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TEXT FLY WITHIN THE BOOK ONLY DO 1620025m OUP -24-4-1-695,0(0 OSMANIA UNIVERSITY LIBRARY Gall No Author 5 1 2* & $ & fi Accession No. Gt ** ^ ^ 'Vr/-v<*^< ^^^.cfok^ ' . This lx> should be returned on or before the date last marked below. Algebra and Trigonometry Algebra and Trigonometry ALVIN Head BETTINGER K. Department and Professor of Mathematics of The Crcif/hton University and JOHN A. ENGLUND Formerly Assistant Professor of Mathematics The Cr eight on University INTERNATIONAL TEXTBOOK COMPANY Scranton, Pennsylvania INTERNATIONAL TEXTBOOKS IN MATHEMATICS L R. W//COX Professor of Mathematics Illinois Institute of Technology CONSULTING EDITOR Second Printing, ] armory 1963 > Copyright I960, by International Textbook Company. All rights reserved. Printed in the United States of America by The Haddon Craftsmen, Inc. at Scranton, Pennsylvania. Library of Congress Card Number: 60-9987. t Preface The authors believe that in this book the basic material of college algebra and trigonometry has been presented with suflicient rigor to provide a firm and coherent groundwork for subsequent courses in mathematics. The material is presented in such a way that it can be grasped by the student without undue assistance. The first chapter consists of a number of introductory topics which are intended to serve as a review of elementary algebra. Actually, something more than a mere review is available in this chapter. Not only are the review topics considered from a more mature point of view than is usual, but the treatment is interwoven with concepts that are basic for an understanding of more advanced mathematical topics. We begin with the algebra of the real-number system. Axioms pertaining to fundamental operations are given, and the various rules for the elementary operations of algebra are derived and logically connected with the bcisic assumptions. We are led naturally to an ordering of the real-number system and to the foundation for a later chapter on inequalities that is easier to understand and more useful than the treatment one customarily finds in textbooks. The second chapter introduces the student to the function concept, which serves as a basis for much of the remaining work of Certain aspects of the discussion become somewhat abstract, but the student is reminded that a proper understanding of the true nature of a function is important for virtually all later courses in mathematics. In line with modern demands, the trigonometric functions are initially introduced in the third chapter as functions of real numbers. Following this presentation, the transition to functions of the book. angles is relatively simple. rest of the volume contains all the usual topics from college algebra and trigonometry. In certain instances a particular devel- The differ somewhat from that usually found. In such the authors believe, the departure is to the advantage of cases, the student. opment may Preface vi We are indebted to our colleague, Professor Morris Dansky, for his valuable suggestions while the manuscript was in preparation. wish particularly to express our deep appreciation to Pro- We fessor L. R. Wilcox for his thorough criticism of the manuscript and his invaluable suggestions for improvement of the text. Finally, a special pany for is due the International Textbook Comand cooperation patience. word its of thanks A. K. BETTINGER J. A. ENGLUND Omaha, Nebraska August, 1960 Contents 1. INTRODUCTORY TOPICS 1-1. The Real-Number System 1-2. Fundamental Assumptions 1 1 1 5 1-4. Operations With Zero Reciprocals 1-5. The Real-Number Scale 6 1-3. 5 Rules of Signs Fundamental Operations on Fractions 1-8. Order Relations for Real Numbers 1-9. Absolute Value 1-10. Inequalities Involving Absolute Values 1-6. 1-7. 1-11. Positive Integral Exponents 1-12. Algebraic Expressions Equations and Identities 1-13. 1-14. Symbols of Grouping Order of Fundamental Operations 1-16. Addition and Subtraction of Algebraic Expressions 1-15. 1-17. Multiplication of Algebraic Expressions 1-18. Special Products Division of Algebraic Expressions 1-19. 1-20. 1-21. 1-22. 1-23. 1-24. 1-25. 1-26. 1-27. 1-28. 1-29. 1-30. 2. Factoring Important Type Forms for Factoring Greatest Common Divisor Least Common Multiple Reduction of Fractions Signs Associated With Fractions Addition and Subtraction of Fractions Multiplication and Division of Fractions Complex Fractions Linear Equations Linear Equations in One Unknown THE FUNCTION CONCEPT 2-1. 2-2. 2-3. 2-4. 2-5. 2-6. 2-7. Rectangular Coordinate Systems Distance Between Two Points Functions Functional Notation Some Special Functions Variation Classification of Functions vii in a Plane 7 9 12 14 14 16 17 18 18 20 21 22 22 23 25 27 30 32 33 34 36 39 40 42 43 49 49 50 52 55 57 57 61 Confenfs viii 3. THE TRIGONOMETRIC FUNCTIONS 3-1. Definitions of the Trigonometric Functions Identities 68 3-4. Tables of Trigonometric Functions Positive and Negative Angles and Standard Position 71 75 76 77 78 3-7. 3-8. 3-9. 3-10. 4-2. 4-3. 4-4. Trigonometric Functions of Angles Tables of Natural Trigonometric Functions of Angles OF EXPONENTS 81 82 86 86 88 DO Positive Integral Exponents Meaning of Negative Exponents Scientific Notation 1)2 1)2 4-6. Rational Exponents The Factorial Symbol 4-7. The Binomial Theorem 1)7 48. General Term 1)9 4-5. in the J)7 Binomial Expansion 101 LOGARITHMS 5-1. 5-2. 5-3. 5-4. 101 Definition of a Logarithm Laws of Logarithms Systems of Logarithms Common Logarithms 5-5. Rules for Characteristic and Mantissa 5-6. How How 5-7. 5-8. 5-9. to to Write Logarithms of Mantissas Use a Table Logarithmic Computation Change of Base RIGHT TRIANGLES AND VECTORS 6-1. Rounding Off Numbers 6-2. 6-3. 6-4. Trigonometric Functions of Acute Angles Procedures for Solving Right Triangles Angles of Elevation and Depression 6-5. Bearing 6-6. Projections Scalar and Vector Quantities Logarithms of Trigonometric Functions Logarithmic Solution of Right Triangles 6-7. 6-8. 6-9. 7. Measurement of Angles The Relation Between Radians and Degrees Arc Length and Area of a Sector THE LAWS 4-1. 6. (54 3-2. 3-6. 5. G3 63 3-3. 3-5. 4. The Point Function P(t) in Navigation arid Surveying TRIGONOMETRIC FUNCTIONS OF SUMS AND DIFFERENCES 7-1. Derivation of the Addition Formulas 7-2. The Double-Angle Formulas 7-3. The Half-Angle Formulas 7-4. Products of Two Functions Expressed as Sums, and Sums Expressed as Products 102 105 105 106 108 108 110 113 115 115 116 117 120 121 122 125 131 133 135 135 139 140 143 Contents 8. GRAPHS OF TRIGONOMETRIC FUNCTIONS; INVERSE FUNCTIONS AND THEIR GRAPHS 8-1. 8-2. 8-3. 117 148 and Phase 8-4. Periodicity, Amplitude, Inverse Functions Inverses of the Trigonometric Functions 14i) ir>r> 156 LINEAR EQUATIONS AND GRAPHS !)-!. 9-2. 1 163 Solutions of Simultaneous Equations Algebraic Solution of Linear Equations in 9-4. Linear Equations in Three Unknown* Graphs of Linear Functions 9-5. Intercepts 9-6. Graphical Solution of Linear Equations in 9-3. 10. 146 146 Variation of the Trigonometric Functions The Graph of the Sine Function The Graphs of the Cosine and Tangent Functions 8-5. 8-6. 9. ix 1(5*5 Two Unknowns... Two Unknowns . . . 173 173 175 177 )ETERMINANTS 10-1 . 10-2. 10-3. 10-4. I )eti'rminants of the Second Order Determinants of the Third Order Properties of Determinants Solution of Three Simultaneous Linear Equations in Three Unknowns 10-5. 10-7. 1 1 . in Three Unknowns 11-5. 183 () 184 185 189 t ^ 189 191 192 105 106 Graphical Representation Trigonometric Representation 11-6. Multiplication and Division in Trigonometric 11-7. DC Moivrc's Theorem Roots of Complex Numbers 11-8. When Homogeneous Equations Sum and Product of Determinants COMPLEX NUMBERS 11-1. The Complex Number System The Standard Notation for Complex Numbers 11-2. 11-3. Operations on Complex Numbers in Standard Form 11-4. 12. 181 Systems of Three Linear Equations D= 10-6. Form 198 199 200 EQUATIONS IN QUADRATIC FORM 12-1. Quadratic Equations in One Unknown 12-9. Solution of Quadratic Equations by Factoring Completing the Square Solution of Quadratic Equations by the Quadratic Formula Equations Involving Radicals Equations in Quadratic Form The Discriminant Sum and Product of the Roots Graphs of Quadratic Functions 12-10. Quadratic Equations 1 2-2. 12-3. 12-4. 12-5. 12-6. 12-7. 12-8. 166 167 169 170 171 in Two Unknowns 204 204 20 1 206 . . . 209 212 214 215 217 218 221 Confenfs x 12-11. 12-12. 12-13. 12-14. 13. THEORY OF EQUATIONS 1 3-1. Introductory Remarks 235 235 235 240 13-7. The Remainder Theorem 241 The Fundamental Theorem of Algebra 243 Pairs of Complex Roots of an Equation 244 The Graph of a Polynomial for Large Values of aRoots Between a and b If /(a) and f(b) Have Opposite Signs 245 13-8. Rational Roots INEQUALITIES 14-3. Introduction Properties of Inequalities Solution of Conditional Inequalities 14-4. Absolute Inequalities 14-1. 14-2. PROGRESSIONS 15-5. Sequences and Series Arithmetic Progressions The General Term of an Arithmetic Progression Sum of the First 71 Terms of an Arithmetic Progression Arithmetic Means 15-6. Harmonic Progressions 15-1 . 15-2. 15-3. 15-4. 1 5-7. 15-8. 15-9. 15-10. 15-11. 15-12. 15-13. Geometric Progression The General Term of a Geometric Progression Sum of the First n Terms of a Geometric Progression Geometric Means Infinite Geometric Progression Repeating Decimals The Binomial Series MATHEMATICAL INDUCTION 16-1. 16-2. 17. 233 Synthetic Division 13-6. 16. 231 13-3. 13-5. 15. 224 227 13-2. 13-4. 14. Graphical Solutions of Systems of Equations Involving Quadratics Algebraic Solutions of Systems Involving Quadratics Exponential and Logarithmic Equations Graphs of Logarithmic and Exponential Functions 248 248 248 249 254 256 256 260 260 261 262 264 265 265 266 267 268 269 271 273 273 Method of Mathematical Induction Proof of the Binomial Theorem for Positive Integral Exponents 275 PERMUTATIONS, COMBINATIONS, AND PROBABILITY Fundamental Principle 17-2. Permutations 17-3. Permutations of n Things Not All Different 17-4. Combinations Binomial Coefficients 17-5. Mathematical Probability 17-6. Most Probable Number and Mathematical Expectation 17-7. 17-1. 17-8. 245 Statistical, or Empirical, Probability 278 278 279 280 281 282 283 284 284 Confenfs 17-9. 17-10. 17-11. 18. xi Mutually Exclusive Events Dependent and Independent Events Repeated Trials 285 286 287 SOLUTION OF THE GENERAL TRIANGLE 18-1. Classes of Problems 18-2. The Law of Sines 18-3. Solution of Case I by the Law 289 289 289 of Sines: Given One Side and Two Angles 290 18-4. Solution of Case II by the Law of Sines: Given the Angle Opposite One of Them 18-5. The Law of Cosines 18-6. Solution of Case III and Case IV by the Law The Law of Tangents The Half-Angle Formulas Area of a Trjangle 18-7. 18-8. 18-9. Two Sides and of Cosines 291 296 297 298 300 302 APPENDIX A. Tables "B. Answers INDEX to Odd-Numbered Problems 307 337 353 1 1-1. Introductory Topics THE REAL-NUMBER SYSTEM The real-number system that we use in the early part of this a development from the original counting numbers, or positive integers, such as 1, 2, and 3. Almost simultaneously with the invention of positive integers, practical problems of measurement gave rise to positive fractions, such as 1/2, 5/6, and 16/7. Much later, in comparatively modern times, the concepts of negative numbers and of other types of numbers were gradually developed. Negative numbers were invented when the problem of subtracting one number from a smaller one presented itself. Thus, the number system was soon enlarged to include the negative integers and course is fractions. These positive and negative numbers, together with zero, are called the rational numbers. Hence, a rational number is defined to be any number that can be expressed as the quotient, or ratio, of two integers. For example, 2/3, 5 (which may be considered as 5/1), and 7 are rational numbers. The number system was then extended to include also numbers which cannot be expressed as the quotient of two integers, namely, the irrational numbers; examples are ^/2 ami rr. The two classes of numbers, rational and irrational, comprise the real numbers. These numbers are so called in contrast to the imaginary or complex numbers considered in Chapter 11. 1-2. FUNDAMENTAL ASSUMPTIONS We shall proceed to introduce the four fundamental operations of addition, subtraction, multiplication, and division into the system of real numbers. The reader has probably been performing these operations in arithmetic and algebra without being conscious that certain basic laws were being obeyed. We shall introduce the four fundamental operations and state, without proof, the laws or assumptions governing them. i 2 Sec. Introductory Topics 1-2 is assumed that there is a mode of combining any numbers a and 6 so as to produce a definite real number called their sum. This mode of combination is called addition. The sum of a and b is denoted by a + b. In this sum a and b are called Addition. It two real terms. Multiplication. It is assumed that there is a mode of combining any two real numbers a and b to produce a definite real number mode of combination is called multiplicaThe product of a and b is denoted by a b or by ab. The individual numbers a and b are called factors of the product. called their product. This tion. Commutative Law for Addition. If a and b are any real numbers, then a (1-1) + b = b + a. Thus the sum of two mumbers is the same regardless in which they are added. For example, 1 , + 3=3 + 2. 2 Associative Law of the order for Addition. If a, b, c are any real numbers, then (1-2) (a + 6) +c =a+ (6 + c). we obtain the same result whether we add the sum of a and 6 to c, or we add a to the sum of b and c. Since the way in which we associate or group these numbers is immaterial, we may write this common value as a + 6 4- c without fear of ambiguity. For That is, example, 2 +3+4 = Commutative Law for (2 + 3) = +4 2 (3 + 4). and b are + Multiplication. If a any real num- bers, then ab (1-3) - ba. is, the product of two numbers is the same regardless of the order in which they are multiplied. For example, That 2-3=3-2. Associative Law for Multiplication. If a, 6, c are , any real num- bers, then (1-4) 1 (o6)c = a(6c). Illustrations of the laws are given here only for the bers, the positive integers. all real numbers. It is understood, most familiar num- however, that the laws apply to 12 Sec. That is, 3 Introductory Topics we we obtain the same result whether multiply the product of a and b by c, or we multiply a by the product of b and c. Since the way in which we associate or group these numbers is immaterial, we may write the result as abc without fear of ambiguity. Thus = 2-:5-4 Distributive Law. If (1 (2-:}) a(b 2- (:;-4). are any real numbers, then 2 a, b, c f>) = i + = r) + ah or. This law, which is usually known as the distributive law for multiplication with respect to addition, effects a connection between addition and multiplication. The distributive law forms the basis for the factoring process in algebra, as will be seen. simple example of the distributive law is A + 2 -(3 4) =2-:5 + 2- 1. This law can be extended to the case where the three or more terms, as sum consists of in the following illustration: For positive integers, multiplication may also be interpreted as repeated addition. Thus, by the distributive law, S 4 ;>,.4 Zero. = = = (1 + + 1) + + + + + + 1 .">(! :$ It is \ I :! ;; -- 1) (1 -4) + (1 4) + (1 4) = (3- 1) + Cl- 1) + (o- 0, - = 4 1) + 4 + C>- 1) + 4, :j. assumed that there and denoted by (1 - 1 is number number a special such that, for every real a (i) + = a. i = called zero , For example, i] It of a a + = + 0' From 2 + o j, + o = o. can be easily shown that only one number with the property can exist. For let 0' be another such number. Then, since = a and b + 0' = b for any numbers r/, b, it follows, by taking and b = 0, that ()' tion, o o=;j, + = the commutative law, The right we agree side of 0', and + = 0. 0'. (1-5) should read (ah) to omit the parentheses performed before any addition. 0' when -f all (ar). However, by conven- multiplications are to be 4 Sec. Introductory Topics 12 Negative of a Number. It is assumed that for every real number a there exists a corresponding number, called the negative of a and designated by a, such that + (- a 7) (1 = a) 0. For example, + 1 (- =0, 1) (- + 2) 2 =0. That each number has but one negative may be shown in the 0. following way: Let x be another negative of a, so that a + x Then By = (- a n) +0 = (- - a ~ ((- a + + <i) + (a x). associativity, o) + n) + x, or In particular, the negative of zero .r. is - - - x = + 0. The Unit. It is assumed that there is a special number called the unit and denoted by 1, such that, for every real number a, (1 a 8) 1 There cannot be a second unit i whence = 1 r = . not is 1'. i, a. we If there were, r- 1 - could say that r, 1'. Reciprocal of a Number. It which = 0, there is is assumed that for every number a an associated number - > a called the rccip- rocal of a, such that a (1-9) The reader may verify the of each number. a x 1, then x ~ Thus, if '7 = L fact that there is only one reciprocal another reciprocal of a, that is, if is ,r a in the definition of important to note the restriction a ^ the reciprocal. In the next section we shall see why this restriction It is is needed. Subtraction. 6, is (1 The difference a 6, of any defined by 10) o - b =a + (- 6). real numbers a and Sec. 14 5 /nfroc/ucfory Topics The operation indicated by the si#n minus which produces for any two real numbers a and b the real number a b is called subtraction. The quotient Division. a f b or or a . b of -r- real any numbers a t) and &, where b ^ 0, is defined by ' The operation associating with quotient numbers a and real b (b =/- 0) their called division. is should be noted that subtraction and division are subordinate and multiplication, in that they are defined in terms of b is that number .r for which these latter. The difference a b + .r - a. Also, tho quotient a b is that number y for which b y ~ a. It should be noted that + (-) = for every number a, and that a 'a a (I/ a) 1 for every number a ~f 0. It to addition OPERATIONS WITH ZERO 1-3. It erty a has already been noted that the special number has the prop- a for every real number a. In particular, we may let (t + = to obtain = + It has already been noted that for every real number a. Next, we prove that for every real a (1-12) a Let x 0. .r () 0. = 0, so that number = a + x. =a 4- a, -r 0. Then, by the distributive law, = a = a (0 + 0) = a - + - a we add -.r, we obtain = + (- .r) = (.r + .r) + (- .r) = x + (x + (Since a: = a 0, (1-12) is established. From this last result, it follows that, for b 0, x If ;r x)) = x + = x. - 013) 1-4. It H RECIPROCALS was noted that every non-zero number has a reciprocal. We can now see why cannot have a reciprocal. If has a reciprocal x, x = 1. Since it has been shown that x = 0, we would have then 6 Sec. Introductory Topics = to conclude that 1. However, = if 1 is allowed, 1-4 then for every number a we have would be the only number Hence, number system. This in the cannot have a situation obviously should be ruled out. Therefore, = a the since a/6 (1/6), reciprocal. Moreover, quotient a/6 is not defined when 6 = The reciprocal 0. way -111 : n 1^ (1-J4)' ^ r b a provided that neither a nor 6 with a r b b To prove I 7 then multiply by ---,* a is 0. 