Pre calculus Grade 11 Learner's Module Senior High School

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Precalculus

Learner’s Material

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Precalculus Learner’s Material First Edition 2016

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Table of Contents

To the Precalculus Learners 1

DepEd Curriculum Guide for Precalculus 2

Unit 1: Analytic Geometry 6

Lesson 1.1: Introduction to Conic Sections and Circles . . . . . . . . 7

1.1.1: An Overview of Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.1.2: Definition and Equation of a Circle . . . . . . . . . . . . . . . . . . . . . . . 8

1.1.3: More Properties of Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.1.4: Situational Problems Involving Circles. . . . . . . . . . . . . . . . . . . . 12

Lesson 1.2: Parabolas. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.2.1: Definition and Equation of a Parabola . . . . . . . . . . . . . . . . . . . . 19

1.2.2: More Properties of Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1.2.3: Situational Problems Involving Parabolas . . . . . . . . . . . . . . . . 26

Lesson 1.3: Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

1.3.1: Definition and Equation of an Ellipse . . . . . . . . . . . . . . . . . . . . . 33

1.3.2: More Properties of Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

1.3.3: Situational Problems Involving Ellipses . . . . . . . . . . . . . . . . . . . 40

Lesson 1.4: Hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

1.4.1: Definition and Equation of a Hyperbola . . . . . . . . . . . . . . . . . . 46

1.4.2: More Properties of Hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

1.4.3: Situational Problems Involving Hyperbolas . . . . . . . . . . . . . . . 54

Lesson 1.5: More Problems on Conic Sections . . . . . . . . . . . . . . . . 60

1.5.1: Identifying the Conic Section by Inspection. . . . . . . . . . . . . . . 60

1.5.2: Problems Involving Different Conic Sections . . . . . . . . . . . . . . 62

iiiAll rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -

electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

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Lesson 1.6: Systems of Nonlinear Equations . . . . . . . . . . . . . . . . . . 67

1.6.1: Review of Techniques in Solving Systems of LinearEquations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

1.6.2: Solving Systems of Equations Using Substitution . . . . . . . . . 69

1.6.3: Solving Systems of Equations Using Elimination. . . . . . . . . . 70

1.6.4: Applications of Systems of Nonlinear Equations . . . . . . . . . . 73

Unit 2: Mathematical Induction 80

Lesson 2.1: Review of Sequences and Series . . . . . . . . . . . . . . . . . . . 81

Lesson 2.2: Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

2.2.1: Writing and Evaluating Sums in Sigma Notation . . . . . . . . . 87

2.2.2: Properties of Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

Lesson 2.3: Principle of Mathematical Induction . . . . . . . . . . . . . . 96

2.3.1: Proving Summation Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

2.3.2: Proving Divisibility Statements. . . . . . . . . . . . . . . . . . . . . . . . . . . 101

?2.3.3: Proving Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

Lesson 2.4: The Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

2.4.1: Pascal’s Triangle and the Concept of Combination. . . . . . . . 109

2.4.2: The Binomial Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

2.4.3: Terms of a Binomial Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . 114

?2.4.4: Approximation and Combination Identities . . . . . . . . . . . . . . . 116

Unit 3: Trigonometry 123

Lesson 3.1: Angles in a Unit Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

3.1.1: Angle Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

3.1.2: Coterminal Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

3.1.3: Arc Length and Area of a Sector . . . . . . . . . . . . . . . . . . . . . . . . . 129

Lesson 3.2: Circular Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

3.2.1: Circular Functions on Real Numbers . . . . . . . . . . . . . . . . . . . . . 136

3.2.2: Reference Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139


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Lesson 3.3: Graphs of Circular Functions and SituationalProblems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

3.3.1: Graphs of y = sinx and y = cosx . . . . . . . . . . . . . . . . . . . . . . . . 145

3.3.2: Graphs of y = a sin bx and y = a cos bx . . . . . . . . . . . . . . . . . . . 147

3.3.3: Graphs of y = a sin b(x− c) + d andy = a cos b(x− c) + d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

3.3.4: Graphs of Cosecant and Secant Functions . . . . . . . . . . . . . . . . 154

3.3.5: Graphs of Tangent and Cotangent Functions . . . . . . . . . . . . . 158

3.3.6: Simple Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

Lesson 3.4: Fundamental Trigonometric Identities. . . . . . . . . . . . .171

3.4.1: Domain of an Expression or Equation . . . . . . . . . . . . . . . . . . . . 171

3.4.2: Identity and Conditional Equation . . . . . . . . . . . . . . . . . . . . . . . 173

3.4.3: The Fundamental Trigonometric Identities . . . . . . . . . . . . . . . 174

3.4.4: Proving Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . 176

Lesson 3.5: Sum and Difference Identities . . . . . . . . . . . . . . . . . . . . .181

3.5.1: The Cosine Difference and Sum Identities . . . . . . . . . . . . . . . . 181

3.5.2: The Cofunction Identities and the Sine Sum andDifference Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

3.5.3: The Tangent Sum and Difference Identities . . . . . . . . . . . . . . . 186

Lesson 3.6: Double-Angle and Half-Angle Identities . . . . . . . . . . . 192

3.6.1: Double-Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

3.6.2: Half-Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195

Lesson 3.7: Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . 201

3.7.1: Inverse Sine Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202

3.7.2: Inverse Cosine Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

3.7.3: Inverse Tangent Function and the Remaining InverseTrigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

Lesson 3.8: Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 220

3.8.1: Solutions of a Trigonometric Equation . . . . . . . . . . . . . . . . . . . . 221

3.8.2: Equations with One Term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224

3.8.3: Equations with Two or More Terms . . . . . . . . . . . . . . . . . . . . . . 227


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Lesson 3.9: Polar Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . 236

3.9.1: Polar Coordinates of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237

3.9.2: From Polar to Rectangular, and Vice Versa . . . . . . . . . . . . . . . 241

3.9.3: Basic Polar Graphs and Applications . . . . . . . . . . . . . . . . . . . . . 244

Answers to Odd-Numbered Exercises in Supplementary Problemsand All Exercises in Topic Tests 255

References 290


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To the Precalculus Learners

The Precalculus course bridges basic mathematics and calculus. This coursecompletes your foundational knowledge on algebra, geometry, and trigonometry.It provides you with conceptual understanding and computational skills that areprerequisites for Basic Calculus and future STEM courses.

Based on the Curriculum Guide for Precalculus of the Department of Edu-cation (see pages 2-5), the primary aim of this Learning Manual is to give youan adequate stand-alone material that can be used for the Grade 11 Precalculuscourse.

The Manual is divided into three units: analytic geometry, summation no-tation and mathematical induction, and trigonometry. Each unit is composedof lessons that bring together related learning competencies in the unit. Eachlesson is further divided into sub-lessons that focus on one or two competenciesfor effective learning.

At the end of each lesson, more examples are given in Solved Examples toreinforce the ideas and skills being developed in the lesson. You have the oppor-tunity to check your understanding of the lesson by solving the SupplementaryProblems. Finally, two sets of Topic Test are included to prepare you for theexam.

Answers, solutions, or hints to odd-numbered items in the SupplementaryProblems and all items in the Topic Tests are provided at the end of the Manualto guide you while solving them. We hope that you will use this feature of theManual responsibly.

Some items are marked with a star. A starred sub-lesson means the discussionand accomplishment of the sub-lesson are optional. This will be decided by yourteacher. On the other hand, a starred example or exercise means the use ofcalculator is required.

We hope that you will find this Learning Manual helpful and convenient touse. We encourage you to carefully study this Manual and solve the exercisesyourselves with the guidance of your teacher. Although great effort has beenput into this Manual for technical correctness and precision, any mistake foundand reported to the Team is a gain for other students. Thank you for yourcooperation.

The Precalculus LM Team


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Unit 1

Analytic Geometry

San Juanico Bridge, by Morten Nærbøe, 21 June 2009,

https://commons.wikimedia.org/wiki/File%3ASan Juanico Bridge 2.JPG. Public Domain.

Stretching from Samar to Leyte with a total length of more than two kilome-ters, the San Juanico Bridge has been serving as one of the main thoroughfaresof economic and social development in the country since its completion in 1973.Adding picturesque effect on the whole architecture, geometric structures aresubtly built to serve other purposes. The arch-shaped support on the main spanof the bridge helps maximize its strength to withstand mechanical resonance andaeroelastic flutter brought about by heavy vehicles and passing winds.



https://commons.wikimedia.org/wiki/File%3ASan_Juanico_Bridge_2.JPG

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Lesson 1.1. Introduction to Conic Sections and Circles

Learning Outcomes of the Lesson

At the end of the lesson, the student is able to:

(1) illustrate the different types of conic sections: parabola, ellipse, circle, hyper-bola, and degenerate cases;

(2) define a circle;

(3) determine the standard form of equation of a circle;

(4) graph a circle in a rectangular coordinate system; and

(5) solve situational problems involving conic sections (circles).

Lesson Outline

(1) Introduction of the four conic sections, along with the degenerate conics

(2) Definition of a circle

(3) Derivation of the standard equation of a circle

(4) Graphing circles

(5) Solving situational problems involving circles

Introduction

We present the conic sections, a particular class of curves which sometimesappear in nature and which have applications in other fields. In this lesson, wefirst illustrate how each of these curves is obtained from the intersection of aplane and a cone, and then discuss the first of their kind, circles. The other conicsections will be covered in the next lessons.

1.1.1. An Overview of Conic Sections

We introduce the conic sections (or conics), a particular class of curves whichoftentimes appear in nature and which have applications in other fields. Oneof the first shapes we learned, a circle, is a conic. When you throw a ball, thetrajectory it takes is a parabola. The orbit taken by each planet around the sunis an ellipse. Properties of hyperbolas have been used in the design of certaintelescopes and navigation systems. We will discuss circles in this lesson, leavingparabolas, ellipses, and hyperbolas for subsequent lessons.

• Circle (Figure 1.1) - when the plane is horizontal

• Ellipse (Figure 1.1) - when the (tilted) plane intersects only one cone to forma bounded curve



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• Parabola (Figure 1.2) - when the plane intersects only one cone to form anunbounded curve

• Hyperbola (Figure 1.3) - when the plane (not necessarily vertical) intersectsboth cones to form two unbounded curves (each called a branch of the hyper-bola)

Figure 1.1 Figure 1.2 Figure 1.3

We can draw these conic sections (also called conics) on a rectangular co-ordinate plane and find their equations. To be able to do this, we will presentequivalent definitions of these conic sections in subsequent sections, and use theseto find the equations.

There are other ways for a plane and the cones to intersect, to form what arereferred to as degenerate conics : a point, one line, and two lines. See Figures 1.4,1.5 and 1.6.


1.1.2. Definition and Equation of a Circle

A circle may also be considered a special kind of ellipse (for the special case whenthe tilted plane is horizontal). As we get to know more about a circle, we willalso be able to distinguish more between these two conics.



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See Figure 1.7, with the point C(3, 1) shown. From the figure, the distanceof A(−2, 1) from C is AC = 5. By the distance formula, the distance of B(6, 5)from C is BC =

√(6− 3)2 + (5− 1)2 = 5. There are other points P such that

PC = 5. The collection of all such points which are 5 units away from C, formsa circle.

Figure 1.7 Figure 1.8

Let C be a given point. The set of all points P having the samedistance from C is called a circle. The point C is called the center ofthe circle, and the common distance its radius.

The term radius is both used to refer to a segment from the center C to apoint P on the circle, and the length of this segment.

See Figure 1.8, where a circle is drawn. It has center C(h, k) and radius r > 0.A point P (x, y) is on the circle if and only if PC = r. For any such point then,its coordinates should satisfy the following.

PC = r√(x− h)2 + (y − k)2 = r

(x− h)2 + (y − k)2 = r2

This is the standard equation of the circle with center C(h, k) and radius r. Ifthe center is the origin, then h = 0 and k = 0. The standard equation is thenx2 + y2 = r2.



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Example 1.1.1. In each item, give thestandard equation of the circle satisfy-ing the given conditions.(1) center at the origin, radius 4

(2) center (−4, 3), radius√

7

(3) circle in Figure 1.7

(4) circle A in Figure 1.9

(5) circle B in Figure 1.9

(6) center (5,−6), tangent to the y-axis Figure 1.9

(7) center (5,−6), tangent to the x-axis

(8) It has a diameter with endpoints A(−1, 4) and B(4, 2).

Solution. (1) x2 + y2 = 16

(2) (x+ 4)2 + (y − 3)2 = 7

(3) The center is (3, 1) and the radius is 5, so the equation is (x−3)2 +(y−1)2 =25.

(4) By inspection, the center is (−2,−1) and the radius is 4. The equation is(x+ 2)2 + (y + 1)2 = 16.

(5) Similarly by inspection, we have (x− 3)2 + (y − 2)2 = 9.

(6) The center is 5 units away from the y-axis, so the radius is r = 5 (you canmake a sketch to see why). The equation is (x− 5)2 + (y + 6)2 = 25.

(7) Similarly, since the center is 6 units away from the x-axis, the equation is(x− 5)2 + (y + 6)2 = 36.

(8) The center C is the midpoint of A and B: C =(−1+4

2, 4+2

2

)=(32, 3). The

radius is then r = AC =√(−1− 3

2

)2+ (4− 3)2 =

√294

. The circle has

equation(x− 3

2

)2+ (y − 3)2 = 29

4. 2

1.1.3. More Properties of Circles

After expanding, the standard equation(x− 3

2

)2

+ (y − 3)2 =29

4

can be rewritten asx2 + y2 − 3x− 6y + 4 = 0,

an equation of the circle in general form.



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If the equation of a circle is given in the general form

Ax2 + Ay2 + Cx+Dy + E = 0, A 6= 0,

orx2 + y2 + Cx+Dy + E = 0,

we can determine the standard form by completing the square in both variables.

Completing the square in an expression like x2 + 14x means determiningthe term to be added that will produce a perfect polynomial square. Since thecoefficient of x2 is already 1, we take half the coefficient of x and square it, andwe get 49. Indeed, x2 + 14x + 49 = (x + 7)2 is a perfect square. To completethe square in, say, 3x2 + 18x, we factor the coefficient of x2 from the expression:3(x2 + 6x), then add 9 inside. When completing a square in an equation, anyextra term introduced on one side should also be added to the other side.

Example 1.1.2. Identify the center and radius of the circle with the given equa-tion in each item. Sketch its graph, and indicate the center.

(1) x2 + y2 − 6x = 7

(2) x2 + y2 − 14x+ 2y = −14

(3) 16x2 + 16y2 + 96x− 40y = 315

Solution. The first step is to rewrite each equation in standard form by complet-ing the square in x and in y. From the standard equation, we can determine thecenter and radius.

(1)

x2 − 6x+ y2 = 7

x2 − 6x+ 9 + y2 = 7 + 9

(x− 3)2 + y2 = 16

Center (3, 0), r = 4, Figure 1.10

(2)

x2 − 14x+ y2 + 2y = −14

x2 − 14x+ 49 + y2 + 2y + 1 = −14 + 49 + 1

(x− 7)2 + (y + 1)2 = 36

Center (7,−1), r = 6, Figure 1.11

(3)

16x2 + 96x+ 16y2 − 40y = 315



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16(x2 + 6x) + 16

(y2 − 5

2y

)= 315

16(x2 + 6x+ 9) + 16

(y2 − 5

2y +

25

16

)= 315 + 16(9) + 16

(25

16

)16(x+ 3)2 + 16

(y − 5

4

)2

= 484

(x+ 3)2 +

(y − 5

4

)2

=484

16=

121

4=

(11

2

)2

Center(−3, 5

4

), r = 5.5, Figure 1.12. 2


In the standard equation (x − h)2 + (y − k)2 = r2, both the two squaredterms on the left side have coefficient 1. This is the reason why in the precedingexample, we divided by 16 at the last equation.

1.1.4. Situational Problems Involving Circles

Let us now take a look at some situational problems involving circles.

?Example 1.1.3. A street with two lanes, each 10 ft wide, goes through asemicircular tunnel with radius 12 ft. How high is the tunnel at the edge of eachlane? Round off to 2 decimal places.



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Solution. We draw a coordinate system with origin at the middle of the highway,as shown. Because of the given radius, the tunnel’s boundary is on the circlex2 + y2 = 122. Point P is the point on the arc just above the edge of a lane, soits x-coordinate is 10. We need its y-coordinate. We then solve 102 + y2 = 122

for y > 0, giving us y = 2√

11 ≈ 6.63 ft. 2

Example 1.1.4. A piece of a broken plate was dug up in an archaeological site.It was put on top of a grid, as shown in Figure 1.13, with the arc of the platepassing through A(−7, 0), B(1, 4) and C(7, 2). Find its center, and the standardequation of the circle describing the boundary of the plate.

Figure 1.13

Figure 1.14

Solution. We first determine the center. It is the intersection of the perpendicularbisectors of AB and BC (see Figure 1.14). Recall that, in a circle, the perpen-dicular bisector of any chord passes through the center. Since the midpoint Mof AB is

(−7+12, 0+4

2

)= (−3, 2), and mAB = 4−0

1+7= 1

2, the perpendicular bisector

of AB has equation y − 2 = −2(x+ 3), or equivalently, y = −2x− 4.



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Since the midpoint N of BC is(1+72, 4+2

2

)= (4, 3), and mBC = 2−4

7−1 = −13,

the perpendicular bisector of BC has equation y − 3 = 3(x− 4), or equivalently,y = 3x− 9.

The intersection of the two lines y = 2x − 4 and y = 3x − 9 is (1,−6) (bysolving a system of linear equations). We can take the radius as the distance ofthis point from any of A, B or C (it’s most convenient to use B in this case). Wethen get r = 10. The standard equation is thus (x− 1)2 + (y + 6)2 = 100. 2

More Solved Examples

1. In each item, give the standard equation of the circle satisying the given con-ditions.

(a) center at the origin, contains (0, 3)

(b) center (1, 5), diameter 8

(c) circle A in Figure 1.15

(d) circle B in Figure 1.15

(e) circle C in Figure 1.15

(f) center (−2,−3), tangent to the y-axis

(g) center (−2,−3), tangent to the x-axis

(h) contains the points (−2, 0) and(8, 0), radius 5

Figure 1.15

Solution:

(a) The radius is 3, so the equation is x2 + y2 = 9.

(b) The radius is 8/2 = 4, so the equation is (x− 1)2 + (y − 5)2 = 16.

(c) The center is (−2, 2) and the radius is 2,so the equation is (x+2)2 +(y−2)2 = 4.

(d) The center is (2, 3) and the radius is 1,so the equation is (x−2)2 +(y−3)2 = 1.

(e) The center is (1,−1) and by thePythagorean Theorem, the radius (seeFigure 1.16) is

√22 + 22 =

√8, so the

equation is (x− 1)2 + (x+ 1)2 = 8.Figure 1.16



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(f) The radius is 3, so the equation is (x+ 2)2 + (y + 3)2 = 9.

(g) The radius is 2, so the equation is (x+ 2)2 + (y + 3)2 = 4.

(h) The distance between (−2, 0) and (8, 0) is 10; since the radius is 5, thesetwo points are endpoints of a diameter. Then the circle has center at(3, 0) and radius 5, so its equation is (x− 3)2 + y2 = 25.

2. Identify the center and radius of the circle with the given equation in eachitem. Sketch its graph, and indicate the center.

(a) x2 + y2 + 8y = 33

(b) 4x2 + 4y2 − 16x+ 40y + 67 = 0

(c) 4x2 + 12x+ 4y2 + 16y − 11 = 0

Solution:

(a)

x2 + y2 + 8y = 33

x2 + y2 + 8y + 16 = 33 + 16

x2 + (y + 4)2 = 49

Center (0,−4), radius 7, see Figure 1.17.

(b)

4x2 + 4y2 − 16x+ 40y + 67 = 0

x2 − 4x+ y2 + 10y = −67

4

x2 − 4x+ 4 + y2 + 10y + 25 = −67

4+ 4 + 25

(x− 2)2 + (y + 5)2 =49

4=

(7

2

)2

Center (2,−5), radius 3.5, see Figure 1.18.

(c)

4x2 + 12x+ 4y2 + 16y − 11 = 0

x2 + 3x+ y2 + 4y =11

4

x2 + 3x+9

4+ y2 + 4y + 4 =

11

4+

9

4+ 4(

x+3

2

)2

+ (y + 2)2 = 9

Center

(−3

2,−2

), radius 3, see Figure 1.19.



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3. A circular play area with radius 3 m isto be partitioned into two sections usinga straight fence as shown in Figure 1.20.How long should the fence be?Solution: To determine the length of thefence, we need to determine the coordi-nates of its endpoints. From Figure 1.20,the endpoints have x coordinate −1 andare on the circle x2 + y2 = 9. Then1 + y2 = 9, or y = ±2

√2. Therefore,

the length of the fence is 4√

2 ≈ 5.66 m. Figure 1.20

4. A Cartesian coordinate system was used to identify locations on a circu-lar track. As shown in Figure 1.21, the circular track contains the pointsA(−2,−4), B(−2, 3), C(5, 2). Find the total length of the track.


Solution: The segment AB is vertical and has midpoint (−2,−0.5), so itsperpendicular bisector has equation y = −0.5. On the other hand, the segmentBC has slope −1/7 and midpoint (1.5, 2.5), so its perpendicular bisector hasequation y − 2.5 = 7(x− 1.5), or 7x− y − 8 = 0.

The center of the circle is the intersection of y = −0.5 and 7x − y − 8 = 0;that is, the center is at

(1514,−1

2

).

The radius of the circle is the distance from the center to any of the points A,

B, or C; by the distance formula, the radius is

√2125

98=

5

14

√170. Therefore,



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the total length of the track (its circumference), is

2× π × 5

14

√170 ≈ 29.26 units.

Supplementary Problems 1.1Identify the center and radius of the circle with the given equation in each item.Sketch its graph, and indicate the center.

1. x2 + y2 =1

4

2. 5x2 + 5y2 = 125

3. (x+ 4)2 +

(y − 3

4

)2

= 1

4. x2 − 4x+ y2 − 4y − 8 = 0

5. x2 + y2 − 14x+ 12y = 36

6. x2 + 10x+ y2 − 16y − 11 = 0

7. 9x2 + 36x+ 9y2 + 72y + 155 = 0

8. 9x2 + 9y2 − 6x+ 24y = 19

9. 16x2 + 80x+ 16y2 − 112y + 247 = 0

Find the standard equation of the circle which satisfies the given conditions.

10. center at the origin, radius 5√

3

11. center at (17, 5), radius 12

12. center at (−8, 4), contains (−4, 2)

13. center at (15,−7), tangent to the x-axis

14. center at (15,−7), tangent to the y-axis

15. center at (15,−7), tangent to the line y = −10

16. center at (15,−7), tangent to the line x = 8

17. has a diameter with endpoints (3, 1) and (−7, 6)



DEPED COPY

18. has a diameter with endpoints

(9

2, 4

)and

(−3

2,−2

)19. concentric with x2 + 20x+ y2 − 14y + 145 = 0, diameter 12

20. concentric with x2 − 2x+ y2 − 2y − 23 = 0 and has 1/5 the area

21. concentric with x2 + 4x+ y2 − 6y + 9 = 0 and has the same circumference asx2 + 14x+ y2 + 10y + 62 = 0

22. contains the points (3, 3), (7, 1), (0, 2)

23. contains the points (1, 4), (−1, 2), (4,−3)

24. center at (−3, 2) and tangent to the line 2x− 3y = 1

25. center at (−5,−1) and tangent to the line x+ y + 10 = 0

26. has center with x-coordinate 4 and tangent to the line −x+ 3y = 9 at (3, 4)

27. A stadium is shaped as in Figure 1.23, where its left and right ends are circulararcs both with center at C. What is the length of the stadium 50 m from oneof the straight sides?

Figure 1.23

28. A waterway in a theme park has asemicircular cross section with di-ameter 11 ft. The boats that aregoing to be used in this waterwayhave rectangular cross sections andare found to submerge 1 ft into thewater. If the waterway is to befilled with water 4.5 ft deep, what isthe maximum possible width of theboats?

Figure 1.24

4



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Lesson 1.2. Parabolas



(1) define a parabola;

(2) determine the standard form of equation of a parabola;

(3) graph a parabola in a rectangular coordinate system; and

(4) solve situational problems involving conic sections (parabolas).

Lesson Outline

(1) Definition of a parabola

(2) Derivation of the standard equation of a parabola

(3) Graphing parabolas

(4) Solving situational problems involving parabolas

Introduction

A parabola is one of the conic sections. We have already seen parabolas whichopen upward or downward, as graphs of quadratic functions. Here, we will seeparabolas opening to the left or right. Applications of parabolas are presentedat the end.

1.2.1. Definition and Equation of a Parabola

Consider the point F (0, 2) and the line ` having equation y = −2, as shown inFigure 1.25. What are the distances of A(4, 2) from F and from `? (The latteris taken as the distance of A from A`, the point on ` closest to A). How aboutthe distances of B(−8, 8) from F and from ` (from B`)?

AF = 4 and AA` = 4

BF =√

(−8− 0)2 + (8− 2)2 = 10 and BB` = 10

There are other points P such that PF = PP` (where P` is the closest point online `). The collection of all such points forms a shape called a parabola.

Let F be a given point, and ` a given line not containing F . The set ofall points P such that its distances from F and from ` are the same, iscalled a parabola. The point F is its focus and the line ` its directrix.



DEPED COPYFigure 1.25

Figure 1.26

Consider a parabola with focus F (0, c) and directrix ` having equation y = −c.See Figure 1.26. The focus and directrix are c units above and below, respectively,the origin. Let P (x, y) be a point on the parabola so PF = PP`, where P` is thepoint on ` closest to P . The point P has to be on the same side of the directrixas the focus (if P was below, it would be closer to ` than it is from F ).

PF = PP`√x2 + (y − c)2 = y − (−c) = y + c

x2 + y2 − 2cy + c2 = y2 + 2cy + c2

x2 = 4cy

The vertex V is the point midway between the focus and the directrix. Thisequation, x2 = 4cy, is then the standard equation of a parabola opening upwardwith vertex V (0, 0).

Suppose the focus is F (0,−c) and the directrix is y = c. In this case, apoint P on the resulting parabola would be below the directrix (just like thefocus). Instead of opening upward, it will open downward. Consequently, PF =√x2 + (y + c)2 and PP` = c − y (you may draw a version of Figure 1.26 for

this case). Computations similar to the one done above will lead to the equation



DEPED COPY

x2 = −4cy.

We collect here the features of the graph of a parabola with standard equationx2 = 4cy or x2 = −4cy, where c > 0.

(1) vertex : origin V (0, 0)

• If the parabola opens upward, the vertex is the lowest point. If theparabola opens downward, the vertex is the highest point.

(2) directrix : the line y = −c or y = c

• The directrix is c units below or above the vertex.

(3) focus : F (0, c) or F (0,−c)

• The focus is c units above or below the vertex.

• Any point on the parabola has the same distance from the focus as ithas from the directrix.

(4) axis of symmetry : x = 0 (the y-axis)

• This line divides the parabola into two parts which are mirror imagesof each other.

Example 1.2.1. Determine the focus and directrix of the parabola with thegiven equation. Sketch the graph, and indicate the focus, directrix, vertex, andaxis of symmetry.



DEPED COPY

(1) x2 = 12y (2) x2 = −6y

Solution. (1) The vertex is V (0, 0) and the parabola opens upward. From 4c =12, c = 3. The focus, c = 3 units above the vertex, is F (0, 3). The directrix,3 units below the vertex, is y = −3. The axis of symmetry is x = 0.

(2) The vertex is V (0, 0) and the parabola opens downward. From 4c = 6, c = 32.

The focus, c = 32

units below the vertex, is F(0,−3

2

). The directrix, 3

2units

above the vertex, is y = 32. The axis of symmetry is x = 0.

Example 1.2.2. What is the standard equation of the parabola in Figure 1.25?

Solution. From the figure, we deduce that c = 2. The equation is thus x2 =8y. 2



DEPED COPY

1.2.2. More Properties of Parabolas

The parabolas we considered so far are “vertical” and have their vertices at theorigin. Some parabolas open instead horizontally (to the left or right), and somehave vertices not at the origin. Their standard equations and properties are givenin the box. The corresponding computations are more involved, but are similarto the one above, and so are not shown anymore.

In all four cases below, we assume that c > 0. The vertex is V (h, k), and itlies between the focus F and the directrix `. The focus F is c units away fromthe vertex V , and the directrix is c units away from the vertex. Recall that, forany point on the parabola, its distance from the focus is the same as its distancefrom the directrix.

(x− h)2 = 4c(y − k) (y − k)2 = 4c(x− h)

(x− h)2 = −4c(y − k) (y − k)2 = −4c(x− h)

directrix `: horizontal directrix `: vertical

axis of symmetry: x=h, vertical axis of symmetry: y=k, horizontal



DEPED COPY

Note the following observations:

• The equations are in terms of x − h and y − k: the vertex coordinates aresubtracted from the corresponding variable. Thus, replacing both h and kwith 0 would yield the case where the vertex is the origin. For instance, thisreplacement applied to (x−h)2 = 4c(y−k) (parabola opening upward) wouldyield x2 = 4cy, the first standard equation we encountered (parabola openingupward, vertex at the origin).

• If the x-part is squared, the parabola is “vertical”; if the y-part is squared,the parabola is “horizontal.” In a horizontal parabola, the focus is on the leftor right of the vertex, and the directrix is vertical.

• If the coefficient of the linear (non-squared) part is positive, the parabolaopens upward or to the right; if negative, downward or to the left.

Example 1.2.3. Figure 1.27 shows the graph of parabola, with only its focusand vertex indicated. Find its standard equation. What is its directrix and itsaxis of symmetry?

Solution. The vertex is V (5,−4) and the focus is F (3,−4). From these, wededuce the following: h = 5, k = −4, c = 2 (the distance of the focus from thevertex). Since the parabola opens to the left, we use the template (y − k)2 =−4c(x− h). Our equation is

(y + 4)2 = −8(x− 5).

Its directrix is c = 2 units to the right of V , which is x = 7. Its axis is thehorizontal line through V : y = −4.

Figure 1.27



DEPED COPY

The standard equation (y+ 4)2 = −8(x− 5) from the preceding example canbe rewritten as y2 + 8x + 8y − 24 = 0, an equation of the parabola in generalform.

If the equation is given in the general form Ax2 +Cx+Dy+E = 0 (A and Care nonzero) or By2+Cx+Dy+E = 0 (B and C are nonzero), we can determinethe standard form by completing the square in both variables.

Example 1.2.4. Determine the vertex, focus, directrix, and axis of symmetryof the parabola with the given equation. Sketch the parabola, and include thesepoints and lines.

(1) y2 − 5x+ 12y = −16

(2) 5x2 + 30x+ 24y = 51

Solution. (1) We complete the square on y, and move x to the other side.

y2 + 12y = 5x− 16

y2 + 12y + 36 = 5x− 16 + 36 = 5x+ 20

(y + 6)2 = 5(x+ 4)

The parabola opens to the right. It has vertex V (−4,−6). From 4c = 5, weget c = 5

4= 1.25. The focus is c = 1.25 units to the right of V : F (−2.75,−6).

The (vertical) directrix is c = 1.25 units to the left of V : x = −5.25. The(horizontal) axis is through V : y = −6.



DEPED COPY

(2) We complete the square on x, and move y to the other side.

5x2 + 30x = −24y + 51

5(x2 + 6x+ 9) = −24y + 51 + 5(9)

5(x+ 3)2 = −24y + 96 = −24(y − 4)

(x+ 3)2 = −24

5(y − 4)

In the last line, we divided by 5 for the squared part not to have any coeffi-cient. The parabola opens downward. It has vertex V (−3, 4).

From 4c = 245

, we get c = 65

= 1.2. The focus is c = 1.2 units below V :F (−3, 2.8). The (horizontal) directrix is c = 1.2 units above V : y = 5.2. The(vertical) axis is through V : x = −3.

Example 1.2.5. A parabola has focus F (7, 9) and directrix y = 3. Find itsstandard equation.

Solution. The directrix is horizontal, and the focus is above it. The parabolathen opens upward and its standard equation has the form (x− h)2 = 4c(y− k).Since the distance from the focus to the directrix is 2c = 9− 3 = 6, then c = 3.Thus, the vertex is V (7, 6), the point 3 units below F . The standard equation isthen (x− 7)2 = 12(y − 6). 2

1.2.3. Situational Problems Involving Parabolas

Let us now solve some situational problems involving parabolas.

Example 1.2.6. A satellite dish has a shape called a paraboloid, where eachcross-section is a parabola. Since radio signals (parallel to the axis) will bounce



DEPED COPY

off the surface of the dish to the focus, the receiver should be placed at the focus.How far should the receiver be from the vertex, if the dish is 12 ft across, and 4.5ft deep at the vertex?

Solution. The second figure above shows a cross-section of the satellite dish drawnon a rectangular coordinate system, with the vertex at the origin. From theproblem, we deduce that (6, 4.5) is a point on the parabola. We need the distanceof the focus from the vertex, i.e., the value of c in x2 = 4cy.

x2 = 4cy

62 = 4c(4.5)

c =62

4 · 4.5= 2

Thus, the receiver should be 2 ft away from the vertex. 2

Example 1.2.7. The cable of a suspension bridge hangs in the shape of aparabola. The towers supporting the cable are 400 ft apart and 150 ft high.If the cable, at its lowest, is 30 ft above the bridge at its midpoint, how high isthe cable 50 ft away (horizontally) from either tower?



DEPED COPY

Solution. Refer to the figure above, where the parabolic cable is drawn withits vertex on the y-axis 30 ft above the origin. We may write its equation as(x− 0)2 = a(y − 30); since we don’t need the focal distance, we use the simplervariable a in place of 4c. Since the towers are 150 ft high and 400 ft apart, wededuce from the figure that (200, 150) is a point on the parabola.

x2 = a(y − 30)

2002 = a(150− 30)

a =2002

120=

1000

3

The parabola has equation x2 = 10003

(y − 30), or equivalently,y = 0.003x2 + 30. For the two points on the parabola 50 ft away from thetowers, x = 150 or x = −150. If x = 150, then

y = 0.003(1502) + 30 = 97.5.

Thus, the cable is 97.5 ft high 50 ft away from either tower. (As expected, weget the same answer from x = −150.) 2

More Solved ExamplesFor Examples 1 and 2, determine the focus and directrix of the parabola with thegiven equation. Sketch the graph, and indicate the focus, directrix, and vertex.

1. y2 = 20x

Solution:Vertex: V (0, 0), opens to the right4c = 20⇒ c = 5Focus: F (5, 0), Directrix: x = −5See Figure 1.28.

2. 3x2 = −12y

Solution: 3x2 = −12y ⇔ x2 = −4yVertex: V (0, 0), opens downward4c = 4⇒ c = 1Focus: F (0,−1), Directrix: y = 1See Figure 1.29.




DEPED COPY

3. Determine the standard equation of theparabola in Figure 1.30 given only itsvertex and focus. Then determine its di-rectix and axis of symmetry.

Solution:

V

(−3

2, 4

), F (−4, 4)

c =5

2⇒ 4c = 10

Parabola opens to the left

Equation: (y − 4)2 = −10

(x+

3

2

)Directrix: x = 1, Axis: y = 4 Figure 1.30

4. Determine the standard equation of theparabola in Figure 1.31 given only itsvertex and diretrix. Then determine itsfocus and axis of symmetry.

Solution:

V

(5,

13

2

), directrix: y =

15

2c = 1 ⇒ 4c = 4Parabola opens downward

Equation:

(y − 13

2

)2

= −4 (x− 5)

Focus:

(5,

11

2

), Axis: x = 5

Figure 1.31

For Examples 5 and 6, determine the vertex, focus, directrix, and axis of sym-metry of the parabola with the given equation. Sketch the parabola, and includethese points and lines.

5. x2 − 6x− 2y + 9 = 0

Solution:

x2 − 6x = 2y − 9

x2 − 6x+ 9 = 2y

(x− 3)2 = 2y

V (3, 0), parabola opens upward

4c = 2⇒ c =1

2, F

(3,

1

2

),

directrix: y = −1

2, axis: x = 3

See Figure 1.32.Figure 1.32



DEPED COPY

6. 3y2 + 8x+ 24y + 40 = 0

Solution:

3y2 + 24y = −8x− 40

3(y2 + 8y) = −8x− 40

3(y2 + 8y + 16) = −8x− 40 + 48

3(y + 4)2 = −8x+ 8

(y + 4)2 = −8

3(x− 1)

V (1,−4), parabola opens to the left

4c =8

3⇒ c =

2

3, F

(1

3,−4

),

directrix: x =5

3, axis: y = −4

See Figure 1.33.

Figure 1.33

7. A parabola has focus F (−11, 8) and directrix x = −17. Find its standardequation.

Solution: Since the focus is 6 units to the right of the directrix, the parabolaopens to the right with 2c = 6. Then c = 3 and V (−14, 8). Hence, theequation is (y − 8)2 = 12(x+ 14).

8. A flashlight is shaped like aparaboloid and the light sourceis placed at the focus so thatthe light bounces off parallel tothe axis of symmetry; this isdone to maximize illumination.A particular flashlight has itslight source located 1 cm fromthe base and is 6 cm deep; seeFigure 1.34. What is the widthof the flashlight’s opening?

Figure 1.34

Solution: Let the base (the vertex) of the flashlight be the point V (0, 0).Then the light source (the focus) is at F (0, 1); so c = 1. Hence, the parabola’sequation is x2 = 4y. To get the width of the opening, we need the x coordinatesof the points on the parabola with y coordinate 6.

x2 = 4(6)⇒ x = ±2√

6

Therefore, the width of the opening is 2× 2√

6 = 4√

6 ≈ 9.8 cm.



DEPED COPY

9. An object thrown from a height of 2m above the ground follows a parabolicpath until the object falls to the ground;see Figure 1.35. If the object reachesa maximum height (measured from theground) of 7 m after travelling a hor-izontal distance of 4 m, determine thehorizontal distance between the object’sinitial and final positions.

Figure 1.35

Solution: Let V (0, 7) be the parabola’s vertex, which corresponds to the high-est point reached by the object. Then the parabola’s equation is of the formx2 = −4c(y − 7) and the object’s starting point is at (−4, 2). Then

(−4)2 = −4c(2− 7)⇒ c =16

20=

4

5.

Hence, the equation of the parabola is x2 = −16

5(y−7). When the object hits

the ground, the y coordinate is 0 and

x2 = −16

5(0− 7) =

112

5⇒ x = ±4

√7

5.

Since this point is to the right of the vertex, we choose x = +4

√7

5. Therefore,

the total distance travelled is 4

√7

5− (−4) ≈ 8.73 m.

Supplementary Problems 1.2Determine the vertex, focus, directrix, and axis of symmetry of the parabola withthe given equation. Sketch the graph, and include these points and lines.

1. y2 = −36x

2. 5x2 = 100y

3. y2 + 4x− 14y = −53

4. y2 − 2x+ 2y − 1 = 0

5. 2x2 − 12x+ 28y = 38

6. (3x− 2)2 = 84y − 112



DEPED COPY

Find the standard equation of the parabola which satisfies the given conditions.

7. vertex (7, 11), focus (16, 11)

8. vertex (−10,−5), directrix y = −1

9. focus

(−10,

23

2

), directrix y = −11

2

10. focus

(−3

2, 3

), directrix x = −37

2

11. axis of symmetry y = 9, directrix x = 24, vertex on the line 3y − 5x = 7

12. vertex (0, 7), vertical axis of symmetry, through the point P (4, 5)

13. vertex (−3, 8), horizontal axis of symmetry, through the point P (−5, 12)

14. A satellite dish shaped like a paraboloid has its receiver located at the focus.How far is the receiver from the vertex if the dish is 10 ft across and 3 ft deepat the center?

15. A flashlight shaped like a paraboloid has its light source at the focus located1.5 cm from the base and is 10 cm wide at its opening. How deep is theflashlight at its center?

16. The ends of a rope are held in place at the top of two posts, 9 m apart andeach one 8 m high. If the rope assumes a parabolic shape and touches theground midway between the two posts, how high is the rope 2 m from one ofthe posts?

17. Radiation is focused to an unhealthy area in a patient’s body using a parabolicreflector, positioned in such a way that the target area is at the focus. If thereflector is 30 cm wide and 15 cm deep at the center, how far should the baseof the reflector be from the target area?

18. A rectangular object 25 m wide is to pass under a parabolic arch that has awidth of 32 m at the base and a height of 24 m at the center. If the vertexof the parabola is at the top of the arch, what maximum height should therectangular object have?

4



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Lesson 1.3. Ellipses



(1) define an ellipse;

(2) determine the standard form of equation of an ellipse;

(3) graph an ellipse in a rectangular coordinate system; and

(4) solve situational problems involving conic sections (ellipses).

Lesson Outline

(1) Definition of an ellipse

(2) Derivation of the standard equation of an ellipse

(3) Graphing ellipses

(4) Solving situational problems involving ellipses

Introduction

Unlike circle and parabola, an ellipse is one of the conic sections that most stu-dents have not encountered formally before. Its shape is a bounded curve whichlooks like a flattened circle. The orbits of the planets in our solar system aroundthe sun happen to be elliptical in shape. Also, just like parabolas, ellipses havereflective properties that have been used in the construction of certain structures.These applications and more will be encountered in this lesson.

1.3.1. Definition and Equation of an Ellipse

Consider the points F1(−3, 0) and F2(3, 0), as shown in Figure 1.36. What is thesum of the distances of A(4, 2.4) from F1 and from F2? How about the sum ofthe distances of B (and C(0,−4)) from F1 and from F2?

AF1 + AF2 = 7.4 + 2.6 = 10

BF1 +BF2 = 3.8 + 6.2 = 10

CF1 + CF2 = 5 + 5 = 10

There are other points P such that PF1 + PF2 = 10. The collection of all suchpoints forms a shape called an ellipse.



DEPED COPYFigure 1.36Figure 1.37

Let F1 and F2 be two distinct points. The set of all points P , whosedistances from F1 and from F2 add up to a certain constant, is calledan ellipse. The points F1 and F2 are called the foci of the ellipse.

Given are two points on the x-axis, F1(−c, 0) and F2(c, 0), the foci, both cunits away from their center (0, 0). See Figure 1.37. Let P (x, y) be a point onthe ellipse. Let the common sum of the distances be 2a (the coefficient 2 willmake computations simpler). Thus, we have PF1 + PF2 = 2a.

PF1 = 2a− PF2√(x+ c)2 + y2 = 2a−

√(x− c)2 + y2

x2 + 2cx+ c2 + y2 = 4a2 − 4a√

(x− c)2 + y2 + x2 − 2cx+ c2 + y2

a√

(x− c)2 + y2 = a2 − cxa2[x2 − 2cx+ c2 + y2

]= a4 − 2a2cx+ c2x2

(a2 − c2)x2 + a2y2 = a4 − a2c2 = a2(a2 − c2)b2x2 + a2y2 = a2b2 by letting b =

√a2 − c2, so a > b

x2

a2+y2

b2= 1

When we let b =√a2 − c2, we assumed a > c. To see why this is true, look at

4PF1F2 in Figure 1.37. By the Triangle Inequality, PF1 + PF2 > F1F2, whichimplies 2a > 2c, so a > c.

We collect here the features of the graph of an ellipse with standard equationx2

a2+y2

b2= 1, where a > b. Let c =

√a2 − b2.



DEPED COPY

(1) center : origin (0, 0)

(2) foci : F1(−c, 0) and F2(c, 0)

• Each focus is c units away from the center.

• For any point on the ellipse, the sum of its distances from the foci is 2a.

(3) vertices : V1(−a, 0) and V2(a, 0)

• The vertices are points on the ellipse, collinear with the center and foci.

• If y = 0, then x = ±a. Each vertex is a units away from the center.

• The segment V1V2 is called the major axis. Its length is 2a. It dividesthe ellipse into two congruent parts.

(4) covertices : W1(0,−b) and W2(0, b)

• The segment through the center, perpendicular to the major axis, is theminor axis. It meets the ellipse at the covertices. It divides the ellipseinto two congruent parts.

• If x = 0, then y = ±b. Each covertex is b units away from the center.

• The minor axis W1W2 is 2b units long. Since a > b, the major axis islonger than the minor axis.

Example 1.3.1. Give the coordinates of the foci, vertices, and covertices of theellipse with equation

x2

25+y2

9= 1.

Sketch the graph, and include these points.



DEPED COPY

Solution. With a2 = 25 and b2 = 9, we have a = 5, b = 3, and c =√a2 − b2 = 4.

foci: F1(−4, 0), F2(4, 0) vertices: V1(−5, 0), V2(5, 0)

covertices: W1(0,−3), W2(0, 3)

Example 1.3.2. Find the (standard) equation of the ellipse whose foci areF1(−3, 0) and F2(3, 0), such that for any point on it, the sum of its distancesfrom the foci is 10. See Figure 1.36.

Solution. We have 2a = 10 and c = 3, so a = 5 and b =√a2 − c2 = 4. The

equation isx2

25+y2

16= 1. 2

1.3.2. More Properties of Ellipses

The ellipses we have considered so far are “horizontal” and have the origin as theircenters. Some ellipses have their foci aligned vertically, and some have centersnot at the origin. Their standard equations and properties are given in the box.The derivations are more involved, but are similar to the one above, and so arenot shown anymore.

In all four cases below, a > b and c =√a2 − b2. The foci F1 and F2 are c

units away from the center. The vertices V1 and V2 are a units away from thecenter, the major axis has length 2a, the covertices W1 and W2 are b units awayfrom the center, and the minor axis has length 2b. Recall that, for any point onthe ellipse, the sum of its distances from the foci is 2a.



DEPED COPY

Center Corresponding Graphs

(0, 0)

x2

a2+y2

b2= 1, a > b

x2

b2+y2

a2= 1, b > a

(h, k)

(x− h)2

a2+

(y − k)2

b2= 1

(x− h)2

b2+

(y − k)2

a2= 1

a > b b > a

major axis: horizontal major axis: vertical

minor axis: vertical minor axis: horizontal

In the standard equation, if the x-part has the bigger denominator, the ellipseis horizontal. If the y-part has the bigger denominator, the ellipse is vertical.

Example 1.3.3. Give the coordinates of the center, foci, vertices, and coverticesof the ellipse with the given equation. Sketch the graph, and include these points.



DEPED COPY

(1)(x+ 3)2

24+

(y − 5)2

49= 1

(2) 9x2 + 16y2 − 126x+ 64y = 71

Solution. (1) From a2 = 49 and b2 = 24, we have a = 7, b = 2√

6 ≈ 4.9, andc =√a2 − b2 = 5. The ellipse is vertical.

center: (−3, 5)

foci: F1(−3, 0), F2(−3, 10)

vertices: V1(−3,−2), V2(−3, 12)

covertices: W1(−3− 2√

6, 5) ≈ (−7.9, 5)

W2(−3 + 2√

6, 5) ≈ (1.9, 5)

(2) We first change the given equation to standard form.

