Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
FOUNDATION MATHEMATICS MAT 110 MODULE 2 A COPPERBELT UNIVERSITY COLLEGE Mathematics Department Copyright © COPPERBELT UNIVERSITY COLLEGE 2010 Acknowledgements The COPPERBELT UNIVERSITY COLLEGE wishes to thank the following for their contribution to this module: EDITORS Directorate of Distance Education (DODE) Ministry of Education, Lusaka AUTHORS 1. Mr. Chikunduzi C. Lyson Lecturer 2. Mr. Kambilombilo Dennis Lecturer 3. Mr. Likando M.Kenneth. Lecturer 4. Mr. Malambo Priestly. Lecturer COPPERBELT UNIVERSITY COLLEGE BOX 20382 KITWE ZAMBIA E-mail: [email protected] Website: www.cce.co.zm Contents About this module 2 1 How this module is structured .......................................................................................... 1 Getting around this module 1 Margin icons ..................................................................................................................... 1 Module 2 overview 3 Is this module 2 for you? .................................................................................................. 3 Module outcomes .............................................................................................................. 3 Timeframe ........................................................................................................................ 3 Study skills ........................................................................................................................ 3 Need help? ........................................................................................................................ 5 Assignments ...................................................................................................................... 6 Assessments ...................................................................................................................... 6 Unit 1 FUNCTIONS 1.0 Introduction ................................................................................................................ 7 1.1 Definition of product set. ............................................................................................ 8 1.2 Relations ..................................................................................................................... 8 1.3 Concept of functions. ............................................................................................... 9 1.3.1 Domain and range of a function ..................................................................... 9 1.3.2 Types of functions ........................................................................................ 11 1.3.2 Even and odd functions ................................................................................ 11 1.3.3 Inverse functions .......................................................................................... 13 1.3.4 Composite functions ..................................................................................... 14 1.3.5 Graphs of quadratic functions ...................................................................... 16 1.3.6 Sketching graphs of quadratic ...................................................................... 17 Unit summary ................................................................................................................. 24 Unit 2 TRIGONOMETRY 25 2.1 Introduction ............................................................................................................... 25 2.2 Trigonometric functions ........................................................................................... 25 2.2.2 Special angles ............................................................................................... 28 2.2.3 Graphs of trigonometric functions ............................................................... 30 2.3 Trigonometric identities ............................................................................................ 35 2.4 Further trigonometric identities ............................................................................. 40 2.4.1 Trigonometric equations............................................................................... 45 2.4.2 Trigonometric equations of the form: .......................................................... 48 Unit summary ................................................................................................................. 53 Unit 3 EXPONENTIAL AND LOGARITHMIC FUNCTIONS ............................................. 53 3.0 Introduction ............................................................................................................... 53 3.1 Exponents.................................................................................................................. 56 3.2 Logarithms ................................................................................................................ 58 3.2.1 Definition...................................................................................................... 58 3.2.2 Laws of logarithms ....................................................................................... 59 3.2.2 The quotient law ........................................................................................... 60 3.2.3 The power law .............................................................................................. 61 3.3 Exponential equations ............................................................................................... 68 3.3.1 Solving exponential equations with the same base. ..................................... 68 3.3.2 Solving quadratic exponential equations ...................................................... 69 3.3.3 Solving exponential equations using the change of base method ................ 70 3.4 logarithmic equations ................................................................................................ 72 log x b c 3.4.1 Solving equations of the form log a b x and ........................... 72 3.4.2 Equations involving natural exponents and natural logarithms ............................. 81 3.5.0 Application of exponential and logarithmic function ............................................ 84 3.5.1 Decay and growth problems ......................................................................... 84 3.5.2 Compound interest........................................................................................ 85 3.5.3 Graphs of exponential and Logarithmic functions ................................................ 87 Graphs of exponential functions............................................................................ 87 3.5.4 The hyperbolic functions ....................................................................................... 95 Answers to activities ..................................................................................................... 101 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 About this module 2 Module 2 of Foundation Mathematics has been produced by Copperbelt University College and is structured in the same way, as outlined below. How this module is structured The course content The course is broken down into units. Each unit comprises: An introduction to the unit content. Unit outcomes. Core content of the unit with a variety of learning activities. A unit summary. Resources For those interested in learning more on this subject, we provide you with a list of additional resources at the end of this module Getting around this module Margin icons While working through this module you will notice the frequent use of margin icons. These icons serve to “signpost” a particular piece of text, a new task or change in activity; they have been included to help you to find your way around. Page 1 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 A complete icon set is shown below. We suggest that you familiarize yourself with the icons and their meaning before starting your study. Activity Assessment Assignment Outcomes Summary Time Help Note it! Page 2 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Module 2 overview Welcome to module 2 MAT 110 This module is based on introductory concepts in Mathematics is part of the foundation mathematics course. Is this module 2 for you? This module 2 is intended for Diploma holder teachers of Mathematics and Science that aim to further teaching skills . Module outcomes During and upon completion of this module you will be able to: Understand and apply knowledge of functions Understand and apply knowledge of Trigonometry. Timeframe The expected study time is 100 hours Study skills As an adult learner your approach to learning will be different to that from your school days: you will choose what you want to study, you will have professional and/or personal motivation for doing so and you will most likely be fitting your study activities around other professional or domestic responsibilities. Essentially you will be taking control of your learning environment. As a Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 consequence, you will need to consider performance issues related to time management, goal setting, stress management, etc. Perhaps you will also need to reacquaint yourself in areas such as essay planning, coping with exams and using the web as a learning resource. Your most significant considerations will be time and space i.e. the time you dedicate to your learning and the environment in which you engage in that learning. We recommend that you take time now—before starting your selfstudy—to familiarize yourself with these issues. There are a number of excellent resources on the web. A few suggested links are: http://www.how-to-study.com/ The “How to study” web site is dedicated to study skills resources. You will find links to study preparation (a list of nine essentials for a good study place), taking notes, strategies for reading text books, using reference sources, test anxiety. http://www.ucc.vt.edu/stdysk/stdyhlp.html This is the web site of the Virginia Tech, Division of Student Affairs. You will find links to time scheduling (including a “where does time go?” link), a study skill checklist, basic concentration techniques, control of the study environment, note taking, how to read essays for analysis, memory skills (“remembering”). http://www.howtostudy.org/resources.php Another “How to study” web site with useful links to time management, efficient reading, questioning/listening/observing skills, getting the most out of doing (“hands-on” learning), memory building, tips for staying motivated, developing a learning plan. The above links are our suggestions to start you on your way. At the time of writing these web links were active. If you want to look for more go to www.google.com and type “self-study basics”, “self-study tips”, “selfstudy skills” or similar. Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Need help? For any help contact the University College Through: 1. University College Secretary – phone +260 212293003 2. Email: [email protected] 3. The Mathematics HOD: +260 955 906714 4. The District Resource Centre Coordinator as advised. All contacts during office hours. Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Assignments Tutor marked assignments will be considered for assessment and to be submitted at stipulated times through the District Resource centres as advised. Assessments There will be a three hour test administered at the University College during a residential school session. Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Unit 1 FUNCTIONS 1.0 Introduction This Unit is aimed at acquainting you with functions which you certainly have met in your earlier study of Mathematics. Without doubt, this is one of the most important concepts in Mathematics and because of that; we shall carry you from the basics and build up the idea so as to prepare you for further Mathematics courses. You will finally learn how to sketch graphs of quadratic function. During and upon completion of this unit you will be able to: Describe a product set Discuss the concept of a relation Discuss the concept of a function Determine domains and ranges of functions Distinguish different types of functions Solve simple problems on linear functions (including their Work with composite functions Determine maximum and minimum turning points of quadra points Sketches graphs of quadratic functions. Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 1.1 Definition of a product set. With your knowledge of sets acquired in Module one, consider a situation where you have two sets A and B. We can define the product set of A and B denoted A B as follows: A B (a, b) : a A, b B. Example 1 Let’s suppose that sets A and B are such that A {3,4,7} and B {1,3} . Using our definition above, the product set of A and B is: A B {( 3,1), (3,3), (4,1), (4,3), (7,1), (7,3)} A product set has ordered pairs for its elements. Activity 1 Given two sets A and B as described in example 1 above, determine the product set B A .What relationship have you noticed between B A and A B ? 1.2 Relations. In simple terms, we can define a relation as a subset of ordered pairs. Let’s now use our knowledge of product sets to amplify and clarify the concept of a relation as follows: Consider the product set A B {( 3,1), (3,3), (4,1), (4,3), (7,1), (7,3)} . Several subsets of this set of ordered pairs A B can be derived such as: {( 3,1), (4,1)} , {( 7,1), (3,3), (4,3)} , {( 7,3), (7,1)} These subsets of ordered pairs are what are referred to as relations and usually the letter R is used to denote a relation. Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Activity 2 (i) Derive relations from the following product set: A B {( 3,1), (3,3), (4,1), (4,3), (7,1), (7,3)} (ii) Construct product sets and determine some of the relations which can be derived from such product sets. 1.3 Concept of functions. You are now ready to be introduced to the concept of a function. Definition: A function (mapping) can be defined as a special relation in which no ordered pairs have the same first entry. What this entails is that a function is a rule which assigns every starting element (input or object), say x X a unique ending element (output or image) say y Y . You could be aware that we usually use the letter f to denote a function, although any other letter can be used. You may equally recall that, functions can be expressed using formulas, diagrammatically or in functional notation. For instance: y f (x) represents a function in formula form where the object is x and the image is f(x). f : X Y represents a function f in functional notation. The meaning here is that ‘the function f maps the elements of set X (objects) onto elements of set Y (images). We shall illustrate the diagrammatic representation of quadratic functions latter in this unit. 1.3.1 DOMAIN AND RANGE OF A FUNCTION A domain of a function f denoted D (f), is a set composed of all possible objects of such a function, whereas a range of f denoted R (f) could be defined as a collection of all images or out puts of f. Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Activity 3 (i) Determine which of the following relations are functions and those which are not: R1 (3,10), (4,10), R2 (7,10), (3,10), (7,13), R3 (10,7), (13, ) (ii) For the functions in part (i), state the domain and range. Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 1.3.2 TYPES OF FUNCTIONS In this section, we are going to study different types of functions. (a) Surjective (onto) function: Let f be a function where y Y , x X and that Y=R (f), X = D (f). Then f is said to be surjective (onto) if every y Y is an image of some x X . Thus, when R (f) is equal to the entire co-domain of f, we say that such a function f is surjective (onto). (b) Injective (one-to-one) function: Suppose that y f (x) is a function and that x1 , x2 D( f ) . The function f can only be said to be injective (one-to-one) if: f ( x1 ) f ( x2 ) x1 x2 . The implication here is that a oneto-one function is one where every image has only one object assigned to it. (c) Many-to-one function: Any function which is not one-to-one is called a many to one. (d) Bijective function: This is a function which is both Surjective (onto) and injective (one-to-one). 1.3.2 EVEN AND ODD FUNCTIONS An even function say y f (x) is one which is such that for all x D( f ) , f ( x) f ( x) . This implies that for a function f to be even, f ( x) f ( x) 0 . On the other hand, a function f (x) can be said to be odd Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 if f ( x) f ( x) for all values of x. Alternatively, f (x) is odd if for all values of x, f ( x) f ( x) 0 . Even functions are symmetrical along the y-axis ( x 0) , whereas odd functions are symmetrical about the origin (0, 0) through 180 0 . Example 2 Show that g ( x) 5 x 6 is a one-to-one function. Solution: Let x1 , x2 D( g ) . g ( x1 ) g ( x2 ) 5x1 6 5x2 6 5 x1 5 x 2 x1 x 2 Since g ( x1 ) g ( x2 ) x1 x2 , it has been shown that g (x ) is a one-to-one function. Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Activity 4 Show that the following functions are one-to-one: f ( x) (i) x 1 , 2x h( x ) 2 x3 Determine which one of the following functions is even and which one is odd: y x 2 , y x3 (ii) Determine whether f ( x) x 1 is odd, even or neither of the two. 1.3.3 INVERSE FUNCTIONS By now you should be familiar that a function maps objects from the domain onto unique images in the range. Nevertheless, let’s now suppose that we require a rule which would map images from the range onto their respective objects in the domain. Such a rule when it exists, and as long as the characteristic of a function is met, is called an inverse function. You may already know that, for functions say f ( x), g ( x), h( x) , their respective inverse functions are denoted by f 1 ( x), g 1 ( x), h 1 ( x) . (i) You should never regard the superscript -1 in an inverse function as an exponent. Thus, f 1 ( x) 1 . f ( x) (ii) Only one-to-one functions are invertible (have inverses). The following example is meant to familiarise you algebraically with a process of finding an inverse of a Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 given one-to-one function. Example 3 Compute the inverse function of y 6 x 3 Solution: For you to find the inverse of this function, you ought to start by making the object variable x the subject of the formula in y 6 x 3 as follows: y 6x 3 y 3 6 x 3 3 (subtracti ng 3 on both sides) y 3 6x y 3 x (dividing by 6 on both sides) 6 y 3 x 6 You should then replace the variable y by x on the right hand side of the preceding equation and use the notation of inverse functions as follows: f 1 ( x) x3 (which is the required inverse function). 6 1.3.4 COMPOSITE FUNCTIONS There are instances when you confront single functions which are a combination of two or more functions. Such functions are called composite functions. Let f (x) and g (x ) be two functions. The composition of f (x) and g (x ) , denoted ( f g )( x) is given by: ( f g )( x) = f [ g ( x )] and the composition of g (x ) and f (x) denoted ( g f )( x) is given by: ( g f )( x) = g[ f ( x )] . You may be wondering what ( f g )( x) Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 and ( g f )( x) means. Let us therefore address this concern: The function ( f g )( x) implies that the function g operates on the object x first which gives an image upon which the function f operates on. Thus, the image of g becomes the object of f in the composite function ( f g )( x) . Similarly, ( g f )( x) means that f operates on the object x and then g operates on the resulting answer. Example 4 Given that f ( x) 2 x 1 and g ( x) 1 x , find the 1 x following functions: (i) ( f g )( x) (ii) ( g f )( x) Solution: 1 x (i) ( f g )( x) = f [ g ( x)] f 1 x 1 x 2 1 , the object replaced by g ( x) 1 x = 2 2x 2 2x 1 x 1 1 x 1 x Hence, ( f g )( x) = 3 x 1 x (ii) ( g f )( x) = g[ f ( x)] g (2 x 1) , replacing the object by f(x). g[ f ( x)] g (2 x 1) = = 1 (2 x 1) 2 x 2 1 (2 x 1) 2x 1 x , factoring out 2 and dividing by 2. x Hence, ( g f )( x) = 1 x x Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 (i) From the preceding example, you may have observed that ( f g )( x) ( g f )( x) . (iii) Given two functions say f ( x) and g ( x) which are one-to-one, the following hold: (a) ( f f 1 )( x) ( f 1 f )( x) x (b) ( f g ) 1 ( x) ( g 1 f 1 )( x) (c) ( g f ) 1 ( x) ( f 1 g 1 )( x) Activity 5 If f ( x) x 4 and g ( x) 5 x 2 , find the following composite functions: (i) ( f f 1 )( x) (ii) ( f 1 f )( x) (iii) ( f g ) 1 (iv) ( g 1 f 1 )( x) (v) ( g f ) 1 ( x) (vi) ( f 1 g 1 )( x) 1.3.5 GRAPHS OF QUADRATIC FUNCTIONS In Module 1, you covered quadratic functions and different methods of finding their roots. You equally studied the concept of a discriminant of quadratic functions. However, in this section, you will learn techniques of sketching graphs of such functions. Additionally, you will study aspects of minimum and maximum values of functions, including the minimum and maximum turning points of quadratic functions. Are you able to remember that a quadratic function is one of the form: Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 f ( x) ax 2 bx c , where a, b and c are constants and a 0 ? If not, you will do well to master this fact. Since we are now interested in graphs of quadratic functions, this is the right stage, to let you know that a graph of a functions like f(x) is called a Parabola. Further, parabolas will either open upwards or downward as shown below: Fig.2 Parabola Opening upwards Fig 1. Parabola opening downward You may be wondering how you can sketch the graph of f(x). This is exactly what we shall study in the following section. 1.3.6 SKETCHING GRAPHS OF QUADRATIC UNCTIONS For you to draw a sketch of the graph of f ( x) ax 2 bx c , where a, b and c are constants and a 0 , the following points must be taken into account: 1. Firstly, determine whether the parabola opens upwards or downwards by establishing the characteristic of the coefficient of x 2 in f ( x) . If a 0 then the parabola opens upwards. However, if a 0 , then the parabola opens downwards. 2. Determine the y-intercept by letting x 0 in f (x) . 3. Your next step should be to establish whether the graph of f (x) cuts the xaxis at two different points, exactly one point or does not touch the x-axis at all. For you to do so, employ the discriminant: b 2 4ac . (Refer to MAT 110). After this, proceed to computation of the Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 x-intercept/s by solving f (x) =0. 