Contentsfaculty.essex.edu/~bannon/precalc/m120.s/notes/mth... · published by McGraw Hill Education, 2013 (ISBN:978-0-07-179559-3).The book is un-der twenty dollars and used copies

Precalculus IIMTH-120

Essex County CollegeDivision of Mathematics

Ron Bannon’sCourse Notes

Contents

0 Introductory Remarks 50.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50.2 Course Web Page . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50.3 This Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50.4 Syllabus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

20 Trigonometric Functions 920.1 Angles and Rotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 920.2 Radian Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 920.3 Two Special Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1120.4 The Circular Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1120.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1320.6 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1820.7 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

21 Graphs of Trigonometric Functions 2321.1 Graphing Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . 2321.2 Calculator Usage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2321.3 Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2721.4 Conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2721.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2721.6 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3521.7 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

22 Angles 4222.1 Arc Length and Radian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4222.2 Angular and Linear Speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4222.3 Area of a Sector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4222.4 Small Details . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4322.5 Degrees, Minutes, and Seconds . . . . . . . . . . . . . . . . . . . . . . . . . . 4322.6 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4322.7 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4622.8 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

23 Trigonometric Identities and Equations 4923.1 Review of The Circular Functions . . . . . . . . . . . . . . . . . . . . . . . . 49

23.1.1 Cautionary Note . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5023.1.2 Fundamental Trigonometric Identities . . . . . . . . . . . . . . . . . . 50

23.2 Simplifying Trigonometric Expressions . . . . . . . . . . . . . . . . . . . . . 5123.3 Proving/Verifying Trigonometric Identities . . . . . . . . . . . . . . . . . . . 5123.4 Solving Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . . 5123.5 Examples: Simplification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

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23.6 Examples: Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5623.7 Examples: Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5823.8 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6323.9 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

23.9.1 Homework: Simplifications . . . . . . . . . . . . . . . . . . . . . . . . 6423.9.2 Homework: Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . 6723.9.3 Homework: Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 68

24 Sum, Difference, Multiple, & Half-Angle Formulas 7324.1 Sum and Difference Identities . . . . . . . . . . . . . . . . . . . . . . . . . . 7324.2 Double Angle Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7424.3 Half-Angle Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7424.4 Product-Sum Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7524.5 Sum-to-Product Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7524.6 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7524.7 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7924.8 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

25 Inverse Trigonometric Functions 8425.1 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8425.2 The Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . 85

25.2.1 The Inverse Sine Function . . . . . . . . . . . . . . . . . . . . . . . . 8625.2.2 The Inverse Cosine Function . . . . . . . . . . . . . . . . . . . . . . . 8925.2.3 The Inverse Tangent Function . . . . . . . . . . . . . . . . . . . . . . 90

25.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9225.4 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9525.5 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

26 Triangles 9826.1 Introductory Information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9826.2 Right Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9826.3 Law of Sines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9826.4 Law of Cosine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

26.4.1 Right Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9926.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10026.6 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10326.7 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

27 Vectors 10627.1 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10627.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10727.3 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11027.4 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

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28 Polar Coordinates; Parametric Equations 11328.1 Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

28.1.1 Technology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11328.1.2 Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11328.1.3 Graphing a Parametric Equation . . . . . . . . . . . . . . . . . . . . 11528.1.4 Easily Tricked? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11628.1.5 Using Software . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

28.2 Polar Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11828.2.1 Cartesian Relationship . . . . . . . . . . . . . . . . . . . . . . . . . . 11828.2.2 Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

28.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12328.4 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12828.5 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

29 Trigonometric Form of Complex Numbers 13529.1 DeMoivre’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13529.2 Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13629.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13729.4 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14129.5 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

37 Loci; Parabolas 14637.1 Loci . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14637.2 Parabola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14737.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14937.4 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15337.5 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

38 Elipses and Hyperbolas 15738.1 The Ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15738.2 Hyperbola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15838.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16138.4 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16838.5 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168

39 Rotation of Axes 17539.1 Rotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

39.1.1 Rotational Invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . 17839.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17939.3 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18239.4 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

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41 Sequences and Series 18741.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18741.2 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18741.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18841.4 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19241.5 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

42 The Principle of Mathematical Induction 19642.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19642.2 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19842.3 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

43 Special Sequences and Series 20043.1 Arithmetic Sequences and Series . . . . . . . . . . . . . . . . . . . . . . . . . 20043.2 Geometric Sequences and Series . . . . . . . . . . . . . . . . . . . . . . . . . 20043.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20143.4 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20543.5 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

44 Binomial Theorem 20944.1 Binomial Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20944.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21044.3 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21244.4 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212

1 Final Exam Review 215

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0 Introductory Remarks

Errors in this document should be reported to Ron Bannon.

[email protected]

This document may be shared with MTH 120 students and instructors only.

0.1 Review

It is absolutely imperative that you have a fairly good grasp of the mathematics covered inprior courses including Precalculus I (MTH 119). You may visit

http://mth119.mathography.org/

to see what materials are covered in MTH 119. Although I may spend some time in classdiscussing material covered in prior mathematics courses, you can not expect that this pre-requisite materials will be covered sufficiently well enough to remediate inadequate prepara-tion. Being prepared is the first step!

0.2 Course Web Page

Please visit http://mth120.mathography.org/ to learn more about how this course will berun. On this website you will be able to view the course outline, section specific syllabi, thisguide1, and information about accessing online homework. Visit this page often!

0.3 This Guide

This guide is being provided as a way to structure MTH 120 across sections and semesters.This guide is not a textbook and it should not be used as such—it just provides structurefor teachers and students.

1. Please come to all classes, these notes are not to be used as an excuse not to attend.You should make every effort to attend all scheduled classes. The notes that follow arenot self-contained and you will need to attend class in order to make sense out of thecontent that follows. The notes just give structure to the classroom lectures, and willkeep you on track with your assignments.

2. These notes will not be read as a script, but teachers/students are encouraged to followthe content of these notes. Your teacher may or may not do all the problems that arein the notes. You should make every attempt to follow what your teacher is doing ineach and every class.

1This is what I’d like my notes to look like If I were a student in this class. You should make someattempt to write your own notes, or at least annotate mine. Please try to follow what is being done in class,and do not do the homework until you understand what was done in class.

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0.4 Syllabus

Essex County CollegeMathematics and Physics Division

Syllabus for MTH 120 §001 — Pre-calculus IISpring 2015 Class Syllabus

Lecturer: Ron Bannon

Office: 2210Phone: 973-877-1886Email: [email protected]

Course Website: http://mth120.mathography.org/

Regular Office Hours: Monday, Wednesday & Friday 2:30 p.m.–3:40 p.m.;Wednesday & Friday 7:50 a.m.–8:20 a.m.;

Appointment Office Hours: Wednesday & Friday 5:15 p.m.–6:35 p.m.;Class Meetings: Monday, Wednesday & Friday 8:30 a.m.–9:50 a.m.,

Room 2101A;First Class: Monday, January 12, 2015;Last Class: Monday, April 27, 2015.

• Course Description: This course, along with MTH 119, prepares students for arigorous treatment of calculus. Topics covered include trigonometric functions; graph-ing trigonometric functions; inverse trigonometric functions; trigonometric identities;trigonometric equations; vectors; analytic geometry; polar coordinates; sequences andseries.

• Optional Materials: Textbook Schaum’s Precalculus, 3rd edition, by Fred Safier;published by McGraw Hill Education, 2013 (ISBN:978-0-07-179559-3).The book is un-der twenty dollars and used copies can be found online for less that ten dollars. Getthe book, it is a bargain, but it is not required.

• Required Materials: Students should have access to a graphics calculator that theyare comfortable using, a notebook, and either a pencil or pen. You may not use acell phone as a calculator. You may not share calculators.

• Course Prerequisites: Grade of “C” or better in MTH 119 or by placement.

• Course Prerequisites & Course Co-requisites: None.

• General Education Goals: MTH 120 is affirmed in the following General Educa-tion Foundation Category: Quantitative Knowledge and Skills. The correspondingGeneral Education Goal is as follows: Students will use appropriate mathematical andstatistical concepts and operations to interpret data and to solve problems.

• Course Goals: Upon successful completion of this course, students should be able todo the following:

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1. demonstrate knowledge of the fundamental concepts and theories from pre-calculusas outlined in this guide;

2. use appropriate technology, such as graphing calculators and computer software,effectively as a tool to solve such problems as those described in this guide.

• Methods of Instruction Instruction will consist of a combination of lectures, classdiscussion, individual study, and computer lab work.

• Class Requirements All students are required to:

1. Follow this guide as presented in class and do the homework problems in eachsection of this guide in a timely manner. Get help if you can not do the homeworkproblems.

2. Be an active participant in all classes. That is, you should behave as if you areinterested in learning the material.

3. Adhere to the in-class assessment schedule.2

• Grading Criteria

1. Online homework will be graded, and is worth up to 10% in bonus credit.3

2. Proctored questions (approximately 33 problems total), equally weighed, is worth75%. If you can follow what we’re doing in class and do the problems in this guideyou should have no problem doing these questions.

3. Cumulative final exam is worth 25%. A final review is provided at the end of thisguide.

Tentative grade breakdown:

A if average ≥ 94%;

B+ if 88% ≤ average < 94%;

B if 82% ≤ average < 88%;

C+ if 76% ≤ average < 82%;

C if 70% ≤ average < 76%;

D if 64% ≤ average < 70%;

F if average < 64%;

• Academic Integrity: Dishonesty disrupts the search for truth that is inherent in thelearning process and so devalues the purpose and the mission of the College. Academicdishonesty includes, but is not limited to, the following:

2I will give students advanced notice in class when an assessment is scheduled.3Information regarding online homework can be found on the course webpage: http://mth120.

mathography.org/.

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Plagiarism: The failure to acknowledge another writer’s words or ideas or to giveproper credit to sources of information.

Cheating: Knowingly obtaining or giving unauthorized information on any test/examor any other academic assignment.

Interference: Any interruption of the academic process that prevents others from theproper engagement in learning or teaching.

Fraud: Any act or instance of willful deceit or trickery.

Violations of academic integrity will be dealt with by imposing appropriate sanctions.Sanctions for acts of academic dishonesty could include the resubmission of an assign-ment, failure of the test/exam, failure in the course, probation, suspension from theCollege, and even expulsion from the College.

• Student Code of Conduct All students are expected to conduct themselves asresponsible and considerate adults who respect the rights of others. Disruptive behaviorwill not be tolerated. All students are also expected to attend and be on time allclass meetings. No cell phones or similar electronic devices are permitted in class.Please refer to the Essex County College student handbook, Lifeline, for more specificinformation about the College’s Code of Conduct and attendance requirements.

• Tentative Class Schedule The course will follow, in order, the entire content ofthis guide. That is, be prepared by having access to this guide during class-time.Electronic or printed copies of this guide are freely available on the course website:http://mth120.mathography.org/.

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20 Trigonometric Functions

20.1 Angles and Rotations

An angle is the union of two rays with a common endpoint called the vertex. In trigonom-

Figure 1: Label this diagram with bold text items!

etry we often think of an angle as a rotation, with the initial ray (called the initial side)positioned on the positive x-axis, and the terminal ray (called the terminal side) rotatedfrom the initial side. Counterclockwise rotations are positive angles, and clockwise rota-tions are negative angles. The angle formed is said to be in standard position.

0

Figure 2: Label this diagram with bold text items!

20.2 Radian Measure

Although degree is a common unit of measure of angles, it is inappropriate for much ofmathematics. The unit that is used in mathematics (actually it’s unit-less) is called radianand here is how it is usually defined: a unit of angle, that is the numerical value of 1, equalto an angle at the center of a circle whose arc length (indicated by s) is equal in length tothe radius (usually indicated by r). So one radian for any circle is roughly 57◦ (Figure 3,page 10). Another way to look at this measure, completely consistent with the above notion,

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0

Figure 3: The illustration of one radian.

is that the arc length, s, is equal to the product of the radius, r, and the angle theta, θ, inradians.4 You should note that both r and s have a unit (length), but that radian is unitless.That is, the relationship between these three variables is:

s = rθ.

Since we know that the circumference for any circle is 2πr, we can determine what theta isfor one complete revolution, or 360◦.

s = rθ

2πr = rθ

2π = θ

So here’s our first concrete relationship between degree, revolution, and radian measure.And you’ll need to know this by heart.

2π = 360◦ ⇒ π

180◦= 1 ⇒ 1 =

180◦

π

2π = 1 rev ⇒ 2π

1 rev= 1 ⇒ 1 =

1 rev

2π

Here we can use the unit, 1, to convert form degree (unit is ◦) to radian (no unit),

90◦ = 90◦ · 1 = 90◦ · π

180◦=π

2;

or for radian (no unit) to degree (unit is ◦),

3π

2=

3π

2· 1 =

3π

2· 180◦

π= 270◦.

4You were told this in grammar school, when your teachers said, C = 2πr.

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20.3 Two Special Triangles

For this class you should be very familiar with two simple triangles that were most likelycovered in a much more basic course.

45◦, 45◦, 90◦ Triangle: In class I will discuss the construction of this triangle, and it isabsolutely important that you learn this basic triangle.

30◦, 60◦, 90◦ Triangle: In class I will discuss the construction of this triangle, and it isabsolutely important that you learn this basic triangle.

30◦, 45◦, 60◦ Reference Angles: Once the above triangles are formed you should be ableto use these triangles to evaluate sixteen points (Figure 4, page 11) along the unitcircle—you should know the point and the angle for all sixteen positions, again we’lldo this together. Some teachers require their students to memorize this, but I believeit is more important to derive this using the above two triangles as references. Yes,you’ll need to use your head!

-2 -1 0 1 2

-1

1

Figure 4: Label 16 points, and their corresponding θ, on this unit circle!

20.4 The Circular Functions

At this point in your mathematical development, you should be well aware of the equation ofa circle that has radius r, and center (h, k). The form you most likely have seen (introducedin MTH 100) before is:

(x− h)2 + (y − k) = r2.

For our introduction to trigonometry, we will take the center to be the origin,5 hence theequation becomes:

x2 + y2 = r2.

5Both h and k become zero.

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The trigonometric (circular) functions are related to the coordinates along this circle as werotate the radius. Of course the radius does not change, but the x and y do as we rotate theradius through an angle, θ. This rotation will be with respect to the positive x axis: as werotate counter-clockwise we say that the radius forms an positive angle with respect to thepositive x-axis; and as we rotate clockwise we say that the radius6 forms a negative anglewith respect to the positive x-axis. Here I will refer to this angle as theta, θ. Here’s how thesix trigonometric functions are defined.

1. The sine function.

sin θ =y

r

2. The cosine function.

cos θ =x

r

3. The tangent function.

tan θ =y

x

4. The cotangent function.

cot θ =x

y

5. The secant function.

sec θ =r

x

6. The cosecant function.

csc θ =r

y

There are many simple thetas to choose from, and everyone should be familiar with degree,radian and revolution measure for these given thetas.7 Furthermore, you should be able torewrite these definitions using other variables. For example, the abscissa could be relabeledwith the variable, a, the ordinate with the variable, b, and the angle could be representedby the variable, t. This is what we would have, where r would now equal

√a2 + b2; just as

in our original definition r was equal to√x2 + y2.

6Some fix the radius to be one unit, hence the unit circle approach, but I prefer the more general casewhere r > 0.

7You should be familiar with the sixteen locations along the unit circle, and their associated angles, thatwas previously discussed.

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sin θ =b

r


cos θ =a

r


tan θ =b

a


cot θ =a

b


sec θ =r

a


csc θ =r

b

Nothing special, but please do take care of what the variables represent: abscissa, ordinateand angle8!

20.5 Examples

1. Take θ = 0◦. First draw the picture of a unit circle, then determine what x, y and r are.Use the result to evaluate sin θ, cos θ, and tan θ.

8On occasion, you will see the angle indicated by the letter x, and that should not be confused with thetypical variable choice for the abscissa.

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Solution: Work will be done in class. Make sure you take notes, or get notes from areliable classmate if you are not present during lecture.

θ = 0◦ = 0 rev = 0

x = 1

y = 0

r = 1

sin 0◦ = 0

cos 0◦ = 1

tan 0◦ = 0

2. Find the point (x, y) on the unit circle that corresponds to the real number t = π/4.Use the result to evaluate sin t, cos t, and tan t.9 Notice that I am using the letter t forthe angle, and you should become comfortable with a variety of letters often used forthe angle.


t =π

4= 45◦ =

1

8rev

x =1√2

y =1√2

r = 1

sinπ

4=

1√2

cosπ

4=

1√2

tanπ

4= 1

3. The point (3, −4) is on the terminal side of an angle in standard position.

• First draw the picture, then determine the values of x, y and r.

• Determine the quadrant that θ is in.

9After doing this problem you should be able to construct a 45◦ reference triangle. Do not memorize.

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• Evaluate all six trigonometric functions for this θ.


x = 3

y = −4

r = 5

θ ∈ IV

sin θ = −4

5

cos θ =3

5

tan θ = −4

3

cot θ = −3

4

sec θ =5

3

csc θ = −5

4

4. If θ = −5π/4, find the values of the of the six trigonometric functions of θ.


θ = −5π

4= −225◦ = −5

8rev

sin θ =1√2

cos θ = − 1√2

tan θ = −1

cot θ = −1

sec θ = −√

2

csc θ =√

2

5. Find the values of the six trigonometric functions of x with the given constraint. Youshould notice that x here refers to the angle and not the x-coordinate. You may want

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to return to the original definitions of the trigonometric functions and choose differentletters for the abscissa (a is often used) and ordinate (b is often used).

cosx = −11

12, 180◦ < x < 270θ


sinx = −√

23

12

cosx = −11

12

tanx =

√23

11

cotx =11√23

secx = −12

11

cscx = − 12√23

6. Take θ = 60◦. First draw the picture, then determine what x and y are. Use the resultto evaluate sin θ, cos θ, and tan θ.


θ = 60◦ =π

3=

1

6rev

x = 1

y =√

3

r = 2

sin θ =√

3/2

cos θ = 1/2

tan θ =√

3

7. Find two solutions of the equation.

csc θ = −2

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Give your answers in degrees (0◦ ≤ θ < 360◦) and radians (0 ≤ θ < 2π). Please do notuse a calculator!


210◦, 330◦, 7π/6, and 11π/6.

8. Find the point (x, y) on the unit circle that corresponds to the real number t = 7π/6.Use the result to evaluate csc t, sec t, and cot t.10


t = 7π/6 = 210◦ = 7/12 rev

x = −√

3/2

y = −1/2

r = 1

cot t =√

3

sec t = −2/√

3

csc t = −2

9. Use the value of the trigonometric function to evaluate the indicated functions.

cos t =7

8,

3π

2< t < 2π

(a) tan t =

Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.

tan t = −√

15/7

(b) sin t =


sin t = −√

15/8

10After doing this problem you should be able to construct a 30◦ reference triangle. Do not memorize.

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10. Take t = 120◦. First draw the picture, then determine what x, y and r are. Use theresult to evaluate csc t, sec t, and cot t.


t = 120◦ = 2π/3 = 1/3 rev

x = −1

y =√

3

r = 2

cot t = −1/√

3

sec t = −2

csc t = 2/√

3

20.6 Supplemental Readings

The material discussed in class covers parts of Chapter 20 of Schaum’s Precalculus. If, forwhatever reason, you missed class or just need additional time with this material, I suggestthat you read over Chapter 20 of Schaum’s Precalculus, and work the Solved Problems beforeattempting the homework assignment.

20.7 Homework

Again, you may need to review Chapter 20 of Schaum’s Precalculus if you are having troublefollowing what we are doing in class, or you have missed the lecture material. It is yourresponsibility to master this material, and you will need to solve all problems in this section.Get help if you can not do these problems.

Directions: Answer each of the following questions. Partial answers are provided, butno work is being provided as a guide. Yes, you need to be able to do these problems and youwill be tested. You need to get help if you cannot do these problems exactly as was done inclass!

1. Take θ = 45◦. First draw the picture, then determine what x, y and r are. Use theresult to evaluate sin θ, cos θ, and tan θ.

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Solution:

θ = 45◦ = π/4 = 1/8 rev

x = 1/√

2

y = 1/√

2

r = 1

sin θ = 1/√

2

cos θ = 1/√

2

tan θ = 1

2. Take θ = 270◦. First draw the picture of a unit circle, then determine what x, y, and rare. Use the result to evaluate sin θ, cos θ, and tan θ.

Solution:

θ = 270◦ = 3π/2 = 3/4 rev

x = 0

y = −1

r = 1

sin θ = −1

cos θ = 0

tan θ = undefined

3. If θ = 5π/3, find the values of the of the six trigonometric functions of θ.


θ = 300◦ = 5π/3 = 5/6 rev

sin θ = −√

3

2

cos θ =1

2

tan θ = −√

3

cot θ = − 1√3

sec θ = 2

csc θ = − 2√3

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4. The point (−1, 2) is on the terminal side of an angle in standard position.

• First draw the picture, then determine the values of x, y and r.

• Determine the quadrant that θ is in.

• Evaluate all six trigonometric functions for this θ.

Solution:

x = −1

y = 2

r =√

5

θ ∈ II

sin θ =2√5

cos θ = − 1√5

tan θ = −2

cot θ = −1

2

sec θ = −√

5

csc θ =

√5

2

5. The terminal side of θ lies on a given line in the specified quadrant. Find the values ofthe six trigonometric functions of θ by finding a point on the line.

y = −3

2x, IV

Solution:

sin θ = − 3√13

cos θ =2√13

tan θ = −3

2

cot θ = −2

3

sec θ =

√13

2

csc θ = −√

13

3

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6. Find the point (x, y) on the unit circle that corresponds to the real number t = −π.Use the result to evaluate sin t, cos t, and tan t.

Solution:

x = −1

y = 0

r = 1

sin t = 0

cos t = −1

tan t = 0

7. Use the value of the trigonometric function to evaluate the indicated functions.

sin t = −4

9, π < t <

3π

2

(a) tan t =

Solution: tan t = 4/√

65

(b) cos t =

Solution: cos t = −√

65/9

8. Find two solutions of the equation.

sec θ = −2

Give your answers in degrees (0◦ ≤ θ < 360◦) and radians (0 ≤ θ < 2π). Please do notuse a calculator!


120◦, 240◦, 2π/3, and 4π/3.

9. Take t = 135◦. First draw the picture, then determine what x, y and r are. Use theresult to evaluate csc t, sec t, and cot t.

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t = 135◦ = 3π/4 = 3/8 rev

x = −1

y = 1

r =√

2

cot t = −1

sec t = −√

2

csc t =√

2

Page 22 of 238




21 Graphs of Trigonometric Functions

21.1 Graphing Trigonometric Functions

The definitions of the trigonometric functions will not change, but now we will need tovisualize many thetas that aren’t so nice.11 Graphs are being provided as a guide and wewill label five key points/characteristics of each—try to use what you already know to followwhat we’re now doing in class. Here’s some questions that you should be able to easilyanswer now.

1. Are there any angles where sine (Figure 5, page 24) and cosine (Figure 6, page 24) areundefined? If so, what are they?

2. What is the domain and range of the sine and cosine?

3. Are there any angles where tangent (Figure 7, page 25) and cotangent (Figure 8, page 25)are undefined? If so, what are they?

4. What is the domain and range of the tangent and cotangent?

5. Are there any angles where secant (Figure 9, page 26) and cosecant (Figure 10, page 26)are undefined? If so, what are they?