9(x2 − 14x) + 16(y2 + 4y) = 71

9(x2 − 14x+ 49) + 16(y2 + 4y + 4) = 71 + 9(49) + 16(4)

9(x− 7)2 + 16(y + 2)2 = 576

(x− 7)2

64+

(y + 2)2

36= 1



DEPED COPY

We have a = 8 and b = 6. Thus, c =√a2 − b2 = 2

√7 ≈ 5.3. The ellipse is

horizontal.

center: (7,−2)

foci: F1(7− 2√

7,−2) ≈ (1.7,−2)

F2(7 + 2√

7,−2) ≈ (12.3,−2)

vertices: V1(−1,−2), V2(15,−2)

covertices: W1(7,−8), W2(7, 4)

Example 1.3.4. The foci of an ellipse are (−3,−6) and (−3, 2). For any pointon the ellipse, the sum of its distances from the foci is 14. Find the standardequation of the ellipse.

Solution. The midpoint (−3,−2) of the foci is the center of the ellipse. Theellipse is vertical (because the foci are vertically aligned) and c = 4. From thegiven sum, 2a = 14 so a = 7. Also, b =

√a2 − c2 =

√33. The equation is

(x+ 3)2

33+

(y + 2)2

49= 1. 2

Example 1.3.5. An ellipse has vertices (2 −√

61,−5) and (2 +√

61,−5), andits minor axis is 12 units long. Find its standard equation and its foci.



DEPED COPY

Solution. The midpoint (2,−5) of the vertices is the center of the ellipse, which ishorizontal. Each vertex is a =

√61 units away from the center. From the length of

the minor axis, 2b = 12 so b = 6. The standard equation is(x− 2)2

61+

(y + 5)2

36=

1. Each focus is c =√a2 − b2 = 5 units away from (2,−5), so their coordinates

are (−3,−5) and (7,−5). 2

1.3.3. Situational Problems Involving Ellipses

Let us now apply the concept of ellipse to some situational problems.

?Example 1.3.6. A tunnel has the shape of a semiellipse that is 15 ft high atthe center, and 36 ft across at the base. At most how high should a passing truckbe, if it is 12 ft wide, for it to be able to fit through the tunnel? Round off youranswer to two decimal places.

Solution. Refer to the figure above. If we draw the semiellipse on a rectangularcoordinate system, with its center at the origin, an equation of the ellipse whichcontains it, is

x2

182+

y2

152= 1.

To maximize its height, the corners of the truck, as shown in the figure, wouldhave to just touch the ellipse. Since the truck is 12 ft wide, let the point (6, n)be the corner of the truck in the first quadrant, where n > 0, is the (maximum)height of the truck. Since this point is on the ellipse, it should fit the equation.Thus, we have

62

182+

n2

152= 1

n = 10√

2 ≈ 14.14 ft 2



DEPED COPY

Example 1.3.7. The orbit of a planet has the shape of an ellipse, and on oneof the foci is the star around which it revolves. The planet is closest to the starwhen it is at one vertex. It is farthest from the star when it is at the other vertex.Suppose the closest and farthest distances of the planet from this star, are 420million kilometers and 580 million kilometers, respectively. Find the equation ofthe ellipse, in standard form, with center at the origin and the star at the x-axis.Assume all units are in millions of kilometers.

Solution. In the figure above, the orbit is drawn as a horizontal ellipse withcenter at the origin. From the planet’s distances from the star, at its closestand farthest points, it follows that the major axis is 2a = 420 + 580 = 1000(million kilometers), so a = 500. If we place the star at the positive x-axis,then it is c = 500 − 420 = 80 units away from the center. Therefore, we getb2 = a2 − c2 = 5002 − 802 = 243600. The equation then is

x2

250000+

y2

243600= 1.

The star could have been placed on the negative x-axis, and the answer wouldstill be the same. 2



DEPED COPY

More Solved Examples1. Give the coordinates of the foci, vertices,

and covertices of the ellipse with equa-

tionx2

169+

y2

144= 1. Then sketch the

graph and include these points.

Solution: The ellipse is horizontal.a2 = 169 ⇒ a = 13, b2 = 144 ⇒ b = 12,c =√

169− 144 = 5Foci: F1(−5, 0), F2(5, 0)Vertices: V1(−13, 0), V2(13, 0)Covertices: W1(0,−12), W2(0, 12)See Figure 1.38. Figure 1.38

2. Find the standard equation of the ellipse whose foci are F1(0,−8) and F2(0, 8),such that for any point on it, the sum of its distances from the foci is 34.

Solution: The ellipse is vertical and has center at (0, 0).

2a = 34⇒ a = 17

c = 8⇒ b =√

172 − 82 = 15

The equation isx2

225+

y2

289= 1.

For Examples 3 and 4, give the coordinates of the center, foci, vertices, andcovertices of the ellipse with the given equation. Sketch the graph, and includethese points.

3.(x− 7)2

64+

(y + 2)2

25= 1

Solution: The ellipse is horizontal.a2 = 64⇒ a = 8, b2 = 25⇒ b = 5c =√

64− 25 =√

39 ≈ 6.24

center: (7,−2)

foci: F1(7−√

39,−2) ≈ (0.76,−2)

F2(7 +√

39,−2) ≈ (13.24,−2)

vertices: V1(−1,−2), V2(15,−2)

covertices: W1(7,−7),W2(7, 3)

See Figure 1.39.

Figure 1.39



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4. 16x2 + 96x+ 7y2 + 14y + 39 = 0

Solution:

16x2 + 96x+ 7y2 + 14y = −39

16(x2 + 6x+ 9) + 7(y2 + 2y + 1) = −39 + 151

16(x+ 3)2 + 7(y + 1)2 = 112

(x+ 3)2

7+

(y + 1)2

16= 1

The ellipse is vertical.a2 = 16 ⇒ a = 4, b2 = 7 ⇒ b =

√7 ≈

2.65c =√

16− 7 = 3

center: (−3,−1)

foci: F1(−3,−4), F2(−3, 2)

vertices: V1(−3,−5), V2(−3, 3)

covertices: W1(−3−√

7,−1) ≈ (−5.65,−1)

W2(−3 +√

7,−1) ≈ (−0.35,−1)

See Figure 1.40.

Figure 1.40

5. The covertices of an ellipse are (5, 6) and (5, 8). For any point on the ellipse,the sum of its distances from the foci is 12. Find the standard equation of theellipse.

Solution: The ellipse is horizontal with center at the midpoint (5, 7) of the

covertices. Also, 2a = 12 so a = 6 while b = 1. The equation is(x− 5)2

36+

(y − 7)2

1= 1.

6. An ellipse has foci (−4−√

15, 3) and (−4 +√

15, 3), and its major axis is 10units long. Find its standard equation and its vertices.

Solution: The ellipse is horizontal with center at the midpoint (−4, 3) of thefoci; also c =

√15. Since the length of the major axis is 10, 2a = 10 and

a = 5. Thus b =√

52 − 15 =√

10. Therefore, the equation of the ellipse is(x+ 4)2

25+

(y − 3)2

10= 1 and its vertices are (−9, 3) and (1, 3).

7. A whispering gallery is an enclosure or room where whispers can be clearlyheard in some parts of the gallery. Such a gallery can be constructed bymaking its ceiling in the shape of a semi-ellipse; in this case, a whisper fromone focus can be clearly heard at the other focus. If an elliptical whispering



DEPED COPY

gallery is 90 feet long and the foci are 50 feet apart, how high is the gallery atits center?

Solution: We set up a Cartesian coordinate system by assigning the center ofthe semiellipse as the origin. The point on the ceiling right above the center is acovertex of the ellipse. Since 2a = 90 and 2c = 50; then b2 = 452−252 = 1400.The height is given by b =

√1400 ≈ 37.4 ft.

8. A spheroid (or oblate spheroid) is the surface obtained by rotating an ellipsearound its minor axis. The bowl in Figure 1.41 is in the shape of the lower halfof a spheroid; that is, its horizontal cross sections are circles while its verticalcross sections that pass through the center are semiellipses. If this bowl is 10in wide at the opening and

√10 in deep at the center, how deep does a circular

cover with diameter 9 in go into the bowl?

Figure 1.41

Solution: We set up a Cartesian coordinate system by assigning the centerof the semiellipse as the origin. Then a = 5, b =

√10, and the equation of

the ellipse is x2

25+ y2

10= 1. We want the y-coordinate of the points on the

ellipse that has x = ±4.5. This coordinate is y = −√

10(1− x2

25

)≈ −1.38.

Therefore, the cover will go 1.38 inches into the bowl.



DEPED COPY

Supplementary Problems 1.3Give the coordinates of the center, foci, vertices, and covertices of the ellipse withthe given equation. Sketch the graph, and include these points.

1.x2

8+y2

4= 1

2.x2

16+

(y − 2)2

25= 1

3. (x− 1)2 + (2y − 2)2 = 4

4.(x+ 5)2

49+

(y − 2)2

121= 1

5. 16x2 − 224x+ 25y2 + 250y − 191 = 0

6. 25x2 − 200x+ 16y2 − 160y = 800

Find the standard equation of the ellipse which satisfies the given conditions.

7. foci (2−√

33, 8) and (2 +√

33, 8), the sum of the distances of any point fromthe foci is 14

8. center (−3,−7), vertical major axis of length 20, minor axis of length 12

9. foci (−21, 10) and (3, 10), contains the point (−9, 15)

10. a vertex at (−3,−18) and a covertex at (−12,−7), major axis is either hori-zontal or vertical

11. a focus at (−9, 15) and a covertex at (1, 10), with vertical major axis

12. A 40-ft wide tunnel has the shape of a semiellipse that is 5 ft high a distanceof 2 ft from either end. How high is the tunnel at its center?

13. The moon’s orbit is an ellipse with Earth as one focus. If the maximumdistance from the moon to Earth is 405 500 km and the minimum distance is363 300 km, find the equation of the ellipse in a Cartesian coordinate systemwhere Earth is at the origin. Assume that the ellipse has horizontal majoraxis and that the minimum distance is achieved when the moon is to the rightof Earth. Use 100 km as one unit.

14. Two friends visit a whispering gallery (in the shape of a semiellipsoid) wherethey stand 100 m apart to be at the foci. If one of them is 6 m from thenearest wall, how high is the gallery at its center?



DEPED COPY

15. A jogging path is in the shape of an ellipse. If it is 120 ft long and 40 ft wide,what is the width of the track 15 ft from either vertex?

16. Radiation is focused to an unhealthy area in a patient’s body using a semiel-liptic reflector, positioned in such a way that the target area is at one focuswhile the source of radiation is at the other. If the reflector is 100 cm wideand 30 cm high at the center, how far should the radiation source and thetarget area be from the ends of the reflector?

4

Lesson 1.4. Hyperbolas



(1) define a hyperbola;

(2) determine the standard form of equation of a hyperbola;

(3) graph a hyperbola in a rectangular coordinate system; and

(4) solve situational problems involving conic sections (hyperbolas).

Lesson Outline

(1) Definition of a hyperbola

(2) Derivation of the standard equation of a hyperbola

(3) Graphing hyperbolas

(4) Solving situational problems involving hyperbolas

Introduction

Just like ellipse, a hyperbola is one of the conic sections that most studentshave not encountered formally before. Its graph consists of two unboundedbranches which extend in opposite directions. It is a misconception that eachbranch is a parabola. This is not true, as parabolas and hyperbolas have verydifferent features. An application of hyperbolas in basic location and navigationschemes are presented in an example and some exercises.

1.4.1. Definition and Equation of a Hyperbola

Consider the points F1(−5, 0) and F2(5, 0) as shown in Figure 1.42. What is theabsolute value of the difference of the distances of A(3.75,−3) from F1 and from



DEPED COPY

F2? How about the absolute value of the difference of the distances of B(−5, 16

3

)from F1 and from F2?

|AF1 − AF2| = |9.25− 3.25| = 6

|BF1 −BF2| =∣∣∣∣16

3− 34

3

∣∣∣∣ = 6

There are other points P such that |PF1 − PF2| = 6. The collection of all suchpoints forms a shape called a hyperbola, which consists of two disjoint branches.For points P on the left branch, PF2 − PF1 = 6; for those on the right branch,PF1 − PF2 = 6.

Figure 1.42

Figure 1.43



DEPED COPY

Let F1 and F2 be two distinct points. The set of all points P , whosedistances from F1 and from F2 differ by a certain constant, is called ahyperbola. The points F1 and F2 are called the foci of the hyperbola.

In Figure 1.43, given are two points on the x-axis, F1(−c, 0) and F2(c, 0), thefoci, both c units away from their midpoint (0, 0). This midpoint is the centerof the hyperbola. Let P (x, y) be a point on the hyperbola, and let the absolutevalue of the difference of the distances of P from F1 and F2, be 2a (the coefficient2 will make computations simpler). Thus, |PF1 − PF2| = 2a, and so∣∣∣√(x+ c)2 + y2 −

√(x− c)2 + y2

∣∣∣ = 2a.

Algebraic manipulations allow us to rewrite this into the much simpler

x2

a2− y2

b2= 1, where b =

√c2 − a2.

When we let b =√c2 − a2, we assumed c > a. To see why this is true, suppose

that P is closer to F2, so PF1 − PF2 = 2a. Refer to Figure 1.43. Suppose alsothat P is not on the x-axis, so 4PF1F2 is formed. From the triangle inequality,F1F2 + PF2 > PF1. Thus, 2c > PF1 − PF2 = 2a, so c > a.

Now we present a derivation. For now, assume P is closer to F2 so PF1 > PF2,and PF1 − PF2 = 2a.

PF1 = 2a+ PF2√(x+ c)2 + y2 = 2a+

√(x− c)2 + y2(√

(x+ c)2 + y2)2

=(

2a+√

(x− c)2 + y2)2

cx− a2 = a√

(x− c)2 + y2

(cx− a2)2 =(a√

(x− c)2 + y2)2

(c2 − a2)x2 − a2y2 = a2(c2 − a2)b2x2 − a2y2 = a2b2 by letting b =

√c2 − a2 > 0

x2

a2− y2

b2= 1

We collect here the features of the graph of a hyperbola with standard equa-tion

x2

a2− y2

b2= 1.

Let c =√a2 + b2.



DEPED COPYFigure 1.44 Figure 1.45

(1) center : origin (0, 0)

(2) foci : F1(−c, 0) and F2(c, 0)

• Each focus is c units away from the center.

• For any point on the hyperbola, the absolute value of the difference ofits distances from the foci is 2a.

(3) vertices : V1(−a, 0) and V2(a, 0)

• The vertices are points on the hyperbola, collinear with the center andfoci.

• If y = 0, then x = ±a. Each vertex is a units away from the center.

• The segment V1V2 is called the transverse axis. Its length is 2a.

(4) asymptotes : y = bax and y = − b

ax, the lines `1 and `2 in Figure 1.45

• The asymptotes of the hyperbola are two lines passing through the cen-ter which serve as a guide in graphing the hyperbola: each branch ofthe hyperbola gets closer and closer to the asymptotes, in the directiontowards which the branch extends. (We need the concept of limits fromcalculus to explain this.)

• An aid in determining the equations of the asymptotes: in the standardequation, replace 1 by 0, and in the resulting equation x2

a2− y2

b2= 0, solve

for y.

• To help us sketch the asymptotes, we point out that the asymptotes`1 and `2 are the extended diagonals of the auxiliary rectangle drawnin Figure 1.45. This rectangle has sides 2a and 2b with its diagonalsintersecting at the center C. Two sides are congruent and parallel tothe transverse axis V1V2. The other two sides are congruent and parallel



DEPED COPY

to the conjugate axis, the segment shown which is perpendicular to thetransverse axis at the center, and has length 2b.

Example 1.4.1. Determine the foci, vertices, and asymptotes of the hyperbolawith equation

x2

9− y2

7= 1.

Sketch the graph, and include these points and lines, the transverse and conjugateaxes, and the auxiliary rectangle.

Solution. With a2 = 9 and b2 = 7, we havea = 3, b =

√7, and c =

√a2 + b2 = 4.

foci: F1(−4, 0) and F2(4, 0)

vertices: V1(−3, 0) and V2(3, 0)

asymptotes: y =√73x and y = −

√73x

The graph is shown at the right. The conju-gate axis drawn has its endpoints b =

√7 ≈

2.7 units above and below the center. 2

Example 1.4.2. Find the (standard) equation of the hyperbola whose foci areF1(−5, 0) and F2(5, 0), such that for any point on it, the absolute value of thedifference of its distances from the foci is 6. See Figure 1.42.

Solution. We have 2a = 6 and c = 5, so a = 3 and b =√c2 − a2 = 4. The

hyperbola then has equationx2

9− y2

16= 1. 2

1.4.2. More Properties of Hyperbolas

The hyperbolas we considered so far are “horizontal” and have the origin as theircenters. Some hyperbolas have their foci aligned vertically, and some have centersnot at the origin. Their standard equations and properties are given in the box.The derivations are more involved, but are similar to the one above, and so arenot shown anymore.

In all four cases below, we let c =√a2 + b2. The foci F1 and F2 are c units

away from the center C. The vertices V1 and V2 are a units away from the center.The transverse axis V1V2 has length 2a. The conjugate axis has length 2b and isperpendicular to the transverse axis. The transverse and conjugate axes bisecteach other at their intersection point, C. Each branch of a hyperbola gets closerand closer to the asymptotes, in the direction towards which the branch extends.The equations of the asymptotes can be determined by replacing 1 in the standardequation by 0. The asymptotes can be drawn as the extended diagonals of theauxiliary rectangle determined by the transverse and conjugate axes. Recall that,



DEPED COPY

for any point on the hyperbola, the absolute value of the difference of its distancesfrom the foci is 2a.

Center Corresponding Hyperbola

(0, 0)

x2

a2− y2

b2= 1

y2

a2− x2

b2= 1

(h, k)

(x− h)2

a2− (y − k)2

b2= 1

(y − k)2

a2− (x− h)2

b2= 1

transverse axis: horizontal transverse axis: vertical

conjugate axis: vertical conjugate axis: horizontal

In the standard equation, aside from being positive, there are no other re-strictions on a and b. In fact, a and b can even be equal. The orientation of thehyperbola is determined by the variable appearing in the first term (the positiveterm): the corresponding axis is where the two branches will open. For example,if the variable in the first term is x, the hyperbola is “horizontal”: the transverse



DEPED COPY

axis is horizontal, and the branches open to the left and right in the direction ofthe x-axis.

Example 1.4.3. Give the coordinates of the center, foci, vertices, and asymp-totes of the hyperbola with the given equation. Sketch the graph, and includethese points and lines, the transverse and conjugate axes, and the auxiliary rect-angle.

(1)(y + 2)2

25− (x− 7)2

9= 1

(2) 4x2 − 5y2 + 32x+ 30y = 1

Solution. (1) From a2 = 25 and b2 = 9, we have a = 5, b = 3, and c =√a2 + b2 =

√34 ≈ 5.8. The hyperbola is vertical. To determine the asymp-

totes, we write (y+2)2

25− (x−7)2

9= 0, which is equivalent to y+ 2 = ±5

3(x− 7).

We can then solve this for y.

center: C(7,−2)

foci: F1(7,−2−√

34) ≈ (7,−7.8) and F2(7,−2 +√

34) ≈ (7, 3.8)

vertices: V1(7,−7) and V2(7, 3)

asymptotes: y = 53x− 41

3and y = −5

3x+ 29

3

The conjugate axis drawn has its endpoints b = 3 units to the left and rightof the center.



DEPED COPY

(2) We first change the given equation to standard form.

4(x2 + 8x)− 5(y2 − 6y) = 1

4(x2 + 8x+ 16)− 5(y2 − 6y + 9) = 1 + 4(16)− 5(9)

4(x+ 4)2 − 5(y − 3)2 = 20

(x+ 4)2

5− (y − 3)2

4= 1

We have a =√

5 ≈ 2.2 and b = 2. Thus, c =√a2 + b2 = 3. The hyperbola

is horizontal. To determine the asymptotes, we write (x+4)2

5− (y−3)2

4= 0

which is equivalent to y − 3 = ± 2√5(x+ 4), and solve for y.

center: C(−4, 3)

foci: F1(−7, 3) and F2(−1, 3)

vertices: V1(−4−√

5, 3) ≈ (−6.2, 3) and V2(−4 +√

5, 3) ≈ (−1.8, 3)

asymptotes: y = 2√5x+ 8√

5+ 3 and y = − 2√

5x− 8√

5+ 3

The conjugate axis drawn has its endpoints b = 2 units above and belowthe center.

Example 1.4.4. The foci of a hyperbola are (−5,−3) and (9,−3). For any pointon the hyperbola, the absolute value of the difference of its of its distances fromthe foci is 10. Find the standard equation of the hyperbola.



DEPED COPY

Solution. The midpoint (2,−3) of the foci is the center of the hyperbola. Eachfocus is c = 7 units away from the center. From the given difference, 2a = 10 soa = 5. Also, b2 = c2− a2 = 24. The hyperbola is horizontal (because the foci arehorizontally aligned), so the equation is

(x− 2)2

25− (y + 3)2

24= 1. 2

Example 1.4.5. A hyperbola has vertices (−4,−5) and (−4, 9), and one of itsfoci is (−4, 2−

√65). Find its standard equation.

Solution. The midpoint (−4, 2) of the vertices is the center of the hyperbola,which is vertical (because the vertices are vertically aligned). Each vertex isa = 7 units away from the center. The given focus is c =

√65 units away from

the center. Thus, b2 = c2 − a2 = 16, and the standard equation is

(y − 2)2

49− (x+ 4)2

16= 1. 2

1.4.3. Situational Problems Involving Hyperbolas

Let us now give an example on an application of hyperbolas.

Example 1.4.6. An explosion was heard by two stations 1200 m apart, locatedat F1(−600, 0) and F2(600, 0). If the explosion was heard in F1 two seconds beforeit was heard in F2, identify the possible locations of the explosion. Use 340 m/sas the speed of sound.

Solution. Using the given speed of sound, we can deduce that the sound traveled340(2) = 680 m farther in reaching F2 than in reaching F1. This is then thedifference of the distances of the explosion from the two stations. Thus, theexplosion is on a hyperbola with foci are F1 and F2, on the branch closer to F1.



DEPED COPY

We have c = 600 and 2a = 680, so a = 340 and b2 = c2 − a2 = 244400.The explosion could therefore be anywhere on the left branch of the hyperbolax2

115600− y2

244400= 1. 2


1. Determine the foci, vertices, and asymptotes of the hyperbola with equationx2

16− y2

33= 1. Sketch the graph, and include these points and lines, the

transverse and conjugate axes, and the auxiliary rectangle.

Solution: The hyperbola is horizontal.a2 = 16⇒ a = 4,b2 = 33⇒ b =

√33,

c =√

16 + 33 = 7

center: (0, 0)

foci: F1(−7, 0), F2(7, 0)

vertices: V1(−4, 0), V2(4, 0)

asymptotes: y =

√33

4x, y = −

√33

4x

The conjugate axis has endpoints(0,−

√33) and (0,

√33). See Figure

1.46.

Figure 1.46

2. Find the standard equation of the hyperbola whose foci are F1(0,−10) andF2(0, 10), such that for any point on it, the absolute value of the difference ofits distances from the foci is 12.

Solution: The hyperbola is vertical and has center at (0, 0). We have 2a = 12,

so a = 6; also, c = 10. Then b =√

102 − 62 = 8. The equation isy2

36− x

2

64= 1.

For Examples 3 and 4, give the coordinates of the center, foci, vertices, andasymptotes of the hyperbola with the given equation. Sketch the graph, and in-clude these points and lines, the transverse and conjugate axes, and the auxiliaryrectangle.

3.(y + 6)2

25− (x− 4)2

39= 1

Solution: The hyperbola is vertical.

a2 = 25⇒ a = 5, b2 = 39⇒ b =√

39, c =√

25 + 39 = 8



DEPED COPY

center: (4,−6)

foci: F1(4,−14), F2(4, 2)

vertices: V1(4,−11), V2(4,−1)

asymptotes:(y + 6)2

25− (x− 4)2

39= 0

⇔ y + 6 = ± 5√39

(x− 4)

The conjugate axis has endpoints b =√

39 units to the left and to the right ofthe center. See Figure 1.47.

Figure 1.47

4. 9x2 + 126x− 16y2 − 96y + 153 = 0

Solution:

9x2 + 126x− 16y2 − 96y = −153

9(x2 + 14x+ 49)− 16(y2 + 6y + 9) = −153 + 9(49)− 16(9)

9(x+ 7)2 − 16(y + 3)2 = 144

(x+ 7)2

16− (y + 3)2

9= 1

The hyperbola is horizontal.

a2 = 16⇒ a = 4, b2 = 9⇒ b = 3, c =√

16 + 9 = 5



DEPED COPY

center: (−7,−3)

foci: F1(−12,−3), F2(−2,−3)

vertices: V1(−11,−3), V2(−3,−3)

asymptotes:(x+ 7)2

16− (y + 3)2

9= 0⇔ y + 3 = ±3

4(x+ 7)

The conjugate axis have endpoints (−7,−6) and (−7, 0). See Figure 1.48.

Figure 1.48

5. The foci of a hyperbola are (−17,−3) and (3,−3). For any point on thehyperbola, the absolute value of the difference of its distances from the foci is14. Find the standard equation of the hyperbola.

Solution: The hyperbola is horizontal with center at the midpoint (−7,−3) ofthe foci. Also, 2a = 14 so a = 7 while c = 10. Then b2 = 102 − 72 = 51. The

equation is(x+ 7)2

49− (y + 3)2

51= 1.

6. The auxiliary rectangle of a hyperbola has vertices (−24,−15), (−24, 9), (10, 9),and (10,−15). Find the equation of the hyperbola if its conjugate axis is hor-izontal.

Solution: The hyperbola is vertical. Using the auxiliary rectangle’s dimen-sions, we see that the length of the transverse axis is 2a = 24 while thelength of the conjugate axis is 2b = 34. Thus, a = 12 and b = 17. Thehyperbola’s vertices are the midpoints (−7,−15) and (−7, 9) of the bottom



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and top sides, respectively, of the auxiliary rectangle. Then the hyperbola’scenter is (−7,−3), which is the midpoint of the vertices. The equation is(y + 3)2

144− (x+ 7)2

289= 1.

7. Two LORAN (long range navigation) stations A and B are situated along astraight shore, where A is 200 miles west of B. These stations transmit radiosignals at a speed 186 miles per millisecond. The captain of a ship travellingon the open sea intends to enter a harbor that is located 40 miles east ofstation A.

Due to the its location, the harbor experiences a time difference in receivingthe signals from both stations. The captain navigates the ship into the harborby following a path where the ship experiences the same time difference as theharbor.

(a) What time difference between station signals should the captain be look-ing for in order the ship to make a successful entry into the harbor?

(b) If the desired time difference is achieved, determine the location of theship if it is 75 miles offshore.

Solution:

(a) Let H represent the harbor on the shoreline. Note that BH−AH = 160−40 = 120. The time difference on the harbor is given by 120÷186 ≈ 0.645milliseconds. This is the time difference needed to be maintained in orderto for the ship to enter the harbor.

(b) Situate the stations A and B on the Cartesian plane so that A (−100, 0)and B (100, 0). Let P represent the ship on the sea, which has coordinates(h, 75). Since PB − PA = 120, then it should follow that h < 0. More-over, P should lie on the left branch of the hyperbola whose equation isgiven by

x2

a2− y2

b2= 1

where 2a = 120⇒ a = 60, and b =√c2 − a2 =

√1002 − 602 = 80.

Therefore,

h2

602− 752

802= 1

h = −

√(1 +

752

802

)602 ≈ −82.24

This means that the ship is around 17.76 miles to the east of station A.



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Supplementary Problems 1.4Give the center, foci, vertices, and asymptotes of the hyperbola with the givenequation. Sketch the graph and the auxiliary rectangle, then include these pointsand lines.

1.x2

100− y2

81= 1

2. y − x =1

y + x

3. 4x2 − 15(y − 5)2 = 60

4.(y − 6)2

64− (x− 8)2

36= 1

5. 9y2 + 54y − 6x2 − 36x− 27 = 0

6. 16x2 + 64x− 105y2 + 840y − 3296 = 0

Find the standard equation of the hyperbola which satisfies the given conditions.

7. foci (−7,−17) and (−7, 17), the absolute value of the difference of the distancesof any point from the foci is 24

8. foci (−3,−2) and (15,−2), a vertex at (9,−2)

9. center (−10,−4), one corner of auxiliary rectangle at (−1, 12), with horizontaltransverse axis

10. asymptotes y = 713− 4

3x and y = 4

3x− 17

3and a vertex at (17, 9)

11. asymptotes y = − 512x+ 19

3and y = 5

12x+ 29

3and a focus at (−4,−5)

12. horizontal conjugate axis, one corner of auxiliary rectangle at (3, 8), and anasymptote 4x+ 3y = 12

13. two corners of auxiliary rectangle at (2, 3) and (16,−1), and horizontal trans-verse axis

14. Two radio stations are located 150 miles apart, where station A is west of sta-tion B. Radio signals are being transmitted simultaneously by both stations,travelling at a rate of 0.2 miles/µsec. A plane travelling at 60 miles aboveground level has just passed by station B and is headed towards the otherstation. If the signal from B arrives at the plane 480 µsec before the signalsent from A, determine the location of the plane.

4



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Lesson 1.5. More Problems on Conic Sections



(1) recognize the equation and important characteristics of the different types ofconic sections; and

(2) solve situational problems involving conic sections.

Lesson Outline

(1) Conic sections with associated equations in general form

(2) Problems involving characteristics of various conic sections

(3) Solving situational problems involving conic sections

Introduction

In this lesson, we will identify the conic section from a given equation. Wewill analyze the properties of the identified conic section. We will also look atproblems that use the properties of the different conic sections. This will allowus to synthesize what has been covered so far.

1.5.1. Identifying the Conic Section by Inspection

The equation of a circle may be written in standard form

Ax2 + Ay2 + Cx+Dy + E = 0,

that is, the coefficients of x2 and y2 are the same. However, it does not followthat if the coefficients of x2 and y2 are the same, the graph is a circle.

General Equation Standard Equation graph

(A) 2x2 + 2y2 − 2x+ 6y + 5 = 0(x− 1

2

)2+(y + 3

2

)2= 0 point

(B) x2 + y2 − 6x− 8y + 50 = 0 (x− 3)2 + (y − 4)2 = −25 empty set

For a circle with equation (x − h)2 + (y − k)2 = r2, we have r2 > 0. This isnot the case for the standard equations of (A) and (B).

In (A), because the sum of two squares can only be 0 if and only if each squareis 0, it follows that x− 1

2= 0 and y + 3

2= 0. The graph is thus the single point(

12,−3

2

).



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In (B), no real values of x and y can make the nonnegative left side equal tothe negative right side. The graph is then the empty set.

Let us recall the general form of the equations of the other conic sections. Wemay write the equations of conic sections we discussed in the general form

Ax2 +By2 + Cx+Dy + E = 0.

Some terms may vanish, depending on the kind of conic section.

(1) Circle: both x2 and y2 appear, and their coefficients are the same

Ax2 + Ay2 + Cx+Dy + E = 0

Example: 18x2 + 18y2 − 24x+ 48y − 5 = 0

Degenerate cases: a point, and the empty set

(2) Parabola: exactly one of x2 or y2 appears

Ax2 + Cx+Dy + E = 0 (D 6= 0, opens upward or downward)

By2 + Cx+Dy + E = 0 (C 6= 0, opens to the right or left)

Examples: 3x2 − 12x+ 2y + 26 = 0 (opens downward)

− 2y2 + 3x+ 12y − 15 = 0 (opens to the right)

(3) Ellipse: both x2 and y2 appear, and their coefficients A and B have the samesign and are unequal

Examples: 2x2 + 5y2 + 8x− 10y − 7 = 0 (horizontal major axis)

4x2 + y2 − 16x− 6y + 21 = 0 (vertical major axis)

If A = B, we will classify the conic as a circle, instead of an ellipse.

Degenerate cases: a point, and the empty set

(4) Hyperbola: both x2 and y2 appear, and their coefficients A and B have dif-ferent signs

Examples: 5x2 − 3y2 − 20x− 18y − 22 = 0 (horizontal transverse axis)

− 4x2 + y2 + 24x+ 4y − 36 = 0 (vertical transverse axis)

Degenerate case: two intersecting lines

The following examples will show the possible degenerate conic (a point, twointersecting lines, or the empty set) as the graph of an equation following a similarpattern as the non-degenerate cases.

(1) 4x2 + 9y2 − 16x+ 18y + 25 = 0 =⇒ (x− 2)2

32+

(y + 1)2

22= 0

=⇒ one point: (2,−1)



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(2) 4x2 + 9y2 − 16x+ 18y + 61 = 0 =⇒ (x− 2)2

32+

(y + 1)2

22= −1

=⇒ empty set

(3) 4x2 − 9y2 − 16x− 18y + 7 = 0 =⇒ (x− 2)2

32− (y + 1)2

22= 0

=⇒ two lines: y + 1 = ±2

3(x− 2)

A Note on Identifying a Conic Sectionby Its General Equation

It is only after transforming a given general equation to standardform that we can identify its graph either as one of the degenerateconic sections (a point, two intersecting lines, or the empty set) or asone of the non-degenerate conic sections (circle, parabola, ellipse, orhyperbola).

1.5.2. Problems Involving Different Conic Sections

The following examples require us to use the properties of different conic sectionsat the same time.

Example 1.5.1. A circle has center at the focus of the parabola y2 + 16x+ 4y =44, and is tangent to the directrix of this parabola. Find its standard equation.

Solution. The standard equation of the parabola is (y + 2)2 = −16(x − 3). Itsvertex is V (3,−2). Since 4c = 16 or c = 4, its focus is F (−1,−2) and its directrixis x = 7. The circle has center at (−1,−2) and radius 8, which is the distancefrom F to the directrix. Thus, the equation of the circle is

(x+ 1)2 + (y + 2)2 = 64. 2

Example 1.5.2. The vertices and foci of 5x2 − 4y2 + 50x + 16y + 29 = 0 are,respectively, the foci and vertices of an ellipse. Find the standard equation ofthis ellipse.

Solution. We first write the equation of the hyperbola in standard form:

(x+ 5)2

16− (y − 2)2

20= 1.

For this hyperbola, using the notations ah, bh, and ch to refer to a, b, and c ofthe standard equation of the hyperbola, respectively, we have ah = 4, bh = 2

√5,

ch =√a2h + b2h = 6, so we have the following points:



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center: (−5, 2)

vertices: (−9, 2) and (−1, 2)

foci: (−11, 2) and (1, 2).

It means that, for the ellipse, we have these points:

center: (−5, 2)

vertices: (−11, 2) and (1, 2)

foci: (−9, 2) and (−1, 2).

In this case, ce = 4 and ae = 6, so that be =√a2e − c2e =

√20. The standard

equation of the ellipse is

(x+ 5)2

36+

(y − 2)2

20= 1. 2


1. Identify the graph of each of the following equations.

(a) 4x2− 8x− 49y2 + 196y− 388 = 0

(b) x2 + 5x+ y2 − y + 7 = 0

(c) y2 − 48x+ 6y = −729

(d) 49x2 + 196x + 100y2 + 1400y +196 = 0

(e) 36x2+360x+64y2−512y+1924 =0

(f) x2 + y2 − 18y − 19 = 0

(g) −5x2 + 60x+ 7y2 + 84y+ 72 = 0

(h) x2 − 16x+ 20y = 136

Solution:

(a) Since the coefficients of x2 and y2 have opposite signs, the graph is ahyperbola or a pair of intersecting lines. Completing the squares, we get

4x2 − 8x− 49y2 + 196y − 388 = 0

4(x2 − 2x)− 49(y2 − 4y) = 388

4(x2 − 2x+ 1)− 49(y2 − 4y + 4) = 388 + 4(1)− 49(4)

(x− 1)2

49− (y − 2)2

4= 1.

Thus, the graph is a hyperbola.

(b) Since x2 and y2 have equal coefficients, the graph is a circle, a point, orthe empty set. Completing the squares, we get

x2 + 5x+ y2 − y + 7 = 0



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x2 + 5x+25

4+ y2 − y +

1

4= −7 +

25

4+

1

4(x+

5

2

)2

+

(y − 1

2

)2

= −1

2.

Since the right hand side is negative, the graph is the empty set.

(c) By inspection, the graph is a parabola.

(d) Since the coefficients of x2 and y2 are not equal but have the same sign,the graph is an ellipse, a point, or the empty set. Completing the squares,we get

49x2 + 196x+ 100y2 + 1400y + 196 = 0

49(x2 + 4x) + 100(y2 + 14y) = −196

49(x2 + 4x+ 4) + 100(y2 + 14y + 49) = −196 + 49(4) + 100(49)

(x+ 2)2

100+

(y + 7)2

49= 1.

Thus, the graph is an ellipse.

(e) Since the coefficients of x2 and y2 are not equal but have the same sign,the graph is an ellipse, a point, or the empty set. Completing the squares,we get

36x2 + 360x+ 64y2 − 512y + 1924 = 0

36(x2 + 10x) + 64(y2 − 8y) = −1924

36(x2 + 10x+ 25) + 64(y2 − 8y + 16) = −1924 + 36(25) + 64(16)

(x+ 5)2

64+

(y − 4)2

36= 0.

Since the right-hand side is 0, the graph is a single point (the point is(−5, 4)).

(f) Since x2 and y2 have equal coefficients, the graph is a circle, a point, orthe empty set. Completing the squares, we get

x2 + y2 − 18y − 19 = 0

x2 + y2 − 18y + 81 = 19 + 81

x2 + (y − 9)2 = 100.

Thus, the graph is a circle.

(g) Since the coefficients of x2 and y2 have opposite signs, the graph is ahyperbola or a pair of intersecting lines. Completing the squares, we get

−5x2 + 60x+ 7y2 + 84y + 72 = 0



DEPED COPY

−5(x2 − 12x) + 7(y2 + 12y) = −72

−5(x2 − 12x+ 36) + 7(y2 + 12y + 36) = −72− 5(36) + 7(36)

(x− 6)2

7− (y + 6)2

5= 0.

Since the right hand side is 0, the graph is a pair of intersecting lines;

these are y + 6 = ±√

5

7(x− 6).

(h) By inspection, the graph is a parabola.

2. The center of a circle is the vertex of the parabola y2 + 24x− 12y + 132 = 0.If the circle intersects the parabola’s directrix at a point where y = 11, findthe equation of the circle.

Solution:

y2 − 12y = −24x− 132

y2 − 12y + 36 = −24x− 132 + 36

(y − 6)2 = −24x− 96

(y − 6)2 = −24(x+ 4)

The vertex of the parabola is (−4, 6) and its directrix is x = 2. Thus, thecircle has center (−4, 6) and contains the point (2, 11). Then its radius is√

(−4− 2)2 + (6− 11)2 =√

61. Therefore, the equation of the circle is (x +4)2 + (y − 6)2 = 61.

3. The vertices of the hyperbola with equation 9x2−72x−16y2−128y−256 = 0are the foci of an ellipse that contains the point (8,−10). Find the standardequation of the ellipse.

Solution:

9x2 − 72x− 16y2 − 128y − 256 = 0

9(x2 − 8x)− 16(y2 + 8y) = 256

9(x2 − 8x+ 16)− 16(y2 + 8y + 16) = 256 + 9(16)− 16(16)

(x− 4)2

16− (y + 4)2

9= 1

The vertices of the hyperbola are (0,−4) and (8,−4). Since these are the fociof the ellipse, the ellipse is horizontal with center C(4,−4); also, the focaldistance of the ellipse is c = 4. The sum of the distances of the point (8,−10)from the foci is√

(8− 0)2 + (−10− (−4))2 +√

(8− 8)2 + (−10− (−4))2 = 16.



DEPED COPY

This sum is constant for any point on the ellipse; so 2a = 16 and a = 8. Thenb2 = 82 − 42 = 48. Therefore, the equation of the ellipse is

(x− 4)2

64+

(y + 4)2

48= 1.

Supplementary Problems 1.5For items 1 to 8, identify the graph of each of the following equations.

1. 9x2 + 72x− 64y2 + 128y + 80 = 0

2. 49x2 − 490x+ 36y2 + 504y + 1225 = 0

3. y2 + 56x− 18y + 417 = 0

4. x2 + 20x+ y2 − 20y + 200 = 0

5. x2 − 10x− 48y + 265 = 0

6. −144x2 − 1152x+ 25y2 − 150y − 5679 = 0

7. x2 + 4x+ 16y2 − 128y + 292 = 0

8. x2 − 6x+ y2 + 14y + 38 = 0

9. An ellipse has equation 100x2 − 1000x + 36y2 − 144y − 956 = 0. Find thestandard equations of all circles whose center is a focus of the ellipse andwhich contains at least one of the ellipse’s vertices.

10. Find all parabolas whose focus is a focus of the hyperbola x2−2x−3y2−2 = 0and whose directrix contains the top side of the hyperbola’s auxiliary rectangle.

11. Find the equation of the circle that contains all corners of the auxiliary rect-angle of the hyperbola −x2 − 18x+ y2 + 10y − 81 = 0.

12. Find the equations of all horizontal parabolas whose focus is the center of theellipse 9x2 + 17y2− 170y+ 272 = 0 and whose directrix is tangent to the sameellipse.

13. Find all values of r 6= 1 so that the graph of

(r − 1)x2 + 14(r − 1)x+ (r − 1)y2 − 6(r − 1)y = 60− 57r

is

(a) a circle,



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(b) a point,

(c) the empty set.

14. Find all values of m 6= −7, 0 so that the graph of

2mx2 − 16mx+my2 + 7y2 = 2m2 − 18m

is

(a) a circle.

(b) a horizontal ellipse.

(c) a vertical ellipse.

(d) a hyperbola.

(e) the empty set.

4

Lesson 1.6. Systems of Nonlinear Equations



(1) illustrate systems of nonlinear equations;

(2) determine the solutions of systems of nonlinear equations using techniquessuch as substitution, elimination, and graphing; and

(3) solve situational problems involving systems of nonlinear equations.

Lesson Outline

(1) Review systems of linear equations

(2) Solving a system involving one linear and one quadratic equation

(3) Solving a system involving two quadratic equations

(4) Applications of systems of nonlinear equations

Introduction

After recalling the techniques used in solving systems of linear equations inGrade 8, we extend these methods to solving a system of equations to systemsin which the equations are not necessarily linear. In this lesson, the equationsare restricted to linear and quadratic types, although it is possible to adapt the



DEPED COPY

methodology to systems with other types of equations. We focus on quadraticequations for two reasons: to include a graphical representation of the solutionand to ensure that either a solution is obtained or it is determined that there isno solution. The latter is possible because of the quadratic formula.

1.6.1. Review of Techniques in Solving Systems of LinearEquations

Recall the methods we used to solve systems of linear equations.There were threemethods used: substitution, elimination, and graphical.

Example 1.6.1. Use the substitution method to solve the system, and sketchthe graphs in one Cartesian plane showing the point of intersection. 4x+ y = 6

5x+ 3y = 4

Solution. Isolate the variable y in the first equation, and then substitute into thesecond equation.

4x+ y = 6

=⇒ y = 6− 4x

5x+ 3y = 4

5x+ 3(6− 4x) = 4

−7x+ 18 = 4

x = 2

y = 6− 4(2) = −2

Example 1.6.2. Use the elimination method to solve the system, and sketch thegraphs in one Cartesian plane showing the point of intersection. 2x+ 7 = 3y

4x+ 7y = 12

Solution. We eliminate first the variable x. Rewrite the first equation whereinonly the constant term is on the right-hand side of the equation, then multiplyit by −2, and then add the resulting equation to the second equation.



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2x− 3y = −7

(−2)(2x− 3y) = (−2)(−7)

−4x+ 6y = 14

−4x+ 6y = 14

4x+ 7y = 12

13y = 26

y = 2

x = −1

2

1.6.2. Solving Systems of Equations Using Substitution

We begin our extension with a system involving one linear equation and onequadratic equation. In this case, it is always possible to use substitution bysolving the linear equation for one of the variables.

Example 1.6.3. Solve the following system, and sketch the graphs in one Carte-sian plane. x− y + 2 = 0

y − 1 = x2

Solution. We solve for y in terms of x in the first equation, and substitute thisexpression to the second equation.

x− y + 2 = 0 =⇒ y = x+ 2

y − 1 = x2

(x+ 2)− 1 = x2

x2 − x− 1 = 0

x =1±√

5

2

x =1 +√

5

2=⇒ y =

1 +√

5

2+ 2 =

5 +√

5

2

x =1−√

5

2=⇒ y =

1−√

5

2+ 2 =

5−√

5

2

Solutions:

(1 +√

5

2,5 +√

5

2

)and

(1−√

5

2,5−√

5

2

)



DEPED COPY

The first equation represents a line with x-intercept −2 and y-intercept 2,while the second equation represents a parabola with vertex at (0, 1) and whichopens upward.

1.6.3. Solving Systems of Equations Using Elimination

Elimination method is also useful in systems of nonlinear equations. Sometimes,some systems need both techniques (substitution and elimination) to solve them.

Example 1.6.4. Solve the following system: y2 − 4x− 6y = 11

4(3− x) = (y − 3)2.

Solution 1. We expand the second equation, and eliminate the variable x byadding the equations.

4(3− x) = (y − 3)2 =⇒ 12− 4x = y2 − 6y + 9 =⇒ y2 + 4x− 6y = 3 y2 − 4x− 6y = 11

y2 + 4x− 6y = 3

Adding these equations, we get

2y2−12y = 14 =⇒ y2−6y−7 = 0 =⇒ (y−7)(y+1) = 0 =⇒ y = 7 or y = −1.

Solving for x in the second equation, we have

x = 3− (y − 3)2

4.



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y = 7 =⇒ x = −1 and y = −1 =⇒ x = −1

Solutions: (−1, 7) and (−1,−1) 2

The graphs of the equations in the preceding example with the points ofintersection are shown below.

Usually, the general form is more convenient to use in solving systems ofequations. However, sometimes the solution can be simplified by writing theequations in standard form. Moreover, the standard form is best for graphing.

Let us again solve the previous example in a different way.

Solution 2. By completing the square, we can change the first equation into stan-dard form:

y2 − 4x− 6y = 11 =⇒ 4(x+ 5) = (y − 3)2. 4(x+ 5) = (y − 3)2

4(3− x) = (y − 3)2

Using substitution or the transitive property of equality, we get

4(x+ 5) = 4(3− x) =⇒ x = −1.

Substituting this value of x into the second equation, we have

4[3− (−1)] = (y − 3)2 =⇒ 16 = (y − 3)2 =⇒ y = 7 or y = −1.

The solutions are (−1, 7) and (−1,−1), same as Solution 1. 2



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Example 1.6.5. Solve the system and graph the curves: (x− 3)2 + (y − 5)2 = 10

x2 + (y + 1)2 = 25.