4. Further, you ought to determine the turning point of f (x) , and this is given b 4ac b 2 . by , 2 a 4 a 5. With the information in the steps above, you can then sketch the graph of f (x) . Let us now use these steps above to sketch the graph of a quadratic function in the following example: Example 5 Sketch the graph of the function f(x) = -2x2 + 5x + 3 Solution: From f(x), a 2 0 . This implies that the curve of function f(x) opens downwards. You now require finding the y-intercept by letting x=0 in f(x) as follows: f (0) 2(0) 2 5(0) 3 =3 (0,3) is the point at which the graph of f(x) meets the y-axis. We shall use the discriminant to discover the behaviour of the graph in relation to the x- axis: From f(x), b 2 4ac = (5) 2 (4 2 3) 1 0 . Since the discriminant is greater than Zero, the curve of f(x) touches the xaxis at two different points. You can find these points by letting y=0 in f(x) as follows: f ( x) 2 x 2 5 x 3 =0 2 x 2 5 x 3 0 (2 x 1)( x 3) 0 (Factorising the left hand side) either 2x 1 0 or x 3 0 x 1 0.5, x 3 2 Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Therefore, the curve touches the x-axis at ( 1 , 0) and (3,0) 2 According to our earlier considered five steps, the turning point is given by: b 4ac b 2 . Now since a 2, b 5, c 3 from f(x), the , 2 a 4 a required turning point 5 4(2)3 52 5 49 , is , . This 2 2 4 2 4 8 information should help you to sketch the graph of f(x) = -2x2 + 5x + 3 as shown below: Fig 3. Graph of f(x) = -2x2 + 5x + 3 Let us consider another example relating to graphs of quadratic functions: Example 6 Sketch the graph of the quadratic function f(x) = x2 + 2x − 3. Solution: By now it should be clear to you that a 1, b 2 and c 3 , from the function f(x). What do you think about the whether the curve opens downwards or upwards? You may have noticed that the curve shall open upwards, since a 0 . Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 The discriminant b 2 4ac 2 [4 1 3] 2 12 14 0 . Do you remember the implication of this value of the discriminant? A positive value implies that the curve will cut the x-axis at two different points. 2 Let us now find those two x-values at which the graph shall cut the horizontal axis. This shall be done by letting y 0 in the function as follows: x 2 2x 3 0 x 2 3x x 3 0 x( x 3) 1( x 3) 0 (Factorising the left hand side) ( x 3)( x 1) 0 x 3 0 or x 1 0 x 3 or x 1 are the two values at which f(x) cuts the x-axis. Before sketching, you also need to ascertain the y-intercept and the turning point of f(x). We shall begin with computing the y-value at which the curve crosses the vertical axis and to do so we let x=0 in f(x) as follows: f (0) (0) 2 2(0) 3 3 y 3 is the sought intercept. b 4a b 2 ? If not, Do you recall that the turning point of f(x ) is given by , 4a 2a revisit the section which deals with this aspect before proceeding. Assuming that you are familiar with turning points, you will notice that the turning point of f(x) 2 (4 1 3) (2) 2 1,4 . is , ( 2 1 ) ( 4 1 ) With all the above information, try to sketch the graph on a piece of paper. Did your graph look like the one below? If not, where are the variations and what could be the justification? If yes, then you have acquired the concepts and skill expected. Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Fig 4. Graph of f(x) = x2 + 2x − 3: (i) A quadratic function whose parabola opens upwards (u-shaped and where a > 0), is said to have a minimum value at the turning point and such a turning point is called a minimum turning point. The turning point of f(x) in example 6, is a minimum turning 4ac b 2 point. Further, the range of f(x) in this case is y . 4a (ii) A quadratic function whose parabola opens downwards (nshaped and where a < 0), has a maximum value at the turning point. A turning point of such a parabola (opening downwards) is called a maximum turning point. Further, the range of f(x) in this respect is y 4ac b 2 . 4a b 4a b 2 is found by completing the (iv) The turning point expression , 4a 2a square for f ( x) ax 2 bx c , where a, b and c are constants and a 0 . In order for you to strengthen your understanding of sketching graphs of quadratic functions, the following questions are provided for that purpose. Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Activity 6 1. Sketch the graphs of the following quadratic functions: (a) f(x) = 2x2 − 8x + 6 (b) f(x) = -x2 + x + 6 (c) f ( x) x 2 5x 2 (d) f ( x) 4 x 2 x 1 (e) f ( x) 2 x 2 5 x 3 2. Determine the minimum and maximum values of the following functions where applicable: Further, indicate the ranges of each given function. f ( x) 2 x 2 5x 3, f ( x) 3x 2 7 x 1, f ( x) 9 x 2 30 x 25 f ( x) 36 48x 9 x 2 Unit Summary In this unit you learnt concepts of product sets, relations and functions. You have also learned how to calculate inverses of linear functions. The idea of composite functions was tackled as well as how to find composite functions. The last aspects which were considered are the sketching of graphs of quadratic functions, minimum and maximum values of quadratic functions and minimum and maximum turning points of graphs of quadratic functions. Page 12 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Unit 2 TRIGONOMETRY 2.1 Introduction In this unit, you will learn of the six trigonometric functions. Curves of sin , cos , tan shall be covered. You will further learn trigonometric identities and how to prove given trigonometric identities. The last section shall be based on how to solve trigonometric equations. During and upon completion of this unit you will be able to: Work with six basic trigonometric functions. Prove trigonometric identities Solve problems involving difference and sum trigonometric formulas. Work with double and half angle formulas. Solve trigonometric equations. 2.2 TRIGONOMETRIC FUNCTIONS To start with, you shall be reminded of the three basic trigonometric functions considered during your earlier study of mathematics. However, you will study three other additional trigonometric functions which shall act as a basis for proofs of identities. You shall in addition learn how to find solutions of trigonometric equations. Let us consider the right angled triangle ABC below in which angle ABC= and angle ACB= 90 0 . Page 25 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Figure 5 We define the three basic trigonometric ratios as follows: opposite side AC adjacent side BC , cos , hypotenuse AB hypotenuse AB opposite side AC and tan adjacent BC sin tan AC sin sin AB AC tan , since cos cos BC BC AB You should now extend your knowledge to the other three trigonometric functions, and these are explained below: 1. The secant of an angle written secant θ. We shall however, be using short form: sec . By definition, you should know that: sec 1 . What identity can you derive from this definition of sec ? cos You may have deduced that sec cos 1. 2. The second of the other three trigonometric functions is the cosecant of an angle written as cosecant θ. You should get acquainted with the short form: csc , where csc 1 . We can derive an identity form this sin definition and this is: csc sin 1 . Page 26 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 3. Lastly, one other trigonometric function is the cotangent of an angle written as cotangent cotangent θ. You require a short form for this function as well, which is cot . In terms of the basic trigonometric functions, cot 1 cos . tan sin consequently, one of the identities which could be derived from this definition is: cot tan 1 . The last three trigonometric functions (cos ecant , cot angent , sec ant ) are reciprocals of the basic three ( sin , cos , tan ) . 2.2.1 ANGLE MEASUREMENT Let us start by stating that angles can either be measured in degrees or radians. It is usually easy to use degrees when talking about angles. However, at this level, you should start using radians as well. What follows therefore, is information meant to acquaint you with the usage of radians. Suppose that the diagram AOBCA below represents a sector of a circle in which the radius is r and the arc ACB is of length k r kr where k is a constant and O is the centre of the circle to which the sector belongs. Figure 6 It follows that the angle which the arc ACB formulates at the centre O is worth k radians. Similarly, when the length of an arc of a circle with radius Page 27 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 r units is r units, such an arc forms an angle of size one (1) radian at the centre. Can you now consider a circle whose radius is r units. You may recall that the circumference of such a circle is given by 2 r 2 r . What do you think then will be the size of the angle, in terms of radians, formed by the circumference of the circle at the centre? Arising from our earlier discussion, the angle will basically be 2 radians. Taking into account that, in terms of degrees, a circle is worth 360 0 , we can consequently conclude that: 360 0 2 Radians………………………………….. *1 . Using *1 , we can always derive relationships between angles in degrees and those measured in radians. The following are some of the several relationships: 180 0 Radians 90 0 45 0 2 Radians Radians etc 4 Can you now derive other equivalences between angles in degrees and those in radians? 2.2.2 SPECIAL ANGLES As you work with trigonometry, you shall meet situations where it will be important for you to make use of angles deemed ‘special’. These angles are special because you naturally do not require to use a calculator or any other technological aid to evaluate a trigonometric functions involving them. Thus, you can easily evaluate trigonometric expressions involving the said ‘special angles’. For our study, we shall only consider three angles: 30 0 or 6 , 45 0 or 4 , and 60 0 or 3 , and these are the ones that are referred to as the ‘special angles’ . Page 28 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Activity 7 Consider the isosceles triangle ABC and the equilateral triangle XYZ below in which AB=1, BC=1, AC= 2 , angle ABC= 90 0 and XY=2, XZ=2, YM=1, MZ=1, XM= 3 , angle XYZ= 60 0 , angle YXM= 30 0 and angle XMZ= 90 0 . Figure 7 Figure 8 Use the two triangles to find the values of the following: sin 450 , cos 450 , tan 450 , sin 60 0 , cos 60 0 , tan 60 0 , sin 30 0 , cos 30 0 tan 30 . Compare your answers with the following: According to the isosceles and equilateral triangles given above, sin 45 1 sin 60 3 1 3 , cos 60 , tan 60 3 and 2 2 1 2 , cos 45 1 1 , tan 45 1 1 2 Page 29 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 sin 30 1 3 1 , cos 30 , tan 30 2 2 3 Where your answers different from the ones given above? If yes, how different and why were they different? You shall be required to use the three angles latter on, but for now, it suffices for you to know how to evaluate trigonometric functions using them. It should be mentioned that the easiest way to remember the values of trigonometric functions involving ‘special angles’ is to master the two triangles given above (isosceles and equilateral). 2.2.3 GRAPHS OF TRIGONOMETRIC FUNCTIONS Have you ever thought of how you can draw the graphs of trigonometric functions? Anyway, this section is intended to give you the knowledge of the graphs of trigonometric functions. However, we shall only consider graphs of the three basic trigonometric functions (i.e. sin , cos , tan ). Zeros of the function y=sin Before we draw the graph of sin , it is important for you to learn certain characteristics of this function. Firstly, you need to know of angles which would give a value of zero for the sine function. From the onset, you should know that the sine function will have a value of zero at radians ( 180 0 ) . Now let us suppose that we write a term like ‘ n ’ where n belongs to the set of integers. Upon investigation, you will notice that sin has a value of Zero (0) at ‘ n (i.e. 0, ±π, ±2π, ±3π, …….), for n Z . Angles like n , for n Z and where the function has a value of zero are called the ‘zeros’ of the function. Alternatively, by the zeros of sin θ, we mean those values of θ for which sin θ will equal 0. The next question which Page 30 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 arises is that of the maximum or minimum value of sin . This concern is addressed in the next point. The function y sin is continuous and will have values lying in the interval 1 sin 1. What this means is that y sin has -1 as the its minimum value whereas its maximum value is 1. 