6. What is the domain and range of the cosecant and secant?

21.2 Calculator Usage

You should also be aware that the angle theta can be given in various ways, and yourcalculator has at least two modes: radian and degree. Make sure that you know how to moveback and forth between radian and degree mode on your calculator—you should always be inradian mode when graphing trigonometric functions. Clearly you should see the sine, cosineand tangent functions on your calculator. However you probably don’t see cotangent, secantand cosecant on your calculator, but you should know that these three functions are relatedto sine, cosine and tangent.

cot θ =1

tan θsec θ =

1

cos θsec θ =

1

sin θ

By now you should be able to use your calculator to graph a variety of functions, and muchof what you learned in MTH 119 regarding your calculator’s use will be applicable here. Youshould, of course, now be able to graph all six trigonometric functions on your calculatorand the graphs should look similar to the ones being presented here.

11Any angle that is co-terminal with the coordinate axis, or those angles whose reference is 30◦, 45◦, or60◦ are considered nice.

Page 23 of 238




0

Figure 5: Partial graph of f (x) = sin x.

0

Figure 6: Partial graph of f (x) = cos x.

Page 24 of 238




0

Figure 7: Partial graph of f (x) = tan x.

0

Figure 8: Partial graph of f (x) = cot x.

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0

Figure 9: Partial graph of f (x) = sec x.

0

Figure 10: Partial graph of f (x) = cscx.

Page 26 of 238




21.3 Graphing

I hope you recall the topic of translation from MTH 119, where you had many simple graphsand you were asked to draw a related graph using simple translation rules. Methods mayvary, but you nonetheless were given a parent, f (x), and then asked to draw the child,A · f (Bx± C)±D. In essence, we will need to translate the five key points/characteristicsof each trigonometric function that were already discussed.

21.4 Conventions

Words may initially interfere with understanding, so let’s not dwell on the following. How-ever, after today you should be able to have an intuitive idea of the following.

Phase Shift: Horizontal shift for a periodic function. It is often hard to see a phase shift ina periodic function, and the term’s definition is not universally agreed upon. Basically,you should have some idea that waves can be shifted left or right, but a left shift canalso be achieved by a right shift, and vice versa.

Amplitude: Half the difference between the minimum and maximum values of the range.Only periodic functions with a bounded range have an amplitude. Essentially, ampli-tude is the radius of the range.

Period of a Periodic Function: The horizontal distance required for the graph of a pe-riodic function to complete one cycle. Formally, a function f is periodic if there existsa number p such that f (x+ p) = f (x) for all x. The smallest possible value of p isthe period. The reciprocal of period is called frequency.

21.5 Examples

1. Graph at least two cycles and find the period and amplitude.

y = −2 sinx

3


Period is 6π, amplitude is 2. You need to label five points on one cycle of the providedgraph (Figure 11, page 28). The five key points are: (0, 0); (3π/2, −2); (3π, 0);(9π/2, 2); (6π, 0).


y = −2 cos(

2x− π

2

)

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0 4 8 12 16 20 24 28 32 36 40

-1.6

-0.8

0.8

1.6

Figure 11: Partial graph of y = −2 sin x3.


Period is π, amplitude is 2. You need to label five points on one cycle of the providedgraph (Figure 12, page 29). The five key points are: (π/4, −2); (π/2, 0); (3π/4, 2);(π, 0); (5π/4, −2).


y = −2 + 3 sin(

2πx− π

4

)


Period is 1, amplitude is 3. You need to label five points on one cycle of the providedgraph (Figure 13, page 29). The five key points are: (1/8, −2); (3/8, 1); (5/8, −2);(7/8, −5); (9/8, −2).


f (x) = −3 cos (π − 2x) + 2

Page 28 of 238




-0.8 0 0.8 1.6 2.4 3.2 4 4.8 5.6

-1.6

-0.8

0.8

1.6

Figure 12: Partial graph of y = −2 cos(2x− π

2

).

-1 -0.5 0 0.5 1 1.5 2

-5

-4

-3

-2

-1

1

Figure 13: Partial graph of y = −2 + 3 sin(2πx− π

4

).

Page 29 of 238





Period is π, amplitude is 3. You need to label five points on one cycle of the providedgraph (Figure 14, page 30). The five key points are: (π/2, −1); (π/4, 2); (0, 5);(−π/4, 2); (−π/2, −1).

0 0.8 1.6 2.4 3.2 4 4.8 5.6 6.4

1

2

3

4

5

Figure 14: Partial graph of f (x) = −3 cos (π − 2x) + 2.

5. Graph at least two cycles.

y = −3 tan 2x


The period is π/2. You need to label five characteristics, three points and two asymp-totes, on one cycle of the provided graph (Figure 15, page 31). The five key character-istics are: x = −π/4; (−π/8, 3); (0, 0); (π/8, −3); x = π/4.


y = −3 cscx

2

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-3.2 -2.4 -1.6 -0.8 0 0.8 1.6 2.4 3.2

-10

-5

5

10

Figure 15: Partial graph of y = −3 tan 2x.


The period is 4π. You need to label five characteristics, two points and three asymp-totes, on one cycle of the provided graph (Figure 16, page 32). The five key character-istics are: x = 0; (π, −3); x = 2π; (3π, 3); x = 4π.

7. Graph at least two cycles and find the period.

f (x) = sec (3πx)


The period is 2/3. You need to label five characteristics, two points and three asymp-totes, on one cycle of the provided graph (Figure 17, page 32). The five key character-istics are: x = −1/6; (0, 1); x = 1/6; (1/3, −1); x = 1/2.


f (x) = cot(πx

3

)

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-10 -5 0 5 10

-10

-5

5

10

Figure 16: Partial graph of y = −3 csc x2.

-2 -1.6 -1.2 -0.8 -0.4 0 0.4 0.8 1.2 1.6 2

-2.4

-1.6

-0.8

0.8

1.6

2.4

Figure 17: Partial graph of f (x) = sec (3πx).

Page 32 of 238





The period is 3. You need to label five characteristics, three points and two asymptotes,on one cycle of the provided graph (Figure 18, page 33). The five key characteristicsare: x = 0; (3/4, 1); (3/2, 0); (9/4, −1); x = 3.

-10 -7.5 -5 -2.5 0 2.5 5 7.5 10

-2.4

-1.6

-0.8

0.8

1.6

2.4

Figure 18: Partial graph of f (x) = cot(πx3

).

9. Use the graph (Figure 19, page 34) of the function to determine whether the function iseven, odd, or neither.

f (x) = −3 secx


Even

10. Use a graphing utility to graph each of the following.

(a) y = |x| · cosx


Your graph should look similar (Figure 20, page 34).

Page 33 of 238




-12.5 -10 -7.5 -5 -2.5 0 2.5 5 7.5 10 12.5

-7.5

-5

-2.5

2.5

5

7.5

Figure 19: Partial graph of f (x) = −3 secx.

-24 -20 -16 -12 -8 -4 0 4 8 12 16 20 24

-16

-12

-8

-4

4

8

12

16

Figure 20: Partial graph of f (x) = |x| · cosx.

Page 34 of 238




(b) y = x · |cosx|



-24 -20 -16 -12 -8 -4 0 4 8 12 16 20 24

-16

-12

-8

-4

4

8

12

16

Figure 21: Partial graph of y = x · |cosx|.



21.7 Homework



Page 35 of 238





y = 3 cos 4x

Solution: Period is π/2, amplitude is 3. You need to label five points on one cycle ofthe provided graph (Figure 22, page 36).

-1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4

-3

-2

-1

1

2

3

Figure 22: Partial graph of y = 3 cos 4x.


y = 3 sin (πx)− 2

Solution: Period is 2, amplitude is 3. You need to label five points on one cycle of theprovided graph (Figure 23, page 36).

-1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

-5

-4

-3

-2

-1

1

Figure 23: Partial graph of y = 3 sin πx− 2.

Page 36 of 238





f (x) = 2 sin(πx− π

2

)− 3

Solution: Period is 2, amplitude is 2. You need to label five points on one cycle of theprovided graph (Figure 24, page 37).

-0.8 -0.4 0 0.4 0.8 1.2 1.6 2 2.4 2.8

-5

-4

-3

-2

-1

Figure 24: Partial graph of f (x) = 2 sin(πx− π

2

)− 3.


f (x) = tan(πx

5

)

Solution: The period is 5. You need to label five characteristics, three points and twoasymptotes, on one cycle of the provided graph (Figure 25, page 37).

-10 -7.5 -5 -2.5 0 2.5 5 7.5 10

-2.4

-1.6

-0.8

0.8

1.6

2.4

Figure 25: Partial graph of f (x) = tan(πx5

).

Page 37 of 238





y = 2 sec 2x+ 3

Solution: The period is π. You need to label five characteristics, two points and threeasymptotes, on one cycle of the provided graph (Figure 26, page 38).

-3.2 -2.4 -1.6 -0.8 0 0.8 1.6 2.4 3.2

-10

-5

5

10

Figure 26: Partial graph of y = 2 sec 2x+ 3.


f (x) = csc (4x)

Solution: The period is π/2. You need to label five characteristics, two points andthree asymptotes, on one cycle of the provided graph (Figure 27, page 38).

-2 -1.6 -1.2 -0.8 -0.4 0 0.4 0.8 1.2 1.6 2

-2.4

-1.6

-0.8

0.8

1.6

2.4

Figure 27: Partial graph of f (x) = csc (4x).

Page 38 of 238




7. Use a graph to approximate (4 decimal places) the solutions to the equation, y = cotx, onthe interval [−2π, 2π]. You will need to do this problem using your graphics calculator,and a graph (Figure 28, page 39) is being provided as a guide.

cotx =√

3, x ∈ [−2π, 2π]

-5 -2.5 0 2.5 5

-2.4

-1.6

-0.8

0.8

1.6

2.4

Figure 28: Partial graph of y = cotx and y =√

3.

Solution: Exact answers are being provided and you should also be able to determinethese values.

x = −11π

6≈ −5.7596

x = −5π

6≈ −2.6180

x =π

6≈ 0.5236

x =7π

6≈ 3.6652

8. Use a graphing utility to graph each of the following.

(a) y = |x · cosx|

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Solution: Your graph should look similar (Figure 29, page 40).

-24 -20 -16 -12 -8 -4 0 4 8 12 16 20 24

-16

-12

-8

-4

4

8

12

16

Figure 29: Partial graph of y = |x · cosx|.

(b) y = |x|+ |cosx|


(c) y = e−x2 · sinx


Page 40 of 238




-24 -20 -16 -12 -8 -4 0 4 8 12 16 20 24

-16

-12

-8

-4

4

8

12

16

Figure 30: Partial graph of y = |x|+ |cosx|.

-1.5 -1 -0.5 0 0.5 1 1.5

-1

-0.5

0.5

1

Figure 31: Partial graph of y = e−x2 · sinx.

Page 41 of 238




22 Angles

22.1 Arc Length and Radian

From the prior sections you should already be familiar with the relationship between theradius of a circle and the arc length subtended by an angle θ in radians.

s = rθ

22.2 Angular and Linear Speed

For objects moving uniformly, the linear speed is just the length traveled divided by the timetaken to travel that distance. We have a similar relationship for uniform rotational speed,where the number of rotations made is divided by the time taken to make those rotations.If we look at the relationship, s = rθ, and divide by time, t, we will get

s

t= r

θ

t⇒ v = rω.

The v is linear velocity, and ω (this is the Greek letter omega) is the angular velocity inradians per unit of time.

22.3 Area of a Sector

The area of a circle is A = πr2. A sector is simply a fractional part (Figure 32, page 42) ofa circle. The area of a sector is just a part of the circle’s whole area, and we can generalize

Figure 32: A pizza cut into eight equal slices.

this (given a radius and a theta) to find the area of any sector, as follows.

A = πr2, Area of the whole circle.θ

2πA =

θ

2ππr2 Area of a fractional part (sector) of a circle.

As =θr2

2, Area of a sector. Again, θ must be in radians!

Page 42 of 238




So the area of a sector is given by this formula.

As =1

2θ r2, θ must be in radians!

22.4 Small Details

Certainly, in your academic career, as in life, you will come across many terms that you maynot be familiar with. You should make some attempt to become familiar with unfamiliarwords, and a dictionary is often quite helpful in this regard. It’s a never ending battle, butnonetheless, an important one to fight. Here’s a short list of simple terms that you will comeacross in the in dealings with trigonometry.

Acute Angle: An acute angle is a positive angle measuring less than 90◦.

Obtuse Angle: An obtuse angle is a positive angle measuring greater than 90◦, but lessthan 180◦.

Complementary: Two acute angles are complementary if their sum is 90◦.

Supplementary: Two positive angles are supplementary if their sum is 180◦.

Coterminal: When two or more angles share the same terminal side. For example, for anyangle θ, the coterminal angles, θk, are given by:

θk = θ + 2πk, k ∈ Z, θ in radian

θk = θ + 360◦k, k ∈ Z, θ in degree

θk = θ + k, k ∈ Z, θ in revolutions

22.5 Degrees, Minutes, and Seconds

Historically, the measure of an angle has been expressed in degrees, minutes and seconds.The notation for degree is a small subscripted circle. For example ten degrees is written as10◦. The notation for minute is a small subscripted tick-mark. For example twelve minutesis written as 12′. The notation for second is two small subscripted tick-marks. For examplefifteen seconds is written as 15′′. The relationships between these units is:

1◦ = 60′ = 3600′′.

Many calculators will allow you to move back and forth between degrees in decimal formand the degree minute second (D◦M’S”) form. You should, of course, be able to to this evenif your calculator can’t.

22.6 Examples

1. Determine the quadrant in which 76.7◦ lies.

Page 43 of 238





First

2. Determine the quadrant in which −223.6◦ lies.


Second

3. Rewrite −288◦ in radian measure as a multiple of π.


−8π

5

4. Rewrite 300◦ in radian measure as a multiple of π.


5π

3

5. Find two positive angles and two negative angles that are co-terminal with 1.125 revo-lutions.


−1.875 rev., −0.875 rev., 0.125 rev., and 2.125 rev., there are other correct answerstoo.

6. Find (if possible) the complement and supplement of 37◦


The complement is 53◦, and the supplement is 143◦.

7. Use your calculator (but show work) to convert 102◦ 15′ 33” to decimal degree form,accurate to the nearest thousandth.

Page 44 of 238





102.259◦

8. Use your calculator (but show work) to convert −17.8567◦ to D◦ M’ S” (degree minuteseconds) form. Only whole numbers should be used.


−17◦ 51′ 24”

9. In a circle with a 120 cm radius, and arc of 132 cm long subtends an angle of how manyradians? How many degrees?


11/10 = 1.1, 198◦/π ≈ 63.025◦

10. Find the radius r of a circle with an arc length s = 3 m and a central angle θ = 60◦.(Round your answer to two decimal places.)


2.86 meters

11. Through how many radians does the minute hand of a clock rotate from 12:40 p.m. and1:30 p.m.?


5π

3

12. A car is traveling at a speed of 60 mph.12 Its tires have an outside diameter of 24 inches.Find the angle through which a tire turns in 10 seconds.

125280 feet in one mile, 12 inches in one foot.

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880, or approximately 50420◦

13. Find the distance between the cities whose latitudes are 25◦ 17′ N and 38◦ 51′ S. Assumethat Earth is a sphere of radius 4000 miles and that the cities are on the same longitude(one city is due north of the other).


4477 miles



22.8 Homework



1. Determine the quadrant in which 245.1◦ lies.

Solution: Third

2. Use your calculator (but show work) to convert −23◦ 39′ 17” to decimal degree form.

Solution: −23.655◦

3. Find the radian measure of the central angle θ of a circle of radius r = 70 km thatintercepts an arc of length s = 156 km. (Round your answer to two decimal places.)

Page 46 of 238




Solution: 2.23

4. Find two positive angles and two negative angles that are coterminal withπ

3.

Solution: −11π

3, −5π

3,

7π

3, and

13π

3.

5. Use your calculator (but show work) to convert 322.4756◦ to D◦ M’ S” (degree minuteseconds) form.

Solution: 322◦ 28′ 32”

6. An earth13 satellite in circular orbit 2600 kilometers above the earth’s surface makes onecomplete revolution every 4 hours. What is its linear speed? Round to the nearest wholenumber.

Solution: 14137 kilometers per hour.

7. If the area of the sector is 15 m2, and the radius is 3 m, what is the angle?

Solution: 10/3 ≈ 3.333 ≈ 190.986◦

8. Find the complement and supplement of 73◦24′17′′.

Solution: 16◦35′43′′, 106◦35′43′′

9. Find the area of the sector, if θ = 50◦ and the diameter is 10 m. Round answer to thenearest hundredth and be sure to use proper units

Solution: 10.91 square meters.

10. The circular blade on a saw has a diameter of 6 inches and rotates at 5400 revolutionsper minute.

(a) Find the angular speed of the blade in radians per second.

13The earth is approximately a sphere with a radius 6400 kilometers.

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Solution: 180π ≈ 565.487, the unit here is simply per second

(b) Find the linear speed of the saw teeth (in feet per second) as they contact the woodbeing cut.

Solution: 45π ≈ 141.372, the unit here is feet per second.

Page 48 of 238




23 Trigonometric Identities and Equations

23.1 Review of The Circular Functions

You should recall that in our introduction to trigonometry, we started with a circle, centeredat the origin, and radius, r > 0:

x2 + y2 = r2.

The trigonometric (circular) functions are related to the coordinates along this circle as werotate the radius. Of course the radius does not change, but the x and y do as we rotatethe radius. This rotation will be with respect to the positive x axis: as we rotate counter-clockwise we say that the radius forms an positive angle with respect to the positive x-axis;and as we rotate clockwise we say that the radius forms a negative angle with respect to thepositive x-axis. Here I will refer to this angle as theta, θ. Here’s how the six trigonometricfunctions they are defined.


sin θ =y

r


cos θ =x

r


tan θ =y

x


cot θ =x

y


sec θ =r

x


csc θ =r

y

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23.1.1 Cautionary Note

An important note on potentially confusing notation.

(sin θ)2 = (sin θ) · (sin θ) = sin θ · sin θ = sin2 θ

Get used to seeing the exponent on the function’s name.

23.1.2 Fundamental Trigonometric Identities

Reciprocal Identities Each will be explained using the definitions that were initially pre-sented. These should all become second nature.

• csc θ =1

sin θ

• sec θ =1

cos θ

• cot θ =1

tan θ

• tan θ =sin θ

cos θ

• cot θ =cos θ

sin θ

Pythagorean Identities Each will be explained using the definitions that were initiallypresented, and their relationship to x2 + y2 = r2. These should all become secondnature.

• sin2 θ + cos2 θ = 1

• tan2 θ + 1 = sec2 θ

• 1 + cot2 θ = csc2 θ

Even-Odd Identities From the graphs you should be able to figure out which functionsare even (symmetric with respect to the y-axis), and which are odd (symmetric withrespect to the origin).

• Sine is odd, sin (−θ) = − sin θ

• Cosine is even, cos (−θ) = cos θ

• Tangent is odd, tan (−θ) = − tan θ

• Cotangent is odd, cot (−θ) = − cot θ

• Secant is even, sec (−θ) = sec θ

• Cosecant is odd, csc (−θ) = − csc θ

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Cofunction Identities You should be able to verify these by using simple graphs. Yes,you know how to graph each of these functions and you should verify that each of thefollowing identities is true. Alternatively, you may prefer using a simple right triangleto verify these, you should note that if θ is acute, then π/2− θ is its compliment.

• sin(π

2− θ)

= cos θ

• cos(π

2− θ)

= sin θ

• tan(π

2− θ)

= cot θ

• cot(π

2− θ)

= tan θ

• sec(π

2− θ)

= csc θ

• csc(π

2− θ)

= sec θ

23.2 Simplifying Trigonometric Expressions

To simplify a trigonometric expression, we use the simple trigonometric identities to rewriteas a simpler expression. Usually it is best to rewrite a trigonometric expression in terms ofsines and cosines. Most simplifications will be quite straight forward and will often mirrorwhat you did in basic algebra: addition of like terms; multiplications; and reductions.

23.3 Proving/Verifying Trigonometric Identities

To verify that a trigonometric equation is an identity, we transform one side of the equationinto the other side by a series of steps, each of which is itself an identity. Initially we willwork with the simple trigonometric identities only. Here’s the general technique.

• Start with one side. It is usually best to choose the side that appears more complicated.

• Use known identities only, and try to use simple obvious steps that bring us closer tothe other side.

• Converting to sines and cosines can in fact be very helpful in most circumstances.

23.4 Solving Trigonometric Equations

Yes, you should be well aware of what it means to solve an equation for x. You’ve solved agood number of equations that have included: linear equations, factorable polynomials equalto zero, non-factorable quadratic equations (completing the square or quadratic formula),rational equations, logarithmic and exponential equations.

Now we will focus our attention on solving equations involving the trigonometric func-tions. The principles will always remain the same! Here’s the general method.

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• Most trigonometric equations—no matter how complicated they initially appear—willreduce to three very simple problems:

sinx = k, or cosx = k, or tanx = k.

Here, it should be noted that x is an unknown angle, and k is a given numerical ratio.For both the sine and cosine the ratio, k, should be on the interval [−1, 1]; but there’sno restriction for tangent’s ratio.

• Once you get one of these forms you should graph one cycle of its related curve andlocate the angles, x, that produce this ratio, k.

• You should note that these functions are periodic and their solutions will also beperiodic. Here, you should note that the sine and cosine have a period of 2π and thetangent has a period of π.

The trouble for most students is getting the equations into this format, but more on thatlatter. Once you see one of these equations it is usually quite straight forward to solve forthe variable x.

23.5 Examples: Simplification

1. Simplify.

sin θ · sec θ


tan θ

2. Simplify.

sec2 θ − 1

sin2 θ


sec2 θ

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3. Simplify.

sin2 x ·(csc2 x− 1

)

Solution: Work will be done in class. Make sure you take notes, or get notes from areliable classmate if you are not present during lecture. Don’t be confused by the useof the variable x here, it just represents some angle.

cos2 x

4. Simplify.

sin (π/2− x)

cos (π/2− x)


cotx

5. Factor the expression and use the fundamental identities to simplify.

csc2 x− 1

cscx− 1


cscx+ 1

6. Perform the indicated operation and use the fundamental identities to simplify.(1

3tanx+

1

3secx

)(2

3tanx− 2

3secx

)

Page 53 of 238





−2

9

7. Use the trigonometric substitution to write the algebraic expression as a trigonometricfunction of θ, where 0 < θ < π/2.

√64− 16x2, x = 2 cos θ


8 sin θ

8. Use the cofunction identities to evaluate the expression without using a calculator.

cos2 49◦ + cos2 83◦ + cos2 7◦ + cos2 41◦


2

9. Use the trigonometric substitution to write the algebraic expression as a trigonometricfunction of θ, where 0 < θ < π/2. Assume a > 0.

√a2 − u2, u = a cos θ


a sin θ

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10. Simplify the trigonometric expression.14

cos θ + tan θ sin θ


sec θ

11. Simplify the trigonometric expression.

sin θ

cos θ+

cos θ

1 + sin θ


sec θ


cos3 β + sin2 β cos β


cos β


secα− cosα

tanα

14θ is the Greek letter theta.15β is the Greek letter beta.16α is the Greek letter alpha.

Page 55 of 238





sinα

23.6 Examples: Identities

1. Verify the trigonometric identity.

cos θ (sec θ − cos θ) = sin2 θ


Make sure you understand the process.


