,
cos
=
, tan 3⇡
= 1, csc 3⇡
= 2, sec 3⇡
=
4
2
4
2
4
4
4
p
3⇡
2, cot 4 = 1
(c) ✓ = 150
p
p
Answer: sin( 150 ) =p 12 , cos( 150 ) = 23 , tan( 150 ) = 33 , csc( 150 ) =
p
2, sec( 150 ) = 2 3 3 , cot( 150 ) = 3
(d) ✓ = 4⇡
3
p
p
3
4⇡
4⇡
1
4⇡
Answer:
sin(
)
=
,
cos(
)
=
,
tan(
)
=
3, csc( 4⇡
)=
3
2
3
2
3
3
p
p
2 3
3
4⇡
4⇡
, sec( 3 ) = 2, cot( 3 ) = 3
3
2. Given a value of one circular function and sign of another function (or the
quadrant where the angle lies), find the value of the indicated function.
p
Answer: 23
Answer: 54
(a) sin ✓ = 12 , ✓ in QI; cos ✓
(b) cos ✓ = 35 , ✓ in QIV; csc ✓
(c) sin ✓ =
(d) cot ✓ =
3
,
7
2
,
9
sec ✓ < 0; tan ✓
Answer:
cos ✓ > 0; csc ✓
Answer:
142
p
3 10
20
p
85
9
3.2.2. Reference Angle
We observe that if ✓1 and ✓2 are coterminal angles, the values of the six circular
or trigonometric functions at ✓1 agree with the values at ✓2 . Therefore, in finding
the value of a circular function at a number ✓, we can always reduce ✓ to a number
between 0 and 2⇡. For example, sin 14⇡
= sin 14⇡
4⇡ = sin 2⇡
. Also, observe
3
3
3
2⇡
⇡
from Figure 3.7 that sin 3 = sin 3 .
Figure 3.7
In general, if ✓1 , ✓2 , ✓3 , and ✓4 are as shown in Figure 3.8 with P (✓1 ) =
(x1 , y1 ), then each of the x-coordinates of P (✓2 ), P (✓3 ), and P (✓4 ) is ±x1 , while
the y-coordinate is ±y1 . The correct sign is determined by the location of the
angle. Therefore, together with the correct sign, the value of a particular circular
function at an angle ✓ can be determined by its value at an angle ✓1 with radian
measure between 0 and ⇡2 . The angle ✓1 is called the reference angle of ✓.
Figure 3.8
143
The signs of the coordinates of P (✓) depends on the quadrant or axis where
it terminates. It is important to know the sign of each circular function in each
quadrant. See Figure 3.9. It is not necessary to memorize the table, since the
sign of each function for each quadrant is easily determined from its definition.
We note that the signs of cosecant, secant, and cotangent are the same as sine,
cosine, and tangent, respectively.
Figure 3.9
Using the fact that the unit circle is symmetric with respect to the x-axis, the
y-axis, and the origin, we can identify the coordinates of all the points using the
coordinates of corresponding points in the Quadrant I, as shown in Figure 3.10
for the special angles.
Figure 3.10
144
Example 3.2.4. Use reference angle and appropriate sign to find the exact value
of each expression.
(1) sin 11⇡
and cos 11⇡
(3) sin 150
6
6
7⇡
6
(2) cos
(4) tan 8⇡
3
Solution. (1) The reference angle of 11⇡
is ⇡6 , and it lies in Quadrant IV wherein
6
sine and cosine are negative and positive, respectively.
⇡
1
=
6
2
p
11⇡
⇡
3
cos
= cos =
6
6
2
sin
11⇡
=
6
sin
(2) The angle 7⇡
lies in Quadrant II wherein cosine is negative, and its refer6
ence angle is ⇡6 .
p
✓
◆
7⇡
⇡
3
cos
= cos =
6
6
2
(3) sin 150 = sin 30 =
(4) tan
8⇡
3
=
tan
⇡
3
=
1
2
⇡
3
⇡
cos
3
sin
p
=
3
2
1
2
=
p
3
2
Seatwork/Homework 3.2.2
Use reference angle and appropriate sign to find the exact value of each expression.
(1) sin 510
(2) tan( 225 )
Answer:
(3) sec 13⇡
3
(4) cot
1
2
Answer:
1
Answer: 2
10⇡
3
Answer:
p
3
3
Exercises 3.2
1. Find the exact value.
(a) sin 600
Answer:
(b) tan( 810 )
p
3
2
Answer: Undefined
p
Answer:
2
(c) sec 585
145
(d) cos( 420 )
1
2
1
2
1
2
Answer:
(e) sin 7⇡
6
Answer:
(f) cos 5⇡
3
Answer:
(g) tan 3⇡
4
Answer:
1
(h) sec 2⇡
3
Answer:
2
(i) csc
(j) cot
11⇡
6
35⇡
6
Answer:
Answer:
4⇡
3
(k) cos
Answer:
(m) cos 7⇡
4
Answer:
(n) sec 19⇡
4
Answer:
4⇡
3
Answer:
(p) sec 23⇡
6
Answer:
(q) csc 13⇡
3
Answer:
(r) tan 5⇡
6
Answer:
2. Find the exact value of each expression.
(b) cos( 30 ) + sin 420
(c) tan( 225 ) + tan 405
csc( 300 )
2⇡
+ sin2 2⇡
3
3
11⇡
5⇡
sin 6 + cos 3
2 cos 5⇡
sin 5⇡
3
2
2 ⇡
tan 4 + 2 cos 8⇡
3
(e) cos2
(f)
(g)
(h)
(i)
(j)
sin(
11⇡
6
⇡
6
)
p
1
2
3
3
p
2
2
p
2
p
3
2
p
2 3
3
p
2 3
3
p
3
3
Answer: 1
Answer: 0
Answer: 0
sin 13⇡
6
1
2
Answer:
tan 2⇡
tan 5⇡
3
6
1+tan 2⇡
tan 5⇡
3
6
sin
3
Teaching Notes
(sin x)2 is denoted
sin2 x.
Answer: 1 by
Similarly, this
p
is used
Answer: 3 notation
with the other
Answer: 0 trigonometric
functions. In
for a
Answer: 0 general,
positive integer n,
sinn x = (sin x)n .
(a) sin2 150 + cos2 150
(d) sec 750
2
Answer:
(l) tan 17⇡
3
(o) sin
p
Answer:
⇡
6
+cos 5⇡
6
cos
p
3
3
Answer: 1
3. Compute P (✓), and find the exact values of the six circular functions.
(a) ✓ =
19⇡
6
Answer: P (✓) =
csc 19⇡
=
6
⇣
p
3
,
2
2, sec 19⇡
=
6
⌘
1
, sin 19⇡
=
2
6
p
2 3
, cot 19⇡
=
3
6
146
1
,
2
p
3
cos 19⇡
=
6
p
3
,
2
tan 19⇡
=
6
p
3
,
3
(b) ✓ =
32⇡
3
Answer: P (✓) =
csc 32⇡
=
3
p
2 3
,
3
⇣
p ⌘
3
1
,
,
2 2
sec 32⇡
=
3
sin 32⇡
=
3
p
3
,
2
p
cos 32⇡
=
3
1
,
2
tan 32⇡
=
3
p
3,
3
3
2, cot 32⇡
=
3
4. Given the value of a particular circular function and an information about the
angle ✓, find the values of the other circular functions.
(a) cos ✓ =
1
2
and
3⇡
2
< ✓ < 2⇡
Answer: sec ✓ = 2, sin ✓ =
8
(b) sin ✓ = 17
and 0 < ✓ < ⇡2
Answer: csc ✓ = 17
, cos ✓ =
8
(c) cos ✓ =
p
2 13
13
and
Answer: sec ✓ =
2
3
p
3
,
2
15
,
17
3⇡
< ✓ < 2⇡
2
p
13
, sin ✓ =
2
tan ✓ =
tan ✓ =
p
3 13
,
13
p
8
,
15
3, csc ✓ =
sec ✓ =
tan ✓ =
3
,
2
17
,
15
p
2 3
,
3
cot ✓ =
cot ✓ =
csc ✓ =
p
3
3
15
8
p
13
,
3
cot ✓ =
4
Lesson 3.3. Graphs of Circular Functions and Situational
Problems
Time Frame: 6 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) determine the domain and range of the di↵erent circular functions;
(2) graph the six circular functions with its amplitude, period, and phase shift;
and
(3) solve situational problems involving circular functions.
Lesson Outline
(1) Domain and range of circular functions
(2) Graphs of circular functions
(3) Amplitude, period, and phase shift
Introduction
There are many things that occur periodically. Phenomena like rotation of
the planets and comets, high and low tides, and yearly change of the seasons
147
follow a definite pattern. In this lesson, we will graph the six circular functions
and we will see that they are periodic in nature.
3.3.1. Graphs of y = sin x and y = cos x
Recall that, for a real number x, sin x = sin ✓ for an angle ✓ with measure x
radians, and that sin ✓ is the second coordinate of the point P (✓) on the unit
circle. Since each x corresponds to an angle ✓, we can conclude that
(1) sin x is defined for any real number x or the domain of the sine function is
R, and
(2) the range of sine is the set of all real numbers between
1 and 1 (inclusive).
From the definition, it also follows that sin(x+2⇡) = sin x for any real number
x. This means that the values of the sine function repeat every 2⇡ units. In this
case, we say that the sine function is a periodic function with period 2⇡.
Table 3.11 below shows the values of y = sin x, where x is the equivalent radian
measure of the special angles and their multiples from 0 to 2⇡. As commented
above, these values determine the behavior of the function on R.
x
0
⇡
6
y
0
1
2
0
0.5
x
7⇡
6
y
1
2
0.5
⇡
4
p
2
2
⇡
3
p
3
2
⇡
2
1
0.71 0.87
1
5⇡
4
p
4⇡
3
p
3
2
1
0.71
0.87
1
2
2
3⇡
2
2⇡
3
p
3
2
3⇡
4
p
2
2
0.87 0.71
5⇡
6
⇡
1
2
0
0.5
0
5⇡
3
p
7⇡
4
p
2
2
1
2
0
0.87
0.71
0.5
0
3
2
11⇡
6
2⇡
Table 3.11
From the table, we can observe that as x increases from 0 to ⇡2 , sin x also
increases from 0 to 1. Similarly, as x increases from 3⇡
to 2⇡, sin x also increases
2
from 1 to 0. On the other hand, notice that as x increases from ⇡2 to ⇡, sin x
decreases from 1 to 0. Similarly, as x increases from ⇡ to 3⇡
, sin x decreases from
2
0 to 1.
To sketch the graph of y = sin x, we plot the points presented in Table 3.11,
and join them with a smooth curve. See Figure 3.12. Since the graph repeats
every 2⇡ units, Figure 3.13 shows periodic graph over a longer interval.
148
Teaching Notes
It is a good
exercise to
construct the
graph of the sine
function using the
height of P (✓).
Put the unit circle
side-by-side with
the coordinate
plane for the
graph, and trace
the height for each
value of x onto the
graph of y = sin x.
Figure 3.12
Figure 3.13
We can make observations about the cosine function that are similar to the
sine function.
• y = cos x has domain R and range [ 1, 1].
• y = cos x is periodic with period 2⇡. The graph of y = cos x is shown in
Figure 3.14.
Figure 3.14
From the graphs of y = sin x and y = cos x in Figures 3.13 and 3.14, respectively, we observe that sin( x) = sin x and cos( x) = cos x for any real
number x. In other words, the graphs of y = cos( x) and y = cos x are the same,
while the graph of y = sin( x) is the same as that of y = sin x.
In general, if a function f satisfies the property that f ( x) = f (x) for all x
in its domain, we say that such function is even. On the other hand, we say that
a function f is odd if f ( x) = f (x) for all x in its domain. For example, the
functions x2 and cos x are even, while the functions x3 3x and sin x are odd.
149
3.3.2. Graphs of y = a sin bx and y = a cos bx
Using a table of values from 0 to 2⇡, we can sketch the graph of y = 3 sin x, and
compare it to the graph of y = sin x. See Figure 3.15 wherein the solid curve
belongs to y = 3 sin x, while the dashed curve to y = sin x. For instance, if x = ⇡2 ,
then y = 1 when y = sin x, and y = 3 when y = 3 sin x. The period, x-intercepts,
and domains are the same for both graphs, while they di↵er in the range. The
range of y = 3 sin x is [ 3, 3].
Figure 3.15
In general, the graphs of y = a sin x and y = a cos x with a > 0 have the same
shape as the graphs of y = sin x and y = cos x, respectively. If a < 0, there is
a reflection across the x-axis. The range of both y = a sin x and y = a cos x is Teaching Notes
Review or teach
[ |a|, |a|].
the reflection
In the graphs of y = a sin x and y = a cos x, the number |a| is called
its amplitude. It dictates the height of the curve. When |a| < 1,
the graphs are shrunk vertically, and when |a| > 1, the graphs are
stretched vertically.
Now, in Table 3.16, we consider the values of y = sin 2x on [0, 2⇡].
0
⇡
6
p
3
2
0
0.87
x
0
y
x
y
⇡
4
7⇡
6
p
3
2
5⇡
4
0.87
1
⇡
3
p
3
2
⇡
2
0
1
0.87
0
3⇡
2
1
4⇡
3
p
3
2
1
0.87
0
0
2⇡
3
p
3
2
1
0.87
1
5⇡
3
p
7⇡
4
5⇡
6
p
⇡
3
2
0
0.87
0
3
2
1
11⇡
6
p
3
2
0.87
1
0.87
Table 3.16
150
3⇡
4
2⇡
0
0
across the x-axis
when the sign of
the function is
changed.
Figure 3.17
Figure 3.17 shows the graphs of y = sin 2x (solid curve) and y = sin x (dashed
curve) over the interval [0, 2⇡]. Notice that, for sin 2x to generate periodic values
similar to [0, 2⇡] for y = sin x, we just need values of x from 0 to ⇡. We then
expect the values of sin 2x to repeat every ⇡ units thereafter. The period of
y = sin 2x is ⇡.
2⇡
.
|b|
If 0 < |b| < 1, the graphs are stretched horizontally, and if |b| > 1, the
graphs are shrunk horizontally.
If b 6= 0, then both y = sin bx and y = cos bx have period given by
To sketch the graphs of y = a sin bx and y = a cos bx, a, b 6= 0, we may proceed
with the following steps:
(1) Determine the amplitude |a|, and find the period 2⇡
. To draw one cycle
|b|
of the graph (that is, one complete graph for one period), we just need to
complete the graph from 0 to 2⇡
.
|b|
(2) Divide the interval into four equal parts, and get five division points: x1 = 0,
x2 , x3 , x4 , and x5 = 2⇡
, where x3 is the midpoint between x1 and x5 (that
|b|
1
is, 2 (x1 + x5 ) = x3 ), x2 is the midpoint between x1 and x3 , and x4 is the
midpoint between x3 and x5 .
(3) Evaluate the function at each of the five x-values identified in Step 2. The
points will correspond to the highest point, lowest point, and x-intercepts
of the graph.
(4) Plot the points found in Step 3, and join them with a smooth curve similar
to the graph of the basic sine curve.
(5) Extend the graph to the right and to the left, as needed.
151
Example 3.3.1. Sketch the graph of one cycle of y = 2 sin 4x.
Solution.
(1) The period is
2⇡
4
= ⇡2 , and the amplitude is 2.
(2) Dividing the interval [0, ⇡2 ] into 4 equal parts, we get the following xcoordinates: 0, ⇡8 , ⇡4 , 3⇡
, and ⇡2 .
8
(3) When x = 0, ⇡4 , and ⇡2 , we get y = 0. On the other hand, when x = ⇡8 , we
have y = 2 (the amplitude), and y = 2 when x = 3⇡
.
8
(4) Draw a smooth curve by connecting the points. There is no need to proceed
to Step 5 because the problem only asks for one cycle.
Example 3.3.2. Sketch the graph of y =
Solution.
(1) The amplitude is |
3 cos x2 .
3| = 3, and the period is
2⇡
1
2
= 4⇡.
(2) We divide the interval [0, 4⇡] into four equal parts, and we get the following
x-values: 0, ⇡, 2⇡, 3⇡, and 4⇡.
(3) We have y = 0 when x = ⇡ and 3⇡, y =
when x = 2⇡.
3 when x = 0 and 4⇡, and y = 3
(4) We trace the points in Step 3 by a smooth curve.
(5) We extend the pattern in Step 4 to the left and to the right.
152
Example 3.3.3. Sketch the graph of two cycles of y = 12 sin
Solution. Since the sine function is odd, the graph of y = 12 sin
as that of y = 12 sin 2x
.
3
(1) The amplitude is 12 , and the period is
2⇡
2
3
2x
3
.
2x
3
is the same
= 3⇡.
(2) Dividing the interval [0, 3⇡] into four equal parts, we get the x-coordinates
of the five important points:
0 + 3⇡
3⇡
=
,
2
2
(3) We get y = 0 when x = 0,
9⇡
.
4
0 + 3⇡
3⇡
2
=
,
2
4
3⇡
,
2
and 3⇡, y =
3⇡
2
1
2
+ 3⇡
9⇡
=
.
2
4
when
3⇡
,
4
and y =
(4) We trace the points in Step 3 by a smooth curve.
(5) We extend the pattern in Step 4 by one more period to the right.
Seatwork/Homework 3.3.2
(1) Sketch the graph of one cycle of y = 12 sin 3x.
Answer:
(2) Sketch the graph of two cycles of y =
Answer:
153
2 cos
x
2
.
1
2
when
(3) Sketch the graph of y =
2 cos 4x.
Answer:
x
3
(4) Sketch the graph of one cycle of y = 3 sin
.
Answer:
3.3.3. Graphs of y = a sin b(x
c) + d and y = a cos b(x
We first compare the graphs of y = sin x and y = sin x
values and the 5-step procedure discussed earlier.
⇡
3
c) + d
using a table of
Teaching Notes
Review or teach
⇡
the horizontal
As x runs from ⇡3 to 7⇡
,
the
value
of
the
expression
x
runs
from
0
to
2⇡.
So
3
3
rule: if
for one cycle of the graph of y = sin x ⇡3 , we then expect to have the graph of translation
x is replaced by
y = sin x starting from x = ⇡3 . This is confirmed by the values in Table 3.18. We x h in the
the
then apply a similar procedure to complete one cycle of the graph; that is, divide equation,
graph is translated
the interval [ ⇡3 , 7⇡
] into four equal parts, and then determine the key values of |h| units to the
3
right if h > 0 and
x in sketching the graphs as discussed earlier. The one-cycle graph of y = sin x to the left if h < 0.
(dashed curve) and the corresponding one-cycle graph of y = sin x ⇡3 (solid
curve) are shown in Figure 3.19.
x
x
sin x
⇡
3
⇡
3
⇡
3
5⇡
6
4⇡
3
11⇡
6
7⇡
3
0
⇡
2
⇡
3⇡
2
2⇡
0
1
0
Table 3.18
154
1
0
Figure 3.19
Observe that the graph of y = sin x ⇡3 shifts ⇡3 units to the right of
y = sin x. Thus, they have the same period, amplitude, domain, and range.
The graphs of
y = a sin b(x
c) and y = a cos b(x
c)
have the same shape as y = a sin bx and y = a cos bx, respectively, but
shifted c units to the right when c > 0 and shifted |c| units to the left
if c < 0. The number c is called the phase shift of the sine or cosine
graph.
Example 3.3.4. In the same Cartesian plane, sketch one cycle of the graphs of
y = 3 sin x and y = 3 sin x + ⇡4 .
Solution. We have sketched the graph of y = 3 sin x earlier at the start of the
lesson. We consider y = 3 sin x + ⇡4 . We expect that it has the same shape as
that of y = 3 sin x, but shifted some units.
Here, we have a = 3, b = 1, and c = ⇡4 . From these constants, we get
the amplitude, the period, and the phase shift, and these are 3, 2⇡, and ⇡4 ,
respectively.
⇡
4
One cycle starts at x =
the important values of x.
⇡
4
+
2
7⇡
4
=
and ends at x =
⇡
4
3⇡
,
4
3⇡
4
⇡
4
x
y = 3 sin x +
+
2
⇡
4
0
155
=
⇡
,
4
⇡
4
+ 2⇡ =
3⇡
4
⇡
4
3⇡
4
3
0
+
2
7⇡
4
5⇡
4
3
7⇡
.
4
=
5⇡
4
7⇡
4
0
We now compute
While the e↵ect of c in y = a sin b(x c) and y = a cos b(x c) is
a horizontal shift of their graphs from the corresponding graphs of
y = a sin bx and y = a cos bx, the e↵ect of d in the equations y =
a sin b(x c) + d and y = a cos b(x c) + d is a vertical shift. That is,
the graph of y = a sin b(x c) + d has the same amplitude, period, and
phase shift as that of y = a sin b(x c), but shifted d units upward
when d > 0 and |d| units downward when d < 0.
Example 3.3.5. Sketch the graph of
⇣
2 cos 2 x
y=
⇡⌘
6
3.
Solution. Here, a = 2, b = 2, c = ⇡6 , and d = 3. We first sketch one cycle of
the graph of y = 2 cos 2 x ⇡6 , and then extend this graph to the left and to
the right, and then move the resulting graph 3 units downward.
The graph of y =
⇡
6
2 cos 2 x
⇡
.
6
⇡
6
Start of one cycle:
End of the cycle:
⇡
6
+ 7⇡
2⇡
6
=
,
2
3
⇡
6
has amplitude 2, period ⇡, and phase shift
⇡
6
+⇡ =
7⇡
6
+ 2⇡
5⇡
3
=
,
2
12
2⇡
3
⇡
6
x
y=
y=
⇡
6
2 cos 2 x
2 cos 2 x
+
2
⇡
6
2
3
5
156
7⇡
6
=
11⇡
12
5⇡
12
2⇡
3
11⇡
12
0
2
0
3
1
3
7⇡
6
2
5
Teaching Notes
Review or teach
the vertical
translation rule: if
the equation
y = f (x) is
changed to
y = f (x) + k, the
graph is translated
|k| units upward if
k > 0 and
downward if k < 0.
Before we end this sub-lesson, we make the following observation, which will
be used in the discussion on simple harmonic motion (Sub-Lesson 3.3.6).
Di↵erent Equations, The Same Graph
1. The graphs of y = sin x and y = sin(x + 2⇡k), k any integer, are
the same.
2. The graphs of y = sin x, y =
y = cos(x + ⇡2 ) are the same.
sin(x + ⇡), y = cos(x
⇡
),
2
and
3. In general, the graphs of
y = a sin b(x
y=
a sin[b(x
y = a cos[b(x
c) + d,
c) + ⇡ + 2⇡k] + d,
c)
⇡
2
+ 2⇡k] + d,
and
y=
a cos[b(x
c) +
⇡
2
+ 2⇡k] + d,
where k is any integer, are all the same.
Similar observations are true for cosine.
Seatwork/Homework 3.3.3
(1) In the same Cartesian plane, sketch one cycle of the graphs of y = 3 cos x
and y = 3 cos x + ⇡3
1.
157
Answer:
(2) In the same Cartesian plane, sketch one cycle of the graphs of y =
and y = 2 14 sin 2 x ⇡4 .
Answer:
(3) Sketch the graph of y = 2 sin
⇡
2
x
Answer:
158
2.
1
4
sin 2x
3.3.4. Graphs of Cosecant and Secant Functions
We know that csc x =
graph of y = csc x.
1
sin x
if sin x 6= 0. Using this relationship, we can sketch the
First, we observe that the domain of the cosecant function is
{x 2 R : sin x 6= 0} = {x 2 R : x 6= k⇡, k 2 Z}.
Table 3.20 shows the key numbers (that is, numbers where y = sin x crosses the
x-axis, attain its maximum and minimum values) and some neighboring points,
where “und” stands for “undefined,” while Figure 3.21 shows one cycle of the
graphs of y = sin x (dashed curve) and y = csc x (solid curve). Notice the
asymptotes of the graph y = csc x.
x
0
⇡
6
⇡
2
5⇡
6
⇡
7⇡
6
3⇡
2
11⇡
6
y = sin x
0
1
2
1
1
2
0
1
2
1
1
2
y = csc x
und
2
1
2
und
2
1
2 und
2⇡
0
Table 3.20
Figure 3.21
We could also sketch the graph of csc x directly from the graph of y = sin x
by observing the following facts:
(1) If sin x = 1 (or
1), then csc x = 1 (or
1).
(2) At each x-intercept of y = sin x, y = csc x is undefined; but a vertical
asymptote is formed because, when sin x is close to 0, the value of csc x will
have a big magnitude with the same sign as sin x.
159
Refer to Figure 3.22 for the graphs of y = sin x (dashed curve) and y = csc x
(solid curve) over a larger interval.
Figure 3.22
Like the sine and cosecant functions, the cosine and secant functions are also
reciprocals of each other. Therefore, y = sec x has domain
{x 2 R : cos x 6= 0} = {x 2 R : x 6=
k⇡
, k odd integer}.
2
Similarly, the graph of y = sec x can be obtained from the graph of y = cos x.
These graphs are shown in Figure 3.23.
Figure 3.23
Example 3.3.6. Sketch the graph of y = 2 csc x2 .
Solution. First, we sketch the graph of y = 2 sin x2 , and use the technique discussed above to sketch the graph of y = 2 csc x2 .
160
The vertical asymptotes of y = 2 csc x2 are the x-intercepts of y = 2 sin x2 :
x = 0, ±2⇡, ±4⇡, . . .. After setting up the asymptotes, we now sketch the graph
of y = 2 csc x2 as shown below.
Example 3.3.7. Sketch the graph of y = 2
sec 2x.
Solution. Sketch the graph of y = cos 2x (note that it has period ⇡), then sketch
the graph of y = sec 2x (as illustrated above), and then move the resulting
graph 2 units upward to obtain the graph of y = 2 sec 2x.
161
Seatwork/Homework 3.3.4
(1) Sketch the graph of y =
sec x on the interval [0, 2⇡].