2.2.4 Graph of the function y=sin Taking into account the discussion in 2.4.4, we shall now draw a sketch of the function y sin , and the following is the required sketch of y sin : Fig 9 2.2.5 Period of the function y=sin You may have noticed from the graph of y sin that the values of this function started repeating themselves after sometime. For instance, the value of y sin at 0 , is same as its value at 2 . When the values of a function regularly repeat themselves, we say that the function is periodic, and the length of the repetitive pattern is what constitutes the period of a function. What do you think is the period of y sin ? You may have discovered that the graph of this function repeated itself after 2 radians and of course this is its period. 2.2.6 Graph of the function y=cos Just like y sin , the function y cos has a period of 2 ( 360 0 ) and is continuous such that 1 cos 1 . The following is a sketch of the Page 31 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 graph of y cos . Fig 10 You may have recognized that the graph of y cos is similar to that of y sin . Apparently, when you translate the curve of y sin to the left by 2 2.2.7 or 90 0 , the result is the curve of y cos . Graph of the function y=tan Do you think that the function y tan is defined for all real values of ? If yes, how and if not, at which points is it discontinuous? Your investigation should have revealed that y tan , is not defined at or at 2 any other angle which a multiple is of . Nevertheless, the range of values 2 for y tan is infinite. What follows now is the sketch of the graph of y tan and certainly this curve reveals the preceding arguments. Sketch of y tan : Page 32 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Fig 11 I, II, III, and IV are referring to the first, second, third and fourth quadrants respectively. Page 33 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Activity 6 1. Determine the domain and range of the function y sin . Expected answers: The domain is (any real value) The range is: 1 y 1 . 2. Show that y sin odd function. 3. Determine the zeros of y cos . 4. What is the period of y tan ? Page 34 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 2.3 TRIGONOMETRIC IDENTITIES You may have heard of the term ‘identity’ used in mathematics. Take a few seconds to reflect on the meaning of trigonometric identities and what it does not mean. What ideas came into your mind upon your reflection? We shall basically describe an identity as any relationship which is true for all permissible values of the variables involved. However, the objective in this unit is to acquaint you with as many identities and proofs as is possible. Let us start by deriving one of the important identities which shall certainly be useful in our proofs of other identities. Consider the diagram below in which POH , PO=r, PH=y and OH=x. Fig 12 Remember that sin and cos y , y r sin .......................1 r x , x r cos .............................................2 r Using the triangle OPH and the Pythagoras theorem, we have: x 2 y 2 r 2 ..........................................................................3 We are now going to use information from (1) and (2) to express (3) as follows: Page 35 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 r cos 2 r sin 2 r 2 r 2 cos 2 r 2 sin 2 r 2 r 2 cos 2 sin 2 r 2 r2 r2 2 2 , cos sin r2 r2 (dividing by r 2 )………………….. (4) From (4) we derive our required and significant identity: cos 2 sin 2 1 …………………………………………………..(5) You can derive several other identities from (5) such as the following: sin 2 1 cos 2 cos 2 1 sin 2 tan 2 1 sec 2 1 cot 2 csc 2 etc. With your knowledge of the above basic identities, you should now be ready to do proofs of trigonometric identities. From the onset, you ought to know that there is no specific method for proving identities. Does this worry you? Be comforted that even when that is the case, you will find proofs on identities very interesting and enriching. Let us now prove some identities: Example Prove the following identities: (a) 1 1 2 sec 2 x 1 sin x 1 sin x (b) 1 1 2 tan x sec x 1 sin x 1 sin x (c) sec x 1 sin 2 x sec x 1 sec 2 x Page 36 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Solutions: (a) 1 1 2 sec 2 x . 1 sin x 1 sin x In order for you to prove a trigonometric identity, you ought to start with the most complicated side and show that, such a side is equal to the other side. For the identity given, it would appear that the most complicated side is the left hand side. We shall therefore start from there. Left hand side: 1 1 1(1 sin x) 1(1 sin x) 1 sin x 1 sin x (1 sin x)(1 sin x) 1 sin x 1 sin x 2 , (1 sin x)(1 sin x) 1 - sin 2 x since 1 sin 2 x (1 sin x)(1 sin x) 2 , cos 2 x 1 sin 2 x 2 cos x 1 2 2 sec 2 x 2 cos x = Right hand side (b) 1 1 2 tan x sec x . 1 sin x 1 sin x Again we shall start from the left hand side as follows: Page 37 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 1 1 1 sin x (1 sin x) , expressing as a single fraction. 1 sin x 1 sin x (1 sin x)(1 sin x) 2 sin x 1 sin 2 x 2 sin x 2 sin x 2 cos x cos x cos x 2 sec x tan x R.H.S. hence proved. (c) sec x 1 sin 2 x . Let us start from the right hand side this time. sec x 1 sec 2 x sin 2 x sin 2 x sec x 1 , multiplyin g by a special one (1) sec x 1 sec x 1 sec x 1 sin 2 x(sec x 1) , because sec 2 x 1 (sec x 1)(sec x 1) 2 sec x 1 sin 2 x(sec x 1) tan 2 x 1 1 sin 2 x(sec x 1) sin 2 x(sec x 1) tan 2 x sin 2 x cos 2 x sin 2 x(sec x 1) cos 2 x 1 (sec x 1) 2 sin x sec 2 x sec x 1 R.H.S. hence proved sec 2 x Page 38 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 The activity which follows is aimed at giving you an opportunity to practice on the proofs of trigonometric identities. Activity 7 1. Prove the following trigonometric identities: (a) cos x 1 sin x sec x tan x (b) sin x cos xtan x cot x sec x csc x cos 2 x 3 sin x 1 1 (c) 2 cos x 2 sin x 2 1 csc x (d) sec x cos x sin x tan x (e) tan 2 x sin 2 x tan 2 x sin 2 x (f) sec 4 x tan 4 x sec 2 x tan 2 x (g) sin x tan x cos x sec x (h) tan x cot x sec x csc x (i) tan y sec y sin y (j) sin y sin y cot 2 y csc y (k) sin x cos x tan x 1 cos 2 x (l) sin sin 7 sin 5 cos cos 7 cos 5 (m) sin sin 2 sin 3 sin 4 4 cos (n) cos cos 2 tan sin sin 2 2 (o) 1 sin 2 sec tan 1 sin (p) tan cot 1 sec csc sin cos (q) sec csc cot sex csc 2 Page 39 2 cos sin 5 2 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 2.4 FURTHER TRIGONOMETRIC IDENTITIES Suppose that A and B are angles measured in either degrees or radians, then you should know the following identity holds: cos( B A) cos B cos A sin B sin A...........................(1) We can derive another identity using (1) as follows: cos( B A) cos[ B ( A)] Now take a close examination of the steps below: cos[ B ( A)] cos B cos( A) sin B sin( A) cos B cos A sin B sin A, since cos( A) cos A (which is an even function ). Additional ly, sin( A) sin A, ( odd function). We have therefore derived our desired idntity, which is : cos( B A) cos B cos A sin B sin A.........................(2) Let us now solve questions using identities (1) and (2): Example Evaluate the following without use of a calculator (a) cos 15 0 (b) cos 7 12 Solutions: (a) We can use the special angles 45 0 and 30 0 to express cos 15 0 Page 40 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 as cos( 45 0 30 0 )............................. *0 . A look at *0 shows that it is similar to the expression in (1), and usage of identity (1) gives us cos(45 30) cos 45 cos 30 sin 45 sin 30............. *2 . Can you remember the values of cos 45, cos 30, sin 45, sin 30 ? In case you have forgotten, go back and restudy the section on ‘special angles’. However, if you understood that section, you will discover that *2 could be written in the following form: 1 3 1 1 cos( 45 30) . We ought to rationalising the 2 2 2 2 denominators, on the right hand side of the preceding equation as shown below: 2 1 3 2 1 1 cos( 45 30) 2 2 2 2 2 2 = 6 2 4 4 6 2 4 Hence, cos15 0 6 2 4 (b) In order for us to find the value of cos of all express 7 , we should first 12 7 either as a sum or difference of ‘special 12 angles’. For instance, cos 7 = 12 3 4 and because of this, 7 = cos . 12 3 4 We can now evaluate cos using identity (2) as 3 4 follows: cos = cos cos sin sin 3 4 3 4 3 4 Page 41 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 1 3 1 1 . At this stage, we ought to rationalise 2 2 2 2 = the denominators and doing so would give us: 3 1 2 1 1 2 cos = 2 2 2 2 3 4 2 2 = 2 6 . 4 4 Consequently, cos 2 6 7 = . 4 4 12 Have you perceived how easy it is to use ‘special angles’ and identities to evaluate trigonometric expressions? If you have not yet mastered the skill, we advise that you go through the above examples again before proceeding. There are other identities which you should know and one of such is developed for you below: Let us suppose that we are given a real angle B. It would then follows that: cos B cos cos B sin sin B...................(3) 2 2 2 Since we know that cos 2 =0 and sin 2 =1, equation (3) becomes: cos B = 0 cos B 1 sin B sin B 2 cos B sin B ………………………….. (4) 2 The activity which follows is intended to help you to strengthen your understanding of the process of deriving similar identities. Attempt it. Page 42 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Activity 8 Follow the steps we have undergone to deduce (4) to show that: (a) sin B cos B 2 (b) tan B cot B 2 Having already considered identities relating to the cosine of a sum and difference of two real angles, let us now derive those for the sine of a difference and sum of two real angles. Suppose A and B are two real angles. Then note that: sin B A cos[ B A] cos[ B A] 2 2 cos B cos A sin B sin A 2 2 sin B cos A cos B sin A. sin B A sin B cos A cos B sin A..............(5) In addition to the above identities, take time to master the following as well: 1. sin B A sin B cos A cos B sin A 2. tan B A tan B tan A 1 tan B tan A 3. tan B A tan B tan A 1 tan B tan A Another category of useful formulas which you require to know, is that of DOUBLE ANGLE FORMULAS. We now state these formulas below for your consideration and understanding: Page 43 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 cos 2 A cos 2 A sin 2 A cos 2 A 1 2 sin 2 A cos 2 A 2 cos 2 A 1 sin 2 A 2 sin A cos A tan 2 A 2 tan A 1 tan 2 A Let us now go through some examples which make use of some the identities we have studied above. Example: Evaluate the following without using a calculator: (a) sin 11 12 (b) sin 195 0 Solutions: (a) Before we can evaluate sin 11 11 , we should express either as a 12 12 a sum or difference of two ‘special angles’. Upon investigation, you will discover that 11 11 7 7 . Consequently, sin . = =sin 12 12 6 4 4 6 With this discovery, we can now use equation (5) to do the evaluation as follows: 7 7 7 cos cos sin = sin sin 6 4 6 4 4 6 1 1 3 1 . = 2 2 2 2 If we rationalise the denominators as we did for our similar calculations, Page 44 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 We would end up with Thus, sin 6 2 . 