2 tanx secx =1

1− sinx− 1

1 + sin x




cosu

1− sinu= secu+ tanu



Page 56 of 238




4. Verify the trigonometric identity.17

1 + cos δ

cos δ=

tan2 δ

sec δ − 1




cosx− cos y

sinx+ sin y+

sinx− sin y

cosx+ cos y= 0




sin5 x = sinx(1− 2 cos2 x+ cos4 x

)



7. Verify the trigonometric identity.18

sin ε

1− cos ε=

1 + cos ε

sin ε

17δ is the Greek letter delta.18ε is the Greek letter epsilon.

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23.7 Examples: Equations

1. Find all solutions of the equation in the interval [0◦, 360◦).

cosx = − 1√2


135◦, 225◦ Your graph should look similar (Figure 33, page 58).

0

Figure 33: Partial graph of y = cosx and y = −1/√

2.


sinx = −√

3

2



Page 58 of 238




0

Figure 34: Partial graph of y = sinx and y = −√

3/2.


tanx = 1



0

Figure 35: Partial graph of y = tanx and y = 1.

4. Find all solutions of the equation in the interval [0, 2π).

secx = −√

2

Page 59 of 238





3π/4, 5π/4


cscx =2√3


π/3, 2π/3

6. Solve for x. Use n as an integer constant.

sinx = − 1√2


x =5π

4+ 2πn

x =7π

4+ 2πn


7. Solve the equation for x. Use n as an integer constant.√

2 cosx− 1 = 0


x =π

4+ 2πn

x =7π

4+ 2πn

Page 60 of 238




0 1.5708 3.1416 4.7124 6.2832

-1

1

Figure 36: Partial graph of y = sinx and y = −1/√

2.


tanx =√

3


x =π

3+ πn


9. Solve for x.

4 cos2 x− 2 = 0


Page 61 of 238




0 1.5708 3.1416 4.7124 6.2832

-2

-1

1

2

Figure 37: Partial graph of y = tanx and y =√

3.

4 cos2 x− 2 = 0

cosx = ± 1√2

x =π

4+ 2πn

x =3π

4+ 2πn

x =5π

4+ 2πn

x =7π

4+ 2πn

This may also be stated simply as

x =π

4+π

2n



2 sin2 x− sinx− 1 = 0

Page 62 of 238




0 1.5708 3.1416 4.7124 6.2832

-2

-1

1

2

Figure 38: Partial graph of y = cosx and y = ±1/√

2.


2 sin2 x− sinx− 1 = 0

(2 sinx+ 1) (sinx− 1) = 0

sinx = −1

2sinx = 1

x =π

2+ 2πn

x =7π

6+ 2πn

x =11π

6+ 2πn




Page 63 of 238




0 1.5708 3.1416 4.7124 6.2832

-2

-1

1

2

Figure 39: Partial graph of y = sinx, y = 1 and y = −1/2.

23.9 Homework


23.9.1 Homework: Simplifications

Directions: Answer each of the following questions. Partial answers are provided, but nowork is being provided as a guide. Yes, you need to be able to do these problems and youwill be tested. You need to get help if you cannot do these problems exactly as was done inclass!


sec4 x− tan4 x

Solution:

sec2 x+ tan2 x

2. Simplify.

cotx · secx

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Solution: cscx


4

3− 8

3sin2 x+

4

3sin4 x

Solution:

4

3

(cos4 x

)

4. Perform the indicated operation and use the fundamental identities to simplify.

5

secx+ 1− 5

secx− 1

Solution:

−10 cot2 x

5. Use the trigonometric substitution to write the algebraic expression as a trigonometricfunction of θ, where 0 < θ < π/2.

√100x2 − 169, 10x = 13 sec θ


13 tan θ


cos2 54◦ + cos2 36◦

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Solution:

1

7. Use the trigonometric substitution to write the algebraic expression as a trigonometricfunction of θ, where 0 < θ < π/2. Assume a > 0.

√u2 − a2, u = a sec θ

Solution:

a tan θ


sin z

csc z+

cos z

sec z

Solution:

1


cos y

sec y + tan y

Solution:

1− sin y

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1 + sin γ

cos γ+

cos γ

1 + sin γ

Solution:

2 sec γ


sin2 61◦ + sin2 87◦ + sin2 29◦ + sin2 3◦

Solution:

2

23.9.2 Homework: Identities


1. Verify that

cscx− sinx = cotx · cosx

is an identity.

Solution: Make sure you understand the process.

2. Verify that

sin4 x− cos4 x = sin2 x− cos2 x

is an identity.

19γ is the Greek letter gamma.

Page 67 of 238





3. Verify that

1

1− cosx+

1

1 + cos x= 2 csc2 x

is an identity.


4. Verify that

(secx+ tanx)2 =1 + sin x

1− sinx

is an identity.


5. Verify that

sin2 x− cos2 x =tanx− cotx

tanx+ cotx

is an identity.


23.9.3 Homework: Equations



tan2 x = 1

Page 68 of 238




Solution: π/4, 3π/4, 5π/4, 7π/4


secx · cscx = 2 csc x

Solution: π/3, 5π/3


sin2 x+ cosx+ 1 = 0

Solution: π

4. Solve for x.

cosx =

√3

2

Solution:

x =π

6+ 2πn

x =11π

6+ 2πn

Your graph, if used as an aid, should look similar (Figure 40, page 70).

5. Solve for x.

tanx = −1

Solution:

x =3π

4+ πn


Page 69 of 238




0 1.5708 3.1416 4.7124 6.2832

-1

1

Figure 40: Partial graph of y = cosx and y =√

3/2.

0 1.5708 3.1416 4.7124 6.2832

-1

1

Figure 41: Partial graph of y = tanx and y = −1.

Page 70 of 238




6. Solve for x.

secx+ 2 = 0

Solution:

secx+ 2 = 0

secx = −2

cosx = −1

2

x =2π

3+ 2πn

x =4π

3+ 2πn


0 1.5708 3.1416 4.7124 6.2832

-2

-1

1

2

Figure 42: Partial graph of y = cosx and y = −1/2.

7. Solve the equation for x. Use n as an integer constant.

3 cot2 x− 1 = 0

Solution:

π/3 + πn, 2π/3 + πn

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8. Solve the equation for x. Use n as an integer constant.

(sinx+ 2) (cosx+ 1) = 0

Solution: π + 2πn

Page 72 of 238




24 Sum, Difference, Multiple, & Half-Angle Formulas

24.1 Sum and Difference Identities

Using a unit circle (radius 1) and two angles A and B, where A > B. You should be able toverify that the points along this circle corresponding to these two angles is: (cosA, sinA)and (cosB, sinB). The distance between these two points is given by:√

(cosA− cosB)2 + (sinA− sinB)2.

If, so inclined, you were to simplify this you would find it is equivalent to:√2− 2 (cosA cosB + sinA sinB).

Now if you were to take a new angle to be A−B, let’s call it S for the time being, it’s pointalong the unit circle would be (cosS, sinS). The distance between this point and the (1, 0)is identical to the distance between points (cosA, sinA) and (cosB, sinB). The distancebetween(cosS, sinS) and (1, 0) is given by:√

(cosS − 1)2 + (sinS − sin 0)2.

Again, if, so inclined, you were to simplify this you would find it is equivalent to:√

2− 2 cosS.

Finally we have this relationship:√

2− 2 cosS =√

2− 2 (cosA cosB + sinA sinB)

2− 2 cosS = 2− 2 (cosA cosB + sinA sinB)

cosS = cosA cosB + sinAsinB

cos (A−B) = cosA cosB + sinA sinB

You need to know this identity:

cos (A−B) = cosA cosB + sinA sinB. (1)

And from this identity we should be able to verify (need to know even odd relationships)this identity:20

cos (A+B) = cosA cosB − sinA sinB. (2)

Now using the cofunction identities we should be able to follow this argument.

sin θ = cos(π

2− θ)

sin (A+B) = cos[π

2− (A+B)

]sin (A+B) = cos

[(π2− A

)−B

]sin (A+B) = cos

(π2− A

)cosB + sin

(π2− A

)sinB

sin (A+B) = sinA cosB + cosA sinB

20We’ll do this in class.

Page 73 of 238




You also need to know this identity:

sin (A+B) = sinA cosB + cosA sinB. (3)

And from this identity we should be able to verify (need to know even odd relationships)this identity:21

sin (A−B) = sinA cosB − cosA sinB. (4)

24.2 Double Angle Formulas

Using these two sum formulas,

cos (A+B) = cosA cosB − sinA sinB

sin (A+B) = sinA cosB + cosA sinB

you should be able to derive the following double angle formulas:22

sin 2x = 2 sinx · cosx

cos 2x = cos2 x− sin2 x

cos 2x = 2 cos2 x− 1

cos 2x = 1− 2 sin2 x

Yes, we’ll do this in class!

24.3 Half-Angle Formulas

Again, make sure you can derive these double angle formulas. Using equations (7) and (8)only, and replacing x with θ/2 we get:

cos θ = 2 cos2θ

2− 1 and cos θ = 1− 2 sin2 θ

2.

Now solving each of these for the functions of half-angles we get:

cosθ

2= ±

√cos θ + 1

2and sin

θ

2= ±

√1− cos θ

2

The choice of the + or − sign depends on the quadrant in which θ/2 lies. These formulasare easy enough to derive, and you’re expected to be able to do so.

tanθ

2=

1− cos θ

sin θ=

sin θ

1 + cos θ

You should notice that there’s no ± sign here, and that’s the tricky part! No need tomemorize either, because you know tangent is related to sine and cosine.

21We’ll do this in class.22Just let, A = B = x.

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24.4 Product-Sum Formulas

Using the sum formulas we can easily show:

sinA cosB =1

2[sin (A+B) + sin (A−B)] (5)

cosA sinB =1

2[sin (A+B)− sin (A−B)] (6)

cosA cosB =1

2[cos (A+B) + cos (A−B)] (7)

sinA sinB =1

2[cos (A−B)− cos (A+B)] (8)

I will show several of these in class.

24.5 Sum-to-Product Formulas

Now, using the formulas just obtained, make these substitutions

A =β + α

2and B =

β − α2

,

and you’ll get,

2 sinβ + α

2cos

β − α2

= sin β + sinα (9)

2 cosβ + α

2sin

β − α2

= sin β − sinα (10)

2 cosβ + α

2cos

β − α2

= cos β + cosα (11)

2 sinβ + α

2sin

β − α2

= cosα− cos β (12)

24.6 Examples

1. Verify.

sin (45◦ + 120◦) 6= sin 45◦ + sin 120◦


√3− 1

2√

26= 2 +

√6

2√

2

Page 75 of 238




2. Write the expression as the sine, cosine, or tangent of an angle.

sin 127◦ cos 19◦ + cos 127◦ sin 19◦


sin 146◦

3. Write the expression as the sine, cosine, or tangent of an angle.23

tan 153◦ − tan 87◦

1 + tan 153◦ tan 87◦


tan 66◦

4. Find the exact value of the trigonometric function given that sinA = 5/13 and cosB =−3/5. (Both A and B are in Quadrant II.)

cos (B − A)

23Here, you are expected to verify that

tan (A−B) =sin (A−B)

cos (A−B)=

tanA− tanB

1 + tanA · tanB.

While you’re at it, you might as well show,

tan (A+B) =sin (A+B)

cos (A+B)=

tanA+ tanB

1− tanA · tanB.

Page 76 of 238





56

65

5. If the sinα = 12/13 and the cos β = 8/17, and both α and β are acute, find:

cos (α + β) .


−140

221

6. Express

sin (3x) · sin (5x)

as a sum of trigonometric functions.24


cos (2x)− cos (8x)

2

7. Verify the identity.25

sin 3x− sinx

cos 3x+ cosx= tanx

24Look back over the notes to find product to sum formulas.25Look back over the notes to find sum to product formulas.

Page 77 of 238





True

8. Rewrite the expression in terms of the first power of the cosine.

3 sin2 x

2· cos2

x

2


3− 3 cos (2x)

8

9. Use the half-angle formulas to determine the exact values.

(a) sin 165◦

Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.√

1−√

3/2

2=

√2−√

3

2=

√6−√

2

4

You may want to verify that all answers are equivalent. The preferred answer isthe middle one.

(b) cos 165◦


−

√1 +√

3/2

2= −

√2 +√

3

2= −√

2 +√

6

4

You may want to verify that all answers are equivalent. The preferred answer isthe middle one.

(c) tan 165◦

Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.√

3− 2

10. Given that tan θ = −3/4 and θ is in the second quadrant, find each of the following.

Page 78 of 238




(a) sin 2θ =


−24/25

(b) cos 2θ =


7/25

(c) tan 2θ =


−24/7

11. Simplify.

sin 2x

2 cos2 x


tanx



24.8 Homework



Page 79 of 238




1. Verify.

cos (45◦ − 120◦) 6= cos 45◦ − cos 120◦

Solution:

√3− 1

2√

26= 2 +

√2

2√

2

2. Find the exact value of the trigonometric function given that sinA = 5/13 and cosB =−3/5. Both A and B are in Quadrant II.

sin (A+B)

Solution:

−63

65

3. Find the exact value of the trigonometric function given that sinA = −15/17 andcosB = −4/5. Both A and B are in Quadrant III. You may want to rewrite the tangentas a ratio.

tan (B − A)

Solution:

−36

77

4. Write

sin (7x) + sin (3x)

as a product of trigonometric functions.26

26Look back over the notes to find sum to product formulas.

Page 80 of 238




Solution:

2 · sin (5x) cos (2x)

5. Find the exact values of sin 2x, cos 2x, and tan 2x using the double-angle formulas.

cosx = −2

7,

π

2< x < π

Solution:

sin 2x = −12√

5

49

cos 2x = −41

49

tan 2x =12√

5

41

6. Rewrite the expression in terms of the first power of the cosine.

5 cos2 2x

Solution:

5 [cos (4x) + 1]

2

7. Use the half-angle formulas to determine the exact values.

(a) sin (π/12)

Solution:

√2−√

3

2=

√6−√

2

4

(b) cos (π/12)

Page 81 of 238




Solution:

√2 +√

3

2=

√2 +√

6

4

(c) tan (π/12)

Solution: 2−√

3

8. Solve for x ∈ [0, 2π).

sin 2x+ cosx = 0

Solution:

π

2,

7π

6,

3π

2

11π

6

9. Find the exact values of sin 2x, cos 2x, and tan 2x using the double-angle formulas.

cotx = −9,3π

2< x < 2π

Solution:

sin 2x = − 9

41

cos 2x =40

41

tan 2x = − 9

40

10. Verify the identity.

tan 2x =2 tanx

1− tan2 x

Solution: True

11. Given that f (x) = sinx, show that

f (x+ h)− f (x)

h= sinx

(cosh− 1

h

)+ cosx

(sinh

h

)

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Solution:

sin (x+ h)− sin (x)

h=

sinx · cosh− sinh · cosx− sinx

h

=sinx · cosh− sinx− sinh · cosx

h

=sinx (cosh− 1)− sinh · cosx

h

=sinx (cosh− 1)

h− sinh · cosx

h

= sin x

(cosh− 1

h

)+ cosx

(sinh

h

)

Page 83 of 238




25 Inverse Trigonometric Functions

25.1 Inverse Functions

First, a partial and simple review of inverses as presented in MTH 119. It is very importantthat you understand much of what you learned about inverses in MTH 119, as it will aidgreatly in your ability to work through trigonometric inverses. Having basic knowledge ofwhat an inverse is will allow you to quickly move forward in your understanding of trigono-metric inverses as presented here.

Definition: A function f is called one-to-one if it never takes on the same value twice;that is

f (x1) 6= f (x2)

whenever x1 6= x2. We can often visually verified that a function is one-to-one by usingthe horizontal line test, which states, “A function is one-to-one if and only if no horizontalline intersects its graph more than once.” For example, the following function, f1 (x), is notone-to-one.

-2 -1 0 1 2

1

2

3

Figure 43: f1 (x) = x2√x3 + 5

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However, this function, f2 (x), is one-to-one.

-2 -1 0 1 2

1

2

3

Figure 44: f2 (x) =√x3 + 1

Some however may be mislead by the horizontal line test, because it is hard to detectfrom the visual alone that f2 (x) =

√x3 + 1 is one-to-one.27 It should be clear though that

f1 (x) = x2√x3 + 5 is not one-to-one from the graph alone.28 To show that f2 (x) =

√x3 + 1

is one-to-one we proceed as follows. Since for all x1, x2 ≥ −1, and x1 6= x2, we have x31 6= x32,which further implies that x31 + 1 6= x32 + 1, and finally that

√x31 + 1 6=

√x32 + 1.

Definition: Let f be a one-to-one function with domain A and range B. Then its inversefunction f−1 has a domain B and range A and is defined by

f−1 (y) = x ⇔ f (x) = y.

If you need more review regarding inverses you should go back and review MTH 119.

25.2 The Inverse Trigonometric Functions

Important: I will now be writing x instead of θ, so please bear in mind that x now representsthe angle! and if I write y = sinx, then y represents the ratio. I know this confuses many

27Note that the horizontal line, y = 1, appears to hit f2 (x) more than once.28Note that the horizontal line, y = 1, appears to hit f1 (x) clearly three times.

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students, but it’s done in almost all books and it’s hard to avoid! The original definitionsthat I gave you back in the beginning have not changed, but here, x and y are now different!

If you’ve learned the previous material in MTH-119 you will immediately notice thatnone of the trigonometric functions are invertible. So we will need to restrict the domain ofall six functions to force the conditions of invertibility—you need to make these functionsone-to-one by restricting their domains. I’ll only do three, starting with the sine function.

0

Figure 45: The sine function and an invertible region.

25.2.1 The Inverse Sine Function

1. What is the functional notation of the red portion of this graph (Figure 45, page 86)?


f (x) = sinx, −π2≤ x ≤ π

2

2. What is this function’s domain?


−π2≤ x ≤ π

2

3. What is this function’s range?

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−1 ≤ y ≤ 1

4. The name of sine’s inverse is arcsine, some also write inverse sine, or y = sin−1 (x).

(a) What is the arcsine’s domain?


−1 ≤ x ≤ 1

(b) What is the arcsine’s range?


−π2≤ y ≤ π

2

5. Label the following graph (Figure 46, page 88). You should indicate the equation ofeach function (there’s two) and important points along each axis.


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0

Figure 46: Please label.

0

Figure 47: The cosine function and an invertible region.

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25.2.2 The Inverse Cosine Function



f (x) = cos x, 0 ≤ x ≤ π



0 ≤ x ≤ π



−1 ≤ y ≤ 1

4. The name of cosine’s inverse is arccosine, some also write inverse cosine, or y =cos−1 (x).

(a) What is the arccosine’s domain?


−1 ≤ x ≤ 1

(b) What is the arccosine’s range?


0 ≤ y ≤ π

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5. Label the following graph (Figure 48, page 90). You should indicate the equation ofeach function (there’s two) and important points/features along each axis.


0

2


25.2.3 The Inverse Tangent Function



f (x) = tan x, −π2< x <

π

2


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0

-2

2

Figure 49: The tangent function and an invertible region.


−π2< x <

π

2



−∞ < y <∞

4. The name of tangent’s inverse is arctangent, some also write inverse tangent, or y =tan−1 (x).

(a) What is the arctangent’s domain?


−∞ < x <∞

(b) What is the arctangent’s range?

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−π2< y <

π

2

5. Label the following graph (Figure 50, page 92). You should indicate the equation ofeach function (there’s two) and important points/features along each axis.


0

-2

2


25.3 Examples

1. Without using your calculator, evaluate arcsin (1/2). What does this number represent?


30◦, or π/6; an angle

2. Without using your calculator, evaluate arccos(−√

3/2). What does this number rep-

resent?

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150◦, or 5π/6; an angle

3. Without using your calculator, evaluate arcsin (3/2).


Not possible. Non-real result.

4. Without using your calculator, evaluate arccos (−1). What does this number represent?


180◦, or π; an angle

5. Without using your calculator, evaluate arctan(−√

3/3). What does this number rep-

resent?


−30◦, or −π/6; an angle

6. Without using your calculator, evaluate arctan (1). What does this number represent?


45◦, or π/4; an angle

7. Using your calculator, evaluate arcsin (0.987). Answer both in radian and degree.29


80.7513◦, or 1.4094

8. Using your calculator, evaluate arccos (−0.0895). Answer both in radian and degree.30

29Approximate to four decimal places.30Approximate to four decimal places.

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95.1348◦, or 1.6604

9. Using your calculator, evaluate arctan (−4.589). Answer both in radian and degree.31


−77.7067◦, or −1.3562

10. Using your calculator, evaluate arccos (−3/4). Answer both in radian and degree.32


138.5904◦, or 2.4189

11. Use the properties of inverse functions to find the exact value of the expression.

arccos

(cos

11π

2

)


90◦, or π/2

12. Write an algebraic expression that is equivalent to the expression. (Hint: Sketch a righttriangle.)

sec [arcsin (x− 1)]


1√2x− x2


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25.5 Homework



1. Without using your calculator, evaluate arcsin(−1/√

2). What does this number repre-

sent?

Solution: −45◦, or −π/4; an angle

2. Without using your calculator, evaluate arccos (0). What does this number represent?

Solution: 90◦, or π/2; an angle

3. Without using your calculator, evaluate arctan(√

3). What does this number represent?

Solution: 60◦, or π/3; an angle

4. Using your calculator, evaluate arcsin (5/7). Answer both in radian and degree.33

Solution: 45.5847◦, or 0.7956

5. Using your calculator, evaluate arctan (5.894). Answer both in radian and degree.34


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Solution: 80.3707◦, or 1.4027


arccos

(tan

17π

4

)

Solution: 0◦, or 0


arccos[sin(−π

4

)]

Solution: 135◦, or 3π/4


cot

(arctan

7

8

)

Solution: 8/7

9. Write an algebraic expression that is equivalent to the expression. (Hint: Sketch a righttriangle.)

cos (arctan x)

Solution:1√

1 + x2

10. Given

y = 3 sin−1 (x− 5)

(a) Find the domain.

(b) Find the range.

(c) Solve for x in terms of y.

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Solution:

(a) 4 ≤ x ≤ 6

(b) −3π

2≤ y ≤ 3π

2

(c) x = 5 + sin(y

3

)

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26 Triangles

26.1 Introductory Information

It is somewhat traditional, when dealing with triangles, to use A, B, and C to indicate theangles; and the lengths opposite these angles are usually indicated by a, b, c. You shouldalready know that the sum of the angles in any triangle is 180◦, that is

A+B + C = 180◦.

26.2 Right Triangle

If you’re given a right triangle, then one of these angles must be 90◦. Usually, but notnecessarily, we take C to be the 90◦ angle. You should be able to verify the followingrelationships when dealing with a right triangle, where C = 90◦.

c2 = a2 + b2

sinA =a

c

cosA =b

c

tanA =a

b

sinB =b

c

cosB =a

c

tanB =b

a

You should also be familiar with the terms: adjacent35, opposite36 and hypotenuse37.

26.3 Law of Sines

The following formula will be discussed and proved in class.

sinA

a=

sinB

b=

sinC

c(13)

a

sinA=

b

sinB=

c

sinC(14)

35From geometry of right triangles, meaning the side next to next to the angle, but not the hypotenuse.36From geometry of right triangles, meaning the side opposite the angle.37The hypotenuse is the longest side of a right triangle, it is opposite the right angle.

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Use the Law of Sines when you know two angles and one side (AAS/ASA), or when we knowtwo sides and an angle opposite one of those sides (SSA). The Law of Sines can be confusingif you don’t have a visual to check. So, in my humble opinion, you should learn to use aprotractor and ruler to draw the triangle before doing any trigonometry. The main reasonthe Law of Sines can be misleading (without a visual) is due to the fact that

sin θ = sin (180◦ − θ)

for all θ. So if you compute θ = 60◦, it could also be θ = 180◦ − 60◦ = 120◦; I seriouslydoubt you’d be confused between 120◦ and 60◦ if you were looking at the angle. So pleasedraw a picture first!