1 11 -.-.&&=a a b We two non-zero numbers can be of the product of expressed in another 'T-O* a I we begin =1. i , this result, , 1 b r to obtain L__ -1 ! /; a b . -. , a a b a b In the preceding proof, free use has been made -! i i a , b of the commutative and associative laws. Finally, if a ^ 0, the reciprocal of the reciprocal of a is a itself. Thus, ^= (1-15) Since a. = ^- (!/) !, multiplication by a gives a - = a 1 -T- I/a 1-5. (I/a) a = -7I/a 1 = . I/a THE REAL-NUMBER SCALE Real numbers may be represented by points on a straight line. such a line select an arbitrary point as origin and lay off equal unit distances in both directions, as shown in Fig. 1-1. (The On i i i i i i i i -7-6-5-4-3-2-10 i i i i i i i I 2 3 4 5 6 7 Fiu. 1-1 unit segment may have any length whatsoever.) Label the points thus far specified as indicated is the origin, 1 is the first point to : Sec. 16 7 Introductory Topics the right, 2 is the second point to the right, 1 is the first point to the left, and so on. Rational numbers that are not integers correspond to certain other points in a natural way. For example, 1/2 corresponds to the midpoint of the segment joining points labeled and 1; and 7/3 represents the point one-third of the distance from the point 2 to the point 3. It is a basic assumption the real numbers that every point corresponds to a concerning unique real number, and that every real number corresponds to The full significance of this assumption cannot be developed in an elementary text. One observation of importance can be made at this time. The non-zero real numbers are divided into two classes. One class consist of numbers representing points to the "right" of 0, and the other consists of numbers representing points to the "left" of 0. The first class consists of positive numbers, and the second of negative numbers. The number may be considered as constituting a third class. It is understood that no two of the three classes zero, exactly one point. positive numbers, a common. Thus and negative numbers number cannot be both have any numbers in and zero, both positive positive and negative, or both negative and zero. The specific desigwhich nation of any negative number will include an explicit sign , prefixed. (This convention, however, does not exclude the possibility of allowing a general symbol, such as x, to stand for a negaPositive numbers do not require such a sign, tive number.) although frequently the sign f is used. is It is to be positive, as is 1-6. assumed that the sum of two positive numbers also the product of two positive numbers. is RULES OF SIGNS To operate effectively with real numbers, a knowledge of the rules of signs and of properties of negative numbers is essential. In each of the following relationships, a and b are any two real numbers, except that the denominator of a fraction - (- (1-16) - (1-17) - (1-18) (1-19) (1-20) (1-21) (- a)b = - (a (a (6); + - a} = a. 6) = - o - 6. 6) = - a + 6. in particular, (-)(- 6) -L = - =<*b. J may not be (- 1)6 = - 6. zero. 8 Sec. Introductory Topics a a (1-22) -b a b a b _ ~ ^T 1-6 a 6' Proo/s o/ (i-i (1-16) (1-17) (1-18) - (- a) = - (- a) = a + (- a) + (- (- a)) = o + = a. - (a + b) = + - (a + V) = - a + a + (- &) + b - (a + 6) = - a - 6 + (a + 6) - (a + 6) = -a-6 + = -a-6 by (1-6), (1-7). - (a - 6) = - (a + (- 6)) = - a - (- 6) = - a + 6 by (1-19) Since (- a) (1-20) = (- a) 0- 6 = (- o b a b - (a (- (- 6) (b , a) -6 a) + + + = = ab ((- - = 0-6 = - = b) a) 6 a) 6) (a = - (- = - 6) - (a = <" a 1 (1-12), b). (1-19), (1-16). by(1 - 20) - = = ,( ~ N a) ^| = d-23) by (a 6)) by /i oo\ (I-22 ) (1-17), (1-16). 1 = ~ - / ( '6 a- 1\ ft) = (- |) | = ~ a i by 6 by /i (1 ' in\ 19) * (1-22), (1-16). has already been observed that non-zero numbers are divided classes, namely, positive and negative. It is assumed that if # is positive, then -a is negative and that if a is negative, then a is positive. All calculations involving negative numbers can be made by performing calculations with positive numbers and applying one or more relationships just given. It follows from (1-17), for example, that the sum of two negative numbers is negative, and It , into two ; is equal to the negative of the sum of the negatives of the given from (1-20) it follows that the product of two negative numbers is positive, and is equal to the product of the numbers. Also, negatives of the given numbers. By (1-19) the product of a positive number and a negative number is negative. Sec. 1-7. 1-7 9 Introductory Topics FUNDAMENTAL OPERATIONS ON FRACTIONS A further study of the algebra of real numbers leads us to the consideration of the fundamental operations as applied to fractions. By definition, a fraction is the quotient obtained by dividing one number a by another number 6, where b is not zero. We a the numerator and b the denominator; and we generally write the fraction a/6, read "a over b" or "a divided by b." call We shall list the following basic relationships for applying the four operations to fractions. In them a, b, c, d are any real numno factor in the denominator of a fraction may be zero. bers, except that ac tt-24) (1 24) a U*. + (1-25) a .b c d (1-27) a + ad __ ~~ 6 _ c ad _ ~~ d c a b I be cd c c (1-28) A _ bc~b be cd ac c d b be c special case of (1-30) is _ ~ / c / d d ' c which states that the reciprocal of a fraction is found by inverting the fraction. Also, by (1-30), dividing by a fraction is equivalent to multiplying by its reciprocal. Proofs of (1-24) = ~ (1-29) Since b by ^ ac 7" to (1-30) * ~5 = d 7" 1 * C b * "1 d (& 1 , , "" * w * 1 * bIT 3 d = a C lac T"~5 == bd = = J.1=J 6c6c66 2 U-24) /^ rp\ (1-25) (1-14), 1 a : a,6 +- = J. l.,l + 6 .- = a .- a + 6 /"'i\l + 6) ,- = ^t- (o T~5 bd by ^ / (1-29). \ v/ie\ .by (1-5). 10 / rt ra-^ f+l-ffz + <'-26> # 6 --- *x (1-27)' v = a ~ c c c - . 6)- ( +- = 1 - a c c + , f , ( v 6)' x -1 = c *->. - / (a v = t"\ t* On\ / , (_ __ _ _ + (- U/ Vx a 6) b) fl(* W/U/ Cfr J. d =3 +9 11 c) (2 + 3) + 7 = (11 + = 2 (3 5). 5 3) + 4) = 5 3 + 5 4. + d) 5(3 c (1-22), (1-5). be (1-29), (1-24). State which of the fundamental assumptions are employed in each of the following equations: + (-5) + (-5) o) 3 6) 1 - ttCt by 1-1. *> c erf Example N 6)' by a & _ a -____ 1-7 Sec. fnfrocfucfory Topics (6 +9. + 6) (3 + 7). Solution: d) 6) c) d) The The The The and commutative laws associative associative law for multiplication. distributive law. Example 1-2. In each of the following, perform the indicated operation: a) 2+3. b) d) (+5) -(+3). e) Solution: 6) c) d) e} /) Each + (-2) -f (-3). (-7) -(+6). + c) 5 /) (-7) -(-6). (-3). case can be treated as an addition. = 5. (-2) +(-3) = 5 + (-3) =2. (+5) -(+3) = (- 7) - (+ 6) = (-7) -(-6) = a) 2 for addition. associative law for addition. 3 -5. +5 + -7 + -7 + (-3) (- 6) <+6) = +2. = - 13. = -1. Example 1-3. By using the fundamental assumptions and rules for and transforming the left side into the right side, justify the equation (a + 6) - (c - d) = (6 - c) + (a + d), operations^ Sec. 1-7 Introductory Topics By Solution: By (1-1) and Example 1 1 (1-18), (a + 6) - (a + 6) - - (c = d) (a + 6) - +A c (1-2), 1-4. + c d = - (b c) + + (a d). using the fundamental assumptions and rules of operations, By justify the equation a Solution: By b (1-29) and (1-24), 6 ac 6 a _ a c be 6 c "~ a _ "~ ac b c (be) (6 6c __ ~~ 6 c) c EXERCISE 1-1 1. Identify the fundamental law or laws that justify each of the following equations: a. x c. 2(3 = (2 e. (a -f 2) (6 - 2. Find the value a. j. (-3) (+7) (- 5) (-5) 3. Evaluate each of the following: a. (+2) (-3). (-7) (+5). b. c. e. 2(-5). f. (-7)0. g. (-!)(- h. (-4) (-5) -(_2)(-l). i. (+2) (-3) -(+7) (-5). 4. Determine the negative of each of the following d. g. + ?/=?/ 5) + = of +(+5). -(+2). - 0. - (+5). 2) d. 5(a 3)5. 3) - (6 3) (a + 2). h. k. - 2-3. J- 3# 5. i ! + cb. ca (- 7) - (- 5). (+32) -(-23) +(-45). 1. (-3) (-5). (-5) (-9). &)] 3. + 2. : d. 2x. 0. c. 2a - 36. k. x - 3. g. - h. a 1. - (x c. b.* 2/3. 1.02. f. _L_. x + y r a +6. ,_ 2? &J * g. - i |^ * j^ * t k A* - y). + 0. * Find the reciprocal of each of the following: a. ], e. (-1) -(-2). 0-(-2). i. +(-3)(0). f. (- c. f. d. e. 2/3. [(a) (a (-5) +(-3). (-8) -(-9). 15 + (- 3). + (-3) -(+4). b. e. b. i- f. each of the following: a. 5. - = sr. + b) - 5a + 56. + b)c = c(a + b) = b. rs x. 3 +|O -- 1 ' d. 2 t h. a o l r - + 5 ' " 0.1 2 - 0.3* 12 6* Sec. Introductory Topics Prove each of the following equations by using rules and for signs 1-7 for operations with fractions: a. a c. - o- - ( [6 6) - +a (a - <J&C ( - c)] = = - c) (a - 6) a - (b -t- b, c). - d. 6 c. - (ac / -- / rt ~~* jr/ <*-< OC ad) h = a[d + c = a. ( c)]. C - f- ORDER RELATIONS FOR REAL NUMBERS 1-8. We to express the fact that a is a a> and the notation a < to indicate that a is negative. The symbol > means is greater than, and < means is less than. These symbols are called order symbols. Assume that a and b are any two given numbers. If a - b > 0, we shall write a > 6, or b < a, and shall read "a is greater than b," or "& is less than a." As can be easily seen, a > b means that a lies shall use the notation positive number, to the right of 6 on the real-number line. a b > tive; b 0, then & which equals a, and conversely. - a < 0. The student is Hence, When (a a> b, that is, when by (1-18), is negaif and only if 6) a>&(or&<a) familiar with the symbol - (for equality), which used to indicate that two quantities are the same. Thus a = b means that the two symbols a and b represent the same matheis matical object. For example, 6-3*2. If a and 6 are two distinct numbers on the scale, we say "a is different from &" or "a does not equal &," and we write symbolically a^b. The symbol means does not equal and = is called the \ inequality symbol. In general, the oblique line or vertical line through any symbol will form a new symbol which is the negation of the original one. Thus, a < b means "a is not less than b." In other words, a = b or a > b (by Property 1 below). For example, 5 < 3. Sometimes we shall find it convenient to combine the symbols < and = or > and =. We write ^ to mean is less than or equal to, and we write ^ to mean is greater than or equal to. We thus have order relations on pairs of real numbers, defined by either of the following equivalent statements : The system tion > (or a > b (or b < a) if and only if a b is positive; a > b (or 6 < a) if and only if b a is negative. numbers is then said to be ordered by the relathe relation <) Assertions of the type a < b or a > b are of real . See. 1-8 13 Introductory Topics The ordering of the called inequalities. real numbers has the fol- lowing properties. Property 1. For every pair of real numbers, a and only one of the following relationships holds: = a < or a 6, 6, or a > 6, one and b. Proof of Property 1: If a = 6, the statement is certainly true. b = 0, so that Saying that a = b is equivalent to saying that a a 6 is either positive or negative. Thus, if a 6 is positive, we have a > b. If, however, a b is neither positive nor zero, then it is negative, and a < b. If two of the three possibilities occurred together, we should have, say, a = b and a > b, or a > b and a < b. b would be both zero and positive, or both positive and Thus, a negative. Since no overlapping may occur among the three classes we of numbers, Property are thus led to a contradiction. For any 2. if numbers real < a b Proof of Property 2: If and a< b b < a, 6, c, it is c, then a and 6 < < true that c. then both 6 c, a and a as (c Let us write c (b 6) a). We have assumed that the sum of two positive numbers is positive. Since c b and 6 a are positive by assumption, their sum, which is c a, is also positive. Hence, c > a, or a < c. c b are positive. Property 3. 4- For any if real by Proof of Property 3: positive. But, by (1-17), (a It a + c) - + (b a, b, c, it is +c>b+ By definition, a> b > a numbers then a true that c. means that a + (-b) + (-c) = a - 6. - (6 + c) is positive, and (a + c) = c) therefore follows that b + c. a + 6 is c that +c> Property 4. For any if real > a b numbers and c > 0, a, 6, c, it is then ac > true that be. We have assumed that the product of two 6 and c are positive, it positive. Since both a Proof of Property 4: positive numbers is follows that their product Therefore, ac Property 5. be is also positive. is positive, For any if a real > b and ac numbers and c < > But (a 6)c = ac be. a, 6, c, it is 0, then ac < be. true that be. 14 Sec. Introductory Topics Proof of Property 5: We numbers with unlike signs is 1-8 have noted that the product of two - b is positive, but negative. Here a = ac 6c, and (a 6) c is negative, it c is negative. Since (a 6) c follows that ac be < 0, or that ac < be. According to Property 3, the order symbol in an inequality is not changed if the same number is added to or subtracted from both sides. It therefore follows that a term on one side of an inequality may be transposed to the other side with its sign changed. For 6 > c, then a > c + b. example, if a to According Property 5, the order symbol in an inequality is reversed if both sides are multiplied or divided by the same negative number. 1-9. ABSOLUTE VALUE As a consequence of the properties of the ordering of real numcan be associated with each number a certain nonnegative number called its absolute value. For any real number a, we define the absolute value of a, denoted by |a|, as follows: bers, there | Thus, 1-10. 3 | = | a = | 3, since 3 a, if > a ^ 0; also 0, and |- 3 | | a = a, if \ = - (- 3) = < a 0. 3, since - 3 < 0. INEQUALITIES INVOLVING ABSOLUTE VALUES We now we let consider some inequalities involving absolute number x be represented by a point P on a number scale, then \x\ is the numerical distance between P and the origin. If we let a be a positive number, then \x\ < a means that the point P is less than a units from the origin; that is, x lies shall If values. between a and a. the We can write this in the form < a < more briefly in the form a < x < a mean exactly the same thing. \x\ < a and A more general inequality which often occurs 6 < a. where a > 0. This is equivalent to a < x a or to each term, ments \x we may b\<a and write 6 6 x a< a. x < x and x < a, Therefore, the statements < a<x<b + a b + a. mean is |# If 6 is 6| < a, added Hence, the stateexactly the same thing. 2<x 3<2 and For example, \x~ 3|<2 may be written means that the distance between x and 3 is less than 2. To solve this inequality for x we add 3 to each term of the inequality, t obtaining 1< x < 5. The following illustrative examples may help to give a better understanding of the processes involved in the solution of the problems in Exercise 1-2. Sec. 1-10 1-5. Example Arrange the following numbers in increasing order: - 2, Since Solution: TT 1-6. Example 3.5, 0, 2, 3.14, Since or |, |-2 1-7. Example - | | -2 = > -2. | integers a |-2 and since - 2 < 2, 6 and such that a Example 1-8. we have the inequality - 2 2, < < \/2 b. be represented approximately by 1.414, the values may satisfy the inequalities. 6 |. such that a 6=2 and 1 as follows: is |-5|. TT, 2 and Find integers a and Solution: Since \/2 = a |. Insert the proper inequality sign (order symbol) between the following numbers: Solution: |-5 3.5, O,TT, 3.14, approximately 3.1416, the desired order is - < |-2 15 Introductory Topics ^ and 1 Express the inequality Thus, 1 < \/2 < 2. Any other pair of would also satisfy the inequalities, 6^2 | < x \ 3 without using the absolute-value symbol. Solution: We know that the statements the same thing. Here a point represented by x and 3. The inequality is x < a and a < x < a mean exactly the positive number 3, and x < 3 means that the 3 less than 3 units from the origin; that is, x is between | \ is | be written may < 3 < x 3. Example 1-9. Explain the meaning of the inequality without using the absolute- value symbol. Solution: The inequality x | x < be written inequalities, we obtain 1 < x 2 | | may 1 \ x | 2 < | 1 and write it b < a is equivalent to a < x b < a. Hence - 1 < x - 2 < 1. If we add 2 to each term of the < 3. \ EXERCISE 1-2 1. a. c. Arrange the numbers in each of the following - e. 3, - 2, 5. 3, 0, 4, - 1, - 1/3, 2, 10, - 2, 1, V3, - 3/2. 2. Insert sets in increasing order: b. d. 4. - - 6, f . 1.4. 0, 1/2, 8, 2, 0, 10, 9, 4, - 2, - 3, 3/8, - V, I - 3/4. - 6/5. 3 |. the proper order symbol between the two numbers in each of the following: ,, and 1/3. - 3 and - a. 3 d. 3. b. 2. Examine each e. f - 3 and 22/7 and | 3 |. w. of the following inequalities, c. \/2 and 1.414. t. 1/8 and 1/6. and determine whether or not it is true. a. d. TT 5 > > 22/7. 3. b. e. | - 3 -f 2 < 0. -2 < 2 | | |. . c. - 3 3 - | f. | | 7 > > | 3. | 5 - 2 |, 16 4. Find the value of each of the following: b.|-3| + |+3|. a.|+2|-|-2|. -6 d. |-7| + |-5|-|+5|. e. 12 - 4 h. 3. g. (-18)* |-9|*|4|. j. 0|4|-|-5|. | 5. 1-10 Sac. Introductory Topics | | c.|+4|-|-4|. f. |5-3|+|3|-|2|. |. 1.0*14. Express each of the following inequalities without using the absolute-value symbol: a. I d. | x < I 2x 2 < | < b. 1. 4. e. | x - c. 1. < 1 | 3. f. | 6. In each of the following, find a pair of integers, a inequalities are satisfied: b. a < 3 < b. a. a < 5 < b. and d, a < TT < 6. 7. If a ^ 3, place the proper order symbol between a 8. If a *z 5, a what can be < < \/3 e. s a f. a &. said about the value of 3a + ^ I - a. 1/2 < | 3/2. such that the given 6, - e. \x < < < 1 1 b. - 2 | < 6 7 and 10. 