4 4 6 2 11 = . 12 4 4 (b) sin 195 0 = sin 150 45 = sin 150 cos 45 cos150 sin 45 1 1 3 1 = 2 2 2 2 = 2 6 4 4 You have now known what double angle formulas are about, but you still need to Learn something about HALF-ANGLE FORMULAS, and these are provided for your masterly below. Thus, given that A is a real angle, then the following hold: 1. cos B 1 cos B 2 2 2. sin B 1 cos B 2 2 3. tan B 1 cos B 2 1 cos B 2.4.1 TRIGONOMETRIC EQUATIONS With all the knowledge of trigonometry we have acquired so far, you are now at a stage where you should use the same to solve trigonometric equations. Do not be worried about the term ‘trigonometric equations’. These are merely equations which are composed of trigonometric terms, and the objective is to find angles which balance such equations. Page 45 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Example Solve the following trigonometric equations: (a) 2 sin 2 sin 0 for 90 0 90 0 (b) 2 tan 2 3 tan 0, for 0 0 360 0 Solutions: (a) 2 sin 2 sin 0 for 90 0 90 0 . Here we shall express the term 2 sin 2 in form where it involves sin and cos . How do you think this can be done? Hopefully, you have recalled the identity sin 2 2 sin cos ................(1) Let us now use (1) in our given equation as follows: 22 sin cos sin 0 If we expand on the left hand side of the preceding equation, we would obtain: 4 sin cos sin 0 ………………(2) Let us now factorise the left hand side of (2), and doing so gives us: sin 4 cos 1 0 . The implication then is that either sin 0 ………………….(3) or 4 cos 1 0 …………………(4) We shall begin by using (3) to solve for as follows: sin 0 sin 1 0 . Arising form the given interval 900 900 , the only value of which satisfies sin 0 is 0 0 . Now let us use equation (4) to find the remaining solutions: From 4 cos 1 =0, we have: 1 cos .........................(5) 4 1 cos 1 75.52 0 4 You may have noticed that 75.52 0 lies within the given interval Page 46 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 900 900 . If we closely examine equation (5), we would discover that the cosine function has a positive value also in the fourth(4th ) quadrant. Therefore, the other value of which lies in the interval 900 900 is 75.52 0 . Hence the required solutions are: 0 0 , 75.52 0 and 75.52 0 . (b) 2 tan 2 3 tan 0, for 0 0 360 0 Let us factorise the left hand side of the given equation as follows: tan 2 2 tan tan 2 0 tan tan 2 1tan 2 0 tan 1tan 2 0 either tan 1 0 .........................(1) or tan 2 0....................................( 2) We shall start by using equation (1) to find the required angles: tan 1 0 tan 1 ………………..(3) Then tan 1 1 450 From (3), we take note that the tangent function is giving a positive value. The angle 45 0 lies in the first quadrant, and but the tangent function tan is also positive in the third quadrant. We therefore require to find the angle which lies in that quadrant. What do you think this angle is? Have you found the angle 225 0 ? If yes, how did you find it? If your answer is different from 225 0 , what might have caused this? Nevertheless, for you to find the value of the angle in the third quadrant ( 225 0 ), you merely need to add 180 0 and the acute angle earlier computed ( 45 0 ). Now that we have used equation (1), let us also use equation (2) to find the other values of . From tan 2 0 , we know that Page 47 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 tan 2 tan 1 2 Thus, 63.4 0 , which lies in the interval 00 3600 . Do you think this is the only solution here? You should take note that tan is has a positive value here also. It therefore follows that we should compute the angle which lies in the third quadrant (where tan is positive). In order for us to find that angle which lies in the third quadrant, we are going to add 180 0 and the acute angle 63.4 0 . Thus, the required angle is 243.4 0 . We have now come to the end of the calculation. Hence solutions are: 450 ,225 0 ,63.4 0 and 243.4 0 2.4.2 TRIGONOMETRIC EQUATIONS OF THE FORM: a cos b sin c For us to find solutions of equations of the form a cos b sin c , we assume that the numerical values a, b and c are appropriately chosen. Additionally, before proceeding to solve such equations, we should confirm that a 2 b 2 c 2 . The following are the steps we ought to go through to solve this type of equations (although any other workable method can be used): 1. Multiply and divide each term on the left hand side of the equation by a 2 b 2 as shown below: a b a2 b2 cos sin ....................(i ) 2 2 a2 b2 a b 2. Secondly, we ought to introduce an auxiliary angle say such that cos a a2 b2 ................................................(ii ) By using information in ( ii ) and Pythagoras theorem we then have: Page 48 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 sin b ..............................................(iii ) a b2 2 3. Using information from ( ii ) and ( iii ), the expression ( i ) transforms into: a 2 b 2 cos cos sin sin ......................(iv ). We are now going to use our knowledge of identities to summarise ( iv ) and once summarised, it becomes: a 2 b 2 cos ...............................................(v) 4. Observe that the expression ( v ) is equal to a cos b sin . Thus, a cos b sin = a cos c 2 a b2 2 b 2 cos =c. Consequently, ......................................(vi) Alternatively, if we let sin a a2 b2 in step ( ii ), where is the auxiliary angle, and go through a similar process, we can deduce that: sin c a2 b2 …………………………….. (vii) 5. With either equation ( vi ) or (vii) , we can then solve for the required values of . (i) The general solutions of an equation cos Z , where Z 1 , are given by : 2n or 360 0 0 , for all n Z (ii) The general solutions of an equation of the form sin Z , where Z 1 are given by: n 1n or (n 180 0 ) 1n 0 , for n Z Page 49 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Example Solve the following trigonometric equations: (a) (b) 3 cos sin 1 , for 0 0 0 360 0 3 cos 4 sin 2.5 , for 180 0 0 180 0 Solutions: (a) Comparing the given equation with a cos b sin c , we conclude that a 3 , b 1 and c 1 . You should observe that a2 b2 3 2 12 3 1 4 c 2 12 1 . Now let us use the steps which we studied earlier to solve our equation: Are you able to deduce that each term of a 2 b 2 2? We should then divide 3 cos sin by 2 and multiply the entire expression by 2 as follows: 3 1 2 cos sin .......................................(i ) 2 2 Equating (i) to 1, which is the right hand side of the question equation we have: 3 1 2 cos sin 1...................................(ii ) 2 2 From (ii) we obtain 3 1 1 cos sin ........(iii ) 2 2 2 At this stage, let us introduce an auxiliary angle such that cos 3 1 . Consequently, sin and then equation (iii) 2 2 transforms into cos cos sin sin 1 1 cos ................(iv ) 2 2 You should take note that using (iv) we would have 1 2 cos 1 60 0 . Let us now use the appropriate expression for finding a general Page 50 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 angle 360 0 n 60 0............................(v) We are looking for the value/s of from (v ) , but we need first of using either cos 3 1 or sin . If we use our basic 2 2 knowledge of trigonometric functions, we shall discover that: 30 0. Replacing by 30 0 in equation (v) and solving for , we have: 30 0 360 0 n 60 0..........................(vi) For n=0 in (vi) = 30 0 60 0 . Here we get two values. That is either 90 0 or 30 0 . However, only 90 0 lies within the given interval 0 0 0 360 0 . This implies that the other value is not among the required solutions. Suppose we now let n=1 in (vi). What do you think will be the solution? Certainly, the only value lying within the given interval is 330 0 . For all the other values of n Z , the values of lie outside 0 0 0 360 0 . Hence, the solutions of 3 cos sin 1 , for 0 0 0 360 0 are: 90 0 and 330 0 . (b) 3 cos 4 sin 2.5 . We shall solve this equation using the steps we studied earlier. This given equation is the same as: 4 3 5 cos sin 2.5 5 5 3 4 2.5 1 cos sin ...................................(i ) 5 5 5 2 We shall now introduce the auxiliary angle such that sin 3 36.87 0 . 5 With the preceding information, (i) become: sin 1 and 30 0..............................(ii ) 2 Using the expression for computing general angles for equations such as (ii) we have: Page 51 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 n 180 0 1n 30 ...................................(iii ) Let us now replace by 36.87 0 and solve for in (iii) to derive 36.87 n 180 0 1 30 0...............(iv ) n for n=0, in (iv) we have 6.87 0 which lies within 180 0 0 180 0 . Additionally, when n=1, in (iv) We obtain 113.130 which also lies within 180 0 0 180 0 . However, you investigate and discover That for all other values of n Z , the values of would lie outside the interval 180 0 0 180 0 . Consequently, the solutions of 3 cos 4 sin 2.5 , for 180 0 0 180 0 are: 113.130 and 6.87 0 . We have now completed the study of unit two. We hope you enjoyed your discoveries. The following activity, however, is intended to accord you an opportunity for further practice on how to solve trigonometric equations. Enjoy your practice. Activity 9 1. Solve the following trigonometric equations for the indicated Intervals: (a) 3 tan 3 3 tan 2 tan 1 0, 0 0 360 0 (b) 1+ cos 2 cos 4 cos 6 0, for 90 0 90 0 (c) 3 cos 2 5 sin 1 0, for 0 0 360 0 (d) 8 cos 2 6 sin 3 0, for - 2 2 (e) 2 sin 3 tan 0, for 90 0 90 0 (f) 7 cos 2 6 sin 2 5, for 0 0 0 180 0 (g) 4 sin 3 cos 2 0, for - 180 0 0 180 0 Page 52 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Unit summary In this unit you have learnt six trigonometric functions, and special angles. You equally studied several kinds of identities and how to prove trigonometric identities. Additionally, you have learnt how to solve trigonometric equation. Page 53 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 ANSWERS TO ACTIVITY 9 (a) 450 ,225 0 30 0 ,210 0 ,150 0 ,330 0 (b) 90 0 ,90 0 ,450 ,450 30 0 ,30 0 (c) 199.47 0 ,340.530 (d) 6 (e) 0 0 ,30 0 ,30 0 (f) 48.880 ,171.72 0 (g) 13.30 ,119.56 0 Page 54 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Unit 3 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 3.0 Introduction This unit will introduce you to exponents and logarithms and will enable you to appreciate the useful role that they play in Algebra. In order to have an easy understanding of this unit, the knowledge of laws of indices which some of you have, will be necessary. During and upon completion of this unit you will be able to: (b)Use the knowledge of exponents and logarithms to simplify and solve algebraic expressions and equations. (c) Appreciate the application of exponents and logarithms in solving real life problems. (d)Sketch, draw and interpret graphs of exponential and logarithmic functions. (e) Identify and sketch hyperbolic graphs of hyperbolic functions . Page 55 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 3.1 Exponents 3.1 Definition An exponent is a power or index of an index notation or number. Let us consider the numbers below 2 x and 73 . For the two numbers x and 3 are called exponents, powers or indices while 2 and 7 are called bases. An index number to base e, that is e x is called the natural exponent. 