Another reason to draw a picture is to see if constructing a triangle is even possible, orif more than one possibility exists. Again, draw a picture first!

26.4 Law of Cosine

The following formulas will be discussed and proved in class.

a2 = b2 + c2 − 2bc cosA (15)

b2 = a2 + c2 − 2ac cosB (16)

c2 = a2 + b2 − 2ab cosC (17)

Use the Law of Cosines when you know two sides and an included angle (SAS), or when weknow three sides (SSS).

26.4.1 Right Triangles

The LOS and LOC applies to all triangles, but right triangles are really very simple cases.For example suppose A = 90◦ and we use equation (15), we get:

a2 = b2 + c2 − 2bc cos 90◦ ⇒ a2 = b2 + c2.

If B = 90◦ and we use equation (16), we get:

b2 = a2 + c2 − 2ac cos 90◦ ⇒ b2 = a2 + c2.

If C = 90◦ and we use equation (17), we get:

c2 = a2 + b2 − 2ab cos 90◦ ⇒ c2 = a2 + b2.

No surprise here. Time to move along.On the other hand, if we use the LOS, where a = 90◦ we get:

sin 90◦

a=

sinB

band

sin 90◦

a=

sinC

c.

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This, of course, gives the following relationships:

sinB =b

aand sinC =

c

a.

You should of course note that angle B is opposite side b, angle C is opposite side c; andangle A is opposite side a which is the hypotenuse.

Once again, no surprise here either. Again, time to move along.

26.5 Examples

1. Solve the right triangle. When necessary, round answers to two decimal places.38

A = 23◦, c = 13 feet


A = 23◦, B = 67◦, C = 90◦, a ≈ 5.08 feet, b ≈ 11.97 feet, c = 13 feet

2. Find, using right triangles, the altitude of the isosceles triangle (two angles/sides equal)where the longest side is 16 kilometers. When necessary, round answers to two decimalplaces.

37◦, 37◦, 106◦


6.03 kilometers

3. Find the angle of depression from the top of a lighthouse 375 feet above water levelto the water line of a ship 3.7 miles39 offshore. When necessary, round answers to twodecimal places.


1.10◦, or 0.02

4. When an airplane leaves the runway, its angle of climb is 5.3◦ and its air speed is 412feet per second. How long will it take the plane to climb to an altitude of 13,500 feet?When necessary, round answers to two decimal places.

38Angles A and B are acute, and c is the hypotenuse.391 mile is equal to 5280 feet.

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354.73 seconds.

5. Use the Law of Sines to solve the triangle. When necessary, round answers to two decimalplaces.

B = 40◦, C = 100◦, c = 13 meters


A = 40◦, a = 8.49 meters, and b = 8.49 meters

6. Use the Law of Sines to solve the triangle. When necessary, round answers to threedecimal places.

A = 21.9◦, C = 51.7◦, b = 0.892 centimeters


B = 106.4◦, a = 0.35, centimeters, and c = 0.73 centimeters

7. Solve the triangle, given a = 11.23 inches, b = 8.24 inches, and B = 29.84◦. Whennecessary, round answers to two decimal places.

Solution: Work will be done in class. Make sure you take notes, or get notes from areliable classmate if you are not present during lecture..

This one is a bit complicated. You should, of course, draw a triangle first. This oneactually gives two distinct and correct answers.

A = 42.70◦, C = 107.46◦, and c = 15.80 inches.

or

A = 137.30◦, C = 12.86◦, and c = 3.69 inches.

8. A forest ranger at an observation point A sights a fire in the direction N27◦10′E. Anotherranger at a observation point B, six miles due east of A, sights the same fire at N52◦40′W.Approximate the distance from each of the observation points to the fire to the nearesttenth of a mile.

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3.7 miles from A, and 5.4 miles from B.

9. The leaning tower of Pisa was originally perpendicular to the ground and 179 feet tall.Because of the sinking into the earth, it now leans at a certain angle θ from the per-pendicular. When the top of the tower is viewed40 from a point 150 from the center ofthe base, the angle of elevation is 53◦. Approximate the angle θ to the nearest wholenumber.


5◦

10. Use the Law of Cosines to solve the triangle.41 When necessary, round answers to twodecimal places.

B = 79.88◦, a = 2.0 feet, c = 6.3 inches


A = 84.96◦, C = 15.16◦, and b = 23.72 inches

11. Use Heron’s Area Formula42 to find the area of the triangle. to find the area of thetriangle. Verify that you get the same answer if you use the Grade School formula.When necessary, round answers to two decimal places.

a = 15 feet, b = 19 feet, c = 6 feet


10√

14 ≈ 37.42 square feet. Both formulas give the same answer. You should note thatHeron’s Area Formula is easier to use in this case. If interested, I suggest you do anInternet search for an algebraic proof of Heron’s Area Formula.

40Tower is leaning towards observer.4112 inches in every foot.42

s =a+ b+ c

2, Area =

√s (s− a) (s− b) (s− c)

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12. A triangular plot of land has sides of lengths 420 feet, 350 feet, and 180 feet. Approximatethe smallest angle between the sides. Round answer to the nearest whole number.


25◦



26.7 Homework



1. Solve the right triangle. Angles A and B are acute, and c is the hypotenuse. Whennecessary, round answers to two decimal places.

B = 17◦, a = 0.98 meters

Solution: A = 73◦, B = 17◦, C = 90◦, a = 0.98 meters, b ≈ 0.30 meters, c ≈ 1.02meters

2. When an airplane leaves the runway, its angle of climb is 11◦ and its ground speed is 312feet per second. How long will it take the plane to climb to an altitude of 21,000 feet?When necessary, round answers to two decimal places.

Solution:

346.27 seconds.

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3. The angle of elevation from the base to the top of a water-slide is 17.3◦. The slide extendshorizontally 43.7 meters. Approximate the height of the water-slide. When necessary,round answers to two decimal places.

Solution: 13 meters

4. When an airplane leaves the runway, its angle of climb is 11◦ and its air speed is 312feet per second. How long will it take the plane to climb to an altitude of 21,000 feet?When necessary, round answers to two decimal places.

Solution: 352.75 seconds.

5. A plane is 76 miles north and 37 miles east of an airport. The pilot wants to fly directlyto the airport. What bearing should be taken? When necessary, round answers to twodecimal places.

Solution: If you are using decimals: South 25.96◦ West, or West 64.04◦ South; orDMS: 25◦ 57′ 32′′ West, or West 64◦ 2′ 28′′ South

6. Use the Law of Sines to solve the triangle. When necessary, round answers to two decimalplaces.

B = 22.3◦, A = 88.98◦, b = 84.7 miles

Solution: Work will be done in class. Make sure you take notes, or get notes from areliable classmate if you are not present during lecture. When necessary, round answersto two decimal places.

C = 68.72◦, a = 223.18 miles, and c = 207.99 miles

7. Find the area of the triangle. Round final answer to three decimal places.

B = 133.6◦, a = 8.7 yards, c = 2.9 yards

Solution: 9.135 square yards

8. To determine the distance between two points A and B, a surveyor chooses a point Cthat is 375 yards from A and 530 yards from B. If the angle BAC has measure 49.5◦,approximate the distance, to the nearest yard, between A and B.

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Solution: 690 yards

9. Use the Law of Cosines to solve the triangle. When necessary, round answers to twodecimal places.

B = 41.8◦, a = 16.4 meters, c = 11.4 meters

Solution: A = 94.32◦, C = 43.88◦, and b = 10.96 meters

10. Use the Law of Cosines to solve the triangle. When necessary, round answers to twodecimal places.

a = 12.3 miles, b = 19.4 miles, c = 7.9 miles

Solution: A = 20.30◦, B = 146.82◦, and C = 12.88◦

11. A jogger runs at a constant rate of one mile every eight minutes in the direction S40◦Efor twenty minutes and then in the direction N20◦E for the next sixteen minutes. Ap-proximate, to the nearest tenth of a mile, the straight-line distance from the startingpoint to the endpoint of the jogger’s course. Round answer to nearest tenth of a mile.

Solution: 2.3 miles

12. To find the distance between two points A and B, a surveyor chooses a point C that is 420yards from A and 540 yards from B. If angle ACB has a measure 63◦10′, approximatethe distance between A and B. Round answer to the nearest yard.

Solution: 513 yards

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27 Vectors

27.1 Vectors

As I hope you now realize, matrices43 are not a single number,44 but rather a collection ofnumbers. Vectors are just particular types of matrices, and are either a single column orsingle row of numbers. In most precalculus books, vectors just have two numbers and arewritten as a row matrix. For example a vector v would look like this:

v =[−7 15

].

However, notation varies, and most people want to differentiate vectors from matrices ingeneral by using this notation instead:

v =⟨−7, 15

⟩.

Here some easily described (we’ll discuss this in class) results related to vectors.

1. The magnitude of the vector v = 〈v1, v2〉, denoted ‖v‖, is given by

‖v‖ = ‖〈v1, v2〉‖ =√v21 + v22.

2. The addition of vectors is the same as the addition of matrices.

u + v = 〈u1, u2〉+ 〈v1, v2〉 = 〈u1 + v1, u2 + v2〉

3. Scalar multiplication is the same as he scalar multiplication of matrices.

kv = k 〈v1, v2〉 = 〈kv1, kv2〉

Two special vectors have reserved names, and they are:

i = 〈1, 0〉 , and j = 〈0, 1〉 ,

and any vector v = 〈v1, v2〉 can be written as a linear combination of i and j as follows:

v = 〈v1, v2〉 = v1i + v2j.

The next thing we need to do is verify that vectors have a visual reference that relatesback to trigonometry. In class we will visually verify that for any vector v = 〈v1, v2〉 that

v1 = ‖v‖ cos θ v2 = ‖v‖ sin θ.

43Covered in MTH 119.44Things that can be described with a single number are called scalar, and are represented by an italic

letter.

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From here we can begin to develop a simple method to determine the angle betweenvectors. Given two non-zero vectors, u and v, the angle between those two vectors can beobtained by using the Law of Cosines as follows. This will be fully explained in class.

‖u− v‖2 = ‖u‖2 + ‖v‖2 − 2 ‖u‖ ‖v‖ cos θ

‖〈u1 − v1, u2 − v2〉‖2 =(u21 + u22

)+(v21 + v22

)− 2 ‖u‖ ‖v‖ cos θ

(u1 − v1)2 + (u2 − v2)2 =(u21 + u22

)+(v21 + v22

)− 2 ‖u‖ ‖v‖ cos θ

u21 + v21 − 2u1v1 + u22 + v22 − 2u2v2 = u21 + u22 + v21 + v22 − 2 ‖u‖ ‖v‖ cos θ

−2u1v1 − 2u2v2 = −2 ‖u‖ ‖v‖ cos θ

u1v1 + u2v2 = ‖u‖ ‖v‖ cos θu1v1 + u2v2‖u‖ ‖v‖

= cos θ

Here we have an expression that pops-up often enough when dealing with vectors that wedefine it as a dot product. That is,

u · v = u1v1 + u2v2.

If we were to use matrix notation, we would have

u · v = uvt = [u1v1 + u2v2] = u1v1 + u2v2.

You should note that the dot product is a scalar (single number) and not a vector (morethan one number). Finally we have a formula for the angle between two non-zero vectors, uand v is given by:

cos θ =u · v‖u‖ ‖v‖

(18)

You should note that if two non-zero vectors are perpendicular or orthogonal (θ = 90◦),then the dot product of these two vectors is zero, because the cos 90◦ is zero.

27.2 Examples

1. Find the component form45 and the magnitude of the vector v. (Express the componentsof v as fractions. Round the magnitude of v to four decimal places.)

Initial Point: (9/2, 0) , Terminal Point: (0, −9/2)

Solution: This will be discussed in class.

v =

⟨−9

2, −9

2

⟩‖v‖ =

9√

2

2≈ 6.3640

45We’ll need to get the vector in standard form first. Pay attention, because this is a very important step.

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2. Given.

u = 〈9, 9〉 , v = 〈−3, 0〉

(a) u + v


〈6, 9〉

(b) u− v


〈12, 9〉

(c) 2u + 3v


〈9, 18〉

(d) 3u− 2v


〈33, 27〉

3. Find a unit vector46 in the direction of the given vector.

u = 〈−3, 0〉


〈−1, 0〉

4. Find the vector v with the given magnitude and the same direction as u.

Magnitude ‖v‖ = 11, Direction u = 9i− 10j


99√181

i− 110√181

j

46A unit vector has length one.

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5. Find the magnitude and direction angle of the vector v.

v = 2 (cos 120◦i + sin 120◦j)


Magnitude is 2 and direction angle is 120◦.

6. Given

u = 2i + 3j, v = −2i + 5j

find

u · v.


11

7. Find s if u = 〈s, −5〉 and v = 〈6, s2〉 are orthogonal.


s = 0 or s = 6/5

8. Find the angle θ between the vectors.

u = i− j, v = −2i + 2j


π or 180◦

9. Use vectors to find the interior angles of the triangle with the given vertices.47

(−5, 0) , (4, 2) , (0, 4)

47Round to nearest hundredth of a degree.

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114.78◦, 26.13◦, 39.09◦



27.4 Homework



1. The initial and terminal points of a vector are given. Write the vector as a linearcombination of the standard unit vectors i and j.

Initial Point (0, −2) , Terminal Point (2, 9)

Solution: 2i + 11j

2. Given.

u = 4i− 2j, v = −3i + 5j

(a) u + v


i + 3j

(b) u− v


7i− 7j

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(c) 2u + 3v


−i + 11j

(d) 3u− 2v


18i− 16j

3. Given

u = 3i− j, v = −2i + 3j

find

u · v.

Solution: −9

4. Find a unit vector in the direction of the given vector.

u = 〈1, −2〉

Solution:

⟨1√5, − 2√

5

⟩

5. Given

u = 〈−1, 3〉 , v = 〈−5, −4〉 , w = 〈3, −7〉 ,

find.

(a) u− 3v

Solution: 〈14, 15〉

(b) ‖v‖

Solution:√

41

(c) 5w + 3v

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Solution: 〈0, −47〉

(d) 3 (5w + 2v)− 7u

Solution: 〈22, −150〉

(e) Rewrite u as a linear combination of i and j.

Solution: −i + 3j

(f) ‖v −w‖

Solution:√

73

6. Find the angle between u = 〈3, 7〉 and v = 〈−4, 2〉.48

Solution: 86.6◦

7. Use the dot product to find the magnitude of u.

u = 2i− 4j

Solution: 2√

5

8. Find the exact angle θ between the vectors.

u = 〈6, 7〉 , v = 〈−3, 1〉

Solution:

θ = arccos

(− 11

5√

34

)

48Approximate to the nearest tenth of a degree.

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28 Polar Coordinates; Parametric Equations

28.1 Parametric Equations

28.1.1 Technology

With few exceptions, we have only dealt with relationships between two variables—somefunctional, others not functional. In those rare cases where we had more than two variablerelationships we invariably found some way to reduce it to two variables. By now youshould have a good idea how to graph simple relationships between two variables includingthe appropriate use of technology. That is, when confronted with a difficult two variablerelationship to graph, you should be able to use a computer to construct a good detailedgraph. So if you’re asked to graph

y sin(x2 − y2

)= x cos (x+ y) ,

you would use technology (Figure 51, page 113) to do so. However, if you were asked to

-2 -1 0 1 2

-1

1

2

Figure 51: Partial graph of y sin (x2 − y2) = x cos (x+ y).

graph y = x2 − 9, it would not be necessary to use technology to do so. In many ways, youshould be able to visualize a relationship between variables, but at times these relationshipsmay prove insurmountable, and in those cases it is highly recommended that technology beemployed. In fact, much insight can be gained by using technology to visualize relationships.

28.1.2 Parameters

Now we are going to intentionally introduce a third variable, often called a parameter intoour two variable problems. Yes, as before we may decide to eliminate this third variable inan attempt to land on familiar ground of two variable relationships. However, no matter howfrustrating this becomes, you should always know that technology can deal with our limitedcapacity to resolve abstractions that are not yet familiar to us. For example, there was a time

Page 113 of 238




when jumping from linear relationships to more complex quadratics relationships provedinsurmountable in your academic careers, so you may have resorted to using a graphingutility. So please consider having a graphic calculator49 nearby for those times your brain istoo stretched to move on.

As an example, suppose we have a simple linear equation,

y = 2x+ 3,

where we now define x as a function of t as follows:

x = f (t) .

Certainly, if x is a function of t, we can also state that y is a functions of t as follows:

y = g (t) .

For this particular example we might say that

x = f (t) = 2t, and y = g (t) = 4t+ 3.

Yes, you probably think this is pointless, and in fact taking a simple problem and making itimpossible. For example, I think it’s really easy to graph y = 2x+ 3 (Figure 52, page 114),but not so easy to graph (f (t) , g (t)) (Figure 53, page 115).

First, the traditional way (Figure 52, page 114).

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

Figure 52: Graph of y = 2x+ 3

49This may mean your computer!

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Now the unfamiliar way (Figure 53, page 115).

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

Figure 53: Graph of (f (t) , g (t))

As you can see there is no difference.

28.1.3 Graphing a Parametric Equation

You should carefully look into using software to graph parametric equations, however, youshould still be able to reason through simple examples. Here’s one

x = f (t) = 2t2 + t, and y = g (t) = 2t, where − 3 ≤ t ≤ 3.

Here, I would like to suggest that you construct a simple table (Table 1, page 116) and thenconnect the dots in order. You’ll see that if you follow t from −3 forward that the graph(Figure 54, page 116). moves in a certain direction. For this example, the initial point is(f (−3) , g (−3)) = (15, −6), and the termial point is (f (3) , g (3)) = (21, 6). Now’s yourchance to do some real work.

1. Verify (Table 1, page 116) is correct. We’ll do this in class.

2. Plot the points, in order, directly on the given graph (Figure 54, page 116). Notice thedirection that these points follow. We’ll do this in class.

3. Try to eliminate the parameter to get a relationship between x and y.50 We’ll do thisin class.

50This relationship between x and y describes a set of points. The parametric equations have a greatadvantage though—if we consider t to be time, we can view these points as location in time and can easilyfollow these points to see how they’re moving.

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4. You should also see that the graph of the relationship between x and y may differ, asin this example.

Table 1: x = 2t2 + t and y = 2t

point t x yNo. 1 −3 15 −6No. 2 −2 6 −4No. 3 −1 1 −2No. 4 0 0 0No. 5 1 3 2No. 6 2 10 4No. 7 3 21 6

-4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

Figure 54: x = 2t2 + t and y = 2t in red, for −3 ≤ t ≤ 3, the black curve is 2x = y2 + y.

28.1.4 Easily Tricked?

Technology will certainly prove helpful, but using your head should be your main goal.Learning this material will require you to not only use your head, but also to explore tech-nology to see why your head may not always produce the correct answer. For example,suppose you’re asked to graph

x = f (t) = cos t y = g (t) = sin2 t, where t ∈ R.

And you decide to eliminate the paramter t, we’ll do this is class. You’ll get

x2 + y = 1,

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which is a very easy equation to graph. Unfortunately, the graphs (Figure 55, page 117)differ. To see why you’ll need to understand that in our parametric form, −1 ≤ x ≤ 1 and0 ≤ y ≤ 1, but in our cartesian form x ∈ R and y ≤ 1. Tricky, I know.

-2 -1 0 1 2

-1

1

Figure 55: Graph of (f (t) , g (t)) in red, and x2 + y = 1 in black.

28.1.5 Using Software

Here’s an interesting problem, graph x = cos (3t) and y = cos (t+ cos (3t)). I really don’tthink you can do this51 using a table and plotting points, so I want to recommend thatyou either use a calculator52 or a computer53. Since we have a site license for Mathematica,here’s the Mathematica code,

ParametricPlot[{Cos[3 t], Cos[t + Cos[3 t]]}, {t, 0, 2 Pi}],

that you should try-out today. Make sure your graph looks like the one in (Figure 56,page 118). You should also play around with the range for t in this example. You may alsowant to visit WolframAlpha54 and just type,

plot (cos(3 t), cos(t + cos(3 t))),

into the compute box.

51At least I can’t. I have literally no patients for this type of problem.52Yes, you may have to read your manual.53I’m use an application called Grapher, which is free on Mac OS X.54http://www.wolframalpha.com

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http://www.wolframalpha.com




-2 -1 0 1 2

-1

1

Figure 56: x = cos (3t) and y = cos (t+ cos (3t))

28.2 Polar Coordinate System

We will now use an alternative coordinate system call the polar coordinate system. Herewe will use the ordered pair (r, θ), where r represents the directed distance from the originto the point P , and θ represents the directed angle, counterclockwise from the polar axis.Here’s a sample of polar graph paper (Figure 57, page 119) to illustrate these concepts.Polar graph paper can be purchased, or you want to simply make copies of the graph paperprovided here (Figure 57, page 119).

In class we will plot the following points: (1, π/3), (2, −π/6), and (−2, −π/2). Youshould note that (−2, −π/2) is the same as (2, π/2). What’s another, possible better, way55

to write (−3, −π/3)? In general, there are an infinite number of representations of any givenpoint in the polar coordinate system.

28.2.1 Cartesian Relationship

By now you should know that we’ve used r and θ in our original definitions of the trigono-metric functions. Where r2 = x2 + y2, and

sin θ =y

r, cos θ =

x

r, tan θ =

y

x.

Initially we said that r > 0, but these relationships hold for r < 0 as well. In the polar system(not in trigonometric functions’ definition though) r can also be zero and this, regardless ofthe value of θ, is always equivalent to the cartesian point (0, 0). Now we can state that forr 6= 0, the following

x = r cos θ and y = r sin θ.

55(3, 2π/3)

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Figure 57: Polar graph paper.

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28.2.2 Graphing

Graphing can be a real challenge, especially if you’re still struggling with evaluating trigono-metric functions at nice values of θ. The general methods that we’ll use in class are: tables,converting to cartesian form, and my all-time favorite, a simple tool56.

Tools: Once tools are mastered you’ll start to be able to do many problems that are con-siderably more difficult than the run of the mill problems you typically see. However,try your level best to do things by hand before using a tool.57

• Visit http://www.wolframalpha.com/ to see a nice free browser-based applica-tion that can graph pretty sophisticated relationships.

• Most graphical calculators can easily do a wide range of graphs in a variety ofcoordinate systems.

• If you have access to Mac OS X you should check out the application calledGrapher—it’s in your Utilities folder.

• If you have an iPad you should install the Wolfram Alpha application.

• If you have a Windows PC you should try WinPlot.

• If you’re using UNIX (or related products such as Linux) you certainly don’t needhelp using software. Congratulations!

Converting: Given

r =4

1 + sin θ.

and the converting (we’ll do this in class) to

y =16− x2

8.

Both equations above will give the following graph Figure 58, page 121), albeit thecartesian form is really simple to graph.

Tables: This is perhaps the most tedious method known, but by plotting points you will geta pretty good idea how to graph relationships between r and θ, for example, supposedyou are asked to graph r = 2 sin θ · cos θ = sin 2θ? Seventeen points in all, but you mayalso decide to graph more points. These values for θ were very easy to evaluate, andthere are many others easy θs too, including 15◦ and 22.5◦. Don’t get carried away,but initially you may have to, especially if you’re having trouble connecting the dots.If you plot these seventeen points (we’ll do this in class ) on polar paper and thenconnect-the-dots in order, you’ll hopefully get a graph (Figure 59, page 121) that lookssimilar to mine.