Answer:
(2) Sketch the graph of y = 2 csc 4x
1 on the interval
Answer:
162
⇥
⇡ ⇡
,
2 2
⇤
.
3.3.5. Graphs of Tangent and Cotangent Functions
sin x
We know that tan x = cos
, where cos x 6= 0. From this definition of the tangent
x
function, it follows that its domain is the same as that of the secant function,
which is
{x 2 R : cos x 6= 0} = {x 2 R : x 6=
k⇡
, k odd integer}.
2
We note that tan x = 0 when sin x = 0 (that is, when x = k⇡, k any integer), and
that the graph of y = tan x has asymptotes x = k⇡
, k odd integer. Furthermore,
2
by recalling the signs of tangent from Quadrant I to Quadrant IV and its values,
we observe that the tangent function is periodic with period ⇡.
To sketch the graph of y = tan x, it will be enough to know its one-cycle
⇡ ⇡
, . See Table 3.24 and Figure 3.25.
Teaching Notes graph on the open interval
2 2
There is also a way
of sketching the
graph of y = tan x
based on the
tangent segment to
the unit circle,
similar to the
construction
described in
sketching the
graph of y = sin x.
But we do not go
anymore into the
details of this
approach.
⇡
2
x
y = tan x
⇡
3
p
und
x
y = tan x
⇡
6
p
3
3
⇡
4
3
1
⇡
6
p
3
3
⇡
4
⇡
3
⇡
2
1
p
3
0
0
und
Table 3.24
Figure 3.25
In the same manner, the domain of y = cot x =
cos x
sin x
is
{x 2 R : sin x 6= 0} = {x 2 R : x 6= k⇡, k 2 Z},
and its period is also ⇡. The graph of y = cot x is shown in Figure 3.26.
163
Figure 3.26
In general, to sketch the graphs of y = a tan bx and y = a cot bx, a 6= 0 and
b > 0, we may proceed with the following steps:
(1) Determine the period ⇡b . Then we draw one cycle of the graph on
for y = a tan bx, and on 0, ⇡b for y = a cot bx.
⇡ ⇡
,
2b 2b
(2) Determine the two adjacent vertical asymptotes. For y = a tan bx, these
⇡
vertical asymptotes are given by x = ± 2b
. For y = a cot bx, the vertical
asymptotes are given by x = 0 and x = ⇡b .
(3) Divide the interval formed by the vertical asymptotes in Step 2 into four
equal parts, and get three division points exclusively between the asymptotes.
(4) Evaluate the function at each of these x-values identified in Step 3. The
points will correspond to the signs and x-intercept of the graph.
(5) Plot the points found in Step 3, and join them with a smooth curve approaching to the vertical asymptotes. Extend the graph to the right and to
the left, as needed.
Example 3.3.8. Sketch the graph of y = 12 tan 2x.
Solution. The period of the function is ⇡2 , and the adjacent asymptotes are x =
⇡ ⇡
± ⇡4 , ± 3⇡
, . . .. Dividing the interval
,
into four equal parts, the key x-values
4
4 4
⇡
⇡
are 8 , 0, and 8 .
x
⇡
8
0
⇡
8
y = 12 tan 2x
1
2
0
1
2
164
Example 3.3.9. Sketch the graph of y = 2 cot x3 on the interval (0, 3⇡).
Solution. The period of the function is 3⇡, and the adjacent asymptotes are x = 0
and x = 3⇡. We now divide the interval (0, 3⇡) into four equal parts, and the
key x-values are 3⇡
, 3⇡
, and 9⇡
.
4
2
4
x
3⇡
4
3⇡
2
y = 2 cot x3
2
0
165
9⇡
4
2
Seatwork/Homework 3.3.5
(1) Sketch the graph of y = cot( x) on the interval [ ⇡, ⇡].
Answer:
(2) Sketch the graph of y = 2 tan x4 on the interval [ 2⇡, 2⇡].
Answer:
3.3.6. Simple Harmonic Motion
Repetitive or periodic behavior is common in nature. The time-telling device
known as sundial is a result of the predictable rising and setting of the sun
everyday. It consists of a flat plate and a gnomon. As the sun moves across the
sky, the gnomon casts a shadow on the plate, which is calibrated to tell the time
of the day.
166
https://commons.wikimedia.org/wiki/File:Sundial 2r.jpg
By liz west (Sundial)
[CC BY 2.0 (http://creativecommons.org/licenses/by/2.0)],
via Wikimedia Commons
Some motions are also periodic. When a weight is suspended on a spring,
pulled down, and released, the weight oscillates up and down. Neglecting resistance, this oscillatory motion of the weight will continue on and on, and its height
is periodic with respect to time.
t = 0 sec
t = 2.8 sec
167
t = 6.1 sec
t = 9 sec
Periodic motions are usually modeled by either sine or cosine function, and are
called simple harmonic motions. Unimpeded movements of objects like oscillation, vibration, rotation, and motion due to water waves are real-life occurrences
that behave in simple harmonic motion.
Equations of Simple Harmonic Motion
The displacement y (directed height or length) of an object behaving
in a simple harmonic motion with respect to time t is given by one of
the following equations:
y = a sin b(t
c) + d
y = a cos b(t
c) + d.
or
In both equations, we have the following information:
• amplitude = |a| = 12 (M m) - the maximum displacement above
and below the rest position or central position or equilibrium, where
M is the maximum height and m is the minimum height;
• period = 2⇡
- the time required to complete one cycle (from one
|b|
highest or lowest point to the next);
• frequency =
|b|
2⇡
- the number of cycles per unit of time;
• c - responsible for the horizontal shift in time; and
• d - responsible for the vertical shift in displacement.
Example 3.3.10. A weight is suspended from a spring and is moving up and
down in a simple harmonic motion. At start, the weight is pulled down 5 cm below
the resting position, and then released. After 8 seconds, the weight reaches its
168
highest location for the first time. Find the equation of the motion.
Solution. We are given that the weight is located at its lowest position at t = 0;
that is, y = 5 when t = 0. Therefore, the equation is y = 5 cos bt.
Because it took the weight 8 seconds from the lowest point to its immediate
highest point, half the period is 8 seconds.
1 2⇡
⇡
⇡t
·
= 8 =) b =
=) y = 5 cos
2
2 b
8
8
? Example 3.3.11. Suppose you ride a Ferris wheel. The lowest point of the
wheel is 3 meters o↵ the ground, and its diameter is 20 m. After it started, the
Ferris wheel revolves at a constant speed, and it takes 32 seconds to bring you
back again to the riding point. After riding for 150 seconds, find your approximate
height above the ground.
Solution. We ignore first the fixed value of 3 m o↵ the ground, and assume that
the central position passes through the center of the wheel and is parallel to the
ground.
Let t be the time (in seconds) elapsed that you have been riding the Ferris
wheel, and y is he directed distance of your location with respect to the assumed
central position at time t. Because y = 10 when t = 0, the appropriate model
is y = 10 cos bt for t 0.
Given that the Ferris wheel takes 32 seconds to move from the lowest point
to the next, the period is 32.
2⇡
⇡
= 32 =) b =
=)
b
16
When t = 150, we get y = 10 cos 150⇡
⇡ 3.83.
16
y=
10 cos
⇡t
16
Bringing back the original condition given in the problem that the riding point
is 3 m o↵ the ground, after riding for 150 seconds, you are approximately located
3.83 + 13 = 16.83 m o↵ the ground.
2
In the last example, the central position or equilibrium may be vertically
shifted from the ground or sea level (the role of the constant d). In the same way,
the starting point may also be horizontally shifted (the role of the constant c).
Moreover, as observed in Sub-Lesson 3.3.3 (see page 157), to find the function
that describes a particular simple harmonic motion, we can either choose
y = a sin b(t
c) + d
y = a cos b(t
c) + d,
or
and determine the appropriate values of a, b, c, and d. In fact, we can assume
that a and b are positive numbers, and c is the smallest such nonnegative number.
169
Example 3.3.12. A signal buoy in Laguna Bay bobs up and down with the
height h of its transmitter (in feet) above sea level modeled by h(t) = a sin bt + d
at time t (in seconds). During a small squall, its height varies from 1 ft to 9 ft
above sea level, and it takes 3.5 seconds from one 9-ft height to the next. Find
the values of the constants a, b, and d.
Solution. We solve the constants step by step.
• The minimum and maximum values of h(t) are 1 ft and 9 ft, respectively.
Thus, the amplitude is a = 12 (M m) = 12 (9 1) = 4.
• Because it takes 3.5 seconds from one 9-ft height to the next, the period is
3.5. Thus, we have 2⇡
= 3.5, which gives b = 4⇡
.
b
7
• Because the lowest point is 1 ft above the sea level and the amplitude is 4,
it follows that d = 5.
2
Example 3.3.13. A variable star is a star whose brightness fluctuates as observed from Earth. The magnitude of visual brightness of one variable star ranges
from 2.0 to 10.1, and it takes 332 days to observe one maximum brightness to
the next. Assuming that the visual brightness of the star can be modeled by the
equation y = a sin b(t c) + d, t in days, and putting t = 0 at a time when the
star is at its maximum brightness, find the constants a, b, c, and d, where a, b > 0
and c the least nonnegative number possible.
Solution.
a=
M
m
2
=
10.1
2.0
2
= 4.05
2⇡
⇡
= 332 =) b =
b
166
d = a + m = 4.05 + 2.0 = 6.05
For the (ordinary) sine function to start at the highest point at t = 0, the least
possible horizontal movement to the right (positive value) is 3⇡
units.
2
bc =
3⇡
2
=)
c=
3⇡
3⇡
=
⇡ = 249
2b
2 · 166
2
? Example 3.3.14. The path of a fast-moving particle traces a circle with equation
(x + 7)2 + (y 5)2 = 36.
It starts at point ( 1, 5), moves clockwise, and passes the point ( 7, 11) for the
first time after traveling 6 microseconds. Where is the particle after traveling 15
microseconds?
170
Solution. As described above, we may choose sine or cosine function. Here, we
choose the sine function to describe both x and y in terms of time t in microseconds; that is, we let
x = a sin b(t
c) + d and y = e sin f (t
g) + h,
where we appropriately choose the positive values for a, b, e, and f , and the least
nonnegative values for c and g.
The given circle has radius 6 and center ( 7, 5). Defining the central position
of the values of x as the line x = 7 and that of the values of y as the line y = 5,
we get a = e = 6, d = 7, and h = 5.
From the point ( 1, 5) to the point ( 7, 11) (moving clockwise), the particle
has traveled three-fourths of the complete cycle; that is, three-fourths of the
period must be 2.
3 2⇡
3 2⇡
·
= ·
=6
4 b
4 f
=)
b=f =
⇡
4
Teaching Notes
As the particle starts at ( 1, 5) and moves clockwise, the values of x start
Here, we need an
at
its
highest value (x = 1) and move downward toward its central position
equation with the
same graph as (x =
7) and continue to its lowest value (x = 13). Therefore, the graph of
y = a sin(bt+ ⇡2 )+d
3⇡
that will fit in the a sin bt + d has to move 2b = 6 units to the right, and so we get c = 6.
equation
y = a sin b(t c)+d,
As to the value of g, we observe the values of y start at its central position
where c is the least
(y
=
5) and go downward to its lowest value (y = 1). Similar to the argument
nonnegative
⇡
possible number. used in determining c, the graph of y = e sin f t + h has to move
= 4 units to
b
Recall the
the
right,
implying
that
g
=
4.
observation made
on page 157.
Hence, We have the following equations of x and y in terms of t:
x = 6 sin ⇡4 (t
6)
7 and y = 6 sin ⇡4 (t
4) + 5.
When t = 15, we get
x = 6 sin ⇡4 (15
and
y = 6 sin ⇡4 (15
6)
7=
p
7+3 2⇡
2.76
p
4) + 5 = 5 + 3 2 ⇡ 9.24.
That is, after traveling for 15 microseconds, the particle is located near the point
( 2.76, 9.24).
2
Seatwork/Homework 3.3.6
? 1. A weight is suspended from a spring and is moving up and down in a simple
harmonic motion. At start, the weight is pushed up 6 cm above the resting
171
position, and then released. After 14 seconds, the weight reaches again to its
highest position. Find the equation of the motion, and locate the weight with
respect to the resting position after 20 seconds since it was released.
Answer: y = 6 cos ⇡7 t or y = 6 sin pi7 (t + 72 ), location of the weight after 20
seconds: about 5.4 cm below the resting position
2. Suppose the lowest point of a Ferris wheel is 1.5 meters o↵ the ground, and its
radius is 15 m. It makes one complete revolution every 30 seconds. Starting at
the lowest point, find a cosine function that gives the height above the ground
of a riding child in terms of the time t in seconds.
⇡
Answer: y = 15 cos 15
t
15
2
+ 16.5
Exercises 3.3
1. Sketch two cycles of the graph (starting from x = 0) of the given function.
Indicate the amplitude, period, phase shift, domain, and range for each function.
(a) y = 4 sin x
Answer: amplitude = 4, period = 2⇡, phase shift = 0, domain = R,
range = [ 4, 4]
(b) y = 3 cos x
Answer: amplitude = 3, period = 2⇡, phase shift = 0, domain = R,
range = [ 3, 3]
(c) y = cos x4
Answer: amplitude = 1, period = 8⇡, phase shift = 0, domain = R,
range = [ 1, 1]
(d) y = sin 2x
Answer: amplitude = 1, period = ⇡, phase shift = 0, domain = R,
range = [ 1, 1]
172
(e) y = 2 + sin 4x
Answer: amplitude = 1, period =
range = [1, 3]
⇡
,
2
phase shift = 0, domain = R,
(f) y = 1 + cos x
Answer: amplitude = 1, period = 2⇡, phase shift = 0, domain = R,
range = [ 2, 0]
(g) y = 12 sin 3x
Answer: amplitude =
range = [ 12 , 12 ]
1
,
2
period =
2⇡
,
3
phase shift = 0, domain = R,
(h) y = 3 sin( x)
Answer: amplitude = 3, period = 2⇡, phase shift = 0, domain = R,
range = [ 3, 3]
(i) y = 3 2 cos x2
Answer: amplitude = 2, period = 4⇡, phase shift = 0, domain = R,
range = [1, 5]
(j) y = sin x ⇡4
Answer: amplitude = 1, period = 2⇡, phase shift =
range = [ 1, 1]
173
⇡
,
4
domain = R,
(k) y = 2 cos x + ⇡3
Answer: amplitude = 2, period = 2⇡, phase shift =
range = [ 2, 2]
⇡
,
3
domain = R,
(l) y = 3 sin(x 4⇡)
Answer: amplitude = 3, period = 2⇡, phase shift = 4⇡, domain = R,
range = [ 3, 3]
(m) y = 2 23 cos x ⇡2
Answer: amplitude =
range = [ 43 , 83 ]
period = 2⇡, phase shift =
⇡
,
2
domain = R,
(n) y = 4 cos x ⇡3 + 2
Answer: amplitude = 4, period = 2⇡, phase shift =
range = [ 2, 6]
⇡
,
3
domain = R,
2
,
3
2. Sketch the graph of the following functions.
(a) y = | sin x|
Answer:
174
(b) y = |4 cos x| + 2
Answer:
(c) y = |2 sin 2(x + ⇡)|
Answer:
1
3. Sketch the graph of each function over two periods, starting from x = 0.
Indicate the period, phase shift, domain, and range of each function.
(a) y = csc( x)
Answer: period = 2⇡, phase shift = 0, domain = {x|x 6= k⇡, k 2 Z},
range = ( 1, 1] [ [1, 1)
(b) y = cot( x)
Answer: period = ⇡, phase shift = 0, domain = {x|x 6= k⇡, k 2 Z},
range = R
(c) y = tan x
Answer: period = ⇡, phase shift = 0, domain = {x|x 6= (2k + 1) ⇡2 , k 2
Z}, range = R
175
(d) y = sec x
Answer: period = 2⇡, phase shift = 0, domain = {x|x 6= (2k + 1) ⇡2 , k 2
Z}, range = ( 1, 1] [ [1, 1)
(e) y = sec 3x
Answer: period = 2⇡
, phase shift = 0, domain = {x|x 6= (2k + 1) ⇡6 , k 2
3
Z}, range = ( 1, 1] [ [1, 1)
(f) y = 3 csc x
Answer: period = 2⇡, phase shift = 0, domain = {x|x 6= k⇡, k 2 Z},
range = ( 1, 3] [ [3, 1)
(g) y = 4 sec 2x
3
Answer: period = 3⇡, phase shift = 0, domain = {x|x 6= (2k + 1) 3⇡
, k2
4
Z}, range = ( 1, 4] [ [4, 1)
(h) y = tan(x + ⇡)
176
Answer: period = ⇡, phase shift =
Z}, range = R
(i) y = tan x ⇡2
Answer: period = ⇡, phase shift =
range = R
⇡, domain = {x|x 6= (2k + 1) ⇡2 , k 2
⇡
,
2
domain = {x|x 6= k⇡, k 2 Z},
(j) y = cot x + ⇡4
Answer: period = ⇡, phase shift = ⇡4 , domain = {x|x 6= (2k 1) ⇡4 , k even integer},
range = R
(k) y = 2 3 csc x
Answer: period = 2⇡, phase shift = 0, domain = {x|x 6= k⇡, k 2 Z},
range = ( 1, 1] [ [5, 1)
(l) y = 4 + sec 3x
Answer: period = 2⇡
, phase shift = 0, domain = {x|x 6= (2k + 1) ⇡6 , k 2
3
Z}, range = ( 1, 3] [ [6, 1)
(m) y = 2 sec x ⇡3
Answer: period = 2⇡, phase shift = ⇡3 , domain = {x|x 6= (2k + 1) 3⇡
, k2
4
Z}, range = ( 1, 2] [ [2, 1)
(n) y = 2 3 sec 2x
3
Answer: period = 3⇡, phase shift = 0, domain = {x|x 6= (2k + 1) 3⇡
, k2
4
Z}, range = ( 1, 2] [ [5, 1)
177
(o) y = 3 csc x 3⇡
2
Answer: period = 2⇡, phase shift =
Z}, range = ( 1, 3] [ [3, 1)
3⇡
,
2
domain = {x|x 6= (2k + 1) ⇡2 , k 2
4. Assuming that there is no vertical shift, find a function that describes a simple
harmonic motion with the following properties.
(a) sine function; displacement zero at time t = 0; moving up initially;
amplitude = 6 cm; period = 4 sec
Answer: y = 6 sin ⇡2 t
(b) cosine function; highest point 4 cm above the equilibrium at time t = 0;
period = 10 sec
Answer: y = 5 cos ⇡5 t
178
(c) cosine function; lowest point 9 cm below the equilibrium at time t = 0;
period = 5 sec
Answer: y = 9 cos 2⇡
(t 52 )
5
5. A point P moving in a simple harmonic motion makes 10 complete revolutions
every 1 second. The amplitude of the motion is 3 m. Assuming that P is at
its minimum displacement with respect to the equilibrium when t = 0 and
there is a vertical shift of 2 m downward, find a sine function that describes
1
the path traced by P in terms of time t.
Answer: y = 3 sin 20⇡(t 40
) 2
? 6. The path of a fast-moving particle (assuming constant speed) traces a circle
with equation
(x 3)2 + (y 4)2 = 25.
It starts at point (3, 1), moves counterclockwise, and passes the point (8, 4)
for the first time after traveling 7 microseconds. Where is the particle after
traveling for 20 microseconds?
Answer: about the point ( 1.87, 5.11)
Hint. The coordinates (x, y) of the location of the particle at time t (in mi⇡
⇡
croseconds) are given by x = 5 cos 14
(t 7) + 3 and y = 5 sin 14
(t 7) + 4.
7. A wooden ball is tied on a string 30 cm long, and is oscillating like a pendulum.
See figure below. It is initially pulled back at 90 angle with the vertical, and
is released with a push so that the ball reaches its maximum height back and
forth. If it reaches its maximum height again after 3 seconds, find its height
10 seconds after it was released.
Answer: 27 cm
Hint. The height h(t) (in cm) of the ball at time t (in seconds) is given by
h(t) = |30 sin ⇡3 (t 32 )| + 12.
8. For what values of k do y = cot x and y = cot(x
k⇡) have the same graph?
Answer: any integer
9. For what values of k do y = sec x and y = sec(x
k⇡) have the same graph?
Answer: any even integer
179
10. Find the least positive value of c such that the graph of y =
coincide with that of y = 2 cos 2x.
2 sin 2(x + c)
Answer: ⇡4
11. Find the largest positive value of c such that the graph of y = 2 cos 3(x c)
coincide with that of y = 2 cos 3(x 2).
Answer: 2 + ⇡3
12. For what values of a do the graphs of y = a cos b(x c) and y = 2 sec ⇡6 (x 6)
never intersect for any values of b and c?
Answer: 2 < a < 2
4
Lesson 3.4. Fundamental Trigonometric Identities
Time Frame: 4 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) determine whether an equation is an identity or a conditional equation;
(2) derive the fundamental trigonometric identities;
(3) simplify trigonometric expressions using fundamental trigonometric identities; and
(4) prove other trigonometric identities using fundamental trigonometric identities.
Lesson Outline
(1) Domain of an equation
(2) Identity and conditional equation
(3) Fundamental trigonometric identities
(4) Proving trigonometric identities
Introduction
In previous lessons, we have defined trigonometric functions using the unit
circle and also investigated the graphs of the six trigonometric functions. This
lesson builds on the understanding of the di↵erent trigonometric functions by
discovery, deriving, and working with trigonometric identities.
3.4.1. Domain of an Expression or Equation
Consider the following expressions:
p
2x + 1,
x2 1,
x2
180
x
3x
4
,
p
x
.
x 1
What are the real values of the variable x that make the expressions defined in
the set of real numbers?
In the first expression, every real value of x when substituted to the expression
makes it defined in the set of real numbers; that is, the value of the expression is
real when x is real.
In the second expression, not every real value of x makes
p the expression defined
in R. For example, when x = 0, the expression becomes
1, which is not a real
number.
p
x2 1 2 R () x2 1 0 () x  1 or x 1
p
Here, for x2 1 to be defined in R, x must be in ( 1, 1] [ [1, 1).
In the third expression, the values of x that make the denominator zero make
the entire expression undefined.
x2
3x
4 = (x
4)(x + 1) = 0
x
3x
()
x = 4 or x =
1
is real when x 6= 4 and x 6= 1.
p
In the fourth expression, because the expression x 1 is in the denominator,
x must be greater than 1. Although the value of the entire expression is 0 when
x = 0, we do not include 0 as allowed value of x because part of the expression
is not real when x = 0.
Hence, the expression
x2
4
In the expressions above, the allowed values of the variable x constitute the
domain of the expression.
The domain of an expression (or equation) is the set of all real values of
the variable for which every term (or part) of the expression (equation)
is defined in R.
In the expressions above, the domains of the first, second, third, and fourth
expressions are R, ( 1, 1] [ [1, 1), R \ { 1, 4}, and (1, 1), respectively.
Example 3.4.1. Determine the domain of the expression/equation.
x2 1
(a) 3
x + 2x2 8x
(b) tan ✓
(c) x2
(d) z
sin ✓
p
p
x+1
1 x
cos 2✓
1 + x2 = p
3
2
x2
cos2 z
= 4 sin z
1 + sin z
1
1
181
Solution. (a) x3 + 2x2
4, or x = 2
p
x + 1 2 R ()
1
()
x=0
8x = x(x + 4)(x
x+1
()
0
x=1
2) = 0
x
()
x = 0, x =
1
Domain = [ 1, 1) \ { 4, 0, 1, 2}
= [ 1, 0) [ (0, 1) [ (1, 2) [ (2, 1)
(b) tan ✓
sin ✓
cos ✓ = 0
cos 2✓ =
()
Domain = R \
✓=
{ k⇡
|
2
sin ✓
cos ✓
k⇡
, k
2
sin ✓
cos 2✓
odd integer
k odd integer}
p
(c) The expression 1+x2 is always positive,
and
so
1 + x2 is defined in R. On
p
3
the other hand, the expression x2 1 is also defined in R, but it cannot
be zero because it is in the denominator. Therefore, x should not be 1
and 1.
Domain = R \ { 1, 1}
(d) 1 + sin z = 0
()
z=
3⇡
2
+ 2k⇡, k 2 Z
Domain = R \ { 3⇡
+ 2k⇡|k 2 Z}
2
2
Seatwork/Homework 3.4.1
Find the domain of the expression/equation.
x+1
2
Answer: R \ { 1, 1}
2
+ 2x + 1 x
1
1
1
(2) 2 sec2 ✓ =
+
Answer: R \ (2k + 1) ⇡2 |k 2 Z
1 sin ✓ 1 + sin ✓
2
(3) 2 tan x = 2 cot x + 1
Answer: R \ k⇡
|k 2 Z
2
1
(4) p
tan x = sin x
Answer: R \ { 1, 1} [ (2k + 1) ⇡2 |k 2 Z
2
1 x
(1)
x
x+1
x2
3.4.2. Identity and Conditional Equation
Consider the following two groups of equations:
Group A
(A1) x2
Group B
1=0
(A2) (x + 7)2 = x2 + 49
x2 4
(A3)
= 2x 1
x 2
(B1) x2
1 = (x
1)(x + 1)
(B2) (x + 7)2 = x2 + 14x + 49
x2 4
(B3)
=x+2
x 2
182
In each equation in Group A, some values of the variable that are in the
domain of the equation do not satisfy the equation (that is, do not make the
equation true). On the other hand, in each equation in Group B, every element
in the domain of the equation satisfies the given equation. The equations in
Group A are called conditional equations, while those in Group B are called
identities.
An identity is an equation that is true for all values of the variable
in the domain of the equation. An equation that is not an identity is
called a conditional equation. (In other words, if some values of the
variable in the domain of the equation do not satisfy the equation,
then the equation is a conditional equation.)