2? POSITIVE INTEGRAL EXPONENTS 1-11. If two or more equal quantities are multiplied by one another, the product of the equal factors is called a power of the repeated factor. Thus 5 2 read "5 squared," means 5 5 5 3 read "5 cubed/' means 5-5-5. In general, an means the product of n factors each ; , , equal to a. We call a the base and n the exponent of the power. It follows from the associative law that a2 Also, if a3 ^ a = (a a a) (a a) = a a a a a = = a5 - ,,5-3 a 2 *3 . 0, a a3 a a a a a a a a = a = a "2 a These and similar results suggest the following laws of expo- m and n are positive integers. The proofs of these them laws are reserved for a later chapter. nents. In Law of Multiplication. To multiply two powers of the same base, add the exponents : am (1-31) Law of Division, power of the same To an divide one = am+n . power of a given base by another base, subtract the exponents : ' (1-32) ^= a"*-*, if a ^ 0, m> n. Sec. 1-12 Law Power of a Power. To for a power, multiply the exponents (1-33) (a For example, Law 17 Introductory Topics (a for a 3 2 ) = Power a3 a3 = m a3 raise a power of a given base to a : = n ) ' 2 = a mn a6 . . To obtain a power of a product, power of a Product. raise each factor of the product to the given (1-34) (aft)* Thus, (3a Law 3 2 ) = 3 2 (a 3 ) 2 = 3 2a6 = = 9a 6 : n a b. . Power of a Quotient. To obtain a power of a quotient, numerator and the denominator to the given power for a raise the : H (n\ I) Thus if inus, n =, nn if 6*0. 6*0 * o, o ALGEBRAIC EXPRESSIONS 1-12. An means algebraic expression is formed by combining numbers by of the fundamental operations of algebra. The distinct parts by plus and minus signs are called The terms of the expression 3x 2 -5xy 2 +7z are 3# 2 -5xy2 Here the numbers 3, 5, and 7 are called numerical coffi- of the expression connected terms. and 7z. , , x 2 xy 2 and z are called the literal parts. An expression containing one or more terms is called a multinomial. A multinomial consisting of one term is a monomial. A binomial is a multinomial consisting of two terms, and a trinomial is a multinomial with three terms. A polynomial is a multinomial are whose terms are of the form ax m y nz p , where m, n, p, cients, or just coefficients; , , positive integers and a is a numerical coefficient, and where one or more of the factors x m , y", z p , may be absent. Thus, 7, 5# 4 , and 3xy + 2 are polynomials, while x + -y is not. The degree of a term of a polynomial is the sum of all the exponents in its literal part. For example, the degree of 3# 2 is 2, the degree of 5xy 2 is 3, and the degree of 7z is 1, because the sums of the exponents are, respectively, 2, 3, and 1. The degree of a polynomial is the degree of its highest-degree term. Thus, in the trinomial 3x 2 -5xy* + 7z, the third-degree 2 2 5xy* + 7z term, -5xy , is its highest-degree term? Therefore, 3# is a polynomial of the third degree. 18 Sec. Introductory Topics By a polynomial in x of degree n we mean an 112 expression of the form a xn + where the coefficients a a T^ 0, and w is a positive aix n~ l + h an, a n are numerical * &i, , , n= coefficients, we agree that the polynomial reduces to a number a which is not 0, and that the degree is zero. The number is regarded as a polynomial also, but as one integer. If 0, , having no degree. If the typical polynomial just given has degree n, the coefficient a is called the leading coefficient. If its leading coefficient is 1, a 2x 2 + 5x + 1 is a monic polypolynomial is called monic. Thus, x* nomial of degree 3. EQUATIONS AND 1-13. IDENTITIES An equation is a statement of equality between two numbers or algebraic expressions. The two expressions are called members, or sides, of the equation. Equations are of two kinds, namely, conditional equations and identities. A conditional equation, or simply an equation, may be true only for certain values (possibly none at all) of the literal quantities appearing. An identity is true for all numerical values that can be substituted for the literal quantities. Illustrations of equations are - 3x 5 = x + 1 5x + 4 = 0. and x2 The x = first 1 or x one is true only - if x = 3, and the second is true only if = 4. Illustrations of identities are - 2) = +4= (x 3(s 3x - 6 and x2 - Each of these equations 1-14. 5x is true for all 1) (x - 4). values of x. SYMBOLS OF GROUPING Parentheses ( and other symbols of grouping which have the ) same meaning as parentheses, namely, brackets [ ], braces { }, and the vinculum are used to associate two or more terms which are to be combined to form a single quantity. The word "parentheses" , Sec. 1-14 19 Introductory Topics is often used to indicate any or all of these symbols of grouping. Removal of the symbols of grouping is accomplished by applying the laws of algebra, such as the laws of signs and the distributive law. The following examples illustrate the procedure. Example 1-10. Solution: The Remove steps - may (2x parentheses from 3). = - 1) (2x - 3) = (-l)(2x) +(-!)(= - 2x + 3. 3) ( 3) from 8x of grouping 2[5y +3 (x y)] basic rules for enclosing a group of terms in parentheses may x [2y Solution: 8x - 3y} and collect terms. One way - 2[5y of obtaining the desired result follows: + 3(z - be stated as follows - y)] = = = The - (2x be indicated as follows: - Remove symbols Example 1-1 1. - Sx Sx 8x {2y - - x 2[5y 2[2y 4y - - Zy] +3x - 3y] - {2y + 3x] - {5y - x} 6x - 5y x + 3y} +x : To write a given expression sign, write the and write + in parentheses preceded by a plus terms as they are given, enclose them in parentheses, in front of the parentheses. Thus, in parentheses preceded by a minus change the sign of each term, write the resulting terms in in front of the parentheses. Thus, parentheses, and write To write a given expression sign, a The - first rule is & = - ( - + 6) = - (6 - a). obvious, and the second follows from the rule of signs (1-18). Example 1-12. a) Enclose the last two terms of 2 4- 3x y within parentheses preceded by a plus sign. 6) Enclose these terms within parentheses preceded by a minus sign. Solution: +, we enclose 3x - y in its given form within the parentheses preceded by a plus sign. In this case the ex* - y). pression becomes 2 -f (3x a) Since the sign before the parentheses is to be 20 Sec. Introductory Topics 1-14 Since the sign before the parentheses is to be , we change the sign of each - y and enclose 3x y within the parentheses preceded by a minus 6) + term of 3x Then the sign. expression becomes 2 3# ( + y). ORDER OF FUNDAMENTAL OPERATIONS 1-15. Parentheses and other symbols of grouping are useful in indicating which operation is to be performed first. We have used them in this way from the outset. In order to avoid using them unnecessarily, as has been already pointed out, the convention is adopted to perform all multiplications first and then the additions (or subtractions). If two or more of these symbols of grouping are used in the same expression, we usually (though not necessarily) remove the innermost pair of symbols first. Illustrations in which symbols of grouping are removed follow : 6) - (6/2) =4-3 = 1. - 6)/2 = (- 2)/2 = (4 c) 6 a) d) 4 + [15/(3 6 + (15/3) 5)] '5 1. + (15/15) =6 + 1=7. = 6 + 5-5 = 6 + 25 = 31. = 6 EXERCISE 1-3 In each problem in the group from 1 to 36, perform the indicated operation or operations. 25 1. 2. . 5. 10'. 6. 34 3. . (- 2) 4. 7. ((- I) 6) 3 . 4. 53 3. 8. - . 10 6 . 4 10. 9. 7*. (-4) 8 11. . ()' "' * 17. i(4 3 ) J 18. f(3 4 ) 19. a(5 8 ). 21. a 3 a 4 . 22. (a 3 ) 4 25. . 2 ( 6 o 3 2 . 26. a ) 34. a8 a - . 2 &YV4/ a 3. 12 - (|) *> (-!)' 23. (a&) 4 27. a 2 a 20. 24. . 4 a 2 6(- a7 28. , ttpV. w/ 36. (W\a / 8 In each problem in the group from 37 to 48, remove symbols of grouping and simplify. 37. 3 - (b - + 39. 4[a (6 a)]. - 36). 41. a - t (2a + 43. (a - 26) - (3a + (y - z). - (3a (a + 26) 38. x 2). - + 6). 40. 42. a 44. - [3 W- + (2a - [3x + (2y b). 4)]. - z 2 )]. Sec. 1-16 - 45. a* 2 46. 3n6 48. + 2bxy 4ac { - 47. 2a - - [(6y - + ab] 3a6 - 6 + a~^~6 - (3a -[ -(a -b -c) -a + { In each problem from 49 to + 52. x +y 55. z 2 58. ax + b - -1. 2 ?/ 2 c. - z2 . 2ax?/ } 2 i/ )]. . (3a 4- 2c) } 4c + a]. -c)]J. (6 two terms in parentheses. First and then use a minus sign. 60, enclose the last 50. a 2 - 2ab 53. 2a + 6 - x2 - 59. 2z - 56. + y*. + (axy 2ac -f [ab [36 -f 4c - 2cx*) use a plus sign before the parentheses, 49. a 21 Introductory Topics + - - 2 - 3y . 54. 3x 3c. ?/ - 62 + c2 - 4y + 20. 51. a 2 6*. z2. 57. 60. 4z. - + a36 x 2 - ab 3 + + &2- 1. In each of the following problems, evaluate the given expression. 61. 16 64. 4 - 67. (3 70. 3 - - 2). 7). (6 - (4 2). 3) 3 - (4 2). 16-6-2. 62. (6 65. 4 6 68. 3 3 + (4/2). (8 + 4)/(2 + 3). 74. (8 76. 77. 8 ADDITION 63. 4 69. 3 2. + 4)/2. + (4/2) + 3. 2 3 Terms, such as 2x y and 5x y 3) 75. 8 78. 8 B , (-4) - - 7. 6) (3 4) 72. 8/(2 AND SUBTRACTION OF 2 (- 66. (4 7. + 4. 71. (8/2) 73. 8 1-16. - (4) (- 2). 2. + 4). + 4(1/2) + 3. + ((4/2) + 3). ALGEBRAIC EXPRESSIONS which have the same literal parts, may be added or subtracted by are called similar or like terms and adding or subtracting their an example. coefficients. To illustrate, let us con- sider Example 1-13. Add 5zy 2 , 7x*y, Solution: Collecting like terms (5 - 2)zt/ 2 + (7 - - 2zy 2 , - and adding 2 9)x y + 4z 2 3 i/ 9x*y, coefficients, we have = 3xy - 2x*y The procedure used in the solution of Example 1-13 follows at once from the distributive law. For example, the sum of the terms 2 5xy* and -2xy 2 is obtained as (5 - 2)xy* or Say This leads at once to the rule for the addition (or subtraction) of . algebraic expressions. In practice, we usually arrange like terms in vertical columns, and then we find the sum of each column by sum of the numerical coefficients procedure may be made clearer by means of the prefixing the in the column. The following examples. 22 Sec. fnfroe/ucfory Topics - 2 Example 1-14. Add 2x 3xy + z, - 2z 2 x - x2 3z2/ -f 3z 2xy - 36 xy +c 2 z. - c2 - 2a 36 -f c2 + - 2 a2 - from 3a 2 3a 2 Solution: +z -5z 2 3x 2 Example 1-15. Subtract 2a 2 + 3z. 2xy 50, 1-16 36 2c 2 c2. . MULTIPLICATION OF ALGEBRAIC EXPRESSIONS 1-17. With the help of the distributive law for multiplication, the product of two algebraic expressions is found by multiplying each term of one by each term of the other and combining like terms. Example 1-16. Multiply 3x 2 3z 2 Solution: - 2xy +y 2xy 2x + y2 - 2 by 2x - 3y 2xy 6z 3 - I3x 2 y 3y. H- 2 Sxy 2 - 3y 3 . SPECIAL PRODUCTS 1-18. The following typical forms of multiplication occur so frequently should learn to recognize them quickly and to obtain the products without resorting to the general process of multiplication. that we They should be learned thoroughly. a(b (1-5) (1-36) (1-37) (1-38) (1-39) (1-40) (1-41) + c) = ab + ac. + b) (a - 6) = a 2 - b 2 2 ab + b 2 = a 3 + 6 3 (a + 6) (a - 6) (a 2 + ab + b 2 = a 3 - 6 3 (a 2 = a 2 + 2a6 + b 2 (a + - b) 2 = a 2 - 2ab + b 2 (a 2 2 (ax + by) (ex + dy) = acx + (ad + bc)xy + bdy (a . . ) . ) . ft) . . See. 1-19 1-19. 23 Introductory Topics DIVISION OF ALGEBRAIC EXPRESSIONS To divide a polynomial by a monomial, divide each term of the polynomial by the monomial and add the results. This rule follows immediately from the rule for fractions expressed by (1-25). Example 1-17. Divide 6a 2 z 2 12a 4 z ~ , ,. 6a 2 z 2 Solution: - 12a% - 30a 6 z 6 by 15a 3 z 2 . 30a 6 x 5 15a 3 z 2 _ 5a To 5x divide a polynomial (the dividend) by a polynomial (the both divisor) , arrange according to descending or ascending powers of some common literal quantity. Then proceed as follows: Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient. Multiply the entire divisor by the first term of the quotient, subtract this product from the dividend. and Use the remainder found by this process as a new dividend, and repeat the process. Continue the work until you obtain a remainder that is of lower degree in the common literal quantity than the divisor. Example 1-18. Divide 6* 3 - 5z 2 2x Solution: - 1 | - 6z 3 6s 2 + + 3x + 3x 2x 2 + 3z 2x 2 4- x 5x 2 3s - by 2x 1 + 1 | - 3x 2 1. x + 1 2 2x 2x + 1 - I 2. The division can be checked by finding the product of and adding 2, proving that 6z Any problem 3 - 5x 2 -f 3z + in division, in general, the relationship dividend 1 = (2x (2s may - - 1) 1) and (3z 2 (3z 2 - x - + x -f 1) 1) -f 2. be checked by means df ? This equation literal is = (divisor) (quotient) an identity that quantities. ; is, it is + remainder. true for all values of the Indeed, this equation supplies the underlying meaning of the process of division, 24 Sec. Introductory Topics 1-19 EXERCISE 1-4 In each problem in the group from separated by commas. - 1. xy, 3xy. + 116, 6a 4. - + 26. 3a 2. 4x 2 y 2 5. 2 4a 26 add the given expressions which are 2xy. , - to 12, 1 2 , - a 2 46 2 . 3. 3x 2 y 2 6. - 3a f 26 2x 2 y*. + 4c, 4a + 36 - 6c. -3a+6~c, -a-6+c. 7. + 2y - z, x + y - 3z, 4x - 3y + 20. - 2xy + y 4xy, - y 2 3z 2 - 4j?/ 2 + 2?/ 3 4z 2 - 2x 2 - y 3 4xy 2 - 2y 8 2x* - 3z + 1, x 2 + 2x - 3, 2z 2 - x s + 4 - 2z. 4az + 3bxy - 4z 2 26z - bx 2 + 2azy, 3z - 2y. 3x 8. x2 9. 10. 2 . , ?/ , 11. 12. . , , In each problem from 13 to 24, subtract the second expression from the - xy, 3xy. 14. 4x 2 y 2 - 2xy. 13. first. , - 15. 3z 2 2/ 2 , - 17. 4a 2 19. - 3a 21. x 2 + 26 2x 2 2/ 2 2 , +6 2a;y 4- - 22. 3 23. . - a2 - c, 2 2/ 2x 46 2 > + . - a 6 + c. 6a + 115, 18. 3a - 26 20. z 3 +x + 4x 4 2 - 3a + 26. + 4c, 4a + 3& - 6c. + x + 1, 3 - z 2 - 16. -1. 3J 4xy. a: 2 - a: 3 + 3z 4 - 2x 5 3x 5 , - a; 3 + 2o: 2 - 2x + 1 . - o; y 4- xy 2 - y 2 , x*y - 4- / In each problem from 25 to 60, perform the indicated operations. 26. (4s) 25. (6*)(-37/). 27. (5**y) 29. (- (- 2sy). (- 2a 3 ). 3a6) (4k) 31. (3z 4- 2y) 37. 4z 2 (z 2 32. (6 (- 2). 39. (a; 41. 4x/(- 2 - 2xy + xy - 2 ?/ y (x ) 36. (3a ). - y). 40. 2 (a; 4c). 3a) (2). 66 2 ) (3a6). - + 26). + 3y - 4?/ ). - y). (x + y) (x 6) (a 2 38. 3xy(2x 2 y 2 (- y)2a. + 34. (4a 2 ) - - 30. (x + 27/ 2 (3xy). + 3y) (x - y). 33. (4xy 35. (2x (-2). 28. (3a) (2b 2 ) y 2 ) 42. 3x 2 y/2xy. 2). 43. 4x*y*z/2xy*z. 44. 20z 6 2/ 4 zyiOz 6 2/2 3 45. 46. (2xy) 2/2xy 2 4a6/(- 4a6). - 2*y)/(- s). - So: 2 2 + 4x*y 2 )/x*y*. - 1). 2 60; + l)/(3x (9x 2 (x y). y) /(x 2 '47. 48. 49. (3xt/ 2 50. (x 2 y 3 (2xy)V2(xy). - fay + 9?/ 2 )/(- 3y). 2 51. (x + 6x + 5)/(x + 1). 53. (x 2 3 57. (x 61. 62. 63. 64. 2/ ' y). 54. (a: ?/ + 3x 2 + 3x + l)/(x + I) 2 -f y )/(x -f y). + y) 8 /( + #) 4 3 2 2 60. (x y )/(x y). (x -f x + 3a; + 2x -h l)/(z + Divide x* - y 2 - (x - y) 2 - (x - y) (x y)byx- y. Divide (a + 6) +6(0+6) + 5 by a + 6 + 5. Divide z 4 + 4z 3 + 6z 2 + 4z + 1 first by x + 1 and then by x 2 + 2z + 1. 2 2x - 3 by x + 4, and divide the result by x + 1. Multiply x 55. (x* 59. - 2 )/(z - y)/(x - 52. . . 3 y). 56. 3 (a; . 58. (x 3 2 2). 1-20 Sec. 65. Divide x 5 3z 4 4- 25 Introductory Topics 2z 2 + x 4 + 3z - 2z - 3 by x - 1 and add the quotient to the excess of - 9z + 7 over 2z + 2x + 7x + 3z + 4. 3 2 4 3 2 Under what conditions will ( - x) n be positive? 6) When Assume first that x is positive and then that x is negative. n For what values of n will ( an ? a) be equal to 66. a) 67. FACTORING Factoring a quantity will it be negative? 1-20. the process of finding quantities which, multiplied together, yield the given quantity. When a quanA is expressed as a product B C, B and C are called factors tity or divisors of A and are said to divide A. Also, A is called a multiple of each of JS and C. These concepts are applicable to numbers or is when algebraic expressions generally, but are most useful when restricted to apply to integers or to polynomials. Such restriction will be adhered to in this book. Thus, when an integer is to be factored, the factors sought are to be integers. And when a polynomial be factored, the desired factors are to be polynomials. Let us is to review the fundamentals of factoring integers (positive, negative, or zero) First, it is clear that every integer n may be expressed as 1 *n or ( !)( n). Such factorizations are called trivial. If an integer n, other than 4-1 or 1, has no factorizations other than trivial ones, then n is called a prime (number). An first . integer having a non-trivial factorization is called composite. Examples of prime integers are 3, 7, and 11; examples of com40. posite integers are 6 and Let n be a composite integer. Then a non-trivial factorization m *p exists in which \m\ and \p\ are less than \n\ and greater than 1. If both m and p are primes, then n is expressed as a product of primes. If not, at least one of m and p, say p, is composite, and so p r-s. Hence, n = m r s. The process begun may be continued if any one of m, r, and s is composite, and additional factors may be found, until the process cannot be continued further, in which case only prime factors are obtained. We may conclude that the factoring process must terminate, since at any stage the new factors introduced are numerically less than their product. The fundamental theorem of arithmetic guarantees that every composite integer is a product of primes, which "are unique except for their signs or the order in which they are written. For example, successive factoring of 156 gives 156 and the = 39 4 = final factors 13, 3, 2, ization of 156 into primes 150 is 13 3 4 = 13 3 2 2, 2 are primes. Another valid factoras follows: = 2-3- (- 13) -(-2). 26 Sec. Introductory Topics 1-20 Let us turn now to polynomials with real coefficients. These may 1 be polynomials in x, such as x 2 or polynomials in x and y, such a3 x 2 + Sxy + 2y 2 or, in fact, polynomials in any number of literal quantities. Every polynomial F has a factorization of the ; ; form for every non-zero number a. And, of course, the factors I/a and aF are themselves polynomials. Such factorizations are called trivial. nomial. polynomial F has no factorizations other than trivial is called a prime polynomial, or an irreducible polypolynomial having a non-trivial factorization is called If a ones, then A F composite or reducible. 