3.2 Laws of exponents From your secondary or high school mathematics, you learnt about laws of indices. We will develop this section with this assumption. Now let’s refresh our memories with the activity below. Activity 1 Simplify the following expressions using laws of indices: a) 3 3 3 3 3 3 = b) a 5 a 3 c) b3 5 Solutions (b) To simplify the expression in a) you will realize that 3 is being multiplied by itself six times. However, this does not imply 3 3 3 3 3 3 3 6 But 3 3 3 3 3 3 31 31 31 31 31 31 in index form And by laws of indices this can be simplified as 3 3 3 3 3 3 31 31 31 31 31 31 3111111 36 with 3 as a common base. (c) In this question you will notice that we have been given two simplified expressions; a5 and a3 , which we can write in expanded form as: Page 56 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 a a a a a a a a a a a a2 as explained in (a) above Now simplifying the same expression using the laws of indices, we will work as follows: a 5 a 3 a 5 3 a 2 This is the same answer we got using the expanded form. Having gone through questions (a) and (b), you can now simplify (c). Hint: Use the law in (a) to assist you in your working. Compare your answer with the one below. c) b15 The three laws of indices described above can be summarized as: 4. Multiplication law of indices 5. Division law of indices 6. power law of indices But since an index is the same as an exponent, the laws above are also called the laws of exponents. Given the exponential functions y a b and x a c , then the following laws hold: 6. xy a b a c xy a b c 7. x ab y ac x a b c y 8. (a b ) n a bn These laws also hold for natural exponents. That is: iv). e a e b e a b v). e a e b e a b vi). (e a ) b e ab Page 57 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 3.2 Logarithms Let’s start our discussion with this activity Activity 2 Evaluate a) 35 b) 107 To find the value of 35 , you can either multiply the base, 3, five times or use a calculator to get 243. Similarly to evaluate 107 , you can either multiply the base, 10, seven times or use a calculator to get 10,000,000. We can therefore deduce from activity 2 that the numbers 5 and 7 determine the value of the index numbers given, that is 35 and 107 . We call such numbers logarithms. 3.2.1 Definition The logarithm (written in short as log or lg) of a positive number to a given base is the power to which the base must be raised to equal the given number. For instance, (a) 35 243 can be expressed in logarithmic form as log3 243 5 read as, the logarithm of 243 in base 3 is equal to 5. 5 is the logarithm to base 3 Now express b) in logarithmic form in the same way. i) Logarithms to base 10 is written in short form as lg and usually with no base attached. Example: log 10 10,000 4 can be written as lg 10,000 4 . ii) Logarithms to base e (2.718 to three decimal places) is written in short form as loge x ln x iii) Logarithms to base 10 are called common logarithms while logarithms to base e are called natural logarithms. Page 58 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 In general, given an equation a b c , this can be expressed in logarithmic form as log a c b . In the same way eb a can be expressed in logarithmic form as loge a b or simply ln a b . However, for this to exist (i) a 1 (ii) a 0 Activity 3 [Self marked. Take 20 minutes on this activity] 1. Identify the logarithm and the base in each of the following statements. a) 1 40 b) 1728 123 1 c) 125 5 3 2. Express the index form given into logarithmic form: a) 63 216 3 b) 27 814 c) 0.001 103 1 d) 4 16 1 2 e) 80 1 3.2.2 Laws of logarithms Having defined the logarithm of a number, we are going to spend some time to discuss some important laws governing logarithms. To do this easily, we will revisit the laws of exponents we started with at the beginning of this unit. Page 59 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Let’s consider the statements a x b and a y c . 2.2.1 The product law The statements a x b and a y c can be expressed in logarithmic form as log a log a Well done. Now write an expression for the product, bc in index form. bc But since a is a common base, we add the exponents and obtain bc a x y Converting bc a x y in logarithmic form we get log a bc Substituting the value of x and the value y in this equation, we obtain log a bc log a b log a c This is the product law of logarithms. 3.2.2 The quotient law To derive this law, we shall begin by expressing the quotient index form. I’m sure this should not be a problem. Let’s do this together. Expressing b in index form gives c b ax This simplifies to c ay b a x y . Since a is a common base. c Now we write this in logarithmic form to obtain Page 60 b in c MODULE 2 FOUNDATION MATHEMATICS – MAT 110 b lo g a c Replacing the values of x and y in the equation above we get b log a log a b log a c c This is the quotient law of logarithms 3.2.3 The power law The rule states that log a b n n log a b , where n is an integer. Proof using the LHS Let n be an integer Then bn a x (From above) n b n a xn (by power law of exponents) Expressing this in logarithmic form, we obtain log a b n xn Substituting the value of x in the equation, we get …………………. as required. The three laws we have discussed above are very important in simplifying expressions and solving equations involving logarithms as you will soon notice. Examples 1 1. Express as single logarithms. a) log 2 log 3 b) log18 log 3 1 d) 3log x log y 1 2 Page 61 1 c) log p log q 3 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Solutions (v) log 2 log 3 To simplify this expression we will use the product law of logarithms explained above. That is log 2 log 3 log(2 3) log 6 b) log18 log 9 To write this as a single logarithm we will apply the quotient law as follows: log18 log 9 log(18 9) log 2 1 c) log p log q 3 To simplify this expression we will need to apply the quotient and the power rules as follows: 1 1 log p log q log p log q 3 using power law 3 1 = log p q 3 using quotient = log p 1 q3 = log p 3 q 1 d) 3log x log y 1 2 To answer the question, we need to apply the three laws of logarithms as follows: 1 1 3log x log y 1 log x3 log y 2 1 2 1 2 = log( x y ) 1 3 Page 62 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 3 x = log10 1 log10 10 2 y log10 = 10 x3 y 2. Evaluate a) log1000 b) 1 log 3 81 2 c) log 2 1 2 d) log 49 log 343 Solutions (b) To answer this question, we take the base to be 10. Hence log1000 log10 1000 = log10 102 = 2. b) 1 log 3 81 2 This question can easily be evaluated by first expressing 81 in exponential form to base 3 as follows: 1 1 log 3 81 log 3 34 2 2 1 = log 3 34 2 ` = log 3 3 2 = 2 log 3 3 =2 c) We will use the power law to evaluate this question 1 1 log 2 log 2 2 2 log 2 1 1 1 log 2 21 2 = log 2 2 =1 Page 63 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 d) log 49 log 343 To answer this question we will first convert the given logarithms to the same base, 7 log 49 log 7 7 2 log 343 log 7 73 = 2 3 The examples lead us to seven important identities: For any base a, the following hold: i) a1 a , if and only if log a a 1 ii) a 0 1 , if and only if log a 1 0 iii) a 1 log a 1 , if and only if a 1 1 a And from identity (i), a more general result can be formulated. That is For any base a, the following is true. a x N , if and only if (iff) log a N x This gives two important results. 9. x log a a x 10. N a loga N 11. e ln x x for all x 0 12. ln e x x for all real x Page 64 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Examples 2 Simplify a) 3log3 4 b) log 2 27 c) d) e) e ln 4.3 ln e 9 e ln 3 f) ln e Solutions a) 3log3 4 4 b) log 2 27 7 c) d) e) e ln 4.3 ln e 9 e ln 3 = 4.3 =9 1 1 = e ln 3 3 1 2 f) 1 ln e ln e = 2 Page 65 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Activity 4 1. Simplify the logarithms below a) 2 3log a 2. Find the value of: a) log8 log 2 b) log 4 1 c) l og a 1 a 2.3 Change of base Changing the base of a logarithmic expression involves taking the same logarithm on both sides. In order to assist you understand this part well. Let’s consider the general statement log a b x , where a is any base other than 10 or e, the natural logarithm. Then a x b in exponential form Now changing the base of the given base a to a new base, c we obtain log c a x log c b Implying that x log c b log c a Finally substituting x log a b , we get log a b log c b log c a In the special case where c b The identity log a b log c b log c b becomes log a b log c c 1 log c a log c a Page 66 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 We usually change logarithms to base 10 or e. Examples 3 1. Evaluate, correct to three significant figures, the following a) log 6 7 b) log8 72.4 2. Show that log a b 2 log b a 3 6 Solutions To answer the questions above, we shall employ the formula log c b log a b . log c a a) log 6 7 . In this logarithmic expression, a = 6 and b = 7. Now changing the base to 10, we have log 6 7 log10 7 lg 7 log10 6 lg 6 To get the value of log 6 7 , we use calculators directly. So this gives log 6 7 1.09 Now try to answer b) on your own using the same approach. If your working is accurate, you should be able to get 2.06 as the answer. Otherwise, go through your work cautiously. 2. To show that the LHS = RHS, let’s work from the LHS as follows log( a b 2 ) log log b a 3 2 log a b 3log b a 2 log a b 3 2 log a b 1 3 2 log a b log a b log a b 3 6 as required log a b Page 67 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Activity 5 [Take 25 minutes on this activity] 1. Simplify a) 3log3 5 b) c) log a a n d) x log x 9 e) log 3 log 4 43 3.0 Exponential and Logarithmic equations Having taken you through the laws of indices and logarithms, its now time to use these laws in solving exponential and logarithmic equations. To do this systematically, we will begin our discussion with exponential equations and then finish off with logarithmic equations. 3.3 Exponential equations An exponential equation is an equation containing a variable in an exponent. For example, 2 x 2 3x 3 . 3.3.1 Solving exponential equations with the same base. Examples 4 Solve for x ii) 34 x 2 3x i) 2 x 64 2 3 Solutions i) To answer this question we are going to express 64 in exponential or index form to base 2 as follows: 2 x 64 26 Now since the base is the same on both sides, it follows that even the exponents are equal. Therefore 2 x 26 Implying that x6 ii) 34 x 2 3x 2 3 Page 68 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 This equation can be solved directly by equating the exponents on both sides since the base is the same. That is 4 x 2 x2 3 Implying x 2 4 x 5 0 This is a quadratic equation. Solving this question gives x 1 or x 5 (Refer to module 1 on how to solve quadratic equations) 3.3.2 Solving quadratic exponential equations Examples 5 i) 42 x 1 7 4 x 3 0 ii) 22 x 2 x 6 0 Solutions We shall go through the first example together and then you will be required to answer question two on your own. So be at tentative. i) 42 x 1 7 4 x 3 0 Step 1: Ensure that the terms expressed in exponential or index form have the same base. If not convert them to the same base. In the question above, the terms expressed in exponential or index notation, that is 42 x 1 and 7(4 x ) are given to the same base, 4. Thus, there is no need to change the base. However, we must express 4 2 x 1 in terms of 4 x as 4 4 x by multiplication and power 2 laws of indices. Step 2: Let some variable, say p, represent the term in index form, in our case 4 x . Now letting p 4 x , we obtain 4 p2 7 p 3 0 Solving this quadratic equation gives p 1 and p 3 4 Page 69 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 But p 4x Therefore for p 1 , we obtain 4 x 1 40 Implying x 0 And for p 4x 3 , we obtain 4 3 4 Expressing both sides in base 10, we have x lg 4 lg 3 4 Implying that x lg 0.75 By use of the calculator, we get lg 4 x 0.21 to 2 decimal places. Now you can work out ii). Compare your answer(s) with this answer, x 1.58 to 2 decimal places. If you fail to get this answer, you should not be discouraged, but try to go through your work carefully. 