Example 3.4.2. Identify whether the given equation is an identity or a conditional equation. For each conditional equation, provide a value of the variable in
the domain that does not satisfy the equation.
(1) x3
p
3
2= x
2
x2 +
p
3
2x +
p
3
4
(2) sin2 ✓ = cos2 ✓ + 1
(3) sin ✓ = cos ✓
p
1
x
1
p =
(4)
1+ x
1
p
2 x+x
1 x
Solution. (1) This is an identity because this is simply factoring of di↵erence of
two cubes.
(2) This is a conditional equation. If ✓ = 0, then the left-hand side of the equation
is 0, while the right-hand side is 2.
(3) This is also a conditional equation. If ✓ = 0, then both sides of the equation
are equal to 0. But if ✓ = ⇡, then the left-hand side of the equation is 0,
while the right-hand side is 2.
(4) This is an identity because the right-hand side of the equation is obtained by
rationalizing the denominator of the left-hand side.
2
Seatwork/Homework 3.4.2
Identify whether the given equation is an identity or a conditional equation. For
each conditional equation, provide a value of the variable in the domain that does
not satisfy the equation.
(1) 1 + x +
x2
1
x
=
1
1
Answer: identity
x
183
cos2 ✓ sin2 ✓
= cos ✓
cos ✓ + sin ✓
(3) tan ✓ = cot ✓
(2)
sin ✓
Answer: identity
Answer: conditional equation, ✓ =
(4) cos2 x = 2 cos x + 3
⇡
2
Answer: conditional equation, x = 0
3.4.3. The Fundamental Trigonometric Identities
Recall that if P (x, y) is the terminal point on the unit circle corresponding to ✓,
then we have
1
y
sin ✓ = y csc ✓ =
tan ✓ =
y
x
1
x
cos ✓ = x sec ✓ =
cot ✓ = .
x
y
From the definitions, the following reciprocal and quotient identities immediately follow. Note that these identities hold if ✓ is taken either as a real number
or as an angle.
Reciprocal Identities
csc ✓ =
1
sin ✓
sec ✓ =
1
cos ✓
cot ✓ =
1
tan ✓
Quotient Identities
tan ✓ =
sin ✓
cos ✓
cot ✓ =
cos ✓
sin ✓
We can use these identities to simplify trigonometric expressions.
Example 3.4.3. Simplify:
tan ✓ cos ✓
(1)
sin ✓
Solution.
(2)
(1)
tan ✓ cos ✓
=
sin ✓
(2)
cos ✓
cot ✓
sin ✓
cos ✓
cos ✓
=1
sin ✓
cos ✓
cos ✓
= cos ✓ = sin ✓
cot ✓
sin ✓
2
If P (x, y) is the terminal point on the unit circle corresponding to ✓, then
x2 + y 2 = 1. Since sin ✓ = y and cos ✓ = x, we get
sin2 ✓ + cos2 ✓ = 1.
184
Teaching Notes By
The assumption in
the division is that
the divisor is
nonzero.
dividing both sides of this identity by cos2 ✓ and sin2 ✓, respectively, we obtain
tan2 ✓ + 1 = sec2 ✓
and 1 + cot2 ✓ = csc2 ✓.
Pythagorean Identities
sin2 ✓ + cos2 ✓ = 1
tan2 ✓ + 1 = sec2 ✓
1 + cot2 ✓ = csc2 ✓
Example 3.4.4. Simplify:
(1) cos2 ✓ + cos2 ✓ tan2 ✓
Solution.
(2)
(2)
1 + tan2 ✓
1 + cot2 ✓
(1) cos2 ✓ + cos2 ✓ tan2 ✓ = (cos2 ✓)(1 + tan2 ✓)
= cos2 ✓ sec2 ✓ = 1
1 + tan2 ✓
sec2 ✓
=
=
1 + cot2 ✓
csc2 ✓
1
cos2 ✓
1
sin2 ✓
=
sin2 ✓
= tan2 ✓
2
cos ✓
2
In addition to the eight identities presented above, we also have the following
identities.
Even-Odd Identities
sin( ✓) =
sin ✓
tan( ✓) =
cos( ✓) = cos ✓
tan ✓
The first two of the negative identities can be obtained from the graphs of
the sine and cosine functions, respectively. (Please review the discussion on page
Teaching Notes 149.) The third identity can be derived as follows:
The corresponding
reciprocal
sin( ✓)
sin ✓
functions follow
tan( ✓) =
=
= tan ✓.
the same
cos( ✓)
cos ✓
Even-Odd
Identities:
The reciprocal, quotient, Pythagorean, and even-odd identities
csc( ✓) = csc ✓
sec( ✓) = sec ✓ what we call the fundamental trigonometric identities.
cot( ✓) = cot ✓.
constitute
We now solve Example 3.2.3 in a di↵erent way.
Example 3.4.5. If sin ✓ =
3
4
and cos ✓ > 0. Find cos ✓.
Solution. Using the identity sin2 ✓ + cos2 ✓ = 1 with cos ✓ > 0, we have
s
✓ ◆2 p
p
3
7
2
cos ✓ = 1 sin ✓ = 1
=
.
4
4
185
2
Example 3.4.6. If sec ✓ = 52 and tan ✓ < 0, use the identities to find the values
of the remaining trigonometric functions of ✓.
Solution. Note that ✓ lies in QIV.
cos ✓ =
sin ✓ =
1
2
=
sec ✓
5
p
cos2 ✓ =
1
1
csc ✓ =
=
sin ✓
sin ✓
tan ✓ =
=
cos ✓
1
cot ✓ =
=
tan ✓
s
1
p
5 21
21
p
21
5
2
5
=
✓ ◆2
2
=
5
p
21
5
p
21
2
p
2 21
21
2
Seatwork/Homework 3.4.3
1. Use the identities presented in this lesson to simplify each trigonometric expression.
(a)
1 + tan x
1 + cot x
Solution.
Answer: tan x
1 + tan x
1 + tan x
=
= tan x
1 + cot x
1 + tan1 x
sin ✓
1 + cos ✓
+
Answer: 2 csc ✓
1 + cos ✓
sin ✓
sin ✓
1 + cos ✓
sin2 ✓
(1 + cos ✓)(1 + cos ✓)
Solution.
+
=
+
1 + cos ✓
sin ✓
sin ✓(1 + cos ✓)
sin ✓(1 + cos ✓)
2
sin ✓ + (1 + 2 cos ✓ + cos2 ✓)
=
sin ✓(1 + cos ✓)
2 + 2 cos ✓
2
=
=
= 2 csc ✓
sin ✓(1 + cos ✓)
sin ✓
tan y + cot y
(c)
Answer: 1
sec y csc y
sin y
y
sin2 y+cos2 y
+ cos
tan y + cot y
cos y
sin y
cos y sin y
Solution.
= 1
=
= sin2 y + cos2 y = 1
1
1
sec y csc y
·
cos y sin y
cos y sin y
(b)
186
(d) 1
cos2 ✓
1 + sin ✓
Solution. 1
Answer: sin ✓
cos2 ✓
1 + sin ✓ cos2 ✓
=
1 + sin ✓
1 + sin ✓
2
sin ✓ + sin ✓
sin ✓(1 + sin ✓)
=
=
= sin ✓
1 + sin ✓
1 + sin ✓
2. Given some initial values, use the identities to find the values of the remaining
trigonometric functions of ✓.
(a) sin ✓ =
2
5
and sec ✓ > 0
Answer: p✓ in QI; cscp✓ = 52 , cos ✓ =
tan ✓ = 2 2121 , cot ✓ = 221
8
3
(b) sec ✓ =
p
1
and tan ✓ > 0
Answer:
✓ in QIII;
cos ✓ =p 83 , sin ✓ =
p
p
4 7
, tan ✓ = 2 3 7 , cot ✓ = 3147
7
(c) tan ✓ = 2 and csc ✓ < 0
p
Answer: ✓pin QIII; cot ✓ = 12 , sec ✓ =
sin ✓ = 4 5 5
(d) csc ✓ =
3
2
and sec ✓ < 0
Answer: ✓ p
in QII; sin ✓ =p 23 , cos ✓ =
tan ✓ = 2 5 5 , cot ✓ = 25
p
sin2 ✓ =
p
1
21
,
5
cos2 ✓ =
5, cos ✓ =
1
p
sin2 ✓ =
p
5
,
5
p
5
,
3
sec ✓ =
p
7
,
4
p
5 21
,
21
csc ✓ =
csc ✓ =
sec ✓ =
p
5
,
4
p
3 5
,
5
3.4.4. Proving Trigonometric Identities
We can use the eleven fundamental trigonometric identities to establish other
identities. For example, suppose we want to establish the identity
csc ✓
cot ✓ =
sin ✓
.
1 + cos ✓
To verify that it is an identity, recall that we need to establish the truth of the
equation for all values of the variable in the domain of the equation. It is not
enough to verify its truth for some selected values of the variable. To prove it, we
use the fundamental trigonometric identities and valid algebraic manipulations
like performing the fundamental operations, factoring, canceling, and multiplying
the numerator and denominator by the same quantity.
Start on the expression on one side of the proposed identity (preferably the
complicated side), use and apply some of the fundamental trigonometric identities
and algebraic manipulations, and arrive at the expression on the other side of the
proposed identity.
187
Expression
Explanation
csc ✓ cot ✓
1
cos ✓
=
sin ✓
sin ✓
Start on one side.
Apply some reciprocal and
quotient identities.
cos ✓
sin ✓
1 cos ✓ 1 + cos ✓
=
·
sin ✓
1 + cos ✓
=
1
1 cos2 ✓
(sin ✓)(1 + cos ✓)
sin2 ✓
=
(sin ✓)(1 + cos ✓)
=
=
Add the quotients.
Multiply the numerator
and denominator by
1 + cos ✓.
Multiply.
Apply a Pythagorean
identity.
sin ✓
1 + cos ✓
Reduce to lowest terms.
Upon arriving at the expression of the other side, the identity has been established. There is no unique technique to prove all identities, but familiarity with
the di↵erent techniques may help.
Example 3.4.7. Prove: sec x
cos x = sin x tan x.
Solution.
sec x
1
cos x
cos x
1 cos2 x
=
cos x
sin2 x
sin x
=
= sin x ·
= sin x tan x
cos x
cos x
cos x =
Example 3.4.8. Prove:
1 + sin ✓
1 sin ✓
2
1 sin ✓
= 4 sin ✓ sec2 ✓
1 + sin ✓
Solution.
1 + sin ✓
1 sin ✓
1 sin ✓
(1 + sin ✓)2 (1 sin ✓)2
=
1 + sin ✓
(1 sin ✓)(1 + sin ✓)
1 + 2 sin ✓ + sin2 ✓ 1 + 2 sin ✓
=
1 sin2 ✓
4 sin ✓
=
= 4 sin ✓ sec2 ✓
cos2 ✓
188
sin2 ✓
2
Seatwork/Homework 3.4.4
Prove each identity.
1. tan x + cot x = csc x sec x
sin x cos x
+
cos x sin x
sin2 x + cos2 x
1
1
1
=
=
=
·
= csc x sec x
sin x cos x
sin x cos x
cos x sin x
Answer: tan x + cot x =
2. sec ✓ + tan ✓ =
1
sec ✓ tan ✓
1
1
Answer:
= 1
sin ✓
sec ✓ tan ✓
cos ✓
cos ✓
cos ✓
=
1 sin ✓
cos ✓
1 + sin ✓
=
·
1 sin ✓ 1 + sin ✓
(cos ✓)(1 + sin ✓)
=
1 sin2 ✓
(cos ✓)(1 + sin ✓)
=
cos2 ✓
1 + sin ✓
1
sin ✓
=
=
+
= sec ✓ + tan ✓
cos ✓
cos ✓ cos ✓
sec y + tan y
3.
= tan y
csc y + 1
sec y + tan y
Answer:
=
csc y + 1
sin y
1
+ cos
cos y
y
1
+
1
sin y
=
1+sin y
cos y
1+sin y
sin y
=
sin y
= tan y
cos y
1
1
+
1 cos ✓ 1 + cos ✓
1
1
1 + cos ✓ + 1 cos ✓
2
Answer:
+
=
=
= 2 csc2 ✓
2
1 cos ✓ 1 + cos ✓
1 cos ✓
sin2 ✓
4. 2 csc2 ✓ =
Exercises 3.4
1. Find the domain of the equation.
p
p
(a) 3 x + 2
x = 2x
Answer: {x|x
0}
Answer: R
(b) sin3 x = sin x + 1
Answer: R \
(c) tan x + cot x = sin x
x+1
(d) 2
+ cos x = csc x
x
1
k⇡
|k
2
2Z
Answer: R \ {{ 1, 1} [ {k⇡|k 2 Z}}
189
2. Simplify each expression using the fundamental identities.
(a)
sin2 ✓
sec2 ✓ 1
Answer: cos2 ✓
sin2 ✓
sin2 ✓
sin2 ✓
=
=
= cos2 ✓
sec2 ✓ 1
tan2 ✓
sin2 ✓
cos2 ✓
1
1
(b)
+
2
1 + tan x 1 + cot2 x
1
1
1
1
Solution.
+
=
+
2
2
2
1 + tan x 1 + cot x
sec x csc2 x
= cos2 x + sin2 x = 1
Solution.
(c) 1
cos2 x
1 + sin x
Answer:
cos2 x
=1
1 + sin x
Solution. 1
=1
=1
(d)
Answer: 1
sin ✓
cos ✓ tan ✓
Solution.
sin x
1 sin2 x
1 + sin x
(1 sin x)(1 + sin x)
1 + sin x
1 sin x = sin x
Answer: 1
sin ✓
1
= tan ✓ ·
=1
cos ✓ tan ✓
tan ✓
3. Given some initial information, use the identities to find the values of the
trigonometric functions of ✓.
(a) csc ✓ =
5
3
and tan ✓ > 0
Answer: ✓ in QI; sin ✓ = 35 , cos ✓ =
cot ✓ = 43
(b) tan ✓ =
12
5
p
1
and cos ✓ < 0
5
Answer: ✓ in QII; cot ✓ = 12
, sec ✓ =
p
12
2
sin ✓ = 1 cos ✓ = 13 , csc ✓ = 13
12
(c) csc ✓ =
3
2
and ⇡ < x <
sin2 ✓ = 45 , sec ✓ = 54 , tan ✓ = 34 ,
p
3⇡
2
Answer:
✓ in QIII;
sin ✓ =p 23 , cos ✓ =
p
p
3 5
, tan ✓ = 2 5 5 , cot ✓ = 25
5
(d) cot ✓ =
7
5
and
3⇡
2
< ✓ < 2⇡
p
1
sin2 ✓ =
p
5
Answer: ✓ in QIV; tan ✓ =
,
sec
✓
=
tan2 ✓ + 1 =
7
p
p
p
74
csc ✓ =
cot2 ✓ + 1 =
, sin ✓ = 5 7474
5
190
13
,
5
tan2 ✓ + 1 =
p
74
,
7
cos ✓ =
p
5
,
3
5
,
13
sec ✓ =
cos ✓ =
p
7 74
,
74
(e) sin ✓ = 1
Answer: ✓ coterminal with
tan ✓ undefined, cot ✓ = 0
3⇡
;
2
csc ✓ =
1, cos ✓ = 0, sec ✓ undefined,
(f) cot ✓ = 1
Answer: ✓ either in QII or QIV
p
p
p
p
✓ in QII: tan ✓ = 1, sin ✓ = 22 , csc ✓ = 2, cos ✓ = 22 , sec ✓ =
2
p
p
p
p
2
2
✓ in QIV: tan ✓ = 1, sin ✓ = 2 , csc ✓ =
2, cos ✓ = 2 , sec ✓ = 2
4. Determine whether the given equation is an identity or a conditional equation.
If it is an identity, prove it; otherwise, provide a value of the variable in the
domain that does not satisfy the equation.
(a) sin x cos x = 1
(b) sin3 x = cos x
(c) (sin x
Answer: conditional equation, x = 0
1
Answer: conditional equation, x =
cos x)2 + (sin x + cos x)2 = 2
⇡
2
Answer: identity
Proof. (sin x cos x)2 + (sin x + cos x)2
= (cos2 x 2 cos x sin x + sin2 x) + (cos2 x + 2 cos x sin x + sin2 x)
=1+1=2
(d) tan( x) cot x =
(e) 2
1
sin2 x = sec x + cos x
Answer: conditional equation, x =
⇡
2
Answer: conditional equation, x =
⇡
2
5. Prove the following identities.
(a) sin3 x = sin x
sin x cos2 x
Solution. sin3 x = sin2 x · sin x = (1
(b) sin4 x
cos4 x = sin2 x
cos x) sin x = sin x
sin x cos2 x
cos2 x
Solution. sin4 x cos4 x = (sin2 x cos2 x)(sin2 x+cos2 x) = sin2 x cos2 x
(c) tan( ✓) sin( ✓) + cos( ✓) = sec( ✓)
Solution. tan( ✓) sin( ✓) + cos( ✓) = tan ✓ sin ✓ + cos ✓
sin2 ✓
+ cos ✓
cos ✓
sin2 ✓ + cos2 ✓
=
cos ✓
1
=
= sec ✓ = sec( ✓)
cos ✓
=
191
(d)
1 + sin u + cos u
1 + cos u
=
1 + sin u cos u
sin u
Solution
1 + sin u + cos u
1 + sin u + cos u 1 + cos u
=
·
1 + sin u cos u
sin u + 1 cos u 1 + cos u
(1 + sin u + cos u)(1 + cos u)
=
sin u + sin u cos u + 1 cos2 u
(1 + sin u + cos u)(1 + cos u)
=
sin u + sin u cos u + sin2 u
(1 + sin u + cos u)(1 + cos u)
=
(sin u)(1 + cos u + sin u)
1 + cos u
=
sin u
sec2 x
in terms of sin x.
sec2 x
1 sec2 x
1
Solution.
=
1 = cos2 x
2
sec x
sec2 x
6. Express
1
7. Express tan x sec x in terms of cos x.
Solution. tan x sec x =
sin x
cos x
·
1
cos x
=
sin x
cos2 x
Teaching Notes
Since you need
1 + cos u to retain
in the numerator
at the end, do not
expand the
numerator.
Answer:
1=
sin2 x
sin2 x
p
± 1 cos2 x
Answer:
cos2 x
p
± 1 cos2 x
=
cos2 x
8. Express all other five trigonometric functions in terms of tan x (allowing ± in
the expression).
tan x
1
1
Answer: sin x = p 2
; cos x = p
; cot x =
; sec x =
2
tan x
± tan x + 1
± tan p
x+1
p
± tan2 x + 1
± tan2 x + 1; csc x =
tan x
Answer:
1
3
(sec ✓ tan ✓)(sec ✓+tan ✓) = 1 =) 3(sec ✓+tan ✓) = 1 =) sec ✓+tan ✓ =
1
3
9. If sec ✓
tan ✓ = 3, what is sec ✓ + tan ✓?
Solution. tan2 ✓ + 1 = sec2 ✓ =) sec2 ✓
tan2 ✓ = 1
4
192
Lesson 3.5. Sum and Di↵erence Identities
Time Frame: 3 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) derive trigonometric identities involving sum and di↵erence of two angles;
(2) simplify trigonometric expressions using fundamental trigonometric identities
and sum and di↵erence identities;
(3) prove other trigonometric identities using fundamental trigonometric identities and sum and di↵erence identities; and
(4) solve situational problems involving trigonometric identities.
Lesson Outline
(1) The sum and di↵erence identities for cosine, sine, and tangent functions
(2) Cofunction identities
(3) More trigonometric identities
Introduction
In previous lesson, we introduced the concept of trigonometric identity, presented the fundamental identities, and proved some identities. In this lesson, we
derive the sum and di↵erence identities for cosine, sine, and tangent functions,
establish the cofunction identities, and prove more trigonometric identities.
3.5.1. The Cosine Di↵erence and Sum Identities
Let u and v be any real numbers with 0 < v  u < 2⇡. Consider the unit circle
with points A = (1, 0), P1 , P2 , P3 , and u and v with corresponding angles shown
in Figure 3.27. Then P1 P2 = AP3 .
Recall that P1 = P (u) = (cos u, sin u), P2 = P (v) = (cos v, sin v), and P3 =
P (u v) = (cos(u v), sin(u v)), so that
p
P1 P2 = (cos u cos v)2 + (sin u sin v)2 ,
while
AP3 =
p
[cos(u
v)
1]2 + [sin(u
v)
0]2 .
Equating these two expressions and expanding the squares, we get
(cos u
cos v)2 + (sin u
sin v)2 = [cos(u
193
v)
1]2 + sin2 (u
v)
Figure 3.27
cos2 u
2 cos u cos v + cos2 v + sin2 u 2 sin u sin v + sin2 v
= cos2 (u v) 2 cos(u v) + 1 + sin2 (u v)
Applying the Pythagorean identity cos2 ✓+sin2 ✓ = 1 and simplifying the resulting
equations, we obtain
(cos2 u + sin2 u) + (cos2 v + sin2 v) 2 cos u cos v 2 sin u sin v
= [cos2 (u v) + sin2 (u v)] 2 cos(u v) + 1
1+1
2 cos u cos v
cos(u
2 sin u sin v = 1
2 cos(u
v) + 1
v) = cos u cos v + sin u sin v.
We have thus proved another identity.
Although we assumed at the start that 0 < v  u < 2⇡, but because
cos( ✓) = cos ✓ (one of the even-odd identities), this new identity is true for
any real numbers u and v. As before, the variables can take any real values or
angle measures.
Cosine Di↵erence Identity
cos(A
Replacing B with
get another identity.
B) = cos A cos B + sin A sin B
B, and applying the even-odd identities, we immediately
194
Cosine Sum Identity
cos(A + B) = cos A cos B
sin A sin B
⇡
Example 3.5.1. Find the exact values of cos 105 and cos 12
.
Solution.
cos 105 = cos(60 + 45 )
= cos 60 cos 45
sin 60 sin 45
p
p p
1
2
3
2
= ·
·
2p 2 p 2
2
2
6
=
4
cos
⇣⇡ ⇡ ⌘
⇡
= cos
12
4
6
⇡
⇡
⇡
⇡
= cos cos + sin sin
4
6
p 4p 6 p
2
3
2 1
=
·
+
·
2
2 2
p2
p
6+ 2
=
4
Example 3.5.2. Given cos ↵ =
QI, find cos(↵ + ).
3
5
and sin
=
12
,
13
Solution. We will be needing sin ↵ and cos .
s
p
sin ↵ =
1 cos2 ↵ =
1
cos
=
q
1
sin
2
=
s
1
where ↵ lies in QIV and
✓ ◆2
3
=
5
✓
12
13
◆2
=
cos(↵ + ) = cos ↵ cos
sin ↵ sin
✓ ◆
3 5
4 12
= ·
5 13
5 13
63
=
65
195
2
in
4
5
5
13
2
Seatwork/Homework 3.5.1
7⇡
1. Find the exact value of cos .
Answer:
12
⇣⇡ ⇡ ⌘
7⇡
⇡
⇡
⇡
⇡
Solution. cos
= cos
+
= cos cos
sin sin
12
4
3
3p p 4 p3
p 4
p
2 1
2
3
2
6
=
·
·
=
2 2
2
2
4
2. Express
cos(5x) cos(2x) + sin(5x) sin(2x)
as a single cosine expression.
p
p
2
6
4
Answer: cos(3x)
Solution. cos(5x) cos(2x) + sin(5x) sin(2x) = cos(5x 2x) = cos(3x)
1
1
3. Given cos ↵ =
and cos =
, where ↵ lies in QI and in QIII, find
3
4
p
1 2 30
cos(↵
).
Answer:
12
p
1
2 2
Solution. cos ↵ = and ↵ in QI =) sin ↵ =
3
3
p
1
15
cos =
and ↵ in QIII =) sin =
4
4
cos (↵
) = cos ↵ cos + sin ↵ sin
p
p !
p
✓ ◆
1
1
2 2
15
1 2 30
=
+
=
3
4
3
4
12
3.5.2. The Cofunction Identities and the Sine Sum and Di↵erence
Identities
In the Cosine Di↵erence Identity, if we let A = ⇡2 , we get
⇣⇡
⌘
⇣⇡ ⌘
⇣⇡ ⌘
cos
B = cos
cos B + sin
sin B
2
2
2
= (0) cos B + (1) sin B
= sin B.
From this identity, if we replace B with ⇡2 B, we have
h⇡ ⇣⇡
⌘i
⇣⇡
⌘
cos
B = sin
B
2
2
2
⇣⇡
⌘
cos B = sin
B .
2
196
As for the tangent function, we have
tan
⇣⇡
2
⌘ sin ⇡
2
B =
cos ⇡2
cos B
=
sin B
= cot B.
B
B
We have just derived another set of identities.
Cofunction Identities
Teaching Notes
The Cofunction
Identities for the
reciprocal
functions will
follow:
csc ⇡2
B =
sec B
sec ⇡2
B =
csc B
cot ⇡2
B =
tan B.
cos
⇣⇡
2
⌘
⇣⇡
⌘
B = sin B
sin
B = cos B
2
⇣⇡
⌘
tan
B = cot B
2
Using the first two cofunction identities, we now derive the identity for sin(A+
B).
i
(A + B)
h⇣2 ⇡
⌘
i
= cos
A
B)
⇣ ⇡2
⌘
⇣⇡
= cos
A cos B + sin
2
2
= sin A cos B + cos A sin B
sin(A + B) = cos
h⇡
⌘
A sin B
Sine Sum Identity
sin(A + B) = sin A cos B + cos A sin B
In the last identity, replacing B with
yield
sin(A
B and applying the even-odd identities
B) = sin[A + ( B)]
= sin A cos( B) + cos A sin( B)
= sin A cos B cos A sin B.
197
Sine Di↵erence Identity
sin(A
B) = sin A cos B
cos A sin B
5⇡
12
Example 3.5.3. Find the exact value of sin
.