2 Examples of prime polynomials are Sx + 2, 2x + 2y, and x + S. Every polynomial in x of the first degree may be shown to be prime; and certain polynomials of higher (even) degree in x also are prime. A development of criteria for primeness of polynomials beyond the scope of this book. lies 2 2 and 4, xy + y Examples of composite polynomials are x factor+ yz + xu + yu, because each of these has a non-trivial ization. Thus, we have xz - x2 + xz + yz + 4 xy y* xu + yu = = = + (x + (x + (x 2) (x - 2), y)y, y) (z + u). Let / be a composite polynomial in x with real coefficients, not all Then it can be shown that there exist polynomials g and h, the degree of each of which is less than that of /, such that of which are zero. / = g. h. g and h are primes, then / is a product of primes. If not, we may proceed to factor (non-trivially) one or both of g and h, and we can continue this process until primes are obtained. The process must terminate eventually, since each non-trivial factorization leads to polynomials of lower degree. The argument just presented applies only to polynomials in one literal quantity x. However, the principle may be extended to If apply to polynomials in any number of "Thus, ?, / = Plp2 literal quantities x, y, z, . where p if p2 , , p n are prime polynomials. The problem of carry- Sec. 1-21 27 Introductory Topics ing out actual factorizations for certain types of polynomials considered in the next section. is A special class of polynomials deserves particular attention. This is the class that consists of polynomials in which all the numerical coefficients are integers. It is possible to prove a factorization theorem such as that just stated, but yielding factors which are polynomials having only integral coefficients. When the general theorem can yield prime factors all of which have integral coefficients, the two theorems give the same result. Otherwise, they will give different results. For example, both theorems applied to x 2 4 factorization (x + 2) (x The polynomial 3# 2 2) trivial factors with integral coefficients. real coefficients are allowed, we have 3x 2 - 4 = (V3x + yield the prime 4 has no nonHowever, when other . - 2) (-x/33 2), which the coefficient \/5 is a perfectly acceptable real number. factors with integral coefficients are desired in such a case as 36# + 24y, we shall agree to remove and factor common in Again, when numerical factors, to obtain 36* + 24y = 3 2 2 (Sx + 2y). Here the prime factors are the numerical primes 3, 2, 2 and the prime polynomial 3x + 2y. In what follows, whenever a polynomial has only integral coefficients, we agree to restrict ourselves to factors of the same kind. In similar fashion, coefficients, we when shall the given polynomial has only rational search for factors having only rational coefficients. 1-21. IMPORTANT TYPE FORMS FOR FACTORING in Section 1-18 applied "in reverse" are formulas Success in factoring a polynomial therefore depends The equations for factoring. on ability to recognize the polynomial as being a particular type of product and as having factors of a definite form. Verify the following type forms by carrying out the indicated multiplications and learn each form, Type 1: (1-42) Common Monomial Factor. From ab + ac = a(b + c). 2 Example 1-19. Factor 4x y Solution: x*y - 6xy* - toy*. = 2xy(2x - 3 (1-5) we have 28 Sec. 1-21 Introductory Topics for actually removing the monomial factor in 1-19 may be put this way Write the common factor, 2xy, Example and in parentheses following 2xy write the algebraic sum of the 2 quotients obtained by dividing successively every term of 4x y &xy 2 by 2xy, in accordance with the distributive law. Instructions : Type 2: Difference of Two Squares. From (1-36) we have - a* (1-43) - 2 Example 1-20. Factor 9z - Solution: 9x 2 Type (1-38) 25y = 2 Sum and 3: 25s/ + (3z = b2 + b) (a - (a b). 2. - 5y) (3x 5y). Two Difference of From Cubes. (1-37) and we have + a3 (1-44) = 63 + (a b) (a - 2 + ab b 2) and - a3 (1-45) = 63 - 3 Example 1-21. Factor &r Solution: - 8x* = = 27 y* + 2 + 3y) [(2x)* + ab b 2 ). (4o: + (2x) (Zy) +6xy + 2 -3y) (2x b) (a 27y*. - (2x - (a 2 (3?/) ] 2 9t/ ). Type 4: Perfect-Square Trinomials. From (1-39) and (1-40), + a2 (1-46) 2ab + b + = 2 (a + = (a 2 b) and a2 - 2a6 2 Example 1-22. Factor 4z + 12a?y (1-47) Solution: Type nomial 4x* 5: may 12z?/ 2 -f 9y* = = (2^) (2x 2 6) . 4 . 9?/ 2 4- 2(2z) (Sz/ + 2 32/ + 2 ) 2 2 (3?/ ) 2 . ) + Ax 2 + Cy (ax + Bxy the two factors is By comparing this + = (ax + by) (ex and by) + (be (ex + + dy). dy) are multiplied ad)xy + bdy 2 . product with the trinomial Ax 2 it is 2 found to be acx 2 note that tri- : together, the product we + a - General Trinomial. Factorization of the general be indicated as follows (1-48) If + 62 + Bxy + Cy 2 , necessary to find four numbers ac = A, be + ad = B, and bd = a, 6, c, d, C. such that Sec. 1-21 29 Introductory Topics The following example illustrates a trial-and-error procedure, which often involves several steps of inspection and testing by multiplication yet it is a method commonly employed in practical work and is recommended here. ; + 2 Example 1-23. Factor 6x Here we wish to Solution: = - llxy find a, l(ty 6, 2 . and d which c, + 6c)zz/ + = satisfy the identity - + llxy 10i/ a c have like and 10, obviously signs, and b and d have 6. Possible 2, db 3, and 1, opposite signs. Possible values for a and c are 10. By trial and error we find the values for 6 and d are db 1, =fc 2, 5, and + (ax 4- by) (ex dy) 6 and bd Since ac acx* + (ad 2 fccfr/ Sx 2 2. = - = = correct selection to be a = 2, c = 3, 6 5, and d = 6z 2 = This selection meets the 2. requirement because + (2x (3s 5t/) - 2y) by Grouping. An expression which does not one of the given type forms can sometimes be reduced to one of these forms by a suitable grouping of terms. Type 6: Factoring fall directly into The following examples illustrate the procedure. Example 1-24. Factor Sax - 56z Solution: First, + - Say 106y. - Sax common group within parentheses the terms having a Thus, + 5bx - Say IGby = common Then, in each group, factor out the - (Sax 5bx) + (Say factor. - quantity. In this case, = - - 5b) - IGby) 56), 2y(3a (Sax x(3a 5bx) to the terms obtained. (This quantity the common factor out Finally, quantity will often be a multinomial factor.) The result is - (Sax x(3a An alternate + - method + 2y(3a - + - = + 56) (x 2y). (3a of grouping would give us the following results 56) 56) First, 3ax - 56x + Say - Then, (3ax + Say) 4- (- = IGby 56x - (3ax IGby) + = Say) 3a(x + + 56x (- 2y) - Finally, 3a(x + 2y) - Example 1-25. Factor x 2 - + 2y) = 56(x 2 ?/ 4- - 6y (x + 2y) (3o - 56(x - : IGby). + 2y). 56). 92 2 . Solution: By grouping the last three terms, we may rewrite the given expression as the difference of two; squares. Thus, 2 x 2 - (y* - 6y X 2 _ y2 ey - 9 + = = x - (y - 3*) 2 = [x - (y - 3z)} [x + (y = (x - y + 3*) (x + y 2 See. 1-21 Introductory Topics 30 4 Example 1-26. Factor 4x + y*. 3x 2 ?/ 4 + 4 rewrite the given expression adding and subtracting x2/ we may as the difference of two squares. Thus, - * 2 2/ 4 2 4 4 4x 4 + 3x 2 j/ 4 + 2/ 8 = 4x + 4* ?y + y* By Solution: , (2x = - + + + 2 [(2x (2x 2 2 - 2 4 ) !/ 4 T/ X 2 T/ 4 2 XT/ ) - 4 ?y 2 XT/ ) ] [(2x (2x + y*) + xy 2 + y* + XT/ ). 2 ] 2 2 EXERCISE 1-5 1 Factor each of the following: 2 2. 6x 3x + Qy 4*. 7. - 2a - 5. ab. - *. + 2cx ax + 9ac - 16. 13. x 2 - 25t/ 4 y 2 _ ai 92/ 4a'x 4 - 81. 16 ' 16x 4 19 22*. 25. 49x 28 . 2 2/ - 4x 2 31. x ty 37. Six" - 2 2/ 17. 49x 2 - - . + 3z) 3 . x 2 3,5. 2 18. . 24. 62 . 64' 67 70*. 72. 74. . - x. - Sly*. + 26) + 36 - . 3 -l/< + 2x - 8 (3c_-t_4d) .__39i _ 42. 45. x 2 T/ 2 15. 48. x 2 61. x 2 54. x 2 - - + 81. + 30. + 4t/ 18* 11* 2 . 4xj/ 3x - 10. + 30x3 +_ 225x. + 5a - /4. 2 x4 63. 2** + 5* - 3. 62. x + 16* + 64. 2x - 11* - 6. 2 66, 4* + 32* + 15. 65. 6x' - 37* + 6. 18x 2 + 16* - 25. 2 69. a* + 3x + 2ay + 6. 18 2 - 1 2x 68. 40x + 35*yz x + 0.36. 12x 2 - 10* + 15. 71. 8x' - 21. x' + 3x 2 - 7x - xz + 6a + xy - 3*. 73. 2x + 3y 3cy. ax + 26x - ex + Say + 66ty - 3. 2 75. x 3 + x - 3x xa6 - xyz - 2aby + 2y*z. oa. & 61 . 33. . 12 y 2 2/a x3. t/ 30. 3x' 16. 2166 8 14. - - 2ay + 3a. x' - 5x 2 + 6*y. 4x* - 9. 25a 2 - 166 2 2 a'x 4 - 9a 4 x 2 - a'x ax'. 27. 36* ' 58. ax 21. 4 4-1 9. 12. . 47. x - 8x 2 + 15. 50. 2x 2 - 3x + 1. 53 x 2 - x - 12. ^' W" *x' 4-4x3+4 T^ 2 - 40x. iQx 5x3 + -5$. _ 6. - 2x - 3. 4ax' 121. 38. 44. 8x 2 4x 6. 2 ty - + " r2 - 35. x ^a at/ + 3. 15. - "" + - . 64. ""41. 216* 12x3 + 2 - 29. 2x' 32. a 3(2?y 3y 14. 26. a 2 . .' 36t + 2 5t/ 23. 0.01 2 &'. z\ + 69 - 11. 20. 1 144a 16x 4 + 3 34. a 3 - 2 Sax 8. 2at 66c. 10. 3a6 + 4x?y. + 2a. I & + 9' - 10x + 9. j -r uu. 57. 1 ' . 60. a 2 2 2 2 . 2/ 1-22. GREATEST COMMON DIVISOR A common divisor of several polynomials (or integers) is a poly- For example, nomial (or integer) which divides each of them. x + y are x and 2 and 3 are common divisors of 12 and 18. Also, common A divisors of greatest common ^ (a* - y-) common and x* (x* + 2xy + y*). known as a highest more polynomials (or integers) divisor (G.C.D.), also factor (H.C.F.), of two or Sec. 1-22 31 Introductory Topics a polynomial (or integer) with the following two properties: a common divisor of the given polynomials (or integers) also it is a multiple of every other common divisor of the given is It is ; polynomials (or integers). follows that, for integers, a G.C.D. is a common divisor of greatest absolute value. It also follows that, for polynomials, a G.C.D. is a common divisor of highest degree. For example, a G.C.D. of 12 and 18 is 4-6 or -6, since 2, 1, It 3, 6 are the only common divisors, and 6 and -6 are those absolute value. This example indicates that a set of and maximum of non-zero integers will have two greatest common divisors, d and d, one being positive and the other negative. For polynomials with real coefficients, if d is a G.C.D., then a d is also a G.C.D. for every real number a not equal to 0. It follows that infinitely many greatest common divisors exist. However, for polynomials with integral coefficients, a G.C.D, should be a polynomial of the same type. Example 1-27, which follows, illustrates the fact that in this case a G.C.D. is uniquely determined except for sign. Since a G.C.D. of given polynomials divides all of them, it must contain as a factor each of the distinct prime factors occurring as common the given polynomials. Since, however, a G.C.D. must divide any common factor of the polynomials, it must contain each of the distinct common primes raised to the highest common power. Thus, a G.C.D. of x* (x y) 2 and # 5 (x y) (x y) must contain the prime factor x to the third power, that is, a factor of all + to the highest common power. + contain (x + y) no further since a G.C.D., Thus, x*(x + y) is common prime factors occur. When no common prime factors occur, to the first power. Example 1-27. Find a G.C.D. Solution: of 3x*y*(x* We shall begin by writing it rftust Similarly, - 4y a G.C.D. 2 ) is 1. - and 6xy*(x* -f 4y*). each of the expressions as the product of prime factors, as follows: 3^3(^2 _ 4xy 40.) = (3 ) ( y*) = (2) (3) (x) (y) (y) (x X ) (X ) (y) (y) (y) ( X + 2 y) ( X - its 2|0, and 6xy*(x* - 4xy + - 2y) (x - 2y). t The different 3, x, y, these and x common G.C.D. 2y), of which only prime factors are 2, 3, x, y, (x + 2y), and (x the product of now We form both are to common 2y polynomials. factors, using for each the maximum common power. Hence, a is 3xy*(x - 2y). 32 Sec. Introductory Topics 1-23. LEAST COMMON 1-23 MULTIPLE A common multiple of two or more polynomials (or integers) one is containing each of the given ones as a factor. Thus, 36 is a common multiple of 6 and 9, and x 2 - y 2 is a common multiple of x y and x + y. A common multiple (L.C.M.) of two or more polynomials a polynomial (or integer) with the following properties: It is a common multiple of the given polynomials (or integers) ; also it is a divisor of every other common multiple of least is (or integers) the given polynomials (or integers). It follows that, for integers, an L.C.M. common multiple of for polynomials, an L.C.M. is a common multiple of lowest degree. For example, an L.C.M. of 6 and -8 is 24 or -24, since the only and 24 and -24 are common multiples are 24, 48, 72, those of minimum absolute value. This example indicates that a least absolute value. It a is also follows that, , m have two least common multiples, and m, one being positive and the other negative. For polynomials with real coefficients, if m is an L.C.M., then a m is also an L.C.M. for every real number a not equal to 0. set of non-zero integers will many least common multiples exist. Howfor with ever, integral coefficients, an L.C.M. should be polynomials a polynomial of the same type. Example 1-28, which follows, illustrates the fact that in this case an L.C.M. is uniquely determined It follows that infinitely except for sign. Since an L.C.M. of given polynomials is a multiple of all of them, it must contain as a factor each of the distinct prime factors occurring as a factor of any one of the given polynomials. Since, however, an L.C.M. must be a multiple of any common multiple of the given polynomials, it must contain each of the various distinct primes to the highest power occurring anywhere. For example, an L.C.M. of x*(x + y) 2 and x*(x -y)(x + y) must contain the various distinct prime factors, which are x, x + y, and x - y. For x, the highest power occurring anywhere is the fifth power for x + y, the highest power is the second f or x - y, the highest power is the 2 5 first. An L.C.M. is therefore x (x + y) (x-y). ; ; Find an L.C.M. of 4z 2 Example 1-28. Solution: We shall first rewrite - 4s = 6z* - 6 = 4*2 - 2 4x, 6z - 6, and 9z 2 18* + 9 = 18x + 9. each of the expressions in factored form. Thus, (2) (2) (x) (x (2) (3) (x + - 1) 1) (x = 2*x(x - 1), 1), and 90 - - (3) (3) (x - 1) (x - 1) = 3*(* - I) 2. Sec. 1-24 The distinct these are 33 Introductory Topics prime factors are and 2, 2, 1, 1, 22 32 x + x 2, 3, #, 2, respectively. (x 1, and x The 1. greatest powers for An L.C.M. is therefore + 1) (x - I) 2 . EXERCISE 1-6 In each problem from + 4. z 7. 9z 3 !/ 2 y, to 12, find a 1 G.C.D. of the given expressions. 3. 4, 7, 39. 2. 9, 21, 33. 1. 4, 14, 36. - y*, x 6. 3a6, 12a 3 6, 6a 3 6 2 6 2 2 21x x x 8. 9. 4x*y*z, Sxifz*, Ux*y*z 2 3) 9, (x 15zV, 3, 10. x + 2, x 2 - 4, x 3 + 2z 2 - 4z - 8. 11. x 8 - s a - 42z, z 4 - 49z 2 z 2 - 36. 12. x 4 + 2z 3 - 3z 2 2z 6 - 5z 4 + 3z 3 z 3 + 3x* - x - 3. - x2 2 y 5. . x3 . 37. . ?/. , . , , , In each problem from 13 to 22, find an L.C.M. of the given expressions. 13. 6, 8, 12. 14. 8, 45, 54. 16. 4z 2 5x 4 20z. y02. 18. 2a 4. 20. 2x 15. xy, 6xz, 8yz. 2 17. 4x ^ 19. x - 4 5 , 2, 9^ + x 21. (s 22. 2x 4 1-24. 6 2/ 22 3 2, , x 6x 4 - 2 , - 49) (x 3 - 8), (x - 7) (x + 7) (x - 22/ 4 6x 2 + 12xy + 6s/ 2 9x 3 + 92/ 3 2) (x 2 - 3, a 2 - 9. 3x - 6, x 2 4- 2x - 8. - 4), (x - 3) (x - 2). a . , , , + 4, + 8, REDUCTION OF FRACTIONS From (1-24) under operations with fractions, it follows that a fraction a/6, in which 6^0, is not changed if both the numerator and the denominator are multiplied or divided by the same = 0, quantity, provided that the quantity is not zero. That is, if k a or ka a 6 For example, = - ~ 2 2 2 2 7 4 57* 2 > ' ~ and A = """ 3 6 6/2 ' The fundamental principles of (1-49) and (1-50) are applied in reducing a fraction to lowest terms and in changing two or more fractions with different denominators into equivalent fractions with a common denominator. The reduction of a fraction to lowest terms, that is, to a form in which all common non-zero factors are removed from both the numerator and the denominator, is accomplished as follows: First, factor both the numerator and the denominator into prime factors. Then, divide both the numerator and the denominator by all their common factors. 34 Sec. Introductory Topics 1-24 be noted that this reduction can also be accomplished It should by dividing the numerator and the denominator by their highest common factor. The following examples will illustrate the reduction of fractions to lowest terms. 30 Example 1-29. Reduce 4< -^ Solution: their to lowest terms. Factoring the numerator and the denominator and dividing both by common factors, we have _2'3-5 30 42 The common 2 3 5 ' 7 7 numerator and the denominator are 2 and factors of the 3. 6xv 2 Example 1-30, Reduce jr-- to lowest terms. Solution: Factoring into prime factors and dividing out 2 3 x y y 3y * common we have factors, = x 2 In this fraction the common x x y factors are 2, x, -- and y. _ Example 1-31. Reduce ^-r2 6z Solution: We - It is 3xy 2 - ^-r~ISy to lowest terms. have 2 6z 2 ~ - 3^ - IS?/ 2 * 3x(x 3(s - - 2y) (x 2y) (2x + 2y) + 3y) 2z 4- important to note that in a fraction of the type 3y a Ct ^ | y i where *C the numerator and the denominator have a common term, any attempt to simplify the fraction by cancelling out this common a y term can lead only to an absurdity. For, quite obviously, "^ *^ ^ i _i_ y does not equal either or - unless a = 1 or 0. For example, if 1 j] x x 2-1-3 we attempt one of these simplifications with the fraction ]" + we reach the obvious contradiction common error, factors, not it is - = - or - - > To avoid this important to remember that only common terms, may be cancelled. One should make common common certain that the 5453 = 6564 quantity is a factor of the entire numera- tor and of the entire denominator. 1-25. SIGNS ASSOCIATED WITH FRACTIONS from (1-49) that we can multiply both the numerator and the denominator of a fraction by 1 without changing the It follows 1-25 Sec. 35 Introductory Joplcs value of the fraction. However, if just one of these is multiplied 1, then the sign of the fraction must be changed in order to keep its value unaltered. Thus, in effect, a minus sign before a fraction can be moved to either the numerator or the denominator without altering the value of the fraction. We have, as previously by stated in (1-22), - ~~ n n 1-11 ~" "~ d , and x - d d - x y y (x i y y) - x We see, then, that any two of the three signs associated with a fraction, namely, the signs of the numerator, the denominator, and the fraction, can be changed without changing the value of the fraction. In general, the rules for changing signs in a fraction are as follows: an even number of factors in the numerathe denominator, or in both, does not change the sign of Changing the signs in tor or in the fraction. Changing the signs in an odd number of factors in the numerator or in the denominator, or in both, does change the sign of the fraction. 2x Example 1-32. Find the missing 7 . . 2x solution: = o ---2x = o - denominator x Solution: - 7 ( ) ) o to an equivalent fraction with fraction y. Since x - y = - - (y x), we make a __ __ the following changes in signs: a y-x~-(y-x)~~x-y - y a a a y x - (y - x) x -y EXERCISE 1-7 1. ( o a or 2x = o 2x - Change the Example 1-33. quantities in Find the missing quantity in each of the following equalities: 36 Sec. introductory Topics 2. Reduce each ** 102010 350470 1-25 of the following fractions to lowest terms. 8a 2 fe 3 . ' ' 6x*y* C* 20* 12a<6 V ' . e* ' 96z 3 2/ 7 * 2 +26 a 2' 4a 2 4 - 15 18x 2 "' a2 ?/) a; [(2x 6x 2 - 5x 4z 2 + lOa; , 2 ' 2 - (a ^- 6)g 4- o& - (a: + 2a)2] d) - 10?