Solution. Expanding both equations, we obtain x2 + y2 − 6x− 10y + 24 = 0

x2 + y2 + 2y − 24 = 0.

Subtracting these two equations, we get

−6x− 12y + 48 = 0 =⇒ x+ 2y − 8 = 0

x = 8− 2y.

We can substitute x = 8− 2y to either the first equation or the second equation.For convenience, we choose the second equation.

x2 + y2 + 2y − 24 = 0

(8− 2y)2 + y2 + 2y − 24 = 0

y2 − 6y + 8 = 0

y = 2 or y = 4

y = 2 =⇒ x = 8− 2(2) = 4 and y = 4 =⇒ x = 8− 2(4) = 0

The solutions are (4, 2) and (0, 4).

The graphs of both equations are circles. One has center (3, 5) and radius√10, while the other has center (0,−1) and radius 5. The graphs with the points

of intersection are show below.



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1.6.4. Applications of Systems of Nonlinear Equations

Let us apply systems of equations to a problem involving modern-day televisionsets.

?Example 1.6.6. The screen size of television sets is given in inches. Thisindicates the length of the diagonal. Screens of the same size can come in differentshapes. Wide-screen TV’s usually have screens with aspect ratio 16 : 9, indicatingthe ratio of the width to the height. Older TV models often have aspect ratio4 : 3. A 40-inch LED TV has screen aspect ratio 16 : 9. Find the length and thewidth of the screen.

Solution. Let w represent the width and h the height of the screen. Then, byPythagorean Theorem, we have the system w2 + h2 = 402 =⇒ w2 + h2 = 1600

w

h=

16

9=⇒ h =

9w

16

w2 + h2 = 1600 =⇒ w2 +

(9w

16

)2

= 1600

337w2

256= 1600

w =

√409 600

337≈ 34.86

h =19x

16≈ 19(34.86)

16= 19.61

Therefore, a 40-inch TV with aspect ratio 16 : 9 is about 35.86 inches wide and19.61 inches high. 2

More Solved ExamplesSolve the system and graph the curves.

1.

x2 − y2 = 21

x+ y = 7

Solution: We can write y in terms of x using the second equation as y = 7−x.



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Substituting this into the first equation, we have

x2 − (7− x)2 = 21

14x− 70 = 0

x = 5.

Thus, the point of intersection is (5, 2).

2.

x2 + y2 − x+ 6y + 5 = 0

x+ y + 1 = 0

Solution: We can write y in terms of x using the second equation as y =−(x+ 1).

Substituting this into the first equation, we have

x2 + (−(x+ 1))2 − x+ 6(−(x+ 1)) + 5 = 0

2x2 − 5x = 0

x(2x− 5) = 0,

which yields x = 0 and x =5

2. Thus, the points of intersection are (0,−1)

and

(5

2,−7

2

).



DEPED COPY

3.

(y − 2)2 = 4(x− 4)

(y − 4)2 = x− 5

Solution: We can rewrite the first equation as

x− 4 =(y − 2)2

4,

which can be substituted into the second equation by rewriting it as

(y − 4)2 = (x− 4)− 1 =(y − 2)2

4− 1

which upon expansion yields

3y2 − 28y + 64 = 0.

This equation has roots y = 16/3 and y = 4, giving us the points (5, 4) and(61

9,16

3

).



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4.

x =1

2(y + 5)2 − 2

y2 + 10y + (x− 2)2 = −9

Solution: We can rewrite the first equation as

(y + 5)2 = 2x+ 4,

which can be substituted into the second equation by completing the squareto get

(y2 + 10y + 25) + (x− 2)2 = −9 + 25

(y + 5)2 + (x− 2)2 = 16

(2x+ 4) + (x− 2)2 = 16

x2 − 2x− 8 = 0,

This equation has roots x = −2 and x = 4, giving us the points (−2,−5),(4,−5−

√12), and

(4,−5 +

√12).

Find the system of equations that represents the given problem and solve.

5. The difference of two numbers is 12, and the sum of their squares is 144. Findthe numbers.

Solution: If x and y are the two numbers, then we have the resulting system{x− y = 12

x2 + y2 = 144,

where the first equation yields x = y + 12. Combining this with the secondequation yields (y + 12)2 + y2 = 144 or equivalently 2y(y + 12) = 0, giving usthe ordered pairs (12, 0) and (0,−12).



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Supplementary Problems 1.6

1. Solve the system and graph the curves:

(a)

x2 + 3x− y + 2 = 0

y − 5x = 1

(b)

(y − 2)2 = 9(x+ 2)

9x2 + 4y2 + 18x− 16y = 0

(c)

(x+ 1)2 + 2(y − 4)2 = 12

y2 − 8y = 4x− 16

(d)

x2 − 2x− 4y2 + 8y − 2 = 0

5x2 − 10x+ 12y2 + 24y − 58 = 0

(e)

x2 + y2 = 2

x− y = 4

2. Ram is speeding along a highway when he sees a police motorbike parked onthe side of the road right next to him. He immediately starts slowing down,but the police motorbike accelerates to catch up with him. It is assumed thatthe two vehicles are going in the same direction in parallel paths.

The distance that Ram has traveled in meters t seconds after he starts toslow down is given by d (t) = 150 + 75t− 1.2t2. The distance that the policemotorbike travels can be modeled by the equation d (t) = 4t2. How long willit take for the police motorbike to catch up to Ram?

3. The square of a certain number exceeds twice the square of another number

by1

8. Also, the sum of their squares is

5

16. Find possible pairs of numbers

that satisfy these conditions.

4. Solve the system of equations x2 + y2 = 41

xy = 20

5. Determine the value(s) of k such that the circle x2 + (y − 6)2 = 36 and theparabola x2 = 4ky will intersect only at the origin.

4



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Topic Test 1 for Unit 1


(a) x2 − x+ y2 + 3y − 32

= 0

(b) x2 + 4x− 14y = 52

(c) 3x2 − 42x− 4y2 − 24y + 99 = 0

(d) 7x2 − 112x+ 2y2 + 448 = 0

2. Determine and sketch the conic with the given equation. Identify the impor-tant parts of the conic and include them in the graph.

(a) 25x2 + 7y2 − 175 = 0

(b) −64x2 + 128x+ 36y2 + 288y − 1792 = 0

3. Find the equation of the conic with the given properties.

(a) parabola; vertex at (−1, 3); directrix x = −7

(b) hyperbola; asymptotes y = 125x − 1

5and y = −12

5x − 49

5; one vertex at

(3,−5)

4. Solve the following system of equations: (x− 1)2 + (y + 1)2 = 5

y = 2(x− 1)2 − 8

5. A doorway is in the shape of a rectangle capped by a semi-ellipse. If therectangle is 1 m wide and 2 m high while the ellipse is 0.3 m high at thecenter, can a cabinet that is 2.26 m high, 0.5 m wide, and 2 m long be pushedthrough the doorway? Assume that the cabinet cannot be laid down on itsside.

6. A point moves so that its distance from the point (0,−1) is twice its distancefrom the line x = 3. Derive the equation (in standard form) of the curve thatis traced by the point, and identify the curve.



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(a) y2 + 8x− 10y = 15

(b) x2 + 10x+ y2 + 18y + 110 = 0

(c) 9x2 + 36x+ 4y2 − 8y + 4 = 0

(d) −11x2 + 132x+ 17y2 − 136y − 124 = 0

2. Determine and sketch the conic with the given equation. Identify the impor-tant parts of the conic and include them in the graph.

(a) x2 − y2 = 64 (b) 4x2 + 24x+ 49y2 − 196y + 36

3. Find the equation of the conic with the given properties.

(a) parabola; directrix y = −2; focus at (7,−12)

(b) ellipse; vertical or horizontal major axis; one vertex at (−5, 12); onecovertex at (−1, 3)

4. Solve the following system of equations: 9x2 − 4y2 + 54x+ 45 = 0

(x+ 3)2 = 4y + 4

5. Nikko goes to his garden to water his plants. He holds the water hose 3 feetabove the ground, with the hoses opening as the vertex and the water flowfollowing a parabolic path. The water strikes the ground a horizontal distanceof 2 feet from where the opening is located. If he were to stand on a 1.5 feetstool, how much further would the water strike the ground?

6. A point moves so that its distance from the point (2, 0) is two-thirds its dis-tance from the line y = 5. Derive the equation (in standard form) of the curvethat is traced by the point, and identify the curve.



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Unit 2

Mathematical Induction

Batad Rice Terraces in Ifugao, by Ericmontalban, 30 September 2012,

https://commons.wikimedia.org/wiki/File%3ABatad rice terraces in Ifugao.jpg. Public Domain.

Listed as one of the United Nations Educational, Scientific and CulturalOrganization (UNESCO) World Heritage sites since 1995, the two-millennium-old Rice Terraces of the Philippine Cordilleras by the Ifugaos is a living testimonyof mankind’s creative engineering to adapt to physically-challenging environmentin nature. One of the five clusters of terraces inscribed in the UNESCO list isthe majestic Batad terrace cluster (shown above), which is characterized by itsamphitheater-like, semicircular terraces with a village at its base.



https://commons.wikimedia.org/wiki/File%3ABatad_rice_terraces_in_Ifugao.jpg

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Lesson 2.1. Review of Sequences and Series



(1) illustrate a series; and

(2) differentiate a series from a sequence.

Lesson Outline

(1) Sequences and series

(2) Different types of sequences and series (Fibonacci sequence, arithmetic andgeometric sequence and series, and harmonic series)

(3) Difference between sequence and series

Introduction

In this lesson, we will review the definitions and different types of sequencesand series.

Lesson Proper

Recall the following definitions:

A sequence is a function whose domain is the set of positive integersor the set {1, 2, 3, . . . , n}.

A series represents the sum of the terms of a sequence.

If a sequence is finite, we will refer to the sum of the terms of thesequence as the series associated with the sequence. If the sequence hasinfinitely many terms, the sum is defined more precisely in calculus.

A sequence is a list of numbers (separated by commas), while a series is asum of numbers (separated by “+” or “−” sign). As an illustration, 1,−1

2, 13,−1

4

is a sequence, and 1− 12

+ 13− 1

4= 7

12is its associated series.

The sequence with nth term an is usually denoted by {an}, and the associatedseries is given by

S = a1 + a2 + a3 + · · ·+ an.



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Example 2.1.1. Determine the first five terms of each defined sequence, andgive their associated series.(1) {2− n}

(2) {1 + 2n+ 3n2}

(3) {(−1)n}

(4) {1 + 2 + 3 + · · ·+ n}

Solution. We denote the nth term of a sequence by an, and S = a1 + a2 + a3 +a4 + a5.

(1) an = 2− nFirst five terms: a1 = 2− 1 = 1, a2 = 2− 2 = 0, a3 = −1, a4 = −2, a5 = −3

Associated series: S = a1 + a2 + a3 + a4 + a5 = 1 + 0− 1− 2− 3 = −5

(2) an = 1 + 2n+ 3n2

First five terms: a1 = 1 + 2 · 1 + 3 · 12 = 6, a2 = 17, a3 = 34, a4 = 57, a5 = 86

Associated series: S = 6 + 17 + 34 + 57 + 86 = 200

(3) an = (−1)n

First five terms: a1 = (−1)1 = −1, a2 = (−1)2 = 1, a3 = −1, a4 = 1,a5 = −1

Associated series: S = −1 + 1− 1 + 1− 1 = −1

(4) an = 1 + 2 + 3 + · · ·+ n

First five terms: a1 = 1, a2 = 1+2 = 3, a3 = 1+2+3 = 6, a4 = 1+2+3+4 =10, a5 = 1 + 2 + 3 + 4 + 5 = 15

Associated series: S = 1 + 3 + 6 + 10 + 15 = 35 2

The sequence {an} defined by an = an−1 + an−2 for n ≥ 3, where a1 =a2 = 1, is called a Fibonacci sequence. It terms are 1, 1, 2, 3, 5, 8, 13, . . ..

An arithmetic sequence is a sequence in which each term after the firstis obtained by adding a constant (called the common difference) to thepreceding term.

If the nth term of an arithmetic sequence is an and the common difference isd, then

an = a1 + (n− 1)d.

The associated arithmetic series with n terms is given by

Sn =n(a1 + an)

2=n[2a1 + (n− 1)d]

2.



DEPED COPY

A geometric sequence is a sequence in which each term after the firstis obtained by multiplying the preceding term by a constant (calledthe common ratio).

If the nth term of a geometric sequence is an and the common ratio is r, then

an = a1rn−1.

The associated geometric series with n terms is given by

Sn =

na1 if r = 1

a1(1− rn)

(1− r)if r 6= 1.

The proof of this sum formula is an example in Lesson 2.3.

When −1 < r < 1, the infinite geometric series

a1 + a1r + a1r2 + · · ·+ a1r

n−1 + · · ·

has a sum, and is given by

S =a1

1− r.

If {an} is an arithmetic sequence, then the sequence with nth termbn = 1

anis a harmonic sequence.


1. How many terms are there in an arithmetic sequence with first term 5, commondifference −3, and last term −76?

Solution: a1 = 5, d = −3, an = −76. Find n.

an = 5 + (n− 1)(−3) = −76

n− 1 =−76− 5

−3= 27, ⇒ n = 28

2. List the first three terms of the arithmetic sequence if the 25th term is 35 andthe 30th term is 5.

Solution: a24 = a1 + 24d = 35 and a30 = a1 + 29d = 5

Eliminating a1 by subtraction, 5d = −30, or d = −6

This implies that a1 = 179, and the first three terms are 179, 173, 167.



DEPED COPY

3. Find the sum of all positive three-digit odd integers.

Solution: Find sn if a1 = 101 = 1 + 50(2), an = 999 = 1 + 499(2).

There are 450 terms from a1 to an, hence n = 450.

sn =450(101 + 999)

2= 247 500

4. The seventh term of a geometric sequence is −6 and the tenth term is 162.Find the fifth term.

Solution: a7 = a1r6 = −6 and a10 = a1r

9 = 162.

Eliminating a1 by division:a10a7

= r3 =162

−6= −27. Thus r = −3

Since a5r2 = a7, a5 =

−6

9= −2

3.

5. Insert three numbers (called geometric means) between 6 and 32/27, so thatthe five numbers form a geometric sequence.

Solution: If a1 = 6 and there are three terms between a1 and 32/27, thena5 = 32/27.

a5 = 6(r)4 =32

27⇒ r4 =

16

81⇒ r = ±2

3One possible set of three numbers is 4, 8/3, 16/9, the other is −4, 8/3,−16/9.

6. A ball dropped from the top of a building 180 m high always rebounds three-fourths the distance it has fallen. How far (up and down) will the ball havetraveled when it hits the ground for the 6th time?

Solution: a1 = 180, r = 3/4, n = 6

s6 =180

[1−

(34

)6]1− 3

4

The ball traveled 2s6 − 180 ≈ 1003.71 meters.

7. The Cantor set is formed as follows. Divide a segment of one unit into threeequal parts. Remove the middle one-third of the segment. From each of thetwo remaining segments, remove the middle third. From each of the remainingsegments, remove the middle third. This process is continued indefinitely. Findthe total length of the segments removed.

Solution: Let an represent the total length removed in the nth iteration. Hencea1 = 1/3, a2 = 2/9, a3 = 4/27, and so on.

This means r =2

3. The sum to infinity is s =

13

1− 23

= 1.

The total length of the segments removed is 1 unit.



DEPED COPY

8. The 7th term of an arithmetic sequence is 25. Its first, third, and 21st termform a geometric sequence. Find the first term and the common difference ofthe sequence.

Solution: a7 = a1 + 6d = 25⇒ a1 = 25− 6da3a1

=a21a3

, or a1a21 = a23.

(25− 6d) (25− 6d+ 20d) = (25− 6d+ 2d)2

d = 0 and an = 25 for all n, or d = 4 and a1 = 1.

9. Let {an} be an arithmetic sequence and {bn} an arithmetic sequence of positiveintegers. Prove that the sequence with nth term abn is arithmetic.

Solution: Let the common difference of {an} be d and of {bn} be c.

abn+1 − abn = [a1 + (bn+1 − 1) d]− [a1 + (bn − 1) d] = bn+1 − bn = c

This proves that the difference between any two consecutive terms of {abn} isa constant independent of n.

10. Let {an} be a geometric sequence. Prove that {a3n} is a geometric sequence.

Solution: Let r be the common ratio of {an}.a3n = (a1r

n−1)3

= a31 (r3)n−1

.

Thus {a3n} is a geometric sequence with first term a31 and common ratio r3.

11. If {an} is a sequence such that its first three terms form both an arithmeticand a geometric sequence, what can be concluded about {an}?Solution: There is a real number r such that a2 = a1r and a3 = a1r

2.

Since a1, a2 and a3 form an arithmetic sequence, then a2 − a1 = a3 − a2, ora3 − 2a2 + a1 = 0.

a3 − 2a2 + a1 = a1r2 − 2a1r + a1 = a1(r − 1)2 = 0 ⇒ a1 = 0 or r = 1.

If a1 = 0, then a1 = a2 = a3 = 0. If r = 1, then a1 = a2 = a3.

In all cases, a1 = a2 = a3.


1. Find the 5th term of the arithmetic sequence whose 3rd term is 35 and whose10th term is 77.

2. Suppose that the fourth term of a geometric sequence is 29

and the sixth termis 8

81. Find the first term and the common ratio.



DEPED COPY

3. The partial sum in the arithmetic series with first term 17 and a commondifference 3 is 30705. How many terms are in the series?

4. An arithmetic sequence a1, a2, . . . , a100 has a sum of 15,000. Find the firstterm and the common difference if the sum of the terms in the sequencea3, a6, a9, . . . , a99 is 5016.

5. The sum of an infinite geometric series is 108, while the sum of the first 3terms is 112. Determine the first term of this series.

6. Evaluate the infinite series32 − 20

51+

33 − 21

52+ · · ·+ 3k+1 − 2k−1

5k+ · · · .

7. Let n = 0.123 = 0.123123 . . . be a nonterminating repeating decimal. Finda rational number that is equal to n by expressing n as an infinite geometricseries. Simplify your answer.

8. An arithmetic sequence whose first term is 2 has the property that its sec-ond, third, and seventh terms are consecutive terms of a geometric sequence.Determine all possible second terms of the arithmetic sequence.

9. Eighty loaves of bread are to be divided among 4 people so that the amountsthey receive form an arithmetic progression. The first two together receiveone-third of what the last two receive. How many loaves does each personreceive?

10. Given a and b, suppose that three numbers are inserted between them sothat the five numbers form a geometric sequence. If the product of the threeinserted numbers between a and b is 27, show that ab = 9.

11. For what values of n will the infinite series (2n− 1) + (2n− 1)2 + . . . +(2n− 1)i + . . . have a finite value?

4

Lesson 2.2. Sigma Notation


At the end of the lesson, the student is able to use the sigma notation torepresent a series.

Lesson Outline

(1) Definition of and writing in sigma notation



DEPED COPY

(2) Evaluate sums written in sigma notation

(3) Properties of sigma notation

(4) Calculating sums using the properties of sigma notation

Introduction

The sigma notation is a shorthand for writing sums. In this lesson, we willsee the power of this notation in computing sums of numbers as well as algebraicexpressions.

2.2.1. Writing and Evaluating Sums in Sigma Notation

Mathematicians use the sigma notation to denote a sum. The uppercase Greekletter Σ (sigma) is used to indicate a “sum.” The notation consists of severalcomponents or parts.

Let f(i) be an expression involving an integer i. The expression

f(m) + f(m+ 1) + f(m+ 2) + · · ·+ f(n)

can be compactly written in sigma notation, and we write it as

n∑i=m

f(i),

which is read “the summation of f(i) from i = m to n.” Here, mand n are integers with m ≤ n, f(i) is a term (or summand) of thesummation, and the letter i is the index, m the lower bound, and nthe upper bound.

Example 2.2.1. Expand each summation, and simplify if possible.

(1)4∑i=2

(2i+ 3)

(2)5∑i=0

2i

(3)n∑i=1

ai

(4)6∑

n=1

√n

n+ 1

Solution. We apply the definition of sigma notation.

(1)4∑i=2

(2i+ 3) = [2(2) + 3] + [2(3) + 3] + [2(4) + 3] = 27



DEPED COPY

(2)5∑i=0

2i = 20 + 21 + 22 + 23 + 24 + 25 = 63

(3)n∑i=1

ai = a1 + a2 + a3 + · · ·+ an

(4)6∑

n=1

√n

n+ 1=

1

2+

√2

3+

√3

4+

2

5+

√5

6+

√6

72

Example 2.2.2. Write each expression in sigma notation.

(1) 1 +1

2+

1

3+

1

4+ · · ·+ 1

100

(2) −1 + 2− 3 + 4− 5 + 6− 7 + 8− 9 + · · · − 25

(3) a2 + a4 + a6 + a8 + · · ·+ a20

(4) 1 +1

2+

1

4+

1

8+

1

16+

1

32+

1

64+

1

128

Solution. (1) 1 +1

2+

1

3+

1

4+ · · ·+ 1

100=

100∑n=1

1

n

(2) −1 + 2− 3 + 4− 5 + · · · − 25

= (−1)1 1 + (−1)2 2 + (−1)3 3 + (−1)4 4

+ (−1)5 5 + · · ·+ (−1)25 25

=25∑j=1

(−1)jj

(3) a2 + a4 + a6 + a8 + · · ·+ a20

= a2(1) + a2(2) + a2(3) + a2(4) + · · ·+ a2(10)

=10∑i=1

a2i

(4) 1 +1

2+

1

4+

1

8+

1

16+

1

32+

1

64+

1

128=

7∑k=0

1

2k2

The sigma notation of a sum expression is not necessarily unique. For ex-ample, the last item in the preceding example can also be expressed in sigmanotation as follows:

1 +1

2+

1

4+

1

8+

1

16+

1

32+

1

64+

1

128=

8∑k=1

1

2k−1.

However, this last sigma notation is equivalent to the one given in the example.



DEPED COPY

2.2.2. Properties of Sigma Notation

We start with finding a formula for the sum of

n∑i=1

i = 1 + 2 + 3 + · · ·+ n

in terms of n.

The sum can be evaluated in different ways. One informal but simple approachis pictorial.

n∑i=1

i = 1 + 2 + 3 + · · ·+ n =n(n+ 1)

2

Another way is to use the formula for an arithmetic series with a1 = 1 andan = n:

S =n(a1 + an)

2=n(n+ 1)

2.

We now derive some useful summation facts. They are based on the axiomsof arithmetic addition and multiplication.

n∑i=m

cf(i) = c

n∑i=m

f(i), c any real number.

Proof.

n∑i=m

cf(i) = cf(m) + cf(m+ 1) + cf(m+ 2) + · · ·+ cf(n)



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= c[f(m) + f(m+ 1) + · · ·+ f(n)]

= c

n∑i=m

f(i) 2

n∑i=m

[f(i) + g(i)] =n∑

i=m

f(i) +n∑

i=m

g(i)

Proof.

n∑i=m

[f(i) + g(i)]

= [f(m) + g(m)] + · · ·+ [f(n) + g(n)]

= [f(m) + · · ·+ f(n)] + [g(m) + · · ·+ g(n)]

=n∑

i=m

f(i) +n∑

i=m

g(i) 2

n∑i=m

c = c(n−m+ 1)

Proof.

n∑i=m

c = c+ c+ c+ · · ·+ c︸︷︷︸n−m+1 terms

= c(n−m+ 1) 2

A special case of the above result which you might encounter more often isthe following:

n∑i=1

c = cn.

Telescoping Sum

n∑i=m

[f(i+ 1)− f(i)] = f(n+ 1)− f(m)



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Proof.

n∑i=m

[f(i+ 1)− f(i)

]= [f(m+ 1)− f(m)] + [f(m+ 2)− f(m+ 1)]

+ [f(m+ 3)− f(m+ 2)] + · · ·+ [f(n+ 1)− f(n)]

Note that the terms, f(m+1), f(m+2), . . . , f(n), all cancel out. Hence, we have

n∑i=m

[f(i+ 1)− f(i)] = f(n+ 1)− f(m). 2

Example 2.2.3. Evaluate:30∑i=1

(4i− 5).

Solution.

30∑i=1

(4i− 5) =30∑i=1

4i−30∑i=1

5

= 430∑i=1

i−30∑i=1

5

= 4(30)(31)

2− 5(30)

= 1710 2

Example 2.2.4. Evaluate:

1

1 · 2+

1

2 · 3+

1

3 · 4+ · · ·+ 1

99 · 100.

Solution.

1

1 · 2+

1

2 · 3+

1

3 · 4+ · · ·+ 1

99 · 100

=99∑i=1

1

i(i+ 1)

=99∑i=1

i+ 1− ii(i+ 1)

=99∑i=1

[i+ 1

i(i+ 1)− i

i(i+ 1)

]



DEPED COPY

=99∑i=1

(1

i− 1

i+ 1

)

= −99∑i=1

(1

i+ 1− 1

i

)

Using f(i) =1

iand the telescoping-sum property, we get

99∑i=1

1

i(i+ 1)= −

(1

100− 1

1

)=

99

100. 2

Example 2.2.5. Derive a formula forn∑i=1

i2 using a telescoping sum with terms

f(i) = i3.

Solution. The telescoping sum property implies that

n∑i=1

[i3 − (i− 1)3

]= n3 − 03 = n3.

On the other hand, using expansion and the other properties of summation,we have

n∑i=1

[i3 − (i− 1)3

]=

n∑i=1

(i3 − i3 + 3i2 − 3i+ 1)

= 3n∑i=1

i2 − 3n∑i=1

i+n∑i=1

1

= 3n∑i=1

i2 − 3 · n(n+ 1)

2+ n.

Equating the two results above, we obtain

3n∑i=1

i2 − 3n(n+ 1)

2+ n = n3

6n∑i=1

i2 − 3n(n+ 1) + 2n = 2n3

6n∑i=1

i2 = 2n3 − 2n+ 3n(n+ 1)



DEPED COPY

= 2n(n2 − 1) + 3n(n+ 1)

= 2n(n− 1)(n+ 1) + 3n(n+ 1)

= n(n+ 1)[2(n− 1) + 3]

= n(n+ 1)(2n+ 1).

Finally, after dividing both sides of the equation by 6, we obtain the desiredformula

n∑i=1

i2 =n(n+ 1)(2n+ 1)

6. 2


1. Expand the following sums and simplify if possible:

(a)5∑i=1

(i2 − i+ 1)

(b)6∑i=3

i2(i+ 1

2

)2

(c)5∑i=0

x3iy15−3i

(d)9∑i=1

x2i+1

(i+ 1)2

(e)∞∑i=1

3−i+22i+1

Solution:

(a)5∑i=1

(i2 − i+ 1) = (12 − 1 + 1) + (22 − 2 + 1) + . . .+ (52 − 5 + 1) = 45 or

5(5 + 1)(2(5) + 1)

6− 5(5 + 1)

2+ 5 = 45

(b)6∑i=3

i2(i+ 1

2

)2

= 32

(4

2

)2

+ . . .+ 62

(7

2

)2

= 802

(c)5∑i=0

x3iy15−3i = x3(0)y15−3(0) + . . . + x3(5)y15−3(5) = xy15 + x3y12 + x6y9 +

x9y6 + x12y3 + x15y



DEPED COPY

(d)9∑i=1

x2i+1

(i+ 1)2=

x3

(1 + 1)2+

x5

(2 + 1)2+ . . .+

x19

(9 + 1)2= 4x3 + 9x5 + 16x7 +

. . .+ 100x19

(e)∞∑i=1

3−i+22i+1 = 27∞∑i=1

(2

3

)i+1

, which is an infinite geometric series with

|r| = 2

3< 1 and a1 =

4

9, giving us

∞∑i=1

3−i+22i+1 = 274/9

1− (2/3)= 36.

2. Evaluate20∑i=1

[2(i− 1) + 2].

Solution:20∑i=1

(2(i− 1) + 2) =20∑i=1

2i = 4(20)(21)

2= 840.

3. Find a formula for1

1(3)+

1

2(4)+

1

3(5)+ · · · + 1

n(n+ 2)given any positive

integer n.

Solution: We have1

1(3)+

1

2(4)+

1

3(5)+· · ·+ 1

n(n+ 2)=

n∑i=1

1

i(i+ 2). Rewriting

yields1

i(i+ 2)=

(2− 1) + (i− i)i(i+ 2)

=1

i− 1

i+ 2− 1

i(i+ 2)

or equivalently1

i+ 2=

1

2

(1

i− 1

i+ 2

).

Expanding the sum term by term,

n∑i=1

1

i(i+ 2)=

n∑i=1

1

2

(1

i− 1

i+ 2

)=

(1− 1

3

)+

(1

2− 1

4

)+

(1

3− 1

5

)· · ·

+

(1

n− 2− 1

n

)+

(1

n− 1− 1

n+ 1

)+

(1

n− 1

n+ 2

)= 1 +

1

2− 1

n+ 1− 1

n+ 2

=n(3n+ 2)

2(n+ 1)(n+ 2).



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4. Determine the value of N such thatN∑i=0

1

i2 + 3i+ 2=

97

98.

Solution: Rewrite the sum as

N∑i=1

1

i2 + 3i+ 2=

N∑i=1

1

(i+ 1)(i+ 2)

=N∑i=1

(1

i+ 1− 1

i+ 2

)

Set f(i) =1

i+ 1and use telescoping sums to get

N∑i=1

1

i2 + 3i+ 2= −

N∑i=1

(1

i+ 2− 1

i+ 1

)= −

(1

N + 2− 1

1

)=N + 1

N + 2.

Since we want the sum to be equal to97

98, N = 96.


1. Expand the following sums:

(a)10∑i=3

√3 · i

2

(b)5∑i=1

x2i

2i

(c)5∑i=2

(−1)i xi−1

2. Write the following in sigma notation.



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(a) (x+ 5)− (x+ 3)2 + (x+ 1)3 − (x− 1)4

(b)1

33+

22

43+

32

53+ . . .+

102

113

(c) a3 + a6 + a9 + . . .+ a81

3. Evaluate the following sums

(a)150∑i=1

(4i+ 2)

(b)120∑i=3

i(i− 5)

(c)50∑i=1

(2i− 1)(2i+ 1)

4. If50∑i=1

f(i) = 20 and50∑i=1

g(i) = 30, what is the value of50∑i=1

g(i) + 3f(i)√2

?

5. If s =200∑i=1

[(i− 1)2 − i2

], express

200∑i=1

i in terms of s.

6. If s =n∑i=1

ai and t =n∑i=1

bi, does it follow thatn∑i=1

aibi

=s

t?

4

Lesson 2.3. Principle of Mathematical Induction



(1) illustrate the Principle of Mathematical Induction; and

(2) apply mathematical induction in proving identities.

Lesson Outline

(1) State the Principle of Mathematical Induction

(2) Prove summation identities using mathematical induction

(3) Prove divisibility statements using mathematical induction

(4) Prove inequalities using mathematical induction



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Introduction

We have derived and used formulas for the terms of arithmetic and geometricsequences and series. These formulas and many other theorems involving positiveintegers can be proven with the use of a technique called mathematical induction.

2.3.1. Proving Summation Identities

The Principle of Mathematical Induction

Let P (n) be a property or statement about an integer n. Supposethat the following conditions can be proven:

(1) P (n0) is true (that is, the statement is true when n = n0).

(2) If P (k) is true for some integer k ≥ n0, then P (k + 1) is true(that is, if the statement is true for n = k, then it is also true forn = k + 1).

Then the statement P (n) is true for all integers n ≥ n0.

The Principle of Mathematical Induction is often compared to climbing aninfinite staircase. First, you need to be able to climb up to the first step. Second,if you are on any step (n = k), you must be able to climb up to the next step(n = k + 1). If you can do these two things, then you will be able to climb upthe infinite staircase.

Part 1 Part 2

Another analogy of the Principle of Mathematical Induction that is used istoppling an infinite line of standing dominoes. You need to give the first dominoa push so that it falls down. Also, the dominoes must be arranged so that if thekth domino falls down, the next domino will also fall down. These two conditionswill ensure that the entire line of dominoes will fall down.



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Standing Domino Tiles, by Nara Cute, 16 October 2015,

https://commons.wikimedia.org/wiki/File:Wallpaper kartu domino.png. Public Domain.

There are many mathematical results that can be proven using mathematicalinduction. In this lesson, we will focus on three main categories: summationidentities, divisibility statements, and inequalities.

Let us now take a look at some examples on the use of mathematical inductionin proving summation identities.

Example 2.3.1. Using mathematical induction, prove that

1 + 2 + 3 + · · ·+ n =n(n+ 1)

2

for all positive integers n.

Solution. We need to establish the two conditions stated in the Principle of Math-ematical Induction.

Part 1.Prove that the identity is true for n = 1.

The left-hand side of the equation consists of one term equal to 1. The right-hand side becomes

1(1 + 1)

2=

2

2= 1.

Hence, the formula is true for n = 1.



https://commons.wikimedia.org/wiki/File:Wallpaper_kartu_domino.png

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Part 2. Assume that the formula is true for n = k ≥ 1:

1 + 2 + 3 + · · ·+ k =k(k + 1)

2.

We want to show that the formula is true for n = k + 1; that is,

1 + 2 + 3 + · · ·+ k + (k + 1) =(k + 1)(k + 1 + 1)

2.

Using the formula for n = k and adding k + 1 to both sides of the equation,we get

1 + 2 + 3 + · · ·+ k + (k + 1) =k(k + 1)

2+ (k + 1)

=k(k + 1) + 2(k + 1)

2

=(k + 1)(k + 2)

2

=(k + 1) [(k + 1) + 1]

2

We have proven the two conditions required by the Principle of MathematicalInduction. Therefore, the formula is true for all positive integers n. 2

Example 2.3.2. Use mathematical induction to prove the formula for the sumof a geometric series with n terms:

Sn =a1 (1− rn)

1− r, r 6= 1.

Solution. Let an be the nth term of a geometric series. From Lesson 2.1, we knowthat an = a1r

n−1.

Part 1.Prove that the formula is true for n = 1.

a1(1− r1)1− r

= a1 = S1

The formula is true for n = 1.

Part 2. Assume that the formula is true for n = k ≥ 1: Sk =a1(1− rk)

1− r. We

want to prove that it is also true for n = k + 1; that is,

Sk+1 =a1(1− rk+1)

1− r.



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We know that

Sk+1 = a1 + a2 + · · ·+ ak︸︷︷︸Sk

+ak+1

= Sk + ak+1

=a1(1− rk

)1− r

+ a1rk

=a1(1− rk

)+ a1r

k (1− r)1− r

=a1(1− rk + rk − rk+1

)1− r

=a1(1− rk+1

)1− r

By the Principle of Mathematical Induction, we have proven that

Sn =a1(1− rn)

1− r

for all positive integers n. 2

Example 2.3.3. Using mathematical induction, prove that

12 + 22 + 32 + · · ·+ n2 =n(n+ 1)(2n+ 1)

6

for all positive integers n.

Solution. We again establish the two conditions stated in the Principle of Math-ematical Induction.

Part 11(1 + 1)(2 · 1 + 1)

6=

1 · 2 · 36

= 1 = 12


Part 2

Assume: 12 + 22 + 32 + · · ·+ k2 =k(k + 1)(2k + 1)

6.

Prove: 12 + 22 + 32 + · · ·+ k2 + (k + 1)2

=(k + 1)(k + 2) [2(k + 1) + 1]

6

=(k + 1)(k + 2)(2k + 3)

6.



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12 + 22 + 32 + · · ·+ k2 + (k + 1)2

=k(k + 1)(2k + 1)

6+ (k + 1)2

=k(k + 1)(2k + 1) + 6(k + 1)2

6

=(k + 1) [k(2k + 1) + 6(k + 1)]

6

=(k + 1) (2k2 + 7k + 6)

6

=(k + 1)(k + 2)(2k + 3)

6

Therefore, by the Principle of Mathematical Induction,

12 + 22 + 32 + · · ·+ n2 =n(n+ 1)(2n+ 1)

6

for all positive integers n. 2

2.3.2. Proving Divisibility Statements

We now prove some divisibility statements using mathematical induction.

Example 2.3.4. Use mathematical induction to prove that, for every positiveinteger n, 7n − 1 is divisible by 6.

Solution. Similar to what we did in the previous session, we establish the twoconditions stated in the Principle of Mathematical Induction.

Part 1

71 − 1 = 6 = 6 · 171 − 1 is divisible by 6.

Part 2

Assume: 7k − 1 is divisible by 6.

To show: 7k+1 − 1 is divisible by 6.

7k+1 − 1 = 7 · 7k − 1 = 6 · 7k + 7k − 1 = 6 · 7k + (7k − 1)

By definition of divisibility, 6 · 7k is divisible by 6. Also, by the hypothesis(assumption), 7k − 1 is divisible by 6. Hence, their sum (which is equal to7k+1 − 1) is also divisible by 6.



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Therefore, by the Principle of Math Induction, 7n − 1 is divisible by 6 for allpositive integers n. 2

Note that 70 − 1 = 1− 1 = 0 = 6 · 0 is also divisible by 6. Hence, a strongerand more precise result in the preceding example is: 7n − 1 is divisible by 6 forevery nonnegative integer n. It does not make sense to substitute negative valuesof n since this will result in non-integer values for 7n − 1.

Example 2.3.5. Use mathematical induction to prove that, for every nonnega-tive integer n, n3 − n+ 3 is divisible by 3.

Solution. We again establish the two conditions in the Principle of MathematicalInduction.

Part 1 Note that claim of the statement is that it is true for every nonnegativeinteger n. This means that Part 1 should prove that the statement is true forn = 0.

03 − 0 + 3 = 3 = 3(1)

03 − 0 + 3 is divisible by 3.

Part 2. We assume that k3− k+ 3 is divisible by 3. By definition of divisibility,we can write k3 − k + 3 = 3a for some integer a.

To show: (k + 1)3 − (k + 1) + 3 is divisible by 3.

(k + 1)3 − (k + 1) + 3 = k3 + 3k2 + 2k + 3

= (k3 − k + 3) + 3k2 + 3k

= 3a+ 3k2 + 3k

= 3(a+ k2 + k)

Since a+k2+k is also an integer, by definition of divisibility, (k+1)3−(k+1)+3is divisible by 3.

Therefore, by the Principle of Math Induction, n3−n+ 3 is divisible by 3 forall positive integers n. 2

?2.3.3. Proving Inequalities

Finally, we now apply the Principle of Mathematical Induction in proving someinequalities involving integers.

Example 2.3.6. Use mathematical induction to prove that 2n > 2n for everyinteger n ≥ 3.



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Solution. Just like the previous example, we establish the two conditions in thePrinciple of Mathematical Induction.

Part 1

23 = 8 > 6 = 2(3)

This confirms that 23 > 2(3).

Part 2

Assume: 2k > 2k, where k is an integer with k ≥ 3

To show: 2k+1 > 2(k + 1) = 2k + 2

We compare the components of the assumption and the inequality we need toprove. On the left-hand side, the expression is doubled. On the right-hand side,the expression is increased by 2. We choose which operation we want to apply toboth sides of the assumed inequality.

Alternative 1. We double both sides.

Since 2k > 2k, by the multiplication property of inequality, we have 2 · 2k >2 · 2k.

2k+1 > 2(2k) = 2k + 2k > 2k + 2 if k ≥ 3.

Hence, 2k+1 > 2(k + 1).

Alternative 2. We increase both sides by 2.

Since 2k > 2k, by the addition property of inequality, we have 2k+2 > 2k+2.

2(k + 1) = 2k + 2 < 2k + 2 < 2k + 2k if k ≥ 3.

The right-most expression above, 2k + 2k, is equal to 2(2k)

= 2k+1.

Hence, 2(k + 1) < 2k+1.

Therefore, by the Principle of Math Induction, 2n > 2n for every integern ≥ 3. 2

We test the above inequality for integers less than 3.

20 = 1 > 0 = 2(0) True

21 = 2 = 2(1) False

22 = 4 = 2(2) False

The inequality is not always true for nonnegative integers less than 3. Thisillustrates the necessity of Part 1 of the proof to establish the result. However,the result above can be modified to: 2n ≥ 2n for all nonnegative integers n.

Before we discuss the next example, we review the factorial notation. Recall



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that 0! = 1 and, for every positive integer n, n! = 1 · 2 · 3 · · ·n. The factorial alsosatisfies the property that (n+ 1)! = (n+ 1) · n!.

Example 2.3.7. Use mathematical induction to prove that 3n < (n + 2)! forevery positive integer n. Can you refine or improve the result?

Solution. We proceed with the usual two-part proof.

Part 1

31 = 3 < 6 = 3! = (1 + 2)! =⇒ 31 < (1 + 2)!

Thus, the desired inequality is true for n = 1.

Part 2

Assume: 3k < (k + 2)!

To show: 3k+1 < (k + 3)!

Given that 3k < (k + 2)!, we multiply both sides of the inequality by 3 andobtain

3(3k)< 3 [(k + 2)!] .

This implies that

3(3k)< 3 [(k + 2)!] < (k + 3) [(k + 2)!] , since k > 0,

and so3k+1 < (k + 3)!.

Therefore, by the Principle of Math Induction, we conclude that 3n < (n+2)!for every positive integer n.

The left-hand side of the inequality is defined for any integer n. The right-hand side makes sense only if n+ 2 ≥ 0, or n ≥ −2.

When n = −2: 3−2 =1

9< 1 = 0! = (−2 + 2)!

When n = −1: 3−1 =1

3< 1 = 1! = (−1 + 2)!

When n = 0: 30 = 1 < 2 = 2! = (0 + 2)!

Therefore, 3n < (n+ 2)! for any integer n ≥ −2. 2



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More Solved ExamplesUse mathematical induction to prove the given statements below.

1. 2 · 3 + 2 · 32 + . . .+ 2 · 3n−1 = 3n − 3 for n ≥ 1

Solution:

Part 1.

2 · 3 = 6 = 32 − 3.


Part 2.

Assume: P = 2 · 3 + 2 · 32 + . . .+ 2 · 3k−1 = 3k − 3.

To show: 2 · 3 + 2 · 32 + . . .+ 2 · 3k = 3k+1 − 3.

2 · 3 + 2 · 32 + . . .+ 2 · 3k = P + 2 · 3k

= 3k − 3 + 2 · 3k

= 3 · 3k − 3

= 3k+1 − 3.

2. 1 + 4 + 42 + . . .+ 4n−1 =1

3(4n − 1) for n ≥ 1

Solution:

Part 1.

1 =1

3(41 − 1).


Part 2.

Assume: P = 1 + 4 + 42 + . . .+ 4k−1 =1

3

(4k − 1

).

To show: 1 + 4 + 42 + . . .+ 4k =1

3

(4k+1 − 1

).

1 + 4 + 42 + . . .+ 4k = P + 4k

=1

3

(4k − 1

)+ 4k

=4

34k − 1

3

=1

3

(4k+1 − 1

).



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3.

(1− 1

22

)·(

1− 1

32

)· · ·(

1− 1

(n− 1)2

)·(

1− 1

n2

)=n+ 1

2nfor n ≥ 2.

Solution:

Part 1.

1− 1

22=

3

4=

2 + 1

2(2).


Part 2.

Assume: P =

(1− 1

22

)·(

1− 1

32

)· · ·(

1− 1

(k − 1)2

)·(

1− 1

k2

)=k + 1

2k.

To show:

(1− 1

22

)· · ·(

1− 1

k2

)·(

1− 1

(k + 1)2

)=

k + 2

2(k + 1).

(1− 1

22

)· · ·(

1− 1

(k + 1)2

)= P ·

(1− 1

(k + 1)2

)=k + 1

2k· (k2 + 2k + 1)− 1

(k + 1)2

=k + 1

2k· k(k + 2)

(k + 1)2

=k + 2

2(k + 1).

4. Prove that 4n+1 + 52n−1 is divisible by 21 for all integers n ≥ 1.

Solution:

Part 1.

41+1 + 52(1)−1 = 21.

The number is divisible by 21 for n = 1.

Part 2.

Assume: 4k+1 + 52k−1 is divisible by 21.

Prove: 4k+2 + 52(k+1)−1 is divisible by 21.

4k+2 + 52(k+1)−1 = 4 · 4k+1 + 25 · 52k−1 = 4(4k+1 + 52k−1)+ 21 · 52k−1

21 ·52k−1 is divisible by 21 and by the hypothesis (assumption), 4k+1 +52k−1 isdivisible by 21. Hence, their sum which is equal to 4k+2 + 52(k+1)−1 is divisibleby 21.



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5. n2 > 2n+ 3 for n ≥ 4.

Solution:

Part 1.

24 = 16 > 7 = 2(2) + 3 The inequality is true for n = 4.

Part 2

Assume: k2 > 2k + 3

Prove: (k + 1)2 > 2(k + 1) + 3

We expand (k + 1)2 and use the inequality in the hypothesis to get

(k + 1)2 = k2 + 2k + 1 > (2k + 3) + 2k + 1 = 4(k + 1) > 2(k + 1) + 3 if k > 0.

Therefore, by the principle of math induction, n2 > 2n+ 3 for n ≥ 4.

6. Prove that 2n+3 < (n+ 3)! for n ≥ 4.

Solution:

Part 1.

24+3 = 27 < 1 · 2 · 3 · · · 7 = (4 + 3)! The inequality is true for n = 1.

Part 2

Assume: 2k+3 < (k + 3)!

Prove: 2k+4 < (k + 4)!

Given that 2k+3 < (k + 3)!, we multiply both sides of the inequality by 2 andobtain

2(2k+3

)< 2 [(k + 3)!].

This implies that

2k+4 < 2 [(k + 3)!] < (k + 4) [(k + 3)!], if k > 0.

Therefore, by the principle of math induction, 2k+3 < (k+3)! for every positiveinteger n.



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Supplementary Problems 2.3Prove the following by mathematical induction:

1.1

2+

2

22+

3

23+ · · ·+ n

2n= 2− n+ 2

2nfor n ≥ 1

2.n∑i=1

−(i+ 1) = −n(n+ 3)

2

3. 1(1!) + 2(2!) + . . .+ n(n!) = (n+ 1)!− 1.

4. The sum of the first n odd numbers is equal to n2.

5.

(1− 1

2

)(1− 1

3

)(1− 1

4

). . .

(1− 1

n

)=

1

2n.

6.n∑i=1

(−1)ii2 =(−1)nn(n+ 1)

2

7. 43n+1 + 23n+1 + 1 is divisible by 7

8. 11n+2 + 122n+1 is divisible by 133

9. 52n+1 · 2n+2 + 3n+2 · 22n+1 is divisible by 19

10. 11n − 6 is divisible by 5

11.10n

3+

5

3+ 4n+2 is divisible by 3

12. n2 < 2n for n ≥ 5.

13.1

13+

1

23+

1

33+ . . .+

1

n3≤ 2− 1

nfor n ≥ 1.