Solution.
sin
✓
5⇡
12
◆
⇡⌘
+
4
6
⇣⇡ ⌘
⇣⇡ ⌘
⇣⇡ ⌘
⇣⇡ ⌘
= sin
cos
+ cos
sin
4
6
p 4p
p6
2
3
2 1
=
·
+
·
2
2
2
2
p
p
6+ 2
=
4
= sin
⇣⇡
3
Example 3.5.4. If sin ↵ = 13
and sin
find sin(↵ + ) and sin(
↵).
= 12 , where 0 < ↵ <
Solution. We first compute cos ↵ and cos .
s
p
cos ↵ = 1 sin2 ↵ = 1
cos
=
q
1
sin2
=
s
1
✓
3
13
◆2
⇡
2
2
and
⇡
2
<
< ⇡,
p
4 10
=
13
✓ ◆2
1
=
2
p
3
2
sin(↵ + ) = sin ↵ cos + cos ↵ sin
p !
p
3
3
4 10 1
=
+
·
13
2
13
2
p
p
4 10 3 3
=
26
sin(
↵) = sin cos ↵ cos sin ↵
p
p !
1 4 10
3 3
= ·
2
13
2
13
p
p
4 10 + 3 3
=
26
198
2
Example 3.5.5. Prove:
sin(x + y) = (1 + cot x tan y) sin x cos y.
Solution.
(1 + cot x tan y) sin x cos y
= sin x cos y + cot x tan y sin x cos y
cos x sin y
= sin x cos y +
sin x cos y
sin x cos y
= sin x cos y + cos x sin y
= sin(x + y)
2
Seatwork/Homework 3.5.2
⇣⇡⌘
1. Find the exact value of sin
.
Answer:
12
⇣⇡ ⇡ ⌘
⇡
⇡
⇡
⇡
⇡
Solution. sin
= sin
= sin cos
cos sin
12
4
6
4
6
p 4p 6 p
2
3
2 1
=
·
·
2
2 2
p2
p
6
2
=
4
2. Find the exact value of sin 20 cos 80
Solution. sin 20 cos 80
sin 80 cos 20 .
sin 80 cos 20 = sin(20
3. Prove:
sin 60 =
sin(x + y)
tan x + tan y
=
.
sin(x y)
tan x tan y
Solution.
sin(x + y)
sin x cos y + cos x sin y
=
sin(x y)
sin x cos y cos x sin y
1
sin x cos y + cos x sin y cos x cos y
=
·
1
sin x cos y cos x sin y
cos x cos y
sin x cos y
cos x sin y
+
tan x + tan y
cos x cos y cos x cos y
=
=
sin x cos y
cos x sin y
tan x tan y
cos x cos y cos x cos y
199
p
6
2
4
Answer:
80 )
= sin( 60 ) =
p
p
3
2
p
3
2
3.5.3. The Tangent Sum and Di↵erence Identities
Recall that tan x is the ratio of sin x over cos x. When we replace x with A + B,
we obtain
sin(A + B)
tan(A + B) =
.
cos(A + B)
Using the sum identities for sine and cosine, and then dividing the numerator
and denominator by cos A cos B, we have
tan(A + B) =
=
=
sin A cos B + cos A sin B
cos A cos B sin A sin B
sin A cos B
cos A sin B
+ cos
cos A cos B
A cos B
cos A cos B
cos A cos B
sin A sin B
cos A cos B
tan A + tan B
.
1 tan A tan B
We have just established the tangent sum identity.
In the above identity, if we replace B with
tan( ✓) = tan ✓, we get
tan(A
B and use the even-odd identity
B) = tan[A + ( B)]
tan A + tan( B)
=
1 tan A tan( B)
tan A tan B
=
.
1 + tan A tan B
This is the tangent di↵erence identity.
Tangent Sum and Di↵erence Identities
tan A + tan B
1 tan A tan B
tan A tan B
B) =
1 + tan A tan B
tan(A + B) =
tan(A
Seatwork/Homework 3.5.3
1. Find the exact values of tan
5⇡
12
, tan
⇡
12
, and tan
7⇡
12
Answer: 2 +
200
.
p
3, 2
p
3, 2 +
p
3
p
⇣⇡ ⇡ ⌘
p
tan ⇡4 + tan ⇡6
1 + 33
5⇡
p = 2+
Solution. tan
= tan
+
=
=
3
3
12
4
6
1 tan ⇡4 tan ⇡6
1
3
p
3
⇣⇡ ⇡ ⌘
p
tan ⇡4 tan ⇡6
1
⇡
3
p
tan
= tan
=
=2
3
⇡
⇡ =
12
4
6
1 + tan 4 tan 6
1 + 33
✓
◆
p
7⇡
5⇡
tan
= tan
=2+ 3
12
12
2. Express tan
⇡
4
+ ✓ and tan(2⇡ ✓) in terms of tan ✓.
⇣⇡
⌘
tan ⇡4 + tan ✓
1 + tan ✓
Solution. tan
+✓ =
=
⇡
4
1 tan 4 tan ✓
1 tan ✓
tan 2⇡ tan ✓
tan(2⇡ ✓) =
= tan ✓
1 + tan 2⇡ tan ✓
cot A cot B 1
3. Prove: cot(A + B) =
.
cot A + cot B
1
1 tan A tan B
Solution. cot(A + B) =
=
tan(A + B)
tan A + tan B
1 tan A tan B cot A cot B
=
·
tan A + tan B cot A cot B
cot A cot B 1
=
cot A + cot B
Exercises 3.5
1. Find the exact value.
(a) cos 255
(b) tan
p
Answer:
⇡
12
2
(c) sin 735
Answer: 2
p
6
Answer:
(d) cot 285
Answer:
(e) cos
(f)
⇡
9
cos
2⇡
9
sin
⇡
9
sin
p
6
4 p
p
3
2
4 p
2+ 3
1
Answer:
2
2⇡
9
tan 20 + tan 25
1 tan 20 tan 25
Answer: 1
2. Given some information about a and b, find sin(a+b), cos(a b), and tan(a+b).
(a) sin a = 35 , cos b =
quadrant
12
,
13
a lies in the third quadrant, and b in the first
56
; cos(a
65
4
5
, sin b = 13
5
Answer: sin(a + b) =
Solution. cos a =
201
b) =
63
56
; tan(a + b) =
65
33
(b) cos a = 12 , tan b =
3
,
2
0 < a < ⇡2 , and ⇡2 < b < ⇡
p
p
p
p
3 13 2 39
3 39 2 13
Answer: sin(a + b) =
; cos(a b) =
; tan(a +
26
26
p
24 13 3
b) =
23
p
p
p
Solution. sin a = 23 , cos b = 2 1313 , sin b = 3 1313
cot b = 35 , a in QII, and b in QIII
p
p
p
p
10 34 3 714
6 34 5 714
Answer: sin(a + b) =
; cos(a b) =
;
170
170
p
375 + 68 28
tan(a + b) =
489
p
p
p
Solution. cos a = 25 , sin a = 521 , cos b = 3 3434 , sin b = 5 3434
(c) sec a =
5
,
2
3. Simplify the following expressions.
(a) cos(⇡
x)
Answer:
(b) tan(x + ⇡)
(c) sin
3⇡
2
(d) cos(x
cos x
Answer: tan x
+x
Answer:
cos x
⇡)
Answer:
cos x
4. Prove each identity.
(a) sin(x y) sin(x + y) = sin2 x sin2 y
Solution. sin(x y) sin(x + y)
= (sin x cos y cos x sin y)(sin x cos y + cos x sin y)
= sin2 x cos2 y cos2 x sin2 y
= (sin2 x)(1 sin2 y) (1 sin2 x) sin2 y
= sin2 x sin2 x sin2 y (sin2 y sin2 x sin2 y)
= sin2 x sin2 y
(b) cos(x
y) = (cot x + tan y) sin x cos y
◆
cos x sin y
Solution. (cot x + tan y) sin x cos y =
+
sin x cos y
sin x cos y
= cos x cos y + sin x sin y = cos(x
csc x csc y
(c) sec(x + y) =
cot x cot y 1
Solution.
sec(x + y) =
✓
1
1
=
cos(x + y)
cos x cos y sin x sin y
=
1
cos x cos y
202
sin x sin y
·
1
sin x sin y
1
sin x sin y
y)
=
=
(d)
1
sin x sin y
cos x cos y sin x sin y
sin x sin y
csc x csc y
cot x cot y 1
cos(x + y)
1 tan x tan y
=
cos(x y)
1 + tan x tan y
cos(x + y)
cos x cos y sin x sin y
Solution.
=
cos(x y)
cos x cos y + sin x sin y
cos x cos y sin x sin y
=
·
cos x cos y + sin x sin y
=
=
1
cos x cos y
1
cos x cos y
cos x cos y sin x sin y
cos x cos y
cos x cos y+sin x sin y
cos x cos y
1 tan x tan y
1 + tan x tan y
5. Let n be an integer. Prove that cos(n⇡ + ✓) = ( 1)n cos ✓ and sin(n⇡ + ✓) =
( 1)n sin ✓.
Solution. cos(n⇡) = ( 1)n and sin(n⇡) = 0 for any integer n.
cos(n⇡ + ✓) = cos(n⇡) cos ✓
sin(n⇡) sin ✓ = ( 1)n cos ✓
sin(n⇡ + ✓) = sin(n⇡) cos ✓ + cos(n⇡) sin ✓ = ( 1)n sin ✓
6. In an alternating current circuit, the instantaneous power P (t) at time t is
given by
P (t) = Im Vm cos ' sin2 (!t)
Im Vm sin ' sin(!t) cos(!t),
where Im and Vm are the maximum current (in amperes) and voltage (in volts),
respectively. Express this function as a product of two sine functions.
Solution. P (t) = Im Vm sin(!t)[cos ' sin(!t)
= Im Vm sin(!t) sin(' !t)
sin ' cos(!t)]
? 7. The force F (in pounds) on the back of a person when he or she bends over
sin(✓+90)
at an acute angle ✓ (in degrees) is given by F = 0.6Wsin
, where W is the
12
weight (in pounds) of the person.
(a) Simplify the formula for F .
(b) Find the force on the back of a person whose weight is 154.32 lbs if he
bends an angle of 40 .
(c) How many pounds should a person weigh for his back to endure a force
of 275 lbs if he bends 38 ?
203
Solution
0.6W sin(✓ + 90
0.6W cos ✓
=
sin 12
sin 12
0.6(154.32) cos 40
(b) F =
⇡ 340.46 lbs
sin 12
F sin 12
275 sin 12
(c) W =
=
⇡ 121.17 lbs
0.6 cos ✓
0.6 cos 38
(a) F (✓) =
8. (a) Prove: sin x + sin y = 2 sin
x+y
2
cos
x y
2
.
(b) A particle is moving according to the equation of motion
⇣
⇣
⇡⌘
⇡⌘
s(t) = sin 4t +
+ sin 4t +
,
3
6
where s(t) centimeters is the directed distance of the particle from the
origin at t seconds.
(i) Express s(t) in the form s(t) = a sin(bt + c).
(ii) Find the amplitude and frequency of the motion. (Here, frequency is
defined as the reciprocal of the period.)
Solution
(a) Adding the identities
sin(A + B) = sin A cos B + cos A sin B
sin(A B) = sin A cos B cos A sin B,
we get
sin(A + B) + sin(A
Let A =
have
x+y
2
and B =
x y
.
2
B) = 2 sin A cos B.
Then A + B = x and A B = y. Thus, we
✓
◆
✓
◆
x+y
x y
sin x + sin y = 2 sin
cos
.
2
2
⇣
⇣
⇡⌘
⇡⌘
(b) (i) s(t) = sin 4t +
+ sin 4t +
6 ✓
✓ 3⇡
◆
⇡◆
4t + 3 + 4t + 6
4t + ⇡3 4t ⇡6
= 2 sin
cos
2
2
⇣
⇣⇡⌘
⇡⌘
= 2 sin 4t +
cos
12
p
p 4⇣
2+ 6
⇡⌘
=
sin 4t +
4
p 2p
p
p
⇣
⇣
2+ 6
⇡⌘
2+ 6
⇡⌘
(ii) s(t) =
sin 4t +
=
sin 4 t +
2 p
4
2
16
p
2+ 6
4
2
Amplitude =
; frequency =
=
2
2⇡
⇡
204
9. The dual tone multi-frequency is the signal information used in touch-tone
phones to identify which digit you touched on the keypad. It works by adding
a pair of sounds, one with a lower frequency and one with a higher frequency.
Refer to the chart below. For example, the sound created by touching 6 is
produced by adding a 770-hertz sound to a 1477-hertz sound. (Note that
“hertz” is a unit of frequency and is equal to 1 cycle per second.) This sound
is modeled by the equation
s(t) = sin(2⇡ · 770t) + sin(2⇡ · 1477t),
where t is time in seconds.
http://cnx.org/contents/XGjYtByD@4/Lab-6-Analog-to-Digital-Conver
(a) Write the equation of the sound created by touching the * (asterisk) key
as a product of sine and cosine functions.
(b) In (a), what is the maximum value of s(t)?
Solution
(a) s(t) = sin(2⇡ · 941t) + sin(2⇡ · 1209t) = 2 sin(2150⇡t) cos(536⇡t)
(b) Max value = 2 occurring at t = 0.75 + k, k nonnegative integer
10. (a) Prove: cos x + cos y = 2 cos
x+y
2
cos
x y
2
.
(b) Two atmospheric waves in space produce pressures of F (t) and G(t) pascals at t seconds, where
✓
◆
3⇡
F (t) = 0.04 cos(2⇡t) and G(t) = 0.04 cos 2⇡t
.
4
Express the total pressure P (t) = F (t) + G(t) in the form
P (t) = a cos(bt + c).
205
Solution
(a) Adding the identities
cos(A + B) = cos A cos B sin A sin B
cos(A B) = cos A cos B + sin A sin B,
we get
cos(A + B) + cos(A
Let A =
have
x+y
2
and B =
B) = 2 cos A cos B.
x y
.
2
Then A + B = x and A B = y. Thus, we
✓
◆
✓
◆
x+y
x y
cos x + cos y = 2 cos
cos
.
2
2
(b) P (t) = F (t) + G(t)
✓
◆
3⇡
= 0.04 cos(2⇡t) + 0.04 cos 2⇡t
4
✓
✓
3⇡ ◆
2⇡t + 2⇡t
2⇡t
4
= 0.04 · 2 cos
cos
2
✓
◆
✓ ◆
3⇡
3⇡
= 0.08 cos 2⇡t
cos
8
8
✓
◆
q
p
3⇡
= 0.04 2
2 cos 2⇡t
8
2⇡t +
2
3⇡
4
◆
11. (a) In the figure, two intersecting lines have equations y = m1 x + b1 and
y = m2 x + b2 , respectively. Let ✓ be the acute angle between them, as
shown. Prove that
m2 m1
tan ✓ =
.
1 + m1 m2
(b) Two non-vertical lines intersect at the point ( 3, 2), and one angle between them measures 30 . If one line is 2y = x + 7, find the equation of
the other line.
206
Solution
(a) Let ↵ and be the angles between each line and the (positive side) x-axis,
as shown in the following diagram:
Then m1 = tan ↵ and m2 = tan , so that
tan ✓ = tan(
↵) =
tan
tan ↵
m2 m1
=
.
1 + tan ↵ tan
1 + m1 m2
(b) Solve for m2 in the equation
tan 30 =
m2 12
,
1 + 12 m2
and then use the point-slope form of the equation of the line to get
p
5 3+8
y=
(x + 3) + 2.
11
12. The length s(✓) of the shadow cast by a vertical pole when the angle of the
sun with the horizontal is given by
s(✓) =
h sin(90
sin ✓
where h is the height of the pole.
207
✓)
,
(a) Express s(✓) as a single trigonometric expression.
(b) At what angle ✓ will give the shortest shadow of the pole? Longest
shadow?
Answer: (a) s(✓) = h cot ✓; (b) Shortest shadow occurs at ✓ = 90 . But the
length of the shadow increases when the value of ✓ approaches 0 ; that is, no
maximum length for the shadow.
13. In 4ABC, prove that
tan A + tan B + tan C = tan A tan B tan C.
Solution
A + B + C = 180
=) tan(A + B + C) = tan 180 = 0
tan A + tan(B + C)
= 0 =) tan A + tan(B + C) = 0
1 tan A tan(B + C)
tan B + tan C
tan A +
=0
1 tan B tan C
tan A tan A tan B tan C + tan B + tan C
=0
1 tan B tan C
=) tan A tan A tan B tan C + tan B + tan C = 0
tan A + tan B + tan C = tan A tan B tan C
4
Lesson 3.6. Double-Angle and Half-Angle Identities
Time Frame: 2 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) derive the double-angle and half-angle identities;
(2) simplify trigonometric expressions using known identities;
(3) prove other trigonometric identities using known identities; and
(4) solve situational problems involving trigonometric identities.
208
Lesson Outline
(1) The double-angle and half-angle identities for cosine, sine, and tangent
(2) More trigonometric identities
Introduction
Trigonometric identities simplify the computations of trigonometric expressions. In this lesson, we continue on establishing more trigonometric identities.
In particular, we derive the formulas for f (2✓) and f 12 ✓ , where f is the sine,
cosine, or tangent function.
3.6.1. Double-Angle Identities
Recall the sum identities for sine and cosine.
sin(A + B) = sin A cos B + cos A sin B
cos(A + B) = cos A cos B
sin A sin B
When A = B, these identities becomes
sin 2A = sin A cos A + cos A sin A = 2 sin A cos A
and
cos 2A = cos A cos A
sin A sin A = cos2 A
sin2 A.
Double-Angle Identities for Sine and Cosine
sin 2A = 2 sin A cos A
cos 2A = cos2 A
sin2 A
The double-identity for cosine has other forms. We use the Pythagorean
identity sin2 ✓ + cos2 ✓ = 1.
cos 2A = cos2 A sin2 A
= cos2 A (1 cos2 A)
= 2 cos2 A 1
cos 2A = cos2 A sin2 A
= (1 sin2 A) sin2 A
= 1 2 sin2 A
209
Other Double-Angle Identities for Cosine
cos 2A = 2 cos2 A
Example 3.6.1. Given sin t =
3
5
1
and
cos 2A = 1
⇡
2
2 sin2 A
< t < ⇡, find sin 2t and cos 2t.
Solution. We first find cos t using the Pythagorean identity. Since t lies in QII,
we have
s
✓ ◆2
p
3
4
2
cos t =
1 sin t =
1
=
.
5
5
cos 2t = 1
sin 2t = 2 sin t cos t
✓ ◆✓ ◆
3
4
=2
5
5
24
=
25
=1
=
7
25
2 sin2 t
✓ ◆2
3
2
5
2
In the last example, we may compute cos 2t using one of the other two doubleangle identities for cosine. For the sake of answering the curious minds, we include
the computations here.
cos 2t = cos2 t sin2 t
✓ ◆ 2 ✓ ◆2
4
3
=
5
5
7
=
25
cos 2t = 2 cos2 t 1
✓ ◆2
4
=2
1
5
7
=
25
In the three cosine double-angle identities, which formula to use depends on
the convenience, what is given, and what is asked.
Example 3.6.2. Derive an identity for sin 3x in terms of sin x.
Solution. We use the sum identity for sine, the double-angle identities for sine
and cosine, and the Pythagorean identity.
sin 3x = sin(2x + x)
= sin 2x cos x + cos 2x sin x
= (2 sin x cos x) cos x + (1 2 sin2 x) sin x
= 2 sin x cos2 x + sin x 2 sin3 x
= 2(sin x)(1 sin2 x) + sin x 2 sin3 x
= 3 sin x 4 sin3 x
210
2
For the double-angle formula for tangent, we recall the tangent sum identity:
tan(A + B) =
tan A + tan B
.
1 tan A tan B
When A = B, we obtain
tan(A + A) =
tan A + tan A
2 tan A
=
.
1 tan A tan A
1 tan2 A
Tangent Double-Angle Identity
tan 2A =
Example 3.6.3. If tan ✓ =
1
3
2 tan A
1 tan2 A
and sec ✓ > 0, find sin 2✓, cos 2✓, and tan 2✓.
Solution. We can compute immediately tan 2✓.
2
2 tan ✓
tan 2✓ =
=
2
1 tan ✓
1
1
3
1 2
3
=
3
4
From the given information, we deduce that ✓ lies in QIV. Using one Pythagorean
identity, we compute cos ✓ through sec ✓. (We may also use the technique discussed in Lesson 3.2 by solving for x, y, and r.) Then we proceed to find cos 2✓.
s
✓ ◆2 p
p
1
10
2
sec ✓ = 1 + tan ✓ = 1 +
=
3
3
p
1
1
3 10
cos ✓ =
= p =
10
sec ✓
10
3
p !2
3
10
4
cos 2✓ = 2 cos2 ✓ 1 = 2
1=
10
5
tan 2✓ =
sin 2✓
=) sin 2✓ = tan 2✓ cos 2✓ =
cos 2✓
3
5
2
Seatwork/Homework 3.6.1
1. If cos ✓ =
2
3
and
3⇡
2
< ✓ < 2⇡, find sin 2✓, cos 2✓, and tan 2✓.
p
p
4 5
1
Answer: sin 2✓ =
, cos 2✓ =
, tan 2✓ = 4 5
9
9
211
2. Express tan 3✓ in terms of tan ✓.
Answer: tan 3✓ =
3 tan ✓ tan3 ✓
1 3 tan2 ✓
2 tan ✓
= sin 2✓.
1 + tan2 ✓
2 tan ✓
2 tan ✓
= 2 sin ✓ cos ✓ = sin 2✓
Solution.
=
1 + tan2 ✓
sec2 ✓
3. Prove:
3.6.2. Half-Angle Identities
Recall two of the three double-angle identities for cosine:
cos 2A = 2 cos2 A
1 and
cos 2A = 1
2 sin2 A.
From these identities, we obtain two useful identities expressing sin2 A and cos2 A
in terms of cos 2A as follows:
cos2 A =
1 + cos 2A
2
sin2 A =
and
1
cos 2A
.
2
Some Useful Identities
cos2 A =
1 + cos 2A
2
sin2 A =
From these identities, replacing A with
cos2
1 + cos 2
A
=
2
2
A
,
2
A
2
1
cos 2A
2
we get
=
1 + cos A
2
and
1 cos 2 A2
A
1 cos A
=
=
.
2
2
2
These are the half-angle identities for sine and cosine.
sin2
Half-Angle Identities for Sine and Cosine
✓ ◆
✓ ◆
A
1 + cos A
A
1 cos A
2
2
cos
=
sin
=
2
2
2
2
Because of the “square” in the formulas, we get
r
r
A
1 + cos A
A
1
cos = ±
and sin = ±
2
2
2
cos A
.
2
The appropriate signs of cos A2 and sin A2 depend on which quadrant
212
A
2
lies.
Example 3.6.4. Find the exact values of sin 22.5 and cos 22.5 .
Solution. Clearly, 22.5 lies in QI (and so sin 22.5 and cos 22.5 are both positive), and 22.5 is the half-angle of 45 .
s
p
p
r
p
2
1
1 cos 45
2
2
2
sin 22.5 =
=
=
2
2
2
s
p
p
r
p
1 + 22
1 + cos 45
2+ 2
cos 22.5 =
=
=
2
2
2
2
✓ ◆
✓
tan ✓ + sin ✓
2
Example 3.6.5. Prove: cos
=
.
2
2 tan ✓
Solution.
✓ ◆
✓
1 + cos ✓
cos
=
2
2
✓
◆
1 + cos ✓ tan ✓
=
2
tan ✓
tan ✓ + cos ✓ tan ✓
=
2 tan ✓
sin ✓
tan ✓ + cos ✓ · cos
✓
=
2 tan ✓
tan ✓ + sin ✓
=
2 tan ✓
2
2
We now derive the first version of the half-angle formula for tangent.
tan
sin A2
A
=
2
cos A2
sin A2
=
cos A2
2 sin A2
2 sin A2
!
2 sin2 A2
=
2 sin A2 cos A2
A
2 · 1 cos
2
=
sin 2 · A2
1 cos A
=
sin A
There is another version of the tangent half-angle formula, and we can derive
it from the first version.
A
1 cos A
tan =
2
sin A
213
=
=
=
=
✓
◆
cos A 1 + cos A
sin A
1 + cos A
2
1 cos A
(sin A)(1 + cos A)
sin2 A
(sin A)(1 + cos A)
sin A
1 + cos A
1
Tangent Half-Angle Identities
tan
A
1 cos A
=
2
sin A
sin A2
A
tan =
2
cos A2
A
sin A
=
2
1 + cos A
✓ ◆
A
1 cos A
tan2
=
2
1 + cos A
tan
⇡
.
Example 3.6.6. Find the exact value of tan 12
Solution.
1 cos ⇡6
1
⇡
tan
=
=
⇡
12
sin 6
Example 3.6.7. If sin ✓ =
tan 2✓ .
2
,
5
p
3
2
1
2
=2
p
3
2
cot ✓ > 0, and 0  ✓ < 2⇡, find sin 2✓ , cos 2✓ , and
Solution. Since sin ✓ < 0 and cot ✓ > 0, we conclude the ⇡ < ✓ < 3⇡
. It follows
2
that
⇡
✓
3⇡
< <
,
2
2
4
which means that 2✓ lies in QII.
s
p
✓ ◆2
p
2
21
cos ✓ =
1 sin2 ✓ =
1
=
5
5
v
⇣ p ⌘
u
p
r
p
u1
21
5
✓
1 cos ✓ t
50 + 10 21
sin =
=
=
2
2
2
10
v
⇣ p ⌘
u
p
r
p
u1 +
21
t
5
✓
1 + cos ✓
50 10 21
cos =
=
=
2
2
2
10
⇣ p ⌘
p
21
1
5
✓
1 cos ✓
5 + 21
tan =
=
=
2
2
2
sin ✓
2
5
214
Seatwork/Homework 3.6.2
1. Find the exact value of tan ⇡8 .
2. If cos ✓ =
3
5
and
3⇡
2
Answer:
< ✓ < 2⇡, find sin 2✓ , cos 2✓ , and tan 2✓ .