/ 2 4- 3sy 2 (a: x*y* ~ 3s 8 . K ' -y x -25 - 3)2 _ - (* - 2 a: - 7x - x*y* , ' + 8a6 - f ' 24*V + 40zV - - 6 1-26. 2 ADDITION +y 13z 49 s - + 14 - z) 81 * x* -f 729 462 (x ' + y) q' (y* 6) (a? - y) (y - x2) (z a6 - 62 - t y) 9a 2 S* u* 3 - 6a 2 8y* xy + ' 24 ' -f - ' 3z ' 9s 2 2 x2 + x + 2y AND SUBTRACTION OF FRACTIONS The sum of two fractions with the same denominator is the whose numerator is the sum of the numerators and whose denominator is the given common denominator. That is, in (1-25), fraction -+=^- (1-25) For example, To add x + 2y x.2y_ + 3 3 3 fractions with different denominators, first change the fractions to equivalent fractions having a common denominator, and then write the sum of the new numerators over this common denominator. Ordinarily, the common denominator which is chosen is a least common multiple of the given denominators, since this leaves the fewest possible common factors in the numerator and denominator However, the same result is obtained, after compilation of all common factors, no matter what common denominator is used. The method of finding an L.C.M. was develof the resultant fractions. oped in Section 1-23. The difference of two fractions, Section 1-2. Thus, by (1-10), when ? > has been defined in a neither 6 nor d o is zero, Sec. 1-26 37 Introductory Topics Applying (1-22), we have a n (1-52) v ' r x T- --c = b d ; The following procedure a c T H--rb d ' suggested for adding (or subtracting) is fractions. common multiple of the given denominators. fraction each to an equivalent fraction having the Change L.C.M. as the denominator in the following way: For each fraction, note which factors are in the L.C.M. but not in the denominator of the given fraction. These factors may be found by dividing the L.C.M. by the denominator of the given fraction. Then multiply the numerator and the denominator of the given 1. Find a least 2. fraction by these factors. Write the sum (or difference) of the numerators of the new fractions found in step 2 over the L.C.M., and reduce the 3. resulting fraction to lowest terms. The following examples will indicate a procedure which should be followed until some skill in working problems has been attained. 357 Example 1-35. Express j - ^ Solution: Step 1. the denominators. Step 2. -f The methods as a single fraction reduced to lowest terms. Q in Section 1-23 give us 36 as the Divide the L.C.M. successively by the denominators 36/4 Change the given = 9, = 36/6 6, = and 36/9 L.C.M. of and 9 to get 4, 6, 4. fractions to fractions having 36 as the denominator in the following way: 3j_9_27 ~ 4 Step 3. Combine the new ' 5^6_30 ~~ 36 9 6 9 __ 6 Note that adding 9 (or subtracting) the common denominator 1. 28 ' 36 30 Since x 2 ~p - u 4 28 25 ~~ " 36 numerators and the denominators of the it is + = Zoo not true that ~ --- 5x 2 -7-75 + 36 leads to absurdities. Thus, 3x Example 1-36. Express X Solution: Step - 36 given fractions leads to absurdities. Thus, dropping the 4 - fractions to obtain 27 3_57_27_3028 + 36 ~ + ~ 36 4 _ ~ 4 7 ' 36 6 -5 A = " (x X + 2) 1 X (x 2 - - - A 4 = 5 it is Also, not true thit as a single simplified fraction. 2) and - - 2 ^ L 2 = X X 5 > an 38 L.C.M. of the denominators as follows: x2 is 3x 2 O i 2 - When the L.C.M. the results are 2. Step (x 4), 5l_ZJ* T+T - I - 5x - x x2 2 ~ *2 4 be written 2 4 -x o ~ X+2) x-2 may expression divided by the denominators (x is -A 2 * The given 4. 1-26 Sec. Introductory Topics *2 "" 4 x - 4 | + ~ 2), (x 2), and 1 lf Multiplying the numerators and the denominators of the given fractions by these we obtain 2 3x (a? - 2) ~ 3a 2 - 6a z 4 + (x 2) (x 2) factors, * ' 2 _ (a (s + 2) + Or 2) 5x a; __ + o;--2 z-h2 - 2 __ 2j-_4 " 2 - x _ ~~ - 5a x2 4 ' 4 2 - 4 Therefore, the desired result is obtained in the following way: 2 - (5x - 2) 5x 2 6g) (2x 4- 4) (3x 3. Step 3x 2 2) a: -2 = -4 2 + x2 2 __ 3o; x2 9x - 4- 6 ~" -4 - 3(x 4 - 1) (x a; 2 - _ 2) 3(x - 1) * t z+2 4 EXERCISE 1-8 Perform each of the indicated operations and express the answer in lowest terms. L 2 1 7 . 5' 1 ~9 ~9* 2 "" 7 5 9 ^_ ~3 + 8* Q 3 R ,2 ^7"" &H"2T' i xZa.^_A ^'S^G Q ^13 3+ 20""Io' 6 12 17 5 5 19 '2~T + T t o 10 Id. - -1 r 1 a; -2,3 + ^ -- x - 2 7^ 1 (x - I) x x /r /* 18. 4. x + 1^ + i x 1 A - 2 +2 x + 3 -f 60; + +2 7^ + ^ ' 12 . a: t 2 x ~r i/ z +2 2 - 4 JK - 8 1 .j Oi 2 zx > _ ^ ^2 + X Q T^ Q a: a a; 6 XV 14. 2 -2 x a; 1yl ~r- - ~ Q 6 < x* + x ^ Q/* *x . 19. x 2 ?'r -y 2 x* -2xy + , y 2 1-27 Sec. 3 20. 4 - 1 _ x2 T x - 2 o^x 2 x - T * _j_ x2 y - 2 1-27. x2 - 2x y 2 MULTIPLICATION - x2 AND +y x ^ Til I/ y j_ 25 ' . y So; 26 * i+?+A. x xy y Til ^ __ y .l*,-<> a -b 23. II _ 24 * 21 . - x -^-=-+4 + 1 x 22. 39 Introductory Topics - x x2 y - . y 2 4xi/ +y 2xy - x3 2 - x 2y f xy 2 -f 3 2/ DIVISION OF FRACTIONS The product of two fractions is the fraction whose numerator is the product of the numerators and whose denominator is the product of the denominators of the given fractions. That is, by (1-29), if neither 6 nor d is 0, a c ac n 9cn (JL4\J) T 1 T~l b d bd . . To illustrate, 26 2-6 4 xx + . 7~~3-7~7 3 + x 1 x-2~ 2 + x2 3 - - 2x 3 _________________ 2 x - 4 A special case of (1-29) is worth noting. To multiply a fraction by a number or expression, we multiply the numerator by that number or expression. Thus, b __ a ~ c For example, 1 b __ a ~~ b c c * 1 _ ~" ab c 55 (-i).Kx = (^lI^. x and The quotient of two 5 if in no factor in a = J/5 b/ d J^ b = *. be c illustrate, /3 __ " 7/ Thus, defined is zero, (1-30) To been has T/^J bf a fractions, Section 1-2 and evaluated in Section 1-7. denominator i\ / 2.| = |, 2 2 5 ' 7 _ ~ 5*2 _ ~ 3 In practice, 7 we 3 10 3a? ' 2 21 / 2/ divide out all 5 /6a 3 7 " 3^ 3 __ e/ 2 ' 6^ 2/ factors j/^_ common 5 _ ~ 3x3 2/ to the * - y 7 Go; _ "" 5 j/^_ ' 2x 3 numerator and denominator before proceeding with the actual multiplication. numerator and the denominator of each given fraction should be factored. If necessary, the T2 __ OT , Example 1-37. Multiply &x*a + &r , O/>2 . -p 9 o Q-y by .u _ r-^V y u and simplify the result 40 Sec. Introductory Topics The work may be Solution: - a2 2x 2x* indicated as follows: - 2z + 5x + 3 x - 3x ' 2 - 2 _ ~ 9 - x(x 9 (x + _ ~ - x2 i * oo ^Example 1-38. Divide i - x(x By Solution: x2 x2 - 5# -f- 4 +4 3x - z3 * 4 - + 2^-1 4x 2 4 a- __ x2 = a; + 3) -2x 3) ' 3) 4x 2 + 3) (x a; 2 -j- - 4s +3 4 ^-^ - 5x + 4) (2x - 1) - 4) (x 3 - 4x 2 + x - 4) (x - 4) (2s - 1) (x -l)(x (x + 1) (x -4) (a -4) (z +l) - 1) (2s - 1) (s - 4) (x + 1) (x + 1) (x a (x - 2 __ + ~~ + - 3 3) (2x 3) (x 2) by (1-30), 5z rr , ^j + (2x 1) - 2) (x (x -f 1) (a n 1-27 3z 2 COMPLEX FRACTIONS 1-28. The fractions we have been discussing so far may be called simple fractions, to distinguish them from the fractions which we now A discuss. fraction which contains other fractions in the numerator or denominator is called a complex fraction. Since the simpli- in the complex fraction is essentially a problem in division, reduce the numerator and the denominator of the complex fraction to simple fractions and then proceed as in division. fication of a we first - 1+ ; r Example 1-39. Simplify- x - First Solution: T fraction I x y The numerator of the given - - complex fraction reduces to the simple 77 I! > and the denominator reduces x +y y -f to - \- 7? Hence, we have xy , + y) xy (y + x) (x x x Alternate Solution: Frequently it may be more convenient to multiply both the numerator and the denominator of the complex fraction by an L.C.M. of the denominators of all simple fractions occurring in the given complex fraction. In this and an L.C.M. of their denomithe fractions are - - and - example, simple nators is xy. > > > by multiplying the numerator and the denominator by xy, we get Therefore, fraction l \ + x)' xy __ xy +y* _ (x + y)y _ of the complex Sec. 1-28 Introductory Topics 41 EXERCISE 1-9 In each of the problems from 1 to 20, perform the indicated operations and express the answer as a simple fraction in lowest terms. 42 Sec. Introductory Topics 8 ^4^- 33. -f - x 1 ~ *v # s +y __ 1 35. 4 g - x2 + 2s 1-2* x _ - - ""L1 ~2. . 4 +g g 1_ 2 34. ____ ; _ x 1-28 2# y x -f 1 x -2y - _ I", 2y __ 07 x -lx + x* l -y* a? Simplify each of the following expressions: LINEAR EQUATIONS 1-29. Introduction. In Section 1-13, an equation was defined as a state- ment of equality between two algebraic expressions. In this section we shall discuss equations in one unknown of the simplest type, called linear equations, typified by the form (1-53) ax , where a and & are specified = &, numbers and a ing to equations in general are needed Solution or Root of an Equation. = 0. Some ideas pertain- first. We shall use the term unknown to designate a literal quantity that appears in an equation and is not regarded as specified at the outset. system of values of the A unknowns which, when substituted for them, makes the equation a true assertion is called also called a root + 3(2) a solution of the equation. A solution is only one unknown is involved. Thus, the 1 define a solution of the equation x = 2 and y = = 2y 4, since the equation + 2(-l) = 4. values 3# when is satisfied for these values ; that is, Sec. 1-30 43 Introductory Topics Two Equivalent Equations. exactly the same solutions. equations are equivalent if they have For example, 2x + 11 = 7x - 4 and 14# - 8 are equivalent, since, as will be seen, each has exactly one root, namely, x = 3. However, 2x + 11 = Ix - 4 and 2# 2 + llz = To; 2 - 4# are not equivalent, because the latter equation 4# + 22 = has, in addition to the root x = 3, the root x = 0. Operations on Equations. Each of the following operations on an equation yields an equivalent equation Adding the same expression to (or subtracting the same expression from) both sides. : Multiplying (or both sides by the same non-zero dividing) number. For any solution of F=G makes also F+H=G+H true, and therefore a solution of this latter equation. Conversely, any = G + is one of F = G, because solution of F + is H H F = F+ H-H = G + H-H = G. Likewise, any solution of Fj= G is one of is a non-zero number, and conversely. aF = aG, provided that a Transposition of Terms. Transposing a term of an equation conmoving the term from one side of the equation to the other and changing its sign. This operation is equivalent to adding the same quantity to both sides (or subtracting the same quantity from both sides) For example, consider the equation sists in . 2x+ 11 = 7x - 4. we transpose 2x from the left side to the -4 from the right to the left, we obtain 11 + 4 = Ix - 2x. If In effect, 1-30. we right, and transpose subtracted 2x from each side and added 4 to each side. LINEAR EQUATIONS IN ONE UNKNOWN An equation in the form shown in (1-53) is called a linear equaunknown. We shall show that such a linear equation has one and only one root, namely, tion in one = a each side of the equation ax = (1-54) If x x b is divided b =a > by a, the result is 44 Introductory Topics which Sec. 1-30 equivalent to the given equation. Therefore, (1-53) and only one solution, namely, b/a. It should be noted that if we allow a to be in (1-53), there are two possibilities. Either no solution exists, when 6^0, since for no number x is it true that (fx = Q x = Q = b^Q; or else every number # is a solution, when 6 = 0, since a*# = 0'# = = & for is clearly has one 9 all values of x. An equation which is not apparently linear may frequently be solved by the theory of linear equations, by replacing the given equation by a linear equation to which it is equivalent. The following steps serve as a guide to the method to be used when there is only one unknown in the equation. 1. Clear the equation of any fractions with numerical denominators by multiplying both sides of the equation by a least common multiple of the denominators of those fractions. Transpose all terms containing the unknown to one side and other terms to the other side. We may collect the terms containing the unknown on either side. 3. Combine like terms. If the equation now assumes the form 2. all in (1-53), it is a linear equation and can be solved sides by the coefficient of the unknown. by dividing both Check the result obtained in step 3 by substituting in the original equation. While it is desirable to include step 4 to show that the number x found in step 3 is actually a solution of the 4. equation, step 4 is not a necessary part of the solution process, since the operations performed in the preceding steps always yield equivalent equations. The purpose of step 4 is to help to make certain that there has been no error. 2, we generally transpose terms containing the unknown whichever side makes solving the equation easier. The following explanations should be noted carefully. Students at times "transpose" coefficients of the unknown. Thus, 2x = 6 takes the erroneous form x = 6 2 by "transposing" 2 to In step to The correct procedure is to remove the coefficient 2 by division, since it is a multiplier of x. Therefore, we should divide both sides of the equation 2x = 6 by 2 and have 2x =6 2 2' or the right side. a; = 3. Errors of this type may be avoided if the student applies the rules of algebra properly and checks his solutions carefully. 1-30 Sec. 45 Introductory Topics Multiplying or dividing both sides of an equation by a polynomial involving the unknown will not necessarily yield an equivalent equation. When the operation is multiplication, the new equation thus obtained may have roots in addition to the roots of the original equation. These extra roots are called extraneous roots. then say that the equation is redundant with respect to the We When both sides of an equation are divided by a polynomial involving the unknown, the new equation may lack some of the original roots. It is then said to be defective with original equation. respect to the original equation. If both sides of an equation are multiplied by a polynomial involving the unknown, the check in step 4 of the recommended procedure is a necessary step in the solution process. An extra- neous root can thus be identified. Dividing both sides of an equaby a polynomial involving the unknown is not a permissible procedure, since roots that are lost cannot be regained. It should be noted that a non-linear equation may sometimes be treated so as to obtain a linear equation which is possibly redundant. For example, the fractions in the equations of Problems 39 to 48 of Exercise 1-10 may be eliminated by multiplying both sides of each equation by a least common multiple of the denomina- tion tors of the fractions in that equation. Since this multiplier contains the unknown, the new equation may have solutions which are not solutions of the given equation. Hence, the solutions must be checked to see whether or not they actually are solutions of the given equation. ST ~Example 1-40. Solve the equation o We Solution: which is x A 7 '= 4 o clear the equation of fractions by multiplying both Collecting the terms containing x on the left side and right, we have 10* Combining like terms, 3* = 60 I3x = 39. + - by 13, by 15, 21. = 3. check, substitute 3 for x in the original equation. 4 o Therefore, 3 is other terms on the we have * | ?l2 - all we obtain Finally, dividing both sides To sides an L.C.M. of the denominators of the fractions, and obtain Wx = 60 - 3x - 21. the root. - iZ o or 2=4-2, The or result is 2=2. 46 Sec. Introductory Topics Example - 1-41. Solve the equation 6x - 3y = 1 5y -f- 2x -f 11 for 1-30 y in terms of x. In this case, regard x as specified. Solution: left. The 6x Combining unknown y on Collect the terms in the other terms on the like terms, - - 2x divide both sides by the 1 - - 12 + by 3y. = %. - 3 check, substitute for in the original equation, ?/ ^ This equation reduces to I2x - to obtain -- ' # unknown _ -IT 2 g To = 11 coefficient of the $x Then all we have 4z Now we the right side, and collect } result is and obtain At 3(x - + - 3) - 2 2 = 5z = 5 (a? - 3) + 4z + 22, which becomes I2x - 3z 9 - 15 + 4^ + 22, or + 9x Since this result is an identity, 7 y = = 9x + 7. is ^ the solution of the original equation, regardless of the value of x. Example 1-42. Find two consecutive integers such that four times the equal to six times the second diminished by first is 20. Solution: Let x be the smaller integer, and x + 1 the next larger integer. Then, from the statement of the problem, we have 4z = 6(x 4- 1) - 20. From this equation, we obtain 4z = 6z -f 6 - 20. Then, 14 Hence x = 7 and x -f 1 = = 2x, or s=7. 8 are the two consecutive integers. The student should carefully check these values by substitution in the original statement of the problem. Example 1-43. The speed of an airplane in still air is 400 miles per hour. If it requires 20 minutes longer to fly from A to B against a wind of 50 miles per hour than it does to fly from B to A with the wind, what is the distance from A to ? Sec. 1-30 47 Introductory Topics Solution: Let x be the distance from A to B. From the data, the speed of the 50 = 350, and the speed of the plane against the wind, in miles per hour, is 400 = 450. in with the miles is 400 50 4wind, plane per hour, Since distance = ;rr ;rrrr OOU = = t. required to fly from A to B, time, in hours, required to fly from B to A. time, or d rate = time, in hours, rt, we have - Hence,, and jrr = Therefore, from the statement of the problem, x x Solving this equation, we have 9z - 7x = 1050, A B to is _ ~450 "~3* 350 So the distance from follows that it 1 x or = 525. 525 miles. EXERCISE 1-10 Solve for the 1. 4. 7. 10. - - 4z unknown = 2 3 - 2. = 2 - 20. 4 - 3z = 6(1 + 2). 60 + 7 = 50 + 6. 30 + problem from in each 2x. 7 5. 3x - - = 6 - 9x 7 x = + 6 1 to 15. 12. - = - 4z 4w + = 3w - i L x - 8 = 2x + 3. - | = | - jx \x 3x 3. Cr - 2 K 6. J i = 3z 4- 4. 9. 11. lOx - 3 = 9z + 4. 12. 4 13. 5z - 7 - 8z = 4x - 17 - fcc. = 2y + 7 + 3y. 15. 6(5 + 4*) 8. 4z - 6 +4 + 6y Solve for y in terms of x in each problem from 16 to 24. 16. 2x - y 19. 40 - 2x 22. 3(x - 4) Solve for 25. 5z - = 3 3. + 2 = 0. + 40 = - all 3. Q> + 20 = + 1 = 1. 2x + 4(0 - 17. 23z 20. - 23. 18. 4. 21. 6z 3(x - + 10 = - 20 = 3. - a: 4 ^ ju 14. Hty 5. 1 40 4) = 0. 0. * 3) = 5. 24. 3x + 7(0 = ij . | values of x which satisfy the equation in each problem from 25 to 48. = 4(s - 2). 26. 8fo - 2^ - 9( - 4) = 13. 27. !^-J> = n. 48 Sec. Introductory Topics 1-30 - 2(x 49. Divide 98 into 2 + 3rc two parts such that one 50. Find three consecutive integers + 9) of 52. If 8 times a certain number is - whose sum is 3 2 ~~ them exceeds the other by two consecutive integers whose squares 51. Find 81 rr 18. 84. differ by 13. 