14. The sequence an =√

2an−1, a1 =√

2 is increasing; that is, an < an+1.

4

Lesson 2.4. The Binomial Theorem



(1) illustrate Pascal’s Triangle in the expansion of (x + y)n for small positiveintegral values of n;



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(2) prove the Binomial Theorem;

(3) determine any term in (x + y)n, where n is a positive integer, without ex-panding; and

(4) solve problems using mathematical induction and the Binomial Theorem.

Lesson Outline

(1) Expand (x+ y)n for small values of n using Pascal’s Triangle

(2) Review the definition of and formula for combination

(3) State and prove the Binomial Theorem

(4) Compute all or specified terms of a binomial expansion

(5) Prove some combination identities using the Binomial Theorem

Introduction

In this lesson, we study two ways to expand (a + b)n, where n is a positiveinteger. The first, which uses Pascal’s Triangle, is applicable if n is not too big,and if we want to determine all the terms in the expansion. The second methodgives a general formula for the expansion of (a + b)n for any positive integer n.This formula is useful especially when n is large because it avoids the process ofgoing through all the coefficients for lower values of n obtained through Pascal’sTriangle.

2.4.1. Pascal’s Triangle and the Concept of Combination

Consider the following powers of a+ b:

(a+ b)1 = a+ b

(a+ b)2 = a2 + 2ab+ b2

(a+ b)3 = a3 + 3a2b+ 3ab2 + b3

(a+ b)4 = a4 + 4a3b+ 6a2b2 + 4ab3 + b4

(a+ b)5 = a5 + 5a4b+ 10a3b2 + 10a2b3 + 5ab4 + b5

We now list down the coefficients of each expansion in a triangular array asfollows:

n = 1 : 1 1

n = 2 : 1 2 1

n = 3 : 1 3 3 1

n = 4 : 1 4 6 4 1

n = 5 : 1 5 10 10 5 1



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The preceding triangular array of numbers is part of what is called the Pas-cal’s Triangle, named after the French mathematician, Blaise Pascal (1623-1662).Some properties of the Triangle are the following:

(1) Each row begins and ends with 1.

(2) Each row has n+ 1 numbers.

(3) The second and second to the last number of each row correspond to therow number.

(4) There is symmetry of the numbers in each row.

(5) The number of entries in a row is one more than the row number (or onemore than the number of entries in the preceding row).

(6) Every middle number after first row is the sum of the two numbers aboveit.

It is the last statement which is useful in constructing the succeeding rows of thetriangle.

Example 2.4.1. Use Pascal’s Triangle to expand the expression (2x− 3y)5.

Solution. We use the coefficients in the fifth row of the Pascal’s Triangle.

(2x− 3y)5 = (2x)5 + 5(2x)4(−3y) + 10(2x)3(−3y)2

+ 10(2x)2(−3y)3 + 5(2x)(−3y)4

+ (−3y)5

= 32x5 − 240x4y + 720x3y2 − 1080x2y3

+ 810xy4 − 243y5 2

Example 2.4.2. Use Pascal’s Triangle to expand (a+ b)8.

Solution. We start with the sixth row (or any row of the Pascal’s Triangle thatwe remember).

n = 6 : 1 6 15 20 15 6 1

n = 7 : 1 7 21 35 35 21 7 1

n = 8 : 1 8 28 56 70 56 28 8 1

Therefore, we get

(a+ b)8 = a8 + 8a7b+ 28a6b2 + 56a5b3

+ 70a4b4 + 56a3b5 + 28a2b6

+ 8ab7 + b8 2



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We observe that, for each n, the expansion of (a + b)n starts with an

and the exponent of a in the succeeding terms decreases by 1, whilethe exponent of b increases by 1. This observation will be shown tobe true in general.

Let us review the concept of combination. Recall that C(n, k) or(nk

)counts

the number of ways of choosing k objects from a set of n objects. It is also usefulto know some properties of C(n, k):

(1) C(n, 0) = C(n, n) = 1,

(2) C(n, 1) = C(n, n− 1) = n, and

(3) C(n, k) = C(n, n− k).

These properties can explain some of the observations we made on the num-bers in the Pascal’s Triangle. Also recall the general formula for the number ofcombinations of n objects taken k at a time:

C(n, k) =

(n

k

)=

n!

k!(n− k)!,

where 0! = 1 and, for every positive integer n, n! = 1 · 2 · 3 · · ·n.

Example 2.4.3. Compute

(5

3

)and

(8

5

).

Solution. (5

3

)=

5!

(5− 3)!3!=

5!

2!3!= 10(

8

5

)=

8!

(8− 5)!5!=

10!

3!5!= 56 2

You may observe that the value of(53

)and the fourth coefficient in the fifth

row of Pascal’s Triangle are the same. In the same manner,(85

)is equal to the

sixth coefficient in the expansion of (a+ b)8 (see Example 2.4.2). These observedequalities are not coincidental, and they are, in fact, the essence embodied in theBinomial Theorem, as you will see in the succeeding sessions.

2.4.2. The Binomial Theorem

As the power n gets larger, the more laborious it would be to use Pascal’s Triangle(and impractical to use long multiplication) to expand (a + b)n. For example,using Pascal’s Triangle, we need to compute row by row up to the thirtieth row



DEPED COPY

to know the coefficients of (a+ b)30. It is, therefore, delightful to know that it ispossible to compute the terms of a binomial expansion of degree n without goingthrough the expansion of all the powers less than n.

We now explain how the concept of combination is used in the expansion of(a+ b)n.

(a+ b)n = (a+ b)(a+ b)(a+ b) · · · (a+ b)︸︷︷︸n factors

When the distributive law is applied, the expansion of (a + b)n consists ofterms of the form ambi, where 0 ≤ m, i ≤ n. This term is obtained by choosinga for m of the factors and b for the rest of the factors. Hence, m + i = n, orm = n − i. This means that the number of times the term an−ibi will appearin the expansion of (a + b)n equals the number of ways of choosing (n − i) or ifactors from the n factors, which is exactly C(n, i). Therefore, we have

(a+ b)n =n∑i=0

(n

i

)an−ibi.

To explain the reasoning above, consider the case n = 3.

(a+ b)3 = (a+ b)(a+ b)(a+ b)

= aaa+ aab+ aba+ abb+ baa+ bab+ bba+ bbb

= a3 + 3a2b+ 3ab2 + b3

That is, each term in the expansion is obtained by choosing either a or b in eachfactor. The term a3 is obtained when a is chosen each time, while a2b is obtainedwhen a is selected 2 times, or equivalently, b is selected exactly once.

We will give another proof of this result using mathematical induction. Butfirst, we need to prove a result about combinations.

Pascal’s Identity

If n and k are positive integers with k ≤ n, then(n+ 1

k

)=

(n

k

)+

(n

k − 1

).

Proof. The result follows from the combination formula.(n

k

)+

(n

k − 1

)=

n!

k!(n− k)!+

n!

(k − 1)!(n− k + 1)!

=n!(n− k + 1) + n!(k)

k!(n− k + 1)!



DEPED COPY

=n!(n− k + 1 + k)

k!(n+ 1− k)!

=n!(n+ 1)

k!(n+ 1− k)!

=(n+ 1)!

k!(n+ 1− k)!

=

(n+ 1

k

)2

Pascal’s identity explains the method of constructing Pascal’s Triangle, inwhich an entry is obtained by adding the two numbers above it. This identityis also an essential part of the second proof of the Binomial Theorem, which wenow state.

The Binomial Theorem

For any positive integer n,

(a+ b)n =n∑i=0

(n

i

)an−ibi.

Proof. We use mathematical induction.

Part 11∑i=0

(1

i

)a1−ibi =

(1

0

)a1b0 +

(1

1

)a0b1 = a+ b

Hence, the formula is true for n = 1.

Part 2. Assume that

(a+ b)k =k∑i=0

(k

i

)ak−ibi.

We want to show that

(a+ b)k+1 =k+1∑i=0

(k + 1

i

)ak+1−ibi.

(a+ b)k+1 = (a+ b)(a+ b)k

= (a+ b)k∑i=0

(k

i

)ak−ibi



DEPED COPY

= a

k∑i=0

(k

i

)ak−ibi + b

k∑i=0

(k

i

)ak−ibi

=k∑i=0

(k

i

)ak−i+1bi +

k∑i=0

(k

i

)ak−ibi+1

=

(k

0

)ak+1b0 +

k∑i=1

(k

i

)ak+1−ibi

+

(k

0

)akb1 +

(k

1

)ak−1b2 +

(k

2

)ak−2b3

+ · · ·+(

k

k − 1

)a1bk +

(k

k

)a0bk+1

= ak+1 +k∑i=1

(k

i

)ak+1−ibi

+k∑i=1

(k

i− 1

)ak+1−ibi + bk+1

=

(k + 1

0

)ak+1b0

+k∑i=1

[(k

i

)+

(k

i− 1

)]ak+1−ibi

+

(k + 1

k + 1

)a0bk+1

=k+1∑i=0

(k + 1

i

)ak+1−ibi

The last expression above follows from Pascal’s Identity.

Therefore, by the Principle of Mathematical Induction,

(a+ b)n =n∑i=1

(n

i

)an−ibi

for any positive integer n. 2

2.4.3. Terms of a Binomial Expansion

We now apply the Binomial Theorem in different examples.

Example 2.4.4. Use the Binomial Theorem to expand (x+ y)6.



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Solution.

(x+ y)6 =6∑

k=0

(6

k

)x6−kyk

=

(6

0

)x6y0 +

(6

1

)x5y1 +

(6

2

)x4y2

+

(6

3

)x3y3 +

(6

4

)x2y4 +

(6

5

)x1y5

+

(6

6

)x0y6

= x6 + 6x5y + 15x4y2 + 20x3y3

+ 15x2y2 + 6xy5 + y6 2

Since the expansion of (a + b)n begins with k = 0 and ends with k = n, theexpansion has n + 1 terms. The first term in the expansion is

(n0

)an = an, the

second term is(n1

)an−1b = nan=1b, the second to the last term is

(nn−1

)abn−1 =

nabn−1, and the last term is(nn

)bn = bn.

The kth term of the expansion is(nk−1

)an−k+1bk−1. If n is even, there is a

middle term, which is the(n2

+ 1)th term. If n is odd, there are two middle

terms, the(n+12

)th and

(n+12

+ 1)th terms.

The general term is often represented by(nk

)an−kbk. Notice that, in any term,

the sum of the exponents of a and b is n. The combination(nk

)is the coefficient

of the term involving bk. This allows us to compute any particular term withoutneeding to expand (a+ b)n and without listing all the other terms.

Example 2.4.5. Find the fifth term in the expansion of(2x−√y

)20.

Solution. The fifth term in the expansion of a fifth power corresponds to k = 4.(20

4

)(2x)20−4 (−√y)4 = 4845

(65536x16

)y2

= 317521920x16y2 2

Example 2.4.6. Find the middle term in the expansion of(x

2+ 3y

)6.

Solution. Since there are seven terms in the expansion, the middle term is thefourth term (k = 3), which is(

6

3

)(x2

)3(3y)3 = 20

(x3

8

)(27y3

)=

135x3y3

2. 2



DEPED COPY

Example 2.4.7. Find the term involving x (with exponent 1) in the expansion

of

(x2 − 2y

x

)8

.

Solution. The general term in the expansion is(8

k

)(x2)8−k (−2y

x

)k=

(8

k

)x16−2k · (−2)kyk

xk

=

(8

k

)(−2)kx16−2k−kyk

=

(8

k

)(−2)kx16−3kyk.

The term involves x if the exponent of x is 1, which means 16 − 3k = 1, ork = 5. Hence, the term is (

8

5

)(−2)5xy5 = −1792xy5. 2

?2.4.4. Approximation and Combination Identities

We continue applying the Binomial Theorem.

?Example 2.4.8. (1) Approximate (0.8)8 by using the first three terms in theexpansion of (1− 0.2)8. Compare your answer with the calculator value.

(2) Use 5 terms in the binomial expansion to approximate (0.8)8. Is there animprovement in the approximation?

Solution.

(0.8)8 = (1− 0.2)8 =8∑

k=0

(8

k

)(1)8−k(−0.2)k

=8∑

k=0

(8

k

)(−0.2)k

(1)2∑

k=0

(8

k

)(−0.2)k =

(8

0

)+

(8

1

)(−0.2) +

(8

2

)(−0.2)2

= 1− 1.6 + 1.12 = 0.52

The calculator value is 0.16777216, so the error is 0.35222784.



DEPED COPY

(2)4∑

k=0

(8

k

)(−0.2)k =

(8

0

)+

(8

1

)(−0.2) +

(8

2

)(−0.2)2

+

(8

3

)(−0.2)3 +

(8

4

)(−0.2)4

= 0.52− 0.448 + 0.112 = 0.184

The error is 0.01622784, which is an improvement on the previous estimate.2

Example 2.4.9. Use the Binomial Theorem to prove that, for any positive in-teger n,

n∑k=0

(n

k

)= 2n.

Solution. Set a = b = 1 in the expansion of (a+ b)n. Then

2n = (1 + 1)n =n∑k=0

(n

k

)(1)n−k(1)k =

n∑k=0

(n

k

). 2

Example 2.4.10. Use the Binomial Theorem to prove that(100

0

)+

(100

2

)+

(100

4

)+ · · ·+

(100

100

)=

(100

1

)+

(100

3

)+

(100

5

)+ · · ·+

(100

99

)Solution. Let a = 1 and b = −1 in the expansion of (a+ b)100. Then

[1 + (−1)

]100=

100∑k=0

(100

k

)(1)100−k(−1)k.

0 =

(100

0

)+

(100

1

)(−1) +

(100

2

)(−1)2 +

(100

3

)(−1)3

+ · · ·+(

100

99

)(−1)99 +

(100

100

)(−1)100

If k is even, then (−1)k = 1. If k is odd, then (−1)k = −1. Hence, we have

0 =

(100

0

)−(

100

1

)+

(100

2

)−(

100

3

)+ · · · −

(100

99

)+

(100

100

)



DEPED COPY

Therefore, after transposing the negative terms to other side of the equation, weobtain (

100

0

)+

(100

2

)+

(100

4

)+ · · ·+

(100

100

)=

(100

1

)+

(100

3

)+

(100

5

)+ · · ·+

(100

99

)2


1. Use the Binomial Theorem to expand (2x4 − 3y2)5.

Solution:(2x4 − 3y2

)5=

5∑k=0

(2x4)5−k (

3y2)k

= 32x20−240x16y2 +720x12y4−

1080x8y6 + 810x4y8 − 243y10

2. Determine the 20th term in the expansion of (x3 − 3y)28

.

Solution: We see that k = 19 should yield the 20th term, yielding−319

(28

19

)x27y19.

3. Find the term containingx2

y2in the expansion of

(x

y− y2

2x2

)20

.

Solution: Setting a =x

y, b = − y2

2x2, the (k + 1)th term in the binomial

expansion is (−1)k(

20

k

)(x

y

)20−k (y2

2x2

)k=

(−1)k

2k

(n

k

)x20−3ky3k−20. To get

x2

y2, we get 20− 3k = 2⇒ k = 6, yielding

1

26

(20

6

)x2y2.

4. Determine the term not involving x in the expansion of

(x3 +

2

x5

)16

.

Solution: Setting a = x3, b =2

x5, the (k+1)th term in the binomial expansion

is

(16

k

)(x3)16−k ( 2

x5

)k= 2k

(16

k

)x48−8k. To get the term without x, we get

48− 8k = 0⇒ k = 6, yielding 26

(16

6

).

5. Determine the coefficient of x9 in the expansion of (1 + 2x)10.



DEPED COPY

Solution: Setting a = 1, b = 2x, the (k + 1)th term in the binomial expansion

of the first factor is

(10

k

)(1)10−k (2x)k = 2k

(10

k

)xk. To get x9, we set k = 9,

yielding 29

(10

9

)x9.

6. Prove thatn∑i=0

(−1)k(n

k

)3n−k = 2n.

Solution: Set a = 3, b = −1.

7. If(√

3 +√

2)5

is written in the form a√

3 + b√

2 where a, b are integers, whatis a+ b?

Solution: We have(√

3 +√

2)5

=5∑

k=0

(5

k

)(√3)5−k (√

2)k

. Note that if

5 − k is odd (or equivalently, k is even), the term has a factor of√

3, while

the rest have a factor of√

2. Thus, a =

(5

0

)+

(5

2

)+

(5

4

)= 16 and

b =

(5

1

)+

(5

3

)+

(5

5

)= 16 yields a+ b = 32.


1. Use the Binomial Theorem to expand the following:

(a) (2x− 3y)5

(b)

(√x

3− 2

x2

)4

(c) (1 +√x)

4

2. Without expanding completely, find the indicated value(s) in the expansion ofthe following:

(a) (2 + x)9, two middle terms

(b)

(p

2+

2

q

)10

, 3rd term

(c) (x2 + y4)21

, last 2 terms

(d)

(1√x

)20

, middle term



DEPED COPY

(e)

(2y4

x3+x5

4y

)15

, term not involving y

(f)

(1

2x2− x2

)13

, term involving x2

(g) (1− 2x)6, coefficient of x3

(h)

(2y7/3 − 1

2y5/3

)30

, coefficient of1

y2

(i) (√x− 3)

8, coefficient of x7/2

(j) (√x+ 2)

6, coefficient of x3/2

3. Approximate (2.1)10 by using the first 5 terms in the expansion of (2 + 0.1)20.Compare your answer with the calculator result.

4. In the expansion of (4x+ 3)34, the kth value and the (k + 1)st terms haveequal coefficients. What is the value of k?

5. Determine the value of(19

0

)−(

19

1

)3 +

(19

2

)32 −

(19

3

)33 + . . .+

(19

18

)318 −

(19

19

)319

4



DEPED COPY


1. Determine if the given sequence is arithmetic, geometric, or neither by writingA, G, or O, respectively.

(a)1

3,1

2,3

4,

9

16,27

32, . . .

(b)1

2,1

7,

1

12,

1

17,

1

21, . . .

(c) 0, 3, 8, 15, 24, . . .

2. Three numbers form an arithmetic sequence, the common difference being 5.If the last number is increased by 1, the second by 2, and the first by 4, theresulting numbers form a geometric sequence. Find the numbers.

3. Evaluate the sum50∑i=1

2i3 + 9i2 + 13i+ 6

i2 + 3i+ 2.

4. Find the indicated terms in the expansion of the given expression.

(a)

(x2 − 1

2

)8

, term involving x8(b) (n3 − 3m)

28, 20th term

5. Prove the statement below for all positive integers n by mathematical induc-tion.

1

1 · 3+

1

3 · 5+ · · ·+ 1

(2n− 1)(2n+ 1)=

n

2n+ 1

6. On his 20th birthday, Ian deposited an amount of 10,000 pesos to a time-deposit scheme with a yearly interest of 4%. Ian decides not to withdraw anyamount of money or earnings and vows to keep it in the same time-depositscheme year after year. Show that the new amounts in Ian’s time-depositaccount in each succeeding birthday represent a geometric sequence, and usethis to determine the value of the money during Ian’s 60th birthday.



DEPED COPY


1. Determine if the given sequence is arithmetic, geometric, or neither by writingA, G, or O, respectively.

(a)2

3,

8

15,32

75,128

375,

512

1875, . . .

(b)1

2,2

3,3

4,4

5,5

6, . . .

(c) 3,11

2, 8,

21

2, 13, . . .

2. The sum of the first two terms of an arithmetic sequence is 9 and the sum ofthe first three terms is also 9. How many terms must be taken to give a sumof −126?

3. Evaluate the following sums.

(a)50∑i=1

(2i+ 1)(i− 3) (b)30∑i=1

√i2 − 2i+ 1

4

4. Find the term not involving x in the expansion of

(x3 +

1

x

)8

.

5. Prove that the following statements are true for all positive integers n bymathematical induction.

(a) 1 + 4 + 7 + . . .+ (3n− 2) =n(3n− 1)

2

(b) 3n + 7n−1 + 8 is divisible by 12.



DEPED COPY

Unit 3

Trigonometry

Puerto Princesa Subterranean River National Park, by Giovanni G. Navata, 12 November 2010,

https://commons.wikimedia.org/wiki/File%3AUnderground River.jpg. Public Domain

Named as one of the New Seven Wonders of Nature in 2012 by the New7WondersFoundation, the Puerto Princesa Subterranean River National Park is world-famous for its limestone karst mountain landscape with an underground river.The Park was also listed as UNESCO World Heritage Site in 1999. The under-ground river stretches about 8.2 km long, making it one of the world’s longestrivers of its kind.



https://commons.wikimedia.org/wiki/File%3AUnderground_River.jpg

DEPED COPY

Lesson 3.1. Angles in a Unit Circle



(1) illustrate the unit circle and the relationship between the linear and angularmeasures of arcs in a unit circle.

(2) convert degree measure to radian measure, and vice versa.

(3) illustrate angles in standard position and coterminal angles.

Lesson Outline

(1) Linear and angular measure of arcs

(2) Conversion of degree to radian, and vice versa

(3) Arc length and area of the sector

(4) Angle in standard position and coterminal angles

Introduction

Angles are being used in several fields like engineering, medical imaging, elec-tronics, astronomy, geography and many more. Added to that, surveyors, pilots,landscapers, designers, soldiers, and people in many other professions heavily useangles and trigonometry to accomplish a variety of practical tasks. In this les-son, we will deal with the basics of angle measures together with arc length andsectors.

3.1.1. Angle Measure

An angle is formed by rotating a ray about its endpoint. In the figure shownbelow, the initial side of ∠AOB is OA, while its terminal side is OB. An angleis said to be positive if the ray rotates in a counterclockwise direction, and theangle is negative if it rotates in a clockwise direction.



DEPED COPY

An angle is in standard position if it is drawn in the xy-plane with its vertexat the origin and its initial side on the positive x-axis. The angles α, β, and θ inthe following figure are angles in standard position.

To measure angles, we use degrees, minutes, seconds, and radians.

A central angle of a circle measures one degree, written 1◦, if it inter-cepts 1

360of the circumference of the circle. One minute, written 1′, is

160

of 1◦, while one second, written 1′′, is 160

of 1′.

For example, in degrees, minutes, and seconds,

10◦30′18′′ = 10◦(

30 +18

60

)′= 10◦30.3′

=

(10 +

30.3

60

)◦= 10.505◦

and

79.251◦ = 79◦(0.251× 60)′

= 79◦15.06′

= 79◦15′(0.06× 60)′′

= 79◦15′3.6′′.

Recall that the unit circle is the circle with center at the origin and radius 1unit.



DEPED COPY

A central angle of the unit circle that intercepts an arc of the circlewith length 1 unit is said to have a measure of one radian, written 1rad. See Figure 3.1.

Figure 3.1

In trigonometry, as it was studied in Grade 9, the degree measure is often used.On the other hand, in some fields of mathematics like calculus, radian measure ofangles is preferred. Radian measure allows us to treat the trigonometric functionsas functions with the set of real numbers as domains, rather than angles.

Example 3.1.1. In the following figure, identify the terminal side of an angle instandard position with given measure.

(1) degree measure: 135◦, −135◦, −90◦, 405◦

(2) radian measure: π4

rad, −3π4

rad, 3π2

rad, −π2

rad



DEPED COPY

Solution. (1) 135◦:−→OC; −135◦:

−−→OD; −90◦:

−−→OE; and 405◦:

−−→OB

(2) radian measure: π4

rad:−−→OB; −3π

4rad:

−−→OD; 3π

2rad:

−−→OE; and −π

2rad:

−−→OE 2

Since a unit circle has circumference 2π, a central angle that measures 360◦

has measure equivalent to 2π radians. Thus, we obtain the following conversionrules.

Converting degree to radian, and vice versa

1. To convert a degree measure to radian, multiply it by π180

.

2. To convert a radian measure to degree, multiply it by 180π

.

Figure 3.2 shows some special angles in standard position with the indicatedterminal sides. The degree and radian measures are also given.

Figure 3.2

Example 3.1.2. Express 75◦ and 240◦ in radians.

Solution.

75( π

180

)=

5π

12=⇒ 75◦ =

5π

12rad

240( π

180

)=

4π

3=⇒ 240◦ =

4π

3rad 2



DEPED COPY

Example 3.1.3. Express π8

rad and 11π6

rad in degrees.

Solution.π

8

(180

π

)= 22.5 =⇒ π

8rad = 22.5◦

11π

6

(180

π

)= 330 =⇒ 11π

6rad = 330◦ 2

3.1.2. Coterminal Angles

Two angles in standard position that have a common terminal side are calledcoterminal angles. Observe that the degree measures of coterminal angles differby multiples of 360◦.

Two angles are coterminal if and only if their degree measures differby 360k, where k ∈ Z.

Similarly, two angles are coterminal if and only if their radian mea-sures differ by 2πk, where k ∈ Z.

As a quick illustration, to find one coterminal angle with an angle that mea-sures 410◦, just subtract 360◦, resulting in 50◦. See Figure 3.3.

Figure 3.3

Example 3.1.4. Find the angle coterminal with −380◦ that has measure

(1) between 0◦ and 360◦, and



DEPED COPY

(2) between −360◦ and 0◦.

Solution. A negative angle moves in a clockwise direction, and the angle −380◦

lies in Quadrant IV.

(1) −380◦ + 2 · 360◦ = 340◦

(2) −380◦ + 360◦ = −20◦ 2

3.1.3. Arc Length and Area of a Sector

In a circle, a central angle whose radian measure is θ subtends an arc that is thefraction θ

2πof the circumference of the circle. Thus, in a circle of radius r (see

Figure 3.4), the length s of an arc that subtends the angle θ is

s =θ

2π× circumference of circle =

θ

2π(2πr) = rθ.

Figure 3.4

In a circle of radius r, the length s of an arc intercepted by a centralangle with measure θ radians is given by

s = rθ.

Example 3.1.5. Find the length of an arc of a circle with radius 10 m thatsubtends a central angle of 30◦.

Solution. Since the given central angle is in degrees, we have to convert it intoradian measure. Then apply the formula for an arc length.

30( π

180

)=π

6rad



DEPED COPY

s = 10(π

6

)=

5π

3m 2

Example 3.1.6. A central angle θ in a circle of radius 4 m is subtended by anarc of length 6 m. Find the measure of θ in radians.

Solution.

θ =s

r=

6

4=

3

2rad 2

A sector of a circle is the portion of the interior of a circle bounded by theinitial and terminal sides of a central angle and its intercepted arc. It is like a“slice of pizza.” Note that an angle with measure 2π radians will define a sectorthat corresponds to the whole “pizza.” Therefore, if a central angle of a sectorhas measure θ radians, then the sector makes up the fraction θ

2πof a complete

circle. See Figure 3.5. Since the area of a complete circle with radius r is πr2, wehave

Area of a sector =θ

2π(πr2) =

1

2θr2.

Figure 3.5

In a circle of radius r, the area A of a sector with a central anglemeasuring θ radians is

A =1

2r2θ.

Example 3.1.7. Find the area of a sector of a circle with central angle 60◦ ifthe radius of the circle is 3 m.

Solution. First, we have to convert 60◦ into radians. Then apply the formula forcomputing the area of a sector.

60( π

180

)=π

3rad



DEPED COPY

A =1

2(32)

π

3=

3π

2m2 2

Example 3.1.8. A sprinkler on a golf course fairway is set to spray water overa distance of 70 feet and rotates through an angle of 120◦. Find the area of thefairway watered by the sprinkler.

Solution.

120( π

180

)=

2π

3rad

A =1

2(702)

2π

3=

4900π

3≈ 5131 ft2 2


1. Find the equivalent degree measure of 548

radians.

Solution: 548

rad = 548

(180π

)=(754π

)◦2. Find the equivalent angle measure in degrees and in radians of an angle tracing

235

revolutions.

Solution: One revolution around a circle is equivalent to tracing 360◦.

23

5rev = 2

3

5rev

(360

1 rev

)= 936◦

936◦ = 936( π

180

)=

26π

5rad

3. Find the smallest positive angle coterminal with −2016◦.

Solution: Add 6 complete revolutions or 6(360◦) = 2160◦ to the given angle(or keep on adding 360◦ until you get a positive angle).

−2016◦ + 2160◦ = 144◦

4. Find the largest negative angle coterminal with 137π5

.

Solution: Subtract 14 complete revolutions or 14(2π) = 28π to the given angle(or keep on subtracting 2π until you get a negative angle).

137π

5− 28π = −3π

5rad



DEPED COPY

5. Find the length of the arc of a circle with radius 15 cm that subtends a centralangle of 84◦.

Solution:

84◦ = 84( π

180

)=

7π

15rad

s = 15

(7π

15

)= 7π cm

6. A central angle θ in a circle of radius 12 inches is subtended by an arc of length27 inches. Find the measure of θ in degrees.

Solution:

s = rθ =⇒ θ =s

r

θ =12

27=

9

4rad

9

4rad =

9

4

(180

π

)=

(405

π

)◦7. Find the area of a sector of a circle with central angle of 108◦ if the radius of

the circle is 15 cm.

Solution:

108◦ = 108( π

180

)=

3π

5rad

A =1

2(15)2

3π

5=

135π

2cm2

8. Given isosceles right triangle ABC with AC as the hypotenuse (as shownbelow), a circle with center at A and radius AB intersects AC at D. What isthe ratio of the area of sector BAD to the area of the region BCD?



DEPED COPY

Solution: Let r be the radius of the circle; that is, r = AB.

∠A =π

4rad =⇒ Area of sector BAD =

1

2r2(π

4

)=πr2

8

Area of region BCD = Area of 4ABC − Area of sector BAD =4r2 − πr2

8

area of sector BAD

area of the region BCD=

πr2

84r2−πr2

8

=π

4− π


1. How many degrees is 115

of a complete revolution?

2. How many radians is 115

of a complete revolution?

3. What is the length of an arc of a circle with radius 4 cm that subtends acentral angle of 216◦?

4. Find the length of an arc of a circle with radius 6π

cm that subtends a centralangle of 99◦.

5. What is the smallest positive angle coterminal with 2110◦?

6. Find the largest negative angle coterminal with 107π6

.

7. Find the area of a sector of a circle with central angle of 7π6

if the diameter ofthe circle is 9 cm.

8. Find the area of a sector of a circle with central angle of 108◦ if the radius ofthe circle is 15 cm.

9. What is the radius of a circle in which a central angle of 150◦ determines asector of area 15 in2?

10. Find the radius of a circle in which a central angle of 5π4

determines a sectorof area 32 in2.

?11. A central angle of a circle of radius 6 inches is subtended by an arc of length 6inches. What is the central angle in degrees (rounded to two decimal places)?

?12. An arc of length π5

cm subtends a central angle θ of a circle with radius 23

cm.What is θ in degrees?



DEPED COPY

13. Two overlapping circles of radii 1 cm are drawn such that each circle passesthrough the center of the other. What is the perimeter of the entire region?

14. The length of arc AB of a circle with center at O is equal to twice the lengthof the radius r of the circle. Find the area of sector AOB in terms of r.

15. The angle of a sector in a given circle is 20◦ and the area of the sector is equalto 800 cm2. Find the arc length of the sector.

16. In Figure 3.6, AE and BC are arcs of two concentric circles with center at D.If AD = 2 cm, BD = 8 cm, and ∠ADE = 75◦, find the area of the regionAECB.

17. In Figure 3.7, AB and DE are diameters. If AB = 12 cm and ∠AOD = 126◦,find the area of the shaded region.


18. A point moves outside an equilateral triangle of side 5 cm such that its distancefrom the triangle is always 2 cm. See Figure 3.8. What is the length of onecomplete path that the point traces?

Figure 3.9

19. The segment of a circle is the region bounded by a chord and the arc subtendedby the chord. See Figure 3.9. Find the area of a segment of a circle with acentral angle of 120◦ and a radius of 64 cm.




20. In Figure 3.10, diameter AB of circle O measures 12 cm and arc BC measures120◦. Find the area of the shaded region.

4

Lesson 3.2. Circular Functions



(1) illustrate the different circular functions; and

(2) use reference angles to find exact values of circular functions.

Lesson Outline

(1) Circular functions

(2) Reference angles

Introduction

We define the six trigonometric function in such a way that the domain ofeach function is the set of angles in standard position. The angles are measuredeither in degrees or radians. In this lesson, we will modify these trigonometricfunctions so that the domain will be real numbers rather than set of angles.



DEPED COPY

3.2.1. Circular Functions on Real Numbers

Recall that the sine and cosine functions (and four others: tangent, cosecant,secant, and cotangent) of angles measuring between 0◦ and 90◦ were defined inthe last quarter of Grade 9 as ratios of sides of a right triangle. It can be verifiedthat these definitions are special cases of the following definition.

Let θ be an angle in standard position and P (θ) = P (x, y) the pointon its terminal side on the unit circle. Define

sin θ = y csc θ =1

y, y 6= 0

cos θ = x sec θ =1

x, x 6= 0

tan θ =y

x, x 6= 0 cot θ =

x

y, y 6= 0

Example 3.2.1. Find the values of cos 135◦, tan 135◦, sin(−60◦), and sec(−60◦).

Solution. Refer to Figure 3.11(a).

(a) (b)

Figure 3.11

From properties of 45◦-45◦ and 30◦-60◦ right triangles (with hypotenuse 1unit), we obtain the lengths of the legs as in Figure 3.11(b). Thus, the coordinatesof A and B are

A =

(−√

2

2,

√2

2

)and B =

(1

2,−√

3

2

).



DEPED COPY

Therefore, we get

cos 135◦ = −√

2

2, tan 135◦ = −1,

sin(−60◦) = −√

3

2, and sec(−60◦) = 2. 2

From the last example, we may then also say that

cos(π

4rad)

=

√2

2, sin

(−π

3rad)

= −√

3

2,

and so on.

From the above definitions, we define the same six functions on real numbers.These functions are called trigonometric functions.

Let s be any real number. Suppose θ is the angle in standard positionwith measure s rad. Then we define

sin s = sin θ csc s = csc θ

cos s = cos θ sec s = sec θ

tan s = tan θ cot s = cot θ

From the last example, we then have

cos(π

4

)= cos

(π4

rad)

= cos 45◦ =

√2

2

and

sin(−π

3

)= sin

(−π

3rad)

= sin(−60◦) = −√

3

2.

In the same way, we have

tan 0 = tan(0 rad) = tan 0◦ = 0.

Example 3.2.2. Find the exact values of sin 3π2

, cos 3π2

, and tan 3π2

.

Solution. Let P(3π2

)be the point on the unit circle and on the terminal side of

the angle in the standard position with measure 3π2

rad. Then P(3π2

)= (0,−1),

and so

sin3π

2= −1, cos

3π

2= 0,

but tan 3π2

is undefined. 2



DEPED COPY

Example 3.2.3. Suppose s is a real number such that sin s = −34

and cos s > 0.Find cos s.

Solution. We may consider s as the angle with measure s rad. Let P (s) = (x, y)be the point on the unit circle and on the terminal side of angle s.

Since P (s) is on the unit circle, we know that x2 + y2 = 1. Since sin s = y =−3

4, we get

x2 = 1− y2 = 1−(−3

4

)2

=7

16=⇒ x = ±

√7

4.

Since cos s = x > 0, we have cos s =√74

. 2

Let P (x1, y1) and Q(x, y) be points on the terminal side of an angle θ instandard position, where P is on the unit circle and Q on the circle of radius r(not necessarily 1) with center also at the origin, as shown above. Observe thatwe can use similar triangles to obtain

cos θ = x1 =x11

=x

rand sin θ = y1 =

y11

=y

r.

We may then further generalize the definitions of the six circular functions.



DEPED COPY

Let θ be an angle in standard position, Q(x, y) any point on the ter-minal side of θ, and r =

√x2 + y2 > 0. Then

sin θ =y

rcsc θ =

r

y, y 6= 0

cos θ =x

rsec θ =

r

x, x 6= 0

tan θ =y

x, x 6= 0 cot θ =

x

y, y 6= 0

We then have a second solution for Example 3.2.3 as follows. With sin s = −34

and sin s = yr, we may choose y = −3 and r = 4 (which is always positive). In

this case, we can solve for x, which is positive since cos s = x4

is given to bepositive.

4 =√x2 + (−3)2 =⇒ x =

√7 =⇒ cos s =

√7

4

3.2.2. Reference Angle

We observe that if θ1 and θ2 are coterminal angles, the values of the six circularor trigonometric functions at θ1 agree with the values at θ2. Therefore, in findingthe value of a circular function at a number θ, we can always reduce θ to a numberbetween 0 and 2π. For example, sin 14π

3= sin

(14π3− 4π

)= sin 2π

3. Also, observe

from Figure 3.12 that sin 2π3

= sin π3.

Figure 3.12



DEPED COPY

In general, if θ1, θ2, θ3, and θ4 are as shown in Figure 3.13 with P (θ1) =(x1, y1), then each of the x-coordinates of P (θ2), P (θ3), and P (θ4) is ±x1, whilethe y-coordinate is ±y1. The correct sign is determined by the location of theangle. Therefore, together with the correct sign, the value of a particular circularfunction at an angle θ can be determined by its value at an angle θ1 with radianmeasure between 0 and π

2. The angle θ1 is called the reference angle of θ.

Figure 3.13

The signs of the coordinates of P (θ) depends on the quadrant or axis whereit terminates. It is important to know the sign of each circular function in eachquadrant. See Figure 3.14. It is not necessary to memorize the table, since thesign of each function for each quadrant is easily determined from its definition.We note that the signs of cosecant, secant, and cotangent are the same as sine,cosine, and tangent, respectively.

Figure 3.14

Using the fact that the unit circle is symmetric with respect to the x-axis, they-axis, and the origin, we can identify the coordinates of all the points using the



DEPED COPY

coordinates of corresponding points in the Quadrant I, as shown in Figure 3.15for the special angles.

Figure 3.15

Example 3.2.4. Use reference angle and appropriate sign to find the exact valueof each expression.

(1) sin 11π6

and cos 11π6

(2) cos(−7π

6

) (3) sin 150◦

(4) tan 8π3

Solution. (1) The reference angle of 11π6

is π6, and it lies in Quadrant IV wherein

sine and cosine are negative and positive, respectively.

sin11π

6= − sin

π

6= −1

2

cos11π

6= cos

π

6=

√3

2

(2) The angle −7π6

lies in Quadrant II wherein cosine is negative, and its refer-ence angle is π

6.

cos

(−7π

6

)= − cos

π

6= −√

3

2

(3) sin 150◦ = sin 30◦ = 12

(4) tan 8π3

= − tan π3

= − sinπ3

cosπ3

= −√3212

= −√

3 2



DEPED COPY


1. If P (θ) is a point on the unit circle and θ = 17π3

, what are the coordinates ofP (θ)?

Solution: 17π3

is coterminal with 5π3

which terminates in QIV. The reference

angle is π3, therefore P

(17π3

)=(

12,−√32

).

2. If P (θ) is a point on the unit circle and θ = −5π6

, find the values of the sixtrigonometric functions of θ.

Solution: The angle −5π6

terminates in QIII, the reference angle is π6, therefore

P(−5π

6

)=(−√32,−1

2

).

cos

(−5π

6

)= −√

3

2sec

(−5π

6

)= − 2√

3= −2

√3

3

sin

(−5π

6

)= −1

2csc

(−5π

6

)= −2

tan

(−5π

6

)=

1√3

=

√3

3cot

(−5π

6

)=

√3

1=√

3

3. Find the six trigonometric functions of the angle θ if the terminal side of θ instandard position passes through the point (5,−12).

Solution: x = 5, y = −12, r =√

(5)2 + (−12)2 = 13.

cos θ =x

r=

5

13sec θ =

r

x=

13

5

sin θ =y

r= −12

13csc θ =

r

y= −13

12

tan θ =y

x= −12

5cot θ =

x

y= − 5

12

4. Given sec θ = −2524

and π ≤ θ ≤ 3π2

, find sin θ + cos θ.

Solution: r = 25, x = −24, y =√

(25)2 − (−24)2 = ±7.

Since θ is in QIII, y = −7.

sin θ + cos θ =−7

25+−24

25= −31

25.

5. If tanA = 45, determine 2 sinA−cosA

3 cosA.



DEPED COPY

Solution:

tanA =4

5=⇒ sinA

cosA=

4

52 sinA− cosA

3 cosA=

2

3

(sinA

cosA

)− 1

3

(cosA

cosA

)=

2

3

(4

5

)− 1

3(1) =

1

5

6. What is the reference angle of −29π6

? Find the value of tan(−29π

6

).

Solution: −29π6

is coterminal with 7π6

in QIII, so its reference angle is π6.

tan

(−29π

6

)= tan

(π6

)=

√3

3

7. For what angle θ in the third quadrant is cos θ = sin 5π3

?

Solution:

sin5π

3= cos θ

cos θ = −√

3

2and θ in QIII =⇒ θ =

7π

6


1. In what quadrant is P (θ) located if θ = 33π4

?

2. In what quadrant is P (θ) located if θ = −17π6

?

3. In what quadrant is P (θ) located if sec θ > 0 and cot θ < 0?

4. In what quadrant is P (θ) located if tan θ > 0 and cos θ < 0 ?

5. If P (θ) is a point on the unit circle and θ = 5π6

, what are the coordinates ofP (θ)?

6. If P (θ) is a point on the unit circle and θ = −11π6

, what are the coordinatesof P (θ)?

7. If cos θ > 0 and tan θ = −23, find sec θ+tan θ

sec θ−tan θ

8. If tan θ = 35

and θ is in QIII, what is sec θ?

9. If csc θ = 2 and cos θ < 0, find sec θ.

10. Find the values of the other trigonometric functions of θ if cot θ = −43

andsin θ < 0.



DEPED COPY

11. Find the values of the other trigonometric functions of θ if csc θ = −4 and θdoes not terminate in QIII.

12. The terminal side of an angle θ in standard position contains the point (7,−1).Find the values of the six trigonometric functions of θ.

13. The terminal side of an angle θ in standard position contains the point (−2, 4).Find the values of the six trigonometric functions of θ.

14. If the terminal point of an arc of length θ lies on the line joining the originand the point (−3,−1), what is cos2 θ − sin2 θ?

15. If the terminal point of an arc of length θ lies on the line joining the originand the point (2,−6), what is sec2 θ − csc2 θ?

16. Determine the reference angle of 35π4

, and find cos 35π4

.

17. If 3π2< θ < 2π, find θ if cos θ = sin 2π

3.

18. Evaluate the sum of sin 30◦ + sin 60◦ + sin 90◦ + · · ·+ sin 510◦ + sin 540◦.

19. If f(x) = sin 2x+ cos 2x+ sec 2x+ csc 2x+ tan 2x+ cot 2x, what is f(7π8

)?

20. Evaluate the sum of sec π6

+ sec 13π6

+ sec 25π6

+ · · ·+ sec 109π6

.

4

Lesson 3.3. Graphs of Circular Functions and SituationalProblems



(1) determine the domain and range of the different circular functions;

(2) graph the six circular functions with its amplitude, period, and phase shift;and

(3) solve situational problems involving circular functions.

Lesson Outline

(1) Domain and range of circular functions

(2) Graphs of circular functions

(3) Amplitude, period, and phase shift



DEPED COPY

Introduction

There are many things that occur periodically. Phenomena like rotation ofthe planets and comets, high and low tides, and yearly change of the seasonsfollow a periodic pattern. In this lesson, we will graph the six circular functionsand we will see that they are periodic in nature.

3.3.1. Graphs of y = sinx and y = cosx

Recall that, for a real number x, sinx = sin θ for an angle θ with measure xradians, and that sin θ is the second coordinate of the point P (θ) on the unitcircle. Since each x corresponds to an angle θ, we can conclude that

(1) sinx is defined for any real number x or the domain of the sine function isR, and

(2) the range of sine is the set of all real numbers between −1 and 1 (inclusive).

From the definition, it also follows that sin(x+2π) = sinx for any real numberx. This means that the values of the sine function repeat every 2π units. In thiscase, we say that the sine function is a periodic function with period 2π.

Table 3.16 below shows the values of y = sinx, where x is the equivalent radianmeasure of the special angles and their multiples from 0 to 2π. As commentedabove, these values determine the behavior of the function on R.

x 0 π6

π4

π3

π2

2π3

3π4

5π6

π

y 0 12

√22

√32

1√32

√22

12

0

0 0.5 0.71 0.87 1 0.87 0.71 0.5 0

x 7π6

5π4

4π3

3π2

5π3

7π4

11π6

2π

y −12

−√22

−√32

−1 −√32

−√22

−12

0

−0.5 −0.71 −0.87 −1 −0.87 −0.71 −0.5 0

Table 3.16

From the table, we can observe that as x increases from 0 to π2, sinx also

increases from 0 to 1. Similarly, as x increases from 3π2

to 2π, sinx also increasesfrom −1 to 0. On the other hand, notice that as x increases from π

2to π, sinx

decreases from 1 to 0. Similarly, as x increases from π to 3π2

, sinx decreases from0 to −1.



DEPED COPY

To sketch the graph of y = sin x, we plot the points presented in Table 3.16,and join them with a smooth curve. See Figure 3.17. Since the graph repeatsevery 2π units, Figure 3.18 shows periodic graph over a longer interval.

Figure 3.17

Figure 3.18

We can make observations about the cosine function that are similar to thesine function.

• y = cosx has domain R and range [−1, 1].

• y = cos x is periodic with period 2π. The graph of y = cos x is shown inFigure 3.19.

Figure 3.19



DEPED COPY

From the graphs of y = sinx and y = cosx in Figures 3.18 and 3.19, re-spectively, we observe that sin(−x) = − sinx and cos(−x) = cosx for any realnumber x. In other words, the graphs of y = cos(−x) and y = cosx are the same,while the graph of y = sin(−x) is the same as that of y = − sinx.

In general, if a function f satisfies the property that f(−x) = f(x) for all xin its domain, we say that such function is even. On the other hand, we say thata function f is odd if f(−x) = −f(x) for all x in its domain. For example, thefunctions x2 and cos x are even, while the functions x3 − 3x and sin x are odd.

3.3.2. Graphs of y = a sin bx and y = a cos bx

Using a table of values from 0 to 2π, we can sketch the graph of y = 3 sin x, andcompare it to the graph of y = sinx. See Figure 3.20 wherein the solid curvebelongs to y = 3 sin x, while the dashed curve to y = sinx. For instance, if x = π

2,

then y = 1 when y = sinx, and y = 3 when y = 3 sin x. The period, x-intercepts,and domains are the same for both graphs, while they differ in the range. Therange of y = 3 sin x is [−3, 3].

Figure 3.20

In general, the graphs of y = a sinx and y = a cosx with a > 0 have the sameshape as the graphs of y = sinx and y = cosx, respectively. If a < 0, there is areflection across the x-axis.

In the graphs of y = a sinx and y = a cosx, the number |a| is calledits amplitude. It dictates the height of the curve. When |a| < 1,the graphs are shrunk vertically, and when |a| > 1, the graphs arestretched vertically.