Answer: sin 2✓ =
p
5
,
5
p
2 5
,
5
cos 2✓ =
p
2
1
tan 2✓ =
1
2
✓ ◆
A
2 2 cos A
3. Prove: sec
=
.
2
sin2 A
A
1
2
2(1 cos A)
2 2 cos A
Solution. sec2 =
=
=
=
A
2
2
2
1 + cos A
1 cos A
sin2 A
cos 2
2
Exercises 3.6
1. Given some information about ✓, find sin 2✓, cos 2✓, and tan 2✓.
1
4
(a) cos ✓ =
and
⇡
2
<✓<⇡
Answer: sin 2✓ =
(b) sec ✓ =
5
2
and sin ✓ > 0
Answer: sin 2✓ =
(c) tan ✓ =
(d) sin ✓ =
2 and
3
5
3⇡
2
p
p
4 21
,
25
15
,
8
cos 2✓ =
< ✓ < 2⇡
Answer: sin 2✓ =
7
,
8
cos 2✓ =
4
,
5
17
,
25
tan 2✓ =
tan 2✓ =
3
,
5
cos 2✓ =
p
15
7
p
4 21
17
tan 2✓ =
4
3
and tan ✓ < 0
Answer: sin 2✓ =
24
,
25
cos 2✓ =
7
,
25
tan 2✓ =
24
7
2. Given the same information as in Item (1), where 0  ✓ < 2⇡, find sin 2✓ , cos 2✓ ,
and tan 2✓ .
Answer:
(a) sin 2✓ =
(b) sin 2✓ =
(c) sin 2✓ =
(d) sin 2✓ =
p
p
6
15
✓
,
tan
=
4
2
3
p
p
cos 2✓ = 1070 , tan 2✓ = 721
p
p
p
p
50 10 5
50+10 5
✓
, cos 2 =
,
10
10
p
p
3 10
, cos 2✓ = 1010 , tan 2✓ = 3
10
10
,
4
p
30
,
10
cos 2✓ =
p
tan 2✓ =
1
p
5
2
3. Express each expression as one trigonometric expression, but do not find the
exact value.
(a) 2 sin 10 cos 10
r
1 cos 7⇡
6
(b)
2
Answer: sin 20
Answer: sin 7⇡
12
215
(c) 1 2 sin2
1 + cos 8
(d)
2
3⇡
10
Answer: cos 3⇡
5
Answer: cos2 4
4. Prove each identity.
(a) tan2
✓
2
= (csc ✓ cot ✓)2
✓
1 cos ✓
1 cos ✓ 1 cos ✓
Solution. tan2 =
=
·
2
1 + cos ✓
1 + cos ✓ 1 cos ✓
(1 cos ✓)2
=
1 cos2 ✓
✓
◆2
1 cos ✓
=
= (csc ✓
sin ✓
cot ✓)2
(b) tan 2✓ + cot 2✓ = 2 csc ✓
✓
✓
1 cos ✓
sin ✓
Solution. tan + cot =
+
2
2
sin ✓
1 cos ✓
(1 cos ✓)2 + sin2 ✓
=
(sin ✓)(1 cos ✓)
1 2 cos ✓ + cos2 ✓ + sin2 ✓
=
(sin ✓)(1 cos ✓)
2 2 cos ✓
2
=
=
= 2 csc ✓
(sin ✓)(1 cos ✓)
sin ✓
(c) sec2
✓
2
= (csc2 ✓)(2 2 cos ✓)
✓
1
2
Solution. sec2 =
=
2
1 + cos ✓
cos2 2✓
2
1 cos ✓
=
·
1 + cos ✓ 1 cos ✓
2 2 cos ✓
=
= (csc2 ✓)(2
sin2 ✓
2 cos ✓)
5. If a = 2 tan ✓, express sin 2✓ and cos 2✓ in terms of a.
Answer: sin 2✓ =
4a
4 a2
,
cos
2✓
=
4 + a2
4 + a2
Solution. sin 2✓ = 2 sin ✓ cos ✓
= 2 tan ✓ cos2 ✓ =
2·
2 tan ✓
2 tan ✓
=
=
2
2
sec ✓
1 + tan ✓
1+
cos 2✓ = 2 cos2 ✓ 1
2
2
=
1=
2
sec ✓
1 + tan2 ✓
216
1=
2
1+
a 2
2
1=
a
2
a 2
2
=
4 a2
4 + a2
4a
4 + a2
6. Find the exact value of cos 36
cos 72 .
Answer:
1
2
Solution
cos 36
cos 72 =
=
=
=
=
=
cos 36
cos 72
2(cos 36 + cos 72 )
1
2(cos 36 + cos 72 )
2
2
2 cos 36
2 cos 72
2(cos 36 + cos 72 )
(2 cos2 36
1) (2 cos2 72
1)
2(cos 36 + cos 72 )
cos 72
cos 144
Half-Angle Identity
2(cos 36 + cos 72 )
cos 72 + cos 36
cos(180
✓) = cos ✓
2(cos 36 + cos 72 )
1
2
·
? 7. The range R of a projectile fired at an acute angle ✓ with the horizontal and
with an initial velocity of v meters per second is given by
R=
v2
sin(2✓),
g
where g is the acceleration due to gravity, which is 9.81 m/sec2 near the Earth’s
surface.
(a) An archer targets an object 100 meters away from her position. If she
positions her arrow at an angle of 32 and releases the arrow at the speed
of 30 m/sec, will she hit her target?
Answer: No
2
30
Solution. R =
· sin(2 · 32 ) ⇡ 82.46 < 100
9.81
(b) If sin ✓ = 25 , solve for v when R = 50.
Answer: v ⇡ 25.86 m/sec
p
2
21
Solution. sin ✓ = and ✓ acute angle =) cos ✓ =
5p
5
v2
2
21
50 =
·2· ·
=) v ⇡ 25.86
9.81
5
5
(c) Given v, find the value of ✓ that gives the largest possible range. At this
v2
✓, what is the range?
Answer: ✓ = 45 , largest R =
g
Solution. To reach the largest R, sin(2✓) must be 1.
8. The figure shows a laser scanner projection system. The optical angle ✓, throw
distance D, and projected image width W are related by the equation
D=
W
2
csc ✓
217
cot ✓
.
Solve for W in terms D and 2✓ .
Answer: W = 2D tan 2✓
https://pangolin.com/userhelp/scanangles.htm
9. The slope of a mountain makes an angle of 45 with the horizontal. At the
base of the mountain, a cannon is fired at an angle ✓ with the horizontal, where
45 < ✓ < 90 , and with initial velocity of v m/sec. Neglecting air resistance,
the distance R (in meters) it drops on the slope of the mountain from the base
is given by
p
2 2v 2
R=
(sin ✓ cos ✓) cos ✓,
g
where g is the acceleration due to gravity in m/sec2 . Express this formula for
R in terms of 2✓.
p
2 2v 2
Solution. R =
(sin ✓ cos ✓) cos ✓
g
p 2
2v
=
(2 sin ✓ cos ✓ 2 cos2 ✓)
g
p 2
2v
=
[sin 2✓ (cos 2✓ + 1)]
g
p 2
2v
=
(sin 2✓ cos 2✓ 1)
g
4
218
Lesson 3.7. Inverse Trigonometric Functions
Time Frame: 3 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) graph the six basic inverse trigonometric functions;
(2) illustrate the domain and range of the inverse trigonometric functions;
(3) evaluate inverse trigonometric expressions; and
(4) solve situational problems involving inverse trigonometric functions.
Lesson Outline
(1) Definitions of the six inverse trigonometric functions
(2) Graphs of inverse trigonometric functions
(3) Domain and range of inverse trigonometric functions
(4) Evaluation of inverse trigonometric expressions
Introduction
In the previous lessons on functions (algebraic and trigonometric), given a
number in the domain of a function, we computed for the value of the function
at that number. Now, given a value in the range of the function, we reverse this
process by finding a number in the domain whose function value is the given
one. Observe that, in this process, the function involved may or may not give a
unique number in the domain. For example, each of the functions f (x) = x2 and
g(x) = cos x do not give a unique number in their respective domains for some
values of each function. Given f (x) = 1, the function gives x = ±1. If g(x) = 1,
then x = 2k⇡, where k is an integer. Because of this possibility, in order that
the reverse process produces a function, we restrict this process to one-to-one
functions or at least restrict the domain of a non-one-to-one function to make it
one-to-one so that the process works. Loosely speaking, a function that reverses
what a given function f does is called its inverse function, and is usually denoted
by f 1 .
More
Teaching Notes
The concept of
inverse function
was studied in
General
Mathematics
and
course.
formally, two functions f and g are inverse functions if
g(f (x)) = x for any x in the domain of f ,
f (g(x)) = x for any x in the domain of g.
We denote the inverse function of a function f by f 1 . The graphs of a function
and its inverse function are symmetric with respect to the line y = x.
219
In this lesson, we first restrict the domain of each trigonometric function
because each of them is not one-to-one. We then define each respective inverse
function and evaluate the values of each inverse trigonometric function.
3.7.1. Inverse Sine Function
All the trigonometric functions that we consider are periodic over their entire
domains. This means that all trigonometric functions are not one-to-one if we
consider their whole domains, which implies that they have no inverses over those
sets. But there is a way to make each of the trigonometric functions one-to-one.
This is done by restricting their respective domains. The restrictions will give us
well-defined inverse trigonometric functions.
The domain of the sine function is the set R of real numbers, and its range is
the closed interval [ 1, 1]. As observed in the previous lessons, the sine function
is not one-to-one, and the first step is to restrict its domain (by agreeing what the
convention is) with the following conditions: (1) the sine function is one-to-one
in that restricted domain, and (2) the range remains the same.
The inverse of the (restricted) sine function
⇥ ⇡ f ⇡(x)
⇤ = sin x, where the
domain is restricted to the closed interval
,
, is called the inverse
2 2
sine function or arcsine function, denoted by f 1 (x) = sin 1 x or
1
f 1 (x) = arcsin ⇥x. Here,
⇤ the domain of f (x) = arcsin x is [ 1, 1],
⇡ ⇡
and its range is
, . Thus,
2 2
y = sin
1
x or y = arcsin x
if and only if
sin y = x,
where
1  x  1 and
⇡
2
 y  ⇡2 .
Throughout the lesson, we interchangeably use sin
the inverse sine function.
1
x and arcsin x to mean
Example 3.7.1. Find the exact value of each expression.
(1) sin 1 12
(3) arcsin 0
(2) arcsin( 1)
(4) sin
1
1
2
Solution. (1) Let ✓ = sin 1 12 . This is equivalent to sin ✓ =⇥ 12 . This
⇤ means that
⇡ ⇡
we are looking for the number ✓ in the closed interval
,
whose sine is
2 2
1 1
1
⇡
⇡
. We get ✓ = 6 . Thus, we have sin 2 = 6 .
2
⇥ ⇡ ⇡⇤
⇡
⇡
(2) arcsin( 1) = ⇡2 because sin
=
1
and
2
, .
2
2
2 2
220
(3) arcsin 0 = 0
(4) sin
1
1
2
⇡
6
=
2
As emphasized
1, sin 1 x is that
⇥ ⇡ ⇡ ⇤in the last example, as long as 1  x 
number y 2
,
such that sin y = x. If |x| > 1, then sin 1 x is not defined in
2 2
R.
We can sometimes find the exact value of sin 1 x (that is, we can find a value
in terms of ⇡), but if no such special value exists, then we leave it in the form
sin 1 x. For example, as shown above, sin 1 12 is equal to ⇡6 . However, as studied
in Lesson 3.2, no special number ✓ satisfies sin ✓ = 23 , so we leave sin 1 23 as is.
Example 3.7.2. Find the exact value of each expression.
(1) sin sin 1 12
(3) arcsin(sin ⇡)
(2) arcsin sin ⇡3
1
(4) sin sin
Solution. (1) sin sin
1 1
2
= sin ⇡6 =
(2) arcsin sin ⇡3 = arcsin
p
3
2
=
1
2
1
2
⇡
3
(3) arcsin(sin ⇡) = arcsin 0 = 0
(4) sin sin
1
1
2
⇡
6
= sin
1
2
=
2
From the last example, we have the following observations:
1. sin(arcsin x) = x for any x 2 [ 1, 1]; and
⇥ ⇡ ⇡⇤
⇥
2. arcsin(sin ✓) = ✓ if and only⇥ if ✓ ⇤2
,
,
and
if
✓
2
6
2 2
⇡ ⇡
arcsin(sin ✓) = ', where ' 2
,
such
that sin ' = sin ✓.
2 2
⇡ ⇡
,
2 2
⇤
, then
To sketch the graph of y = sin 1 x, Table 3.28 presents the tables of values
for y = sin x and y = sin 1 x. Recall that the graphs of y = sin x and y = sin 1 x
are symmetric with respect to the line y = x. This means that if a point (a, b) is
on y = sin x, then (b, a) is on y = sin 1 x.
y = sin x
y = sin
1
x
x
⇡
2
y
1
x
1
y
⇡
2
⇡
3
p
3
2
⇡
4
p
2
2
p
p
2
2
1
2
0
1
2
⇡
3
⇡
4
⇡
6
0
⇡
6
3
2
⇡
6
0
⇡
6
1
2
0
1
2
Table 3.28
221
⇡
4
p
2
2
⇡
3
p
3
2
p
p
3
2
1
⇡
4
⇡
3
⇡
2
2
2
⇡
2
1
The graph (solid thick curve) of the restricted sine function y = sin x is shown
in Figure 3.29(a), while the graph of inverse sine function y = arcsin x is shown
in Figure 3.29(b).
(a) y = sin x
(b) y = sin
1
x
Figure 3.29
Example 3.7.3. Sketch the graph of y = sin 1 (x + 1).
Solution 1. In this solution, we use translation of graphs.
Because y = sin 1 (x + 1) is equivalent to y = sin 1 [x ( 1)], the graph of
y = sin 1 (x + 1) is 1-unit to the left of y = sin 1 x. The graph below shows
y = sin 1 (x + 1) (solid line) and y = sin 1 x (dashed line).
222
Solution 2. In this solution, we graph first the corresponding sine function, and
then use the symmetry with respect to y = x to graph the inverse function.
y = sin 1 (x + 1) () sin y = x + 1 () x = sin y
1
The graph below shows the process of graphing of y = sin 1 (x + 1) from y =
Teaching Notes sin x
1 with ⇡2  x  ⇡2 , and then reflecting it with respect to y = x.
Keep in mind that,
because of the
restriction in the
domain, we have
the following:
sin(sin 1 x) = x
for all x 2 [ 1, 1].
But sin 1 (sin x) is
not always x. We
have
sin 1 (sin x) = x
only if
⇡
 x  ⇡2 .
2
Seatwork/Homework 3.7.1
1. Find the exact value of each expression.
(a) sin
(b)
(c)
(d)
(e)
1
1
⇣ p ⌘
2
arcsin
2
⇣p ⌘
arcsin 23
⇣ p ⌘
3
sin 1
2
⇣
p ⌘
1 2
sin sin
2
Answer:
Answer:
Answer:
Answer:
Answer:
(f) arcsin sin ⇡3
(g) arcsin cos
(h) sin
1
⇡
3
sin 4⇡
3
⇡
2
⇡
3
⇡
3
p
2
2
Answer:
⇡
3
Answer:
⇡
6
Answer:
223
⇡
2
⇡
3
2. Sketch the graph of each equation.
(a) y = sin 1 (x
2)
(b) y = sin 1 (2x)
224
3.7.2. Inverse Cosine Function
The development of the other inverse trigonometric functions follows similarly
from that of the inverse sine function.
y = cos
Teaching Notes
Observe that this
definition of
cos 1 x is
equivalent to
cos 1 x =
⇡
sin 1 x.
2
1
x or y = arccos x
means
cos y = x,
where
1  x  1 and 0  y  ⇡.
The graph (solid thick curve) of the restricted cosine function y = cos x is
shown in Figure 3.30(a), while the graph of inverse cosine function y = arccos x
is shown in Figure 3.30(b).
(a) y = cos x
(b) y = cos
Figure 3.30
Example 3.7.4. Find the exact value of each expression.
(1) cos 1 0
(4) cos 1 cos 3⇡
4
⇣ p ⌘
3
(2) arccos
(5) arccos cos 7⇡
2
6
⇣
(3) cos cos
1
⇣
p
3
2
⌘⌘
Solution. (1) cos 1 0 =
⇣ p ⌘
3
(2) arccos
= 5⇡
2
6
⇡
2
⇣
(6) sin cos
because cos ⇡2 = 0 and
225
⇡
2
1
p
2
2
2 [0, ⇡].
⌘
1
x
⇣
(3) cos cos
(4) cos
1
1
⇣
p
3
2
cos 3⇡
=
4
(5) arccos cos 7⇡
6
⇣
(6) sin cos
1
p
2
2
⌘
⌘⌘
p
3
2
=
p
3
2
because
2 [ 1, 1]
3⇡
4
because 3⇡
2 [0, ⇡].
4
⇣ p ⌘
3
= arccos
= 5⇡
2
6
=
p
2
2
2
Example 3.7.5. Simplify: sin arcsin 23 + arccos 12 .
Solution. We know that arccos 12 = ⇡3 . Using the Sine Sum Identity, we have
sin arcsin 23 + arccos 12
= sin arcsin 23 +
⇡
3
= sin arcsin 23 cos ⇡3 + cos arcsin 23 sin ⇡3
=
=
2
3
1
3
· 12 + cos arcsin 23 ·
+
p
3
2
p
3
2
cos arcsin 23 .
We compute cos arcsin 23 . Let ✓ = arcsin 23 . By definition, sin ✓ = 23 , where
✓ lies in QI. Using the Pythagorean identity, we have
p
p
cos arcsin 23 = cos ✓ = 1 sin2 ✓ = 35 .
Going back to the original computations above, we have
p
3
1
+
3
2
p
1
+ 23
3
p
2+ 15
.
6
sin arcsin 23 + arccos 12 =
=
=
Example 3.7.6. Simplify: sin 2 cos
1
4
5
cos arcsin 23
·
p
5
3
2
.
4
Solution. Let ✓ = cos 1
. Then cos ✓ = 45 . Because cos ✓ < 0 and range
5
of inverse
cosine function is [0, ⇡], we know that ✓ must be within the interval
⇤
⇡
, ⇡ . Using the Pythagorean Identity, we get sin ✓ = 35 .
2
Using the Sine Double-Angle Identity, we have
sin 2 cos
1
4
5
= sin 2✓
= 2 sin ✓ cos ✓
= 2 · 35
= 24
.
25
226
4
5
2
Example 3.7.7. Sketch the graph of y = 14 cos 1 (2x).
Solution.
y=
1
1
cos 1 (2x) () 4y = cos 1 (2x) () x = cos(4y)
4
2
We graph first y = 12 cos(4x). The domain of this graph comes from the restriction
of cosine as follows:
⇡
0  4x  ⇡ =) 0  x  .
4
Then reflect this graph with respect to y = x, and we finally obtain the graph of
y = 14 cos 1 (2x) (solid line).
In the last example, we may also use the following technique. In graphing
y = 14 cos 1 (2x), the horizontal length of cos 1 x is reduced to half, while the
vertical height is reduced to quarter. This comparison technique is shown in
the graph below with the graph of y = cos 1 x in dashed line and the graph of
y = 14 cos 1 (2x) in solid line.
227
Seatwork/Homework 3.7.2
1. Find the exact value of each expression.
(a) cos 1 ( 1)
⇣ p ⌘
2
(b) arccos
2
Answer: ⇡
Answer:
(c) arccos sin 5⇡
2
(d) cos cos
(e) cos
1
1
tan
(f) arccos cos
3⇡
4
Answer: 0
2
5
3⇡
4
13⇡
3
2
5
Answer:
Answer: 0
Answer:
⇡
3
p
4 2
5
2. Simplify each expression.
(a) cos cos
1 2
3
sin
1
1
3
Answer:
p
9
2
Solution. Let ↵ = cos 1 23 . This implies that cos ↵
=
and
↵ in
3
p
5
QI. Using a Pythagorean identity, one gets sin ↵ = 3 . Similarly, let
1
= sin 1
. This implies that sin = 13 and in QIV. Then,
3 p
we get cos = 2 3 2 .
We now use Cosine Di↵erence Identity, and the given and computed
values to simplify the expression.
✓
✓ ◆◆
1
1 2
1
cos cos
sin
= cos(↵
)
3
3
p !
p !✓ ◆
p
p
✓ ◆
2
2 2
5
1
4 2
5
= cos ↵ cos + sin ↵ sin =
+
=
.
3
3
3
3
9
p
p
17 3 4 21
(b) tan arcsin
+ arccos
Answer:
9
1
⇡
Solution. Note that arcsin
= 6 . Second, we let ✓ =p arccos 25 ,
2
which means that cos ✓ = 25 and ✓ is in QI. We get tan ✓ = 221 .
✓
✓ ◆
◆
⇣ ⇡
⌘
1
2
tan arcsin
+ arccos
= tan
+✓
2
5
6
tan( ⇡6 ) + tan ✓
=
1 (tan( ⇡6 ))(tan ✓)
1
2
2
5
p
=
=
228
3
1
3
p
21
2
p
3
21
)(
)
3
2
+
p
(
p
p
17 3 4 21
9
3. Sketch the graph of each equation.
(a) y = 2 cos 1 (3x)
(b) y = 12 cos 1 (x + 2)
3.7.3. Inverse Tangent Function and the Remaining Inverse
Trigonometric Functions
The inverse tangent function is similarly defined as inverse sine and inverse cosine
functions.
y = tan
1
x or y = arctan x
means
tan y = x,
where x 2 R and
⇡
2
< y < ⇡2 .
229
The graph (solid thick curve) of the restricted function y = tan x is shown
in Figure 3.31(a), while the graph of inverse function y = arctan x is shown in
Figure 3.31(b).
(a) y = tan x
(b) y = tan
1
x
Figure 3.31
Example 3.7.8. Find the exact value of each expression.
(1) tan 1 1
(4) tan 1 tan
p
(2) arctan
3
(5) tan 1 tan 7⇡
6
(3) tan tan
5
2
1
1
p
(4) tan
1
.
⇡
3
3 =
1
5
2
=
5
2
tan
⇡
6
=
⇡
6
(3) tan tan
⇡ ⇡
,
2 2
⇡
4
1=
(2) arctan
19⇡
6
(6) arctan tan
Solution. Note the range of arctan is the open interval
(1) tan
⇡
6
(5) Here, note that
tan 7⇡
= tan ⇡6 .
6
7⇡
6
62
tan
because
⇡ ⇡
,
2 2
1
✓
⇡
6
2
⇡ ⇡
,
2 2
.
. Use the idea of reference angle, we know that
7⇡
tan
6
◆
= tan
1
⇣
⇡⌘ ⇡
tan
=
6
6
(6) Here, we cannot use the idea of reference angle, but the idea can help in a
way. The number (or angle) 19⇡
is in QII, wherein tangent is negative, and
6
230
its reference angle is ⇡6 .
✓
✓
◆◆
⇣
⇣ ⇡ ⌘⌘
19⇡
arctan tan
= arctan tan
6
6
⇡
=
6
Example 3.7.9. Find the exact value of each expression.
8
(1) sin 2 tan 1
(2) tan sin 1 35 tan 1
3
2
1
4
8
Solution. (1) Let ✓ = tan 1
. Then tan ✓ = 83 . Following the notations in
3
Lesson 3.2 and the definition of inverse tangent
that ✓ lies
p function, we know
p
2
2
in QIV, and x = 3 and y = 8. We get r = 3 + ( 8) = 73.
Applying the Sine Double-Angle Identity (page 209) gives
✓
✓ ◆◆
8
1
sin 2 tan
= sin 2✓
3
= 2 sin ✓ cos ✓
y x
=2· ·
✓r r ◆ ✓
◆
8
3
p
p
=2
73
73
48
.
=
73
(2) Using the Tangent Di↵erence Identity, we obtain
✓
◆
1 3
1 1
tan sin
tan
5
4
1 3
tan sin 5
tan tan 1 14
=
1 + tan sin 1 35 tan tan 1 14
tan sin 1 35
=
1 + tan sin 1
1
4
3
5
·
1
4
.
We are left to compute tan sin 1 35 . We proceed as in (1) above. Let
✓ = sin 1 35 . Then sin ✓ = 35 . From the definition of inverse sine function and
the notations used p
in Lesson 3.2, we know that ✓ lies in QI, and y = 3 and
r = 5. We get x = 52 32 = 4, so that tan ✓ = xy = 34 .
✓
◆
1
tan sin 1 35
1 3
1 1
4
tan sin
tan
=
1 3
5
4
1 + tan sin 5 · 14
=
3
4
1
4
1 + 34 ·
8
=
19
231
1
4
2
? Example 3.7.10. A student is
viewing a painting in a museum.
Standing 6 ft from the painting,
the eye level of the student is 5 ft
above the ground. If the painting is 10 ft tall and its base is
4 ft above the ground, find the
viewing angle subtended by the
painting at the eyes of the student.
Solution. Let ✓ be the viewing angle, and let ✓ = ↵ +
We observe that
tan ↵ =
1
6
and
as shown below.
9
= .
6
tan
Using the Tangent Sum Identity, we have
tan ↵ + tan
1 tan ↵ tan
1
+9
= 6 169
1 6·6
20
= .
9
Using a calculator, the viewing angle is ✓ = tan
tan ✓ = tan(↵ + ) =
1 20
9
⇡ 65.8 .
We now define the remaining inverse trigonometric functions.
Define
⇡
tan 1 x.