9 more than 5 times the same number, what is the number? 53. A is 54. rectangular plot of ground is four times as long as feet, what is its area? it is wide. If its perimeter 4,800 How many pounds of coffee at 90 cents per 98 cents per pound will it take to pound and how many pounds at make 100 pounds of a mixture costing 96 cents per pound? 55. At a was 25 cents for a child and 75 cents for an adult. from 500 admissions, how many children and how many adults were admitted? If college play, admission $210 was taken 56. A man 57. What has $365 in 41 bills of $5 and $10 denominations. each denomination does he have? and 58. in are the angles of a triangle, if one angle is How many bills of three times the second angle six times the third angle? One man, same job together? X y can do a certain job in 7 days, and another man, Y, can do the How long would it take them to do the job working in 15 days. 2 2-1. The Function Concept RECTANGULAR COORDINATE SYSTEMS IN A PLANE In Section 1-5 we saw how we can associate a real number with every point on a number scale. The real number attached to a given point is called the coordinate of the point. This representation suggests the assumption that to any real number there corresponds precisely one point on the scale, and to any point of the scale there corresponds precisely one real number. This one-to-one correspondence between the set of real numbers and points on the number scale is known as a one-dimensional coordinate system. We shall now extend the concept of a one-dimensional coordinate system to a system of coordinates in a plane in which two number scales are perpendicular to each other. The two perpendicular lines, which we shall call coordinate axes, divide the plane into four parts, or quadrants^ numbered as shown in Fig. 2-1. The horizontal and vertical lines are designated as the x-axis and the y-axis, respectively, and their point of intersection is called the origin and On is each of these axes, number scale D labeled O. by we selecting construct a an arbitrary unit of length and the origin as the zero point. As in Section 1-5, a coordinate on the x-axis will be considered positive if it is to the right of 0, that is, to the right of the #-axis, and will be negative if it is to the left. A coordinate on the IV p IGt 2-1. be considered positive if it is above the #-axis, and negabelow. Just as the real-number scale of Fig. 1-1 gave us a system of one-dimensional coordinates by which we could set up a one-to-one correspondence between points on a line and real numbers, so the 2/-axis will tive if it is 49 50 Sec. 2-1 The Function Concept system of coordinates with respect to two mutually perpendicular axes sets up a one-to-one correspondence between points in a plane and ordered number pairs. We use the designation "ordered" pairs for the following reason. To designate any point, we shall agree to give its directed distance frdm the i/-axis first, and call it the abscissa or x-coordinate, and then the directed distance from the and call it the ordinate or y-coordinate. The abscissa and ordinate of a point constitute its rectangular coordinates. They are written in parentheses as an ordered number pair, as in the nota#-axis tion (x, y), the abscissa always being written first. By this scheme we assign to each point of the plane a definite ordered pair (x, y) of real numbers and, conversely, to each ordered pair (x, y) of real numbers there PO.2) T \ \ FlG 2-2 \ assigned a definite point of the plane. Thus, the abscissa of the point P in Fig. 2-2 is 3, and its ordinate is 2, and we say that the coordinates of the point P are (3, 2). Similarly, the coordinates of Q are (2, 0), the coordinates of R are (0, +-x 3), and those of the origin are (0, 0). Marking in the plane the position of a point designated by its coordinates is called plotting the point. The coordinate system we have constf ucted is a particular case of cartesian is coordinates, so called in honor of Rene Descartes (1596-1650), who first introduced a coordinate system in 1637. It is called a rectangular system, since the axes intersect in a right angle. (Actually the axes may intersect at any angle, but it is usually simpler to take them perpendicular to each other. When the two axes are not perpendicular, the coordinate system is called an oblique coordinate system. Oblique systems will not be used in this book.) 2-2. It DISTANCE BETWEEN was seen TWO POINTS in Section 1-10 that |a 6| equals the distance between two points on the number scale represented by the real numbers a and 6. It follows that \x 2 - a?i| represents the distance between the points A (xi, 0) and B(x 2 6) on the #-axis of Fig. 2-3. Let us now consider the two points PI and P2 with coordinates (#i> #1) and (# 2 yi). The points have the same ^/-coordinate, which means that they lie on the same horizontal line. Hence, the distance between the points PI(XI, yi) and P2 (tf2> 2/i) is the same as , , Sec. 2-2 The Function Concept 51 *K *-JT FIG. 2-3. FIG. 2-4. between A(#i, 0) and Z?(# 2 ,0). This distance is dis#i|- Similarly, we can show that [2/2 2/i| represents the tance between two points (#2 2/1) and (# 2 2/2) on a vertical line. We have thus arrived at the following two important properties. If two points PI (#1,1/1) and P 2 (# 2 2/i) have the same ^-coordinate, then the distance between them, or |PiP 2 is given by the distance | #2 ~ , , , |, = |PiP2 (2-1) - |*2 | xi\. two points Qi (#1,2/1) and Q 2 (#1,2/2) have the same #-coordinate then the distance between them is given by If (2-2) - IQiQal ~ (2/2 l/i|. The concept of the distance between any two points in a plane is so important that we shall now develop a formula for it. Let us denote by d the distance between the points PI (#1,2/1) and P 2 (#2, 2/2). That is, d is the length of the line segment PiP2 2-4. Let P 3 be the point (# 2 yi) as shown. Since the angle at P3 is a right angle, we have, by the in Fig. , Pytha- gorean theorem, |PlP 2 2 Therefore, |PiP 2 = = 2 | - | |PlP3 2 | + - #i| 2 + - #i) 2 + (#2 |*2 |P 3 P2| It/2 (2/2 ~ 2 . 2 yi\ 2 2/i) . the distance d between any two points PI (#1,2/1) P2(#2, 2/2) in the plane is given by That is, = V(#2 - d (2-3) This formula is known 2 #i) + (2/2 - and 2 2/i) - as the distance formula. Example 2-1. Find the distance between the points ( 3 2) and (5, 2). Solution: It makes no difference which point is labeled Pi. Let us label the first one Pi and the second one P 2 Since the two points have the same ^-coordinate, . (2-1) applies and |PiP 3 = 1 |5 - (- 3)| = |5 + 3| = |8| = 8. 52 The Function Concept Sec. Example 2-2. Find the distance between the points Solution: Again let us label the first point PI points have the same ^-coordinate, (2-2) applies Solution: and (3, 7). P2 and the second one and |PiP2| = |7 Example 2-3. Find the distance between the points - Let us designate the points as Pi(2, the distance |PiP2| (3, 1) - (2, 1) Since the . = 1| and and /Vo, 1) 2-2 6. (5, 3). By 3). (2-3), is -2)* +(3 -(-I)) 2 = V9 + We 16 = \/25 = 5. now consider a special application of the distance to a point formula (2-3). The line segment OP from the origin P(x,y) is called the radius vector to P. By (2-3), the distance between O(0, 0) and any point P(x,y), or the length of the radius shall vector to P, is d or d (2-4) = V(* ~ = V* 2 + O) y 2 + (y - 2 O) 2 - Thus, for the point P(3, 2), the radius vector has the length V3 2 + 2 2 = VIST EXERCISE 2-1 1. Plot the following points: b. a. (3,5). d. (-3, -6). e. g. (0,2). h. In each of the following between them: a. (2, 3) and (7, 3). c. (1, 4) and (1, 0). e. (2, 1) and (5, 6). g. (3/2) and (5, 7). 2. 3. (-4,7). (1, -3). (-5,0). c. f. i. cases, plot the pair of points b. (5, - d. ( f. (0, h. (- 3, 4, and (5, -2). (-8, -6) (3,0). find the distance and ( - 1, - 2). - 3, 4). 2) and ( and 8) 3). (5, - 6) and (- 8, - 6). 2) - Find the length of the radius vector to each of the following points : - a. (4, 3). b. (12, 5). c. (1, d. (5, e. (7,0). f. (-3, -2). h. (-1,2). i. (a, 6). k. (- \/, - L (m, n). gj- (0, (1, 2-3. -12). 4)._ V3). 1). 1). FUNCTIONS So far we have been concerned with single numbers and pairs of numbers. Now we shall consider mathematical relations, known Sec. 2*3 The Function Concept 53 as functions, between two sets of numbers. To distinguish between the two sets, we shall call one of these sets the domain of definition of the function, and the other the range set Y of the function. We begin by defining a variable to be a symbol which may take any value in a given set of numbers. If # is a symbol which is used to denote any number of the domain and y is a symbol which denotes any number of the set Y, then x is called an independent variable and y is called a dependent variable. If a set contains only a single number, the symbol used to represent that number is called a constant. To set forth a function, the domain should be explicitly specified ; that is, it is necessary to determine definitely just what elements or numbers the domain contains. The same is true of the range set Then, as soon as a definite rule of correspondence is given which assigns to each number x of the domain one or more numbers y of the range set, the function is specified completely. We thus have a set of ordered pairs of numbers (x, y), where x is any number of and y is a number of Y. The set of ordered pairs of numbers (x,y) is called the function. The rule of correspondence which determines the collection of pairs (x,y) is often expressed by a formula involving algebraic or other processes. In .such cases we usually find it convenient to refer to the formula as though it were the function. For example, we often 2 3x + 5 when actually we mean the speak of the function y = x X X X 3x + 5. If just determined by y = x 2 one number of Y is paired with each value of x, the function is said to be single-valued. If more than one number of Y is paired with some value of x, the function is said to be multiple-valued. We frequently find it possible to deal with multiple-valued functions by set of ordered pairs (x,y) separating them into distinct single-valued functions. The range of values of the function consists of those numbers y in the range set Y which actually correspond to some number x of the domain. When the range of a function has been determined, it is always possible to replace the original range set, which may include numbers in addition to those of the range, by the range itself. We shall now consider several examples of functions. The constant function y c associates the same number with every number x of the domain X. Hence, the range Y of the function consists of just one number c. Since y has the same value for all pairs (x, y), the function is evidently singleIllustration 1. c valued. The identity function y = x associates with every number x the number itself. In other words, the numbers Illustration 2. real 54 The Function Concept Sec. 2-3 are y = 1, 2, 3, respectively. The corresponding to x = 1, 2, 3, domain is the set of all real numbers, and the range Y is also this entire set. The function y = x is single-valued, because it associates just one number of Y with each value of x. X Illustration 3. Let us consider the linear function defined by the is the set of all real numbers, 2. The domain equation y = 3x and the range Y X is mines a unique y sponding to x = 1, also this entire set. In this case a given x deterwhich is equal to 3x 2. For example, corre2, 3, we have y = 1, 4, 7. The function is single- valued. Illustration 4. Let the rule of correspondence be given by the 2 be the set of all real numbers, and let Y Also, let equation y = x be the set of all non-negative real numbers. Then, to the number x 2 there corresponds the number y = ( 2) 2 4; to the number x = 3 there corresponds the number y = 9 and so on. Hence, y = x 2 associates just one number of Y with a given value of x, and defines y as a single-valued function of x. Although this is not obvious, the X . ; range of the function is the given range set Y. Illustration 5. In this case let the rule of correspondence be given by the equation y 2 = x. Here X is the set of all non-negative real numbers, and Y is the set of all real numbers. Then to the number x - 2 there correspond the two numbers y \/2 and y = \/2; to 3 and the number x = 9 there correspond the numbers y = V9 2 = = = defines as a two-valued x on. ~3 and so y Thus, y y V9 is there a single value of y, namely, function of x. Only for x = y = 0. ; The function y = \A&2 #2 with domain X cona ^ x ^ a, is a singlesisting of all real numbers x such that valued function with range set Y consisting of the numbers ^ y ^ a. Here the range is Y, as may be proved. x 2 is negative. x 2 has no meaning when a 2 Note that \fa2 2 x 2 < 0, Hence, those values of x must be excluded for which a and we must restrict the value of x to the interval a ^ x ^ a. (We shall consider the meaning of the square root of a negative number in Chapter 11.) The function y = -\/a2 x 2 is single- valued Illustration 6. > because \/a2 x 2 represents the non-negative square root of a 2 a2 ~~ # 2 -) x 2 is denoted by (The other square root of a2 x2 . V Illustration 7. The function y = l/x is defined for all real num0. For x = 0, l/x is not defined, since division bers different from by zero is not permissible. In other words, although there are Sec. 2-4 values of y for values of x neat* when x 55 The Function Concept 0, there is no possible value for y actually equals 0. Usually the functions which we are about to consider are defined all values of x 9 with the following two exceptions Values of x must be excluded which involve even roots of for : negative numbers, since these are not defined as real numbers. Values of x must be excluded for which a denominator is zero, since division by zero is not a permissible operation. use to be made of a function will restrict the occasion, the values of x for which it is to be regarded as defined. On 2-4. FUNCTIONAL NOTATION Since functions are mathematical entities, they may be given such as /, #, <>. To designate the number, or numbers, y corresponding to a given number x according to the rule letter notations, specified /, we us6 the notation f(x). the function / be defined by the equation by a given function illustration, let y Then /(O) = =* x2 - 2x + As an 3. = 6, and so on. Frequently, the symbol f(x) used to designate the function rather than the functional values. The context will make the meaning cleai*. It should be remembered that the notation y = f(x) does not mean that y is a number / multiplied by another number x. Instead 3, /(-I) is an abbreviation for "/ of x." The set X does not have to be as simple as it is trations. If in the preceding illus- X should consist of a set of ordered pairs of numbers, the rule of correspondence would then determine a value, or values, of y for each ordered pair of X. We would then have a function of two independent variables. For example, the area of a triangle is given by the relationship A = /(a,6)=|a&. Here X is the set of ordered pairs, (a, 6), of positive real numbers, where a is the length of the altitude of the triangle and 6 is the length of its base and Y is the set of numbers A, each of which represents an area corresponding to a given pair (a, 6). Similarly, ; a function of three variables, f(x,y,z), of ordered triples (x,y,z). Thus, is X 2 2 /(2, 3) = 2 + 3 = 13. = 3-2 + 2(5) =11. Also, if f(x, y> z) defined in terms of a set if ~ x = x 2 + y* then 2*, then / (3, 2, 5) f(x, y) -y+ 9 56 T/ra Function Sec. Concept 2-4 EXERCISE 2-2 By using the phrase "a function of" in each problem from 1 to 8, express each given quantity, which is regarded as a dependent variable, as a function of one or more independent variables. Where possible write the relationship both in words and in symbols. 7. The The The The 8. A 9. Given f(x) = 2x - 10. Given = 3x + 1. 3. 5. 11. = area of a 2. circle. area of a trapezoid. volume of a cylinder. annual premium for a 4. 6. life The area of a triangle. The volume of a sphere. The retail price of food in a grocery store. insurance policy. person's height. g(x) 3, find /(O), /(- 1), 5, find 0(1), g(- 3), /(3), /(1/2), /(V), 3/(l), /(3)/4, (0(2)), 0(4), 20(4), g(z Using f(x) and 0(x) as defined in problems 9 and 1). 2) /( 10, find - j^- In problems 12 to 26, let /(a;, j/) = 2xy + 3.t - 2y, and a 2 + 6 2 + c 2 Evaluate or simplify each given expression. let /(6)0(3), > 0(a, 6, c) . 12. /(I, 2). 13. /(O, 0). 14. 0(0, 0, 0). 15. 0(1, 2, 3). 16. /(i, 1/x). 17. p(p, g, r). 19- 20. 18. M 25. g(- if) + ffe a, 6, c) V, -). - g(a, - 26. 0(a, 6, c). 6, - c). A and circumference of C of a circle as functions of the radius r. r, express A as a function of C, and also express C as a func- 27. Express the area By eliminating A. tion of 28. Express the volume of a sphere as a function of its surface area. 29. Express the surface arfea of a sphere as a function of its volume. 30. Suppose that is U is a function of V and that V is a function of W. Show that U a function of W. Determine the maximal domain of values of x for which y is defined as a function from 31 to 61. Assume that x and y are real numbers. of x in each problem 31. y = x. 32. y + = z2 = #(2z 34. y == 3z 37. y 35. y 1. 38. y . 40. y 43. y=x(x-l)(x+l). -f- 1). 41. y 44. = 3z. = 2x - 3. = x - 2. = (3x- 1) 42. y = | = 4z + 5. = x + 1. = (2x - 3) 45. y = 33. y 36. y 2 39. y (a? 2/=z2+i- 2 = 9. 4- 1). 2 X\X 3/ 46. x 2 49. x 2 + y 2 = 9. + = 0. 47. x 2 50. 3* 2 2/ + tf = 0. l 48. x 2 " - 2 2/ L) = *2 0. (2x + 1). 2-6 Sec. 57 The Function Concept 54.,= 55 - y = 58. y = Vr^2. 2-5. SOME x -T7 T- i ' 56. z = 4i/2. 59. = V4 - y a: 2 . 57. s + 60. = y 1 = Zy*. SPECIAL FUNCTIONS The following additional illustrations of two rather unusual, but functions are given here to help us become better very useful, with the function concept. acquainted The absolute-value function is defined by associating with x its absolute value x\. The functional relation is given by j/=|a?|. Thus, the domain comprises the set of all real numbers, while the range comprises the set of all non-negative real numbers. For this function we have the values 2, 3, 0, TT, \/2 corresponding, respectively, to x = 2, -3, 0, 77, -\/2. The bracket function or greatest-integer function, represented by the notation y = [x] is defined as the largest integer which does not exceed x. Its domain is the set of all real numbers, and its range is the set of all integers. Thus, if f(x) = [>], then /(-3.5) , = -4, /(-I) - -1, /(O) - 0, /(2.5) - 2, and /(5) = 5. EXERCISE 2-3 /W = 1. Given 2. Given /(x) 3. Given /(x) 4. Given /(x) = = = = fa? , | r x | x find/(- 3),/(2.3), find/(3.2),/(2), [x], \ [x], find/(- 3) and/(2.5). 