3.3.3 Solving exponential equations using the change of base method This method was introduced earlier in the unit, but we will now demonstrate how to use it to solve exponential equations. Examples 6 Solve for x i) 53 x 7 ii) 2 x1 5 Solutions Here is how to go about solving these types of equations. i) 53 x 7 Page 70 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 To solve this equation, you must convert both sides to logarithm to base 10. That is lg 53 x lg 7 Implying that 3x lg 7 1.209 lg 5 Giving x 0.40 to 2 decimal places ii) 2 x1 5 To solve this equation, we will use the same approach as in (i). That is lg 2 x 1 lg 5x , by power rule, this becomes x 1 lg 2 Expanding the LHS and collecting like terms we obtain lg 2 Now factorizing the RHS and dividing with the LHS, we get lg 2 x which is the same as lg 5 lg 3 x lg 2 By use of calculators we obtain the value of x as lg 5 lg 3 x 0.76 to 2 decimal places. I’m sure you didn’t find problems in filling the missing parts. If you didn’t have no problems then you are ready to do the activity below. Activity 6 1. Solve the exponential equations a) 2 x 8 b) x 3 729 Page 71 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 c) 52 x 25 d) 83t 32 4t 1 2. Solve the equations: a) 3x 6 b) 22 x 5 c) 42 x1 3 3. Solve the equations a) 5 x 5 x1 10 b) 2 22 x 5 2 x 2 0 c) 32 x 1 26 3x 9 0 d) 4 x 6 2 x 16 0 3.4 logarithmic equations A logarithmic equation is an equation which involves a logarithmic function of some variable. For example, 7 ln x 15 . Like exponential equations, there are several methods used to solve logarithmic equations as you will see from the worked examples below. 3.4.1 Solving equations of the form log a b x and log x b c where the unknown is represented by x. Examples 7 Solve for x a) log3 81 x b) log x 8 2 Solutions a) log3 81 x Page 72 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 To solve this equation, we must express the given equation in index form as follows: 3x 81 Now we express 81 in exponential form to base 3. So we have 3 x 34 Implying that x4 b) log x 8 2 The approach used in solving the previous equation will be employed to solving this equation. That is x2 8 Implying that x 8. 3.2.2 Solving logarithmic equation with the same base. Examples 8 a) l og3 x 3 log3 x 3 3 b) log2 x log2 x 2 log2 9 2 x Solutions a) l og3 x 3 log3 x 3 3 To solve this question, we will begin by simplifying the LHS by applying the multiplication law of logarithms. That is log3 x 3 x 3 3 Converting in index form we obtain 33 x 3 x 3 Implying that 27 x 2 9 Thus x 2 36 And x 6 is the only answer. Page 73 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 b) log2 x x 2 log2 9 2 x Using the approach as the one above, we simplify the equation as follows: log2 x x 2 log2 9 2 x But the LHS and RHS are both given in base two, we can cancel or ignore the logarithms and rewrite the equation, log2 x x 2 log2 9 2 x as follows: x( x 2) 9 2 x Simplifying the equation we obtain x2 9 0 And x 3 or x 3 as answers. 3.2.3 Solving logarithmic equations with different bases Examples 9 Solve for y a) log 2 y log8 y 4 b) log 5 y 2 log y 5 3 Solutions To solve equations in example 8 above, we utilize the change of base method which we have already discussed. a) log 2 y log8 y 4 We will write the equation in base two as follows: log 2 y log 2 y 4 , simplifying gives log 2 8 3log 2 y log 2 y 12 , since ( log 2 8 log 2 23 3 ) Now applying the quotient law of logarithms and simplifying the given equation, we obtain y3 log 2 12 y Expressing in index form, we get Page 74 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 212 y 2 2 y 1 12 2 1 2 2 2 6 y or y 64 b) log 5 y 2 log y 5 3 Changing the base to 5, we obtain log5 52 log5 y 3 log5 y This simplifies to log5 y 2 2 3log5 y Letting p log5 y , we get a new equation in p as p 2 2 3 p , giving us p2 3 p 2 0 Solving for p, we obtain p 1 or p 2 But p log5 y Now For p 1 Implies that log5 y 1 This gives y 51 5 For p 2 Implies log5 y 2 This gives Page 75 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 y 52 Implying that y 25 Activity 7 [Take 40 minutes on this task] 1. Solve for x a) log5 x 2 b) log x 125 3 c) log 5 x 2 4 x 1 d) x log2 log4 256 e) log3 3 2 x 2 2. Solve the following equations a) log7 x log7 18x 61 1 b) log 5 4 x 2 1 2 log 5 2 x 1 c) log8 x 2 9 log8 x 3 2 3. Solve for x a) log3 x 4log x 3 3 0 b) log 2 x log x 2 2 c) log2 x log 4 x 6 d) log3 x 2log x 3 1 3.3 Simultaneous equations involving Exponential and Logarithmic functions Apart from the equations we have discussed, that is, equations in one variable, exponential and logarithmic equation can be presented in two variables as we shall seen. Examples 10 Solve for x and y a) log x y 2 ; xy 8 b) 2 lg y lg 2 lg x ; 2 y 4x c) y log 2 8 x ; 2 x 8 y 8192 d) ln 6 ln x 3 2ln y ; 2 y x 3 e) 5x 2 7 y 1 3468 ; 7 y 5x 76 Page 76 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 log3 x y log9 2x 1 f) Solutions log x y 2 ; xy 8 a) To solve this equation, we start by converting log x y 2 in index form as follows x2 y Now replacing x 2 y in the equation xy 8 and solving for x we obtain x x2 8 Implying that x3 8 x2 Substituting x 2 in the equation x 2 y , we find the value of y as y 22 4 b) 2 lg y lg 2 lg x ; 2 y 4x To answer this question, it’s easier to work in index form. Therefore, we will begin by converting the equation 2 lg y lg 2 lg x in index form as follows: lg y 2 lg( 2 x) Can you state the laws used to simplify the equation? Now from the equation above, you will notice that, both sides of the equation have been expressed in the same base, hence we will cancel the logarithms and obtain an equation in x and y only as y 2 2x . Expressing x in terms of y we have x y2 2 y2 Substituting the value of x, that is x in the equation 2 y 4x 2 and solving for y, we get Page 77 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 2 4 y y2 2 Implying that 2 2 y y2 2 2 , this simplifies to 2 y 2 y . Since the base is the same, then the powers also equal. 2 Implying y y 2 .Hence y 1 or y 0 Replacing y in the equation, x y2 we obtain two values of x as 2 For x 1 x 12 1 2 2 And for x 0 x 02 0 2 y log 2 8 x ; 2 x 8 y 8192 c) This equation is similar to the one we have just solved. As such I will ask you to express y log 2 8 x in index form. 2x Now substituting x in the equation 2 x 8 y 8192 , we obtain 2 8 y 8192 Now expressing the LHS in index form to base 8, we get 8 y 8 y 8192 Implying that 8 4096 and expressing both sides in same base, 8, we have 8 8 2 8 y 8192 , dividing both sides by 2, get y y 4 And Page 78 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 y4 Substituting y 4 and solving for x, we obtain x 12 d) ln 6 ln x 3 2ln y 2 y x 3 ; The same method we used to answer b) and c) above will be utilized to solve equation d). Again we start solving the equations by expressing ln 6 ln x 3 2ln y in index form as follows ln[ 6x 3] ln y 2 by multiplication and power laws of logarithms Since the LHS and RHS are expressed to the same base, then this equation simplifies to 6x 3 y 2 . Expanding and collecting like terms gives y 2 6 x 18 0 Now expressing x in terms of in the equation 2 y x 3 gives x 2y 3 And substituting this x value in the equation y 2 6 x 18 0 we obtain a quadratic equation in y as y 2 12 y 36 0 Solving this equation gives y6 Implying that x 9 .That is by substituting y 6 in the equation x 2 y 3 . e) 5x 2 7 y 1 3468 ; 7 y 5x 76 Since the two equations have expressed in index form, we straight away use one of the methods of solving simultaneous equations to work out the solutions. In this case we shall apply substitution method. y x x2 y 1 We shall substitute 7 5 76 in the equation 5 7 3468 y since 7 is the subject. This gives Page 79 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 5 x 2 7 y 1 3468 By laws of indices this equation can written as 5 2 (5 x ) 7(7 y ) 3468 Replacing the value of 7 y , we obtain 5 2 (5 x ) 7 5 x 76 3468 This reduces to 25(5 x ) 7 5 x 532 3468 Implying 32 5 x 4000 . Dividing by 32, we get 5 x 125 Therefore x5 Now substituting x 5 in the equation 7 y 5 x 76 , we obtain 7 y 5 3 76 Implying that 7 y 49 . Rewriting 49 in base 7 and solving for y gives y 2. f) log3 x y log9 2x 1 The equations given above are log 3 x y ; y log 9 2 x 1 Now expressing the equations in index form, we obtain 3 y x ; 9 y 2x 1 Substituting x 3 y in the equation 9 y 2 x 1 , we get 9 y 2(3 y ) 1 (32 ) y 2(3 y ) 1 And from laws of indices this gives (3 y ) 2 2(3 y ) 1 This exponential equation is of the form ax 2 bx c 0 which we have already discussed above. Therefore y 0 and x 1. I’m sure getting these values would not be a problem. Page 80 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 3.4.2 Equations involving natural exponents and natural logarithms 3.4.2.1 Solving equations involving natural exponents Examples 11 1. Solve for x a) e ln(2 x 1) 5 b) e 2 x 1 5 c) e x ex 2 2 Solutions a) e ln(2 x 1) 5 To solve this equation, we will use the identity e ln x x . So in our case x 2 x 1 Implying 2x 1 5 And x 3. b) e 2 x 1 5 To answer this question, we will begin by writing the equation e 2 x 1 5 in logarithmic form as log e 5 2 x 1 But you will recall that log e x ln x from our discussions on page 14. Now using this, we rewrite the equation as 2x 1 ln 5 x 1 ln 5 1 . Using the calculator we get 2 x 1.305 to three decimal places. Page 81 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 c) e x ex 2 2 To solve this equation, we first multiply both sides of the equation by 2 as follows 2 e x ex 2 2 giving us 2 e x ex 4 Multiplying both sides by e x we then obtain (e x ) e x (e e ) e x 4e x . Simplifying gives e 2 x 1 4e x this gives the quadratic equation (e x ) 2 4e x 1 0 To solve this equation, we let a variable say P replace e x in the equation. That is p2 4 p 1 0 And solving for p gives p 2 5 But p e x Hence ex 2 5 Since e x 0 for all x, we only use the positive solution for e x ; That is e x 2 5 Now taking the natural log on both sides we obtain ln e x ln( 2 5) Implying that x ln (2 5 ) 3.4.2 Solving equations involving natural logarithms Examples 12 Solve for x a) ln( x 1) 1 ln x Page 82 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 To solve this equation, we first collect like terms as follows ln( x 1) ln x 1 Applying the quotient law of logarithms, we express ln( x 1) ln x 1 as ln x 1 1 . Now expressing this in exponential form, we obtain e1 x x 1 , using the fact that ln x log e x x Implying that xe x 1 xe x 1 And x 1 0.582 to three decimal places. e 1 Self marked activity 8 [Take 20 minutes on this task] Solve for x 1. a) e 3 x 5 100 b) e 0.01x 27 3 d) e ln1 x 2 x c) e x e x e 4 2 e) e ln6 x 2. 2 4 5x a) ln x ln 2 1 c) ln x b) ln( x 1) 0 1 2 ln 4 ln 8 2 3 d) ln x 2 x 2 ln x ln x 1 Page 83 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 3.5.0 Application of exponential and logarithmic function Exponential and logarithmic functions are very essential in real life situations. In this section we will look at how exponential and logarithmic functions are applied in business and science. 