2
It follows that the domain of y = cot 1 x is R and its range is (0, ⇡).
cot
y = sec
1
x=
1
x or y = arcsec x
means
where |x|
⇥
1 and y 2 0, ⇡2
sec y = x,
⇥
[ ⇡, 3⇡
.
2
232
2
Teaching Notes
Keep in mind that
the domain
restrictions are
conventions we set.
Other books and
sources might have
di↵erent domain
restrictions. The
restrictions we
made aim to make
calculus
computations
easier in the future.
Define
⇡
sec 1 x.
2
This means that the⇤domain⇤of y = csc 1 x is ( 1, 1] [ [1, 1) and
its range is
⇡, ⇡2 [ 0, ⇡2 .
csc
1
x=
The graphs of these last three inverse trigonometric functions are shown in
Figures 3.32, 3.33, and 3.34, respectively.
(a) y = cot x
(b) y = cot
1
x
(b) y = sec
1
x
Figure 3.32
(a) y = sec x
Figure 3.33
233
(a) y = csc x
(b) y = csc
1
x
Figure 3.34
Observe that the process in getting the value of an inverse function is the
same to all inverse functions. That is, y = f 1 (x) is the same as f (y) = x. We
need to remember the range of each inverse trigonometric function. Table 3.35
summarizes all the information about the six inverse trigonometric functions.
Function
Domain
sin
1
x
[ 1, 1]
cos
1
x
[ 1, 1]
tan
1
x
cot
1
sec
csc
Range
⇥
⇡ ⇡
,
2 2
Graph
⇤
Figure
3.29(b)
[0, ⇡]
Figure
3.30(b)
R
⇡ ⇡
,
2 2
Figure
3.31(b)
x
R
(0, ⇡)
Figure
3.32(b)
1
x
{x : |x|
1}
1
x
{x : |x|
1}
⇥
⇥
0, ⇡2 [ ⇡, 3⇡
2
⇡,
Table 3.35
⇡
2
⇤
[ 0, ⇡2
Figure
3.33(b)
⇤
Example 3.7.11. Find the exact value of each expression.
p
(1) sec 1 (⇣ 2) ⌘
(3) cot ⇣1
3
⇣
p
2 3
1
3
1
1
(2) csc
(4) sin sec
csc
3
2
234
Figure
3.34(b)
p ⌘⌘
2 3
3
Solution. (1) sec 1 ( 2) =
⇣ p ⌘
2 3
1
(2) csc
= 2⇡
3
3
(3) cot
1
p
3 =
4⇡
3
because sec 4⇡
=
3
5⇡
6
(4) From (2), we know that csc
3
.
2
1
⇣
p ⌘
2 3
3
=
2 and
2⇡
.
3
4⇡
3
⇥
2 ⇡, 3⇡
2
Let ✓ = sec
1
3
2
. Then
sec ✓ =
From defined range of inverse secant function and the notations
in Lesson
3.2,
✓ lies in QIII,
and r = 3 and x = 2.p Solving for y, we get
p
p
2
2
y=
3
( 2) =
5. It follows that sin ✓ = 35 and cos ✓ = 23 .
We now use the Sine Sum Identity.
p !!
✓ ◆
✓
✓
◆◆
3
2
3
2⇡
1
1
sin sec
csc
= sin ✓
2
3
3
✓
◆
2⇡
= sin ✓ +
3
2⇡
2⇡
= sin ✓ cos
+ cos ✓ sin
3
3
p !✓ ◆ ✓ ◆
2
5
1
=
+
3
2
3
p
p
5 2 3
=
6
p !
3
2
2
Seatwork/Homework 3.7.3
1. Find the exact value of each expression.
(a) sec 1 ( 1)
Answer: ⇡
(b) arctan( 1)
Answer:
(c) arccsc csc 5⇡
2
Answer:
1
(d) cot (cot ( 10))
(e) sec
1
(f) csc csc
tan
1
Answer:
3⇡
4
3
8
⇡
4
⇡
2
10
Answer: 0
Answer:
3
8
2. Simplify each expression.
p
p
6
10
+
210
(a) cos arcsec 52 arccot 3
Answer:
50
Solution. Let ↵ = arcsec 52 and = arccot 3. These imply that sec ↵ =
5
and cot = 3, where
↵ and are
both in QI. Wepobtain the following:
2
p
p
21
3 10
2
cos ↵ = 5 , sin ↵ = 5 , cos = 10 , and sin = 1010 .
235
Using Cosine Di↵erence Identity and the above values, we simplify the
expression as follows:
cos(arcsec
5
2
arccot 3) = cos ↵ cos + sin ↵ sin
p !
p ! p !
✓ ◆
2
3 10
21
10
=
+
5
10
5
10
p
p
6 10 + 210
=
50
✓
✓ ◆
◆
1
5
7
(b) tan tan 1
+ tan 1
Answer:
2
3
11
Solution
✓
✓ ◆
◆
1
tan tan 1
+ tan tan 1 53
1
1
1 5
2
tan tan
+ tan
=
1
2
3
1 tan tan 1
tan tan 1 53
2
=
1
2
1
+
1
2
5
3
5
3
=
7
11
Exercises 3.7
1. Find the exact value of each expression.
(a) sin
1
(b) cos
1
(c) tan
(d) csc
1
2
0
p
1
1
Answer:
Answer:
⇡
3
⇡
Answer: 2
Answer: 4⇡
3
3⇡
Answer: 4
3
Answer:
1
1
(e) sec ( 2)
(f) cot 1 ( 1)
(g) csc
⇡
6
⇡
2
1 1
2
Answer: Undefined
? 2. Find the value of each expression using a calculator. Round your answer to
two decimal places.
(a) sin 1 (1/3)
Answer: 0.34
(b) cos 1 ( 2/5)
Answer: 1.98
1
Answer: 1.56
1
(d) csc (11/9)
Answer: 0.96
(e) sec 1 ( 20/3)
Answer: 1.72
(f) cot 1 (5/7)
Answer: 0.95
(c) tan (100)
236
3. Simplify each expression.
(a) cos
1
cos ⇡3
(b) csc
1
tan ⇡6
(c) tan
1
tan 5⇡
4
(d) sin
(e) cos
1
1
csc
Answer: undefined
⇡
4
⇡
4
⇡
6
Answer:
⇡
4
cos
⇡
3
Answer:
Answer:
⇡
3
Answer:
4. Simplify each expression.
(a) sin sin 1 12 + cos 1 12
⇣
⇣
p
1
1
(b) cos tan
3 + sin
p
3
2
(c) tan (2 tan 1 ( 1))
5
(d) cos tan 1 43 + cos 1 13
⇣
(e) sin 2 sin 1 12 3 tan 1
5. Solve for t in terms of x.
p
3
3
Answer: 1
⌘⌘
Answer: 1
Answer: undefined
⌘
Answer: t = 13 sin
(b) x = 2 tan(t + 1)
Answer: t = tan
(c) x = 12 cos(2t + 1)
Answer: t =
(e) x =
1
2
3
2
sec(1
cot(2
1
2
Answer:
(a) x = sin 3t
(d) x = 2
33
65
Answer:
t)
Answer: t = 1
3t)
Answer: t =
6. Sketch the graph of each function.
(a) y = cos 1 (x + 1)
237
sec
2
3
1
3
1
cot
cos
1
1 x
2
x
1
1 (2x)
2
2
(2
3
1 1
2
x)
x
1
(b) y = sin 1 (x
(c) y = sin
1
2x
(d) y = cos
1
x
2
2)
238
(e) y = 2 cos 1 (x
(f) y =
1
2
1)
sin 1 (2x)
(g) y = 2 sin 1 (2x + 2)
239
(h) y =
2 cos 1 (2x
1)
7. Solve for x in the equation sin 1 (x2
2x) =
8. Solve for x in the equation tan 1 (4x2 + 5x
⇡
.
2
7) =
Answer: x = 1
⇡
.
4
Answer:
2, 34
9. A woman is standing x ft from a wall with a billboard nailed on it. The
billboard is 15 ft tall, and its base is 6 ft above the eye level of the woman.
Find the viewing angle subtended on the eyes of the woman from the base to
15x
the top of the billboard.
Answer: tan 1 2
x + 126
? 10. During a leap year, the number of hours of daylight in a city can be modeled
by D(t) = 12 + 2.4 sin(0.017t 1.377), where t is the day of the year (that is,
t = 1 means January 1, t = 60 is February 29, and so on).
(a) Give one day of that year whose number of hours of daylight is about
14.4.
Answer: ⇡ 173 days, so the day would be June 21
(b) Find another day of that year whose number of hours of daylight is the
same as that of February 29.
Answer: ⇡ 287 days, so the day would be October 13
? 11. After getting a job, a man started saving a percentage of his annual income,
which can be modeled by
P (t) = 2.5 cos(0.157t) + 5.2,
where P (t) is the percentage of his annual income that he was able to save on
year t after he got a job.
(a) What percentage of his annual income did he save on the second year?
Answer: 7.58% of his annual income
240
(b) On what year right after getting a job did he save the least?
Answer: 20 years after getting a job
(c) On what year right after getting a job did he save the most? When would
it happen again?
Answer: 40 years after getting a job
(d) If he got his job at the age of 20, how much will he save on the year of
his retirement (that is, when he is 60)?
Answer: 7.7% of his annual
income
12. Prove each identity.
p
2
1+x
(a) cos (tan 1 x) = 1+x
2 , x 2 R
Solution. Let ✓ = tan 1 x, where x is any real number. This implies that
tan ✓ = x. One can think of a right triangle, with acute angle ✓ whose
opposite side is x and adjacent side as 1. Solving for the hypotenuse,
p
p
adjacent side
1
1+x2
we get 1 + x2 . Thus cos ✓ =
= p1+x
. But
2 =
1+x2
hypotenuse
p
1+x2
✓ = tan 1 x, therefore, cos(tan 1 x) = 1+x
2 .
(b) sin (tan
1
2x) =
p 2x
,
1+4x2
1
x2R
Solution. Let ✓ = tan 2x, where x is any real number. This implies that
tan ✓ = 2x. One can think of a right triangle, with acute angle ✓ whose
opposite side is 2x and adjacent side as 1. Solving for the hypotenuse, we
p
opposite side
2x
1
get 1 + 4x2 . Thus sin ✓ =
= p1+4x
2x,
2 . But ✓ = tan
hypotenuse
2x
therefore, sin(tan 1 2x) = p1+4x
2.
(c) tan 1 x + tan 1 x1 = ⇡2 , x > 0
Solution. To prove the identity, one can prove an equivalent identity;
that is, to show that sin(tan 1 x + tan 1 x1 ) = 1, x > 0.
Let ↵ = tan 1 x and = tan 1 x1 . The same techniques as above are
x
1
p 1
applied to get the following: sin ↵ = p1+x
, sin = p1+x
2 , cos ↵ =
2,
1+x2
x
p
and cos = 1+x2 .
We now do the following manipulations:
1
)
x
= sin(↵ + ) = sin ↵ cos + cos ↵ sin
✓
◆✓
◆ ✓
◆✓
◆
x
x
1
1
p
p
= p
+ p
1 + x2
1 + x2
1 + x2
1 + x2
x2 + 1
=
= 1.
1 + x2
sin(tan
1
It follows that tan
x + tan
1
1
x + tan
1 1
x
= ⇡2 .
241
(d) sin 1 x + cos 1 x = ⇡2 , x 2 [ 1, 1]
Solution. Same as in (c), we show that sin(sin 1 x + cos 1 x) = 1, where
x 2 [ 1, 1].
Let ↵ = sin 1 x and = cos 1 x. It follows that sin ↵ = x andp
cos = x.
Using fundamental identities, we obtain the following: cos ↵ = 1 x2 =
sin .
Then, we do the following manipulations:
sin(sin
Therefore, sin
1
1
x + cos
x + cos
1
1
x) =
=
=
=
=
sin(↵ + )
sin ↵ cos + cos ↵ sin
p
p
x(x) + ( 1 x2 )( 1
x2 + 1 x2
1.
x2 )
x = ⇡2 .
4
Lesson 3.8. Trigonometric Equations
Time Frame: 3 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) solve trigonometric equations; and
(2) solve situational problems involving trigonometric equations.
Lesson Outline
(1) Definition of a trigonometric equation
(2) Solution to a trigonometric equation
(3) Techniques of solving a trigonometric equation
Introduction
We have studied equations in Lesson 3.4. We di↵erentiated an identity from
a conditional equation. Recall that an identity is an equation that is true for all
values of the variable in the domain of the equation, while a conditional equation
is an equation that is not an identity.
In this lesson, we mostly study conditional trigonometric equations. We
have started it unnoticeably in the preceding lesson. For example, the equation sin x = 12 has the unique solution x = sin 1 12 = ⇡6 in the closed interval
242
⇥
⇡ ⇡
,
2 2
⇤
. However, if we consider the entire domain (not the restricted domain)
of the sine function, which is the set R of real numbers, there are solutions (other
than ⇡6 ) of the equation sin x = 12 . This current lesson explores the techniques of
solving (conditional) trigonometric equations.
We divide the lesson into two groups of equations: the ones with a basic way
of solving and those that use more advanced techniques.
3.8.1. Solutions of a Trigonometric Equation
Any equation that involves trigonometric expressions is called a trigonometric
Teaching Notes equation. Recall that a solution or a root of an equation is a number in the domain
The word
of the equation that, when substituted to the variable, makes the equation true.
“solution” has two
meanings in our The set of all solutions of an equation is called the solution set of the equation.
discussion. One is
a process of solving
Technically, the basic method to show that a particular number is a solution
a problem, and the
other is a number of an equation is to substitute the number to the variable and see if the equation
that makes an becomes true. However, we may use our knowledge gained from the previous
equation true. The
intended meaning lessons to do a quicker verification process by not doing the manual substitution
depends on the and checking. We use this technique in the example.
context of its
usage.
Example 3.8.1. Which numbers in the set 0, ⇡6 , ⇡4 , ⇡3 , ⇡2 , 2⇡
, 3⇡
, 5⇡
, ⇡, 2⇡ are
3
4
6
Teaching Notes solutions to the following equations?
In the process of
1
showing that a (1) sin x =
(7) cos2 x = cos 2x + sin2 x
2
number is a
solution of an (2) tan x = 1
(8) sin x + cos 2x = 0
equation, note that
p
we cannot assume (3) 3 sec x =
2 3
(9) 2 sin x + tan x 2 cos x = 2
yet that it is a
p
2
2
solution. This
(4)
means that we
cannot use the (5)
equality sign yet in
the process. (6)
3| cot x| = 1
sec2 x
tan2 x = 1
sin x + cos x = 0
(10) sin x + cos x = 2
(11) sin 2x = sin x
(12) 2 tan x + 4 sin x = 2 + sec x
Solution. Note that the choices (except 2⇡) are numbers within the interval [0, ⇡].
To quickly determine which numbers among the choices are solutions to a parTeaching Notes ticular equation, we use some distinctive properties of the possible solutions.
In the actual
classroom
discussion, you
(1)
may only choose
some of these
items.
The sine function is positive on (0, ⇡). From Lesson 3.2, we recall that ⇡6 is
an obvious solution. We may imagine the graph of y = sin x. We may also
use the idea of reference angle. Thus, among the choices, only ⇡6 and 5⇡
are
6
1
the only solutions of sin x = 2 .
(2) Since tan x = 1 > 0, any solution of the equation among the choices must
be in the interval 0, ⇡2 (that is, in QI). Again, among the choices, the only
solution to tan x = 1 is ⇡4 .
243
p
(3) Here, the given equation is equivalent to sec x = 2 3 3 . Among the choices, Teaching Notes
p
For convenience in
the only solution of the equation 3 sec x = 2 3 is 5⇡
.
showing or finding
6
a solution of an
(4) Eliminating
the absolute
value sign, the given equation is equivalent topcot x = equation, we may
p
p
use an equivalent
3
3
3
⇡
or
cot
x
=
.
Among
the
choices,
the
only
solution
of
cot
x
=
is
,
3
3
3
3 equation. By
p
definition, the
2⇡
while the other equation has 3 . Thus, the only solutions of 3| cot x| = 1 solutions of the
equivalent equation
from the given set are ⇡3 and 2⇡
.
3
(5)
are exactly the
same as the
The given equation is one of the Pythagorean Identities (page 185). It means solutions of the
that all numbers in the domain of the equation are solutions. The domain original equation.
of the equation is R \ {x : cos x = 0}. Thus, all except
sec2 x tan2 x = 1.
⇡
2
are solutions of
(6) For the sum of sin x and cos x to be 0, they must have equal absolute values
but di↵erent signs. Among the choices, only 3⇡
satisfies these properties, and
4
it is the only solution of sin x + cos x = 0.
(7) This equations is one of the Double-Angle Identities for Cosine. This means
that all numbers in the domain of the equation are its solutions. Because the
domain of the given equation is R, all numbers in the given set are solutions
of cos2 x = cos 2x + sin2 x.
(8) We substitute each number in the choices to the expression on the left-side
of the equation, and select those numbers that give resulting values equal to
1.
Teaching Notes
x = 0: sin 0 + cos 2(0) = 0 + 1 = 1
x = ⇡6 : sin ⇡6 + cos 2( ⇡6 ) =
x = ⇡4 : sin ⇡4 + cos 2( ⇡4 ) =
x = ⇡3 : sin ⇡3 + cos 2( ⇡3 ) =
1
+ 12 = 1
2
p
p
2
2
+
0
=
2
2
p
p
3
3 1
1
=
2
2
2
x = ⇡2 : sin ⇡2 + cos 2( ⇡2 ) = 1
x=
2⇡
:
3
sin 2⇡
+ cos 2( 2⇡
)=
3
3
x=
3⇡
:
4
5⇡
:
6
sin 3⇡
+ cos 2( 3⇡
)=
4
4
x=
sin 5⇡
+ cos 2( 5⇡
)=
6
6
1=0
p
p
3
3 1
1
=
2
2
2
p
p
2
+ 0 = 22
2
1
+ 12 = 1
2
x = ⇡: sin ⇡ + cos 2⇡ = 0 + 1 = 1
x = 2⇡: sin 2⇡ + cos 2(2⇡) = 0 + 1 = 1
From these values, the only solution of sin x + cos 2x = 0 among the choices
is ⇡2 .
(9) We again substitute the numbers in the given set one by one, and see which
resulting values are equal to 1.
244
If one side of an
equation is
constant and if the
non-constant
expression is a bit
complicated, the
basic method of
showing that a
number is a
solution of the
equation is more
appropriate; that
is, to start on the
non-constant side,
then substitute the
number to the
variable, simplify
the expression, and
end on the value of
the constant on the
other side.
x = 0: 2 sin 0 + tan 0
2 cos 0 =
2
p
3 2 3
3
x = ⇡6 : 2 sin ⇡6 + tan ⇡6
2 cos ⇡6 =
x = ⇡4 : 2 sin ⇡4 + tan ⇡4
2 cos ⇡4 = 1
p
2 cos ⇡3 = 2 3
x = ⇡3 : 2 sin ⇡3 + tan ⇡3
1
x = ⇡2 : Since tan ⇡2 is undefined, this value of x cannot be a solution of the
equation.
x=
2⇡
:
3
2 sin 2⇡
+ tan 2⇡
3
3
x=
3⇡
:
4
2 sin 3⇡
+ tan 3⇡
4
4
p
2 cos 3⇡
=2 2
4
5⇡
:
6
2 sin 5⇡
+ tan 5⇡
6
6
2 cos 5⇡
=
6
x=
x = ⇡: 2 sin ⇡ + tan ⇡
2 cos 2⇡
=1
3
1
p
3+2 3
3
2 cos ⇡ = 2
x = 2⇡: 2 sin 2⇡ + tan 2⇡
2 cos 2⇡ =
Thus, the only solution of 2 sin x + tan x
2
2 cos x = 2 from the given set is ⇡.
(10) This equation has no solution because one of the Pythagorean Identities says
sin2 x + cos2 x = 1.
(11) We substitute each number in the given set to the expression of each side of
the equation, and see which resulting values are equal.
Teaching Notes
If both sides of an
equation are both
non-constant
expressions, one
method of showing
that a number is a
solution of the
equation is to
substitute the
number to both
expressions (but
never join them
with equality sign
because they are
not yet equal
logically), and
check if the
resulting values are
equal.
x = 0: sin 2(0) = 0; sin 0 = 0
x = ⇡6 : sin 2( ⇡6 ) =
p
3
;
2
sin ⇡6 =
x = ⇡4 : sin 2( ⇡4 ) = 1; sin ⇡4 =
x = ⇡3 : sin 2( ⇡3 ) =
p
3
;
2
1
2
p
2
2
sin ⇡3 =
p
3
2
x = ⇡2 : sin 2( ⇡2 ) = 0; sin ⇡2 = 1
x=
3⇡
:
4
sin 2( 3⇡
)=
4
x=
5⇡
:
6
sin 2( 5⇡
)=
6
1; sin 3⇡
=
4
p
3
;
2
sin ⇡3 =
p
2
2
1
2
x = ⇡: sin 2⇡ = 0; sin ⇡ = 0
x = 2⇡: sin 2(2⇡) = 0; sin 2⇡ = 0
Thus, among the numbers in the given set, the solutions of sin 2x = sin x are
0, ⇡3 , ⇡, and 2⇡.
245
(12) We employ the same technique used in the previous item.
x=0:
2 tan 0 + 4 sin 0 = 0
2 + sec 0 = 3
x=
⇡
6
:
2 tan ⇡6 + 4 sin ⇡6 =
2 + sec ⇡6 =
x=
⇡
4
:
x=
⇡
3
:
p
2 3+6
3
p
2 3+6
3
p
2 tan ⇡4 + 4 sin ⇡4 = 2 2 + 2
p
2 + sec ⇡4 = 2 + 2
p
2 tan ⇡3 + 4 sin ⇡3 = 4 3
2 + sec ⇡3 = 4
x=
⇡
2
x=
2⇡
3
:
Both tan ⇡2 and sec ⇡2 are undefined.
: 2 tan 2⇡
+ 4 sin 2⇡
=0
3
3
2 + sec 2⇡
=0
3
x=
3⇡
4
x=
5⇡
6
p
: 2 tan 3⇡
+ 4 sin 3⇡
=2 2
4
4
p
2 + sec 3⇡
=
2
2
4
: 2 tan 5⇡
+ 4 sin 5⇡
=
6
6
2 + sec 5⇡
=
6
x=⇡:
p
6 2 3
3
2
p
6 2 3
3
2 tan ⇡ + 4 sin ⇡ = 0
2 + sec ⇡ = 1
x = 2⇡ : 2 tan 2⇡ + 4 sin 2⇡ = 0
2 + sec 2⇡ = 3
After checking the equal values, the solutions of 2 tan x + 4 sin x = 2 + sec x
among the given choices are ⇡6 , 2⇡
, and 5⇡
.
2
3
6
Seatwork/Homework 3.8.1
In each equation, list down its solutions from the set
(1)
p
3 sec ✓ = 2
⇡
,
3
⇡ ⇡ ⇡ 2⇡
, , , , ⇡, 3⇡
4 6 4 3
2
.
Answer:
⇡
6
(2) (sin x)(tan x + 1) = 0
Answer:
⇡
,
4
⇡
(3) 2 + cos ✓ = 1 + 2 sin2 ✓
Answer:
⇡
,
3
⇡
(4) cos ✓ tan2 ✓ = 3 cos ✓
Answer:
2⇡ 3⇡
, 2
3
246
3.8.2. Equations with One Term
From the preceding discussion, you may observe that there may be more solutions
of a given equations outside the given set. We now find all solutions of a given
equation.
We will start with a group of equations having straightforward techniques
of finding their solutions. These simple techniques involve at least one of the
following ideas:
(1) equivalent equations (that is, equations that have the same solutions as the
original equation);
(2) periodicity of the trigonometric function involved;
(3) inverse trigonometric function;
(4) values of the trigonometric function involved on the interval [0, ⇡] or [0, 2⇡]
(depending on the periodicity of the function); and
(5) Zero-Factor Law: ab = 0 if and only if a = 0 or b = 0.
To “solve an equation” means to find all solutions of the equation. Here,
unless stated as angles measured in degrees, we mean solutions of the equation
that are real numbers (or equivalently, angles measured in radians).
Example 3.8.2. Solve the equation 2 cos x
1 = 0.
Solution. The given equation is equivalent to
1
cos x = .
2
On the interval [0, 2⇡], there are only two solutions of the last equation, and these
are x = ⇡3 (this is in QI) and x = 5⇡
(in QIV).
3
Because the period of cosine function is 2⇡, the complete solutions of the
equation are x = ⇡3 + k(2⇡) and x = 5⇡
+ k(2⇡) for all integers k.
2
3
In the preceding example, by saying that the “complete solutions are x =
+ k(2⇡) and x = 5⇡
+ k(2⇡) for all integers k,” we mean that any integral
3
Teaching Notes value of k will produce a solution to the given equation. For example, when
Any particular
k = 3, x = ⇡3 + 3(2⇡) = 19⇡
is a solution of the equation. When k = 2,
3
solution in a family
5⇡
7⇡
is another solution of 2 cos x 1 = 0. The family of
of solutions can be x = 3 + ( 2)(2⇡) =
3
⇡
used as a “seed
solutions
x
=
+
k(2⇡)
can
be equivalently enumerated as x = 19⇡
+ 2k⇡, while
3
3
solution” to
5⇡
7⇡
produce the other the family x =
+ k(2⇡) can also be stated as x = 3 + 2k⇡.
3
⇡
3
solutions in the
family.
Example 3.8.3. Solve: (1 + cos ✓)(tan ✓
247
1) = 0.
Solution. By the Zero-Factor Law, the given equation is equivalent to
1 + cos ✓ = 0
cos ✓ =
or
tan ✓
1
1=0
tan ✓ = 1
✓ = ⇡ + 2k⇡, k 2 Z
✓=
⇡
4
+ k⇡, k 2 Z.
Therefore, the solutions of the equation are ✓ = ⇡ + 2k⇡ and ✓ =
k 2 Z.