Now, in Table 3.21, we consider the values of y = sin 2x on [0, 2π].



DEPED COPY

x 0 π6

π4

π3

π2

2π3

3π4

5π6

π

y 0√32

1√32

0 −√32

−1 −√32

0

0 0.87 1 0.87 0 −0.87 −1 −0.87 0

x 7π6

5π4

4π3

3π2

5π3

7π4

11π6

2π

y√32

1√32

0 −√32

−1 −√32

0

0.87 1 0.87 0 −0.87 −1 −0.87 0

Table 3.21

Figure 3.22

Figure 3.22 shows the graphs of y = sin 2x (solid curve) and y = sinx (dashedcurve) over the interval [0, 2π]. Notice that, for sin 2x to generate periodic valuessimilar to [0, 2π] for y = sinx, we just need values of x from 0 to π. We thenexpect the values of sin 2x to repeat every π units thereafter. The period ofy = sin 2x is π.

If b 6= 0, then both y = sin bx and y = cos bx have period given by2π

|b|.

If 0 < |b| < 1, the graphs are stretched horizontally, and if |b| > 1, thegraphs are shrunk horizontally.

To sketch the graphs of y = a sin bx and y = a cos bx, a, b 6= 0, we may proceedwith the following steps:

(1) Determine the amplitude |a|, and find the period 2π|b| . To draw one cycle

of the graph (that is, one complete graph for one period), we just need tocomplete the graph from 0 to 2π

|b| .



DEPED COPY

(2) Divide the interval into four equal parts, and get five division points: x1 = 0,x2, x3, x4, and x5 = 2π

|b| , where x3 is the midpoint between x1 and x5 (that

is, 12(x1 + x5) = x3), x2 is the midpoint between x1 and x3, and x4 is the

midpoint between x3 and x5.

(3) Evaluate the function at each of the five x-values identified in Step 2. Thepoints will correspond to the highest point, lowest point, and x-interceptsof the graph.

(4) Plot the points found in Step 3, and join them with a smooth curve similarto the graph of the basic sine curve.

(5) Extend the graph to the right and to the left, as needed.

Example 3.3.1. Sketch the graph of one cycle of y = 2 sin 4x.

Solution. (1) The period is 2π4

= π2, and the amplitude is 2.

(2) Dividing the interval [0, π2] into 4 equal parts, we get the following x-

coordinates: 0, π8, π

4, 3π

8, and π

2.

(3) When x = 0, π4, and π

2, we get y = 0. On the other hand, when x = π

8, we

have y = 2 (the amplitude), and y = −2 when x = 3π8

.

(4) Draw a smooth curve by connecting the points. There is no need to proceedto Step 5 because the problem only asks for one cycle.

Example 3.3.2. Sketch the graph of y = −3 cos x2.

Solution. (1) The amplitude is | − 3| = 3, and the period is2π12

= 4π.

(2) We divide the interval [0, 4π] into four equal parts, and we get the followingx-values: 0, π, 2π, 3π, and 4π.



DEPED COPY

(3) We have y = 0 when x = π and 3π, y = −3 when x = 0 and 4π, and y = 3when x = 2π.

(4) We trace the points in Step 3 by a smooth curve.

(5) We extend the pattern in Step 4 to the left and to the right.

Example 3.3.3. Sketch the graph of two cycles of y = 12

sin(−2x

3

).

Solution. Since the sine function is odd, the graph of y = 12

sin(−2x

3

)is the same

as that of y = −12

sin 2x3

.

(1) The amplitude is 12, and the period is

2π23

= 3π.

(2) Dividing the interval [0, 3π] into four equal parts, we get the x-coordinatesof the five important points:

0 + 3π

2=

3π

2,

0 + 3π2

2=

3π

4,

3π2

+ 3π

2=

9π

4.

(3) We get y = 0 when x = 0, 3π2

, and 3π, y = −12

when 3π4

, and y = 12

when9π4

.

(4) We trace the points in Step 3 by a smooth curve.

(5) We extend the pattern in Step 4 by one more period to the right.



DEPED COPY

3.3.3. Graphs of y = a sin b(x − c) + d and y = a cos b(x − c) + d

We first compare the graphs of y = sinx and y = sin(x− π

3

)using a table of

values and the 5-step procedure discussed earlier.

As x runs from π3

to 7π3

, the value of the expression x− π3

runs from 0 to 2π. Sofor one cycle of the graph of y = sin

(x− π

3

), we then expect to have the graph of

y = sinx starting from x = π3. This is confirmed by the values in Table 3.23. We

then apply a similar procedure to complete one cycle of the graph; that is, dividethe interval [π

3, 7π

3] into four equal parts, and then determine the key values of

x in sketching the graphs as discussed earlier. The one-cycle graph of y = sinx(dashed curve) and the corresponding one-cycle graph of y = sin

(x− π

3

)(solid

curve) are shown in Figure 3.24.

x π3

5π6

4π3

11π6

7π3

x− π3

0 π2

π 3π2

2π

sin(x− π

3

)0 1 0 −1 0

Table 3.23

Figure 3.24

Observe that the graph of y = sin(x− π

3

)shifts π

3units to the right of

y = sinx. Thus, they have the same period, amplitude, domain, and range.

The graphs of

y = a sin b(x− c) and y = a cos b(x− c)

have the same shape as y = a sin bx and y = a cos bx, respectively, butshifted c units to the right when c > 0 and shifted |c| units to the leftif c < 0. The number c is called the phase shift of the sine or cosinegraph.



DEPED COPY

Example 3.3.4. In the same Cartesian plane, sketch one cycle of the graphs ofy = 3 sin x and y = 3 sin

(x+ π

4

).

Solution. We have sketched the graph of y = 3 sinx earlier at the start of thelesson. We consider y = 3 sin

(x+ π

4

). We expect that it has the same shape as

that of y = 3 sin x, but shifted some units.

Here, we have a = 3, b = 1, and c = −π4. From these constants, we get

the amplitude, the period, and the phase shift, and these are 3, 2π, and −π4,

respectively.

One cycle starts at x = −π4

and ends at x = −π4

+ 2π = 7π4

. We now computethe important values of x.

−π4

+ 7π4

2=

3π

4,−π

4+ 3π

4

2=π

4,

3π4

+ 7π4

2=

5π

4

x −π4

π4

3π4

5π4

7π4

y = 3 sin(x+ π

4

)0 3 0 −3 0

While the effect of c in y = a sin b(x − c) and y = a cos b(x − c) isa horizontal shift of their graphs from the corresponding graphs ofy = a sin bx and y = a cos bx, the effect of d in the equations y =a sin b(x− c) + d and y = a cos b(x− c) + d is a vertical shift. That is,the graph of y = a sin b(x−c)+d has the same amplitude, period, andphase shift as that of y = a sin b(x − c), but shifted d units upwardwhen d > 0 and |d| units downward when d < 0.



DEPED COPY

Example 3.3.5. Sketch the graph of

y = −2 cos 2(x− π

6

)− 3.

Solution. Here, a = −2, b = 2, c = π6, and d = −3. We first sketch one cycle of

the graph of y = −2 cos 2(x− π

6

), and then extend this graph to the left and to

the right, and then move the resulting graph 3 units downward.

The graph of y = −2 cos 2(x− π

6

)has amplitude 2, period π, and phase shift

π6.

Start of one cycle: π6

End of the cycle: π6

+ π = 7π6

π6

+ 7π6

2=

2π

3,

π6

+ 2π3

2=

5π

12,

2π3

+ 7π6

2=

11π

12

x π6

5π12

2π3

11π12

7π6

y = −2 cos 2(x− π

6

)−2 0 2 0 −2

y = −2 cos 2(x− π

6

)− 3 −5 −3 −1 −3 −5

Before we end this sub-lesson, we make the following observation, which willbe used in the discussion on simple harmonic motion (Sub-Lesson 3.3.6).



DEPED COPY

Different Equations, The Same Graph

1. The graphs of y = sinx and y = sin(x + 2πk), k any integer, arethe same.

2. The graphs of y = sinx, y = − sin(x + π), y = cos(x − π2), and

y = − cos(x+ π2) are the same.

3. In general, the graphs of

y = a sin b(x− c) + d,

y = −a sin[b(x− c) + π + 2πk] + d,

y = a cos[b(x− c)− π2

+ 2πk] + d,

andy = −a cos[b(x− c) + π

2+ 2πk] + d,

where k is any integer, are all the same.

Similar observations are true for cosine.

3.3.4. Graphs of Cosecant and Secant Functions

We know that csc x = 1sinx

if sinx 6= 0. Using this relationship, we can sketch thegraph of y = cscx.

First, we observe that the domain of the cosecant function is

{x ∈ R : sinx 6= 0} = {x ∈ R : x 6= kπ, k ∈ Z}.

Table 3.25 shows the key numbers (that is, numbers where y = sinx crosses thex-axis, attain its maximum and minimum values) and some neighboring points,where “und” stands for “undefined,” while Figure 3.26 shows one cycle of thegraphs of y = sinx (dashed curve) and y = cscx (solid curve). Notice theasymptotes of the graph y = cscx.

x 0 π6

π2

5π6

π 7π6

3π2

11π6

2π

y = sinx 0 12

1 12

0 −12−1 −1

20

y = cscx und 2 1 2 und −2 −1 −2 und

Table 3.25



DEPED COPY

Figure 3.26

We could also sketch the graph of cscx directly from the graph of y = sinxby observing the following facts:

(1) If sinx = 1 (or −1), then cscx = 1 (or −1).

(2) At each x-intercept of y = sinx, y = csc x is undefined; but a verticalasymptote is formed because, when sinx is close to 0, the value of csc x willhave a big magnitude with the same sign as sin x.

Refer to Figure 3.27 for the graphs of y = sinx (dashed curve) and y = cscx(solid curve) over a larger interval.

Figure 3.27



DEPED COPY

Like the sine and cosecant functions, the cosine and secant functions are alsoreciprocals of each other. Therefore, y = secx has domain

{x ∈ R : cosx 6= 0} = {x ∈ R : x 6= kπ

2, k odd integer}.

Similarly, the graph of y = secx can be obtained from the graph of y = cosx.These graphs are shown in Figure 3.28.

Figure 3.28

Example 3.3.6. Sketch the graph of y = 2 csc x2.

Solution. First, we sketch the graph of y = 2 sin x2, and use the technique dis-

cussed above to sketch the graph of y = 2 csc x2.

The vertical asymptotes of y = 2 csc x2

are the x-intercepts of y = 2 sin x2:

x = 0,±2π,±4π, . . .. After setting up the asymptotes, we now sketch the graphof y = 2 csc x

2as shown below.



DEPED COPYExample 3.3.7. Sketch the graph of y = 2− sec 2x.

Solution. Sketch the graph of y = − cos 2x (note that it has period π), then sketchthe graph of y = − sec 2x (as illustrated above), and then move the resultinggraph 2 units upward to obtain the graph of y = 2− sec 2x.



DEPED COPY

3.3.5. Graphs of Tangent and Cotangent Functions

We know that tanx = sinxcosx

, where cosx 6= 0. From this definition of the tangentfunction, it follows that its domain is the same as that of the secant function,which is

{x ∈ R : cosx 6= 0} = {x ∈ R : x 6= kπ

2, k odd integer}.

We note that tanx = 0 when sinx = 0 (that is, when x = kπ, k any integer), andthat the graph of y = tanx has asymptotes x = kπ

2, k odd integer. Furthermore,

by recalling the signs of tangent from Quadrant I to Quadrant IV and its values,we observe that the tangent function is periodic with period π.

To sketch the graph of y = tanx, it will be enough to know its one-cyclegraph on the open interval

(−π

2, π2

). See Table 3.29 and Figure 3.30.

x −π2

−π3

−π4−π

60

y = tanx und −√

3 −1 −√33

0

x π6

π4

π3

π2

y = tanx√33

1√

3 und

Table 3.29

Figure 3.30

In the same manner, the domain of y = cotx = cosxsinx

is

{x ∈ R : sinx 6= 0} = {x ∈ R : x 6= kπ, k ∈ Z},

and its period is also π. The graph of y = cotx is shown in Figure 3.31.




In general, to sketch the graphs of y = a tan bx and y = a cot bx, a 6= 0 andb > 0, we may proceed with the following steps:

(1) Determine the period πb. Then we draw one cycle of the graph on

(− π

2b, π2b

)for y = a tan bx, and on

(0, π

b

)for y = a cot bx.

(2) Determine the two adjacent vertical asymptotes. For y = a tan bx, thesevertical asymptotes are given by x = ± π

2b. For y = a cot bx, the vertical

asymptotes are given by x = 0 and x = πb.

(3) Divide the interval formed by the vertical asymptotes in Step 2 into fourequal parts, and get three division points exclusively between the asymp-totes.

(4) Evaluate the function at each of these x-values identified in Step 3. Thepoints will correspond to the signs and x-intercept of the graph.

(5) Plot the points found in Step 3, and join them with a smooth curve ap-proaching to the vertical asymptotes. Extend the graph to the right and tothe left, as needed.

Example 3.3.8. Sketch the graph of y = 12

tan 2x.

Solution. The period of the function is π2, and the adjacent asymptotes are x =

±π4,±3π

4, . . .. Dividing the interval

(−π

4, π4

)into four equal parts, the key x-values

are −π8, 0, and π

8.

x −π8

0 π8

y = 12

tan 2x −12

0 12



DEPED COPY

Example 3.3.9. Sketch the graph of y = 2 cot x3

on the interval (0, 3π).

Solution. The period of the function is 3π, and the adjacent asymptotes are x = 0and x = 3π. We now divide the interval (0, 3π) into four equal parts, and thekey x-values are 3π

4, 3π

2, and 9π

4.

x 3π4

3π2

9π4

y = 2 cot x3

2 0 −2

3.3.6. Simple Harmonic Motion

Repetitive or periodic behavior is common in nature. As an example, the time-telling device known as sundial is a result of the predictable rising and settingof the sun everyday. It consists of a flat plate and a gnomon. As the sun movesacross the sky, the gnomon casts a shadow on the plate, which is calibrated totell the time of the day.



DEPED COPY

Sundial, by liz west, 29 March 2007,

https://commons.wikimedia.org/wiki/File:Sundial 2r.jpg. Public Domain.

Some motions are also periodic. When a weight is suspended on a spring,pulled down, and released, the weight oscillates up and down. Neglecting resis-tance, this oscillatory motion of the weight will continue on and on, and its heightis periodic with respect to time.

t = 0 sec t = 2.8 sec



https://commons.wikimedia.org/wiki/File:Sundial_2r.jpg

DEPED COPYt = 6.1 sec t = 9 sec

Periodic motions are usually modeled by either sine or cosine function, and arecalled simple harmonic motions. Unimpeded movements of objects like oscilla-tion, vibration, rotation, and motion due to water waves are real-life occurrencesthat behave in simple harmonic motion.

Equations of Simple Harmonic Motion

The displacement y (directed height or length) of an object behavingin a simple harmonic motion with respect to time t is given by one ofthe following equations:

y = a sin b(t− c) + d

ory = a cos b(t− c) + d.

In both equations, we have the following information:

• amplitude = |a| = 12(M − m) - the maximum displacement above

and below the rest position or central position or equilibrium, whereM is the maximum height and m is the minimum height;

• period = 2π|b| - the time required to complete one cycle (from one

highest or lowest point to the next);

• frequency = |b|2π

- the number of cycles per unit of time;

• c - responsible for the horizontal shift in time; and

• d - responsible for the vertical shift in displacement.

Example 3.3.10. A weight is suspended from a spring and is moving up anddown in a simple harmonic motion. At start, the weight is pulled down 5 cm belowthe resting position, and then released. After 8 seconds, the weight reaches its



DEPED COPY

highest location for the first time. Find the equation of the motion.

Solution. We are given that the weight is located at its lowest position at t = 0;that is, y = −5 when t = 0. Therefore, the equation is y = −5 cos bt.

Because it took the weight 8 seconds from the lowest point to its immediatehighest point, half the period is 8 seconds.

1

2· 2π

b= 8 =⇒ b =

π

8=⇒ y = −5 cos

πt

82

?Example 3.3.11. Suppose you ride a Ferris wheel. The lowest point of thewheel is 3 meters off the ground, and its diameter is 20 m. After it started, theFerris wheel revolves at a constant speed, and it takes 32 seconds to bring youback again to the riding point. After riding for 150 seconds, find your approximateheight above the ground.

Solution. We ignore first the fixed value of 3 m off the ground, and assume thatthe central position passes through the center of the wheel and is parallel to theground.

Let t be the time (in seconds) elapsed that you have been riding the Ferriswheel, and y is he directed distance of your location with respect to the assumedcentral position at time t. Because y = −10 when t = 0, the appropriate modelis y = −10 cos bt for t ≥ 0.

Given that the Ferris wheel takes 32 seconds to move from the lowest pointto the next, the period is 32.

2π

b= 32 =⇒ b =

π

16=⇒ y = −10 cos

πt

16

When t = 150, we get y = 10 cos 150π16≈ 3.83.

Bringing back the original condition given in the problem that the riding pointis 3 m off the ground, after riding for 150 seconds, you are approximately located3.83 + 13 = 16.83 m off the ground. 2

In the last example, the central position or equilibrium may be verticallyshifted from the ground or sea level (the role of the constant d). In the same way,the starting point may also be horizontally shifted (the role of the constant c).Moreover, as observed in Sub-Lesson 3.3.3 (see page 154), to find the functionthat describes a particular simple harmonic motion, we can either choose

y = a sin b(t− c) + d

ory = a cos b(t− c) + d,

and determine the appropriate values of a, b, c, and d. In fact, we can assumethat a and b are positive numbers, and c is the smallest such nonnegative number.



DEPED COPY

Example 3.3.12. A signal buoy in Laguna Bay bobs up and down with theheight h of its transmitter (in feet) above sea level modeled by h(t) = a sin bt+ dat time t (in seconds). During a small squall, its height varies from 1 ft to 9 ftabove sea level, and it takes 3.5 seconds from one 9-ft height to the next. Findthe values of the constants a, b, and d.

Solution. We solve the constants step by step.

• The minimum and maximum values of h(t) are 1 ft and 9 ft, respectively.Thus, the amplitude is a = 1

2(M −m) = 1

2(9− 1) = 4.

• Because it takes 3.5 seconds from one 9-ft height to the next, the period is3.5. Thus, we have 2π

b= 3.5, which gives b = 4π

7.

• Because the lowest point is 1 ft above the sea level and the amplitude is 4,it follows that d = 5. 2

Example 3.3.13. A variable star is a star whose brightness fluctuates as ob-served from Earth. The magnitude of visual brightness of one variable star rangesfrom 2.0 to 10.1, and it takes 332 days to observe one maximum brightness tothe next. Assuming that the visual brightness of the star can be modeled by theequation y = a sin b(t − c) + d, t in days, and putting t = 0 at a time when thestar is at its maximum brightness, find the constants a, b, c, and d, where a, b > 0and c the least nonnegative number possible.

Solution.

a =M −m

2=

10.1− 2.0

2= 4.05

2π

b= 332 =⇒ b =

π

166

d = a+m = 4.05 + 2.0 = 6.05

For the (ordinary) sine function to start at the highest point at t = 0, the leastpossible horizontal movement to the right (positive value) is 3π

2units.

bc =3π

2=⇒ c =

3π

2b=

3π

2 · π166

= 249 2

?Example 3.3.14. The path of a fast-moving particle traces a circle with equa-tion

(x+ 7)2 + (y − 5)2 = 36.

It starts at point (−1, 5), moves clockwise, and passes the point (−7, 11) for thefirst time after traveling 6 microseconds. Where is the particle after traveling 15microseconds?



DEPED COPY

Solution. As described above, we may choose sine or cosine function. Here, wechoose the sine function to describe both x and y in terms of time t in microsec-onds; that is, we let

x = a sin b(t− c) + d and y = e sin f(t− g) + h,

where we appropriately choose the positive values for a, b, e, and f , and the leastnonnegative values for c and g.

The given circle has radius 6 and center (−7, 5). Defining the central positionof the values of x as the line x = −7 and that of the values of y as the line y = 5,we get a = e = 6, d = −7, and h = 5.

From the point (−1, 5) to the point (−7, 11) (moving clockwise), the particlehas traveled three-fourths of the complete cycle; that is, three-fourths of theperiod must be 2.

3

4· 2π

b=

3

4· 2π

f= 6 =⇒ b = f =

π

4

As the particle starts at (−1, 5) and moves clockwise, the values of x startat its highest value (x = −1) and move downward toward its central position(x = −7) and continue to its lowest value (x = −13). Therefore, the graph ofa sin bt+ d has to move 3π

2b= 6 units to the right, and so we get c = 6.

As to the value of g, we observe the values of y start at its central position(y = 5) and go downward to its lowest value (y = −1). Similar to the argumentused in determining c, the graph of y = e sin ft + h has to move π

b= 4 units to

the right, implying that g = 4.

Hence, We have the following equations of x and y in terms of t:

x = 6 sin π4(t− 6)− 7 and y = 6 sin π

4(t− 4) + 5.

When t = 15, we get

x = 6 sin π4(15− 6)− 7 = −7 + 3

√2 ≈ −2.76

andy = 6 sin π

4(15− 4) + 5 = 5 + 3

√2 ≈ 9.24.

That is, after traveling for 15 microseconds, the particle is located near the point(−2.76, 9.24). 2


1. Find the period of the function y = 4 sin(x−π4

)− 3.

Solution: y = 4 sin(x−π4

)− 3 =⇒ y = 4 sin 1

3(x− π)− 3 =⇒ P = 2π

13

= 6π



DEPED COPY

2. In the function y = 3 tan(2kx− π), the period is 4π. Find the value of k andthe phase shift of the graph of the function.

Solution: The period of the tangent function is P = πb.

y = 3 tan(2kx− π) =⇒ y = 3 tan 2k(x− π

2k

)=⇒ P =

π

2k= 4π.

1

2k= 4 =⇒ k =

1

8and Phase shift =

π

2k=

π

2(18

) = 4π

3. Sketch the graph of function y = 12

sin 12

(x+ π

6

)+2 over one period. Determine

the domain and range of the function.

Solution: The graph is a vertical translation of y = 12

sin 12

(x+ π

6

)by 2 units

upward. The period of the given function is 2π12

= 4π. One complete cycle may

start at x = −π6

and end at x = −π6

+ 4π = 23π6

.

The critical points for the graph are

x = −π6, x = −5π

6, x =

11π

6, x =

17π

6, and x =

23π

6.

The domain of the function is R and its range is[52, 32

].

4. Sketch the graph of the function y = −2 cos(x− π2) + 3 over two periods. Find


Solution: The graph of the given function is a vertical translation of y =−2 cos(x − π

2) by 3 units upward. The period of the function is 2π. One

complete cycle may start and end at x = π2

and x = 5π2

, respectively. Thenext complete cycle starts at x = 5π

2and ends at x = 9π

2.

critical points:π

2, π,

3π

2, 2π,

5π

2, 3π,

7π

2, 4π,

9π

2



DEPED COPYThe domain of the given cosine function is R, and its range is [1, 5].

5. Sketch the graph of the function y = 14

tan(x− π

4

)over three periods. Find


Solution: The period of the function is π. One complete cycle may start atx = π

4and end at x = 5π

4.

The domain of the function is {x|x 6= 3π4

+ kπ, k ∈ Z}, and its range is R.

6. Sketch the graph of the function y = −3 cot(12x+ π

12

)+ 2 over three periods.

Find the domain and range of the function.

Solution: y = −3 cot(12x+ π

12

)+ 2 = −3 cot 1

2

(x+ π

6

)+ 2 =⇒ P = 2π

One complete cycle may start at x = −π6

and end at x = 11π6

.



DEPED COPY

The domain of the function is {x|x 6= −π6

+ 2kπ, k ∈ Z}, and its range is R.

7. The graph of the function g(x) is the same as that of f(x) = 3 sin(x− π

3

)but

shifted 2 units downward and π2

units to the right. What is g(−π)?

Solution: The function f(x) = 3 sin(x− π

3

)when shifted 2 units downward

and π2

units to the right is

g(x) = 3 sin(x− π

3− π

2

)− 2 = 3 sin

(x− 5π

6

)− 2.

g(−π) = 3 sin

(−π − 5π

6

)− 2 = −1

2

8. The graph of the function h(x) is the same as that of f(x) = 3 sin(2x−3π)+1but shifted 3 units upward and π

2units to the left. What is h(5π

6)?

Solution: h(x) = 3 sin[2(x+ π

2

)− 3π

]+ 1 + 3 = 3 sin(2x− 2π) + 4

h

(5π

6

)= 3 sin

[2

(5π

6

)− 2π

]+ 4 =

8− 3√

3

2

9. Sketch the graph of y = 2 sec 12

(x− π

4

)over two periods. Find the domain

and range of the function.

Solution: The period of the function is 4π. One complete cycle may start atx = π

4and end at x = 17π

4.



DEPED COPY

The domain of the function is {x|x 6= 5π4

+ 2kπ, k ∈ Z}, and its range is(−∞,−2] ∪ [2,∞).

10. Sketch the graph of y = − csc(x+ π

3

)+ 2 over two periods. Find the domain

and range of the function.

Solution: The period of the function is 2π. One complete cycle may start andend at x = −π

3and x = 5π

3, respectively.

The domain of the function is {x|x 6= −π3

+ kπ, k ∈ Z}, and its range is(−∞, 1] ∪ [3,∞).



DEPED COPY


1. What is the period of the function y = −2 cos 14

(x− π

2

)?

2. The amplitude and period of the function y = 4− a2

cos(bx3− π

)are 3 and 4π,

respectively. Find |a|+ b.

3. In the function y = 2π− 3π cot 4πk

(x− 2), the period is 2. Find the value of k.

4. What are the minimum and maximum values of the function y = 3 sin 34

(x+ 2π

3

)−

5?

5. Given the function y = 3 sin 34

(x+ 2π

3

)− 5, find the value of y when x = 8π

9.

6. Given the function y = −2 cot 43

(x− π

6

)+ 3, find the value of y when x = 7π

6.

7. Find the domain and range of the function y = −23

sin 13

(x− 3π

4

)+ 2?

8. Find the range of the function y = 3 sec(2x3

).

9. Find the equation of the secant function whose graph is the graph of y =3 sec 2x shifted π units to the right and 3 units downward.

10. Find the equation of the sine function whose graph is the graph of y =−2 sin 2

(x− π

4

)+ 1 shifted π

2units to the left and 3 units upward.

11. Given the tangent function y = 1 − 3 tan(2x−π

4

), find the equations of all its

vertical asymptotes.

12. Given the cosecant function y = csc(x2− π

3

), find the equations of all its

vertical asymptotes.

13. Sketch the graph over one period, and indicate the period, phase shift, domain,and range for each.

(a) y = 2 sin 14

(x+ π

4

)− 1

(b) y = tan 12

(2x+ π

3

)− 2

(c) y = 12

csc 34(2x− π)− 1

(d) y = sec 12

(4x+ 2π

3

)+ 2

14. A point P in simple harmonic motion has a frequency of 12

oscillation perminute and amplitude of 4 ft. Express the motion of P by means of anequation in the form d = a sin bt.

?15. A mass is attached to a spring, and then pulled and released 8 cm below itsresting position at the start. If the simple harmonic motion is modeled byy = a cos 1

10(t − c), where a > 0, c the least nonnegative such number, and t

in seconds, find the location of the mass 10 seconds later.

4



DEPED COPY

Lesson 3.4. Fundamental Trigonometric Identities



(1) determine whether an equation is an identity or a conditional equation;

(2) derive the fundamental trigonometric identities;

(3) simplify trigonometric expressions using fundamental trigonometric identi-ties; and

(4) prove other trigonometric identities using fundamental trigonometric identi-ties.

Lesson Outline

(1) Domain of an equation

(2) Identity and conditional equation

(3) Fundamental trigonometric identities

(4) Proving trigonometric identities

Introduction

In previous lessons, we have defined trigonometric functions using the unitcircle and also investigated the graphs of the six trigonometric functions. Thislesson builds on the understanding of the different trigonometric functions bydiscovery, deriving, and working with trigonometric identities.

3.4.1. Domain of an Expression or Equation

Consider the following expressions:

2x+ 1,√x2 − 1,

x

x2 − 3x− 4,

x√x− 1

.

What are the real values of the variable x that make the expressions defined inthe set of real numbers?

In the first expression, every real value of x when substituted to the expressionmakes it defined in the set of real numbers; that is, the value of the expression isreal when x is real.

In the second expression, not every real value of xmakes the expression definedin R. For example, when x = 0, the expression becomes

√−1, which is not a real

number.√x2 − 1 ∈ R ⇐⇒ x2 − 1 ≥ 0 ⇐⇒ x ≤ −1 or x ≥ 1



DEPED COPY

Here, for√x2 − 1 to be defined in R, x must be in (−∞,−1] ∪ [1,∞).

In the third expression, the values of x that make the denominator zero makethe entire expression undefined.

x2 − 3x− 4 = (x− 4)(x+ 1) = 0 ⇐⇒ x = 4 or x = −1

Hence, the expressionx

x2 − 3x− 4is real when x 6= 4 and x 6= −1.

In the fourth expression, because the expression√x− 1 is in the denominator,

x must be greater than 1. Although the value of the entire expression is 0 whenx = 0, we do not include 0 as allowed value of x because part of the expressionis not real when x = 0.

In the expressions above, the allowed values of the variable x constitute thedomain of the expression.

The domain of an expression (or equation) is the set of all real values ofthe variable for which every term (or part) of the expression (equation)is defined in R.

In the expressions above, the domains of the first, second, third, and fourthexpressions are R, (−∞,−1] ∪ [1,∞), R \ {−1, 4}, and (1,∞), respectively.

Example 3.4.1. Determine the domain of the expression/equation.

(a)x2 − 1

x3 + 2x2 − 8x−√x+ 1

1− x

(b) tan θ − sin θ − cos 2θ

(c) x2 −√

1 + x2 =2

3√x2 − 1

(d) z − cos2 z

1 + sin z= 4 sin z − 1

Solution. (a) x3 + 2x2 − 8x = x(x + 4)(x − 2) = 0 ⇐⇒ x = 0, x =−4, or x = 2√x+ 1 ∈ R ⇐⇒ x+ 1 ≥ 0 ⇐⇒ x ≥ −1

1− x = 0 ⇐⇒ x = 1

Domain = [−1,∞) \ {−4, 0, 1, 2}= [−1, 0) ∪ (0, 1) ∪ (1, 2) ∪ (2,∞)



DEPED COPY

(b) tan θ − sin θ − cos 2θ = sin θcos θ− sin θ − cos 2θ

cos θ = 0 ⇐⇒ θ = kπ2, k odd integer

Domain = R \ {kπ2| k odd integer}

(c) The expression 1+x2 is always positive, and so√

1 + x2 is defined in R. Onthe other hand, the expression 3

√x2 − 1 is also defined in R, but it cannot

be zero because it is in the denominator. Therefore, x should not be −1and 1.

Domain = R \ {−1, 1}

(d) 1 + sin z = 0 ⇐⇒ z = 3π2

+ 2kπ, k ∈ ZDomain = R \ {3π

2+ 2kπ|k ∈ Z} 2

3.4.2. Identity and Conditional Equation

Consider the following two groups of equations:

Group A Group B

(A1) x2 − 1 = 0 (B1) x2 − 1 = (x− 1)(x+ 1)

(A2) (x+ 7)2 = x2 + 49 (B2) (x+ 7)2 = x2 + 14x+ 49

(A3)x2 − 4

x− 2= 2x− 1 (B3)

x2 − 4

x− 2= x+ 2

In each equation in Group A, some values of the variable that are in thedomain of the equation do not satisfy the equation (that is, do not make theequation true). On the other hand, in each equation in Group B, every elementin the domain of the equation satisfies the given equation. The equations inGroup A are called conditional equations, while those in Group B are calledidentities.

An identity is an equation that is true for all values of the variablein the domain of the equation. An equation that is not an identity iscalled a conditional equation. (In other words, if some values of thevariable in the domain of the equation do not satisfy the equation,then the equation is a conditional equation.)

Example 3.4.2. Identify whether the given equation is an identity or a condi-tional equation. For each conditional equation, provide a value of the variable inthe domain that does not satisfy the equation.

(1) x3 − 2 =(x− 3√

2) (x2 + 3

√2x+ 3

√4)



DEPED COPY

(2) sin2 θ = cos2 θ + 1

(3) sin θ = cos θ − 1

(4)1−√x

1 +√x

=1− 2

√x+ x

1− x

Solution. (1) This is an identity because this is simply factoring of difference oftwo cubes.

(2) This is a conditional equation. If θ = 0, then the left-hand side of the equationis 0, while the right-hand side is 2.

(3) This is also a conditional equation. If θ = 0, then both sides of the equationare equal to 0. But if θ = π, then the left-hand side of the equation is 0,while the right-hand side is −2.

(4) This is an identity because the right-hand side of the equation is obtained byrationalizing the denominator of the left-hand side. 2

3.4.3. The Fundamental Trigonometric Identities

Recall that if P (x, y) is the terminal point on the unit circle corresponding to θ,then we have

sin θ = y csc θ =1

ytan θ =

y

x

cos θ = x sec θ =1

xcot θ =

x

y.

From the definitions, the following reciprocal and quotient identities immedi-ately follow. Note that these identities hold if θ is taken either as a real numberor as an angle.

Reciprocal Identities

csc θ =1

sin θsec θ =

1

cos θcot θ =

1

tan θ

Quotient Identities

tan θ =sin θ

cos θcot θ =

cos θ

sin θ

We can use these identities to simplify trigonometric expressions.



DEPED COPY

Example 3.4.3. Simplify:

(1)tan θ cos θ

sin θ(2)

cos θ

cot θ

Solution. (1)tan θ cos θ

sin θ=

sin θcos θ

cos θ

sin θ= 1

(2)cos θ

cot θ=

cos θcos θsin θ

= sin θ 2

If P (x, y) is the terminal point on the unit circle corresponding to θ, thenx2 + y2 = 1. Since sin θ = y and cos θ = x, we get

sin2 θ + cos2 θ = 1.

By dividing both sides of this identity by cos2 θ and sin2 θ, respectively, we obtain

tan2 θ + 1 = sec2 θ and 1 + cot2 θ = csc2 θ.

Pythagorean Identities

sin2 θ + cos2 θ = 1

tan2 θ + 1 = sec2 θ 1 + cot2 θ = csc2 θ

Example 3.4.4. Simplify:

(1) cos2 θ + cos2 θ tan2 θ (2)1 + tan2 θ

1 + cot2 θ

Solution. (1) cos2 θ + cos2 θ tan2 θ = (cos2 θ)(1 + tan2 θ)

= cos2 θ sec2 θ

= 1

(2)1 + tan2 θ

1 + cot2 θ=

sec2 θ

csc2 θ=

1cos2 θ

1sin2 θ

=sin2 θ

cos2 θ= tan2 θ 2

In addition to the eight identities presented above, we also have the followingidentities.

Even-Odd Identities

sin(−θ) = − sin θ cos(−θ) = cos θ

tan(−θ) = − tan θ



DEPED COPY

The first two of the negative identities can be obtained from the graphs ofthe sine and cosine functions, respectively. (Please review the discussion on page147.) The third identity can be derived as follows:

tan(−θ) =sin(−θ)cos(−θ)

=− sin θ

cos θ= − tan θ.

The reciprocal, quotient, Pythagorean, and even-odd identities constitutewhat we call the fundamental trigonometric identities.

We now solve Example 3.2.3 in a different way.

Example 3.4.5. If sin θ = −34

and cos θ > 0. Find cos θ.

Solution. Using the identity sin2 θ + cos2 θ = 1 with cos θ > 0, we have

cos θ =√

1− sin2 θ =

√1−

(−3

4

)2

=

√7

4. 2

Example 3.4.6. If sec θ = 52

and tan θ < 0, use the identities to find the valuesof the remaining trigonometric functions of θ.

Solution. Note that θ lies in QIV.

cos θ =1

sec θ=

2

5

sin θ = −√

1− cos2 θ = −

√1−

(2

5

)2

= −√

21

5

csc θ =1

sin θ= −5

√21

21

tan θ =sin θ

cos θ=−√215

25

= −√

21

2

cot θ =1

tan θ= −2

√21

212

3.4.4. Proving Trigonometric Identities

We can use the eleven fundamental trigonometric identities to establish otheridentities. For example, suppose we want to establish the identity

csc θ − cot θ =sin θ

1 + cos θ.



DEPED COPY

To verify that it is an identity, recall that we need to establish the truth of theequation for all values of the variable in the domain of the equation. It is notenough to verify its truth for some selected values of the variable. To prove it, weuse the fundamental trigonometric identities and valid algebraic manipulationslike performing the fundamental operations, factoring, canceling, and multiplyingthe numerator and denominator by the same quantity.

Start on the expression on one side of the proposed identity (preferably thecomplicated side), use and apply some of the fundamental trigonometric identitiesand algebraic manipulations, and arrive at the expression on the other side of theproposed identity.

Expression Explanation

csc θ − cot θ Start on one side.

=1

sin θ− cos θ

sin θApply some reciprocal andquotient identities.

=1− cos θ

sin θAdd the quotients.

=1− cos θ

sin θ· 1 + cos θ

1 + cos θMultiply the numeratorand denominator by1 + cos θ.

=1− cos2 θ

(sin θ)(1 + cos θ)Multiply.

=sin2 θ

(sin θ)(1 + cos θ)Apply a Pythagoreanidentity.

=sin θ

1 + cos θReduce to lowest terms.

Upon arriving at the expression of the other side, the identity has been estab-lished. There is no unique technique to prove all identities, but familiarity withthe different techniques may help.

Example 3.4.7. Prove: secx− cosx = sinx tanx.

Solution.

secx− cosx =1

cosx− cosx

=1− cos2 x

cosx=

sin2 x

cosx= sinx · sinx

cosx= sinx tanx 2

Example 3.4.8. Prove:1 + sin θ

1− sin θ− 1− sin θ

1 + sin θ= 4 sin θ sec2 θ



DEPED COPY

Solution.

1 + sin θ

1− sin θ− 1− sin θ

1 + sin θ=

(1 + sin θ)2 − (1− sin θ)2

(1− sin θ)(1 + sin θ)

=1 + 2 sin θ + sin2 θ − 1 + 2 sin θ − sin2 θ

1− sin2 θ

=4 sin θ

cos2 θ= 4 sin θ sec2 θ 2


1. Express each of the other circular functions of θ in terms of cos θ.

Solution:

• sec θ =1

cos θ

• sin2 θ + cos2 θ = 1 =⇒ sin2 θ = 1− cos2 θ =⇒ sin θ = ±√

1− cos2 θ

• csc θ =1

sin θ=

1

±√

1− cos2 θ

• cot θ =cos θ

sin θ=

cos θ

±√

1− cos2 θ

• tan θ =sin θ

cos θ=±√

1− cos2 θ

cos θ

2. If tan θ = a, express cos2 θ in terms of a.

Solution:

a =sin θ

cos θ=⇒ a2 =

sin2 θ

cos2 θ=⇒ a2 =

1− cos2 θ

cos2 θa2 cos2 θ = 1− cos2 θ

a2 cos2 θ + cos2 θ = 1 =⇒ cos2 θ(a2 + 1) = 1 =⇒ cos2 θ =1

a2 + 1

3. Given a = cosx, simplify and express sin4 x− cos4 x in terms of a.

Solution: sin4 x− cos4 x = (sin2 x+ cos2 x)(sin2 x− cos2 x)

= sin2 x− cos2 x

= 1− cos2 x− cos2 x

= 1− 2 cos2 x = 1− 2a2



DEPED COPY

4. Simplify (cscx− secx)2 + (cscx+ secx)2.

Solution: (cscx− secx)2 + (cscx+ secx)2

= (csc2 x− 2 cscx secx+ sec2 x) + (csc2 x+ 2 cscx secx+ sec2 x)

= 2 csc2 x+ 2 sec2 x

=2

sin2 x+

2

cos2 x

=2(cos2 x+ sin2 x)

sin2 x cos2 x=

2

sin2 x cos2 x= 2 csc2 x sec2 x

5. Verify the identitycsc θ

tan θ + cot θ= cos θ.

Solution:

csc θ

tan θ + cot θ=

1

sin θsin θ

cos θ+

cos θ

sin θ

=

1

sin θsin2 θ + cos2 θ

cos θ sin θ

=1

sin θ· cos θ sin θ

1= cos θ

6. Establish the identitycsc θ + cot θ − 1

cot θ − csc θ + 1=

1 + cos θ

sin θ.

Solution:csc θ + cot θ − 1

cot θ − csc θ + 1

=csc θ + cot θ − 1

cot θ − csc θ + 1· csc θ + cot θ

csc θ + cot θ

=(csc θ + cot θ − 1)(csc θ + cot θ)

(cot θ − csc θ)(csc θ + cot θ) + (csc θ + cot θ)


cot2 θ − csc2 θ + csc θ + cot θ


−1 + csc θ + cot θ

= csc θ + cot θ =1

sin θ+

cos θ

sin θ=

1 + cos θ

sin θ


1. Using fundamental identities, simplify the expressiontanx− sinx

sinx.

2. Using fundamental identities, simplify the expression1

cscx− cotx.

3. Simplify sinA+cos2A

1 + sinA.



DEPED COPY

4. Simplify (1− cos2A)(1 + cot2A).

5. Expresscscx+ secx

cotx+ tanxin terms of sine and cosine.

6. Expresstanx− cotx

tanx+ cotxin terms of sine and cosine.

7. Expresstanx+ sinx

cscx+ cotxin terms cosine only.

8. Express1

1 + tan2 xin terms sine only.

9. If cot θ = a, express sin θ cos θ in terms of a.

10. If sec θ = a > 0 and sin θ > 0, express sin θ cos θ in terms of a.

For numbers 11 - 20, establish the identities.

11.csc a+ 1

csc a− 1=

1 + sin a

1− sin a

12.1 + sin a

1− sin a− 1− sin a

1 + sin a= 4 tan a sec a

13.cos a

sec a+ tan a= 1− sin a

14.csc a+ sec a tan a

csc2 a= tan a sec a

15.1

1− cos a+

1

1 + cos a= 2 csc2 a

16.sin3 α− cos3 α

sinα− cosα= 1 + sinα cosα

17.tanα

1− tan2 α=

sinα cosα

2 cos2 α− 1

18.tan2 α + secα + 1

tanα + cotα= tanα + sinα

19.cotα− sinα secα

secα cscα= cos2 α− sin2 α

20. tan2 α sec2 α− sec2 α + 1 = tan4 α

4



DEPED COPY

Lesson 3.5. Sum and Difference Identities



(1) derive trigonometric identities involving sum and difference of two angles;

(2) simplify trigonometric expressions using fundamental trigonometric identitiesand sum and difference identities;

(3) prove other trigonometric identities using fundamental trigonometric identi-ties and sum and difference identities; and

(4) solve situational problems involving trigonometric identities.

Lesson Outline

(1) The sum and difference identities for cosine, sine, and tangent functions

(2) Cofunction identities

(3) More trigonometric identities

Introduction

In previous lesson, we introduced the concept of trigonometric identity, pre-sented the fundamental identities, and proved some identities. In this lesson, wederive the sum and difference identities for cosine, sine, and tangent functions,establish the cofunction identities, and prove more trigonometric identities.

3.5.1. The Cosine Difference and Sum Identities

Let u and v be any real numbers with 0 < v ≤ u < 2π. Consider the unit circlewith points A = (1, 0), P1, P2, P3, and u and v with corresponding angles asshown below. Then P1P2 = AP3.



DEPED COPY

Recall that P1 = P (u) = (cosu, sinu), P2 = P (v) = (cos v, sin v), and P3 =P (u− v) = (cos(u− v), sin(u− v)), so that

P1P2 =√

(cosu− cos v)2 + (sinu− sin v)2,

whileAP3 =

√[cos(u− v)− 1]2 + [sin(u− v)− 0]2.

Equating these two expressions and expanding the squares, we get

(cosu− cos v)2 + (sinu− sin v)2 = [cos(u− v)− 1]2 + sin2(u− v)

cos2 u− 2 cosu cos v + cos2 v + sin2 u− 2 sinu sin v + sin2 v

= cos2(u− v)− 2 cos(u− v) + 1 + sin2(u− v)

Applying the Pythagorean identity cos2 θ+sin2 θ = 1 and simplifying the resultingequations, we obtain

(cos2 u+ sin2 u) + (cos2 v + sin2 v)− 2 cosu cos v − 2 sinu sin v

= [cos2(u− v) + sin2(u− v)]− 2 cos(u− v) + 1

1 + 1− 2 cosu cos v − 2 sinu sin v = 1− 2 cos(u− v) + 1

cos(u− v) = cosu cos v + sinu sin v.

We have thus proved another identity.

Although we assumed at the start that 0 < v ≤ u < 2π, but becausecos(−θ) = cos θ (one of the even-odd identities), this new identity is true forany real numbers u and v. As before, the variables can take any real values orangle measures.

Cosine Difference Identity

cos(A−B) = cosA cosB + sinA sinB

Replacing B with −B, and applying the even-odd identities, we immediatelyget another identity.

Cosine Sum Identity

cos(A+B) = cosA cosB − sinA sinB

Example 3.5.1. Find the exact values of cos 105◦ and cos π12

.



DEPED COPY

Solution.

cos 105◦ = cos(60◦ + 45◦)

= cos 60◦ cos 45◦ − sin 60◦ sin 45◦

=1

2·√

2

2−√

3

2·√

2

2

=

√2−√

6

4

cosπ

12= cos

(π4− π

6

)= cos

π

4cos

π

6+ sin

π

4sin

π

6

=

√2

2·√

3

2+

√2

2· 1

2

=

√6 +√

2

42

Example 3.5.2. Given cosα = 35

and sin β = 1213

, where α lies in QIV and β inQI, find cos(α + β).

Solution. We will be needing sinα and cos β.

sinα = −√

1− cos2 α = −

√1−

(3

5

)2

= −4

5

cos β =

√1− sin2 β =

√1−

(12

13

)2

=5

13

cos(α + β) = cosα cos β − sinα sin β

=3

5· 5

13−(−4

5

)12

13

=63

652

3.5.2. The Cofunction Identities and the Sine Sum and DifferenceIdentities

In the Cosine Difference Identity, if we let A = π2, we get

cos(π

2−B

)= cos

(π2

)cosB + sin

(π2

)sinB



DEPED COPY

= (0) cosB + (1) sinB

= sinB.

From this identity, if we replace B with π2−B, we have

cos[π

2−(π

2−B

)]= sin

(π2−B

)cosB = sin

(π2−B

).

As for the tangent function, we have

tan(π

2−B

)=

sin(π2−B

)cos(π2−B

)=

cosB

sinB= cotB.

We have just derived another set of identities.