56Visit: http://www.wolframalpha.com/ to see an example free tool.57Many people don’t know the value of gasoline and think its way too expensive—I just tell them to push

a car full of people for 30 miles and you will understand the value of gasoline. Cheap energy, like cheap food,has made us unfit to function. Please don’t resort to technology as the only way!

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http://www.wolframalpha.com/

http://www.wolframalpha.com/




Figure 58: Partial graph of r =4

1 + sin θ.

Figure 59: Graph of r = 2 sin θ cos θ = sin 2θ.

Page 121 of 238




point θ rNo. 1 0◦ = 0 0

No. 2 30◦ =π

6

√3

2≈ 0.87

No. 3 45◦ =π

61

No. 4 60◦ =π

3

√3

2≈ 0.87

No. 5 90◦ =π

20

No. 6 120◦ =2π

3−√

3

2≈ −0.87

No. 7 135◦ =3π

4−1

No. 8 150◦ =5π

6−√

3

2≈ −0.87

No. 9 180◦ = π 0

No. 10 210◦ =7π

6

√3

2≈ 0.87

No. 11 225◦ =5π

41

No. 12 240◦ =4π

3

√3

2≈ 0.87

No. 13 270◦ =3π

20

No. 14 300◦ =5π

3−√

3

2≈ −0.87

No. 15 315◦ =7π

4−1

No. 16 330◦ =11π

6−√

3

2≈ −0.87

No. 17 360◦ = π 0

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28.3 Examples

1. Consider the following.

x = t, y =t

7

(a) Graph for −∞ ≤ t ≤ ∞.


Your graph should look similar to mine (Figure 104, page 235).

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

Figure 60: Partial graph of x = t, y = t7.

(b) Eliminate the parameter and write the corresponding rectangular equation whosegraph represents the curve.


y =x

7

(c) Adjust and find the domain/range of the resulting rectangular equation so its graphis the same as the parametric form.


All real numbers.


x = |t− 1| , y = t+ 4

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-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

-2

-1

1

2

3

4

5

6

7

8

9

Figure 61: Partial graph of x = |t− 1| , y = t+ 4.



x = |y − 5|



x ≥ 0, −∞ < y <∞


x = e2t, y = 2et




Page 124 of 238




-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

-2

-1

1

2

3

4

5

6

7

8

9

Figure 62: Partial graph of x = e2t, y = 2et.



y = 2√x



x > 0, y > 0

4. A point in polar coordinates (r, θ) is given. Find the corresponding rectangular coordi-nates for the point.(

4,3π

2

)


(0, −4)

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5. Use a calculator to find the rectangular coordinates (x, y) of the point given in polarcoordinates (r, θ) . Round your results to two decimal places.

(7.80, 2.65)


(−6.88, 3.68)

6. Plot the point given in rectangular coordinates. Find two sets of polar coordinates forthe point for 0 ≤ θ < 2π.

(0, −9)


(9, 3π/2) ; (−9, π/2)

7. Convert the rectangular equation to polar form. Assume r > 0.

x2 + y2 = 64


r = 8

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8. Convert the rectangular equation to polar form. Assume r > 0.

7x+ 9y − 5 = 0


r =5

7 cos θ + 9 sin θ

9. Graph the polar equation.

r = 9 cos θ


This equation is equivalent to x2 + y2 = 9x, which is a circle. You should use software(Figure 63, page 128) to help you construct this graph.


r = 3 + 6 cos θ


You should use software (Figure 64, page 129) to help you construct this graph.


r = 2 cos (2θ)

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Figure 63: Using wolframalpha to graph r = 9 cos θ.


You should use software (Figure 65, page 130) to help you construct this graph.



28.5 Homework


Directions: Answer each of the following questions. Partial answers are provided, butno work is being provided as a guide. Yes, you need to be able to do these problems and you

Page 128 of 238




Figure 64: Using wolframalpha to graph r = 3 + 6 cos θ.

will be tested. You need to get help if you cannot do these problems exactly as was done inclass!


x = t, y = 3t3


Solution: Your graph should look similar to mine (Figure 66, page 130).


Solution:

y = 3x3


Solution: All real numbers.


x = 3 + 4 cos t, y = 5 + 4 sin t

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Figure 65: Using wolframalpha to graph r = 2 cos (2θ).

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

Figure 66: Partial graph of x = t, y = 3t3.

Page 130 of 238






-2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

1

2

3

4

5

6

7

8

9

Figure 67: Complete graph of x = 3 + 4 cos t, y = 5 + 4 sin t.


Solution:

(x− 3)2 + (y − 5)2 = 16


Solution: 1 ≤ y ≤ 9, −1 ≤ x ≤ 7

3. A point in polar coordinates (r, θ) is given. Find the corresponding rectangular coordi-nates for the point.(

2, −π4

)

Solution:(√2, −

√2)

4. Plot the point given in rectangular coordinates. Find two sets of polar coordinates forthe point for 0 ≤ θ < 2π.(

8√

3, −8)

Page 131 of 238




Solution:

(16, 11π/6) ; (−16, 5π/6)

5. Convert the polar equation to rectangular form. Drawing the picture here may be worth1000 words.

θ =2π

3

Solution:

y = −√

3x

6. Convert the polar equation to rectangular form.

r =1

1 + sin θ

Solution:

2y = 1− x2


r2 = 4 sin θ

Solution: You should use software (Figure 68, page 133) to help you construct thisgraph.


r = 8− sec θ

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Figure 68: Using wolframalpha to graph r2 = 4 sin θ.



r = 4 (1− 2 sin θ)


Page 133 of 238




Figure 69: Using wolframalpha to graph r = 8− sec θ.

Figure 70: Using wolframalpha to graph r = 4 (1− 2 sin θ).

Page 134 of 238




29 Trigonometric Form of Complex Numbers

Fact is, you were probably told that the unit imaginary number is necessary mainly becausewe need them to solve some quadratic equations. However, the fact remains that theirrise to importance is really not due to quadratics but more to do with an Italian namedGerolamo Cardano, and he is remembered here because he showed that complex numberswere absolutely necessary in solving a certain type cubic equation.

As I hope you recall the unit imaginary is denoted by i, where i =√−1. You should fur-

ther recall that a complex number is written as a linear combination of a real and imaginarypart, and is usually denoted by the letter z, where z = a+ bi.

We’re are going to visualize these complex numbers as an ordered pair (a, b). Here ais the position along the real axis (similar to the x-axis, and often labeled Re) and b is theposition along the imaginary axis (similar to the y-axis, and often labeled Im).

Here if we visualize the point (a, b), we can easily find many relationships to trigonometryand geometry. For example, in class we will illustrate the following relationships.

1. The modulus (or absolute value) of the complex number z = a+ bi is

|z| =√a2 + b2.

2. The complex number z = a+ bi has a trigonometric form (or polar form)

z = r (cos θ + i sin θ)

where r = |z| =√a2 + b2 and tan θ = b/a. The number r is the modulus of z, and θ

is an argument of z.

3. If two complex number z1 and z2 have polar forms

z1 = r1 (cos θ1 + i sin θ1) and z2 = r2 (cos θ2 + i sin θ2)

then

z1z2 = r1r2 [cos (θ1 + θ2) + i sin (θ1 + θ2)] multiplicationz1z2

=r1r2

[cos (θ1 − θ2) + i sin (θ1 − θ2)] , z2 6= 0 division

29.1 DeMoivre’s Theorem

If we were to take our multiplication rule, where z1 = z2 = z, we would get:

zz = rr [cos (θ + θ) + i sin (θ + θ)] ;

z2 = r2 [cos 2θ + i sin 2θ] .

Page 135 of 238




Of course if we keep repeating this n times for the same z we would get:

zn = rn [cosnθ1 + i sinnθ1] .

Basically DeMoivre’s Theorem tells us that to take the nth power of any complex number,we take the nth power of the modulus and multiple the argument by n.

In MTH-100 you were asked to raise complex numbers to integral powers, and this wasgenerally easy to do if the powers were small. For example:

(2 + 3i)2 = 4 + 12i+ 9i2 = 4 + 12i− 9 = −5 + 12i.

However, suppose you were asked to do(1

2+

1

2i

)10

?

Multiplying this out, even if you are familiar with expanding binomials, would be difficultfor most students. Using DeMoivre’s Theorem makes it a lot simpler. First write in polarform

1

2+

1

2i =

1√2

(cos

π

4+ i sin

π

4

),

then we have(1

2+

1

2i

)10

=

(1√2

)10(cos

10π

4+ i sin

10π

4

).

If you simplify you’ll find:(1

2+

1

2i

)10

=1

32

(cos

5π

2+ i sin

5π

2

)=

1

32i.

29.2 Roots

We can extend this to take nth roots of a complex number by letting w = n√z or wn = z. To

find the nth root of z, we basically need to find a complex number w such that wn = z. If

z = r (cos θ + i sin θ) ,

then

w = n√r

(cos

θ

n+ i sin

θ

n

)However, you may have noticed that the argument θ can be replaced by any θ+ 2πk, k ∈ Z.So we have

wk = n√r

(cos

θ + 2πk

n+ i sin

θ + 2πk

n

)where k = 0, 1, 2, . . . , n− 1.

As an example, let’s find the six sixth roots of z = −64. Here’s the steps:

Page 136 of 238




• Write the polar form of z.

z = 64 (cos π + i sin π)

• Use the formula.

wk =6√

64

(cos

π + 2πk

6+ i sin

π + 2πk

6

)• Now compute.

w0 =6√

64(

cosπ

6+ i sin

π

6

)=√

3 + i

w1 =6√

64

(cos

π + 2π

6+ i sin

π + 2π

6

)= 2i

w2 =6√

64

(cos

π + 4π

6+ i sin

π + 4π

6

)= −√

3 + i

w3 =6√

64

(cos

π + 6π

6+ i sin

π + 6π

6

)= −√

3− i

w4 =6√

64

(cos

π + 8π

6+ i sin

π + 8π

6

)= −2i

w5 =6√

64

(cos

π + 10π

6+ i sin

π + 10π

6

)=√

3− i

If were dealing with real numbers, the solution to

x6 + 64 = 0

would be the empty set. However, if you are dealing with complex numbers the solution tothis same equation would be{√

3 + i, 2i, −√

3 + i, −√

3− i, −2i,√

3− i}.

This is exactly what we learned in MTH 119 (there are six roots).

29.3 Examples

1. Find the modulus of the complex number 3 + 4i.

Solution: Work will be done in class. Make sure you take notes, or get notes from areliable classmate if you are not present during lecture.√

32 + 42 = 5

2. Write the complex number 1 + i in trigonometric form.

Page 137 of 238




Solution: Work will be done in class. Make sure you take notes, or get notes from areliable classmate if you are not present during lecture.√

2 (cos 45◦ + i sin 45◦)

3. Write the given trigonometric form of a complex number in the form a+ bi.

2

(cos

2π

3+ i sin

2π

3

)


−1 +√

3i

4. Let

z1 = 2(

cosπ

4+ i sin

π

4

)and z2 = 5

(cos

π

3+ i sin

π

3

),

find both z1 · z2 and z2/z1.


z1 · z2 = 10

(cos

7π

12+ i sin

7π

12

)z2z1

=5

2

(cos

π

12+ i sin

π

12

)5. Repeat the last problem using the form a+ bi and verify that it is the same.


They are the same.

z1 · z2 =5(√

2−√

6)

2+

5(√

2 +√

6)

2i

z2z1

=5(√

2 +√

6)

8+

5(√

6−√

2)

8i

You may want to consider learning how to do complex arithmetic on your calculatorstoo. You should be able to find i and your calculators are quite capable of doingarithmetic with both real and complex numbers. They’re certainly better than mosthumans when it comes to computational speed.

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6. Represent graphically the powers z, z2, z3, and z4 on the provided graph (Figure 72,page 143) paper.

z =1

2

(1 +√

3i)


z = 1 · (cos 60◦ + i sin 60◦)

z2 = 12 · (cos 120◦ + i sin 120◦)

z3 = 13 · (cos 180◦ + i sin 180◦)

z4 = 14 · (cos 240◦ + i sin 240◦)

-2 -1 0 1 2

-1

-0.5

0.5

1

00.4 0.8 1.2 1.6 2

0.1667π

0.3333π

0.5π

0.6667π

0.8333π

π

1.1667π

1.3333π

1.5π

1.6667π

1.8333π

Figure 71: Complex Plane

7. Use DeMoivre’s Theorem to find the indicated power of the complex number. Write theresult in standard form.[

2(

cosπ

3+ i sin

π

3

)]12

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z = 2(

cosπ

3+ i sin

π

3

)z12 = 212 (cos 4π + i sin 4π)

z12 = 4096

8. Find the three cube roots of 1.


z = 1 + 0 · iz = 1 · [cos 0◦ + i sin 0◦]

w3 = z

w0 =3√

1

[cos

0◦

3+ i sin

0◦

3

]= 1

w1 =3√

1

[cos

0◦ + 360◦

3+ i sin

0◦ + 360◦

3

]= −1

2+

√3

2i

w2 =3√

1

[cos

0◦ + 720◦

3+ i sin

0◦ + 720◦

3

]= −1

2−√

3

2i

9. Solve for x, where x is complex.

x4 + 81 = 0


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x4 + 81 = 0

x4 = −81

z = −81 + 0 · iz = 81 · [cos 180◦ + i sin 180◦]

x4 = z

x0 =4√

81

[cos

180◦

4+ i sin

180◦

4

]=

3√

2

2(1 + i)

x1 =4√

81

[cos

180◦ + 360◦

4+ i sin

180◦ + 360◦

4

]=

3√

2

2(−1 + i)

x2 =4√

81

[cos

180◦ + 720◦

4+ i sin

180◦ + 720◦

4

]=

3√

2

2(−1− i)

x3 =4√

81

[cos

180◦ + 1080◦

4+ i sin

180◦ + 1080◦

4

]=

3√

2

2(1− i)



29.5 Homework



1. Use DeMoivre’s Theorem to find the indicated power of the complex number. Write theresult in standard form.(

1 +√

3i)3

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Solution:

z = 1 +√

3i

z = 2 (cos 60◦ + i sin 60◦)

z3 = 8 (cos 180◦ + i sin 180◦)

z3 = −8

2. Verify using polar form that i9 is i.

Solution:

You should be able to due this using simple algebra too!

i = cos 90◦ + i sin 90◦

i9 = cos 810◦ + i sin 810◦

i = i9

3. Represent graphically the powers z, z2, z3, z4, z5, z6, and z7 on the provided graph(Figure 72, page 143) paper.

z =1

2

(1 +√

3i)

Solution:

z = 1 · (cos 60◦ + i sin 60◦)

z2 = 12 · (cos 120◦ + i sin 120◦)

z3 = 13 · (cos 180◦ + i sin 180◦)

z4 = 14 · (cos 240◦ + i sin 240◦)

z5 = 15 · (cos 300◦ + i sin 300◦)

z6 = 16 · (cos 360◦ + i sin 360◦)

z7 = 17 · (cos 420◦ + i sin 420◦) = z = 1 · (cos 60◦ + i sin 60◦)

Page 142 of 238




-2 -1 0 1 2

-1

-0.5

0.5

1

00.4 0.8 1.2 1.6 2

0.1667π

0.3333π

0.5π

0.6667π

0.8333π

π

1.1667π

1.3333π

1.5π

1.6667π

1.8333π

Figure 72: Complex Plane

4. Let

z1 =√

2 +√

2i and z2 =5

2+

5√

3

2i,

find both z1 · z2 and z2/z1.

Solution:

First, find trigonometric form.

z1 = 2(

cosπ

4+ i sin

π

4

)and z2 = 5

(cos

π

3+ i sin

π

3

),

then use DeMoivre’s Theorem.

z1 · z2 = 10

(cos

7π

12+ i sin

7π

12

)z2z1

=5

2

(cos

π

12+ i sin

π

12

)5. Find the three cube roots of z = 2 + 2i.


Page 143 of 238




z = 2 + 2i

z = 2√

2 [cos 45◦ + i sin 45◦]

w3 = z

w0 =3

√2√

2

[cos

45◦

3+ i sin

45◦

3

]=

√3 + 1

2+

√3− 1

2i

w1 =3

√2√

2

[cos

45◦ + 360◦

3+ i sin

45◦ + 360◦

3

]= −1 + i

w2 =3

√2√

2

[cos

45◦ + 720◦

3+ i sin

45◦ + 720◦

3

]=

1−√

3

2−√

3 + 1

2i

Again, you should be able to write the trigonometric form quickly, but the algebraicform may prove difficult to compute.


x5 − 1 = 0

Solution:

x5 − 1 = 0

x5 = 1

z = 1 + 0 · iz = 1 · [cos 0◦ + i sin 0◦]

x5 = z

x0 =5√

1

[cos

0◦

5+ i sin

0◦

5

]x1 =

5√

1

[cos

0◦ + 360◦

5+ i sin

0◦ + 360◦

5

]x2 =

5√

1

[cos

0◦ + 720◦

5+ i sin

0◦ + 720◦

5

]x3 =

5√

1

[cos

0◦ + 1080◦

5+ i sin

0◦ + 1080◦

5

]x4 =

5√

1

[cos

0◦ + 1440◦

5+ i sin

0◦ + 1440◦

5

]

Page 144 of 238





x5 +√

3 + i = 0

Solution:

x5 +√

3 + i = 0

x5 = −√

3− iz = −

√3− i

z = 2 · [cos 210◦ + i sin 210◦]

x5 = z

x0 =5√

2

[cos

210◦

5+ i sin

210◦

5

]x1 =

5√

2

[cos

210◦ + 360◦

5+ i sin

210◦ + 360◦

5

]x2 =

5√

2

[cos

210◦ + 720◦

5+ i sin

210◦ + 720◦

5

]x3 =

5√

2

[cos

210◦ + 1080◦

5+ i sin

210◦ + 1080◦

5

]x4 =

5√

2

[cos

210◦ + 1440◦

5+ i sin

210◦ + 1440◦

5

]

Page 145 of 238




37 Loci; Parabolas

37.1 Loci

In geometry, a locus (plural: loci) is a set of points whose location satisfies or is determinedby one or more specified conditions. For example, a circle is the set of points for which thedistance from a single point (the center) is constant (the radius). The equation,58

x2 + y2 − 2x+ 2y − 2 = 0,

basically describes a set of points that are all the same distance (r = 2) from the point(1, −1), it’s graph (Figure 73, page 146) illustrates the infinite number of points in thislocus. In general you should also understand that a circle is a set of points that is equidistant

-2 -1 0 1 2 3 4 5 6 7

-4

-3

-2

-1

1

Figure 73: Graph of x2 + y2 − 2x+ 2y − 2 = 0, with important features indicated.

from the center, so we can generalize the equations of a circle using the distance formula.Here we will let r represent the radius, and (h, k) the center. If we let (x, y) represent anarbitrary point on the circle we get59√

(x− h)2 + (y − k)2 = r.

Squaring both sides we get the standard from of a circle

(x− h)2 + (y − k)2 = r2

58Thinking back to MTH 100, you would have rewritten this equation in standard form,

(x− 1)2

+ (y + 1)2

= 2

59We’re using the distance formula.

Page 146 of 238




Again, the radius is r, the center is (h, k), and the point on the circle is (x, y). For example,if we have

(x− 3)2 + (y + 4)2 = 25,

the center is (3, −4), and the radius is 5. If you’re asked to graph the problem you shouldbe able to find four easy points along the circle’s circumference:

(3, −9) , (3, 1) , (−2, −4) , (8, −4) .

If the equation given is not in standard form you will need to use the method of completingthe square that was taught in MTH 100. For example:

x2 + y2 + 10x− 6y + 25 = 0

x2 + 10x+ y2 − 6y = −25

x2 + 10x+ 25 + y2 − 6y + 9 = −25 + 25 + 9

(x+ 5)2 + (y − 3)2 = 9

The center is (−5, 3), and the radius is 3. If you’re asked to graph the problem you shouldalso be able to find four easy points along the circle’s circumference:

(−5, 0) , (−5, 6) , (−8, 3) , (−2, 3) .

37.2 Parabola

-6 -5 -4 -3 -2 -1 0 1 2 3

-4

-3

-2

-1

1

2

Figure 74: Partial graph of f (x) = y = x2 + 2x− 3, with important features indicated.

Page 147 of 238




You should be able to graph a simple case of the general form of a parabola,

y = Ax2 +Bx+ C.

I strongly suggest you start by creating a table with simple points, at least six. I do expectthat you are capable to finding the following key features though:

• x-intercepts by setting y = 0. They’re not always there, but in general they’re worthlooking for. In our example above we would do the following.

y = x2 + 2x− 3

0 = x2 + 2x− 3

0 = (x+ 3) (x− 1)

So the x-intercepts are: (−3, 0) and (1, 0).

• y-intercept by setting x = 0. That’s just too easy! The y-intercept in the exampleabove is: (0, −3).

• The vertex, which is the point (at least in our examples) that will either be the highest(maximum) or lowest (minimum) point on our graph. This point is dead-center of thex values of the x-intercepts. If you use the quadratic formula and take the average(dead center) you’ll get a nice formula for the vertex.(

− B

2A, f

(− B

2A

))So in our example above we have (−1, −4) as the vertex.

• The axis-of-symmetry which is the dashed-line in our graph above. It should be notedthat this line is a folding line of symmetry. The equation of this line is x = −1.

Now we will move beyond this to a more formal approach. For example we can describe aparabola as a collection of points (locus) satisfying the following property:

• these points (x, y) on the parabola are equidistant from a fixed line (directrix), andfixed point (focus). The focus is not on the directrix. The midpoint between thedirectrix and focus is the vertex. The line passing through the focus and vertex iscalled the axis of symmetry (AOS) of the parabola.

The standard form of the equation of a parabola with vertex (h, k) is:

Horizontal Directrix: (x− h)2 = 4p (y − k), p 6= 0 with directrix y = k − p;

Vertical Directrix: (y − k)2 = 4p (x− h), p 6= 0 with directrix x = h− p.

Page 148 of 238




The focus lies on the AOS p units (directed distance) from the vertex.If you like60 you may attempt to derive these formulas by placing the vertex at zero,

using (x, y) as a point on the parabola, and the selecting point along the x or y axis to actas a focus. By selecting a focal point you should then be able to select a directrix. Oncedone you’ll need use simple translation to get the forms above. It’s really not difficult.

Here is a simple example (Figure 75, page 149) that should illustrate the key featuresof a parabola with equation, (x− 1)2 = 4 · 2 (y + 3), You should be able to find and labelthe vertex, AOS, focus, directrix. The other points, at x = 3 and x = 5, indicated on theparabola/directrix are there for you to verify the equidistance property.

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

Figure 75: Partial graph of (x− 1)2 = 4 · 2 (y + 3).

37.3 Examples

1. Find the standard form of the equation of the circle with the given characteristics.

Center: (−2, 6) ; point on circle: (−3, −2)


(x+ 2)2 + (y − 6)2 = 65

60Your teacher may do this as well.

Page 149 of 238




2. Write the equation of the circle in standard form. Then identify its center and radius.

x2 + y2 + 18x− 4y + 49 = 0


The standard form of the equation of the circle is (x+ 9)2 + (y − 2)2 = 36; center(−9, 2); radius is 6.

3. Find the x and y-intercepts of the graph of the circle.

x2 − 10x+ y2 + 8y = −16


The y-intercept is (0, −4); the x-intercepts are (2, 0) and (8, 0).