⇡
4
+ k⇡ for all
2
Example 3.8.4. Find all values of x in the interval [ 2⇡, 2⇡] that satisfy the
equation (sin x 1)(sin x + 1) = 0.
Solution.
sin x
1=0
or
sin x + 1 = 0
sin x = 1
x=
⇡
2
or
sin x =
3⇡
2
x=
⇡
,
2
Solutions:
3⇡
2
3⇡ 3⇡
, 2,
2
1
⇡
2
or
⇡
2
2
Example 3.8.5. Solve: cos x = 0.1.
Solution. There is no special number whose cosine is 0.1. However, because
0.1 2 [ 1, 1], there is a number whose cosine is 0.1. In fact, in any one-period
interval, with cos x = 0.1 > 0, we expect two solutions: one in QI and another in
QIV. We use the inverse cosine function.
From Lesson 3.7, one particular solution of cos x = 0.1 in QI is x = cos 1 0.1.
We can use this solution to get a particular solution in QIV, and this is x =
2⇡ cos 1 0.1, which is equivalent to x = cos 1 0.1.
From the above particular solutions, we can produce all solutions of cos x =
0.1, and these are x = cos 1 0.1+2k⇡ and x = cos 1 0.1+2k⇡ for all k 2 Z. 2
Example 3.8.6. Solve: 3 tan ✓ + 5 = 0.
Solution.
3 tan ✓ + 5 = 0
=)
tan ✓ =
5
3
We expect only one solution in any one-period interval.
tan ✓ =
5
3
=)
✓ = tan
1
5
3
+ k⇡, k 2 Z
2
? Example 3.8.7. The voltage V (in volts) coming from an electricity distributing company is fluctuating according to the function V (t) = 200 + 170 sin(120⇡t)
at time t in seconds.
248
(1) Determine the first time it takes to reach 300 volts.
(2) For what values of t does the voltage reach its maximum value?
Solution. (1) We solve for the least positive value of t such that V (t) = 300.
200 + 170 sin(120⇡t) = 300
100
sin(120⇡t) =
170
120⇡t = sin
t=
1
100
170
sin 1 100
170
⇡ 0.00167 seconds
120⇡
(2) The maximum value of V (t) happens when and only when the maximum
value of sin(120⇡t) is reached. We know that the maximum value of sin(120⇡t)
is 1, and it follows that the maximum value of V (t) is 370 volts. Thus, we
need to solve for all values of t such that sin(120⇡t) = 1.
sin(120⇡t) = 1
⇡
120⇡t = + 2k⇡, k nonnegative integer
2
⇡
+ 2k⇡
t= 2
120⇡
1
+ 2k
t= 2
⇡ 0.00417 + 0.017k
120
This means that the voltage is maximum when t ⇡ 0.00417 + 0.017k for each
nonnegative integer k.
2
Seatwork/Homework 3.8.2
1. Solve each equation.
(a) tan x =
(b) sin x =
(c) (cos x
1
⇡
4
Answer:
1
2
Answer:
1)(tan x + 1) = 0
⇡
6
+ 2k⇡,
Answer: 2k⇡,
5⇡
6
+ k⇡, k 2 Z
+ 2k⇡, k 2 Z
⇡
4
+ k⇡, k 2 Z
2. Find all values of the variable in the interval [ 2⇡, 2⇡] that satisfy the given
equation.
(a) (sin ✓ + 1)(tan ✓) = 0
(b) sec ✓ + 2 = 0
Answer:
249
⇡ 3⇡
, , 0, ⇡, 2⇡, ⇡,
2 2
Answer: 2⇡
, 4⇡
, 4⇡
,
3
3
3
2⇡
2⇡
3
3.8.3. Equations with Two or More Terms
We will now consider a group of equations having multi-step techniques of finding their solutions. Coupled with the straightforward techniques discussed in the
preceding discussion, these more advanced techniques involve factoring of expressions and trigonometric identities. The primary goal is to reduce a given equation
into equivalent one-term equations.
Example 3.8.8. Solve: 2 cos x tan x = 2 cos x.
Solution.
2 cos x tan x = 2 cos x
2 cos x tan x 2 cos x = 0
(2 cos x)(tan x 1) = 0
2 cos x = 0
or
tan x
1=0
cos x = 0
tan x = 1
x = ⇡2 + 2k⇡ or
x = 3⇡
+ 2k⇡,
2
k2Z
x = ⇡4 + k⇡,
k2Z
Solutions:
⇡
2
+ 2k⇡,
3⇡
2
⇡
4
+ 2k⇡,
+ k⇡, k 2 Z
2
Example 3.8.9. Solve for x 2 [0, 2⇡): sin 2x = sin x.
Solution.
sin 2x = sin x
sin 2x
sin x = 0
2 sin x cos x
(sin x)(2 cos x
sin x = 0
Sine Double-Angle Identity
1) = 0
sin x = 0
or
2 cos x
x = 0 or x = ⇡
1=0
cos x =
x=
Solutions: 0, ⇡,
250
⇡
3
⇡ 5⇡
, 3
3
1
2
or x =
5⇡
3
2
Teaching Notes
The method for
solving
trigonometric
equations follows
the usual way of
solving nonlinear
equations; that is,
transform the
equation so that
one side is 0, and
then factor.
Tips in Solving Trigonometric Equations
(1) If the equation contains only one trigonometric term, isolate that
term, and solve for the variable.
(2) If the equation is quadratic in form, we may use factoring, finding
square roots, or the quadratic formula.
(3) Rewrite the equation to have 0 on one side, and then factor (if
appropriate) the expression on the other side.
(4) If the equation contains more than one trigonometric function,
try to express everything in terms of one trigonometric function.
Here, identities are useful.
(5) If half or multiple angles are present, express them in terms of a
trigonometric expression of a single angle, except when all angles
involved have the same multiplicity wherein, in this case, retain
the angle. Half-angle and double-angle identities are useful in
simplification.
Example 3.8.10. Solve for x 2 [0, 2⇡): 2 cos2 x = 1 + sin x.
Solution.
2 cos2 x = 1 + sin x
sin2 x) = 1 + sin x
2(1
2 sin2 x + sin x
(2 sin x
Pythagorean Identity
1=0
1)(sin x + 1) = 0
2 sin x
1=0
sin x =
x=
⇡
6
Factoring
or
sin x + 1 = 0
1
2
or x =
sin x =
5⇡
6
x=
Solutions:
1
3⇡
2
⇡ 5⇡ 3⇡
, 6, 2
6
Example 3.8.11. Solve for x 2 [0, 2⇡) in the equation 3 cos2 x + 2 sin x = 2.
Solution.
3 cos2 x + 2 sin x = 2
3(1
sin2 x) + 2 sin x = 2
(3 sin x + 1)(sin x
1) = 0
251
Pythagorean Identity
Factoring
2
3 sin x + 1 = 0
sin x =
or
sin x
1
3
1=0
sin x = 1
x = sin 1 ( 13 ) + 2⇡
or
x = ⇡ sin 1 ( 13 )
x=
⇡
2
sin 1 ( 13 )+, ⇡ + sin 1 ( 13 ),
Solutions: 2⇡
⇡
2
2 Teaching Notes
Using the
Odd-Even Identity,
sin 1 ( 13 ) =
sin 1 ( 13 ).
One part of the last solution needs further explanation. In the equation
sin x = 13 , we expect two solutions in the interval [0, 2⇡): one in (⇡, 3⇡
) (which
2
3⇡
is QIII), and another in ( 2 , 2⇡) (which is QIV). Since no special number satisfies
sin x = 13 , we use inverse sine function. Because the range of sin 1 is [ ⇡2 , ⇡2 ], we Teaching Notes
the reference
know that ⇡2 < sin 1 ( 13 ) < 0. From this value, to get the solution in ( 3⇡
, 2⇡), Using
2
angle of sin 1 ( 13 ),
we simply add 2⇡ to this value, resulting to x = sin 1 ( 13 ) + 2⇡. On the other we get two
(QIII and
hand, to get the solution in (⇡, 3⇡
), we simply add sin 1 ( 13 ) to ⇡, resulting to solutions
2
QIV), and these
are ⇡ + sin 1 ( 13 )
x = ⇡ sin 1 ( 13 ).
and 2⇡
x
Example 3.8.12. Solve: sin2 x + 5 cos2 = 2.
2
Solution.
x
sin2 x + 5 cos2 = 2
2 ◆
✓
1 + cos x
2
sin x + 5
=2
2
Cosine Half-Angle Identity
2 sin2 x + 5 cos x + 1 = 0
2(1
cos2 x) + 5 cos x + 1 = 0 Pythagorean Identity
2 cos2 x
5 cos x
3=0
(2 cos x + 1)(cos x
3) = 0
2 cos x + 1 = 0
cos x =
or
cos x
1
2
cos x = 3
x = 2⇡
+ 2k⇡ or
3
4⇡
x = 3 + 2k⇡,
k2Z
Solutions:
3=0
2⇡
3
no solution
+ 2k⇡,
4⇡
3
+ 2k⇡, k 2 Z
Example 3.8.13. Solve for x 2 [0, 2⇡) in the equation tan 2x
252
2
2 cos x = 0.
sin
1 ( 1 ).
3
Solution.
tan 2x 2 cos x = 0
sin 2x
2 cos x = 0
cos 2x
sin 2x 2 cos x cos 2x = 0
Apply the Double-Angle Identities for Sine and Cosine, and then factor.
2 sin x cos x 2(cos x)(1 2 sin2 x) = 0
(2 cos x)(2 sin2 x + sin x 1) = 0
(2 cos x)(2 sin x 1)(sin x + 1) = 0
2 cos x = 0
or
2 sin x
1=0
or
sin x + 1 = 0
1
2
cos x = 0
sin x =
x = ⇡2 or
x = 3⇡
2
x = ⇡6 or
x = 5⇡
6
sin x =
x=
1
3⇡
2
These values of x should be checked in the original equation because tan 2x may
not be defined. Upon checking, this is not the case for each value of x obtained.
The solutions are ⇡2 , 3⇡
, ⇡6 , 5⇡
, and 3⇡
.
2
2
6
2
? Example 3.8.14. A weight is suspended from a spring and vibrating vertically
according to the equation
4
⇡
5
f (t) = 20 cos
t
5
6
,
where f (t) centimeters is the directed distance of the weight from its central
position at t seconds, and the positive distance means above its central position.
(1) At what time is the displacement of the weight 5 cm below its central
position for the first time?
(2) For what values of t does the weight reach its farthest point below its central
position?
Solution. (1) We find the least positive value of t such that f (t) =
20 cos
cos
4
⇡
5
4
⇡
5
5
6
t
t
5
6
=
=
5
1
4
There are two families of solutions for this equation.
253
5.
•
4
⇡
5
t
t=
5
6
+
5
1
= cos 1
6
4
cos 1 ( 14 )+2k⇡
4
⇡
5
+ 2k⇡, k 2 Z
In this family of solutions, the least positive value of t happens when
k = 0, and this is
t=
•
4
⇡
5
t
t=
5
6
+
5
= 2⇡
6
2⇡ cos 1 (
4
⇡
5
5 cos
+
6
+ 2(0)⇡
⇡ 1.5589.
+ 2k⇡, k 2 Z
1
4
cos 1
1
+2k⇡
4)
1
4
4
⇡
5
1
Here, the least positive value of t happens when k =
5 2⇡
+
6
t=
cos
1
1
4
+ 2( 1)⇡
4
⇡
5
1, and this is
⇡ 0.1078.
Therefore, the first time that the displacement of the weight is 5 cm below
its central position is at about 0.1078 seconds.
(2) The minimum value of f (t) happens when and only when the minimum
value of cos 45 ⇡ t 56 is reached. The minimum value of cos 45 ⇡ t 56 is
1, which implies that the farthest point the weight can reach below its
central position is 20 cm. Thus, we need to solve for all values of t such that
cos 45 ⇡ t 56 = 1.
cos 45 ⇡ t
4
⇡
5
4
⇡
5
t
t
5
6
5
6
5
6
=
1
= cos 1 ( 1) + 2k⇡, k
0
= ⇡ + 2k⇡
t=
5
6
+
⇡+2k⇡
4
⇡
5
=
25
12
+ 52 k
Therefore, the weight reaches its farthest point (which is 20 cm) below its
central position at t = 25
+ 52 k for every integer k 0.
2
12
Seatwork/Homework 3.8.3
1. Solve each equation.
(a) 2 sin2 ✓ = sin ✓ + 1
Answer:
+ 2k⇡,
⇡
2
+ 2k⇡, k 2 Z
(b) tan2 x + tan x = 6
Answer: tan 1 ( 3) + k⇡, tan
1
2 + k⇡, k 2 Z
(c) sin x = 1 + cos x
7⇡
6
+ 2k⇡,
11⇡
6
Answer:
254
⇡
2
+ 2k⇡, ⇡ + 2k⇡, k 2 Z
2. Find the solutions in the interval [0, 2⇡).
Answer: ⇡2 , 3⇡
, ⇡ , 5⇡
2 6 6
Answer: 0, ⇡, sin 1 13 , ⇡ sin 1 13
(a) sin 2✓ = cos ✓
(b) 3 cos2 x + sin x = 3
Exercises 3.8
1. Solve each equation.
(a) 2 sin x + 1 = 0
⇡
6
Answer:
(c) tan x + 1 = 0
Answer:
(f) sec x
1=0
Answer:
(g) sec2 x + 6 tan x + 4 = 0
(h) cos 2x + 3 = 5 cos x
(j) 6 sec2
x
2
x
2
=1
Answer:
Answer:
+ 3 = 7 tan x2
Answer: 2 tan
1
3
1
1
⇡(4k
16
1), k 2 Z
+ 1), k 2 Z
+ k⇡, tan 1 ( 5) + k⇡, k 2 Z
⇡
4
Answer:
+ k⇡, k 2 Z
Answer: k⇡, k 2 Z
1=0
(i) cos2 x + sin2
3⇡
4
Answer: 16 ⇡(2k
(d) cos 3x = 0
2
+ 2k⇡, k 2 Z
Answer: k⇡, k 2 Z
(b) sin x tan x = 0
(e) tan 4x
7⇡
6
+ 2k⇡,
2⇡
3
⇡
3
+ 2k⇡,
+ 2k⇡,
4⇡
3
⇡
3
+ 2k⇡, k 2 Z
+ 2k⇡, 2k⇡, k 2 Z
+ 2k⇡, 2 tan
1 3
2
+ 2k⇡, k 2 Z
2. Find the solutions in the interval [0, 2⇡).
(a) 4 sin2 x
1=0
(b) 2 cos2 x + 3 cos x
2=0
(c) tan x
cot x = 0
p
(d) 2 sin 2x = 3
(e) sec2 x
2
(f) 2 sin x
⇡ 5⇡ 7⇡ 11⇡
, 6, 6, 6
6
Answer: ⇡3 , 5⇡
3
⇡ 5⇡
Answer: 4 , 4
Answer: ⇡6 , ⇡3 , 7⇡
, 4⇡
6
3
⇡ 2⇡ 4⇡ 5⇡
Answer: 3 , 3 , 3 , 3
Answer: 7⇡
, 11⇡
6
6
3⇡
Answer: 2
⇡, cos 1 34 , 2⇡ cos 1 34
Answer: 0, ⇡3 , 5⇡
3
11⇡
Answer: 6
Answer:
4=0
5 sin x = 3
(g) tan x + sec x = 0
(h) 2 sin 2x = 3 sin x
Answer: 0,
(i) tan2 x = 1 + sec x
p
(j) tan x + 3 = sec x
3. Find the solutions in the interval [0 , 180 ).
(a) sin x
cos x = 0
(b) cot 4x
1=0
Answer: 45
Answer: 11.25 , 56.25 , 101.25 , 146.25
255
(c) 3 cos 2x
3 cos x = 0
Answer: 0 , 120
? (d) 6 sec2 x + 3 = 7 tan x
? (e) tan2 x + tan x = 6
Answer: 161.6 , 56.3
Answer: 63.4 , 108.4
? 4. A weight is suspended from a spring and vibrating vertically according to the
equation
f (t) = 25.2 sin(3.8t 2.1),
where f (t) centimeters is the directed distance of the weight from its central position at t seconds, and the positive distance means above its central
position.
(a) Find the times when the weight is at its central position.
Solution. We solve the equation f (t) = 0.
25.2 sin(3.8t 2.1) = 0
sin(3.8t 2.1) = 0
3.8t 2.1 = k⇡ k nonnegative integer
2.1 + k⇡
t =
3.8
t ⇡ 0.55 + 0.83k
Therefore, the weight is at its central position at t ⇡ 0.55 + 0.83k seconds
(where k is a nonnegative integer). In other words, it is at central position
when t = 0.55 s, t = 1.38 s, t = 2.21 s, etc.
(b) For what values of t does the weight reach its farthest point below its
central position?
Solution. The weight reaches it farthest point below the central position
when sin(3.8t 2.1) = 1. Solving for t, we get
sin(3.8t
3.8t
2.1) =
2.1 =
t =
t =
t ⇡
1
3⇡
+ 2k⇡ where k is a whole number
2
3⇡+4k⇡
+ 2.1
2
where k is a whole number
3.8
(3 + 4k)⇡ + 4.2
where k is a whole number
7.6
1.79 + 1.65k where k is a whole number
Therefore, the weight reaches it farthest point below the central position
at t ⇡ 1.79 + 1.65k seconds (where k is a whole number). For instance,
at t = 1.79 s, t = 3.44 s, t = 5.09 s, etc.
256
? 5. The finance department of a car company conducted a study of their weekly
sales in the past years, and came out with the following approximating function:
s(t) = 12.18 cos(0.88t 7.25) + 20.40, t 0,
where s(t) represents weekly car sales in million pesos at week t (t = 0 represents the start of the study).
(a) Find the weekly sales at the start of the study.
Solution.
s(0) =
=
=
=
12.18 cos(0.88(0) 7.25) + 20.40
12.18 cos( 7.25) + 20.40
(12.18)(0.5679) + 20.40
27.32
The weekly sales of the car company at the start of the study is approximately 27.32 million pesos.
(b) Find the projected maximum and minimum weekly sales of the company.
Solution. The projected maximum and minimum weekly sales of the
company are attained when the cosine values are 1 and 1, respectively.
Thus, the maximum weekly sales is 12.18 + 20.40 = 32.58 million pesos,
and the minimum weekly sales is 12.18 + 20.40 = 8.22 million pesos.
(c) If the company were able to reach its maximum sales this week, when
will the next projected maximum weekly sales and upcoming projected
minimum weekly sales be?
Solution. The next projected maximum weekly sales will be attained after
2⇡
one period. That is, P = 0.88
⇡ 7.14. Hence, if the company were able
to reach its maximum sales this week, then the next projected maximum
weekly sales will be after about 7 weeks.
On the other hand, the upcoming minimum weekly sales is projected after
2⇡
half the period. That is, 12 P = 12 0.88
⇡ 3.57. Hence, if the company were
able to reach its maximum sales this week, then the upcoming projected
minimum weekly sales will be after about 3.5 weeks.
(d) After the start of the study, when did the company experience a weekly
sales of only 10 million for the first time?
Solution. Here, we want to solve s(t) = 10 for the least nonnegative value
of t.
12.18 cos(0.88t
cos(0.88t
257
7.25) + 20.40 = 10
520
7.25) =
609
We get
0.88t
7.25 = cos
1
✓
520
609
◆
+ 2k⇡, k 2 Z
t ⇡ 11.19 + 7.14k
=)
or
0.88t
7.25 =
cos
1
✓
520
609
◆
+ 2k⇡
=)
t ⇡ 5.29 + 7.14k.
Among these solutions, the least nonnegative value of t is t = 11.19 +
7.14( 1) ⇡ 4.05. Thus, about 4 weeks after the start of the study, the
company experienced a weekly sales of only 10 million for the first time.
? 6. After many years in business, the financial analyst of a shoe company projected
that the monthly costs of producing their products and monthly revenues from
the sales of their products are fluctuating according to the following formulas:
C(t) = 2.6 + 0.58 sin(0.52t
7.25)
R(t) = 2.6 + 1.82 cos(0.52t
7.25),
and
where C(t) and R(t) are the costs and revenues in million of pesos at month t
(t = 0 represents January 2010). About how many months after January 2010
did the company experience a zero profit for the first time?
Solution. The profit is zero when the revenue is the same as the cost.
2.6 + 0.58 sin(0.52t
7.25) = 2.6 + 1.82 cos(0.52t 7.25)
91
tan(0.52t 7.25) =
✓ ◆ 29
91
0.52t 7.25 = tan 1
+ k⇡, k 2 Z
29
7.25 + tan 1 91
+ k⇡
29
t=
⇡ 16.37 + 6.04k
0.52
The company experienced zero profit for the first time about 16.37+6.04( 2) ⇡
4.29 or 4 months after January 2010.
7. If x be a real number such that
cos 2x =
2
3
and
what is cos x + sin x?
258
cos x
1
sin x = ,
2
Solution.
cos 2x =
cos2 x
sin2 x =
(cos x + sin x)(cos x
sin x) =
✓ ◆
1
(cos x + sin x)
=
2
cos x + sin x =
2
3
2
3
2
3
2
3
4
3
8. Solve for x 2 [ ⇡, ⇡): 16 sin4 x + 1 = 8 sin2 x.
Solution.
16 sin4 x + 1
16 sin4 x 8 sin2 x + 1
(4 sin2 x 1)2
4 sin2 x 1
8 sin2 x
0
0
0
1
sin x = ±
2
5⇡ ⇡
x = ± ,±
6
6
=
=
=
=
9. Find the smallest positive value of ✓ that satisfies the equation
⇣⇡
⌘
⇣⇡
⌘ 3
sin
+ ✓ + sin
✓ = .
3
3
8
Solution.
⇣⇡
⌘
⇣⇡
⌘
3
3
3
8
⇡
⇡
⇡
⇡
3
sin cos ✓ + cos sin ✓ + sin cos ✓ cos sin ✓ =
3
3
3
3
8
⇡
3
2 sin cos ✓ =
3
8
p
3
3 cos ✓ =
8
sin
+ ✓ + sin
✓
=
✓ = cos
259
1
p !
3
8
10. In 4ABC, angles A and B (in degrees) satisfy
3 sin A + cos B = 3 and
sin B + 3 cos A = 2.
Prove that C = 30 .
Solution.
(
3 sin A + cos B = 3
3 cos A + sin B = 2
Square both equations:
(
9 sin2 A + 6 sin A cos B + cos2 B = 9
9 cos2 A + 6 cos A sin B + sin2 B = 4
Add and solve:
9 + 6 sin(A + B) + 1 = 13
6 sin(A + B) = 3
1
sin(A + B) =
2
A + B = 30 , 150
If A + B = 150 , then C = 30 . If A + B = 30 , then A < 30 , and
3 sin A + cos B < 3 · 12 + 1 = 52 < 3, contradicting the first equation. Therefore,
we only have C = 30 .
4
Lesson 3.9. Polar Coordinate System
Time Frame: 3 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) locate points in polar coordinate system;
(2) convert the coordinates of a point from rectangular to polar system and vice
versa; and
(3) solve situational problems involving polar coordinate system.
Lesson Outline
(1) Polar coordinate system: pole and polar axis
260
(2) Polar coordinates of a point and its location
(3) Conversion from polar to rectangular coordinates, and vice versa
(4) Simple graphs and applications
Introduction
Two-dimensional coordinate systems are used to describe a point in a plane.
We previously used the Cartesian or rectangular coordinate system to locate a
point in the plane. That point is denoted by (x, y), where x is the signed distance of the point from the y-axis, and y is the signed distance of the point
from the x-axis. We sketched the graphs of equations (lines, circles, parabolas,
ellipses, and hyperbolas) and functions (polynomial, rational, exponential, logarithmic, trigonometric, and inverse trigonometric) in the Cartesian coordinate
plane. However, it is often convenient to locate a point based on its distance
from a fixed point and its angle with respect to a fixed ray. Not all equations
can be graphed easily using Cartesian coordinates. In this lesson, we also use
another coordinate system, which can be presented in dartboard-like plane as
shown below.
3.9.1. Polar Coordinates of a Point
We now introduce the polar coordinate system. It is composed of a fixed point
called the pole (which is the origin in the Cartesian coordinate system) and a
fixed ray called the polar axis (which is the nonnegative x-axis).
261
In the polar coordinate system, a point is described by the ordered pair (r, ✓),
where the radial coordinate r refers to the directed distance of the point from the
pole and the angular coordinate ✓ refers to a directed angle (usually in radians)
from the polar axis to the segment joining the point and the pole.
Because a point in polar coordinate system is described by an order pair of
radial coordinate and angular coordinate, it will be more convenient to geometrically present the system in a polar plane, which serves just like the Cartesian
plane. In the polar plane shown below, instead of rectangular grids in the Cartesian plane, we have concentric circles with common center at the pole to identify
easily the distance from the pole (radial coordinate) and angular rays emanating
from the pole to show the angles from the polar axis (angular coordinate).
262
Example 3.9.1. Plot the following points in one polar plane: A(3, ⇡3 ), B(1, 5⇡
),
6
7⇡
19⇡
7⇡
17⇡
17⇡
5⇡
C(2, 6 ), D(4, 12 ), E(3, ⇡), F (4, 6 ), G(2.5, 4 ), H(4, 6 ), and I(3, 3 ).
Solution.
As seen in the last example, unlike in Cartesian plane where a point has a
unique Cartesian coordinate representation, a point in polar plane have infinitely
many polar coordinate representations. For example, the coordinates (3, 4) in
the Cartesian plane refer to exactly one point in the plane, and this particular
point has no rectangular coordinate representations other than (3, 4). However,
the coordinates (3, ⇡3 ) in the polar plane also refer to exactly one point, but
this point has other polar coordinate representations. For example, the polar
coordinates (3, 5⇡
), (3, 7⇡
), (3, 13⇡
), and (3, 19⇡
) all refer to the same point as
3
3
3
3
⇡
that of (3, 3 ).