Cofunction Identities

cos(π

2−B

)= sinB sin

(π2−B

)= cosB

tan(π

2−B

)= cotB

Using the first two cofunction identities, we now derive the identity for sin(A+B).

sin(A+B) = cos[π

2− (A+B)

]= cos

[(π2− A

)−B)

]= cos

(π2− A

)cosB + sin

(π2− A

)sinB

= sinA cosB + cosA sinB

Sine Sum Identity

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

In the last identity, replacing B with −B and applying the even-odd identitiesyield

sin(A−B) = sin[A+ (−B)]



DEPED COPY

= sinA cos(−B) + cosA sin(−B)

= sinA cosB − cosA sinB.

Sine Difference Identity

sin(A−B) = sinA cosB − cosA sinB

Example 3.5.3. Find the exact value of sin(5π12

).

Solution.

sin

(5π

12

)= sin

(π4

+π

6

)= sin

(π4

)cos(π

6

)+ cos

(π4

)sin(π

6

)=

√2

2·√

3

2+

√2

2· 1

2

=

√6 +√

2

42

Example 3.5.4. If sinα = 313

and sin β = 12, where 0 < α < π

2and π

2< β < π,

find sin(α + β) and sin(β − α).

Solution. We first compute cosα and cos β.

cosα =√

1− sin2 α =

√1−

(3

13

)2

=4√

10

13

cos β = −√

1− sin2 β = −

√1−

(1

2

)2

= −√

3

2

sin(α + β) = sinα cos β + cosα sin β

=3

13

(−√

3

2

)+

4√

10

13· 1

2

=4√

10− 3√

3

26

sin(β − α) = sin β cosα− cos β sinα

=1

2· 4√

10

13−

(−√

3

2

)3

13

=4√

10 + 3√

3

262



DEPED COPY

Example 3.5.5. Prove:

sin(x+ y) = (1 + cotx tan y) sinx cos y.

Solution.

(1 + cot x tan y) sinx cos y = sinx cos y + cotx tan y sinx cos y

= sinx cos y +cosx

sinx

sin y

cos ysinx cos y

= sinx cos y + cosx sin y = sin(x+ y) 2

3.5.3. The Tangent Sum and Difference Identities

Recall that tanx is the ratio of sinx over cos x. When we replace x with A+B,we obtain

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

cos(A+B).

Using the sum identities for sine and cosine, and then dividing the numeratorand denominator by cosA cosB, we have

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

cosA cosB − sinA sinB

=sinA cosBcosA cosB

+ cosA sinBcosA cosB

cosA cosBcosA cosB

− sinA sinBcosA cosB

=tanA+ tanB

1− tanA tanB.

We have just established the tangent sum identity.

In the above identity, if we replace B with −B and use the even-odd identitytan(−θ) = − tan θ, we get

tan(A−B) = tan[A+ (−B)] =tanA+ tan(−B)

1− tanA tan(−B)=

tanA− tanB

1 + tanA tanB.

This is the tangent difference identity.

Tangent Sum and Difference Identities

tan(A+B) =tanA+ tanB

1− tanA tanB

tan(A−B) =tanA− tanB

1 + tanA tanB



DEPED COPY


1. Given cosA = −35, π < A < 3π

2, and tanB = 7

24, B in QI, find: (a) sin(A+B),

(b) cos(A+B), and (c) tan(A+B).

Solution: cosA = −3

5and A in QIII =⇒ sinA = −4

5

tanB =7

24and B in QIII =⇒ sinB = − 7

25and cosB = −24

25

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

=

(−4

5

)(−24

25

)+

(−3

5

)(− 7

25

)=

117

125

(b) cos(A+B) = cosA cosB − sinA sinB

=

(−3

5

)(−24

25

)−(−4

5

)(− 7

25

)=

44

125

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

cos(A+B)=

11712544125

=117

44

2. Find the exact value of cos5π

12.

Solution: cos5π

12= cos

(2π

3− π

4

)= cos

2π

3cos

π

4+ sin

2π

3sin

π

4

=

(−1

2

)(√2

2

)+

(√3

2

)(√2

2

)

=

√6−√

2

4

3. If A+B = π2

+ 2kπ, k ∈ Z, prove that sinA = cosB.

Solution: sinA = sin(π

2+ 2kπ −B

)= sin

(π2

+ 2kπ)

cosB − cos(π

2+ 2kπ

)sinB = cosB

4. Find the value of

(tan 10◦)(tan 15◦)(tan 20◦)(tan 15◦) · · · (tan 65◦)(tan 70◦)(tan 75◦)(tan 80◦).

Solution: From the previous item, we know that sin θ = cos(90◦ − θ). Wewrite each tangent in terms of sine and cosine.

(tan 10◦)(tan 15◦)(tan 20◦)(tan 15◦) · · · (tan 65◦)(tan 70◦)(tan 75◦)(tan 80◦)

=

(sin 10

cos 10

)(sin 15

cos 15

)(sin 20

cos 20

)· · ·(

sin 70

cos 70

)(sin 75

cos 75

)(sin 80

cos 80

)= 1



DEPED COPY

5. If A, B and C are the angles of a triangle and

(tanA)(tanB)(tanC) = −√

3

3,

find tanA+ tanB + tanC.

Solution:

A+B + C = 180◦ =⇒ tan(A+B + C) = tan 180◦ = 0

tanA+ tan(B + C)

1− tanA tan(B + C)= 0 =⇒ tanA+ tan(B + C) = 0

tanA+tanB + tanC

1− tanB tanC= 0

tanA− tanA tanB tanC + tanB + tanC

1− tanB tanC= 0

=⇒ tanA− tanA tanB tanC + tanB + tanC = 0

tanA+ tanB + tanC = tanA tanB tanC = −√

3

3

6. Establish the identitysin(A+B)

cos(A−B)=

tanA+ tanB

1 + tanA tanB.

Solution:

sin(A+B)

cos(A−B)=

sinA cosB + cosA sinB

cosA cosB + sinA sinB

=sinA cosB + cosA sinB

cosA cosB + sinA sinB·

1

cosA cosB1

cosA cosB

=

sinA cosB

cosA cosB+

cosA sinB

cosA cosBcosA cosB

cosA cosB+

sinA sinB

cosA cosB

=tanA+ tanB

1 + tanA tanB


1. If 3π2< θ < 2π, find the radian measure of θ if cos θ = sin 2π

3.

2. For what angle θ in QIV is sin θ = cos 4π3

?

3. If A+B = π2

+ kπ, k ∈ Z, prove that tanA = cotB.



DEPED COPY

4. What is the exact value of cot(−5π

12

)?

5. What is the exact value of sin 105◦ − cos 15◦?

6. What is the exact value of tan 1875◦?

7. Let α and β be acute angles such that cotα = 7 and csc β =√

10. Findcos(α + β).

8. Given sinα = − 817

and sin β = −12, find cos(α+β) if both α and β are in QIV.

9. If 3 sinx = 2, find sin(x− π) + sin(x+ π).

10. Simplify: cos(x+ π

2

)+ cos

(π2− x).

11. Given sinA = 45, π

2≤ A ≤ π, and cosB = 4

5, B not in QI, find: (a) sin(A−B),

(b) cos(A − B), and (c) tan(A − B). Also, determine the quadrant in whichA−B terminate.

12. Given cscA =√

3, A in QI, and secB =√

2, sinB < 0, find: (a) sin(A− B),(b) cos(A − B), and (c) tan(A − B). Also, determine the quadrant in whichA−B terminate.

13. Given sinα = 45

and cos β = 513

, find sin(α + β) + sin(α− β).

14. Given sinα = 23, α in QII, and cos β = 3

4, find cos(α + β) + cos(α− β).

15. If A and B are acute angles (in degrees) such that cscA =√

17 and cscB =√343

, what is A+B?

16. If tan(x+ y) = 13

and tan y = 12, what is tanx?

17. Evaluate:tan π

9+ tan 23π

36

1− tan π9

tan 23π36

.

18. Establish the identity:

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

+ cosA cosB sinC − sinA sinB sinC.

19. Prove: sin 2θ = 2 sin θ cos θ.

20. Prove: cot 2θ =cot2 θ − 1

2 cot θ.

4



DEPED COPY


1. A central angle in a circle of radius 6 cm measures 37.5◦. Find: (a) length ofthe intercepted arc and (b) the area of the sector.

2. The point (−1,−2) lies on the terminal side of the angle θ in standard position.Find sin θ + cos θ + tan θ.

3. Given sinA = 1213

, where A is not in QI, and cscB = −53, where B is not in

QIII, find: (a) cos(A−B) and (b) tan(A−B).

4. Find the exact value oftan 57◦ + tan 78◦

1− tan 57◦ tan 78◦.

5. If sinx = a and cos x ≥ 0, expresscosx tanx+ sinx

tanxin terms of a.

6. Prove the identity cos6 x+ sin6 x = 3 cos4 x− 3 cos2 x− 1.

7. A regular hexagon of side length 1 unit is inscribed in a unit circle such thattwo of its vertices are located on the x-axis. Determine the coordinates of thehexagon.

8. Determine the amplitude, period and phase shift of the graph of

y = 2 sin(x

2+π

3

)− 1,

and sketch its graph over one period. Find the range of the function.



DEPED COPY


1. The area of a sector of a circle formed by a central angle of 30◦ is dπ3

cm2.Find the length of the intercepted arc.

2. The point (8,−6) lies on the terminal side of the angle θ in standard position.Find (sin θ + cos θ)2.

3. Given sinA = − 8

17and cosA > 0, evaluate sin

(π2− A

)+ cos

(π2− A

).

4. Find the exact value of sin 160◦ cos 35◦ − sin 70◦ cos 55◦.

5. Find the exact value of tan7π

12.

6. Given cosA = −35, where A is not in QII, and tanB = 24

7, where B is not in

QI, find: (a) sin(A+B) and (b) cot(A+B).

7. Establish the identitytan2 x

tanx+ tan3 x= sinx cosx.

8. If sinx− cosx =1

3, find

sinx

secx.

9. Determine the period and phase shift of the graph of y = tan( π

18− x

3

)+ 2,

and sketch its graph over two periods.



DEPED COPY

Lesson 3.6. Double-Angle and Half-Angle Identities



(1) derive the double-angle and half-angle identities;

(2) simplify trigonometric expressions using known identities;

(3) prove other trigonometric identities using known identities; and

(4) solve situational problems involving trigonometric identities.

Lesson Outline

(1) The double-angle and half-angle identities for cosine, sine, and tangent

(2) More trigonometric identities

Introduction

Trigonometric identities simplify the computations of trigonometric expres-sions. In this lesson, we continue on establishing more trigonometric identities.In particular, we derive the formulas for f(2θ) and f

(12θ), where f is the sine,

cosine, or tangent function.

3.6.1. Double-Angle Identities

Recall the sum identities for sine and cosine.

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

cos(A+B) = cosA cosB − sinA sinB

When A = B, these identities becomes

sin 2A = sinA cosA+ cosA sinA = 2 sinA cosA

andcos 2A = cosA cosA− sinA sinA = cos2A− sin2A.

Double-Angle Identities for Sine and Cosine

sin 2A = 2 sinA cosA cos 2A = cos2A− sin2A

The double-identity for cosine has other forms. We use the Pythagoreanidentity sin2 θ + cos2 θ = 1.

cos 2A = cos2A− sin2A



DEPED COPY

= cos2A− (1− cos2A)

= 2 cos2A− 1

cos 2A = cos2A− sin2A

= (1− sin2A)− sin2A

= 1− 2 sin2A

Other Double-Angle Identities for Cosine

cos 2A = 2 cos2A− 1 cos 2A = 1− 2 sin2A

Example 3.6.1. Given sin t = 35

and π2< t < π, find sin 2t and cos 2t.

Solution. We first find cos t using the Pythagorean identity. Since t lies in QII,we have

cos t = −√

1− sin2 t = −

√1−

(3

5

)2

= −4

5.

sin 2t = 2 sin t cos t

= 2

(3

5

)(−4

5

)= −24

25

cos 2t = 1− 2 sin2 t

= 1− 2

(3

5

)2

=7

252

In the last example, we may compute cos 2t using one of the other two double-angle identities for cosine. For the sake of answering the curious minds, we includethe computations here.

cos 2t = cos2 t− sin2 t

=

(−4

5

)2

−(

3

5

)2

=7

25

cos 2t = 2 cos2 t− 1

= 2

(−4

5

)2

− 1

=7

25

In the three cosine double-angle identities, which formula to use depends onthe convenience, what is given, and what is asked.

Example 3.6.2. Derive an identity for sin 3x in terms of sinx.



DEPED COPY

Solution. We use the sum identity for sine, the double-angle identities for sineand cosine, and the Pythagorean identity.

sin 3x = sin(2x+ x)

= sin 2x cosx+ cos 2x sinx

= (2 sin x cosx) cosx+ (1− 2 sin2 x) sinx

= 2 sin x cos2 x+ sinx− 2 sin3 x

= 2(sinx)(1− sin2 x) + sin x− 2 sin3 x

= 3 sin x− 4 sin3 x 2

For the double-angle formula for tangent, we recall the tangent sum identity:

tan(A+B) =tanA+ tanB

1− tanA tanB.

When A = B, we obtain

tan(A+ A) =tanA+ tanA

1− tanA tanA=

2 tanA

1− tan2A.

Tangent Double-Angle Identity

tan 2A =2 tanA

1− tan2A

Example 3.6.3. If tan θ = −13

and sec θ > 0, find sin 2θ, cos 2θ, and tan 2θ.

Solution. We can compute immediately tan 2θ.

tan 2θ =2 tan θ

1− tan2 θ=

2(−1

3

)1−

(−1

3

)2 = −3

4

From the given information, we deduce that θ lies in QIV. Using one Pythagoreanidentity, we compute cos θ through sec θ. (We may also use the technique dis-cussed in Lesson 3.2 by solving for x, y, and r.) Then we proceed to find cos 2θ.

sec θ =√

1 + tan2 θ =

√1 +

(−1

3

)2

=

√10

3

cos θ =1

sec θ=

1√103

=3√

10

10

cos 2θ = 2 cos2 θ − 1 = 2

(3√

10

10

)2

− 1 =4

5

tan 2θ =sin 2θ

cos 2θ=⇒ sin 2θ = tan 2θ cos 2θ = −3

52



DEPED COPY

3.6.2. Half-Angle Identities

Recall two of the three double-angle identities for cosine:

cos 2A = 2 cos2A− 1 and cos 2A = 1− 2 sin2A.

From these identities, we obtain two useful identities expressing sin2A and cos2Ain terms of cos 2A as follows:

cos2A =1 + cos 2A

2and sin2A =

1− cos 2A

2.

Some Useful Identities

cos2A =1 + cos 2A

2sin2A =

1− cos 2A

2

From these identities, replacing A with A2, we get

cos2A

2=

1 + cos 2(A2

)2

=1 + cosA

2

and

sin2 A

2=

1− cos 2(A2

)2

=1− cosA

2.

These are the half-angle identities for sine and cosine.

Half-Angle Identities for Sine and Cosine

cos2(A

2

)=

1 + cosA

2sin2

(A

2

)=

1− cosA

2

Because of the “square” in the formulas, we get

cosA

2= ±

√1 + cosA

2and sin

A

2= ±

√1− cosA

2.

The appropriate signs of cos A2

and sin A2

depend on which quadrant A2

lies.

Example 3.6.4. Find the exact values of sin 22.5◦ and cos 22.5◦.

Solution. Clearly, 22.5◦ lies in QI (and so sin 22.5◦ and cos 22.5◦ are both posi-tive), and 22.5◦ is the half-angle of 45◦.

sin 22.5◦ =

√1− cos 45◦

2=

√1−

√22

2=

√2−√

2

2

cos 22.5◦ =

√1 + cos 45◦

2=

√1 +

√22

2=

√2 +√

2

22



DEPED COPY

Example 3.6.5. Prove: cos2(θ

2

)=

tan θ + sin θ

2 tan θ.

Solution.

cos2(θ

2

)=

1 + cos θ

2

=1 + cos θ

2

(tan θ

tan θ

)=

tan θ + cos θ tan θ

2 tan θ

=tan θ + cos θ · sin θ

cos θ

2 tan θ

=tan θ + sin θ

2 tan θ2

We now derive the first version of the half-angle formula for tangent.

tanA

2=

sin A2

cos A2

=sin A

2

cos A2

(2 sin A

2

2 sin A2

)

=2 sin2

(A2

)2 sin A

2cos A

2

=2 · 1−cosA

2

sin(2 · A

2

)=

1− cosA

sinA

There is another version of the tangent half-angle formula, and we can deriveit from the first version.

tanA

2=

1− cosA

sinA

=1− cosA

sinA

(1 + cosA

1 + cosA

)=

1− cos2A

(sinA)(1 + cosA)

=sin2A

(sinA)(1 + cosA)

=sinA

1 + cosA



DEPED COPY

Tangent Half-Angle Identities

tanA

2=

1− cosA

sinAtan

A

2=

sinA

1 + cosA

tanA

2=

sin A2

cos A2

tan2

(A

2

)=

1− cosA

1 + cosA

Example 3.6.6. Find the exact value of tan π12

.

Solution.

tanπ

12=

1− cos π6

sin π6

=1−

√32

12

= 2−√

3 2

Example 3.6.7. If sin θ = −25, cot θ > 0, and 0 ≤ θ < 2π, find sin θ

2, cos θ

2, and

tan θ2.

Solution. Since sin θ < 0 and cot θ > 0, we conclude the π < θ < 3π2

. It followsthat

π

2<θ

2<

3π

4,

which means that θ2

lies in QII.

cos θ = −√

1− sin2 θ = −

√1−

(−2

5

)2

= −√

21

5

sinθ

2=

√1− cos θ

2=

√√√√1−(−√215

)2

=

√50 + 10

√21

10

cosθ

2= −

√1 + cos θ

2= −

√√√√1 +(−√215

)2

= −√

50− 10√

21

10

tanθ

2=

1− cos θ

sin θ=

1−(−√215

)−2

5

= −5 +√

21

22


1. If cos θ = − 513

with 0 < θ < π, find sin 2θ and cos 2θ.



DEPED COPY

Solution: In this problem, we use the Pythagorean identity sin2 θ+cos2 θ = 1.Because cos θ = − 5

13, we must have

sin2 θ = 1− cos2 θ = 1− 25

169=

144

169.

Moreover, since 0 < θ < π, we take the square root of both sides of the aboveequation to get

sin θ =12

13.

Now, using the double-angle identities we get

sin 2θ = 2 sin θ cos θ and cos 2θ = cos2 θ − sin2 θ

= 2(1213

) (− 5

13

)= 25

169− 144

169

= −120169

= −119169.

2. Derive an identity for cos 3x in terms of cosx.

Solution: We use the sum identity for cosine, the double-angle identities forsine and cosine, and the Pythagorean identity.

cos 3x = cos(2x+ x)

= cos 2x cosx− sin 2x sinx

= (2 cos2 x− 1) cosx− (2 sinx cosx) sinx

= 2 cos3 x− cosx− 2 sin2 x cosx

= 2 cos3 x− cosx− 2(1− cos2 x) cosx

= 4 cos3 x− 3 cosx.

3. Derive the identity for tan 3t in terms of tan t.

Solution: Using the sum identity for tangent, we obtain

tan 3t = tan(2t+ t) =tan 2t+ tan t

1− tan 2t tan t.

Now, using the tangent double-angle identity, we have

tan 3t =2 tan t

1−tan2 t + tan t

1− 2 tan t1−tan2 t tan t

.

Upon simplifying the terms on the right side of the equation, we finally obtain

tan 3t =3 tan t− tan3 t

1− 3 tan2 t.



DEPED COPY

4. Find the exact value of sin π12

.

Solution: To find the value of sin π12

, we use the identity sin2 12y = 1−cos y

2.

With y = π6, we obtain

sin2 π

12= sin2 1

2

(π6

)=

1− cos π6

2=

1−√32

2=

2−√

3

4.

Now, since 0 < π12< π, sin π

12must be positive, and so

sinπ

12=

√2−√

3

2.

5. Prove: sin2

(θ

2

)=

sec θ − 1

2 sec θ.

Solution:

sin2

(θ

2

)=

1− cos θ

2

=1− cos θ

2

(sec θ

sec θ

)=

sec θ − cos θ sec θ

2 sec θ

=sec θ − cos θ · 1

cos θ

2 sec θ

=sec θ − 1

2 sec θ.

6. Use the half-angle identity to find the exact value of tan 75◦.

Solution: tan 75◦ = tan(12· 150◦

)= 1−cos 150◦

sin 150◦=

1+√32

12

= 2 +√

3.

7. A ball is thrown following a projectile motion. It is known that the horizontaldistance (range) the ball can travel is given by

R =v20g

sin 2θ,

where R is the range (in feet), v0 is the initial speed (in ft/s), θ is the angleof elevation the ball is thrown, and g = 32 ft/s2 is the acceleration due togravity.

(a) Express the new range in terms of the original range when an angle θ(0 < θ ≤ 45◦) is doubled?

(b) If a ball travels a horizontal distance of 20 ft when kicked at an angleof α with initial speed of 20

√2 ft/s, find the horizontal distance it can

travel when you double α. Hint: Use the result of item (a)



DEPED COPY

Solution:

(a) Let R =v20g

sin 2θ be the original range. When the angle is doubled, the

new range will become R′ =v20g

sin 4θ. Now, we solve sin 4θ in terms ofthe original range.

Note that sin 2θ = gRv20

. So, as a consequence of the fundamental identity,

we obtain

cos 2θ =

√1− g2R2

v40=

√v40 − g2R2

v20.

Since sin 4θ = 2 sin 2θ cos 2θ, it follows that

R′ =v20g

sin 4θ =v20g

(2 · gR

v20·√v40 − g2R2

v20

)=

2R√v40 − g2R2

v20.

(b) Using the result in (a), if α is doubled, then the new range is given by

R′ =2R√v40 − g2R2

v20=

40√

640000− 409600

800= 24.

Therefore, the new horizontal distance is 24 ft.


1. Let θ be an angle in the first quadrant and sin θ = 13. Find

(a) sin 2θ

(b) cos 2θ

(c) tan 2θ

(d) sec 2θ

(e) csc 2θ

(f) cot 2θ

2. Find the approximate value of csc 46◦ and sec 46◦, given that sin 23◦ ≈ 0.3907.

3. If cos t = 34, what is cos 2t?

4. Derive a formula for sin 4x in terms of sinx and cos x.

5. Let −π4< x < 0. Given that tan 2x = −2, solve for tanx.

6. Obtain an identity for tan 4θ in terms of tan θ.

7. Solve for the exact value of cot 4θ if tan θ = 12.

8. Use half-angle identities to find the exact value of (a) sin2 15◦ and (b) cos2 15◦.

9. Use half-angle identities to find the exact value of (a) sin2 5π8

and (b) cos2 5π8

.



DEPED COPY

10. Find the exact value of cos π8.

11. Prove thattan 1

2y − 1

tan 12y + 1

=sin y − cos y − 1

sin y + cos y + 1.

12. Verify that the following equation is an identity: cot 12t = cot t(sec t+ 1).

13. Use half-angle identities to find the exact value of (a) cos 105◦ and (b) tan 22.5◦.

4

Lesson 3.7. Inverse Trigonometric Functions



(1) graph the six basic inverse trigonometric functions;

(2) illustrate the domain and range of the inverse trigonometric functions;

(3) evaluate inverse trigonometric expressions; and

(4) solve situational problems involving inverse trigonometric functions.

Lesson Outline

(1) Definitions of the six inverse trigonometric functions

(2) Graphs of inverse trigonometric functions

(3) Domain and range of inverse trigonometric functions

(4) Evaluation of inverse trigonometric expressions

Introduction

In the previous lessons on functions (algebraic and trigonometric), we com-puted for the value of a function at a number in its domain. Now, given a valuein the range of the function, we reverse this process by finding a number in thedomain whose function value is the given one. Observe that, in this process,the function involved may or may not give a unique number in the domain. Forexample, each of the functions f(x) = x2 and g(x) = cosx do not give a uniquenumber in their respective domains for some values of each function. Givenf(x) = 1, the function gives x = ±1. If g(x) = 1, then x = 2kπ, where k is aninteger. Because of this possibility, in order for the reverse process to produce afunction, we restrict this process to one-to-one functions or at least restrict thedomain of a non-one-to-one function to make it one-to-one so that the process



DEPED COPY

works. Loosely speaking, a function that reverses what a given function f doesis called its inverse function, and is usually denoted by f−1.

More formally, two functions f and g are inverse functions if

g(f(x)) = x for any x in the domain of f ,

andf(g(x)) = x for any x in the domain of g.

We denote the inverse function of a function f by f−1. The graphs of a functionand its inverse function are symmetric with respect to the line y = x.

In this lesson, we first restrict the domain of each trigonometric functionbecause each of them is not one-to-one. We then define each respective inversefunction and evaluate the values of each inverse trigonometric function.

3.7.1. Inverse Sine Function

All the trigonometric functions that we consider are periodic over their entiredomains. This means that all trigonometric functions are not one-to-one if weconsider their whole domains, which implies that they have no inverses over thosesets. But there is a way to make each of the trigonometric functions one-to-one.This is done by restricting their respective domains. The restrictions will give uswell-defined inverse trigonometric functions.

The domain of the sine function is the set R of real numbers, and its range isthe closed interval [−1, 1]. As observed in the previous lessons, the sine functionis not one-to-one, and the first step is to restrict its domain (by agreeing what theconvention is) with the following conditions: (1) the sine function is one-to-onein that restricted domain, and (2) the range remains the same.

The inverse of the (restricted) sine function f(x) = sinx, where thedomain is restricted to the closed interval

[−π

2, π2

], is called the inverse

sine function or arcsine function, denoted by f−1(x) = sin−1 x orf−1(x) = arcsinx. Here, the domain of f−1(x) = arcsinx is [−1, 1],and its range is

[−π

2, π2

]. Thus,

y = sin−1 x or y = arcsinx

if and only ifsin y = x,

where −1 ≤ x ≤ 1 and −π2≤ y ≤ π

2.

Throughout the lesson, we interchangeably use sin−1 x and arcsinx to meanthe inverse sine function.



DEPED COPY

Example 3.7.1. Find the exact value of each expression.(1) sin−1 1

2

(2) arcsin(−1)

(3) arcsin 0

(4) sin−1(−1

2

)Solution. (1) Let θ = sin−1 1

2. This is equivalent to sin θ = 1

2. This means that

we are looking for the number θ in the closed interval[−π

2, π2

]whose sine is

12. We get θ = π

6. Thus, we have sin−1 1

2= π

6.

(2) arcsin(−1) = −π2

because sin(−π

2

)= −1 and −π

2∈[−π

2, π2

].

(3) arcsin 0 = 0

(4) sin−1(−1

2

)= −π

62

As emphasized in the last example, as long as −1 ≤ x ≤ 1, sin−1 x is thatnumber y ∈

[−π

2, π2

]such that sin y = x. If |x| > 1, then sin−1 x is not defined in

R.

We can sometimes find the exact value of sin−1 x (that is, we can find a valuein terms of π), but if no such special value exists, then we leave it in the formsin−1 x. For example, as shown above, sin−1 1

2is equal to π

6. However, as studied

in Lesson 3.2, no special number θ satisfies sin θ = 23, so we leave sin−1 2

3as is.

Example 3.7.2. Find the exact value of each expression.(1) sin

(sin−1 1

2

)(2) arcsin

(sin π

3

) (3) arcsin(sin π)

(4) sin(sin−1

(−1

2

))Solution. (1) sin

(sin−1 1

2

)= sin π

6= 1

2

(2) arcsin(sin π

3

)= arcsin

√32

= π3

(3) arcsin(sin π) = arcsin 0 = 0

(4) sin(sin−1

(−1

2

))= sin

(−π

6

)= −1

22

From the last example, we have the following observations:

1. sin(arcsin x) = x for any x ∈ [−1, 1]; and

2. arcsin(sin θ) = θ if and only if θ ∈[−π

2, π2

], and if θ 6∈

[−π

2, π2

], then

arcsin(sin θ) = ϕ, where ϕ ∈[−π

2, π2

]such that sinϕ = sin θ.

To sketch the graph of y = sin−1 x, Table 3.32 presents the tables of valuesfor y = sinx and y = sin−1 x. Recall that the graphs of y = sinx and y = sin−1 xare symmetric with respect to the line y = x. This means that if a point (a, b) ison y = sinx, then (b, a) is on y = sin−1 x.



DEPED COPY

y = sinxx −π

2−π

3−π

4−π

60 π

6π4

π3

π2

y −1 −√32−√22−1

20 1

2

√22

√32

1

y = sin−1 xx −1 −

√32−√22−1

20 1

2

√22

√32

1

y −π2−π

3−π

4−π

60 π

6π4

π3

π2

Table 3.32

The graph (solid thick curve) of the restricted sine function y = sinx is shownin Figure 3.33(a), while the graph of inverse sine function y = arcsin x is shownin Figure 3.33(b).

(a) y = sinx (b) y = sin−1 x

Figure 3.33

Example 3.7.3. Sketch the graph of y = sin−1(x+ 1).

Solution 1. In this solution, we use translation of graphs.

Because y = sin−1(x + 1) is equivalent to y = sin−1[x − (−1)], the graph ofy = sin−1(x + 1) is 1-unit to the left of y = sin−1 x. The graph below showsy = sin−1(x+ 1) (solid line) and y = sin−1 x (dashed line).



DEPED COPYSolution 2. In this solution, we graph first the corresponding sine function, and

then use the symmetry with respect to y = x to graph the inverse function.

y = sin−1(x+ 1) ⇐⇒ sin y = x+ 1 ⇐⇒ x = sin y − 1

The graph below shows the process of graphing of y = sin−1(x + 1) from y =sinx− 1 with −π

2≤ x ≤ π

2, and then reflecting it with respect to y = x.

3.7.2. Inverse Cosine Function

The development of the other inverse trigonometric functions is similar to thatof the inverse sine function.



DEPED COPY

y = cos−1 x or y = arccosx

meanscos y = x,

where −1 ≤ x ≤ 1 and 0 ≤ y ≤ π.

The graph (solid thick curve) of the restricted cosine function y = cosx isshown in Figure 3.34(a), while the graph of inverse cosine function y = arccosxis shown in Figure 3.34(b).

(a) y = cosx (b) y = cos−1 x

Figure 3.34

Example 3.7.4. Find the exact value of each expression.(1) cos−1 0

(2) arccos(−√32

)(3) cos

(cos−1

(−√32

))(4) cos−1

(cos 3π

4

)(5) arccos

(cos 7π

6

)(6) sin

(cos−1

√22

)Solution. (1) cos−1 0 = π

2because cos π

2= 0 and π

2∈ [0, π].

(2) arccos(−√32

)= 5π

6

(3) cos(

cos−1(−√32

))= −

√32

because −√32∈ [−1, 1]

(4) cos−1(cos 3π

4

)= 3π

4because 3π

4∈ [0, π].



DEPED COPY

(5) arccos(cos 7π

6

)= arccos

(−√32

)= 5π

6

(6) sin(

cos−1√22

)=√22

2

Example 3.7.5. Simplify: sin(arcsin 2

3+ arccos 1

2

).

Solution. We know that arccos 12

= π3. Using the Sine Sum Identity, we have

sin(arcsin 2

3+ arccos 1

2

)= sin

(arcsin 2

3+ π

3

)= sin

(arcsin 2

3

)cos π

3+ cos

(arcsin 2

3

)sin π

3

= 23· 12

+ cos(arcsin 2

3

)·√32

= 13

+√32

cos(arcsin 2

3

).

We compute cos(arcsin 2

3

). Let θ = arcsin 2

3. By definition, sin θ = 2

3, where

θ lies in QI. Using the Pythagorean identity, we have

cos(arcsin 2

3

)= cos θ =

√1− sin2 θ =

√53.

Going back to the original computations above, we have

sin(arcsin 2

3+ arccos 1

2

)= 1

3+√32

cos(arcsin 2

3

)= 1

3+√32·√53

= 2+√15

6. 2

Example 3.7.6. Simplify: sin(2 cos−1

(−4

5

)).

Solution. Let θ = cos−1(−4

5

). Then cos θ = −4

5. Because cos θ < 0 and range

of inverse cosine function is [0, π], we know that θ must be within the interval(π2, π]. Using the Pythagorean Identity, we get sin θ = 3

5.

Using the Sine Double-Angle Identity, we have

sin(2 cos−1

(−4

5

))= sin 2θ

= 2 sin θ cos θ

= 2 · 35

(−4

5

)= −24

25. 2

Example 3.7.7. Sketch the graph of y = 14

cos−1(2x).

Solution.

y =1

4cos−1(2x) ⇐⇒ 4y = cos−1(2x) ⇐⇒ x =

1

2cos(4y)



DEPED COPY

We graph first y = 12

cos(4x). The domain of this graph comes from the restrictionof cosine as follows:

0 ≤ 4x ≤ π =⇒ 0 ≤ x ≤ π

4.

Then reflect this graph with respect to y = x, and we finally obtain the graph ofy = 1

4cos−1(2x) (solid line).

In the last example, we may also use the following technique. In graphingy = 1

4cos−1(2x), the horizontal length of cos−1 x is reduced to half, while the

vertical height is reduced to quarter. This comparison technique is shown inthe graph below with the graph of y = cos−1 x in dashed line and the graph ofy = 1

4cos−1(2x) in solid line.

3.7.3. Inverse Tangent Function and the Remaining InverseTrigonometric Functions

The inverse tangent function is similarly defined as inverse sine and inverse cosinefunctions.



DEPED COPY

y = tan−1 x or y = arctanx

meanstan y = x,

where x ∈ R and −π2< y < π

2.

The graph (solid thick curve) of the restricted function y = tan x is shownin Figure 3.35(a), while the graph of inverse function y = arctanx is shown inFigure 3.35(b).

(a) y = tanx (b) y = tan−1 x

Figure 3.35

Example 3.7.8. Find the exact value of each expression.(1) tan−1 1

(2) arctan(−√

3)

(3) tan(tan−1

(−5

2

))(4) tan−1

(tan(−π

6

))(5) tan−1

(tan 7π

6

)(6) arctan

(tan(−19π

6

))Solution. Note the range of arctan is the open interval

(−π

2, π2

).

(1) tan−1 1 = π4

(2) arctan(−√

3)

= −π3

(3) tan(tan−1

(−5

2

))= −5

2

(4) tan−1(tan(−π

6

))= −π

6because −π

6∈(−π

2, π2

).



DEPED COPY

(5) Here, note that 7π66∈(−π

2, π2

). Use the idea of reference angle, we know that

tan 7π6

= tan π6.

tan−1(

tan7π

6

)= tan−1

(tan

π

6

)=π

6

(6) Here, we cannot use the idea of reference angle, but the idea can help in away. The number (or angle) −19π

6is in QII, wherein tangent is negative, and

its reference angle is π6.

arctan

(tan

(−19π

6

))= arctan

(tan(−π

6

))= −π

62

Example 3.7.9. Find the exact value of each expression.(1) sin

(2 tan−1

(−8

3

))(2) tan

(sin−1 3

5− tan−1 1

4

)Solution. (1) Let θ = tan−1

(−8

3

). Then tan θ = −8

3. Following the notations in

Lesson 3.2 and the definition of inverse tangent function, we know that θ liesin QIV, and x = 3 and y = −8. We get r =

√32 + (−8)2 =

√73.

Applying the Sine Double-Angle Identity (page 192) gives

sin

(2 tan−1

(−8

3

))= sin 2θ

= 2 sin θ cos θ

= 2 · yr· xr

= 2

(− 8√

73

)(3√73

)= −48

73.

(2) Using the Tangent Difference Identity, we obtain

tan

(sin−1

3

5− tan−1

1

4

)=

tan(sin−1 3

5

)− tan

(tan−1 1

4

)1 + tan

(sin−1 3

5

)tan(tan−1 1

4

)=

tan(sin−1 3

5

)− 1

4

1 + tan(sin−1 3

5

)· 14

.

We are left to compute tan(sin−1 3

5

). We proceed as in (1) above. Let

θ = sin−1 35. Then sin θ = 3

5. From the definition of inverse sine function and



DEPED COPY

the notations used in Lesson 3.2, we know that θ lies in QI, and y = 3 andr = 5. We get x =

√52 − 32 = 4, so that tan θ = y

x= 3

4.

tan

(sin−1

3

5− tan−1

1

4

)=

tan(sin−1 3

5

)− 1

4

1 + tan(sin−1 3

5

)· 14

=34− 1

4

1 + 34· 14

=8

192

?Example 3.7.10. A student isviewing a painting in a museum.Standing 6 ft from the painting,the eye level of the student is 5 ftabove the ground. If the paint-ing is 10 ft tall, and its base is4 ft above the ground, find theviewing angle subtended by thepainting at the eyes of the stu-dent.

Solution. Let θ be the viewing angle, and let θ = α + β as shown below.

We observe that

tanα =1

6and tan β =

9

6.

Using the Tangent Sum Identity, we have

tan θ = tan(α + β) =tanα + tan β

1− tanα tan β

=16

+ 96

1− 16· 96

=20

9.

Using a calculator, the viewing angle is θ = tan−1 209≈ 65.8◦. 2

We now define the remaining inverse trigonometric functions.



DEPED COPY

Definecot−1 x =

π

2− tan−1 x.

It follows that the domain of y = cot−1 x is R and its range is (0, π).

y = sec−1 x or y = arcsecx

meanssec y = x,

where |x| ≥ 1 and y ∈[0, π

2

)∪[π, 3π

2

).

Definecsc−1 x =

π

2− sec−1 x.

This means that the domain of y = csc−1 x is (−∞,−1] ∪ [1,∞) andits range is

(−π,−π

2

]∪(0, π

2

].

The graphs of these last three inverse trigonometric functions are shown inFigures 3.36, 3.37, and 3.38, respectively.

(a) y = cotx (b) y = cot−1 x

Figure 3.36



DEPED COPY(a) y = secx (b) y = sec−1 x

Figure 3.37

(a) y = cscx (b) y = csc−1 x

Figure 3.38

Observe that the process in getting the value of an inverse function is thesame to all inverse functions. That is, y = f−1(x) is the same as f(y) = x. Weneed to remember the range of each inverse trigonometric function. Table 3.39summarizes all the information about the six inverse trigonometric functions.



DEPED COPY

Function Domain Range Graph

sin−1 x [−1, 1][−π

2, π2

] Figure3.33(b)

cos−1 x [−1, 1] [0, π]Figure3.34(b)

tan−1 x R(−π

2, π2

) Figure3.35(b)

cot−1 x R (0, π)Figure3.36(b)

sec−1 x {x : |x| ≥ 1}[0, π

2

)∪[π, 3π

2

) Figure3.37(b)

csc−1 x {x : |x| ≥ 1}(−π,−π

2

]∪(0, π

2

] Figure3.38(b)

Table 3.39

Example 3.7.11. Find the exact value of each expression.

(1) sec−1(−2)

(2) csc−1(−2√3

3

) (3) cot−1(−√

3)

(4) sin(

sec−1(−3

2

)− csc−1

(−2√3

3

))Solution. (1) sec−1(−2) = 4π

3because sec 4π

3= −2 and 4π

3∈[π, 3π

2

)(2) csc−1

(−2√3

3

)= −2π

3

(3) cot−1(−√

3)

= 5π6

(4) From (2), we know that csc−1(−2√3

3

)= −2π

3. Let θ = sec−1

(−3

2

). Then

sec θ = −32. From defined range of inverse secant function and the notations

in Lesson 3.2, θ lies in QIII, and r = 3 and x = −2. Solving for y, we gety = −

√32 − (−2)2 = −

√5. It follows that sin θ = −

√53

and cos θ = −23.

We now use the Sine Sum Identity.

sin

(sec−1

(−3

2

)− csc−1

(−2√

3

3

))

= sin

(θ −

(−2π

3

))= sin

(θ +

2π

3

)214


DEPED COPY

= sin θ cos2π

3+ cos θ sin

2π

3

=

(−√

5

3

)(−1

2

)+

(−2

3

)(√3

2

)

=

√5− 2

√3

62


1. Find the exact values of the following, if they exist.

(a) sin−1√22

(b) arcsin(−1

2

)(c) sin−1 2

Solution: Note that the range of f(x) = sin−1 x is [−π2, π2]. Thus, if we let

y = sin−1 x, then we are looking for y ∈ [−π2, π2] such that sin y = x. Hence,

(a) sin−1√22

= π4,

(b) arcsin(−1

2

)= −π

6, and

(c) sin−1 2 is undefined because sin y ≤ 1.

2. Find the exact value of each expression.

(a) sin(

sin−1√22

)(b) cos

[arcsin

(−1

2

)](c) sin−1

(sin 11π

2

)Solution:

(a) sin(

sin−1√22

)= sin(π

4) =

√22

(b) cos[arcsin

(−1

2

)]= cos(−π

6) =

√32

(c) sin−1(sin 11π

2

)= sin−1(−1) = −π

2

3. Answer the following.

(a) What is the domain of y = sin−1 2x?

(b) What is the range of y = sin−1 2x?

(c) What is the x−intercept of y = sin−1 2x?

Solution:

(a) Consider the function f(θ) = sin−1 θ. The domain of sin−1 θ is [−1, 1].So, θ = 2x ∈ [−1, 1]. Therefore, the domain of sin−1 2x is [−1/2, 1/2].



DEPED COPY

(b) [−π/2, π/2]

(c) (0, 0)

4. From the concept of projectile motion, if an object is directed at an angle θ

(with θ ∈ [0, π/2]), then the range will be R =v20g

sin 2θ (in feet) where v0 (in

ft/s) is the initial speed and g = 32 ft/s2 is the acceleration due to gravity.At what angle shall the object be directed so that the range will be 100 ft,given that the initial speed is v0 = 80 ft/s?

Solution: From the formula of the range, we get

100 =802

32sin 2θ =⇒ 1

2= sin 2θ

Since θ must be from 0 to π2

(i.e. 0 ≤ 2θ ≤ π), this is equivalent to finding 2θsuch that 2θ = sin−1 1

2. Hence,

2θ =π

6=⇒ θ =

π

12.

Therefore, the object must be directed at an angle of π12rad (or 15◦), to have

a projectile range of 100 ft.


(a) cos−1√22

(b) cos(cos−1

(−1

2

)) (c) arccos(cos π)

(d) arccos π

Solution:

(a) cos−1√22

= π4

(b) cos(cos−1

(−1

2

))= cos 2π

3= −1

2

(c) arccos(cos π) = arccos(−1) = π

(d) Let y = arccosπ. Since cos y ≤ 1, we have y is undefined because π > 3.

6. Simplify: (a) cos(

cos−1√32− cos−1 1

3

)Solution: We know that cos−1

√32

= π6. Let θ = cos−1 1

3. Which is equivalent

to cos θ = 13

with 0 ≤ θ ≤ π. Using the Cosine Difference Identity, we have

cos

(cos−1

√3

2− cos−1

1

3

)= cos

(π6− θ)

= cosπ

6cos θ + sin

π

6sin θ

=

√3

2· 1

3+

1

2· sin θ.



DEPED COPY

Now, we solve for sin θ using the Pythagorean Identity which gives

sin2 θ = 1− (1/3)2 = 1− (1/9) = 8/9.

Thus, sin θ = 2√2

3because θ ∈ [0, π]. Finally, we obtain

cos

(cos−1

√3

2− cos−1

1

3

)=

√3

2· 1

3+

1

2· 2√

2

3

=

√3

6+

2√

2

6

=

√3 + 2

√2

6.

7. Simplify: (a) cos(2 cos−1 2

5

); (b) sin

(cos−1 2

5

)Solution: Let θ = cos−1 2

5. Which is equivalent to cos θ = 2

5with 0 ≤ θ ≤

π. Using the Double-Angle Identity for Cosine and one of the FundamentalIdenity, we have

cos

(2 cos−1

2

5

)= cos(2θ) = 2 cos2 θ − 1 = 2

(2

5

)2

− 1 =8

25− 1 = −17

25

and

sin

(cos−1

2

5

)= sin θ =

√1− cos2 θ =

√1− 4

25=

√21

5.

Here, sin θ ≥ 0 because θ ∈ [0, π].

8. Graph: y = 1 + cos−1 x

Solution: The graph can be obtained by translating the graph of the inversecosine function one unit upward.



DEPED COPY

9. Find the exact value of each expression.

(a) arctan(tan 4π3

) (b) tan(tan−1 45)

Solution:

(a) arctan(tan 4π3

) = arctan√

3 = π/3

(b) tan(tan−1 45) = 4

5

10. Find the exact value of tan(tan−1 76

+ tan−1 12).

Solution:

tan

(tan−1

7

6+ tan−1

1

2

)=

tan(tan−1 76) + tan(tan−1 1

2)

1− tan(tan−1 76) · tan(tan−1 1

2)

=76

+ 12

1− 76· 12

= 4


(a) sec−1√

2

(b) csc−1 1

(c) cot−1√33

(d) arcsec−1(cot(−π4))

(e) cos(arccsc−1 2)

(f) arccot−1(sin 20π3

)

Solution:

(a) sec−1√

2 = π4

(b) csc−1 1 = π2

(c) cot−1√33

= π3

(d) arcsec(cot(−π4)) = arcsec(−1) = −π

(e) cos(arccsc(2)) = cos π6

=√32

(f) arccot(sin 20π3

) = arccot√32

. Let θ = arccot√32

. Then,

cot θ =

√3

2=⇒ tan θ =

2√3

=⇒ θ = tan−12√3

(≈ 0.8571).

Here, we needed to use a calculator to solve for the approximate value,since 2√

3is not a special value for tangent function.



DEPED COPY


1. Find the exact value of the following.

(a) sin[sin−1(1/2)]

(b) cos[cos−1(−√

2/2)]

(c) tan[tan−1(−√

3)]

(d) sin[arctan(√

3)]

(e) cos[arccos(√

2)]

(f) tan[arcsin(1/4)]

(g) cos[sin−1(√

3/2)]


(a) sin−1[sin(25π/6)]

(b) arccos[cos(23π/4)]

(c) tan−1[tan(−1)]

(d) arcsin[cos(13π/4)]

(e) cos−1[sec(23π)]

(f) arctan[sin(−π/12)]

3. Solve the exact value of the following.

(a) sin[2 cos−1(−4/5)]

(b) cos[2 sin−1(5/13)]

(c) sin(sin−1(3/5) + cos−1(−5/13))

(d) cos[sin−1(1/2)− cos−1(8/17)]

4. Consider the function f(x) = tan−1(x+ 1). Do the following.

(a) Find the domain of f .

(b) Find the range of f .

(c) Find the x− and y−intercept of f , if there are any.

(d) Graph f .