4. Find the standard form of the equation of the parabola with the given characteristic andvertex at the origin.

Focus:

(25

2, 0

)


The vertex is (0, 0); p = 25/2; the standard form of the equation of the parabola is(y − 0)2 = 50 (x− 0) or y2 = 50x.

5. Find the standard form of the equation of the parabola with the given characteristic andvertex at the origin.

Directrix: y = −2


The vertex is (0, 0); p = 2; the standard form of the equation of the parabola is(x− 0)2 = 8 (y − 0) or x2 = 8y.

Page 150 of 238




6. Find the vertex, focus, and directrix of the parabola and sketch its graph.

x = −7y2


x = −7y2

y2 = −1

7x

(y − 0)2 = 4 ·(− 1

28

)· (x− 0)

The vertex is (0, 0); the directrix is x = 1/28, p = −1/28; the focus is (−1/28, 0).Your graph (Figure 76, page 151) should look similar to mine.

-1 0

Figure 76: Partial graph of x = −7y2.


−x+ y2 = 0

Page 151 of 238





−x+ y2 = 0

y2 = x

(y − 0)2 = 4 ·(

1

4

)· (x− 0)

The vertex is (0, 0); the directrix is x = −1/4, p = 1/4; the focus is (1/4, 0). Yourgraph (Figure 77, page 152) should look similar to mine.

0 1 2 3 4

-1

1

Figure 77: Partial graph of −x+ y2 = 0.


y2 − 12y + 20− 16x = 0


Page 152 of 238




y2 − 12y + 20− 16x = 0

y2 − 12y + 20 = 16x

y2 − 12y + 20 + 16 = 16x+ 16

(y − 6)2 = 4 · (4) · (x+ 1)

The vertex is (−1, 6); the directrix is x = −5, p = 4; the focus is (3, 6). Your graph(Figure 78, page 153) should look similar to mine.

-15 -10 -5 0 5 10 15 20 25 30 35 40 45 50 55 60

-15

-10

-5

5

10

15

20

25

30

Figure 78: Partial graph of y2 − 12y + 20− 16x = 0.



37.5 Homework


Page 153 of 238





1. Find the equation of the circle given its graph (Figure 79, page 154).

-2 -1 0 1 2 3 4 5 6 7

-4

-3

-2

-1

1

Figure 79: Graph of a circle.

Solution:

(x− 1)2 + (y + 1)2 = 4

x2 + y2 − 2x+ 2y − 2 = 0

2. Find the center and radius of the circle whose equation is

x2 + y2 − 6x+ 6y + 9 = 0

Solution: r = 3; (3, −3)

3. Find the vertex and AOS of the parabola whose equation is

(y − 3)2 = 16 (x+ 2)

Page 154 of 238




Solution: y = 3; (−2, 3)

4. Find the equation of the parabola given its graph (Figure 80, page 155).

-2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

-6

-5

-4

-3

-2

-1

1

2

3

4

Figure 80: Partial graph of of a parabola.

Solution:

(y + 2)2 = 4 · 2 (x− 3)

5. Show that y2 − 8x + 2y + 9 = 0 is an equation of a parabola by rewriting the equationin standard form.

Solution:

(y + 1)2 = 8 (x− 1)

6. Find the focus, directrix, vertex, AOS, for the parabola whose equation is

(y + 1)2 = 8 (x− 1) .

Graph this parabola.

Page 155 of 238




Solution: The vertex is (1, −1); the directrix is x = −1, p = 2; the focus is (3, −1);AOS is y = −1. Your graph (Figure 81, page 156) should look similar to mine.

-5 0 5 10 15 20 25 30

-10

-5

5

10

Figure 81: Partial graph of of a (y + 1)2 = 8 (x− 1).

7. Find the equation of a parabola with focus (3, 4) and directrix is y-axis.

Solution:

(y − 4)2 = 4 · 3

2·(x− 3

2

)y2 − 8y + 25 = 6x

8. Find the equation of a parabola with focus (−4, 3) and directrix is y = 4.

Solution:

(x+ 4)2 = 4 · (−1) · (y − 3)

x2 + 8x+ 4 = −4y

Page 156 of 238




38 Elipses and Hyperbolas

38.1 The Ellipse

An ellipse is the set of all points (x, y) where the sum of the distances from two distinctpoints (foci) is constant. The line through the foci intersects the ellipse at two points calledthe vertices. The line segment joining the two vertices is called the major axis, and themidpoint on this line segment is called the center of the ellipse. The chord through themidpoint that is perpendicular to the major axis is called the minor axis. Here’s a picture(Figure 82, page 157) that we’ll label in class.

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12

-7

-6

-5

-4

-3

-2

-1

1

2

3

Figure 82: Graph of an ellipse whose equation is 4x2 + y2 − 8x+ 4y − 8 = 0.

The standard equations of is ellipse, with center (h, k) and major and minor axes oflengths 2a and 2b respectively, where 0 < b < a, where the major axis is:

Horizontal:

(x− h)2

a2+

(y − k)2

b2= 1

Vertical:

(x− h)2

b2+

(y − k)2

a2= 1

The foci lie on the major axis, c units from the center, with c2 = a2 − b2. For example inthe graph above (Figure 82, page 157) we can rewrite (we’ll do this in class) the equation,4x2 + y2 − 8x+ 4y − 8 = 0, as:

(x− 1)2

22+

(y + 2)2

42= 1

Page 157 of 238




The center here is (1, −2); the major axis is 2 · 4 = 8 units in length; the vertices are (1, 2)and (1, −6); the foci are

(1, −2− 2

√3)

and(1, −2 + 2

√3).

You’ll see terms related to the ellipse in the textbook61 that you will probably not befamiliar with. There is no need to memorize these terms!

Apogee: is the point in the orbit of the moon or a satellite at which it is furthest from theearth. These orbits may appear circular but they are really elliptical.

Perigee: the point in the orbit of the moon or a satellite at which it is nearest to the earth.It is opposite of the apogee.

Eccentricity: is a technical term indicating the deviation from circularity. The eccentricitye of an ellipse is given by the ratio

e =c

a.

Again, a is the distance from the center to the vertices, and c is the distance fromthe center to the foci. Since a > c we have 0 < e < 1 for every ellipse, and as e getscloser to 1 the ellipse becomes more elongated (egg shaped), and as e gets closer to 0the ellipse becomes more circular. For example, the moon’s orbit about the earth hase ≈ 0.05 (nearly circular), whereas Mercury’s orbit about the sun has e ≈ 0.2 (circular,but not as circular as the moon’s orbit.).

38.2 Hyperbola

A hyperbola is the set of points (x, y), the difference of whose distances from two distinctfixed points (foci) is a positive constant. The standard form of a hyperbola with center(h, k) is:

Vertical Transverse:

(x− h)2

a2− (y − k)2

b2= 1

Horizontal Transverse:

(y − k)2

a2− (x− h)2

b2= 1

The vertices are a units from the center, and the foci are c (c2 = a2+b2) units from the center.Two graphic examples follow and we will label them in class. Here is a graph (Figure 86,page 161) of the hyperbola whose equation is 9x2 − 4y2 − 18x − 16y = 43. In class we will

61I also occasionally see these terms in common news stories about the moon. Yes, I often have to lookupthe definitions when I come across these terms.

Page 158 of 238




-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12

-7

-6

-5

-4

-3

-2

-1

1

2

3

Figure 83: Partial graph of 9x2 − 4y2 − 18x− 16y = 43.

also show that 9x2 − 4y2 − 18x− 16y = 43 can be rewritten as

(x− 1)2

22− (y + 2)2

32= 1.

The center here is (1, −2); the vertical transverse x = 1; the vertices are (−1, −2) and(3, −2); the foci are

(1−√

13,−2)

and(1 +√

13, −2).

Here is another graph (Figure 84, page 159) of the hyperbola whose equation is 9y2 −4x2 + 54y + 32x = 19. In class we will also show that 9y2 − 4x2 + 54y + 32x = 19 can be

-5 0 5 10

-5

Figure 84: Partial graph of 9y2 − 4x2 + 54y + 32x = 19.

rewritten as

(y + 3)2

22− (x− 4)2

32= 1.

Page 159 of 238




The center here is (4, −3); the horizontal transverse y = −3; the vertices are (4, −1) and(4, −5); the foci are

(4,−3−

√13)

and(4, −3 +

√13).

Certainly graphing hyperbolas will be more of a challenge and I want to suggest lookingat asymptotic behavior as an aid. By asymptotic behavior, I mean as x and y go towardsinfinity, the graph of the hyperbola starts to look linear. We’ll cover the following linearequation in class and you should be clear on how to derive these.

Vertical Transverse: The asymptotes are:

y − k = ± ba

(x− h)

Horizontal Transverse: The asymptotes are:

y − k = ±ab

(x− h)

For example, the hyperbola whose equation is

(y + 3)2

22− (x− 4)2

32= 1,

would yield the following asymptotes

y + 3 = ±2

3(x− 4) .

Here’s the graph (Figure 85, page 160) .

-5 0 5 10

-5

Figure 85: Partial graph of 9y2 − 4x2 + 54y + 32x = 19.

And the hyperbola whose equation is

(x− 1)2

22− (y + 2)2

32= 1,

Page 160 of 238




would yield the following asymptotes

y + 2 = ±3

2(x− 1) .

Here’s the graph (Figure 86, page 161).

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12

-7

-6

-5

-4

-3

-2

-1

1

2

3

Figure 86: Partial graph of 9x2 − 4y2 − 18x− 16y = 43.

38.3 Examples


49x2 + 36y2 − 98x− 360y − 815 = 0

(a) Find the standard form of the equation of the ellipse.


(x− 1)2

62+

(y − 5)2

72= 1

(b) Find the center, vertices, foci, and eccentricity of the ellipse.


The center: (1, 5); the vertices: (1, −2), (1, 12);the foci:

(1, 5−

√13),(1, 5 +

√13); eccentricity

√13/7. To graph you should

also find these two points: (−5, 5), (7, 5).

Page 161 of 238




(c) Graph.


Your graph (Figure 101, page 228) should look similar to mine.

-18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

-8-7-6-5-4-3-2-1

12345678910111213141516

Figure 87: Partial graph of 49x2 + 36y2 − 98x− 360y − 815 = 0.

2. Find the standard form of the equation of the ellipse with the given characteristics andcenter at the origin.

Vertices: (0, ±8) ; foci: (0, ±7)


(x− 0)2(√15)2 +

(y − 0)2

82= 1

or

x2

15+y2

64= 1

Page 162 of 238





Foci: (±6, 0) ; major axis of length 14


x2

49+y2

13= 1

4. Find the standard form of the equation of the ellipse with the given characteristics.

Foci: (0, 0) , (8, 0) ; major axis of length 18


(x− 4)2

81+y2

65= 1


Center: (7, 4) ; a = 3c; foci: (1, 4) , (13, 4)


(x− 7)2

324+

(y − 4)2

288= 1

Page 163 of 238




6. Consider the following hyperbola.

y2

4− x2

16= 1

(a) Find the center.


(0, 0)

(b) Find the vertices.


(0, −2) and (0, 2)

(c) Find the foci.

Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.(0, −2

√5)

and(0, 2√

5)

(d) Find the asymptotes.


y = ±1

2x

(e) Graph.




(x+ 3)2

25− (y − 1)2

16= 1



(−3, 1)

Page 164 of 238




-20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

-12-11-10-9-8-7-6-5-4-3-2-1

123456789101112

Figure 88: Partial graph ofy2

4− x2

16= 1.



(2, 1) and (−8, 1)

(c) Find the foci.

Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.(−3−

√41, 1

)and

(−3 +

√41, 1

)(d) Find the asymptotes.


y = 1± 4

5(x+ 3)

(e) Graph.



Page 165 of 238




-20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

-12-11-10-9-8-7-6-5-4-3-2-1

123456789101112

Figure 89: Partial graph of(x+ 3)2

25− (y − 1)2

16= 1.


16x2 − 25y2 + 50y − 425 = 0

(a) Find the standard form.


x2

25− (y − 1)2

16= 1

(b) Find the center.


(0, 1)

(c) Find the vertices.


(−5, 1) and (5, 1)

(d) Find the foci.

Page 166 of 238




Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.(−√

41, 1)

and(√

41, 1)

(e) Find the asymptotes.


y = 1± 4

5x

(f) Graph.



-20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

-12-11-10-9-8-7-6-5-4-3-2-1

123456789101112

Figure 90: Partial graph of 16x2 − 25y2 + 50y − 425 = 0.

9. Find the standard form of the equation of the hyperbola with the given characteristicsand center at the origin.

Foci: (±8, 0) ; vertici: (±4, 0)

Page 167 of 238





x2

16− y2

48= 1


Vertices: (0, ±9) ; asymptotes: y = ±3

4x


y2

81− x2

144= 1



38.5 Homework




Foci: (0, 0) , (8, 0) ; major axis of length 12

Page 168 of 238




Solution:

(x− 4)2

36+y2

20= 1


Vertices: (0, ±9) ; asymptotes: y = ±3

5x

Solution:

y2

81− x2

225= 1


Foci: (±6, 0) ; major axis of length 14

Solution:

x2

49+y2

13= 1


y2

4− x2

16= 1


Solution: (0, 0)


Page 169 of 238




Solution: (0, −2) and (0, 2)

(c) Find the foci.

Solution:(0, −√

20)

and(0,√

20)

(d) Find the asymptotes.

Solution: y = ±1

2x

(e) Graph.

Solution: Your graph (Figure 91, page 170) should look similar to mine.

-20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

-12-11-10-9-8-7-6-5-4-3-2-1

123456789101112

Figure 91: Partial graph ofy2

4− x2

16= 1.


49x2 + 36y2 − 98x− 360y − 815 = 0


Solution:(x− 1)2

36+

(y − 5)2

49= 1

Page 170 of 238





Solution: The center: (1, 5); the vertices: (1, −2), (1, 12); the foci:(1, 5−

√13),(

1, 5 +√

13); eccentricity

√13/7.

(c) Graph.


-18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

-8-7-6-5-4-3-2-1

12345678910111213141516



Vertices: (0, ±8) ; foci: (0, ±7)

Solution:

x2

15+y2

64= 1


16x2 − 25y2 + 50y − 425 = 0

Page 171 of 238





Solution:x2

25− (y − 1)2

16= 1


Solution: (0, 1)


Solution: (−5, 1) and (5, 1)

(d) Find the foci.

Solution:(−√

41, 1)

and(√

41, 1)


Solution: y = 1± 4

5x

(f) Graph.


-20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

-12-11-10-9-8-7-6-5-4-3-2-1

123456789101112



Center: (7, 4) ; a = 3c; foci: (1, 4) , (13, 4)

Page 172 of 238




Solution:

(x− 7)2

324+

(y − 4)2

288= 1


Foci: (±8, 0) ; vertici: (±4, 0)

Solution:

x2

16− y2

48= 1


(x+ 3)2

25− (y − 1)2

16= 1


Solution: (−3, 1)


Solution: (2, 1) and (−8, 1)

(c) Find the foci.

Solution:(−3−

√41, 1

)and

(−3 +

√41, 1

)(d) Find the asymptotes.

Solution: y = 1± 4

5(x+ 3)

(e) Graph.


Page 173 of 238




-20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

-12-11-10-9-8-7-6-5-4-3-2-1

123456789101112

Figure 94: Partial graph of(x+ 3)2

25− (y − 1)2

16= 1.

Page 174 of 238




39 Rotation of Axes

39.1 Rotations

The graph62 of Ax2 +Bxy + Cy2 +Dx+ Ey + F = 0 is one of the following.

Circle: A = C, A 6= 0

Parabola: AC = 0, but A and C are not both zero.

Ellipse: AC > 0, A and C have the same signs.

Hyperbola: AC < 0, A and C have the different signs.

As you should know by now, none of our prior examples involved an xy term. In fact, allour examples so far had lines of symmetry that were perpendicular/parallel to the coordinateaxes. What we want to do now is eliminate the xy term by rotating θ counterclockwise froma point (x′, y′) to a point (x, y) that’s on the graph of Ax2 +Bxy+Cy2 +Dx+Ey+F = 0.This certainly makes the problem a lot more difficult. Once we eliminate the xy term wewill transform Ax2 +Bxy + Cy2 +Dx+ Ey + F = 0 into a more manageable form:

A′ (x′)2

+ C ′ (y′)2

+D′x′ + E ′y′ + F ′ = 0.

First, let the point (x, y) be rewritten as (r cos (θ + β) , r sin (θ + β)), and the point (x′, y′)be rewritten as (r cos (β) , r sin (β)). This may be visualized as a point (x′, y′) being rotatedθ to a point (x, y). A simple expansion and substitution will yield this transformation:

x = x′ cos θ − y′ sin θ and y = x′ sin θ + y′ cos θ.

Then, although challenging, you need to substitute these expressions into Ax2+Bxy+Cy2+Dx+Ey+F = 0 , expand and then find the θ that makes the xy term disappear. If you dothis you will find that the solution to this is:

cot 2θ =A− CB

.

Here’s the steps you need to take.

1. Make sure the problem is of this form: Ax2 + Bxy + Cy2 + Dx + Ey + F = 0. Thiswill only be done if B 6= 0. If B = 0 we have a much simpler problem.

2. Find the values if A, B and C

3. Solve for a normally acute angle θ using

cot 2θ =A− CB

.

62There are cases where the graph will not be any of these.

Page 175 of 238




4. Evaluate these expressions using the θ found above.

x = x′ cos θ − y′ sin θ and y = x′ sin θ + y′ cos θ

5. Substitute these expressions into Ax2 + Bxy + Cy2 + Dx + Ey + F = 0, and expandand you’ll end up with a new equation that is easier to graph.

6. Graph this new equation and then simply rotate the axes θ and you’ll have the graphof Ax2 +Bxy + Cy2 +Dx+ Ey + F = 0.

Let’s start with a very simple case, supposed you’re told to identify the key features ofthe hyperbola whose equation is xy − 1 = 0. Your first step is to identify A, B and C.

A = C = 0, and B = 1

Then

cot 2θ =A− CB

cot 2θ = 0

2θ =π

2

θ =π

4

Now, using this θ we have:

x = x′ cosπ

4− y′ sin π

4

x =x′√

2− y′√

2

y = x′ sinπ

4+ y′ cos

π

4

y =x′√

2+

y′√2

Now substitute this back into xy − 1 = 0 and we get:(x′√

2− y′√

2

)(x′√

2+

y′√2

)− 1 = 0

(x′)2 − (y′)2

2= 1

(x′ − 0)2(√2)2 − (y′ − 0)2(√

2)2 = 1

So just graph the hyperbola as written, and then rotate 45◦.

Page 176 of 238




Here, I certainly want to suggest that graphing the above hyperbola using this techniqueis painful to say the least. If I were asked to do this I would have simple graphed

y =1

x

and that’s it! However, let’s say you were asked to graph:

7x2 − 6√

3xy + 13y2 − 16 = 0

Let’s proceed as before. Your first step is to identify A, B and C.

A = 7, C = 13, and B = −6√

3

Then

cot 2θ =A− CB

cot 2θ =1√3

tan 2θ =√

3

2θ =π

3+ πk

θ =π

6+π

2k

θ =π

6

Now, using this θ we have:

x = x′ cosπ

6− y′ sin π

6

x =

√3x′

2− y′

2

y = x′ sinπ

6+ y′ cos

π

6

y =x′

2+

√3y′

2

Now substitute this back into 7x2 − 6√

3xy + 13y2 − 16 = 0 and we get:

0 = 7

(√3x′

2− y′

2

)2

− 6√

3

(√3x′

2− y′

2

)(x′

2+

√3y′

2

)+ 13

(x′

2+

√3y′

2

)2

− 16

16 = 4 (x′)2

+ 16 (y′)2

1 =(x′ − 0)2

22+

(y′ − 0)2

12

So just graph the ellipse as written,

1 =(x′)2

4+

(y′)2

1

Page 177 of 238




and then rotate 30◦. That is the ellipse is centered at the origin with vertices ±2 in thex′y′-system.

Here, I certainly want to state that the above transformation was difficult, so I also wantto suggest that you learn to use software to graph equations.

-2 -1 0 1 2

-1

1

Figure 95: Graph of 7x2 − 6√

3xy + 13y2 − 16 = 0, with axis rotated 30◦ using software.

39.1.1 Rotational Invariants

The rotation of the coordinate axes through an angle θ that transforms the equation Ax2 +Bxy + Cy2 +Dx+ Ey + F = 0 into the form A′ (x′)2 + C ′ (y′)2 +D′x′ + E ′y′ + F ′ = 0 hasthe following rotational invariants.

1. F = F ′

2. A+ C = A′ + C ′

3. B2 − 4AC = (B′)2 − 4A′C ′ or 4AC −B2 = 4A′C ′ because B′ = 0

So now just find the sign of 4AC − B2 to determine the sign of A′C ′. The graph63 isdetermined by the sign of A′C ′. (Discriminant method)

Parabola: 4AC = B2

Ellipse or Circle: 4AC > B2, a circle if B = 0, and A = C.

Hyperbola: 4AC < B2

63There are cases where the graph will not be any of these.

Page 178 of 238




39.2 Examples

1. The x′y′-coordinate system has been rotated θ degrees from the xy-coordinate system.The coordinates of a point in the xy-coordinate system are given. Find the coordinatesof the point in the rotated coordinate system.

θ = 45◦, (7, −7)


Using simple geometry you’ll get:(0, −7

√2). Or you may use:

7 = x′ cos 45◦ − y′ sin 45◦

7 =x′√

2− y′√

2

7√

2 = x′ − y′

−7 = x′ sin 45◦ + y′ cos 45◦

−7 =x′√

2+

y′√2

−7√

2 = x′ + y′

And solve the system:

7√

2 = x′ − y′

−7√

2 = x′ + y′

for x′ and y′ and you’ll get:(0, −7

√2).

2. Rotate the axes to eliminate the xy-term in the equation. Then write the equation instandard form. (Rotate the coordinate axes through an angle θ with 0 ≤ θ ≤ π/2.)

2x2 + 4√

3xy − 2y2 + 20 = 0

Sketch the graph of the resulting equation, showing both sets of axes.


To begin you should note that A = 2, B = 4√

3 and C = −2.

Page 179 of 238




cot 2θ =1√3

tan 2θ =√

3

2θ = 60◦ + 180◦k

θ = 30◦ + 90◦k

θ = 30◦

x =

√3x′ − y′

2

y =x′ +√

3y′

2

4 (x′)2 − 4 (y′)

2+ 20 = 0

(y′)2

5− (x′)2

5= 1


-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

Figure 96: Partial graph of 2x2 + 4√

3xy − 2y2 + 20 = 0.

Again, to do this graph, I suggest you graph 4x2 − 4y2 + 20 = 0 and then rotate 30◦

counterclockwise and you’ll have the graph of 2x2 + 4√

3xy − 2y2 + 20 = 0.

3. Rotate the axes to eliminate the xy-term in the equation. Then write the equation instandard form. (Rotate the coordinate axes through an angle θ with 0 ≤ θ ≤ π/2.)

13x2 − 6√

3xy + 7y2 − 80 = 0

Sketch the graph of the resulting equation, showing both sets of axes.

Page 180 of 238





cot 2θ = − 1√3

tan 2θ = −√

3

2θ = −60◦ + 180◦k

θ = −30◦ + 90◦k

θ = 60◦

x =x′ −

√3y′

2

y =

√3x′ + y′

2

4 (x′)2

+ 16 (y′)2 − 80 = 0

(x′)2

20+

(y′)2

5= 1


-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

Figure 97: Partial graph of 13x2 − 6√

3xy + 7y2 − 80 = 0.