The polar coordinates (r, ✓ + 2k⇡), where k 2 Z, represent the same
point as that of (r, ✓).
In polar coordinate system, it is possible for the coordinates (r, ✓) to have
a negative value of r. In this case, the point is |r| units from the pole in the
opposite direction of the terminal side of ✓, as shown in Figure 3.36.
263
Figure 3.36
Example 3.9.2. Plot the following points in one polar plane: A( 3, 4⇡
), B( 4, 11⇡
),
3
6
C( 2, ⇡), and D( 3.5, 7⇡
).
4
Solution. As described above, a polar point with negative radial coordinate lies
on the opposite ray of the terminal side of ✓.
264
Points in Polar Coordinates
1. For any ✓, the polar coordinates (0, ✓) represent the pole.
2. A point with polar coordinates (r, ✓) can also be represented by
(r, ✓ + 2k⇡) or ( r, ✓ + ⇡ + 2k⇡) for any integer k.
Seatwork/Homework 3.9.1
1. Plot the following points in one polar plane:
A(2,
⇡
)
2
B(3, 7⇡
)
4
C(4,
D(2,
4⇡
)
3
E( 4, ⇡4 )
F (1, 41⇡
)
12
G( 3,
2⇡
)
3
H( 4,
⇡
)
12
I( 2,
⇡
)
6
11⇡
)
2
Answer:
2. Give the polar coordinates (r, ✓) with indicated properties that represent the
same point as the given polar coordinates.
265
(a)
(b)
(c)
(d)
(e)
(f)
(2, ⇡); r > 0, 2⇡ < ✓  0
(5, 3⇡
); r < 0, 0 < ✓  2⇡
4
4⇡
( 5, 3 ); r < 0, 2⇡ < ✓  0
(1, 0); r < 0, 0 < ✓  2⇡
(2, sin 1 (0.6)); r < 0, 2⇡ < ✓  0
( 3, cos 1 ( 0.4)); r > 0, 0 < ✓  2⇡
Answer: (2, ⇡)
Answer: ( 5, 7⇡
)
4
2⇡
Answer: ( 5, 3 )
Answer: ( 1, ⇡)
Answer: ( 2, sin 1 (0.6) ⇡)
Answer: (3, cos 1 ( 0.4) + ⇡)
3.9.2. From Polar to Rectangular, and Vice Versa
We now have two ways to describe points on a plane – whether to use the Cartesian coordinates (x, y) or the polar coordinates (r, ✓). We now derive the conversion from one of these coordinate systems to the other.
We superimpose the Cartesian and polar planes, as shown in the following
diagram.
Figure 3.37
Suppose a point P is represented by the polar coordinates (r, ✓). From Lesson
3.2 (in particular, the boxed definition on page 142), we know that
x = r cos ✓
and
y = r sin ✓.
Conversion from Polar to Rectangular Coordinates
8
< x = r cos ✓
(r, ✓) !
! (x, y)
: y = r sin ✓
Given one polar coordinate representation (r, ✓), there is only one
rectangular coordinate representation (x, y) corresponding to it.
266
Example 3.9.3. Convert the polar coordinates (5, ⇡) and ( 3, ⇡6 ) to Cartesian
coordinates.
Solution.
(5, ⇡)
Teaching Notes
One can also easily
convert the polar
coordinates (5, ⇡)
to its
corresponding
rectangular
coordinates ( 5, 0)
by simply plotting
the point.
( 3, ⇡6 )
8
< x = 5 cos ⇡ = 5
!
: y = 5 sin ⇡ = 0
8
< x=
!
: y=
3 cos ⇡6 =
3 sin
⇡
6
=
p
3 3
2
3
2
! ( 5, 0)
! (
p
3 3
,
2
3
)
2
2
As explained on page 263 (right after Example 3.9.1), we expect that there
are infinitely many polar coordinate representations that correspond to just one
given rectangular coordinate representation. Although we can actually determine
all of them, we only need to know one of them and we can choose r 0.
Suppose a point P is represented by the rectangular coordinates (x, y). Referring back to Figure 3.37, the equation of the circle is
p
x2 + y 2 = r 2
=)
r = x2 + y 2 .
We now determine ✓. If x = y = 0, then r = 0 and the point is the pole. The
pole has coordinates (0, ✓), where ✓ is any real number.
If x = 0 and y 6= 0, then we may choose ✓ to be either ⇡2 or
equivalents) depending on whether y > 0 or y < 0, respectively.
3⇡
2
(or their
Now, suppose x 6= 0. From the boxed definition again on page 142, we know
that
y
tan ✓ = ,
x
where ✓ is an angle in standard position whose terminal side passes through the
point (x, y).
Conversion from Rectangular to Polar Coordinates
! (r, ✓) = (0, ✓), ✓ 2 R
8
< (y, ⇡ )
if y > 0
2
! (r, ✓) =
: (|y|, 3⇡ ) if y < 0
(x, y) = (0, 0)
(0, y)
y6=0
(x, 0)
x6=0
2
8
< (x, 0) if x > 0
! (r, ✓) =
: (|x|, ⇡) if x < 0
267
! (r, ✓)
(x, y)
x6=0, y6=0
r=
p
x2 + y 2
tan ✓ =
y
x
✓ same quadrant as (x, y)
Given one rectangular coordinate representation (x, y), there are
many polar coordinate representations (r, ✓) corresponding to it. The
above computations just give one of them.
Example 3.9.4. Convert each Cartesian coordinates to polar coordinates (r, ✓),
where r 0.
(4) (6, 2)
(1) ( 4, 0)
(2) (4, 4)
(3) ( 3,
(5) ( 3, 6)
p
3)
(6) ( 12, 8)
Solution. (1) ( 4, 0)
! (4, ⇡)
Teaching Notes
Plotting the points
on the
superimposed
Cartesian and
polar planes is a
quicker approach
in converting
rectangular
coordinates to
polar.
(2) The point (4, 4) is in QI.
p
p
p
r = x2 + y 2 = 4 2 + 4 2 = 4 2
tan ✓ =
y
x
=
4
4
= 1 =)
p ⇡
4 2, 4
✓=
!
p
(3) ( 3,
3) in QIII
q
p
p
r = ( 3)2 + (
3)2 = 2 3
(4, 4)
p
p
tan ✓ = 33 = 33 =) ✓ =
p
p
( 3,
3) ! 2 3, 7⇡
6
(4) (6, 2) in QIV
p
p
r = 62 + ( 2)2 = 2 10
tan ✓ =
(6, 2)
2
6
=
!
1
3
=)
p
2 10, tan
⇡
4
7⇡
6
✓ = tan
1
1
1
3
(5) ( 3, 6) in QII
p
p
r = ( 3)2 + 62 = 3 5
tan ✓ =
( 3, 6)
6
3
=
!
1
3
2 =) ✓ = ⇡ + tan 1 ( 2)
p
3 5, ⇡ + tan 1 ( 2)
268
Teaching Notes
Recall that
tan 1 ( 2) is in
QIV.
Teaching Notes
We may also use
✓ = tan 1 23 ⇡.
(6) ( 12, 8) in QIII
p
p
r = ( 12)2 + ( 8)2 = 4 13
tan ✓ =
8
12
=
2
3
!
( 12, 8)
=) ✓ = ⇡ + tan
p
4 13, ⇡ + tan 1 23
1 2
3
2
Seatwork/Homework 3.9.2
1. Convert each polar coordinates to Cartesian coordinates.
(a) (2, ⇡)
(b)
(c)
(d)
(e)
Answer: ( 2, 0)
p p
Answer: ( 2 2, 2 2)
(4, 3⇡
)
4
(6, 3⇡
)
2
( 2, 2⇡
)
3
( 4, 7⇡
)
6
(f) ( 3,
Answer: (0, 6)
p
Answer: (2,
3)
p
Answer: (2 3, 2)
p
Answer: ( 32 , 3 2 3 )
2⇡
)
3
(g) (1, sin 1 (
(h) ( 2, tan
p
Answer: ( 2 3 2 ,
1
))
3
1 4
)
3
Answer: (
2. Convert each Cartesian coordinates to polar coordinates (r, ✓), where r
6
,
5
1
)
3
8
)
5
0.
Answer: (6, ⇡2 )
p
Answer: (3 2, ⇡4 )
(a) (0, 6)
(b) (3, 3)
p
(c) ( 3 3, 3)
p
(d) ( 1,
3)
Answer: (6, 5⇡
)
6
Answer: (2, 4⇡
)
3
p
Answer: ( 17, tan 1 4)
p
Answer: (2 5, ⇡ + tan 1 ( 2))
p
Answer: (2 10, ⇡ + tan 1 13 )
p
Answer: ( 2, 3⇡
)
4
(e) (1, 4)
(f) ( 2, 4)
(g) ( 6, 2)
(h) ( 1, 1)
3.9.3. Basic Polar Graphs and Applications
From the preceding session, we learned how to convert polar coordinates of a
point to rectangular and vice versa using the following conversion formulas:
y
r2 = x2 + y 2 , tan ✓ = , x = r cos ✓, and y = r sin ✓.
x
Because a graph is composed of points, we can identify the graphs of some equations in terms of r and ✓.
269
Graph of a Polar Equation
The polar graph of an equation involving r and ✓ is the set of all
points with polar coordinates (r, ✓) that satisfy the equation.
As a quick illustration, the polar graph of the equation r = 2 2 sin ✓ consists
of all points (r, ✓) that satisfy the equation. Some of these points are (2, 0), (1, ⇡6 ),
(0, ⇡2 ), (2, ⇡), and (4, 3⇡
).
2
Example 3.9.5. Identify the polar graph of r = 2, and sketch its graph in the
polar plane.
Solution. Squaring the equation, we get r2 = 4. Because r2 = x2 + y 2 , we have
x2 + y 2 = 4, which is a circle of radius 2 and with center at the origin. Therefore,
the graph of r = 2 is a circle of radius 2 with center at the pole, as shown below.
In the previous example, instead of using the conversion formula r2 = x2 + y 2 ,
we may also identify the graph of r = 2 by observing that its graph consists of
points (2, ✓) for all ✓. In other words, the graph consists of all points with radial
distance 2 from the pole as ✓ rotates around the polar plane. Therefore, the
graph of r = 2 is indeed a circle of radius 2 as shown.
Example 3.9.6. Identify and sketch the polar graph of ✓ =
5⇡
.
4
Solution. The graph of ✓ = 5⇡
consists of all points (r, 5⇡
) for r 2 R. If
4
4
5⇡
r > 0, then points (r, 4 ) determine a ray from the pole with angle 5⇡
from
4
the polar axis. If r = 0, then (0, 5⇡
)
is
the
pole.
If
r
<
0,
then
the
points
4
(r, 5⇡
)
determine
a
ray
in
opposite
direction
to that of r > 0. Therefore, the
4
270
graph of ✓ = 5⇡
is a line passing through the pole and with angle
4
respect to the polar axis, as shown below.
5⇡
4
with
Example 3.9.7. Identify (and describe) the graph of the equation r = 4 sin ✓.
Solution.
r
r2
x2 + y 2
x2 + y 2 4y
x2 + (y 2)2
= 4 sin ✓
= 4r sin ✓
= 4y
=0
=4
Therefore, the graph of r = 4 sin ✓ is a circle of radius 2 and with center at (2, ⇡2 ).
271
? Example 3.9.8. Sketch the graph of r = 2
2 sin ✓.
Solution. We construct a table of values.
x
0
⇡
6
r
2
1
x
7⇡
6
r
3
⇡
4
⇡
3
⇡
2
0.59 0.27
5⇡
4
4⇡
3
3.41 3.73
0
3⇡
2
4
2⇡
3
3⇡
4
5⇡
6
⇡
1
2
0.27 0.59
5⇡
3
7⇡
4
11⇡
6
2⇡
3
2
3.73 3.41
This heart-shaped curve is called a cardioid.
2
? Example 3.9.9. The sound-pickup capability of a certain brand of microphone
is described by the polar equation r = 4 cos ✓, where |r| gives the sensitivity of
the microphone to a sound coming from an angle ✓ (in radians).
(1) Identify and sketch the graph of the polar equation.
(2) Sound coming from what angle ✓ 2 [0, ⇡] is the microphone most sensitive
to? Least sensitive?
Solution. (1)
r
r2
x2 + y 2
x2 + 4x + y 2
(x + 2)2 + y 2
= 4 cos ✓
= 4r cos ✓
= 4x
=0
=4
272
This is a circle of radius 2 and with center at (2, ⇡).
(2) We construct a table of values.
x
r
0
4
⇡
6
3.46
⇡
4
2.83
⇡
3
2
⇡
2
2⇡
3
0
2
3⇡
4
5⇡
6
2.83 3.46
⇡
4
From the table, the microphone is most sensitive to sounds coming from
angles ✓ = 0 and ✓ = ⇡, and least sensitive to sound coming from an angle
✓ = ⇡2 .
2
Seatwork/Homework 3.9.3
1. Identify (and describe) the graph of each polar equation.
(a) ✓ = 2⇡
3
Answer: Line passing through the pole with angle
polar axis
2⇡
3
with respect to the
(b) r = 3
Answer: Circle with center at the pole and of radius 3
(c) r = 2 sin ✓
Answer: Circle of radius 1 and with center at (1, ⇡2 )
273
(d) r = 3 cos ✓
Answer: Circle of radius 1.5 and with center at (1.5, 0)
(e) r = 2 + 2 cos ✓
Answer: A cardioid
2. Sketch the graph of each polar equation.
(a) r =
3
(b) r =
2 sin ✓
274
(c) r = 2 + 2 sin ✓
(d) r = 4 cos ✓
3. The sound-pickup capability of a certain brand of microphone is described by
the polar equation
r = 1.5(1 + cos ✓),
275
where |r| gives the sensitivity of the microphone of a sound coming from an
angle ✓ (in radians).
(a) Identify and sketch the graph of the polar equation.
Answer: A cardioid
(b) Sound coming from what angle ✓ 2 [0, 2⇡) is the microphone most sensitive to? Least sensitive?
Answer: Most sensitive at ✓ = 0; least sensitive at ✓ = ⇡
Exercises 3.9
1. Plot the following points in one polar plane:
A( 2, ⇡2 )
D( 3,
G(4,
2⇡
)
3
8⇡
)
3
B(1,
7⇡
)
3
C( 2, ⇡4 )
E(4,
⇡
)
4
F ( 3, 7⇡
)
12
H( 2,
Answer:
276
11⇡
)
12
I(1, 15⇡
)
2
2. Give the polar coordinates (r, ✓) with indicated properties that represent the
same point as the given polar coordinates.
(a)
(b)
(c)
(d)
(e)
( 3, 2⇡); r > 0, 0 < ✓  2⇡
(10, 4⇡
); r < 0, 0 < ✓  2⇡
3
3⇡
( 4, 2 ); r < 0, 2⇡ < ✓  0
( 1, ⇡); r < 0, 0 < ✓  2⇡
( 2, cos 1 23 ); r > 0, 2⇡ < ✓  0
Answer: (3, ⇡)
Answer: ( 10, 5⇡
)
3
⇡
Answer: ( 4, 2 )
Answer: ( 1, ⇡)
Answer: (2, ⇡ + cos 1 23 )
3. Convert each polar coordinates to Cartesian coordinates.
(a) (4, ⇡)
(b) ( 4, 7⇡
)
4
Answer: ( 4, 0)
p
p
Answer: (2 2, 2 2)
p
Answer: ( 1,
3)
Answer: (5, 0)
p
Answer: (4 3, 4)
(c) (2, 2⇡
)
3
(d) ( 5, 3⇡)
(e) (8, 11⇡
)
6
4. Convert each Cartesian coordinates to polar coordinates (r, ✓), where r
and 0  ✓2⇡.
0
Answer: (6, 3⇡
)
2
p 5⇡
Answer: (5 2, 4 )
(a) (0, 6)
(b) ( 5, 5)
p
Answer: (2 10, ⇡ + tan 1 ( 3))
p
Answer: ( 17, 2⇡ + tan 1 ( 4))
(c) ( 2, 6)
(d) (1, 4)
p
(e) (1,
3)
Answer: (2, 5⇡
)
3
5. Identify and sketch the graph of each polar equation.
(a) ✓ = ⇡3
Answer: A line passing through the pole and with angle
to the polar axis
277
⇡
3
with respect
(b) r = 3 sin ✓
Answer: A circle tangent to the x-axis with center at (0, 1.5)
(c) r = cos ✓
Answer: A circle tangent to the y-axis with center at (0.5, 0)
(d) r = 2 2 cos ✓
Answer: A cardioid
278
(e) r = 1 + sin ✓
Answer: A cardioid
6. The graph of the polar equation r = 2 cos 2✓ is a four-petaled rose. Sketch its
graph.
Answer:
? 7. A comet travels on an elliptical orbit that can be described by the polar
equation
1.164
r=
1 + 0.967 sin ✓
with respect to the sun at the pole. Find the closest distance between the sun
and the comet.
Answer: Closest distance occurs when sin ✓ = 1, so r =
1.164
1.967
⇡ 0.59 units.
? 8. Polar equations are also used by scientists and engineers to model motion of
satellites orbiting the Earth. One satellite follows the path
r=
36210
,
6 cos ✓
279
where r is the distance in kilometers between the center of the Earth and the
satellite, and ✓ is the angular measurement in radians with respect to a fixed
predetermined axis.
(a) At what value of ✓ 2 [0, 2⇡) is the satellite closest to Earth, and what is
the closest distance?
Answer: The satellite is closest to Earth when cos ✓ = 1, and this occurs
when ✓ = ⇡. The closest distance is, therefore, r = 636210
⇡ 5182.86
( 1)
kilometers.
(b) How far away from Earth can the satellite reach?
Answer: The satellite can reach as far as r = 36210
⇡ 7242 km away from
6 1
the Earth.
9. The graph of the polar equation
15
3 2 cos ✓
is a conic section. Identify and find its equation in rectangular coordinate
system.
Answer: Ellipse, 5x2 + 9y 2 60x 225 = 0
r=
Solution
3r 2r cos ✓ = 15
p
x
3 x2 + y 2 2r · = 15
r
⇣ p
⌘2
3 x2 + y 2 = (2x + 15)2
5x2 + 9y 2
60x
225 = 0 an ellipse
10. The graph of the polar equation
6
3 + 3 sin ✓
is a parabola. Find its equation in rectangular coordinate system.
r=
Answer: y =
1 2
x
4
+1
11. For what values of ✓ 2 [0, 2⇡) will the graphs of r = 4 cos ✓ and r cos ✓ = 1
intersect?
Answer: ⇡3 , 2⇡
, 4⇡
, 5⇡
3
3
3
12. Convert the polar equation
r=
2 sin 2✓
✓ sin3 ✓
cos3
Answer: x3 = y 3 + 4xy
into Cartesian equation.
4
280
References
[1] R.N. Aufmann, V.C. Barker, and R.D. Nation, College Trigonometry, Houghton
Mi✏in Company, 2008.
[2] E.A. Cabral, M.L.A.N. De Las Peñas, E.P. De Lara-Tuprio, F.F. Francisco,
I.J.L. Garces, R.M. Marcelo, and J.F. Sarmiento, Precalculus, Ateneo de
Manila University Press, 2010.
[3] R. Larson, Precalculus with Limits, Brooks/Cole, Cengage Learning, 2014.
[4] L. Leithold, College Algebra and Trigonometry, Addison Wesley Longman
Inc., 1989, reprinted by Pearson Education Asia Pte. Ltd., 2002.
[5] M.L. Lial, J. Hornsby, and D.I. Schneider, College Algebra and Trigonometry
and Precalculus, Addison-Wesley Educational Publisher, Inc., 2001.
[6] J. Stewart, L. Redlin, and S. Watson, Precalculus: Mathematics for Calculus,
Brooks/Cole, Cengage Learning, 2012.
[7] M. Sullivan, Algebra & Trigonometry, Pearson Education, Inc., 2012.
[8] C. Young, Algebra and Trigonometry, John Wiley & Sons, Inc., 2013.
281
Biographical Notes
PRECALCULUS TEAM
Leader
Ian June L. Garces, Ph.D.
Dr. Garces is currently an Associate Professor at the Ateneo de Manila University. He finished his Bachelor of Science in Mathematics at Mindanao State
University in Marawi City as Magna Cum Laude. He was granted a straight program to the Doctor of Philosophy in Mathematics degree at the Ateneo de Manila
University as a DOST scholar. He was the leader and head coach of the Philippine
Team to the International Mathematical Olympiad for three years, the convenor
of the Math Learning Area Team of the K-12 Program of the Department of Education, and a member of the Technical Panel of the Metrobank-MTAP-DepEd
Math Challenge. He has published books, articles, and research papers in integration theory, graph theory, mathematical problem solving and recreation,
and mathematical competitions. He was also the Team Leader of the Grade 11
Precalculus Learning Manual commissioned by the Department of Education.
Writers
Richard B. Eden, Ph.D.
Dr. Eden is currently an Assistant Professor at the Ateneo de Manila University,
where he has been teaching Probability and Calculus at the undergraduate and
graduate levels. He finished his bachelor’s degree in Mathematics at Ateneo de
Manila as Cum Laude, and his master’s and doctorate degrees in Mathematics
at Purdue University in Indiana, USA. He also presented papers at Purdue University, University of Kansas, and Duke University - focusing on Statistics, Malliavin Calculus, Stochastic Analysis, and Probability. As a mathematical problem
solver, he is currently involved in the Philippine Mathematical Olympiad and the
International Mathematical Olympiad.
Glenn Rey A. Estrada, M.S.
Mr. Estrada is teaching Mathematics at the Leyte Normal University (LNU).
He finished Cum Laude with a degree of Elementary Education and Bachelor in
Secondary Education major in Mathematics at the LNU. He finished his master’s
degree in Mathematics at the University of San Carlos as a CHED scholar. Mr.
Estrada is a member of international and national mathematics organizations
such as the Asian Qualitative Research Association, Mathematical Society of the
Philippines, and the Philippine Council of Mathematics Teacher Educators. An
upholder of the K-12 education, Mr. Estrada has also served as Trainer in several
teacher training workshops on K-12 Basic Education Curriculum.
282
Mark-Anthony J. Vidallo
Mr. Vidallo is a Mathematics teacher and trainer at the Makati Science High
School. He graduated magna cum laude with a degree in Secondary Education
major in Mathematics at the University of Makati and is pursuing graduate
studies at the Philippine Normal University. He received citations including the
MTAP Exemplary Secondary Mathematics Teacher Award and won first place
at the 1st Casio-DepEd NCR Mathematics Teachers Olympics. Mr. Vidallo coauthored textbooks on Calculus, Analytic Geometry, and Math Ed magazines
for junior high school. He is a trainer for mathematics competitions such as the
Philippine Mathematical Olympiad, Metrobank-MTAP-DepEd Math Challenge,
and Australian Mathematics Competition.
Jerico B. Bacani, Ph.D.
Dr. Bacani is currently the Chairman of the Department of Mathematics and
Computer Science at University of the Philippines Baguio, and the Regional Coordinator of the Philippine Mathematical Olympiad. He finished his doctorate
degree at Karl-Franzens Universität Graz in Austria, through the Austria’s Science and Technology Grant for Southeast Asia. He finished his master’s degree
in Mathematics at UP Baguio as a CHED scholar, and his bachelor’s degree in
Mathematics at UP Diliman as a DOST scholar. Besides publishing numerous
academic articles both locally and internationally, he also wrote learning resources
for Grade 7 students and teaching modules for Grade 11 Mathematics teachers.
Flordeliza F. Francisco, Ph.D.
Dr. Francisco is a current member of the CHED Technical Panel for General
Education and the DepEd Mathematics Group of Consultants for K-12 Basic
Education Program. She is an Assistant Professor at the Ateneo de Manila University, where she is teaching Advanced Calculus, Real Analysis, Finite Mathematics, and related courses. She completed her doctorate degree in Mathematics
and bachelor’s degree in Mathematics both at the Ateneo. Dr. Francisco is also
member of the Philippine Council of Mathematics Teacher Educators. She coauthored and edited books on Mathematics, including Precalculus (Ateneo Press)
and Math Challenge Questions (Anvil Publications).
Technical Editors
Reginaldo M. Marcelo, Ph.D.
Dr. Marcelo is currently a member of the CHED Technical Panel for Science and
Mathematics. He is an Assistant Professor at the Ateneo de Manila University.
He finished his doctorate degree in Mathematics at Sophia University through a
scholarship from the Japan’s Ministry of Education and his master’s degree in
Mathematics at the University of the Philippines Diliman. He graduated with a
bachelor’s degree in Mathematics at the Ateneo as a DOST scholar. Dr. Marcelo
is a member of the Mathematical Society of the Philippines and the Southeast
Asian Mathematics Society, and co-authored and edited publications such as
Precalculus and College Algebra textbooks.
283
Maria Alva Q. Aberin, Ph.D.
Dr. Aberin is an Assistant Professor at the Ateneo de Manila University, where
she is teaching undergraduate and graduate Mathematics courses. She has also
served as National Trainer for a number of teacher training activities initiated by
the DepEd. Dr. Aberin was Vice President of the National Board of Directors for
the Philippine Council of Mathematics Teacher Educators. She completed her
doctorate degree in Mathematics Education, her master’s and bachelor’s degrees
in Mathematics at the University of the Philippines Diliman.
Copyreader
Naomi L. Tupas
Ms. Tupas holds various technical writing and editing positions for both private
and public institutions. She is the technical writer at the provincial government
of Camarines Sur, an editor at Operation Blessing Foundation Philippines Inc., a
copywriter at Coastal Training Technologies, Inc., and the editor-in-chief for the
Marketing Communication Office at the De La Salle University Manila. She is
currently taking Master of Arts in Language Education at the University of the
Philippines Diliman, where she finished her degree in Bachelor of Arts in English,
major in Creative Writing.
284