5. Evaluate and simplify the following, if they exist.

(a) arcsec(−√

2)

(b) arccsc(−2)

(c) arccot√

3

(d) [sec−1(−1)] · [cos−1(−1)]

(e) 2 cot−1√

3 + 3 csc−1 2

(f) csc−1 0

6. Evaluate and simplify the following, if they exist.

(a)cos(sec−1 3 + tan−1 2)

cos(tan−1 2)

(b) tan(2 arcsin(1/6))

(c) cos2(

sin−1(1/2)2

)(d) arcsec(sin(100π/3))



DEPED COPY

7. A trough is in the shape of an inverted triangular prism whose cross sectionhas the shape of an inverted isosceles triangle (see Figure 3.40). If the lengthof the base of the cross section is 2

√3 m. and the length of the trough is

100√

3 m., find the size of the vertex angle so that the volume is 900 m3.Hint: V = bhl/2.

Figure 3.40

4

Lesson 3.8. Trigonometric Equations



(1) solve trigonometric equations; and

(2) solve situational problems involving trigonometric equations.

Lesson Outline

(1) Definition of a trigonometric equation

(2) Solution to a trigonometric equation

(3) Techniques of solving a trigonometric equation

Introduction

We have studied equations in Lesson 3.4. We differentiated an identity froma conditional equation. Recall that an identity is an equation that is true for all



DEPED COPY

values of the variable in the domain of the equation, while a conditional equationis an equation that is not an identity.

In this lesson, we mostly study conditional trigonometric equations. Thoughnot explicitly, we have started it in the preceding lesson. For example, the equa-tion sinx = 1

2has the unique solution x = sin−1 1

2= π

6in the closed interval[

−π2, π2

]. However, if we consider the entire domain (not the restricted domain)

of the sine function, which is the set R of real numbers, there are solutions (otherthan π

6) of the equation sinx = 1

2. This current lesson explores the techniques of

solving (conditional) trigonometric equations.

We divide the lesson into two groups of equations: the ones using a basic wayof solving, and those using more advanced techniques.

3.8.1. Solutions of a Trigonometric Equation

Any equation that involves trigonometric expressions is called a trigonometricequation. Recall that a solution or a root of an equation is a number in the domainof the equation that, when substituted to the variable, makes the equation true.The set of all solutions of an equation is called the solution set of the equation.

Technically, the basic method to show that a particular number is a solutionof an equation is to substitute the number to the variable and see if the equationbecomes true. However, we may use our knowledge gained from the previouslessons to do a quicker verification process by not doing the manual substitutionand checking. We use this technique in the example.

Example 3.8.1. Which numbers in the set{

0, π6, π4, π3, π2, 2π

3, 3π

4, 5π

6, π, 2π

}are

solutions to the following equations?

(1) sinx = 12

(2) tanx = 1

(3) 3 secx = −2√

3

(4)√

3| cotx| = 1

(5) sec2 x− tan2 x = 1

(6) sinx+ cosx = 0

(7) cos2 x = cos 2x+ sin2 x

(8) sinx+ cos 2x = 0

(9) 2 sinx+ tanx− 2 cosx = 2

(10) sin2 x+ cos2 x = 2

(11) sin 2x = sinx

(12) 2 tanx+ 4 sinx = 2 + secx

Solution. Note that the choices (except 2π) are numbers within the interval [0, π].To quickly determine which numbers among the choices are solutions to a par-ticular equation, we use some distinctive properties of the possible solutions.

(1) The sine function is positive on (0, π). From Lesson 3.2, we recall that π6

isan obvious solution. We may imagine the graph of y = sin x. We may alsouse the idea of reference angle. Thus, among the choices, only π

6and 5π

6are

the only solutions of sin x = 12.



DEPED COPY

(2) Since tan x = 1 > 0, any solution of the equation among the choices mustbe in the interval

(0, π

2

)(that is, in QI). Again, among the choices, the only

solution to tanx = 1 is π4.

(3) Here, the given equation is equivalent to secx = −2√3

3. Among the choices,

the only solution of the equation 3 sec x = −2√

3 is 5π6

.

(4) Eliminating the absolute value sign, the given equation is equivalent to cot x =√33

or cotx = −√33

. Among the choices, the only solution of cot x =√33

is π3,

while the other equation has 2π3

. Thus, the only solutions of√

3| cotx| = 1from the given set are π

3and 2π

3.

(5) The given equation is one of the Pythagorean Identities (page 175). It meansthat all numbers in the domain of the equation are solutions. The domainof the equation is R \ {x : cosx = 0}. Thus, all except π

2are solutions of

sec2 x− tan2 x = 1.

(6) For the sum of sinx and cos x to be 0, they must have equal absolute valuesbut different signs. Among the choices, only 3π

4satisfies these properties, and

it is the only solution of sinx+ cosx = 0.

(7) This equations is one of the Double-Angle Identities for Cosine. This meansthat all numbers in the domain of the equation are its solutions. Because thedomain of the given equation is R, all numbers in the given set are solutionsof cos2 x = cos 2x+ sin2 x.

(8) We substitute each number in the choices to the expression on the left-sideof the equation, and select those numbers that give resulting values equal to1.

x = 0: sin 0 + cos 2(0) = 0 + 1 = 1

x = π6: sin π

6+ cos 2(π

6) = 1

2+ 1

2= 1

x = π4: sin π

4+ cos 2(π

4) =

√22

+ 0 =√22

x = π3: sin π

3+ cos 2(π

3) =

√32− 1

2=√3−12

x = π2: sin π

2+ cos 2(π

2) = 1− 1 = 0

x = 2π3

: sin 2π3

+ cos 2(2π3

) =√32− 1

2=√3−12

x = 3π4

: sin 3π4

+ cos 2(3π4

) =√22

+ 0 =√22

x = 5π6

: sin 5π6

+ cos 2(5π6

) = 12

+ 12

= 1

x = π: sin π + cos 2π = 0 + 1 = 1

x = 2π: sin 2π + cos 2(2π) = 0 + 1 = 1



DEPED COPY

From these values, the only solution of sin x+ cos 2x = 0 among the choicesis π

2.

(9) We again substitute the numbers in the given set one by one, and see whichresulting values are equal to 1.

x = 0: 2 sin 0 + tan 0− 2 cos 0 = −2

x = π6: 2 sin π

6+ tan π

6− 2 cos π

6= 3−2

√3

3

x = π4: 2 sin π

4+ tan π

4− 2 cos π

4= 1

x = π3: 2 sin π

3+ tan π

3− 2 cos π

3= 2√

3− 1

x = π2: Since tan π

2is undefined, this value of x cannot be a solution of the

equation.

x = 2π3

: 2 sin 2π3

+ tan 2π3− 2 cos 2π

3= 1

x = 3π4

: 2 sin 3π4


4= 2√

2− 1

x = 5π6

: 2 sin 5π6


6= 3+2

√3

3

x = π: 2 sinπ + tan π − 2 cosπ = 2

x = 2π: 2 sin 2π + tan 2π − 2 cos 2π = −2

Thus, the only solution of 2 sin x+ tanx−2 cosx = 2 from the given set is π.

(10) This equation has no solution because one of the Pythagorean Identities sayssin2 x+ cos2 x = 1.

(11) We substitute each number in the given set to the expression of each side ofthe equation, and see which resulting values are equal.

x = 0: sin 2(0) = 0; sin 0 = 0

x = π6: sin 2(π

6) =

√32

; sin π6

= 12

x = π4: sin 2(π

4) = 1; sin π

4=√22

x = π3: sin 2(π

3) =

√32

; sin π3

=√32

x = π2: sin 2(π

2) = 0; sin π

2= 1

x = 3π4

: sin 2(3π4

) = −1; sin 3π4

=√22

x = 5π6

: sin 2(5π6

) = −√32

; sin π3

= 12

x = π: sin 2π = 0; sin π = 0

x = 2π: sin 2(2π) = 0; sin 2π = 0

Thus, among the numbers in the given set, the solutions of sin 2x = sinx are0, π

3, π, and 2π.



DEPED COPY

(12) We employ the same technique used in the previous item.

x = 0 : 2 tan 0 + 4 sin 0 = 0

2 + sec 0 = 3

x = π6

: 2 tan π6

+ 4 sin π6

= 2√3+63

2 + sec π6

= 2√3+63

x = π4

: 2 tan π4

+ 4 sin π4

= 2√

2 + 2

2 + sec π4

=√

2 + 2

x = π3

: 2 tan π3

+ 4 sin π3

= 4√

3

2 + sec π3

= 4

x = π2

: Both tan π2

and sec π2

are undefined.

x = 2π3

: 2 tan 2π3

+ 4 sin 2π3

= 0

2 + sec 2π3

= 0

x = 3π4

: 2 tan 3π4

+ 4 sin 3π4

= 2√

2− 2

2 + sec 3π4

= 2−√

2

x = 5π6

: 2 tan 5π6

+ 4 sin 5π6

= 6−2√3

3

2 + sec 5π6

= 6−2√3

3

x = π : 2 tan π + 4 sinπ = 0

2 + sec π = 1

x = 2π : 2 tan 2π + 4 sin 2π = 0

2 + sec 2π = 3

After checking the equal values, the solutions of 2 tan x + 4 sin x = 2 + sec xamong the given choices are π

6, 2π

3, and 5π

6. 2

3.8.2. Equations with One Term

From the preceding discussion, you may observe that there may be more solutionsof a given equation outside the given set. We now find all solutions of a givenequation.

We will start with a group of equations having straightforward techniquesin finding their solutions. These simple techniques involve at least one of thefollowing ideas:



DEPED COPY

(1) equivalent equations (that is, equations that have the same solutions as theoriginal equation);

(2) periodicity of the trigonometric function involved;

(3) inverse trigonometric function;

(4) values of the trigonometric function involved on the interval [0, π] or [0, 2π](depending on the periodicity of the function); and

(5) Zero-Factor Law: ab = 0 if and only if a = 0 or b = 0.

To “solve an equation” means to find all solutions of the equation. Here,unless stated as angles measured in degrees, we mean solutions of the equationthat are real numbers (or equivalently, angles measured in radians).

Example 3.8.2. Solve the equation 2 cos x− 1 = 0.

Solution. The given equation is equivalent to

cosx =1

2.

On the interval [0, 2π], there are only two solutions of the last equation, and theseare x = π

3(this is in QI) and x = 5π

3(in QIV).

Because the period of cosine function is 2π, the complete solutions of theequation are x = π

3+ k(2π) and x = 5π

3+ k(2π) for all integers k. 2

In the preceding example, by saying that the “complete solutions are x =π3

+ k(2π) and x = 5π3

+ k(2π) for all integers k,” we mean that any integralvalue of k will produce a solution to the given equation. For example, whenk = 3, x = π

3+ 3(2π) = 19π

3is a solution of the equation. When k = −2,

x = 5π3

+ (−2)(2π) = −7π3

is another solution of 2 cosx − 1 = 0. The family ofsolutions x = π

3+ k(2π) can be equivalently enumerated as x = 19π

3+ 2kπ, while

the family x = 5π3

+ k(2π) can also be stated as x = −7π3

+ 2kπ.

Example 3.8.3. Solve: (1 + cos θ)(tan θ − 1) = 0.

Solution. By the Zero-Factor Law, the given equation is equivalent to

1 + cos θ = 0 or tan θ − 1 = 0

cos θ = −1 tan θ = 1

θ = π + 2kπ, k ∈ Z θ = π4

+ kπ, k ∈ Z.

Therefore, the solutions of the equation are θ = π + 2kπ and θ = π4

+ kπ for allk ∈ Z. 2



DEPED COPY

Example 3.8.4. Find all values of x in the interval [−2π, 2π] that satisfy theequation (sin x− 1)(sinx+ 1) = 0.

Solution.

sinx− 1 = 0 or sin x+ 1 = 0

sinx = 1 sinx = −1

x = π2

or − 3π2

x = 3π2

or − π2

Solutions: π2, −3π

2, 3π

2, −π

22

Example 3.8.5. Solve: cosx = 0.1.

Solution. There is no special number whose cosine is 0.1. However, because0.1 ∈ [−1, 1], there is a number whose cosine is 0.1. In fact, in any one-periodinterval, with cos x = 0.1 > 0, we expect two solutions: one in QI and another inQIV. We use the inverse cosine function.

From Lesson 3.7, one particular solution of cosx = 0.1 in QI is x = cos−1 0.1.We can use this solution to get a particular solution in QIV, and this is x =2π − cos−1 0.1, which is equivalent to x = − cos−1 0.1.

From the above particular solutions, we can produce all solutions of cosx =0.1, and these are x = cos−1 0.1+2kπ and x = − cos−1 0.1+2kπ for all k ∈ Z. 2

Example 3.8.6. Solve: 3 tan θ + 5 = 0.

Solution.3 tan θ + 5 = 0 =⇒ tan θ = −5

3

We expect only one solution in any one-period interval.

tan θ = −53

=⇒ θ = tan−1(−5

3

)+ kπ, k ∈ Z 2

?Example 3.8.7. The voltage V (in volts) coming from an electricity distribut-ing company is fluctuating according to the function V (t) = 200 + 170 sin(120πt)at time t in seconds.

(1) Determine the first time it takes to reach 300 volts.

(2) For what values of t does the voltage reach its maximum value?

Solution. (1) We solve for the least positive value of t such that V (t) = 300.

200 + 170 sin(120πt) = 300

sin(120πt) =100

170



DEPED COPY

120πt = sin−1100

170

t =sin−1 100

170

120π≈ 0.00167 seconds

(2) The maximum value of V (t) happens when and only when the maximumvalue of sin(120πt) is reached. We know that the maximum value of sin(120πt)is 1, and it follows that the maximum value of V (t) is 370 volts. Thus, weneed to solve for all values of t such that sin(120πt) = 1.

sin(120πt) = 1

120πt =π

2+ 2kπ, k nonnegative integer

t =π2

+ 2kπ

120π

t =12

+ 2k

120≈ 0.00417 + 0.017k

This means that the voltage is maximum when t ≈ 0.00417 + 0.017k for eachnonnegative integer k. 2

3.8.3. Equations with Two or More Terms

We will now consider a group of equations having multi-step techniques of findingtheir solutions. Coupled with the straightforward techniques we learned in thepreceding discussion, these more advanced techniques involve factoring of expres-sions and trigonometric identities. The primary goal is to reduce a given equationinto equivalent one-term equations.

Example 3.8.8. Solve: 2 cos x tanx = 2 cos x.

Solution.

2 cosx tanx = 2 cos x

2 cosx tanx− 2 cosx = 0

(2 cosx)(tanx− 1) = 0

2 cosx = 0 or tanx− 1 = 0

cosx = 0 tan x = 1

x = π2

+ 2kπ orx = 3π

2+ 2kπ,

k ∈ Z

x = π4

+ kπ,k ∈ Z

Solutions: π2

+ 2kπ, 3π2

+ 2kπ, π4

+ kπ, k ∈ Z 2



DEPED COPY

Example 3.8.9. Solve for x ∈ [0, 2π): sin 2x = sinx.

Solution.

sin 2x = sinx

sin 2x− sinx = 0

2 sinx cosx− sinx = 0 Sine Double-Angle Identity

(sinx)(2 cosx− 1) = 0

sinx = 0 or 2 cosx− 1 = 0

x = 0 or x = π cosx = 12

x = π3

or x = 5π3

Solutions: 0, π, π3, 5π

32

Tips in Solving Trigonometric Equations

(1) If the equation contains only one trigonometric term, isolate thatterm, and solve for the variable.

(2) If the equation is quadratic in form, we may use factoring, findingsquare roots, or the quadratic formula.

(3) Rewrite the equation to have 0 on one side, and then factor (ifappropriate) the expression on the other side.

(4) If the equation contains more than one trigonometric function,try to express everything in terms of one trigonometric function.Here, identities are useful.

(5) If half or multiple angles are present, express them in terms of atrigonometric expression of a single angle, except when all anglesinvolved have the same multiplicity wherein, in this case, retainthe angle. Half-angle and double-angle identities are useful insimplification.

Example 3.8.10. Solve for x ∈ [0, 2π): 2 cos2 x = 1 + sinx.

Solution.

2 cos2 x = 1 + sinx

2(1− sin2 x) = 1 + sinx Pythagorean Identity



DEPED COPY

2 sin2 x+ sinx− 1 = 0

(2 sinx− 1)(sinx+ 1) = 0 Factoring

2 sinx− 1 = 0 or sin x+ 1 = 0

sinx = 12

sinx = −1

x = π6

or x = 5π6

x = 3π2

Solutions: π6, 5π

6, 3π

22

Example 3.8.11. Solve for x ∈ [0, 2π) in the equation 3 cos2 x+ 2 sinx = 2.

Solution.

3 cos2 x+ 2 sinx = 2

3(1− sin2 x) + 2 sinx = 2 Pythagorean Identity

(3 sinx+ 1)(sinx− 1) = 0 Factoring

3 sinx+ 1 = 0 or sin x− 1 = 0

sinx = −13

sinx = 1

x = sin−1(−13) + 2π

orx = π − sin−1(−1

3)

x = π2

Solutions: 2π − sin−1(13)+, π + sin−1(1

3), π

22

One part of the last solution needs further explanation. In the equationsinx = −1

3, we expect two solutions in the interval [0, 2π): one in (π, 3π

2) (which

is QIII), and another in (3π2, 2π) (which is QIV). Since no special number satisfies

sinx = −13, we use inverse sine function. Because the range of sin−1 is [−π

2, π2], we

know that −π2< sin−1(−1

3) < 0. From this value, to get the solution in (3π

2, 2π),

we simply add 2π to this value, resulting to x = sin−1(−13) + 2π. On the other

hand, to get the solution in (π, 3π2

), we simply add − sin−1(−13) to π, resulting to

x = π − sin−1(−13).

Example 3.8.12. Solve: sin2 x+ 5 cos2 x2

= 2.

Solution.

sin2 x+ 5 cos2 x2

= 2

sin2 x+ 5(1+cosx

2

)= 2 Cosine Half-Angle Identity

2 sin2 x+ 5 cosx+ 1 = 0



DEPED COPY

2(1− cos2 x) + 5 cosx+ 1 = 0 Pythagorean Identity

2 cos2 x− 5 cosx− 3 = 0

(2 cosx+ 1)(cosx− 3) = 0

2 cosx+ 1 = 0 or cos x− 3 = 0

cosx = −12

cosx = 3

x = 2π3

+ 2kπ orx = 4π

3+ 2kπ,

k ∈ Z

no solution

Solutions: 2π3

+ 2kπ, 4π3

+ 2kπ, k ∈ Z 2

Example 3.8.13. Solve for x ∈ [0, 2π) in the equation tan 2x− 2 cosx = 0.

Solution.

tan 2x− 2 cosx = 0

sin 2x

cos 2x− 2 cosx = 0

sin 2x− 2 cosx cos 2x = 0

Apply the Double-Angle Identities for Sine and Cosine, and then factor.

2 sinx cosx− 2(cosx)(1− 2 sin2 x) = 0

(2 cosx)(2 sin2 x+ sin x− 1) = 0

(2 cosx)(2 sinx− 1)(sinx+ 1) = 0

2 cosx = 0 or 2 sinx− 1 = 0 or sinx+ 1 = 0

cosx = 0 sinx = 12

sinx = −1

x = π2

orx = 3π

2

x = π6

orx = 5π

6

x = 3π2

These values of x should be checked in the original equation because tan 2x maynot be defined. Upon checking, this is not the case for each value of x obtained.The solutions are π

2, 3π

2, π

6, 5π

6, and 3π

2. 2

?Example 3.8.14. A weight is suspended from a spring and vibrating verticallyaccording to the equation

f(t) = 20 cos(45π(t− 5

6

)),

where f(t) centimeters is the directed distance of the weight from its centralposition at t seconds, and the positive distance means above its central position.



DEPED COPY

(1) At what time is the displacement of the weight 5 cm below its centralposition for the first time?

(2) For what values of t does the weight reach its farthest point below its centralposition?

Solution. (1) We find the least positive value of t such that f(t) = −5.

20 cos(45π(t− 5

6

))= −5

cos(45π(t− 5

6

))= −1

4

There are two families of solutions for this equation.

• 45π(t− 5

6

)= cos−1

(−1

4

)+ 2kπ, k ∈ Z

t = 56

+cos−1(− 1

4)+2kπ

45π

In this family of solutions, the least positive value of t happens whenk = 0, and this is

t =5

6+

cos−1(−1

4

)+ 2(0)π

45π

≈ 1.5589.

• 45π(t− 5

6

)= 2π − cos−1

(−1

4

)+ 2kπ, k ∈ Z

t = 56

+2π−cos−1(− 1

4)+2kπ

45π

Here, the least positive value of t happens when k = −1, and this is

t =5

6+

2π − cos−1(−1

4

)+ 2(−1)π

45π

≈ 0.1078.

Therefore, the first time that the displacement of the weight is 5 cm belowits central position is at about 0.1078 seconds.

(2) The minimum value of f(t) happens when and only when the minimumvalue of cos 4

5π(t− 5

6

)is reached. The minimum value of cos 4

5π(t− 5

6

)is

−1, which implies that the farthest point the weight can reach below itscentral position is 20 cm. Thus, we need to solve for all values of t such thatcos 4

5π(t− 5

6

)= −1.

cos 45π(t− 5

6

)= −1

45π(t− 5

6

)= cos−1(−1) + 2kπ, k ≥ 0

45π(t− 5

6

)= π + 2kπ

t = 56

+ π+2kπ45π

= 2512

+ 52k

Therefore, the weight reaches its farthest point (which is 20 cm) below itscentral position at t = 25

12+ 5

2k for every integer k ≥ 0. 2



DEPED COPY


1. Give a particular solution of the following equation.

(a) sin2 x− 1 = 0

(b) cot x =√

3

(c) sec 3x = −1

(d) csc2 x− cscx− 2 = 0

(e) cos2 2x = sin2 x

(f) 2 cosx− 3 = 0

Solution:

(a) x = −π2

is a solution because sin2(−π2)− 1 = (−1)2 − 1 = 0.

(b) Note that cos π6

=√32

and sin π6

= 12. Thus, cot π

6=√

3. So, x = π6

is asolution.

(c) Since sec θ = −1 if and only if cos θ = −1, a particular solution of theequation in 3x is π, that is, 3x = π. Hence, x = π

3is a solution.

(d) Note that csc 3π2

= −1. So, csc2 3π2

= 1. As a consequence, csc2 3π2−

csc 3π2− 2 = 1− (−1)− 2 = 0.

(e) x = π2

is a solution.

(f) Because cosx must not be more than 1, then the equation has no solution.

2. What is the solution set of the following trigonometric equation sin2 x +cos2 x = 1?

Solution: The equation is the Pythagorean Identity, meaning any element ofthe domain of sin x and cosx satisfies the equation. The domain of both sinxand cosx is R. Therefore, the solution set of this trigonometric equation is R.One may try the numbers −π

6, 0, and π

4for illustration.

(a) x = −π6

sin2 x+ cos2 x = sin2(−π

6

)+ cos2

(−π

6

)=(

sin(−π

6

))2+(

cos(−π

6

))2=

(−1

2

)2

+

(√3

2

)2

=1

4+

3

4= 1.

(b) x = 0

sin2 x+ cos2 x = sin2 0 + cos2 0 = 02 + 12 = 0 + 1 = 1.

(c) x = π4

sin2 x+ cos2 x = sin2 π

4+ cos2

π

4

=

(√2

2

)2

+

(√2

2

)2

=1

4+

1

4= 1.



DEPED COPY

3. Find the solution set of the trigonometric equation tan2 x+ 1 = sec2 x.

Solution: Notice that this is a fundamental identity. Thus, the solution ofthis equation is any number common to the domain of the tangent and secantfunction. That is, the solution set is

{x ∈ R | cosx 6= 0} = {x ∈ R | x 6= ±π2,±3π

2,±5π

2, ...}

= {x ∈ R | x 6= (2k+1)π2

, k ∈ Z}.

4. Find all solutions of√

3 tanx+ 1 = 0.

Solution: The equation is equivalent to tan x = − 1√3. This is true only if

x = 5π6

+ kπ where k ∈ Z.

5. What are the solutions of√

3 tanx+ 1 = 0, where x ∈ [0, 2π].

Solution: The solutions are x = 5π6

and x = 11π6

.

6. Determine all solutions of 4 cos2 x− 1 = 0.

Solution: Note that the equation is quadratic in form, so we can apply tech-niques in solving quadratic equations. For this case, we factor the expressionon the left and obtain, (2 cosx − 1)(2 cosx + 1) = 0. Consequently, we havecosx = 1/2 or cosx = −1/2. The first equation have solutions of the form(π/3 + 2kπ) or (5π/3 + 2kπ) where k ∈ Z, while the second equation havesolutions of the form (2π/3 + 2kπ) or (4π/3 + 2kπ). Combining the two solu-tions, one observes that the solution set of the original equation may be givenby {

π

3,2π

3,4π

3,5π

3,7π

3, ...

}.

We can write this in a more compact form as{kπ

3: k 6= 3j, where j ∈ Z

}.

7. Find the solutions of 4 cos2 x− 1 = 0 within the closed interval [0, 2π].

Solution: Similar to Example 6, the solution of the above equation is{kπ

3: k 6= 3j, where j ∈ Z

}.

Since we are to find solutions in [0, 2π], we take k = 1, 2, 4, and 5 to obtainthe solutions π/3, 2π/3, 4π/3, and 5π/3.



DEPED COPY

8. If x ∈ [0, 2π), solve the equation 2 sin2 x =√

3 sinx.

Solution: First, we write the equation as 2 sin2 x −√

3 sinx = 0. Then, wefactor out sinx and get

sinx(2 sinx−√

3) = 0 =⇒ sinx = 0 or sinx =√

3/2.

The first of these equations has solutions x = 0 and x = π, while the secondhas solutions x = π/3 and 2π/3. The solutions of the original equation is theunion of the two, i.e., the solution set is {0, π, π

3, 2π

3}.

9. Solve 2 cos2 x+ 5 cosx− 3 = 0, where x ∈ [0, 2π).

Solution: By factoring the left hand side of the given equation, we get (2 cosx−1)(cosx+ 3) = 0. This gives us two equations, namely

cosx =1

2and cos x = −3.

First, we remark that the second equation does not have a solution becausecosx should be more than or equal to -1. Hence, the solution of the firstequation is the solution of the original equation. Thus, the solution set is{π3, 5π

3}.

10. Determine the solution set of the equation cos 2x = sinx on [0, 2π).

Solution: Combining the equation cos 2x = sinx with the cosine double-angleidentity cos 2x = 1− 2 sin2 x, we get

sinx = 1− 2 sin2 x.

This is equivalent to

2 sin2 x+ sinx− 1 = 0 =⇒ (2 sinx− 1)(sinx+ 1) = 0

=⇒ sinx = 1/2 and sinx = −1.

The solutions of the first equation is x = π/6 and x = 5π/6. The numberx = 3π/2 is the solution of the second equation. Therefore, the solution set ofthe original equation is {π

6, 5π

6, 3π

2}.

11. Solve cosx = cos 2x, for x ∈ [0, 2π).

Solution:

cosx = cos 2x

⇒ cosx = 2 cos2 x− 1

⇒ 0 = 2 cos2 x− cosx− 1

⇒ 0 = (2 cos x+ 1)(cosx− 1).



DEPED COPY

The given trigonometric equation is equivalent to solving 2 cos x + 1 = 0 andcosx = 1. For 2 cosx+ 1 = 0 which is the same as cosx = −1/2, the solutionsin the given interval are x = 2π/3, 4π/3. For cosx = 1, the solution is x = 0.Therefore, the solution set of the original equation is {0, 2π

3, 4π

3}.

12. A lighthouse at sea level is 34 mi from a boat. It is known that the top ofthe lighthouse is 42.5 mi from the boat and that x = r cos θ, where x is thehorizontal distance, r is the distance of the top of the lighthouse from theboat, and θ is the angle of depression from the top of lighthouse. Find θ.

Solution:

x = r cos θ =⇒ cos θ =x

r=

34

42.5=

4

5

=⇒ θ = cos−14

5≈ 0.6435 (or 36.87◦).

For this case, we used a calculator to find the value of the unknown variableθ since 4

5is not a special value for cosine.

13. Three cities, A,B, and C, are positioned in a triangle as seen in the figurebelow.

It is known that City A is 140 mi from City C, while City B is 210 mi fromCity C. Cities A and B are 70

√7 mi apart. Also, by the Cosine Law, we have

z2 = x2 + y2 − 2xy cos γ

where x, y, and z are the respective distances of BC,AC, AB, and γ =m∠ACB. Find γ.

Solution: Substituting the corresponding values of x, y, and z, the problem isnow equivalent to solving the equation

34300 = 44100 + 19600− 58800 cos γ

⇒ −29400 = −58800 cos γ

⇒ 12

= cos γ

⇒ π3

= γ.



DEPED COPY


1. Find all solutions of the equation 2 cos x− cosx sinx = 0.

2. Determine the solution set of the equation csc2 x+ 1 = 0.

3. What are the solutions of sec2 x+ secx− 2 = 0.

4. Find the solutions of the equation 4 sin2 x− 1 = 0 on [0, 2π).

5. Find the values of x ∈ [0, 2π) for which csc 2x =√

2.

6. What is the solution set of the equation sin θ = csc θ?

7. Solve t = sin−1(cos 2t).

8. Let x ∈ [0, 2π). Find the solutions of the equation cos2 4x+ sin2 2x = 1.

9. If a projectile, such as a bullet, is fired into the air with an initial velocity v atan angle of elevation θ, then the height h of the projectile at time t is given byh(t) = −16t2 + vt cos θ meters. If the initial velocity is 109 meters per second,at what angle should the bullet be fired so that its height is 45 meters abovethe floor in 2 seconds.

10. In a baseball field, a pitcher throws the ball at a speed of 60 km/h to thecatcher who is 100 m away. When the ball leaves a starting point at an angleof elevation of θ , the horizontal distance the ball travels is determined byd = v2

32sin θ, where d is measured in meters and velocity in kilometers per

hour. At what angle of elevation (in degrees) is the ball thrown?

4

Lesson 3.9. Polar Coordinate System



(1) locate points in polar coordinate system;

(2) convert the coordinates of a point from rectangular to polar system and viceversa; and

(3) solve situational problems involving polar coordinate system.



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Lesson Outline

(1) Polar coordinate system: pole and polar axis

(2) Polar coordinates of a point and its location

(3) Conversion from polar to rectangular coordinates, and vice versa

(4) Simple graphs and applications

Introduction

Two-dimensional coordinate systems are used to describe a point in a plane.We previously used the Cartesian or rectangular coordinate system to locate apoint in the plane. That point is denoted by (x, y), where x is the signed dis-tance of the point from the y-axis, and y is the signed distance of the pointfrom the x-axis. We sketched the graphs of equations (lines, circles, parabolas,ellipses, and hyperbolas) and functions (polynomial, rational, exponential, log-arithmic, trigonometric, and inverse trigonometric) in the Cartesian coordinateplane. However, it is often convenient to locate a point based on its distancefrom a fixed point and its angle with respect to a fixed ray. Not all equationscan be graphed easily using Cartesian coordinates. In this lesson, we also useanother coordinate system, which can be presented in dartboard-like plane asshown below.

3.9.1. Polar Coordinates of a Point

We now introduce the polar coordinate system. It is composed of a fixed pointcalled the pole (which is the origin in the Cartesian coordinate system) and a



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fixed ray called the polar axis (which is the nonnegative x-axis).

In the polar coordinate system, a point is described by the ordered pair (r, θ).The radial coordinate r refers to the directed distance of the point from the pole.The angular coordinate θ refers to a directed angle (usually in radians) from thepolar axis to the segment joining the point and the pole.

Because a point in polar coordinate system is described by an order pair ofradial coordinate and angular coordinate, it will be more convenient to geomet-rically present the system in a polar plane, which serves just like the Cartesianplane. In the polar plane shown below, instead of rectangular grids in the Carte-sian plane, we have concentric circles with common center at the pole to identifyeasily the distance from the pole (radial coordinate) and angular rays emanatingfrom the pole to show the angles from the polar axis (angular coordinate).



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Example 3.9.1. Plot the following points in one polar plane: A(3, π3), B(1, 5π

6),

C(2, 7π6

), D(4, 19π12

), E(3,−π), F (4,−7π6

), G(2.5, 17π4

), H(4, 17π6

), and I(3,−5π3

).

Solution.

As seen in the last example, unlike in Cartesian plane where a point has aunique Cartesian coordinate representation, a point in polar plane have infinitelymany polar coordinate representations. For example, the coordinates (3, 4) inthe Cartesian plane refer to exactly one point in the plane, and this particularpoint has no rectangular coordinate representations other than (3, 4). However,the coordinates (3, π

3) in the polar plane also refer to exactly one point, but

this point has other polar coordinate representations. For example, the polarcoordinates (3,−5π

3), (3, 7π

3), (3, 13π

3), and (3, 19π

3) all refer to the same point as

that of (3, π3).

The polar coordinates (r, θ + 2kπ), where k ∈ Z, represent the samepoint as that of (r, θ).

In polar coordinate system, it is possible for the coordinates (r, θ) to havea negative value of r. In this case, the point is |r| units from the pole in theopposite direction of the terminal side of θ, as shown in Figure 3.41.




Example 3.9.2. Plot the following points in one polar plane: A(−3, 4π3

), B(−4, 11π6

),C(−2,−π), and D(−3.5,−7π

4).

Solution. As described above, a polar point with negative radial coordinate lieson the opposite ray of the terminal side of θ.



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Points in Polar Coordinates

1. For any θ, the polar coordinates (0, θ) represent the pole.

2. A point with polar coordinates (r, θ) can also be represented by(r, θ + 2kπ) or (−r, θ + π + 2kπ) for any integer k.

3.9.2. From Polar to Rectangular, and Vice Versa

We now have two ways to describe points on a plane – whether to use the Carte-sian coordinates (x, y) or the polar coordinates (r, θ). We now derive the conver-sion from one of these coordinate systems to the other.

We superimpose the Cartesian and polar planes, as shown in the followingdiagram.

Figure 3.42

Suppose a point P is represented by the polar coordinates (r, θ). From Lesson3.2 (in particular, the boxed definition on page 139), we know that

x = r cos θ and y = r sin θ.

Conversion from Polar to Rectangular Coordinates

(r, θ) −→

x = r cos θ

y = r sin θ−→ (x, y)

Given one polar coordinate representation (r, θ), there is only onerectangular coordinate representation (x, y) corresponding to it.



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Example 3.9.3. Convert the polar coordinates (5, π) and (−3, π6) to Cartesian

coordinates.

Solution.

(5, π) −→

x = 5 cos π = −5

y = 5 sin π = 0−→ (−5, 0)

(−3, π6) −→

x = −3 cos π6

= −3√3

2

y = −3 sin π6

= −32

−→ (−3√3

2,−3

2) 2

As explained on page 239 (right after Example 3.9.1), we expect that thereare infinitely many polar coordinate representations that correspond to just onegiven rectangular coordinate representation. Although we can actually determineall of them, we only need to know one of them and we can choose r ≥ 0.

Suppose a point P is represented by the rectangular coordinates (x, y). Re-ferring back to Figure 3.42, the equation of the circle is

x2 + y2 = r2 =⇒ r =√x2 + y2.

We now determine θ. If x = y = 0, then r = 0 and the point is the pole. Thepole has coordinates (0, θ), where θ is any real number.

If x = 0 and y 6= 0, then we may choose θ to be either π2

or 3π2

(or theirequivalents) depending on whether y > 0 or y < 0, respectively.

Now, suppose x 6= 0. From the boxed definition again on page 139, we knowthat

tan θ =y

x,

where θ is an angle in standard position whose terminal side passes through thepoint (x, y).

Conversion from Rectangular to Polar Coordinates

(x, y) = (0, 0) −→ (r, θ) = (0, θ), θ ∈ R

(0, y)y 6=0

−→ (r, θ) =

(y, π2) if y > 0

(|y|, 3π2

) if y < 0

(x, 0)x6=0

−→ (r, θ) =

(x, 0) if x > 0

(|x|, π) if x < 0



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(x, y)x6=0, y 6=0

−→ (r, θ)

r =√x2 + y2

tan θ = yx

θ same quadrant as (x, y)

Given one rectangular coordinate representation (x, y), there aremany polar coordinate representations (r, θ) corresponding to it. Theabove computations just give one of them.

Example 3.9.4. Convert each Cartesian coordinates to polar coordinates (r, θ),where r ≥ 0.(1) (−4, 0)

(2) (4, 4)

(3) (−3,−√

3)

(4) (6,−2)

(5) (−3, 6)

(6) (−12,−8)

Solution. (1) (−4, 0) −→ (4, π)

(2) The point (4, 4) is in QI.

r =√x2 + y2 =

√42 + 42 = 4

√2

tan θ = yx

= 44

= 1 =⇒ θ = π4

(4, 4) −→(4√

2, π4

)(3) (−3,−

√3) in QIII

r =√

(−3)2 + (−√

3)2 = 2√

3

tan θ = −√3

−3 =√33

=⇒ θ = 7π6

(−3,−√

3) −→(2√

3, 7π6

)(4) (6,−2) in QIV

r =√

62 + (−2)2 = 2√

10

tan θ = −26

= −13

=⇒ θ = tan−1(−1

3

)(6,−2) −→

(2√

10, tan−1(−1

3

))(5) (−3, 6) in QII

r =√

(−3)2 + 62 = 3√

5

tan θ = 6−3 = −2 =⇒ θ = π + tan−1(−2)

(−3, 6) −→(3√

5, π + tan−1(−2))243


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(6) (−12,−8) in QIII

r =√

(−12)2 + (−8)2 = 4√

13

tan θ = −8−12 = 2

3=⇒ θ = π + tan−1 2

3

(−12,−8) −→(4√

13, π + tan−1 23

)2

3.9.3. Basic Polar Graphs and Applications

From the preceding session, we learned how to convert polar coordinates of apoint to rectangular and vice versa using the following conversion formulas:

r2 = x2 + y2, tan θ =y

x, x = r cos θ, and y = r sin θ.

Because a graph is composed of points, we can identify the graphs of some equa-tions in terms of r and θ.

Graph of a Polar Equation

The polar graph of an equation involving r and θ is the set of allpoints with polar coordinates (r, θ) that satisfy the equation.

As a quick illustration, the polar graph of the equation r = 2−2 sin θ consistsof all points (r, θ) that satisfy the equation. Some of these points are (2, 0), (1, π

6),

(0, π2), (2, π), and (4, 3π

2).

Example 3.9.5. Identify the polar graph of r = 2, and sketch its graph in thepolar plane.

Solution. Squaring the equation, we get r2 = 4. Because r2 = x2 + y2, we havex2 + y2 = 4, which is a circle of radius 2 and with center at the origin. Therefore,the graph of r = 2 is a circle of radius 2 with center at the pole, as shown below.



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In the previous example, instead of using the conversion formula r2 = x2 +y2,we may also identify the graph of r = 2 by observing that its graph consists ofpoints (2, θ) for all θ. In other words, the graph consists of all points with radialdistance 2 from the pole as θ rotates around the polar plane. Therefore, thegraph of r = 2 is indeed a circle of radius 2 as shown.

Example 3.9.6. Identify and sketch the polar graph of θ = −5π4

.

Solution. The graph of θ = −5π4

consists of all points (r,−5π4

) for r ∈ R. Ifr > 0, then points (r,−5π

4) determine a ray from the pole with angle −5π

4from

the polar axis. If r = 0, then (0,−5π4

) is the pole. If r < 0, then the points(r,−5π

4) determine a ray in opposite direction to that of r > 0. Therefore, the

graph of θ = −5π4

is a line passing through the pole and with angle −5π4

withrespect to the polar axis, as shown below.

Example 3.9.7. Identify (and describe) the graph of the equation r = 4 sin θ.

Solution.

r = 4 sin θ

r2 = 4r sin θ

x2 + y2 = 4y

x2 + y2 − 4y = 0

x2 + (y − 2)2 = 4

Therefore, the graph of r = 4 sin θ is a circle of radius 2 and with center at (2, π2).



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?Example 3.9.8. Sketch the graph of r = 2− 2 sin θ.

Solution. We construct a table of values.

x 0 π6

π4

π3

π2

2π3

3π4

5π6

π

r 2 1 0.59 0.27 0 0.27 0.59 1 2

x 7π6

5π4

4π3

3π2

5π3

7π4

11π6

2π

r 3 3.41 3.73 4 3.73 3.41 3 2

This heart-shaped curve is called a cardioid. 2



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?Example 3.9.9. The sound-pickup capability of a certain brand of microphoneis described by the polar equation r = −4 cos θ, where |r| gives the sensitivity ofthe microphone to a sound coming from an angle θ (in radians).

(1) Identify and sketch the graph of the polar equation.

(2) Sound coming from what angle θ ∈ [0, π] is the microphone most sensitiveto? Least sensitive?

Solution. (1) r = −4 cos θ

r2 = −4r cos θ

x2 + y2 = −4x

x2 + 4x+ y2 = 0

(x+ 2)2 + y2 = 4

This is a circle of radius 2 and with center at (2, π).

(2) We construct a table of values.

x 0 π6

π4

π3

π2

2π3

3π4

5π6

π

r −4 −3.46 −2.83 −2 0 2 2.83 3.46 4

From the table, the microphone is most sensitive to sounds coming fromangles θ = 0 and θ = π, and least sensitive to sound coming from an angleθ = π

2. 2



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1. Locate in the polar plane the following polar points: M(1, π/3), A(0, π), T (π, 0),and H(4, 5π/3).

Solution:

2. Locate in the polar plane the following polar points: W (−1, 7π/4), X(2,−π/6), Y (4,−5π/6)and Z(−3,−11π/3).

Solution:



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3. Convert the following polar points to Cartesian coordinates.

(a) (5, 5π/4) to Cartesian coordinates

(b) (−2, 3π/4) to Cartesian coordinates

(c) (π, π)

(d) (0, 10)

Solution: (a) Using the conversion formulas with r = 5 and θ = 5π/4, we get

x = r cos θ = 5 cos(5π/4) = −5√

2/2

andy = r sin θ = 5 sin(5π/4) = −5

√2/2.

Therefore, (5, 5π/4) in Cartesian coordinate is(−5√

2/2,−5√

2/2).

(b) Using the conversion formulas with r = −2 and θ = 5 = 3π/4, we get

x = r cos θ = −2 cos(3π/4) = 2√

2/2

andy = r sin θ = −2 sin(3π/4) = −2

√2/2.

Therefore, (−2, 3π/4) in Cartesian coordinate is (2√

2/2,−2√

2/2).

(c) Notice here that π is used in two different ways. First is π, with numericalvalue approximately equal to −3.14, is used as a radius and second, as anangle equivalent to 180◦. That is, the point is in the negative x-axis π unitsaway from the origin. Hence, the Cartesian coordinate of (π, π) is (−π, 0).

(d) Since the radius is 0, then the polar point (0, 10) is the origin with Cartesiancoordinate (0, 0).

4. Convert the following Cartesian points to polar coordinates.

(a) (5,−5)

(b) (−3,√

3)

(c) (−5√

3,−15)

(d) (8, 0)

Solution: (a) The point (x, y) = (5,−5) is in the fourth quadrant. Using theconversion formulas, we get

r =√x2 + y2 =

√52 + (−5)2 = 5

√2

andθ = tan−1(y/x) = tan−1(−1) = −π/4.

Therefore, (5,−5) in polar coordinate is (5√

2,−π/4).



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(b) Similarly, we use the conversion formulas to get

r =

√(−3)2 + (

√3)2 =

√12 = 2

√3

andθ = tan−1[

√3/(−3)] = −π/6.

Note that the point is in the second quadrant so we must use −pi6

+π. There-

fore, (−3,√

3) in polar coordinate is (2√

3, 5π/6).

(c) The point is in the third quadrant.

r =

√(−5√

3)2 + (−15)2 =√

300 = 10√

3

andtan θ = −15/(−5

√3)⇒ θ = 4π/3.

Therefore, (−5√

3,−15) in polar coordinate is (10√

3, 4π/3).

(d) Using the conversion formula, one can show that the point (8, 0) in polarcoordinate is also (8, 0).

5. Identify (and describe) the graph of the equation r = 4 sin θ. Using a graphingsoftware, graph the following equations.

(a) r = 2 sin θ

(b) r = −5

(c) θ = 2r

(d) r = 2− 2 cos θ

Solution: (a) r = 2 sin θ is a circle with radius 1 centered at (1, π2).

(b) r = −5 is a standard circle with radius 5.

(c) Notice that as θ increases, the r also increases. The graph of θ = 2r is aspiral rotating counter-clockwise from the pole.

(d) The graph of r = 2− 2 cos θ is a cardioid.

(a) (b)



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(c) (d)

6. A boy is flying a kite with an angle of elevation of 60◦ from where he stands.What is the direct distance of the kite from him, if the the kite is 6 ft abovethe ground?

Solution: The problem can be illustrated as follows:

Here, r (in ft) is the distance of the kite from the boy and θ is the angle ofdepression. To solve for r, we apply the formula y = r sin θ. Thus,

r = y/ sin θ = 6/ sin(60◦) = 6/(√

3/2) = 12/√

3 = 4√

3.

Therefore, the kite is 4√

3 ft away from the boy.



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1. Give two more pairs of coordinates that describe the same point.

(a) (13, π/3) (b) (0, 0) (c) (15, 15π/4)

2. Locate the following points in the polar coordinate plane:

(a) P (3,−π)

(b) Q(−3, 7π/4)

(c) R(5/2, 5π/2)

(d) S(−8,−23π/6)

3. Transform the following to Cartesian coordinates:

(a) (3,−π)

(b) (−3, 7π/4)

(c) (5/2, 5π/2)

(d) (−8,−23π/6)

4. Transform the following to polar coordinates:

(a) (−9, 40)

(b) (15, 20)

(c) (5/2, 5π/2)

(d) (14,−14)

5. Consider the equation in polar form r = 4 cos 2θ.

(a) Complete the table

θ 0 π/6 π/4 π/3 π/2 2π/3 3π/4 5π/6 π 7π/6 5π/4 4π/3 3π/2

r

(b) Plot the points obtained in part (a) in a polar coordinate system.

6. A helicopter is hovering 800 feet above a road. A truck driver observes thehelicopter at a horizontal distance of 600 feet. Find the angle of elevation ofthe helicopter from the truck driver.

4



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1. Let θ be an angle in QIII such that cos θ = −12

13. Find the values of the six

trigonometric functions of 2θ.

2. Prove that cot(2x) =cot2 x− 1

2 cotx.

3. Using half-angle identities to find the exact values of the following.

(a) tan 15◦ (b) tan 7.5◦


(a) tan−1(

cot103π

6

)(b) cos

(sin−1

40

41

)5. Let y ∈ [0, 2π). Find the solutions of the equation

sin−1(cos2 y − cos y − 1) = −π/2.

6. Let θ ∈ [0, 2π]. Find all the solutions of the equation

4 cos2 θ sin θ = 3 sin θ.