4. Sketch the graph of the degenerate conic.

x2 + y2 + 10x+ 8y + 41 = 0

Page 181 of 238





It is a single point, (−5, −4)

5. Consider the following equation.

x2 + xy + 4y2 + x+ y − 6 = 0

(a) Use the discriminant to classify the graph of the equation.


Ax2 +Bxy + Cy2 +Dx+ Ey + F = 0

Parabola: 4AC = B2

Ellipse or Circle: 4AC > B2

Hyperbola: 4AC < B2

Ellipse

(b) Use the Quadratic Formula to solve for y.


0 = x2 + xy + 4y2 + x+ y − 6

0 = 4y2 + xy + y + x2 + x− 6

0 = 4y2 + (x+ 1) y +(x2 + x− 6

)y =

− (x+ 1)±√

(x+ 1)2 − 4 · 4 · (x2 + x− 6)

2 · 4

y =−x− 1±

√97− 14x− 15x2

8

(c) Use a graphing utility to graph the equation.




The material discussed in class covers parts of Chapter 39 of Schaum’s Precalculus. If, forwhatever reason, you missed class or just need additional time with this material, I suggest

Page 182 of 238




-5 -4 -3 -2 -1 0 1 2 3 4 5

-2

-1

1

2

Figure 98: Partial graph of x2 + xy + 4y2 + x+ y − 6 = 0.

that you read over Chapter 39 of Schaum’s Precalculus, and work the Solved Problems beforeattempting the homework assignment.

39.4 Homework



1. Sketch the graph of the degenerate conic.

x2 − y2 + 2x+ 1 = 0

Solution: This will graph as two lines, y = x+ 1 and y = −x− 1.

2. Given.

3x2 − 2√

3xy + y2 + 2x+ 2√

3y = 0.

(a) This equation is of the form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0. Identify thecoefficients.

Page 183 of 238




Solution:

A = 3

B = −2√

3

C = 1

D = 2

E = 2√

3

F = 0

(b) The quantity, B2 − 4AC is called the discriminant. Compute the discriminant forthis problem.

Solution:

B2 − 4AC = 0

(c) In general if the discriminant is: 0, the graph is a parabola; negative, the graph isan ellipse or circle (B = 0, A = C); positive, the graph is a hyperbola. What is thisgraph?

Solution: Parabola

(d) Solve

cot 2θ =A− CB

,

for θ, where 0 < θ < π/2.

Solution:

cot 2θ =3− 1

2√

3

tan 2θ = −√

3

2θ = −π3

+ πk

θ = −π6

+π

2k

To meet the condition that 0 < θ < π/2, we take k = 1 and get θ = 60◦ = π/3.

(e) Use computer software to graph 3x2 − 2√

3xy + y2 + 2x+ 2√

3y = 0.

Solution: Your graph should look like mine (Figure 99, page 185).

(f) Transform 3x2 − 2√

3xy + y2 + 2x+ 2√

3y = 0 using,

x = x′ cos 60◦ − y′ sin 60◦ and y = x′ sin 60◦ + y′ cos 60◦.

Solution: Might be helpful to know how to use a computer algebra system (CAS),

Page 184 of 238




-6 -5 -4 -3 -2 -1 0 1 2 3

-4

-3

-2

-1

1

Figure 99: 3x2 − 2√

3xy + y2 + 2x+ 2√

3y = 0; y =√

3x in red; and y2 = −x in blue.

one is actually built into your graphics calculator.

x =x′ − y′

√3

2

y =x′√

3 + y′

2

0 = 3x2 − 2√

3xy + y2 + 2x+ 2√

3y

0 = 3

(x′ − y′

√3

2

)2

− 2√

3

(x′ − y′

√3

2

)(x′√

3 + y′

2

)

+

(x′√

3 + y′

2

)2

+ 2

(x′ − y′

√3

2

)+ 2√

3

(x′√

3 + y′

2

)0 = 4y′2 + 4x′

y′2 = −x′

Again, I strongly suggest you learn how to use your calculator’s built in CAS, asthis example clearly indicates.

3. Solve the system of quadratic equations algebraically.{x2 − 4y2 − 10x − 64y − 247 = 0

16x2 + 4y2 − 160x + 64y + 400 = 0

Then verify your results by using a graphing utility to graph the equations and find anypoints of intersection of the graphs.

Page 185 of 238




Solution: Just add the equations together and you’ll get after simplification: x2 −10x+ 9 = 0. Solving this equation for x, yields x = 1 or x = 9. Substitute these valuesinto either equation and you’ll get: (1, −8) and (9, −8). Your graph (Figure 100,page 186) should look similar to mine, and is in full support of the answers above.

-26 -25 -24 -23 -22 -21 -20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53

-21-20-19-18-17-16-15-14-13-12-11-10-9-8-7-6-5-4-3-2-1

12345678

Figure 100: Partial graph of the system, indicating the points of intersection of the ellipseand hyperbola.

Page 186 of 238




41 Sequences and Series

41.1 Sequences

An infinite sequence is a function having for its domain the set of positive integers,{1, 2, 3, . . . }. A finite sequence is a function having for its domain a set of finite pos-itive integers, {1, 2, 3, . . . , n}. Although we normally talk about positive integers for thedomain, it is often convenient to start at a non-positive integer and you should be especiallycareful to note the starting number in a sequence’s domain. The terms of the sequence aredenoted as follows,

a1, a2, . . . , an,

where

an = f (n) .

These functions can be defined in a variety of ways. For example:

Explicitly: We will discuss this is class.

an = f (n) =(−1)n

3n2 + 1

Recursively: We will discuss this is class.

an = f (n) =(n+ 1)

an−1, a1 = 5

41.2 Series

We will also add the terms of a sequence together, typically called a sum or series. Initiallywe will do this the old-fashioned way—that is, by adding each terms to the next until weget to the end. For example,

1 + 2 + 3 + · · ·+ 100 = 5050,

may take some time to compute, but it is doable. This old-fashioned way of adding willbecome quite tedious and we will eventually move towards using formulas (with reasonablepatterns) to find sums. Notationally a finite sum looks like this

a1 + a2 + a3 + · · ·+ an =n∑i=1

ai,

and notationally infinite sum looks like this

a1 + a2 + a3 + · · · =∞∑i=1

ai.

Page 187 of 238




Again, initially we will have a reasonable number of terms and we should be able to do thisby hand. However, you may also want to investigate your calculator’s ability to add. Almostall graphics calculator have the ability to do sums.

You may be wondering if you can add an infinite number of terms together, and I wantto remind that you already have many times in the past—even in grade school.

3

10+

3

100+

3

1000+ · · · =

∞∑i=1

3

10i=

1

3

41.3 Examples

1. Write the first five terms of the sequence.64

an =

(1

2

)n+2


a1 = 1/8

a2 = 1/16

a3 = 1/32

a4 = 1/64

a5 = 1/128


an =2n

n+ 3


a0 = 0

a1 = 1/2

a2 = 4/5

a3 = 1

a4 = 8/7

64Assume n begins with 1.65Assume n begins with 0.

Page 188 of 238





an = 8 (−1)2n−1(

n

n− 1

)


a2 = −16

a3 = −12

a4 = −32/3

a5 = −10

a6 = −48/5

4. Write an expression for the apparent nth term of the sequence.67

−2, 1, 4, 7, 10, . . .


an = 3n− 5


an = n an−1, where a1 = 1.


66Assume n begins with 2.67Assume n begins with 1.68Assume n begins with 2.

Page 189 of 238




a2 = 2

a3 = 6

a4 = 24

a5 = 120

a6 = 720

This sequence is in fact a sequence of factorial, where an = n! = n (n− 1) (n− 2) · · · 1.

a1 = 1

a2 = 2 · 1a3 = 3 · 2 · 1a4 = 4 · 3 · 2 · 1a5 = 5 · 4 · 3 · 2 · 1a6 = 6 · 5 · 4 · 3 · 2 · 1

Furthermore, a0 = 1. As you can imagine, this sequence grows rather quickly, forexample,

a15 = 15 · 14 · 13 · . . . · 3 · 2 · 1 = 1307674368000

Once again, your calculator has a factorial function built in, and you may want to seehow high your calculator can go.


2

3,

3

7,

4

11,

1

3,

6

19, . . .


an =n+ 1

4n− 1

7. Find the sum.

5∑i=0

6i2

69Assume n begins with 1.

Page 190 of 238





330

8. Find the sum.

4∑i=0

1

1 + i


137/60

9. Find the sum.

5∑k=1

(k − 1) (2k + 1)


90

10. Use sigma notation to write the sum.

7

11− 1+

7

11− 2+

7

11− 3+ · · ·+ 7

11− 10


10∑i=1

7

11− i

Page 191 of 238




11. Given the sequence

1, 2, 2,4

3,

2

3,

4

15, . . .

find a formula for the nth term.70


Where n begins with 0.

an =2n

n!



41.5 Homework




an = 3 (−2)n

70Hint: Its has something to do with 2n and n!.71Assume n begins with 1.

Page 192 of 238




Solution:

a1 = −6

a2 = 12

a3 = −24

a4 = 48

a5 = −96


an =n

2n− 1

Solution:

a0 = 0

a1 = 1

a2 = 2/3

a3 = 3/5

a4 = 4/7


an = (−1)n−1 sin(πn

4

)

Solution:

a1 = 1/√

2

a2 = −1

a3 = 1/√

2

a4 = 0

a5 = −1/√

2

72Assume n begins with 0.73Assume n begins with 1.

Page 193 of 238





−1, 2, −4, 8, −16, . . .

Solution:

an = (−1)n 2n−1


an =an−1

4, where a1 = 12.

Solution:

a2 = 3

a3 = 3/4

a4 = 3/16

a5 = 3/64

a6 = 3/256


1

2,

2

5,

3

10,

4

17,

5

26, . . .

Solution:

an =n

n2 + 1

74Assume n begins with 1.75Assume n begins with 2.76Assume n begins with 1.

Page 194 of 238




7. Find the sum.

4∑i=0

i

i2 + 2

Solution: 115/99

8. Find the sum.

10∑i=1

2i+ 1

Solution: 120

9. Find the sum.

4∑k=1

(k − 1) (2k − 1)

Solution: 34

10. Use sigma notation to write the sum.

1

7 + 2+

2

7 + 4+

3

7 + 6+ · · ·+ 15

7 + 30

Solution:

15∑i=1

i

7 + 2i

Page 195 of 238




42 The Principle of Mathematical Induction

Let Pn be a statement involving positive integers n. If

1. P1 is true, and

2. the truth of Pk implies the truth of Pk+1 for every positive integer k,

then Pn must be true for all positive integers n.This is a very simple proof technique, and I think everyone should make an attempt to

understand how to construct a proof using the Principle of Mathematical Induction (PMI).Better still, everyone should try to reason why this method is called a proof technique.

The following very important formulas can easily be proved using Principle of Mathe-matical Induction (PMI)77.

n∑i=1

i = 1 + 2 + 3 + · · ·+ n =n (n+ 1)

2

n∑i=1

i2 = 12 + 22 + 32 + · · ·+ n2 =n (n+ 1) (2n+ 1)

6

n∑i=1

i3 = 13 + 23 + 33 + · · ·+ n3 =

[n (n+ 1)

2

]2n∑i=1

i4 = 14 + 24 + 34 + · · ·+ n4 =n (n+ 1) (2n+ 1) (3n2 + 3n− 1)

30

n∑i=1

i5 = 15 + 25 + 35 + · · ·+ n5 =n2 (n+ 1)2 (2n2 + 2n− 1)

12

No, you do not need to memorize these formulas, but when you get to calculus you shouldbe able to use them without too much trouble.

42.1 Examples

1. Given

Pk =6

(k + 4) (k + 6),

find Pk+1.


Pk+1 =6

(k + 5) (k + 7)

77We’ll do two in class in the Examples section.

Page 196 of 238




2. Find Pk+1 for the given Pk. (Give your answer in terms of Pk)

Pk = 11 + 13 + 15 + · · ·+ [2 (k − 1) + 9] + [2k + 9]


Pk+1 = Pk + [2k + 11]

3. Use PMI to prove.

1 + 2 + 3 + · · ·+ n =n (n+ 1)

2


Verify for n = 1.

P1 : 1 =1 (1 + 1)

21 = 1

Assume Pk and show that Pk → Pk+1.

Pk : 1 + 2 + 3 + · · ·+ k =k (k + 1)

2

Add (k + 1) to both sides.

Pk : 1 + 2 + 3 + · · ·+ k =k (k + 1)

2

1 + 2 + 3 + · · ·+ k + (k + 1) =k (k + 1)

2+ (k + 1)

1 + 2 + 3 + · · ·+ k + (k + 1) =(k + 1) (k + 2)

2

This last line is exactly what we wanted, Pk+1. Q.E.D.


13 + 23 + 33 + · · ·+ n3 =

[n (n+ 1)

2

]2

Page 197 of 238





Verify for n = 1.

P1 : 13 =

[1 (1 + 1)

2

]21 = 1


Pk : 13 + 23 + 33 + · · ·+ k3 =

[k (k + 1)

2

]2Add (k + 1)3 to both sides.

Pk : 13 + 23 + 33 + · · ·+ k3 =

[k (k + 1)

2

]213 + 23 + 33 + · · ·+ k3 + (k + 1)3 =

[k (k + 1)

2

]2+ (k + 1)3

13 + 23 + 33 + · · ·+ k3 + (k + 1)3 = (k + 1)2[k2

4+ (k + 1)

]13 + 23 + 33 + · · ·+ k3 + (k + 1)3 = (k + 1)2

[k2 + 4k + 4

4

]13 + 23 + 33 + · · ·+ k3 + (k + 1)3 = (k + 1)2

[(k + 2)2

4

]

13 + 23 + 33 + · · ·+ k3 + (k + 1)3 =

[(k + 1) (k + 2)

2

]2This last line is exactly what we wanted, Pk+1. Q.E.D.



Page 198 of 238




42.3 Homework




n∑i=1

i2 = 12 + 22 + 32 + · · ·+ n2 =n (n+ 1) (2n+ 1)

6

Solution: Make sure you understand the method to construct a PMI proof.

2. Use PMI to prove.78

n∑i=1

i4 = 14 + 24 + 34 + · · ·+ n4 =n (n+ 1) (2n+ 1) (3n2 + 3n− 1)

30


3. Use PMI to prove.79

n∑i=1

i5 = 15 + 25 + 35 + · · ·+ n5 =n2 (n+ 1)2 (2n2 + 2n− 1)

12


78Here you may want to use your calculator’s ability to do symbolic algebra.79Here you may want to use your calculator’s ability to do symbolic algebra.

Page 199 of 238




43 Special Sequences and Series

43.1 Arithmetic Sequences and Series

Definition: A sequence is arithmetic if the difference between consecutive terms are thesame. So the sequence is of this form

a1, a1 + d, a1 + 2d, . . . .

You should note that the nth term is simply

an = a1 + (n− 1) d.

In class we will discuss how to sum together a finite arithmetic sequence. You should tryto make sense out of this—memorization will fail you in the long run. Please try to reason!

To sum the integers from 1 to n, we have80

1 + 2 + 3 + · · ·+ n =n (n+ 1)

2.

To sum an arithmetic series, we have81

a1 + a2 + a3 + · · ·+ an = an + [a1 + d] + [a1 + 2d] + · · ·+ [a1 + (n− 1) d]

= n · a1 + d · [1 + 2 + 3 + · · ·+ (n− 1)]

= n · a1 + d · n (n− 1)

2

=n · [2a1 + d · (n− 1)]

2

=n · (a1 + an)

2

43.2 Geometric Sequences and Series

Definition: A sequence is geometric if the ratio between consecutive terms are the same.So the sequence is of this form

a1, a1 · r, a1 · r2, . . . .

You should note that the nth term is simply

an = a1 · rn−1.

It should be noted that r 6= 0.

80This is really very simple to derive and I will cover this in class. Everyone should be able to do this ontheir own!

81Nothing to memorize, but you will need to follow what I do in class in order to make sense out of theseformulas.

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In class we will discuss how to sum together a finite/infinite geometric sequence. Youshould try to make sense out of this—memorization will fail you in the long run. Please tryto reason! To sum a geometric series, for finite sums we have82

a1 + a2 + a3 + · · ·+ an = an + a1 · r + a1 · r2 + · · ·+ a1 · rn−1

=a1 − r · a1

1− r

=

(1− rn

1− r

)a1

If −1 < r < 1 and we let n→∞, we have

a1 + a2 + a3 + . . . = an + a1 · r + a1 · r2 + · · ·=

a11− r

43.3 Examples

1. Find a formula for an for the arithmetic sequence.

a1 = 11, d = 5


an = 6 + 5n


19, 14, 9, 4, −1, . . .


an = 24− 5n


a1 = 110, a10 = 35

82Nothing to memorize, but you will need to follow what I do in class in order to make sense out of theseformulas.

Page 201 of 238





an =355− 25n

3

4. Write the first five terms of the arithmetic sequence. Find the common difference andwrite the nth term of the sequence as a function of n.

a1 = 390, ak+1 = ak − 10


a1 = 390

a2 = 380

a3 = 370

a4 = 360

a5 = 350

d = −10

an = 400− 10n

5. Find the sum of the first 190 positive odd integers.


a1 = 1

a2 = 3

a3 = 5

a4 = 7

a5 = 9

d = 2

an = 2n− 1

a190 = 379190∑i=1

(2i− 1) = 36100

Page 202 of 238




6. Find the sum of the integers from −30 to 80.


a1 = −30

a2 = −29

a3 = −28

a4 = −27

a5 = −26

d = 1

an = n− 31

80 = n− 31

n = 111111∑i=1

(n− 31) = 2775 =111 (−30 + 80)

2

7. Write the first five terms of the geometric sequence.

a1 = 6, r = 3


a1 = 6

a2 = 18

a3 = 54

a4 = 162

a5 = 486

8. Write the first five terms of the geometric sequence. Find the common ratio and writethe nth term of the sequence as a function of n.

a1 = 256, ak+1 =ak4

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a1 = 256

a2 = 64

a3 = 16

a4 = 4

a5 = 1

r =1

4

an =256

4n−1= 210−2n

9. Find the sum. (Round your answer to two decimal places.)

13∑n=0

2

(2

7

)n


Geometric series.14∑i=1

2

(2

7

)n−1=

14(1− (2/7)14

)5

≈ 2.80

You may want to try to compute this sum as n→∞ and you will see that this sum isactually 14/5 = 2.8.

10. Find the rational number representation of the repeating decimal.

3.864


Infinite geometric series.

Method 1:

x = 3.864

1000x = 3864.864

999x = 3861

x =3861

999=

143

37

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Method 2:

3.864 = 3 +864

1000+

864

1000000+ · · ·

= 3 +864/1000

1− 1/1000

= 3 +32

37=

143

37

3861

999=

143

37



43.5 Homework



1. Identify the sequence as arithmetic, geometric, or neither.

3

4,

9

8,

27

16, . . .

Solution: Geometric, r = 3/2


7, 5, 3, . . .

Page 205 of 238




Solution: Arithmetic, d = −2


1

2,

1

3,

1

4, . . .

Solution: Neither


2, 4, 8, . . .

Solution: Geometric, r = 2

5. Write the first five terms of the arithmetic sequence.

a4 = 27, a10 = 63

Solution: an = 3 + 6n, {9, 15, 21, 27, 33}

6. Find the indicated nth partial sum of the arithmetic sequence.

a1 = 32, a180 = 310, n = 100

Solution:

a1 = 32

a180 = 310

d =278

179

an =278n+ 5450

179

a100 =33250

179100∑i=1

278n+ 5450

179=

1948900

179=

100 (32 + 33250/179)

2

Page 206 of 238




7. Find a formula for the nth term of the geometric sequence. Then find the indicated nthterm of the geometric sequence.

21st term : 4, 8, 16, . . .

Solution: an = 4 · 2n−1 = 2n+1; a21 = 222 = 4194304

8. Given a positive term geometric sequence with a3 = 16/3 and a5 = 64/27, find r, a1,and a10 Also find the sum of the first 10 terms. If possible find the infinite sum startingat a1, that is, add all the terms in the infinite sequence together.

Solution:

Geometric.

r = 2/3

a1 = 12

a10 =2048

656110∑i=1

12

(2

3

)i−1= 36− 36

(2

3

)10

=232100

6561

∞∑i=1

12

(2

3

)i−1= 36

9. Find the sum. Both exact and approximate to four decimal places.

15∑n=0

(−3)n−1

Solution: Geometric series.10761680

3≈ 3587226.6667

10. Find the rational number representation of the repeating decimal.

2.098

Page 207 of 238




Solution: Infinite geometric series.

2078

990=

1039

495

Page 208 of 238




44 Binomial Theorem

In the expansion of (a+ b)n, where n ∈ Z+, we have this very simple pattern:

(a+ b)n = an + k1an−1b+ k2a

n−2b2 + k3an−3b3 + · · ·+ kn−2a

2bn−2 + kn−1abn−1 + bn.

The pattern should be easy to follow, but the constants ki may in fact be difficult to compute.Here’s a short list to get us started finding the coefficients.

(a+ b)0 = 1

(a+ b)1 = a+ b

(a+ b)2 = a2 + 2ab+ b2

(a+ b)3 = a3 + 3a2b+ 3ab2 + b3

(a+ b)4 = a4 + 4a3b+ 6a2b2 + 4ab3 + b4

You could certainly write down the expansion for (a+ b)7, but you may have a difficult timewith the coefficients. Here’s what you should be able to do right now:

(a+ b)7 = a7 + k1a6b+ k2a

5b2 + k3a4b3 + k4a

3b4 + k5a2b5 + k6ab

6 + b7.

There are many ways to compute ki, and the first way for many is to construct a triangle(called Pascal’s Triangle) of coefficients that has a very simple pattern once it is pointedout. We’ll do this in class, but I want to suggest that this is not the way to proceed indetermining he coefficients of a binomial expansion.

44.1 Binomial Coefficients

The coefficients of the binomial expansion can be computed83 using this formula:

nCr =

(nr

)=

n!

(n− r)!r!,

where n represents the degree (row of Pascal’s Triangle) and r represents the the position ofeach term starting with 0 and ending with n in each expansion (row of Pascal’s Triangle).Using this formula, you should be able to generate any row you want without having tocompute prior rows. These coefficients actually indicate the number of district arrangementsof the binomial entry’s variable part. For example, if you look at the term a2b in theexpansion of (a+ b)3 you notice that its coefficient is 3 and this is the actual number ofarrangements84 of a2b. Now let’s say we were asked to find the number of arangments ofa7b5, we would get a staggering number of

12C7 =12 C5 =

(127

)=

(125

)=

12!

7!5!= 792 arrangements.

83As a reminder, n! = 1 · 2 · 3 · · · · · n.84The arrangements of a2b are: aab, aba, baa.

Page 209 of 238




There’s no way that I could have obtained that by listing all the distinct orderings of a7b5,nor could I have created the thirtieth row (n = 12) of Pascal’s Triangle either. So, you’llneed to be able to generate coefficients using this formula instead.

Here’s the binomial formula using this new notation. In the expansion of (a+ b)n, wheren ∈ Z+, we now have:

(a+ b)n =n∑r=0

(nr

)an−rbr

44.2 Examples

1. Evaluate.(1614

)


(1614

)= 120

2. Evaluate 27C23.


27C23 = 17550

3. Use the Binomial Theorem to expand and simplify the expression.

(x− y)5


(x− y)5 = (x+ [−y])5 = x5 − 5x4y + 10x3y2 − 10x2y3 + 5xy4 − y5

Page 210 of 238




4. Find the fifth term in from the left in the expansion of the binomial.

(3x− 10y)7


(74

)(3x)3 (−10y)4 = 9450000x3y4

5. Solve for x.85

4∑k=0

(4k

)x4−k5k = 81


Rewrite as a binomial to a power.