3.5.1 Decay and growth problems Decay and growth problems are solved using the formula f ( x) Ae kx , where A represents the initial amount of substance, K is the constant and x represents the time. Examples 13 1) A radioactive substance is decaying according to the formula y Ae 0.2 x where y is the amount of substance remaining after x years. 2 3 If the initial amount A = 80 grams, how much is left after 3 years? The half-life of a radioactive substance is the time it takes for half of it to decompose. Find the half-life of this substance in which A = 80 grams. Solution i. In this question, we have been given the initial amount of substance (A = 80) and the time (x = 3) Thus substituting these values in the general formula, that is y Ae kx , we obtain y 80e 0.2(3) .This simplifies to y 80e 0.6 Page 84 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 This gives us y 43.9 grams to 1 decimal place ii. This question calls for the time x which only half of the initial amount of substance is left. Thus half of the substance gives us 40 grams. Implying 1 e 0.2 x . Now taking the natural logarithms on both sides 2 we get 0.2 x ln 1 2 0.2x ln 1 ln 2 Meaning x 3.466 Years to 3 decimal places 3.5.2 Compound interest The formula used in the evaluation of none compounding interest is also the application of exponential growth. The formula for none compound interest is given as At p1 r n nt where p = initial investment r = annual interest rate (as a decimal) n = number of annual interest periods t = number of years of investment At = final value after t years Page 85 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Examples 14 A $5000 investment earns interest at the annual rate of 8.4% compounded monthly. a) What is the investment worth after one year? b) What is it worth after 10 years? c) How much interest was earned in 10 years? Solutions From the data given, p = $5000 r = 8.4% = 0.084 n = monthly Therefore interest per month = r 8.4% 0.084 0.007 n 12 12 to three decimal places. Now with this data we can find the solutions to the given question as follows: a) A1 50001 0.007 112 = 50001.007 12 = $5437 to the nearest dollar b) A10 50001 0.007 1210 = 50001.007 120 = $11,548 to the nearest dollar c) To find the interest earned in 10 years, we find the difference of the final amount and the initial investment. That is Interest after 10 years = $11,548 - $5000 = $6548 Page 86 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 3.5.3 Graphs of exponential and Logarithmic functions Graphs of exponential functions. An Exponential function is a function of the form f ( x) a x , where a is a positive constant. That is a 0 and a 1 In other words an exponential function is one where the variable appears as an exponent. Let’s consider the exponential function f ( x) 3 x for 3 x 3 x f ( x) 3 x -3 -2 -1 0 1 2 3 0.037 0.111 0.333 1 3 9 27 From the table above, we observe that: i) f ( x) 0 for all real values of x, ii) As x increases f (x) increases rapidly, iii) f (x) =1 when x = 0 iv) As x decreases (That is x 2,3,..., ) , f(x) becomes numerically smaller. We say that as x approaches minus infinity, f (x) approaches the value zero. This is written in short as x , f ( x) 0 v) The domain of f (x) is the set of positive real numbers. Now let’s sketch the graph of f ( x) 3 x for 3 x 3 Page 87 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 From the graph of f ( x) 3 x , you will observe that the graph does not cross the line y = 0 or the X- axis. Hence we say that the X – axis is the horizontal asymptote. In general, all exponential functions of the form y a x , for a > 0, have graphs of the same shape and all pass through the point (0 , 1) and the X axis or the line y = 0 is the horizontal asymptote. Self marked Activity 9 Draw the graphs of: a) y 2 x b) y 1.5 x c) y 4 x , for 3 x 3 From the basic graphs of exponential functions discussed above, we can also draw other graphs of exponential functions which are quite challenging. Examples Draw graphs of: f ( x) 2 x3 . and f ( x) 2 x 1 for 3 x 3 . Page 88 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 The first one f ( x) 2 x3 has been done for you. Study it and then draw the graph of f ( x) 2 x 1 . Now the table of values of f ( x) 2 x3 are shown below: x -3 -2 -1 0 f(x) 0.0156 0.0313 0.0625 0.125 1 2 3 0.25 0.5 1 Graph of f ( x) 2 x3 Now try to draw the graph of f ( x) 2 x 1 on your own. 5.2 Logarithmic functions A function of the form f ( y) log b x , where b 0 and b 1, is called a logarithmic function with base b. Let’s consider the logarithmic function f ( x) log 2 x Page 89 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 x 0.125 0.25 0.5 1 2 4 f ( x) log 2 x -3 -2 -1 0 1 2 From the table, the following we can observe the following: i) When x 1, f (1) does not exist. Put it in a different way, there is no real value for f (x) when x = -1.Meaning f ( x) log b x does not exist for negative values of x. ii) For x 1, f ( x) 0 and as x approaches positive infinity, f (x) approaches positive infinity. iii) For x 0 , f (0) is undefined. As x decreases to zero, f (x) approaches negative infinity. And that for 0 x 1 , f ( x) 0 . iv) the graph is concave down wards v) it is an increasing function Now we draw the graph of f ( x) log 2 x for 3 x x . Page 90 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 The graph of f ( x) log 2 x , does not cross the line x = 0 or the Yaxis. Hence we say that the Y – axis is the vertical asymptote All graphs of the form f ( x) log a x , a 1 and a > 0, have the same shape as the basic function, and they all pass through the point (1,0) and the Y – axis or the line x = 0 is the vertical asymptote. Self marked activity 10 Draw graphs of: a) y log x 3 b) y log x 4 Now by comparing or drawing the graphs of y 4 x and y log x 4 , you will observe that: (iii)the two graphs are inverses of each other. In other words the graph of y 4 x is the inverse of the graph of y log x 4 . In general the functions f ( x) a x and g ( x) log x a are inverse functions. (iv) the line y x is the mirror line (line of reflection) of the two graphs. That is to say you can draw the graph of y 4 x by reflecting this graph in the line y x and the reverse is true. Apart from the basic graphs demonstrated above, we also draw higher graphs of logarithmic functions. Example Graph the functions y log 2 (2 x 4) and y log 3 ( x 2) for 1 x 2 Solution Since logarithms can be taken only of positive numbers, the domain of the function is the solution of 2x 4 0 ; that is , all x 2 . Furthermore, the asymptote is x = -2. Now to locate points on the curve, first we rewrite the logarithmic function in exponential form. That is Page 91 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 y log 2 (2 x 4) 2 x 4 2 y Now substituting the values of x in 2 y 2 x 4 , we get x f(x) -1 1 0 2 1 2 3 Now graph of y log 3 ( x 2) . Graphs of natural exponents and logarithms behave in the same way as those of common exponents and logarithms. For instance the graph of the natural exponential function, y e x passes through the point (0, 1) and has the line y = 0 as its horizontal asymptote. Similarly, the graph of the natural logarithmic function, y ln x passes through the point (0, 1) and has the line x = 0 as its vertical asymptote. Refer to the graphs below. It follows from the fore going discussion that graphs of natural exponential functions and logarithmic functions have the same properties as the earlier graphs discussed above. Page 92 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 The graph of y e x The graph of y ln x Self marked activity 11 Sketch the following graphs a) y = e power x-2 b) y = 3e power 2x minus 1 c) y = ln (x-2) d) y = ln2x Page 93 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Unit Summary In this unit you have learnt the laws of exponents and logarithms and how they can be used to simplify and solve equations involving exponential and logarithmic functions. You have also learnt how to draw and sketch graphs of exponential and logarithmic functions as well as hyperbolic functions. Page 94 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 3.5.4 The hyperbolic functions These are functions that have been generated by the use of (e x e x ) is 2 called the hyperbolic sine of x and is abbreviated sinh x . Similarly, exponential functions. For example the function e ex is the hyperbolic cosine of x and is 2 abbreviated cosh x . The other hyperbolic functions worth mentioning are; the function x e x ex tanh x x e e x csc hx 2 e e x sec hx 2 e e x coth x e x ex e x e x x x The graph of sinh x Page 95 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 The graph of cosh x The graph of tanh x Page 96 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 The graph of csc hx The graph of sec hx Page 97 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 The graph of coth x Hyperbolic functions are not periodic. 2. sinh x varies from to and cosh x varies between 1 and tanh x 1 3. 0 sec hx 1 1. However, it is not our intention to go into detailed discussions about the hyperbolic functions in this unit, but rather to show you the differences between these graphs and the exponential and logarithmic functions. You will discuss the hyperbolic functions in details in higher courses. Assessment 1. Express as a single logarithmic a) 2 log a 5 log a 4 log a 10 b) log 5 7 log 5 x c) 3 2 lg 5 2. Simplify a) log y 2 log xy b) log x 3 log x log x 2 log x Page 98 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 c) log(log y 2 ) log(log y) 3. Evaluate a) ln 10 ln 5 b) ln 8 ln 0.2 c) ln 25 3 ln 5 4. Solve for x a) 2 x 45 c) 80e x 120 b) 100(2 x ) 2000 d) 4 2 x 1 6 x e) e 2 x 1 3 x 1 5. Solve the equations a) log 2 (2x 1) 3 b) 2 log 2 x log 2 ( x 1) 2 c) log 3 x 2 log x 3 1 d) 4 2 x 1 5 x 2 61 x e) 4 x 2 x 1 3 0 f) 4 32 x 1 17 3 x 7 0 6. Solve the simultaneous equations a) log y x 2 , 5 y x 12 log x y b) log 2 x log 2 y 3 , log y x 2 c) log x 2 log 2 2 log y , log x 3 y 3 0 d) Show that log 16 xy 1 1 log 4 x log 4 y . Hence or otherwise, 2 2 solve the simultaneous equations 1 log 4 x log 16 xy 3 , 8 2 (log 4 y) 7. A radioactive material is decaying according to the formula y Ae kx , where x is the time in years. The initial amount A = 10 grams, and 8 grams remain after 5 years. a) Find k. Leave your answer in terms of natural logarithms. b) Estimate the amount remaining after 10 years. c) Find the half – life to the nearest tenth of a year. 8. Suppose that a $10,000 investment earns compound interest at the annual interest rate of 9%. If the time of deposit of the investment is one year (t = 1), find the value of the investment for each of the following types of compounding. Page 99 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 a) n = 4 (quarterly) b) n = 52 (weekly) c) n = 365 (daily) 9. Graph each of the following, state the domain, intercepts, equations of the asymptotes, if: a) f ( x) 0.2 x b) f ( x) log 4 x 1 c) f ( x) e x2 d) f ( x) ln x 4 e) f ( x) ln x 3 Page 100 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 Answers to activities Activity five a) 5 b) n c) 9 Activity six d) 1 1. a) 3 b) 9 d) c) 0 5 11 2. a) 1.6309 b) 1.1610 c) -0.1038 3. a) 1.2153 c) 2 Activity seven 1. a) 25 b) 5 c) 5; -1 d) 2 e) -3 1 2. a) 9 c) 67 1 3. a) 81 ; 3 c) 3 Activity eight 1 3 4 1. a) c) ; e) 2 2 3 e 2. a) c) 64 2 Assessment 1. a) log a 10 b) log 5 7 x c) log 10 40 y 2. a) log c) 2 x 2 3. a) b) -1.292 c) 3 4. a) 5.492 c) -0.405 1 9 5. a) c) ; 9 e) 3 2 6. a) x 4 or x 9 and y 2 or y 3 c) x 4 or 10 and y 4 or and y 2 1 4 7. a) ln b) 6.4 grams c) 15.5 year 5 5 END OF MODULE 2 Page 101 MODULE 2 FOUNDATION MATHEMATICS – MAT 110 References 10 3