5. Given f(x) Let/(x) be the function whose domain the definition is 3). + [x] - x, find/( - 3.5) and/(4). + [2 - x] - 1, find/(0),/(- l/2),/(l),/(3/2), 6. [x] an and/(- as follows and/(2). X is the set of all real numbers for which : if x < 0, then/(aO \ix ;>0, then/(x) = = x. x\ Find/(3)and/(-2). 2-6. VARIATION A particularly important example of a simple type of function often occurring in the physical sciences is given by the formula y If k > 0, this equation = kx. shows that y increases a$ x increases, and We usually say that y varies directly that y decreases as x decreases. The Function Concept 58 Sec. is directly proportional to x. If x also be written as follows as x, or that y ship may where k ^ 0, 2-6 the relation- : the constant of proportionality. This relationship to saying that the ratio of y to x remains constant for equivalent all non-zero values of x. is called is The value of the constant k in any particular problem may be determined from a known pair of values of x and y in the problem. Thus, if the given relationship is y = kx, and if we know that y = 6 when x = 2, then k = 3. Th formula then becomes y = 3x. We say that y varies inversely as tO a, if x, or y is inversely proportional r. V = (**<>> IC This relationship shows that ?/ decreases as x increases, and that y increases as x decreases. But, when x = 0, the following two formulas are equivalent ^ xy = fc and y = : *c Therefore, the relationship between a? and # is such that the product of x and y is constant. Several types of variations may be combined in a single equation to express a certain law. For example, when y varies directly as x and z, we say that y varies jointly with x and z 9 and we write y = kx&. Direct variation and inverse variation are often combined in applications. Thus, according to Newton's law of gravitation, the force F of attraction between two bodies of masses m\ and 2 varies w directly as the product of their masses and inversely as the square of the distance d between them. The equation is _ * Example 2-4, If a __ kmim,2 ~~d?~' man is paid $15 for an 8-hour day, how much would he make in a 35-hour week? Solution: Since this is The wages a man earns vajy directly as the amount of time he works. a problem in direct variation, we have w = kt. In this formula, w represents the total wages, in dollars; t represents the time worked, in hours; and the constant k represents the wage rate in dollars per hour. Substituting w = 15 and t = 8 determine^ the constant k = 1.875. The general formula then becomes w = 1.875 t. Sec. 2-6 The Function Concept Therefore, when Hence, the man t = 35, 59 we have w = = (1.875) (35) 65.62. earns $65.62 in a week. A motorist traveling at an average rate of 50 miles per hour made Example 2-5. a trip in 5 hours. How long would rate of 60 miles per hour? it take him to make the same trip at an average Solution: Since the time required Varies inversely as the speed, we have '=* In this case, k represents the distance traveled, in miles; r is the speed, in miles t is the time required, in hours. Substituting t = 5 and r = 50 per hour; and determines k = 250. Hence, the general formula , t when Therefore, r = _250 r we obtain 60, ' _ 25 ~ 250 ~ 60 Hence, the time required Example 2-6. is Under is ' 6 4 hours 10 minutes. suitable conditions the electric current / in a conductor E and inversely as the resistance R of the R = 10 ohms, / = 11 amperes. Find what varies directly as the electromotive force When E = conductor. voltage is ohms necessary to cause 2 amperes to flow through 60 Solution: is 110 volts and From the statement of the problem, given by the formula we see that the of resistance. combined variation fejjj 1 = 11, E 110, and Substituting I Therefore, the formula becomes ~~W' R = 10 determines the constant k to be 1. /= R This relationship may also be expressed as follows: The required voltage E equal to the current / flowing through the conductor multiplied by the resistance R, or E = IR. is For the specified values, In this problem k It is = 1. E= (2) (60) = 120. Hence, The formula obtained is E= 120 volts. commonly known widely applied to entire and partial circuits through which elefctric as Ohm's law. currents flow. EXERCISE 2-4 y varies directly with x, and y = 15 wh&i x = 7, find a formula for y in terms of x. If x varies directly with y, and x *= 32 when y = 4, find x When y = 3. If y is directly proportional to x 2 and y = 112 when x = 4, find y when x = 9. 1. If 2. 3. 4. If , y is inversely proportional to x, and y = - % when x = 1, find y when x = -3. 60 Sec. 2-6 = 6. = 12 and when x =9 The Function Concept 5. If y varies inversely with x, and y = 10 = 3, when x # varies directly with y and inversely with 2=2, find x when ?/ = 16 and 2=4. 6. If 2, 22, y varies directly with \/x and inversely with and 2=2, find / when z = 25 and 2=6. 8. If y is directly proportional to x when # 9. If = 1 and 2 = 1, find y 2=2, find y when a: 4 when y and y = 18 and inversely proportional to = 2 and 2=4. \/z, and 2/ = 4 = 1 when x 3 y varies directly with x and inversely with and = and # 7. If y when x find = - 1 two spheres have radii r\ and and 2 respectively, show that 10. If and r2 , = - 2 22 , 1 = and y 2 when # 2. diameters d\ and ^2, and surface areas Si , ri* 11. If d Si.* Sa the two spheres in problem 10 have volumes V\ and ~" r2 2 "~ d22 F2 , respectively, show that 12. If the radii of two spheres are 3 units and 1 unit, respectively, and b) the ratio of their volumes. find a) the ratio of their surface areas 13. What are the ratio of the surface areas spheres if the ratio of their radii is and the ratio of the volumes of two 3/2? The diameter 14. of the planet Jupiter is approximately 10.9 times the diameter of Earth. Assuming that both planets are spheres, find a) the ratio of their surface areas and 6) the ratio of their volumes. 15. The diameter of the sun Compare the volumes and that the sun and 16. approximately 109 times the diameter of Earth. the surface areas of the sun and Jupiter, both planets arc spheres. By how much must whose surface area 17. is is we assume the diameter of a sphere be multiplied to give a sphere 25 times that of a given sphere? When the volume V of a gas remains absolute temperature T. so-called absolute zero, which its if constant, the pressure (Absolute temperature - is P varies directly as measured from the - 460 F or is approximately 273 C.) If gas enclosed in a tank having a volume of 1,000 cubic feet and the pressure is_ 54 pounds per square inch at a temperature of 27 C, what will be the temperature when the pressure is raised to 108 pounds per square inch? is T of a gas remains constant, the pressure P varies inversely A gas at a pressure of 50 pounds per square inch has a volume 18. If the temperature as the volume V. of 1,000 cubic feet. If the pressure is increased to 150 while the temperature remains constant, 19* The weight of a of the distance what pounds per square inch the volume? body above the earth's surface varies inversely as the Square from the center of the earth. If a certain body weighs 100 is 4,000 miles from the center of the earth, hdw much will it pounds when it weigh when it is 4,010 miles from the center center? is . arid when 4,100 miles from the Sec. 2-7 The Function Concept 61 20. The electrical resistance of a wire varies directly as its length and inversely as the square of its diameter. A copper wire 10 inches long and 0.04 inches in diameter has a resistance of 0.0656 ohms, approximately. What is the resistance of a copper wire 1 inch long and 0.01 inches in diameter? 21. What is is the diameter of a copper wire 1,000 inches long whose resistance ohms? 10 22. According to Kepler's third law, the square of the time it takes a planet to make one circuit about the sun varies as the cube of its mean distance from the The mean sun. distance of the earth distance of Jupiter make one circuit is 92.9 million miles, and the mean Find the time it takes Jupiter to is 475.5 million miles. about the sun. j 2-7. CLASSIFICATION OF FUNCTIONS It is often desirable to immediate purpose it group functions into a will suffice to consider classes. For our classification into algebraic functions and non-algebraic, or transcendental, functions. Let us first give a more precise definition of a polynomial function and then define algebraic functions and give some illustrations of both. A polynomial function of # is a function given by the relationship n~ n + + a n_iz + a y = a G x + aix an where a a an are real constants, a ^ 0, and n is a posil n, - lf , , -i, The polynomial function is said to be of degree n. The function which makes the number correspond to x zero number is also called the but this polyevery polynomial, nomial has no degree. A rational function of x is a function which either is a polynomial function or can be expressed as a quotient of two polynomials. Thus, a polynomial is often referred to as a rational tive integer or zero. integral function of x. A polynomial, or a rational integral function, of x, y, sum of terms of the form z, , is defined to be the algebraic kxa y bz c where k is - , a constant coefficient and each of the exponents a, b, c, either a positive integer or zero*. The degree of such a function is the degree of the highest-degree term which is present. - - - is and 5x 2 - Ixy* + 62 define polynomial functions of the second degree and third degree, respec- For example, the expressions 3x 2 - tively. These and the expressions x x 7 + 5 y and 2x y - x \/7 + x*9 + t . are l examples of rational functions. * Zero exponents will be defined in Chapter 1 for any non-zero number u. that u = 4. For now, one needs only note 62 The Function Concept A number an algebraic number is equation of the which the 2-7 a root of a polynomial form a,QX in if it is Sec. n -{- aix n~~* 4~ a n ix H~ -}- an = 0, an are integers, not coefficients Oo, 0i, all zero. term algebraic number, we have the term Analogous A function y = f(x) is called an algebraic funcalgebraic function. tion of x if y is a solution of an equation of the form to the Po(x)y n + Pi(x)y*~ + l - - + Pn~i(x)y + Pn (x) = 0, Pn (x) are polynomials in where the coefficients PQ(X), PI(X), and n is a x, positive integer. Polynomials and rational functions are special types of algebraic functions. The functions that we have considered so far were illus- - , trations y = \/# ft case, y = y x According to our definition, algebraic functions. an algebraic function of x because y 2 - x = 0. In this P (a?)=l, Pi(x)=0, and P2 (#) = x. Similarly, 2, of is = "" is an algebraic function of & because #7/ 2 - #2 Here n = 2, P (a;) = x, PI(X) = 0, and P2 (z) = - 2 + 1. An irrational function is an algebraic function which = + 1 is not a 0. a: rational function. A number a number which is not algebraic, and a transcendental function is a function which is not algebraic. Functions like the trigonometric functions, which we shall take up in Section 3-2, belong to the class of transcendental functions. Later we shall consider other types of transcendental functions, namely, the logarithmic and exponential functions such as log x and transcendental is w 3-1. Tfie Trigonometric Functions THE POINT FUNCTION P(t) The trigonometric functions that we are about to define are functions in the sense previously described in Section 2-3 ; that is, they are relations between two sets of numbers. The student who is familiar with the trigonometric functions from his high-school work cautioned to note that we are not, for the present, discussing angles with these functions. We shall see that the concept of a trigonometric function need not be associated with an angle; in fact, many of the most important applications of mathematics in modern science and engineering are concerned with trigonometric functions of pure numbers. Hence, we shall adopt the numerical point of view, leaving the study in terms of angles as a is in connection secondary consideration. Consider a circle with a radius of one unit placed at the origin of a rectangular-coordinate system. See Fig. 3-1. Let t be any real D ffi) FIG. 3-2. FIG. 3-1. number. Starting at the point with coordinates (1,0), we lay off on the unit circle an arc of length \t\. If t> 0, we measure the arc in a counterclockwise direction. If t < 0, we measure in the clockwise direction. If t = 0, the arc consists only of the point (1,0). By this procedure, there is associated with each real number t a definite end-point P(t) of the arc whose 63 initial point is (1, 0). There- 64 The Trigonometric Functions Sec. 31 number t, we have a definite ordered pair (x, y) of numbers which are the coordinates of the endpoint of the arc. Since P(t) lies on the unit circle, it is at one unit distance from the origin. Hence, it follows from the distance formula that fore, corresponding to every real x2 (3-1) + y = 2 1. By means of this equation we can find the second coordinate of the point P(t), except for sign, if one of the coordinates is known. To determine the number pair (x, y) for the point P(t) corresponding to a given value of t, we shall take note of the fact that the circumference of the unit circle is 2?r = 6.2832 units (approxiin Fig. 3-2 is one-half of the mately) For example, since arc is equal TT units in it is length, and arc complete circumference, ABC . AB to 7T/2. now becomes apparent that P(0) is the initial point (1, 0) ; P(TT) is the point (-1, 0) P(?r/2) is the point (0, 1) and P(37r/2) and P(-ir/2) both represent the point (0, -1). It ; ; 3-2. DEFINITIONS OF THE TRIGONOMETRIC FUNCTIONS The correspondence between the of ordered pairs (x, y) set of real numbers t and the leads to definitions of the six set common trigonometric functions, namely, the sine, cosine, tangent, cotangent, secant and cosecant. General Relationships. We shall define the cosine of the real t to be x, and the sine of the real number t to be y. Thus, number we have (3-2) x = cos t, y = sin t. and (3-3) of each of these functions (the sine function and the is the set of all real numbers t. Since, however, every point P(t) lies on the unit circle, neither of its coordinates (x, y) can exceed 1 in absolute value. Therefore, The domain cosine function) and |cosJ|<;i (3-4) |sin|<;i. In other words, the ranges of cos t and sin t are restricted by the requirements 1 ^ cos t ^ 1 and 1 ^ sin t ^ 1, respectively, for all values of t. It may be shown that the range of each of these functions is the set of all real numbers u such that 1 = u ^ 1. The other four functions may be defined in terms of the cosine and sine, as follows : (3-5) tan t = ^ cos t (cos * * 0), See. 3-2 65 The Trigonometric Functions (3-6) cot t = (3-7) sec t = -^- 55i| sin t cos esc (3-8) = t (sin t * 0), (cos < 5* 0), (sin * 5^ 0). t -r Since the cosine and sine are defined in terms of the coordinates it is also possible to express the other functions same coordinates. From the definitions of the cosine and the sine given by (3-2) and (3-3) and from the definitions of the other functions given by (3-5), (3-6), (3-7), and of the point P(t), in terms of these (3-8), * we have ^ =~ =l' <"-> ^ <""' i (3-9) t (3-10) <" (3-11)' v sec*=-J = = cos esc (3-12) t = -4- = sin We note here that cos = i t x - \ G*?*0), /> ' y (y ^ 0). or x, appears in the denominators of both tan t and sec t. Hence, tan t and sec t are not defined when t is a number for which the ^-coordinate of P(t) equals zero. For t, example, since the ^-coordinate of P(7r/2) or P(37r/2) is 0, it follows that tan 7r/2, sec 7r/2, tan 3?r/2, and sec 377/2 are not defined. Similarly, it can be shown that cot 0, esc 0, cot TT, and esc TT are not defined. We functions tan conclude, therefore, that the domain of each of the t, cot t, sec t, and esc t is the set of all real numbers for which the denominator is not zero. follows from (3-7) and (3-8) that numerical values of or of esc t can never be less than 1. Hence, the ranges of It also sec t these two functions are restricted by the requirements (3-13) |sec t\ 1 and |csc *| ^ 1. From (3-5) and (3-6) we obtain an idea of the behavior of the tangent and cotangent functions. It may be shown that the range of each of these funptions is the set of all real numbers. The Trigonometric Functions of ?r/6, Tr/4, and Tr/3. The computa- tion of the numerical values of the trigonometric functions in general is beyond the scope of this book. However, we shall use the methods of plane geometry to find sin t, cos t, and tan t for ,= 7r/6, 66 The Trigonometric Functions 3-2 FIG. 3-5. FIG. 3-4. FIG. 3-3. Sec. and 77/8, in order to show that for certain values of t the trigonometric functions can be found exactly without tables. To compute the functions for the real value t 77/8, we construct the unit circle of Fig. 3-3, Arc AB is given to be equal to 7T/3, which is one-sixth of the complete circumference. Triangle OAB is inscribed in the circle, as shown, with side BO extended through the origin to C. Our problem now is to find the values of x = cos (7T/3) and y = sin (77/8) as coordinates of the point B. Since OA = OB, triangle AOB is isosceles. Hence, 77/4, OAB = angle OBA. angle We note also that arc CAB ~ 77 and arc CA = arc CAB - arc AB = 77 - 77/8 = 277/3. Therefore, arc Also, angle Furthermore, angle Hence, it equals the and OBA 9 or angle 2 angle AOB sum COA = is 2 arc AB. 2 angle AOB. an exterior angle of triangle AOB. two remote interior angles OAB of the COA = angle OAB + angle OBA. Thus, and triangle COA CA = AOB = angle OAB + angle OBA, is equilateral. we draw BD perpendicular to OA, it will bisect OA. Then ~ 3/4 and = 1/2 and from x 2 + y 2 = 1 it follows that y 2 y V5/2. Hence, cos (77/8) =1/2, sin (77/8) = V5/2, and tan (77/8) = \/3 For i = 77/6, place the equilateral triangle AOB of Fig. 3-3 in the unit circle as shown in Fig. 3-4, where E is the mid-point of arc AB. Then arc EB = 77/6 and OE is the perpendicular bisector of chord AB. If x = ; . Sec. 3-2 67 The Trigonometric Functions = 1/2. From x* + y 2 = 1 it Since DB is one4ialf of chord^AB, y 2 = = follows that x 3/4 and x VS/2. Hence, cos (TT/G) = \/5/2 and sin (7T/6) = 1/2. We then have tan (77/6) = 1/V3 = yff/3. For = 7T/4, construct a unit circle as shown in Fig. 3-5 with arc AB = Tr/4. Since arc AC = 7r/2, arc 5C = ir/2 - Tr/4 = ir/4. Hence, arc AB = arc BC, and angle A 05 = angle BOC. Draw BZ) perpendicular to OA. Since the two parallels are cut by the transversal OB, angle BOC = angle angle A0# = angle OBD, Hence, OC and and Thus, y 2x 2 = = x. 2 2 Substituting this value of y in x + y = 1, we have = 1/2. Therefore, x - cos (Tr/4) = \/2/2, and # = V2/2. It follows-that tan (ir/4) = 1. and x 2 1 sin (77/4) = Other Special Values. In a similar fashion we can compute rt_ C/jf exactly the trigonometric functions of such values of 7 t 00 as > Q > and - -~ Functions of multiples of ?r/2 may also be -5- > computed in this fashion, if one considers a straight line as a right triangle in which one angle is and, hence, one side has zero length. Example 3-1. Calculate the values Solution: (0, 1). As explained of the six trigonometric functions of t = ?r/2. in Section 3-1, the coordinates of the point P(ir/2) are Hence, by (3-2), (3-3), (3-9), (3-10), (3-11), and (3-12), sin (?r/2) = 0, = 1, tan (7T/2) is cos (7T/2) undefined, sec (?r/2) is esc (7T/2) = = cot (w/2) undefined, 1, 0. EXERCISE 3-1 1. Determine the coordinates of each of the following points: a. P(2ir). d. P( - y) b. P(-T). e. P(4ir). c. P(5x/2;. f. P(- 7ir). In each of the following cases, carefully draw a unit circle and estimate the coordinates of P(t). c. P(3). b. P(2). a. P(l). 