(x+ 5)4 = 81

x+ 5 = ±3

x = −5± 3

So x = −2 or x = −8.

6. Find the term of(3x2

2− 1

3x

)12

that does not contain x.

85Hint: It’s a binomial!

Page 211 of 238





The general term has this form.(12k

)(3x2

2

)k (− 1

3x

)12−k

We’re looking for x to have a zero exponent.

x2k

x12−k= x3k−12 ⇒ k = 4

Substitute k = 4 in the general term and simplify.(124

)(3x2

2

)4(− 1

3x

)8

=

(124

)(1

6

)4

=55

144



44.4 Homework



1. Write the first three terms in the binomial expansion.(2x2 − 3y

)5

Page 212 of 238




Solution:

32x10 − 240x8y + 720x6y2

2. Write the fourth term in the expansion.(√x− 1√

x

)5

Solution:

10√x

3. Use the Binomial Theorem to expand and simplify the expression.

(2x− 3y)4

Solution:

(2x− 3y)4 = 16x4 − 96x3y + 216x2y2 − 216xy3 + 81y4

4. Verify.(nr

)=

(n

n− r

)

Solution: It is true.

5. Find the middle term in the expansion.(x3 + 2y2

)10

Page 213 of 238




Solution:

8064x15y10

Page 214 of 238




1 Final Exam Review

These questions are being provided as review for the final exam. They will not be covered inclass, and if you need help with these problems I suggest you get help during my scheduledoffice hours.

1. Use csc θ = 3 and sec θ =3√

2

4to find the exact value of each of the following.

(a) The quadrant that θ is in.

Solution: First (I) quadrant

(b) sin θ

Solution: sin θ = 1/3

(c) tan θ

Solution: tan θ =√

2/4

(d) cos θ

Solution: cos θ = 2√

2/3

(e) sec (90◦ − θ)

Solution: sec (90◦ − θ) = csc θ = 3

2. Evaluate (exact values) all six trigonometric functions for x = −120◦.

(a) sinx =

Solution: sinx = −√

3/2

(b) cos x =

Solution: cosx = −1/2

(c) tan x =

Solution: tanx =√

3

(d) cot x =

Solution: cotx = 1/√

3

(e) secx =

Solution: secx = −2

Page 215 of 238




(f) cscx =

Solution: cscx = −2/√

3

3. Use your calculator to evaluate the trigonometric function. Round your answers to fivedecimal places.

(a) sin 11.67◦ =

Solution: sin 11.67◦ ≈ 0.20227

(b) cos 0.345

Solution: cos 0.345 ≈ 0.94108

(c) tan

(−8π

9

)Solution: tan

(−8π

9

)≈ 0.36397

(d) cot 2.379

Solution: cot 2.379 = 1/ tan 2.379 ≈ −1.04668

(e) csc (−2.689)

Solution: csc (−2.689) = 1/ sin (−2.689) ≈ −2.28677

(f) sec 45

Solution: sec 45 = 1/ cos 45 ≈ 1.90359

(g) arcsin 0.564

Solution: arcsin 0.564 ≈ 0.59922 or arcsin 0.564 ≈ 34.33288◦

(h) arccos (−0.367)

Solution: arccos (−0.367) ≈ 1.94658 or arccos (−0.367) ≈ 111.53072◦

(i) arctan 1.113

Solution: arctan 1.113 ≈ 0.83883 or arctan 1.113 ≈ 48.06117◦

4. Write an expression for the nth term.

2, 1,8

9, 1,

32

25, . . .

Page 216 of 238




Solution: By inspection.

an =2n

n2

5. Convert each of the following angle measures to radian measure.

(a) 60◦ =

Solution:

60◦ = 60◦ ·( π

180◦

)=π

3

(b) 90◦ =

Solution:

90◦ = 90◦ ·( π

180◦

)=π

2

(c) 50◦ =

Solution:

50◦ = 50◦ ·( π

180◦

)=

5π

18

6. Verify the identity.

sin2 θ =sec2 θ − 1

sec2 θ

Solution: Select the right side.

sec2 θ − 1

sec2 θ=

sec2 θ

sec2 θ− 1

sec2 θ= 1− cos2 θ

= sin2 θ

Q.E.D.

7. Find the rational number representation of the given repeating decimal.

1.287

Page 217 of 238




Solution: Let x = 1.287.

10x = 12.87

1000x = 1287.87

1000x− 10x = 1287.87− 12.87

990x = 1275.0000

x =1275

990=

85

66

8. Given

u = 〈1, −5〉 , v = 〈5, −3〉 , w = 〈−3, −6〉 ,

find.

(a) The angle between v and u. Both exact and approximate to four decimal places.

Solution:

cos θ =v · u‖v‖ ‖u‖

cos θ =20√

26√

34=

10√221

θ = arccos

(10√221

)≈ 47.7263◦ or 0.8330

(b) ‖u−w‖

Solution:

‖u−w‖ = ‖〈4, 1〉‖ =√

16 + 1 =√

17

(c) 5w + 3v

Solution:

5w + 3v = 5 〈−3, −6〉+ 3 〈5, −3〉 = 〈−15, −30〉+ 〈15, −9〉 = 〈0, −39〉

(d) w · v

Solution:

w · v = 〈−3, −6〉 · 〈5, −3〉 = −15 + 18 = 3

9. Use summation notation to write the given sum.

3 + 9 + 27 + 81 + · · ·+ 729

Page 218 of 238




Solution: Here’s one possible answer.

3 + 9 + 27 + 81 + · · ·+ 729 =6∑

n=1

3n

10. Find the exact value of the following in both degree and radian.

(a)

arccos

(1√2

)Solution:

arccos

(1√2

)=

π

4

= 45◦

(b)

arccos

[sin

(5π

3

)]Solution:

arccos

[sin

(5π

3

)]= arccos

(−√

3

2

)=

5π

6= 150◦

11. Approximate, to the nearest tenth, the length of arc on a circle of radius 14 inches anda central angle of 60◦.

Solution:

S = 14 · π3

So, the length is 14.7 inches.

12. Perform the indicated addition and use the fundamental identities to simplify.

1

1− sinx+

1

1 + sin x

Page 219 of 238




Solution:

1

1− sinx+

1

1 + sin x=

1

1− sinx· 1 + sin x

1 + sin x+

1

1 + sin x· 1− sinx

1− sinx

=1 + sin x

1− sin2 x+

1− sinx

1− sin2 x

=1 + sin x+ 1− sinx

1− sin2 x

=2

1− sin2 x

=2

cos2 x= 2 sec2 x

13. If the sec θ = 7 and 270◦ < θ < 360◦, find the following.

(a) sin θ =

Solution: From the information given we can conclude that x = 1, r = 7, andthat y < 0. To find the value of y, solve

72 = 12 + y2 ⇒ y = ±√

48 = ±4√

3.

So, y = −4√

3, so sin θ = −4√

3/7.

(b) cos θ =

Solution: cos θ = 1/7

(c) tan θ =

Solution: tan θ = −4√

3

(d) cot θ =

Solution: cot θ = −√

3/12

(e) csc θ =

Solution: csc θ = −7√

3/12

14. Solve the triangle, given a = 11.23 inches b = 8.24 inches, and B = 29.84◦.

Solution: You should, of course, draw a triangle first. This one actually gives twodistinct answers.

A = 42.70◦, C = 107.46◦ and c = 15.80 inches.

or

A = 137.30◦, C = 12.86◦ and c = 3.69 inches.

Page 220 of 238




15. Find the supplement of 83◦.

Solution:

180◦ − 83◦ = 97◦

16. Find all solutions.

2 sin2 x+ 3 cosx− 3 = 0

Solution: You’ll need to write it in terms of cosines first.

2 sin2 x+ 3 cosx− 3 = 0

2(1− cos2 x

)+ 3 cosx− 3 = 0

−2 cos2 x+ 3 cosx− 1 = 0

Now multiply both sides of this equation by −1 and factor.

2 cos2 x− 3 cosx+ 1 = 0

(2 cosx− 1) (cosx− 1) = 0

Set each factor equal to zero and solve.

cosx =1

2⇒ x =

π

3+ 2πk, k ∈ Z

=5π

3+ 2πk, k ∈ Z

cosx = 1 ⇒ x = 2πk, k ∈ Z.

17. If z = −1− 1i, find the trigonometric form of z and z9, also find z9 in standard form.

Solution: Here θ = 225◦ or θ =5π

4, and r =

√2.

z =√

2 cos 225◦ +(√

2 sin 225◦)i

z9 = 16√

2 cos 2025◦ +(

16√

2 sin 2025◦)i

z9 = −16− 16i

Page 221 of 238




18. Rewrite the expression so that it is not in fractional form

tan2 x

cscx+ 1.

Solution:

tan2 x

cscx+ 1=

tan2 x1

sinx+ 1· sinx

sinx

=tan2 x sinx

1 + sin x· 1− sinx

1− sinx

=tan2 x sinx · (1− sinx)

1− sin2 x

=tan2 x sinx · (1− sinx)

cos2 x= tan2 x sinx · (1− sinx) · sec2 x

= sin3 x · (1− sinx) · sec4 x

19. Verify the identity.√1 + sin x

1− sinx=

1 + sin x

|cosx|

Solution: Select the left side.√1 + sin x

1− sinx=

√1 + sin x

1− sinx· 1 + sin x

1 + sin x

=

√(1 + sin x)2

1− sin2 x

=

√(1 + sin x)2

cos2 x

=1 + sin x

|cosx|

Q.E.D.

Here it might be nice to mention that 1 + sinx ≥ 0 so√

(1 + sin x)2 = 1 + sinx.

However, since −1 ≤ cosx ≤ 1, the√

cos2 x = |cosx|.

Page 222 of 238




20. Find the sum.

2π

3+π

4+

π

12=

Solution:

2π

3+π

4+

π

12=

8π + 3π + π

12= π

21. If

sinα = − 5

13and cos β =

3

5,

with both α and β are in the fourth quadrant. Find the exact values of the following.

Solution: Preamble: Since α is in the fourth quadrantα

2will be in the second

quadrant, which will determine the signs of the half-angle formulas.

sinα = − 5

13and cosα =

12

13

sin β = −4

5and cos β =

3

5

(a) sin (α− β)

Solution:

sin (α− β) = sinα cos β − sin β cosα

=

(− 5

13

)(3

5

)−(−4

5

)(12

13

)=

33

65

(b) cos (α− β)

Solution:

cos (α− β) = cosα cos β + sinα sin β

=

(12

13

)(3

5

)+

(− 5

13

)(−4

5

)=

56

65

Page 223 of 238




(c) tan (α− β)

Solution:

tan (α− β) =sin (α− β)

cos (α− β)

=

33

6556

65

=33

56

(d) sin(α

2

)Solution:

α

2is in the second quadrant, so the sine will be positive.

sin(α

2

)=

√1− cosα

2

=

√1− 12/13

2

=1√26

(e) cos(α

2

)Solution:

α

2is in the second quadrant, so the cosine will be negative.

cos(α

2

)= −

√1 + cosα

2

= −√

1 + 12/13

2

= − 5√26

(f) tan(α

2

)Solution: Just take the ratio of sine over cosine.

tan(α

2

)=

sin(α

2

)cos(α

2

)= −1

5

22. Determine two coterminal angles in radian measure (one positive, one negative) for

θ =π

3.

Page 224 of 238




Solution:

π

3+ 2π =

7π

3π

3− 2π = −5π

3

23. Find all solutions.

2 cos2 x+ 3 sinx− 3 = 0

Solution: You’ll need to write it in terms of cosines first.

2 cos2 x+ 3 sinx− 3 = 0

2(1− sin2 x

)+ 3 sinx− 3 = 0

−2 sin2 x+ 3 sinx− 1 = 0

Now multiply both sides of this equation by −1 and factor.

2 sin2 x− 3 sinx+ 1 = 0

(2 sinx− 1) (sinx− 1) = 0

Set each factor equal to zero and solve.

sinx =1

2⇒ x =

π

6+ 2πk, k ∈ Z

=5π

6+ 2πk, k ∈ Z

sinx = 1 ⇒ x =π

2+ 2πk, k ∈ Z.

24. Find the equation of a parabola with focus (3, 4) and directrix is y-axis.

Solution:

(y − 4)2 = 4 · 3

2·(x− 3

2

)y2 − 8y + 25 = 6x

Page 225 of 238




25. Find the complement of π/6.

Solution:

π

2− π

6=π

3

26. Given that

Pk =k

2· [5k − 3] ,

find Pk+1.

Solution: Just replace k with (k + 1).

Pk+1 =(k + 1)

2· [5 (k + 1)− 3]

27. Solve the triangle, given a = 13.25 inches b = 7.28 inches, and c = 9.02 inches.

Solution: You should, of course, draw a triangle first.

A = 108.28◦, B = 31.45◦ and C = 40.27◦.

28. Given.

3x2 − 2√

3xy + y2 + 2x+ 2√

3y = 0.

(a) This equation is of the form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0. Identify thecoefficients.

Solution:

A = 3

B = −2√

3

C = 1

D = 2

E = 2√

3

F = 0

(b) The quantity, B2 − 4AC is called the discriminant. Compute the discriminant forthis problem.

Page 226 of 238




Solution:

B2 − 4AC = 0

(c) In general if the discriminant is: 0, the graph is a parabola; negative, the graph isan ellipse or circle (B = 0, A = C); positive, the graph is a hyperbola. What is thisgraph?

Solution: Parabola

(d) Solve

cot 2θ =A− CB

,

for θ, where 0 < θ < π/2.

Solution:

cot 2θ =3− 1

2√

3

tan 2θ = −√

3

2θ = −π3

+ πk

θ = −π6

+π

2k

To meet the condition that 0 < θ < π/2, we take k = 1 and get θ = 60◦ = π/3.

29. A ramp 21 feet in length rises to a loading platform that is 5 feet off the ground.Assuming that the ground is level, what is the angle (to the nearest whole degree)between the ramp and the ground?

Solution:

sin θ =5

21

arcsin

(5

21

)≈ 14◦


49x2 + 36y2 − 98x− 360y − 815 = 0


Solution:(x− 1)2

62+

(y − 5)2

72= 1

Page 227 of 238





Solution: The center: (1, 5); the vertices: (1, −2), (1, 12);the foci:

(1, 5−

√13),(1, 5 +

√13); eccentricity

√13/7. To graph you should

also find these two points: (−5, 5), (7, 5).

(c) Graph.


-18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

-8-7-6-5-4-3-2-1

12345678910111213141516



Vertices: (0, ±8) ; foci: (0, ±7)

Solution:

x2

15+y2

64= 1

32. Determine the exact values of the nine indicated points on the graph (Figure 102,page 229) of

f (x) = 2 sin (πx)− 1

Page 228 of 238




0

Figure 102: Partial graph of f (x) = 2 sin (πx)− 1.

Solution:

(0, −1) ;

(1/2, 1) ;

(1, −1) ;

(3/2, −3) ;

(2, −1) ;

(5/2, 1) ;

(3, −1) ;

(7/2, −3) ;

(4, −1)


16x2 − 25y2 + 50y − 425 = 0


Page 229 of 238




Solution:x2

25− (y − 1)2

16= 1


Solution: (0, 1)


Solution: (−5, 1) and (5, 1)

(d) Find the foci.

Solution:(−√

41, 1)

and(√

41, 1)


Solution: y = 1± 4

5x

(f) Graph.


-20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

-12-11-10-9-8-7-6-5-4-3-2-1

123456789101112


34. Evaluate the given infinite geometric sum.

∞∑n=1

0.9n

Page 230 of 238




Solution:

∞∑n=1

0.9n = 0.9 · 1

1− 0.9=

0.9

0.1= 9

35. Simplify the factorial expression.

(2n+ 2)!

(2n)!

Solution:

(2n+ 2)!

(2n)!=

(2n+ 2) (2n+ 1) (2n)!

(2n)!

= (2n+ 2) (2n+ 1)

= 4n2 + 6n+ 2

36. Find the sum.

70∑n=48

(2n− 1)

Solution: Expand first. You should also notice that there are 70− 48 + 1 = 23 terms.

70∑n=48

(2n− 1) = (2 · 48− 1) + (2 · 49− 1) + (2 · 50− 1) + · · ·+ (2 · 70− 1)

= 2 · (48 + 49 + 50 + · · ·+ 70)− (1 + 1 + 1 + · · ·+ 1)

= 2 · 23

2· (70 + 48)− 23 = 2691

37. When an airplane leaves the runway, its angle of climb is 19◦ and its air speed is 300feet per second. Find the plane’s altitude after 30 seconds.

Page 231 of 238




Solution: The plane will travel 30 · 300 = 9, 000 feet (this is the hypotenuse of a righttriangle), so the plane’s altitude (this is the side opposite the 19◦ angle) is 9000·sin 19◦ ≈2930 feet.

38. Use mathematical induction to prove the formula for every positive integer n.

2(1 + 3 + 32 + 33 + · · ·+ 3n−1

)= 3n − 1

Solution: Verify for n = 1.

P1 : 2 (1) = 3− 1

2 = 2


Pk : 2(1 + 3 + 32 + 33 + · · ·+ 3k−1

)= 3k − 1

Add 2 · 3k to both sides.

Pk : 2(1 + 3 + 32 + 33 + · · ·+ 3k−1

)= 3k − 1

2(1 + 3 + 32 + 33 + · · ·+ 3k−1

)+ 2 · 3k = 3k − 1 + 2 · 3k

2(1 + 3 + 32 + 33 + · · ·+ 3k−1 + 3k

)= 3 · 3k − 1

2(1 + 3 + 32 + 33 + · · ·+ 3k−1 + 3k

)= 3k+1 − 1

This last line is exactly what we wanted, Pk+1. Q.E.D.

39. Find the first five terms of the recursively defined sequence.

a1 = 15, ak+1 = 3ak − 2.

Solution:

a1 = 15

a2 = 3a1 − 2 = 3 · 15− 2 = 43

a3 = 3a2 − 2 = 3 · 43− 2 = 127

a4 = 3a3 − 2 = 3 · 127− 2 = 379

a5 = 3a4 − 2 = 3 · 379− 2 = 1135

Page 232 of 238




40. Write an algebraic expression that is equivalent to

sec [arcsin (x− 1)] .

Solution: You should draw a triangle first and use the Pythagorean Theorem to de-termine the missing side.

sec [arcsin (x− 1)] =1√

2x− x2

41. Use your calculator to approximate each of the following to three decimal places in bothdegree and radian.

(a) arcsin (−0.987)

Solution:

arcsin (−0.987) ≈ −1.409

arcsin (−0.987) ≈ −80.751◦

(b) arctan (−89.456)

Solution:

arctan (−89.456) ≈ −1.560

arctan (−89.456) ≈ −89.360◦

(c) arccos (−0.009)

Solution:

arccos (−0.009) ≈ 1.580

arccos (−0.009) ≈ 90.516◦

42. Solve for x.

(a) sin x+√

2 = − sinx in the interval [0, 2π).

Solution:

sinx+√

2 = − sinx

2 sinx = −√

2

sinx = −√

2

2= − 1√

2

Page 233 of 238




The reference is 45◦ and the solutions occur in the third and fourth quadrant, so

x = 45◦ + 180◦ = 225◦ =5π

4

x = 360◦ − 45◦ = 315◦ =7π

4

(b) 2 sin2 x− sinx− 1 = 0 in the interval [0, 2π).

Solution:

2 sin2 x− sinx− 1 = 0

(2 sinx+ 1) (sinx− 1) = 0

Which gives two equations to solve.

sinx = −1

2sinx = 1

The reference for the first equation is 30◦ and the solutions occur in the third andfourth quadrant, so

x = 30◦ + 180◦ = 210◦ =7π

6

x = 360◦ − 30◦ = 330◦ =11π

6,

and the solution to the second equation is x = π/2.

43. Find the infinite geometric sum.

8 + 6 +9

2+

27

8+ · · ·

Solution: a1 = 8 and r =6

8=

3

4.

S∞ = 8 · 1

1− 3/4= 32

44. Let

an = n2n

(a) Write the first five terms of the sequence, starting with a1.

Page 234 of 238




Solution:

a1 = 1 · 21 = 2

a2 = 2 · 22 = 8

a3 = 3 · 23 = 24

a4 = 4 · 24 = 64

a5 = 5 · 25 = 160

(b) Is this sequence arithmetic, geometric, or neither.

Solution: Neither

45. Find θ in degrees (0◦ < θ < 90◦) and radians (0 < θ < π2), if

cot θ =1√3.

Solution: Draw a triangle first, clearly θ = 60◦ = π/3.


x = t, y =t

7



-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

Figure 104: Partial graph of x = t, y = t7.

Page 235 of 238





Solution:

y =x

7


Solution: All real numbers.

47. A deposit of P dollars is made at the beginning of each month in an account earning anannual interest rate r, compounded monthly. The balance after t years is given by

A =12t∑i=1

P(

1 +r

12

)i.

Show that A is also equal to

A = P

[(1 +

r

12

)12t− 1

](1 +

12

r

).

Solution: Expand12t∑i=1

P(

1 +r

12

)iand you should notice that it is geometric. Just

simplify to show that it’s the same as A = P

[(1 +

r

12

)12t− 1

](1 +

12

r

).

A = P(

1 +r

12

)·[1 +

(1 +

r

12

)+(

1 +r

12

)2+(

1 +r

12

)3+ · · ·+

(1 +

r

12

)12t−1]= P

(1 +

r

12

)·

1−(1 + r

12

)12t1−

(1 + r

12

)= P

(1 +

r

12

)·

1−(1 + r

12

)12t− r

12

= P(

1 +r

12

)·[1−

(1 +

r

12

)12t]·(−12

r

)= P

(1 +

r

12

)·(

12

r

)·[1−

(1 +

r

12

)12t]· (−1)

= P

(12

r+ 1

)·[−1 +

(1 +

r

12

)12t]= P

[(1 +

r

12

)12t− 1

](1 +

12

r

)

Page 236 of 238




Q.E.D.

48. Find the numerical coefficient of the term whose variable part is x6y3 in the expansionof (x− 2y)9.

Solution:(96

)(x)6 (−2y)3 = 84 · (−8)x6y3 = −672x6y3

The numerical coefficient is −672.

49. Use the properties of inverse functions to find the exact value of

arccos

(cos

7π

2

).

Solution:

arccos

(cos

7π

2

)= arccos (0) =

π

2

50. Find the sum.

100∑n=51

6n

Solution:

100∑n=51

6n = 6 · 51 + 6 · 52 + 6 · 53 + · · ·+ 6 · 100

= 6 · (51 + 52 + 53 + · · ·+ 100)

= 6 · 50

2· (100 + 51) = 22650

Page 237 of 238




51. Use the Binomial Theorem to expand and simplify(3

3√x2 − 2 3

√y)3.

Solution: First, rewrite,(3

3√x2 − 2 3

√y)3

=(3x2/3 − 2y1/3

)3,

then expand(33

)(3x

23

)3+

(32

)(3x

23

)2 (−2y

13

)+

(31

)(3x

23

)(−2y

13

)2+

(30

)(−2y

13

)3,

finally the simplification

27x2 − 54x 3√xy + 36 3

√x2y2 − 8y.

52. Write an expression for the nth term.

1,5

2,25

6,

125

24,

625

120, . . .

Solution:

an =5n−1

n!

Page 238 of 238

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