7. Let r = −2−2 sin θ. Complete the table and plot the points (r, θ) in the samepolar coordinates.

θ 0 π/6 π/4 π/3 π/2 2π/3 3π/4 5π/6 π 7π/6 5π/4 4π/3 3π/2

r

8. Transform the following points from Cartesian to polar coordinates.

(a) (−42,−56)

(b) (100, 100)

(c) (0, 7)

(d) (7, 0)

(e) (2π, 2π)

(f) (5, 12)

9. Transform the following points from polar to Cartesian coordinates.

(a) (3, π/3)

(b) (45, 7π/4)

(c) (−1,−π)

(d) (5, 0)

(e) (2π, 2π)

(f) (9, 17π/6)



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1. Let θ be an angle in the 2nd quadrant such that cos θ = − 7

25. Find the

following.

(a) cos(2θ) (b) sin(2θ) (c) tan(2θ)

2. Given that cos 48◦ ≈ 0.6691. Find the approximate value of the following.

(a) cos2 24◦ (b) sin2 24◦ (c) tan2 24◦

3. Using half-angle identities to find the exact values of the following.

(a) tan(π/12) (b) tan(π/24)

4. Find the exact value of cos

(cos−1

1

7+ cos−1

3

5

).

5. Let x ∈ [0, 2π). Find the solutions of the equation

4 sin2 x+ (2√

3− 2√

2) sinx−√

6 = 0.

6. Let θ ∈ [0, 2π]. Find all the solutions of the equation

2 sin2(2θ)− sin(2θ)− 1 = 0.

7. Let r = 2 + 2 cos θ. Complete the table and plot the points (r, θ) in the samepolar coordinates.

θ 0 π/6 π/4 π/3 π/2 2π/3 3π/4 5π/6 π 7π/6 5π/4 4π/3 3π/2

r

8. Transform the following points from Cartesian to polar coordinates.

(a) (21,−28)

(b) (−100,−100)

(c) (0,−5)

(d) (−5, 0)

(e) (π, π)

(f) (15, 8)

9. Transform the following points from polar to Cartesian coordinates.

(a) (4, π/6)

(b) (100, 5π/4)

(c) (1, π)

(d) (−5, 0)

(e) (π, π)

(f) (15, 8π/3)



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Answers to

Odd-Numbered Exercisesin Supplementary Problems

and

All Exercises in Topic Tests

4


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1. center (0, 0), r =1

2

3. center

(−4,

3

4

), r = 1

5. center (7,−6), r = 11

256

Supplementary Problems 1.1 (page 17)


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7. center (−2,−4), r =5

3

9. center

(−5

2,7

2

), r =

7

4

11. (x− 17)2 + (y − 5)2 = 144

13. (x− 15)2 + (y + 7)2 = 49

15. (x− 15)2 + (y + 7)2 = 9

17. (x+ 2)2 + (y − 3.5)2 = 31.25

19. (x+ 10)2 + (y − 7)2 = 36

21. (x+ 2)2 + (y − 3)2 = 12

23. (x− 2.5)2 + (y − 0.5)2 = 14.5

25. (x+ 5)2 + (y + 1)2 = 8

27. Set up a Cartesian coordinate system by assigning C as the origin. Then thecircle on the left end has radius 100 and has equation x2 + y2 = 10000. Aradius of the circle on the right end can be drawn from C to the upper rightcorner of the figure; this radius has length (by the Pythagorean theorem)



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√3002 + 1002 =

√100000. Then the circle on the right end has equation x2 +

y2 = 100000. We want the length of the segment at y = 50. In this case, theleft endpoint has x = −

√10000− 502 = −

√7500 and the right endpoint has

x =√

100000− 502 =√

97500. Then the total length is√

97500− (−√

7500)= 50

√3 + 50

√39 m ≈ 398.85 m.

1. vertex (0, 0), focus (−9, 0), directrix x = 9, axis y = 0

3. vertex (−1, 7), focus (−2, 7), directrix x = 0, axis y = 7

258



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5. vertex (3, 2), focus

(3,−3

2

), directrix y =

11

2, axis x = 3

7. (y − 11)2 = 36(x− 7)

9. (x+ 10)2 = 34(y − 3)

11. (y − 9)2 = −80(x− 4)

13. (y − 8)2 = −8(x+ 3)

15. ≈ 4.17 cm

17. 3.75 cm

1. center: (0, 0)

foci: F1(−2, 0), F2(2, 0)

vertices: V1(−2√

2, 0), V2(2√

2, 0)


3. center: (1, 1)

foci: F1(1−√

3, 1),

F2(1 +√

3, 1)

vertices: V1(−1, 1), V2(3, 1)

covertices: W1(1, 0),W2(1, 2)

259



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5. center: (7,−5)

foci: F1(1,−5), F2(13,−5)

vertices: V1(−3,−5), V2(17,−5)


7.(x− 2)2

49+

(y − 8)2

16= 1

9. The center is (−9, 10) and c = 12. We see that the given point (−9, 15) isa covertex, so b = 5. Then a =

√52 + 122 = 13. Therefore, the equation is

(x+ 9)2

169+

(y − 10)2

25= 1.

11. Since the major axis is vertical, the center has the same x coordinate as thefocus and the same y coordinate as the covertex; that is, the center is (−9, 10).

Then c = 5, b = 10, and a2 = 125. Therefore, the equation is(x+ 9)2

100+

(y − 10)2

125= 1.

13. Recall that the unit is 100 km. The vertices of the ellipse are at (3633, 0) and(−4055, 0). Then the center of the ellipse is at (−211, 0). Then a = 3844

and c = 211. It follows that b2 = 14731815. The equation is(x+ 211)2

14776336+

y2

14731815= 1.

15. Set up a coordinate system with the center of the ellipse at the origin. Thena = 60 and b = 20. We want the length of the segment with endpoints (on the

ellipse) having x = 45 (or −45). The y coordinates are given by 452

602+ y2

202= 1,

or y = ±√

202(1− 452

602

)≈ ±13.23. Hence, the desired width is 26.46 ft.



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1. center: (0, 0)

foci: F1(−√

181, 0), F2(√

181, 0)

vertices: V1(−10, 0), V2(10, 0)

asymptotes: y = ± 9

10x

3. center: (0, 5)

foci: F1(−√

19, 5), F2(√

19, 5)

vertices: V1(−√

15, 5), V2(√

15, 5)

asymptotes: y − 5 = ± 2√15x

5. center: (−3,−3)

foci: F1(−3,−3−√

15),

F2(−3,−3 +√

15)

vertices: V1(−3,−3−√

6),

V2(−3,−3 +√

6)

asymptotes: y + 3 = ±√

6

3(x+ 3)

261



DEPED COPY

7.y2

144− (x+ 7)2

145= 1 9.

(x+ 10)2

81− (y + 4)2

256= 1

11. The intersection (−4, 8) of the two asymptotes is the center of the hyperbola.Then the hyperbola is vertical and c = 13. Since the slopes of the asymptotesare ± 5

12, we have a

b= 5

12.

Since c = 13, we have a2 + b2 = 169 and b =√

169− a2. It follows that

a√169− a2

=5

12=⇒ a = 5 and b = 12 =⇒ (y − 8)2

25− (x+ 4)2

144= 1.

13. The midpoint (9, 1) of the two given corners is the center of the hyperbola.Since the transverse axis is horizontal, a = 7 and b = 2. Therefore, the

equation is(x− 9)2

49− (y − 1)2

4= 1.

1. pair of intersecting lines

3. parabola

5. parabola

7. empty set

9. The standard equation of the ellipse is(x− 5)2

36+

(y − 2)2

100= 1; so its foci

are (5, 10) and (5,−6) while its vertices are (5, 12) and (5,−8). The equationsof the circles are (x − 5)2 + (y − 10)2 = 4, (x − 5)2 + (y − 10)2 = 324,(x− 5)2 + (y + 6)2 = 4, and (x− 5)2 + (y + 6)2 = 324.

11. The standard equation of the hyperbola is(y + 5)2

25− (x+ 9)2

25= 1. Its auxil-

iary rectangle has corners (−14, 0), (−4, 0), (−4,−10), (−14,−10). The equa-tion of the circle is (x+ 9)2 + (y + 5)2 = 50.

13. The equation simplifies to

(x+ 7)2 + (y − 3)2 =r + 2

r − 1.

Its graph

(a) is a circle ifr + 2

r − 1> 0; that is, when r ∈ (−∞,−2) ∪ (1,+∞).

(b) is a point ifr + 2

r − 1= 0; that is, when r = −2.

(c) is the empty set ifr + 2

r − 1< 0; that is, when r ∈ (−2, 1).

262



DEPED COPY

1. (a) (1, 6)

(b)

(−4

3,− 9

20

),

(−4

3,89

20

)

(c) (1, 6), (1, 2)

263



DEPED COPY

(d)

(1,

3

2

), (1−

√15,−1), (1 +

√15,−1)

(e) No solution

3. Let (x, y) be the ordered pair that satisfies the conditions. The resultingsystem of equations is

x2 = 2y2 +1

8

x2 + y2 =5

16

Solving yields

(1

2,1

4

),

(−1

2,1

4

),

(1

2,−1

4

), and

(−1

2,−1

4

).



DEPED COPY

5. We have the system x2 + (y − 6)2 = 36

x2 = 4ky,

where the first equation is a circle above the x-axis, tangent to the x-axis atx = 0, and the second equation is a parabola facing up/down, depending onk.

Substituting the second equation in the first equation yields y2+(4k−12)y = 0.Note that y = 0 is already a root.

We now consider two cases.

If k > 0, the system might have one or two solutions. To ensure that thesolution is unique, we set the discriminant to be nonpositive: 4k − 12 ≤ 0⇒k ≥ 3.

If k ≤ 0, the system will always have a unique solution.

Thus k ∈ (−∞, 0] ∪ (3,+∞).

1. (a) Since the coefficients of x2 and y2 are equal, the graph is a circle, a point,or the empty set. Completing the squares, we see that the equation isequivalent to (

x− 1

2

)2

+

(y +

3

2

)2

= 4.

Hence, the graph is a circle with center (0.5,−1.5) and radius 2.

(b) By inspection, the graph is a parabola. Completing the squares, we seethat the equation is equivalent to (x+ 2)2 = 14(y+ 4). Hence, the graphhas vertex at (−2,−4) and is opening upward.

(c) Since the coefficients of x2 and y2 are of opposite signs, the graph is ahyperbola or a pair of intersecting lines. Completing the squares, we seethat the equation is equivalent to

(x− 7)2

4− (y + 3)2

3= 1.

Hence, the graph is a horizontal hyperbola with center at (7,−3).

(d) Since the coefficients of x2 and y2 have the same sign and are unequal,the graph is an ellipse, a point, or the empty set. Completing the squares,we see that the equation is equivalent to

(x− 8)2

2+y2

7= 0.

Hence, the graph is the point (8, 0).

265

Topic Test 1 for Unit 1 (page 78)


DEPED COPY

2. (a) The equation is equivalent tox2

7+y2

25= 1. This is a vertical ellipse.

center: (0, 0)

foci: F1(0,−3√

2), F2(0, 3√

2)

vertices: V1(0,−5), V2(0, 5)

covertices: W1(−√

7, 0),W2(√

7, 0)

(b) The equation is equivalent to(y + 4)2

64− (x− 1)2

36= 1. This is a vertical

hyperbola.

center: C(1,−4)

foci: F1(1,−14), F2(1, 6)

vertices: F1(1,−12), F2(1, 6)

asymptotes: y + 4 = ±43(x− 1)

3. (a) The parabola opens to the right and has focal distance c = 6. Its equationis (y − 3)2 = 24(x+ 1).

(b) The intersection (−2,−5) of the two asymptotes is the center of thehyperbola. Then the hyperbola is horizontal and a = 5. Using the slopesof the asymptotes, we have b

a= 12

4. It follows that b = 12 and the

equation is(x+ 2)2

25− (y + 5)2

144= 1.

4. Multiplying the first equation by 2, we get 2(x − 1)2 + 2(y + 1)2 = 10. Bysubtracting the second equation from this new equation, we get the equation2(y + 1)2 + 8 = 10− y. This has solutions y = 0 and y = −5/2.

When y = 0, the corresponding x values are 3 and −1. When y = −5/2,

the corresponding x values are ±√112

+ 1. Therefore, the solutions are (−1, 0),

(3, 0),(±√112

+ 1,−52

).



DEPED COPY

5. Set up a coordinate system by making the center of the door’s base the origin.Then the ellipse has center (0, 2) with a = 1/2 and b = 0.3; then its is equationx2

0.52+ (y−2)2

0.32= 1.

To determine if the cabinet can be pushed through the doorway, we determinethe height of the doorway when x = 0.25 (or −0.25). We solve for y from the

equation 0.252

0.52+ (y−2)2

0.32= 1. Solving for the y coordinate, we see that the height

is ≈ 2.2598 m. Hence, the cabinet cannot be pushed through the doorway.

6. Let (x, y) be the coordinates of the point. This point satisfies√x2 + (y + 1)2 = 2|x− 3|.

Manipulating this equation gives us

x2 + (y + 1)2 = 4(x2 − 6y + 9)

−3(x2 − 8x+ 16) + (y + 1)2 = 36− 48

−3(x− 4)2 + (y + 1)2 = −12

(x− 4)2

4− (y + 1)2

12= 1.

Therefore, the point traces a horizontal hyperbola with center at (4,−1).

1. (a) By inspection, the graph is a parabola. Completing the squares, we seethat the equation is equivalent to (y−5)2 = −8(x−5). Hence, the graphhas vertex at (5, 5) and is opening to the left.

(b) Since the coefficients of x2 and y2 are equal, the graph is a circle, a point,or the empty set. Completing the squares, we see that the equation isequivalent to (x+ 5)2 + (y + 9)2 = −4. Hence, the graph is the emptyset.

(c) Since the coefficients of x2 and y2 have the same sign and are unequal,the graph is an ellipse, a point, or the empty set. Completing the squares,

we see that the equation is equivalent to(x+ 2)2

4+

(y − 1)2

9= 1. Hence,

the graph is a vertical ellipse with center (−2, 1).

(d) Since the coefficients of x2 and y2 are of opposite signs, the graph is ahyperbola or a pair of intersecting lines. Completing the squares, we

see that the equation is equivalent to(y − 4)2

11− (x− 6)2

17= 0. Hence,

the graph is a pair of intersecting lines given by the equations y − 4 =±11

17(x− 6).

267



DEPED COPY

2. (a) The equation is equivalent tox2

64− y

2

64= 1. This is a horizontal hyperbola.

center: (0, 0)

foci: F1(−8√

2, 0),

F2(8√

2, 0)

vertices: V1(−8, 0), V2(8, 0)

asymptotes: y = ±x

(b) The equation is equivalent to(x+ 3)2

49+

(y − 2)2

4= 1. This is a horizontal

ellipse.

center: C(−3, 2)

foci: F1(−3− 3√

5, 2), F2(−3 + 3√

5, 2)

vertices: F1(−10, 2), F2(4, 2)

covertices: W1(−3, 0),W2(−3, 4)

3. (a) The parabola opens downward and has focal distance c = 5. Its equationis (x− 7)2 = −20(y + 7).

(b) Since the ellipse has vertical or horizontal major axis, the center is ateither (−1, 12) or (−5, 3). Since the major axis is longer than the minoraxis, the center must be at (−5, 3). Then the ellipse is vertical with a = 9and b = 4. Its equation is

(x+ 5)2

16+

(y − 3)2

81= 1.



DEPED COPY

4. Completing the squares, we see that the first equation is equivalent to 9(x +3)2 = 4y2 + 36. On the other hand, the second equation is equivalent to9(x + 3)2 = 36y + 36. Subtracting the second equation from the first, we get4y2 − 36y = 0, which has solutions y = 0 and y = 9.

When y = 0, the corresponding x values are −5 and −1. When y = 9, thecorresponding x values are −3 ± 2

√10. Therefore, the solutions are (−5, 0),

(−1, 0),(−3± 2

√10, 9

).

5. Set up a coordinate system so that the opening of the hose (the parabola’svertex) is at (0, 3) and that the water flows towards the positive x-axis. Thenthe x-axis (y = 0) corresponds to the ground; it follows the parabola passesthrough the point (2, 0). Hence, the equation of the parabola is x2 = −4

3(y−3).

If Nikko stands on a 1.5-ft stool and the vertex remains at (0, 3), the liney = −1.5 will correspond to the ground. Hence, the water will strike the

ground when y = −1.5. This gives x =√−4

3(−1.5− 3) =

√6. Therefore, the

water will travel√

6− 2 ft further before striking the ground.

6. Let (x, y) be the coordinates of the point. This point satisfies√(x− 2)2 + y2 =

2

3|y − 5|.

Manipulating this equation gives us

(x− 2)2 + y2 =4

9(y2 − 10y + 25)

9(x− 2)2 + 5(y2 + 8y) = 100

9(x− 2)2 + 5(y + 4)2 = 100 + 80

(x− 2)2

20+

(y + 4)2

36= 1.

Therefore, the point traces a vertical ellipse with center at (2,−4).

1. a3 = a1+(3−1)d = 35; a10 = a1+(10−1)d = 77⇒ d = 6, a1 = 23⇒ a5 = 47.

3. sn =n (2(17) + (n− 1)3)

2= 30705 ⇒ 3n2 + 31n − 61410 = 0. Using the

quadratic formula and noting that n must be a whole number, we have n =138.

5. We have s = 108 =a1

1− rand s3 = 112 =

a1 (1− r3)1− r

= 108 (1− r3) ⇒ r =

− 1

27⇒ a1 = 144.

269



DEPED COPY

7. Note that 0.123123 . . . = 0.123 + 0.000123 + 0.000000123 + . . . = 0.123 +0.123(0.001) + 0.123(0.001)2 + . . ., which is an infinite geometric series with

r = 0.001. Thus, 0.123 =0.123

1− 0.001=

41

333.

9. We have s4 =4 (2a1 + (4− 1)d)

2= 80⇒ 2a1 + 3d = 40. Since the sum of the

first two numbers are one-third of the sum of the last two numbers, we have1

3(a1 + a2) = a3 + a4 ⇒ 4a1 + 14d = 0. Combining yields d = 10, and thus

a1 = 5, a2 = 15, a3 = 25, a4 = 35.

11. Note that this is a geometric series with common ratio 2n− 1. Thus, the sumwill have a finite value if |2n− 1| < 1 ⇒ −1 < 2n − 1 < 1 ⇒ 0 < n < 1.Thus, n ∈ (0, 1).

1. (a)10∑i=3

√3 · i

2=

√3

2

(3

2+ . . .+

10

2

)= 26

√3

(b)5∑i=1

x2i

2i=x2

2+x4

4+x6

8+x8

16+x10

32

(c)5∑i=2

(−1)i xi−1 = x− x2 + x3 − x4

3. (a)150∑i=1

(4i+ 2) = 4150∑i=1

i+150∑i=1

2 = 4150(151)

2+ 2(150) = 45, 600

(b)120∑i=3

i(i−5) =120∑i=1

(i2−5i)−1(1−5)−2(2−5) =120(121)(2(120) + 1)

6+10 =

583, 230

(c)130∑i=1

(2i−3)(2i+3) =130∑i=1

(4i2−9) = 4130∑i=1

i2−130∑i=1

9 =130(131)(2(130) + 1)

6+

9(130) = 741, 975

5. s =200∑i=1

[(i− 1)2 − i2

]=

200∑i=1

(1− 2i)⇒200∑i=1

i =200− s

2

270



DEPED COPY

1. Part 1.1

2= 2− 1 + 2

21


Part 2.

Assume:k∑i=1

i

2i= 2− k + 2

2k.

To show:k+1∑i=1

i

2i= 2− k + 3

2k+1.

k+1∑i=1

i

2i=

k∑i=1

i

2i+k + 1

2k+1

= 2− k + 2

2k+k + 3

2k+1

= 2− k + 3

2k+1.

3. Part 1.1(1!) = (1 + 1)!− 1


Part 2.

Assume:k∑i=1

i(i!) = (k + 1)!− 1.

To show:k+1∑i=1

i(i!) = (k + 2)!− 1.

k+1∑i=1

i(i!) =k∑i=1

i(i!) + (k + 1)[(k + 1)!]

= (k + 1)!− 1 + (k + 1)[(k + 1)!]

= (k + 2)(k + 1)!− 1

= (k + 2)!− 1.

271



DEPED COPY

5. Part 1.

1− 1

2=

1

2=

1

2(1)


Part 2.

Assume: P =

(1− 1

2

)·(

1− 1

3

)· · ·(

1− 1

k − 1

)·(

1− 1

k

)=

1

2k.

To show:

(1− 1

2

)· · ·(

1− 1

k

)·(

1− 1

k + 1

)=

1

2(k + 1).

(1− 1

2

)· · ·(

1− 1

k + 1

)= P ·

(1− 1

k + 1

)=

1

2k· k

k + 1

=1

2(k + 1).

7. Part 1.43(1)+1 + 23(1)+1 + 1 = 273 = 7(39)


Part 2.

Assume: 43k+1 + 23k+1 + 1 is divisible by 21.

Prove: 43(k+1)+1 + 23(k+1)+1 + 1 is divisible by 21.

43(k+1)+1+23(k+1)+1+1 = 64·43k+1+8·23k+1+1 = 56·43k+1+8(43k+1 + 23k+1 + 1

)−

7

9. Part 1.52(1)+1 · 21+2 + 31+2 · 22(1)+1 = 1216 = 19(64)


Part 2.

Assume: 52k+1 · 2k+2 + 3k+2 · 22k+1 is divisible by 19.

Prove: 52(k+1)+1 · 2(k+1)+2 + 3(k+1)+2 · 22(k+1)+1 is divisible by 19.

52(k+1)+1 · 2(k+1)+2 + 3(k+1)+2 · 22(k+1)+1 = 50 · 52k+1 · 2k+2 + 12 · 3k+2 · 22k+1 =12 (52n+1 · 2n+2 + 3n+2 · 22n+1) + 38 · 52n+1 · 2n+2



DEPED COPY

11. Part 1.101

3+

5

3+ 41+2 = 1029 = 3(343)


Part 2.

Assume:10k

3+

5

3+ 4k+2 is divisible by 3.

Prove:10k+1

3+

5

3+ 4k+3 is divisible by 3.

10k+1

3+

5

3+4k+3 = 10· 10k

3+

5

3+4·4k+2 = 10

(10k

3+

5

3+ 4k+2

)−6·4n+2−9· 5

3

13. Part 1. 1 ≤ 2− 1

1= 1

Part 2

Assume:k∑i=1

1

i3≤ 2− 1

k

Prove:k+1∑i=1

1

i3≤ 2− 1

k + 1

k+1∑i=1

1

i3≤ 2 − 1

k+

1

(k + 1)3= 2 − (k + 1)3 − k

(k + 1)3. Note that 0 < (k + 1)2 ⇒

(k + 1)2 < (k + 1)3 − k, thus 2− (k + 1)3 − k(k + 1)3

< 2− (k + 1)2

(k + 1)3= 2− 1

k + 1.

1. (a) (2x− 3y)5 = 32x5 − 240x4y + 720x3y2 − 1080x2y3 + 810xy4 − 243y5

(b)

(√x

3− 2

x2

)4

= −32

3x−11/2 + 16x−8 +

8

3x−3 +

1

81x2 − 8

27x−1/2

(c) (1 +√x)

4= 4x3/2 + x2 + 6x+ 4

√x+ 1

3. Approximating yields (2.1)10 ≈4∑

k=0

(10

k

)210−k(0.1)k = 1667.904, which has

an approximate error of −0.08.

5. In sigma notation we have19∑k=0

(19

k

)(−3)k = (1− 3)19 = (−2)19.

273



DEPED COPY

1. (a) G, r = 3/2 (b) O (c) O

2. First, a3 = a2 + 5 = a1 + 10. Also,a3 + 1

a2 + 2=

a2 + 2

a1 + 4. Thus, a1 = 5,

a2 = 10, and a3 = 15.

3. We have50∑i=1

2i3 + 9i2 + 13i+ 6

i2 + 3i+ 2=

(2i+ 3)(i+ 1)(i+ 2)

(i+ 1)(i+ 2)= 2i+ 3.

Thus,50∑i=1

2i3 + 9i2 + 13i+ 6

i2 + 3i+ 2=

50∑i=1

(2i+ 3) = 2700.

4. (a)

(8

k

)x16−2k

(−1

2

)k=⇒ 16− 2k = 8 =⇒ k = 4

=⇒(

8

4

)x8(−1

2

)4

=35

8x8

(b) k = 19 =⇒(

28

19

)(n3)

28−19(−3m)19 = −

(28

19

)319n27m19

5. For n = k + 1:

1

1 · 3+

1

3 · 5+ · · ·+ 1

(2(k + 1)− 1)(2(k + 1) + 1)

=k

2k + 1+

1

(2k + 1)(2k + 3)=

2k2 + 3k + 1

(2k + 1) · (2k + 3)=

k + 1

2k + 3

6. a1 = 10, 000, r = 1.04, 60− 20 = 40

s40 = 10, 000 · 1− (1.04)40

1− (1.04)≈ 499, 675.83 pesos

1. (a) G, r = 4/5 (b) O (c) A, d = 5/2

2. We have a1 + a2 = 2a1 + d = 9 and a1 + a2 + a3 = 3a1 + 2d = 9 yieldinga1 = 9, d = −9. Using sn = −126, we get n = 7.

3. (a)50∑i=1

(2i+ 1)(i− 3) =50∑i=1

2i2 − 5i− 3 = 79, 472

(b)30∑i=1

√i2 − 2i+ 1

4=

30∑i=1

i− 1

2=

435

2

274




DEPED COPY

4.

(8

k

)(x3)

8−k(

1

x

)k=

(8

k

)x24−4k =⇒ 24− 4k = 0 =⇒ k = 6 =⇒

(8

6

)= 28

5. (a) For n = k + 1:

1 + 4 + 7 + · · ·+ (3(k + 1)− 2)

=k(3k − 1)

2+ (3k + 1) =

3k2 + 5k + 2

2=

(k + 1)(3k + 2)

2

(b) For n = k + 1: 3(n+1) + 7(n+1)−1 + 8 = 7 (3n + 7n−1 + 8) − 4 · 3n − 6 · 8,where 4 · 3n is divisible by 12 for n ≥ 1, and 6 · 8 = 48 = 12(4).

1.6

5rev =

6

5rev

(360

1 rev

)= 432◦

3. 216◦ = 216( π

180

)=

6π

5rad; s = 4

(6π

5

)=

24π

5cm

5. 2110◦ − 5(360◦) = 310◦

7. θ =7π

6; r =

9

2cm; A =

1

2

(9

2

)27π

6=

189π

16cm2

9. θ = 150◦ =5π

6; A = 15 in2; r =

√2(15)

5π6

=6√π

=6√π

πin

11. r = 6 in; s = 6 in; θ =s

r=

6

6= 1 rad; 1 rad = 1

(180◦

π

)≈ 57.30◦

13.8π

3cm

15. θ = 20◦ =π

9rad; A = 800 cm2

r =

√√√√2(800)(π9

) =120√π

=120√π

πcm; s =

(120√π

π

)(π9

)=

40√π

3cm

17. r = 6 cm; θ = 54◦ =3π

10

Area of shaded region = 2×area of sector AOE = 2

[1

2(6)2

(3π

10

)]=

54π

5cm2

19. Asegment = Asector − Atriangle =1

2

(2π

3

)(6)2 − 1

2(3)(6

√3) = (12π − 9

√3) cm2

275



DEPED COPY

1.33π

4is coterminal with

33π

4− 8π =

π

4, and

33π

4terminates in QI.

3. The secant function is positive in QI and QIV. The cotangent function isnegative in QII and QIV. Therefore, the angle θ is in QIV.

5.5π

6is in QII. The reference angle is

π

6, and therefore P

(5π

6

)=

(−√

3

2,1

2

).

7. tan θ = −2

3, cos θ > 0 =⇒ sec θ =

√13

3

sec θ + tan θ

sec θ − tan θ=

√13

3− 2

3√13

3+

2

3

=17− 4

√13

9

9. csc θ = 2, cos θ < 0; r = 2, y = 1, x = −√

3; sec θ =r

x=

2

−√

3= −2

√3

3

11. csc θ = −4 and θ not in QIII =⇒ θ in QIV

csc θ =4

−1=⇒ r = 4, y = −1

x =√

(4)2 − (−1)2 = ±√

15, θ is in Quadrant IV, x =√

15

cos θ =x

r=

√15

4sec θ =

r

x=

4√

15

15sin θ =

y

r= −1

4

csc θ =r

y= −4 tan θ =

y

x= −√

15

15cot θ =

x

y= −√

15

13. x = −2, y = 4 =⇒ r =√

(−2)2 + (4)2 = 2√

5

cos θ =x

r= −√

5

5sec θ =

r

x= −√

5 sin θ =y

r=

2√

5

5

csc θ =r

y=√

52 tan θ =y

x= −2 cot θ =

x

y= −1

2

15. x = 2, y = −6, r =√

(2)2 + (−6)2 = 2√

10; sec θ =√

10, csc θ = −√

10

3

sec2 θ − csc2 θ = 10− 10

9=

80

9

17. cos θ = sin 2π3

=√32

and 3π2< θ < 2π =⇒ θ =

11π

6

276



DEPED COPY

19. f(x) = sin 2x+ cos 2x+ sec 2x+ csc 2x+ tan 2x+ cot 2x

f

(7π

8

)= −√

2

2+

√2

2+√

2−√

2− 1− 1 = −2

1. P =2π14

= 8π

3.π4πk

= 2 =⇒ k = 8

5. y = 3 sin 34

(8π9

+ 2π3

)− 5 = −13

2

7. domain = R; range =[43, 83

]9. y = 3 sec 2(x− π)− 3

11. Asymptotes: x = 3π2

+ 2kπ, k ∈ Z

13. (a) P = 8π, phase shift = −π4, domain = R, range = [−3, 1]

(b) P = π, phase shift = −π6, domain =

{x|x 6= π

3+ kπ, k ∈ Z

}, range = R

277



DEPED COPY

(c) P = 4π3

, phase shift = π2, domain =

{x|x 6= π

2+ 2kπ

3, k ∈ Z

}, range =(

−∞,−32

]∪[−1

2,∞)

(d) P = π, phase shift = −π6, domain =

{x|x 6= π

12+ kπ

2, k ∈ Z

}, range =

(−∞, 1] ∪ [3,∞)

15. y = 8 cos 110

(t− 10π); at t = 10, y ≈ −4.32 (that is, the mass is located about4.32 cm below the resting position)



DEPED COPY

1.tanx− sinx

sinx=

sinx

cosxsinx

− sinx

sinx=

sinx

cosx· 1

sinx− 1 = secx− 1

3. sinA+cos2A

1 + sinA=

sinA+ sin2A cos2A

1 + sinA=

sinA+ 1

1 + sinA= 1

5.cscx+ secx

cotx+ tanx=

1

sinx+

1

cosxcosx

sinx+

sinx

cosx

=

cosx+ sinx

sinx cosxcos2 x+ sin2 x

sinx cosx

= cosx+ sinx

7.tanx+ sinx

cscx+ cotx=

sinx

cosx− sinx

1

sinx+

cosx

sinx

=

sinx+ sinx cosx

cosx1 + cos x

sinx

=

sinx(1 + cos x)

cosx1 + cos x

sinx

=sin2 x

cosx=

1− cos2 x

cosx

9. sin θ cos θ = sin2 θ · cos θ

sin θ=

cot θ

csc2 θ=

cot θ

1 + cot2 θ=

a

1 + a2

11.csc a+ 1

csc a− 1=

1

sin a+ 1

1

sin a− 1

=

1 + sin a

sin a1− sin a

sin a

=1 + sin a

1− sin a

13.cos a

sec a+ tan a=

cos a1

cos a+

sin a

cos a

=cos2 a

1 + sin a=

1− sin2 a

1 + sin a= 1− sin a

15.1

1− cos a+

1

1 + cos a=

1 + cos a+ 1− cos a

(1− cos a)(1 + cos a)=

2

1− cos2 a=

2

sin2 a= 2 csc2 a

17.tanα

1− tan2 α=

sinα

cosα

1− sin2 α

cos2 α

=

sinα

cosαcos2 α− sin2 α

sin2 α

=sinα

cosα

(cos2 α

cos2 α− sin2 α

)=

sinα cosα

cos2 α− (1− cos2 α)=

sinα cosα

2 cos2 α− 1

279



DEPED COPY

19.cotα− sinα secα

secα cscα=

cosα

sinα− sinα

cosα1

cosα· 1

sinα

=cos2 α− sin2 α

sinα cosα· cosα sinα = cos2 α− sin2 α

1. cos θ = sin2π

3=

√3

2and θ in QIV =⇒ θ =

11π

6

3. tanA = tan(π

2+ kπ −B

)=

sin(π2

+ kπ −B)

cos(π2

+ kπ −B)

=sin(π2

+ kπ)

cosB − cos(π2

+ kπ)

sinB

cos(π2

+ kπ)

cosB + sin(π2

+ kπ)

sinB

=sin(π2

+ kπ)

cosB

sin(π2

+ kπ)

sinB

= cotB

5. sin 105◦ − cos 15◦ = sin(90◦ + 15◦)− cos 15◦ = cos 15◦ − cos 15◦ = 0

7. cotα = 7, csc β =√

10, and α and β are acute

=⇒ cosα =7√

2

10, sinα =

√2

10, sin β =

√10

10, cos β =

3√

2

10

cos(α + β) = cosα cos β − sinα sin β

=

(7√

2

10

)(3√

2

10

)−

(√2

10

)(√10

10

)=

2√

5

5

9. 3 sinx = 2 =⇒ sinx =2

3

sin(x− π) + sin(x+ π)

= sinx cos π − cosx sin π + sinx cos π + cosx sin π

= 2 sin x cos π = 2

(2

3

)(−1) = −4

3

11. sinA =4

5and A in QII =⇒ cosA = −3

5

cosB =4

5and B in QIV =⇒ sinB = −3

5

(a) sin(A−B) =

(4

5

)(4

5

)−(−3

5

)(−3

5

)=

7

25

280



DEPED COPY

(b) cos(A−B) =

(−3

5

)(4

5

)+

(4

5

)(−3

5

)= −24

25

(c) tan(A−B) =

7

25

−24

25

= − 7

24

cos(A−B) < 0 and sin(A−B) > 0 =⇒ A−B in QII

13. Given: sinα =4

5and cos β =

5

13

sin(α + β) + sin(α− β) = sinα cos β + cosα sin β + sinα cos β − cosα sin β

= 2 sinα cos β = 2

(4

5

)(5

13

)=

8

13

15. cscA =√

17, A in QI =⇒ tanA =1

4

cscB =√343, B in QI =⇒ tanB =

3

5

tan(A+B) =

1

4+

3

5

1− 1

4

(3

5

) = 1 =⇒ A+B = 45◦

17.tan

π

9+ tan

23π

36

1− tanπ

9tan

23π

36

= tan

(π

9+

23π

36

)= tan

3π

4= −1

19. sin 2θ = sin(θ + θ) = sin θ cos θ + cos θ sin θ = 2 sin θ cos θ

1. r = 6 cm, θ = 37.5◦ = 37.5( π

180

)=

5π

24rad

(a) s = 6

(5π

24

)=

5π

4cm

(b) A =1

2(6)2

(5π

24

)=

15π

4cm2

2. x = −1, y = −2, r =√

(−1)2 + (−2)2 =√

5

sin θ + cos θ + tan θ =−2√

5

5+−√

5

5+ 2 =

10− 3√

5

5

281



DEPED COPY

3. sinA =12

13, A is in QII =⇒ cosA = − 5

13, tanA = −12

5

cosB = −5

3, B is in QIV =⇒ sinB = −3

5, cosB =

4

5, tanA = −3

4

(a) cos(A−B) = cosA cosB + sinA sinB

=

(− 5

13

)(4

5

)+

(12

13

)(−3

5

)= −56

65

(b) tan(A−B) =tanA+ tanB

1 + tanA tanB=

−12

5+

(−3

4

)1 +−12

5

(−3

4

) = −33

56

4.tan 57◦ + tan 78◦

1− tan 57◦ tan 78◦= tan(57 + 78)◦ = tan 135◦ = −1

5.cosx tanx+ sinx

tanx= cosx+

sinx

tanx= 2 cos x = 2

√1− sin2 x = 2

√1− a2

6. cos6 x+ sin6 x = (cos2 x)3 + (sin2 x)3

= (cos2 x+ sin2 x)(cos4 x− cos2 x sin2 x+ sin4 x)

= cos4 x− cos2 x sin2 x+ sin4 x

= cos4 x− cos2 x(1− cos2 x) + (1− cos2 x)2

= cos4 x− cos2 x+ cos4 x+ 1− 2 cos2 x+ cos4 x

= 3 cos4 x− 3 cos2 x− 1

7. Connect the three diagonals of the hexagon. In doing this, the hexagon is

divided into 6 equilateral triangles. Hence, B

(1

2,

√3

2

). Same coordinates

for C, E and F , except that they will just vary in signs depending on thequadrant.

8. y = 2 sin(x

2+π

3

)− 1 =⇒ y = 2 sin

1

2

(x+

2π

3

)− 1

P = 4π, Phase Shift =2π

3, Amplitude = 2, Range = [−3, 1]



DEPED COPY

1. Asector =π

3cm2, θ = 30◦ = 30

( π

180

)=π

6rad

π

3=

1

2

(π6

)r2 =⇒ r = 2 cm =⇒ 2

(π6

)=π

3cm

2. x = 8, y = −6, r =√

(8)2 + (−6)2 = 10

(sin θ + cos θ)2 = sin2 θ + 2 sin θ cos θ + cos2 θ

= 1 + 2 sin θ cos θ = 1 + 2

(−6

10

)(8

10

)=

1

25

3. sinA = − 8

17

sin(π

2− A

)+ cos

(π2− A

)= cosA+ sinA =

15

17+−8

17=

7

17

4. sin 160 cos 35− sin 70 cos 55

= sin 20 cos 35− cos 20 sin 35

= sin(20− 35) = − sin(45− 30) =

√2−√

6

4

5. tan7π

12= tan

(π4

+π

3

)=

tanπ

4+ tan

π

3

1− tanπ

4tan

π

3

=1 +√

3

1−√

3= −2−

√3

6. cosA = −3

5, A is in QIII =⇒ sinA = −4

5, tanA =

4

3

tanB =24

7, B is in QIII =⇒ sinB = −24

25, cosB = − 7

25,

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

=

(−4

5

)(− 7

25

)+

(−3

5

)(−24

25

)=

4

5

(b) cot(A+B) =1− tanA tanB

tanA+ tanB=

1− 4

3

(24

7

)4

3+

(24

7

) = −3

4

7.tan2 x

tanx+ tan3 x=

tan2 x

tanx(1 + tan2 x)=

tanx

sec2 x= sinx cosx

283



DEPED COPY

8.sinx

secx= sinx cosx

sinx− cosx =1

3

(sinx− cosx)2 =

(1

3

)2

=⇒ sin2 x− 2 sinx cosx+ cos2 x =1

9

=⇒ 1− 2 sinx cosx =1

9=⇒ −2 sinx cosx = −8

9=⇒ sinx cosx =

4

9

9. y = tan( π

18− x

3

)+ 2 = − tan

1

3

(x− π

6

)+ 2

P = 3π, phase shift =π

6

√2

9

(b) cos 2θ = 79

(c) tan 2θ = 4√2

7

(d) sec 2θ = 97

(e) csc 2θ = 9√2

8

(f) cot 2θ = 7√2

8

3. cos(2t) = 18

5. tanx = 1−√5

2

7. cot 4θ = 1/(tan 4θ) = −7/24

9. sin2 5π8

= 2+√2

4and cos2 5π

8= 2−

√2

4

11.tan 1

2y − 1

tan 12y + 1

=

1− cos y

sin y− 1

sin y

1 + cos y+ 1

=

1− cos y − sin y

sin ysin y + 1 + cos y

1 + cos y

=1− cos2 y − sin y − sin y cos y

sin2 y + sin y + sin y cos y

=sin2 y − sin y − sin y cos y

sin2 y + sin y + sin y cos y

=sin y − 1− cos y

sin y + 1 + cos y.

284


1. (a) sin 2θ = 4


DEPED COPY

13. (a) cos 105◦ =

√2+√3

2(b) tan 22.5◦ =

√2− 1

1. (a) sin[sin−1(1/2)] = 1/2

(b) cos[cos−1(−√

2/2)] = −√

2/2

(c) tan[tan−1(−√

3)] = −√

3

(d) sin[arctan(√

3)] = −√

3/2

(e) cos[arccos(√

2)] does not exist

(f) tan[arcsin(1/4)] =√

15/15

(g) cos(sin−1√

3/2) = 1/2

3. (a) sin[2 cos−1(−4/5)] = −24/25

(b) cos[2 sin−1(5/13)] = 119/169

(c) sin[sin−1(3/5) + cos−1(−5/13)] = 33/65

(d) cos[sin−1(1/2)− cos−1(8/17)] = (15 + 8√

3)/34

5. (a) arcsec(−√

2) = 3π/4

(b) arccsc(−2) = −π/6(c) arccot

√3 = π/6

(d) [sec−1(−1)] · [cos−1(−1)] = π · π = π2

(e) 2 cot−1√

3 + 3 csc−1 2 = 2(π/6) + 3(π/6) = 5π/6

(f) csc−1 0 does not exist

7. Vertex angle θ should be π/3.

1. Solution set: {π/2, 3π/2, 5π/2, 7π/2, ...} = {(2k + 1)π/2 | k ∈ Z}

3. Solution set: {2kπ/3 | k ∈ Z}

5. Solution set: {π/8, 3π/8, 9π/8, 11π/8}

7. Solution set: {−π/2, π/6}

9. The bullet should be fired with an angle of θ = 60◦.

285




DEPED COPY

1. (a) (13, 7π/3), (13, 13π/3)

(b) (0, 2π), (0, π/4)

(c) (15, 7π/4), (15, 23π/4)

3. (a) (−3, 0)

(b) (−3√

2/2, 3√

2/2)

(c) (0, 5/2)

(d) (−4√

3,−4)

5. (a) r = 4 cos 2θ

θ 0 π/6 π/4 π/3 π/2 2π/3 3π/4 5π/6 π 7π/6 5π/4 4π/3 3π/2

r 4 2 0 −2 −4 −2 0 2 4 2 0 −2 −4

(b)

169

(b) sin(2θ) = −120169

(c) tan(2θ) = −120119

(d) sec(2θ) = 169119

(e) csc(2θ) = −169120

(f) cot(2θ) = −119120

2. Hint: Use the double-angle identity for tangent tan(2x) =2 tanx

1− tan2 x.

3. (a) tan 15◦ = 2−√

3 (b) tan 7.5◦ =4−√

6−√

2√6−√

2

4. (a) tan−1(

cot103π

6

)=π

3(b) cos

(sin−1

40

41

)=

9

41

5. Solution Set =

{0,π

2,3π

2

}286



1. (a) cos(2θ) = 119


DEPED COPY

6. Solution Set =

{0, π, 2π,

π

6,5π

6,7π

6,11π

6

}7. r = −2− 2 sin θ

θ 0 π/6 π/4 π/3 π/2 2π/3 3π/4

r −2 −3 −2−√

2 −2−√

3 −4 −2−√

3 −2−√

2

θ 5π/6 π 7π/6 5π/4 4π/3 3π/2

r −3 −2 −1 −2 +√

2 −2 +√

3 0

8. (a) (r, θ) = (−70, tan−1 43)

(b) (r, θ) = (100√

2, π4)

(c) (r, θ) = (7, π2)

(d) (r, θ) = (7, 0)

(e) (r, θ) = (2π√

2, π4)

(f) (r, θ) = (13, tan−1 125

)

9. (a) (x, y) = (32, 3√3

2)

(b) (x, y) = (45√2

2,−45

√2

2)

(c) (x, y) = (1, 0)

(d) (x, y) = (5, 0)

(e) (x, y) = (2π, 0)

(f) (x, y) = (−9√3

2, 92)



DEPED COPY

625(b) sin(2θ) = −336

625(c) tan(2θ) = 336

527

2. (a) cos2 24◦ ≈ 0.8346 (b) sin2 24◦ ≈ 0.1655 (c) tan2 24◦ ≈ 0.1983

3. (a) tan(π/12) = 2−√

3 (b) tan(π/24) =4−√

6−√

2√6−√

2

4. cos

(cos−1

1

7+ cos−1

3

5

)=

3− 16√

3

35

5. Solution set =

{π

4,3π

4,4π

3,5π

3

}6. Solution set =

{5π

8,7π

8,13π

8,15π

8,π

4,5π

4

}7. r = 2 + 2 cos θ

θ 0 π/6 π/4 π/3 π/2 2π/3 3π/4

r 4 2 +√

3 2 +√

2 3 2 1 2−√

2

θ 5π/6 π 7π/6 5π/4 4π/3 3π/2

r 2−√

3 0 2−√

3 2−√

2 1 2

288


1. (a) cos(2θ) = −527


DEPED COPY

8. (a) (r, θ) = (35, tan−1(−43))

(b) (r, θ) = (100√

2, 5π4

)

(c) (r, θ) = (5,−π2)

(d) (r, θ) = (5, π)

(e) (r, θ) = (π√

2, π4)

(f) (r, θ) = (17, tan−1( 815

))

9. (a) (x, y) = (2√

3, 2)

(b) (x, y) = (−50√

2,−50√

2)

(c) (x, y) = (−1, 0)

(d) (x, y) = (−5, 0)

(e) (x, y) = (−π, 0)

(f) (x, y) = (−152, 15√3

2)



DEPED COPY

References

[1] R.N. Aufmann, V.C. Barker, and R.D. Nation, College Trigonometry, HoughtonMifflin Company, 2008.

[2] E.A. Cabral, M.L.A.N. De Las Penas, E.P. De Lara-Tuprio, F.F. Francisco,I.J.L. Garces, R.M. Marcelo, and J.F. Sarmiento, Precalculus, Ateneo deManila University Press, 2010.

[3] R. Larson, Precalculus with Limits, Brooks/Cole, Cengage Learning, 2014.

[4] L. Leithold, College Algebra and Trigonometry, Addison Wesley LongmanInc., 1989, reprinted by Pearson Education Asia Pte. Ltd., 2002.

[5] M.L. Lial, J. Hornsby, and D.I. Schneider, College Algebra and Trigonometryand Precalculus, Addison-Wesley Educational Publisher, Inc., 2001.

[6] J. Stewart, L. Redlin, and S. Watson, Precalculus: Mathematics for Calculus,Brooks/Cole, Cengage Learning, 2012.

[7] M. Sullivan, Algebra & Trigonometry, Pearson Education, Inc., 2012.

[8] C. Young, Algebra and Trigonometry, John Wiley & Sons, Inc., 2013.