2. d. P(-2). e. P(4). f.-P(-8). 68 3. The Trigonometric Functions Sec. 3-2 Evaluate each of the following: K b. cos -r- o d. g. 4. cot- e. sin csc(-). u h. -- llTT / sm (- , f. e. cos t t = 1/2. = - 1/2 b. cot t sin t = - t 1. being positive. Complete the following table, secy 7r \ ). In each of the following cases assume that approximately the appropriate arc (or arcs). a. sin 5. . f 2ir, c. tan t = f. cot t = and draw a figure d. esc 1. 1, sec i t showing = - 1. being negative. which shows the algebraic signs of the trig- onometric functions in the four quadrants. 6. Use the equation x 2 where ^ J g 27r. +y = 1 7. Use the equation x 2 + = 1 2 2 1/ trigonometric functions of a. sin t d. esc t g. cot t 8. e. sec h. tan = - sin = sec ( ( 9. For each of the following t) given condition b. t which tan f = 3/5. = - 2. = - 6/7. Prove that each of the following equations sec for t = cot t, in each of the following cases to find the other b. cos c. t t. = 1/2. = - 3/2. = 2. a. sin a. to find all values of is t. '^ . c. sec f. tan i. sin J t J = 13/12. = 4/3. = - 3/5. correct. b. cos ( i) d. tan ( t) cos f. tan . cases, state the quadrant, or quadrants, in which the is satisfied. The sine and cosine have the same signs. The tangent and cosine have opposite signs. 3-3. IDENTITIES As an immediate consequence of the definitions of the six trigonometric functions, we can establish certain relationships among Sec. 3-3 69 The Trigonometric Functions them which hold for every value of Since P(t) t. lies on the unit 2 2 (3-1) holds that is, x + y = 1. But, according to (3-2) and (3-3) x = cos t and y = sin t. Therefore, we have the equation circle, ; , cos 2 (3-14) t + sin 2 = t 1. This states that "the square of the cosine of t plus the square of the sine of t equals unity." Since (3-14) holds for every value of t, 2 it is an identity. Note that we use the symbol cos t instead of 2 This simplified notation is used for all positive exponents, (cos t) but is never used in the case of a negative exponent. Thus, cos' 1 t does not mean the same as (cos t)~ l as we shall see in Chapter 8. . , Similarly, we can prove that for each value of t for which the functions are defined, (3-15) 1 (3-16) 1 Proof of (3-15): By + tan 2 = + cot 2 = t sec 2 , t esc 2 t. tan definition, = t ^ cos and sec t = cos t t However, these relationships have no meaning if cos t equals zero, that is, if the ^-coordinate T)f P(t) equals zero. When cos 1 7*0, we may divide both sides of the identity cos 2 1 + sin 2 1 = 1 by cos2 * and obtain sin , 2 1 * ~~~ j cos 2 c cos 2 t Iherefore, 1 Proof of (3-16): By In this case of sin 2 1 + we assume cos 2 1 1 by + tan 2 = t sec 2 definition, cot that sin sin 2 1, t sin t. = t ~~ nf\o f sin t ^ 0. When we we obtain cos 2 _ ~~ t 1 t sin 2 t 2 and esc t = 1 sin t divide both sides t Therefore, 1 + cot 2 t = esc 2 *. We thus have established the three identities which here for easier reference we restate : (3-14) cos 2 (3-15) 1 (3-16) 1 + + t + sin 2 tan 2 t cot 2 t = = 1 = sec 1, 2 esc 2 t, t. These fundamental! identities are very important in working with trigonometric identities and should be remembered. Our present work with identities will consist of reducing given trigonometric expressions to other forms. Unfortunately, no specific rule of procedure can be given for making these reductions. Profit- 70 The Trigonometric Functions Sec. 3-3 is a matter of both practice and Generally speaking, when we want to reduce a given expression to some other form, it is helpful first to perform any indicated algebraic operations and then to use some form of one of the fundamental relationships to simplify the expression. To prove an identity, we may proceed in any one of the following ways 1) We may work on the more complicated member of the identity and attempt to reduce it to the simpler member. 2) We may work with both sides of the identity and show that they making such reductions ciency in experience. : induce to the same expression. We may form the difference of the two sides of the identity and prove that difference to be equal to zero. It is frequently desirable to express both sides of the given identity in terms of sines and cosines alone, and then use (3-14) if needed. We shall consider a few examples to illustrate the procedure in 3) reducing expressions. Example 3-2. Show that cos From Solution: = sec t -f sin t t tan = t cos -f sin t , Adding, we have 1 + sin = 2 1 Since sin 2 1 + t sec tan cot + cos t 2 1 = t tan t sin tan t cos + cot = Example 3-4. t t = : t to sin t cost cos t sin cos t = sec L t. t cost sin and cot sin 2 cos t t - sin Therefore, t cos + 1 =- t cos 2 sin sin 2 1 t sin t + By 1 t cos + cot t sin t = sin t cos t. t Establish the identity (sec t - cos O2 = tan 2 t(l reducing both sides to the same expression. Solution: t cos 2 we obtain 1, t t 2J t, cos 1 tan cos t - definition,' + we have t ^ , t 111 By J t cos = cos 3-3. Reduce Example * tan t and 1 cos Solution: -f sin t cog2 = t cos Since cos 2 t. (3-5), or the definition of the tangent, cos .,,. tan -f sin t t definition, sec f I [ Vcos t = cos T t - Hence, the left side t .\* COS / = /I - cos I \ cos 1 t 2 1\* ) ) becomes - cos 2 t) by Sec. By 3-4 The Trigonometric Functions we have (3-14) cos 2 *\ 2 /I \ The cos _ ~~ ) t = t sin J(l = o i\ 1) cos 2 _ . 1 <?os 1 By definition, t sin 2 . - o 1/1 tan 2 cos 22 be reduced as follows. may Hence, we have x sin 4 _ "" Vcos t) right side of the given identity tan /sin 2 *\ 2 71 . . . sin 2 1 . 2 Since both sides reduce to the same expression, = t cos 2 1 ~- the identity > cos . , sin 4 is established. t EXERCISE 3-2 Prove each of the following - . - . 1. sin cos t tan identities: ~ 0. = sin rt , 2. H esc sin sec t 3. : cos 5. sin 9. + 13. I - (1 esc sin ) + = cos 2 L= cos 2 tan + t 14. 2 J 2 J 1 1 ^ sec 2 t - tan 2 = 27. 1 -f tan esc 2 <(sec 8 2 < = (sec rinf-coBJ + 1 cos - 1 - J cos = t 1 2 sin 1 2 cos 2 J. 2 1 J 2 1 1 2 J. t 2 2 J. 2 1 2 1 < t. 2 1 J + cos t i J 2 1 4 J t t 4 J. *. t) 1 J . ^ 2. 1 t ^ cos ^ t . t. + cot O = sec esc - sin = tan sin t(l + tan 2 28. ^ 32. 2 * 2 2 1 < cosM sin3 sin : 1 3-4. 2 ... + tan = sin 4- cos 26. (tan . sin t 2 l)csc i. cot 2 f. = t) - J t ; . sec 2 cos sin 1 cos 2 t 2 J -I- __ sin ~~ 1). tan|-l tan + 1 t cos 2 2 esc J 1. = cos (csc = 1. cos tan cos - 1) (sec + 1) = tan 2 (sec 2 = sec sec sin tan -f cos 2 = tan sin sec esc esc 2 + sec = sec esc - cos 4 = sin - cos sin tan + cot 2 = sec csc ? - sin t sec 1 - , Ojl 31. cot 2 22. t. ^ sm 24. ^ 20. J. 1 1 cos sin 18. * 2^ 29. 16. . J. 1 . 25. sec 2 12. ^. cot 4 cos 2 10. 1. 2 tan - = = . . , 1 esc 2 1 t t 8. sin . t) cos sec 6. (sin t. 2 tan cos 2 + cos 1 H- = 2 - cos sec = sec esc - sec = tan 4 + tan 2 sec sm tan = Sec 2 * 21. = = = t (1 cos 2 A 4. esc - sin -~ - 15. sin 19. + 2 tan tan 2 J(cot 2 11. (1 17. t (cot 7. sec 2 ~ 0. = ; esc t " 1+ -f t cos f*os ^ + cos i . . = = x ^ i (esc t - . 3 1 cos J. ^ ^^ f cot O2 TABLES OF TRIGONOMETRIC FUNCTIONS Exact numerical values of trigonometric functions in general cannot be found. However, by use of methods beyond the scope of this book, the values can be computed to as many decimal places as desired. The results of such computations have been tabulated an<l are included in this text in the form of tables of trigonometric functions. Table I at the end of the text contains the values of the The Tr/gonomefr/c Funcfions Sec. 3-4 P(t) FIG. 3-6. ^ t ^ ir/2. six functions corresponding to numbers t such that = contains table since 1.5708 the ir/2 Actually, approximately, values of between and 1.60. be the point corresponding to a given value of t, Fig. 3-6, and let ti denote the length of the shorter arc which joins P(t) to the o>axis. In each case in Fig. 3-6, the point P(ti) is located by measuring the arc t\ counterclockwise from the positive half of the #-axis. We shall call ti the reference number, or ^ t\ ^ rr/2. Since t v is a real related number, for t. Note that number, there is associated with it a point P(t\) = (#1, 2/1). Also, since ti lies between and 77/2, P(ti) must lie in the first quadrant. In each case in Fig. 3-6 the coordinates of P(t^) must be numerthat is, \x\ = Xi and \y\ = y\. Since ically equal to those of P(t) are defined in terms of x and y, we functions all the trigonometric Let P(t) (x, y) ; can say that (3-17) | any function of = t j same function of t\. not have the same algebraic sign, and all functions of t\ have in any quadrant and the lie whereas P(t) may positive values, functions of t do not necessarily have positive values. It is important to see that the proper sign is chosen to make the equation a true one. The algebraic sign in each case depends on the quadrant in which P(t) lies. The following examples and Fig. 3-7 will illustrate the method of reducing a function of any positive or negative t to the same These functional values may since P(ti) lies in the first quadrant f unctipn of the reference We number ti. our discussion for the present to direct use of Table I and consider ipnly values of ti which are shown there. The process of using the table for values of ti which are not shown will be treated in Section 3-10 when we discuss interpolation. For shall limit simplicity at this time, we shall use the approximate value TT = 3.14. Sec. 3-4 Example 3-5. Reduce the functions of t As shown number t\ is Solution: reference of 1.14 73 The Trigonomefric Functions may = 2 to functions of its reference number. in Fig. 3-7(a), P(2) is in the second quadrant, be found from Table by noting that only the Thus, we have The numerical 2, or 1.14. TT I. The and the values of the functions signs of the functions of 2 are determined in the second quadrant. and cosecant are positive sine : cos 2 sin 2 tan 2 cot 2 sec 2 esc 2 = - cos (T - 2) = - cos 1.14 = - 0.4176, = sin (w - 2) = sin 1.14 = 0.9086, = - tan (w - 2) = - tan 1.14 = - 2.176, = - cot (T - 2) = - cot 1.14 = - 0.4596, = - sec (T - 2) = - sec 1.14 = - 2.395, = esc (TT - 2) = esc 1.14 = 1.101. Example 3-6. Find tan Solution: As shown Since the tangent is in Fig. 3-7(6), the reference Solution: = Solution: 4 = ir = .86. 1.162. 5. Here, as shown in Fig. 3-7(c), t\ = 2?r 5 = 1.28; and cos 5 = To 20. locate the point P(20), a positive direction from revolution is .86 t\ = tan 0.2867. Example 3-8. Find cot circle in number positive in the third quadrant, tan 4 Example 3-7. Find cos cos 1.28 4. is 2ir = 6.28, find we we must proceed 20 units around the The number of units in one complete (1, 0). find that 20 = 3(6.28) + 1.16. Therefore, to locate P(20), we must proceed three times around the unit circle and then continue for 1.16 additional units in a counterclockwise direction. This means that *! t may = = t be taken as 1.16. From 1.16. the table, Since P(20) or P(1.16) lies in the first quadrant, 0.4356. we have cot 20 cot 1.16 = = 74 The Trigonometric Funcf/ons Example 3-9. Find sin = - Sec. (-2). = shown in Fig. 3-8 that IT -f 2| = 1.14. Hence, t\ is the same as t\ in Example negative in the third quadrant, we have For Solution: t 2, it is i = - (- 2) = - sin 2 = - sin 1.14 = - - (- |- * Since sin 3-5. | sin 3-4 2)| t is 0.9086. should be noted that in Fig. 3-8 the points P(-t) andP(t) are located on opposite sides of the #-axis, and are joined by a It P(2) segment which is bisected perpendicularly by the axis. Hence the coordinates of the two points are numerically equal, but the ^-coordinates have opposite signs. Therefore, by the definitions of the funcline tions given in Section 3-2, sin Hence, ( cos ( tan ( if t (3-18) > ) t) = sin t, = cos t, = -tan t, it follows that sec (t) = esc t, (t) = sec t, cot ( esc t) = cot t. 0, (any function of (-t) = | |same function of t\. However, the algebraic sign of the function is changed for all functions except the cosine and the secant. Because cos t and sec t remain unchanged when t is replaced by its negative, these functions are called even functions. The remaining functions are called odd functions, since their values change sign when t is replaced by its negative. EXERCISE 3-3 1. Construct a figure, locating each of the following points. its related number ti. (Use TT = 3.14). Show the point P(t) and a. P(l). 2. With the e. P(- 4). d. P(- 5). c. P(10). aid of Table I find each of the following values, using b. P(3). a. sin 1.45. b. cos 3.5. d. tan 5. e. esc g. sin 28. h. cos 60. (- c. 2.41). 3-5. POSITIVE We P(3/2). = 3.14. f. cot (- i. tan ( - 4.50). 30). 227T k. cot sec TT sec 4.75. Sir j. f. 1. 4 AND NEGATIVE ANGLES AND STANDARD sin POSITION have defined each of the six trigonometric functions as a relation between two sets of numbers, employing as the independent Sec. 3-5 The Trigonometric Functions number whose absolute value variable a real t represents the length of an arc of a unit circle. Now we return to the shall tradi- viewpoint and tional consider trigonometric functions of angles. Although the student is probably familiar with the idea of FIG. 3-9. angle from the study of geometry, we shall try to make the definition in a plane and draw the halfmore precise. Let us select a point line or ray a emanating from O, as shown in Fig. 3-9. We shall call the vertex of the ray. Finally, we let A be a point on the ray in its initial position. Now rotate the ray a about^O to some terminal position 6, so that moves along the arc indicated by the curved arrow AB. the point A The ray may be rotated in the counterclockwise sense, as in Fig. 3-9 (a) or in the clockwise sense, as in view (b). Moreover, it may be turned through one or more complete revolutions, as in view (c) We shall speak of the position b as the terminal ray b. We have , . then an ordered pair of half -lines consisting of the initial ray a and the terminal ray b. We can now define an angle as follows An angle is a geometric figure consisting of two ordered rays : emanating from a common vertex. With each angle is associated a number, called the measure of the angle, which indicates the sense and amount of rotation required to turn from the initial ray of the angle to the terminal ray. This usually represented graphically by a curved arrow. Its evaluation will be considered in Section 3-6. rotation is We may designate the angle in Fig. 3-9 as angle AOB; or we may use a Greek letter, such as 0, $, a, /3, or y, as the designation. The line OA is called the initial side of angle AOB f and OB is the terminal side. Counterclockwise rotation, as in Fig. 3-9 (a) 01 3-9 (c), gives rise tb a positive angle, while clockwise rotation, such as the one in Fig. 3-9(6), gives rise to a negative angle. Finally, we shall say that an angle is in standard position with respect to a rectangular coordinate system when its vertex is at the origin and its initial side coincides with the positive #-axis. See The Trigonometric Functions 76 Sec. 3-5 III FIG. 8-11. FIG. 3-10. Fig. 3-10. When an angle is placed in standard position, the terminal side determines the quadrant to which an angle is said to belong. Thus, angle XOP in Fig. 3-10 is positive because it is generated in a counterclockwise direction, and is a second-quadrant angle because the terminal side OP lies in the second quadrant. We note that the definition of angle does not specify that the rotation should stop at the first arrival at the terminal side OP, Fig. 3-10. In fact, angles of any size may be generated, since any number of angles which end at the terminal side OP of a given angle may be obtained simply by adding a number of complete rotations, positive or negative, to the given angle. For example, the same terminal side may also be reached by rotation in the opposite direction. All angles which are in standard position and have the same terminal sides are called coterminal angles. In Fig. 3-11 the angle a is generated by rotation of OX counterclockwise to the position OP. The angle /3, which is coterminal with a, is generated by adding to a one complete rotation of OX. The angle y is a negative angle, which is coterminal with a and is generated by rotating OX in the clockwise direction to the position OP. 3-6. MEASUREMENT OF ANGLES The problem of measuring an angle is equivalent to that of finding the measure of the associated arc. One should, therefore, apply the discussion of Section 3-1 and construct a unit circle as shown in Fig. 3-12. FIG. 3-12. measuring the angle Let be an angle in standard position. Since the initial and terminal sides of the angle intersect the circle in the points P (0) an(j P(t), respectively, the problem of reduces to that of measuring the appropriate $ec. 37 77 The Trigonometric Functions arc length t. Thus, the measure of the angle can be found in terms of a real number in any one of several ways, depending on the unit of measure chosen. We shall consider first the circular system, or natural system, of measuring angles, which is used almost exclusively in the calculus and its applications. Its fundamental unit is the radian. This unit may be defined as follows A radian is the measure of an angle which, if placed at the center of a circle, intercepts an arc on the circumference equal in length to the radius of the circle. In Fig. 3-13 the angle AOB is 1 radian, and the length of the subtended arc AB is equal to the : / radius r. If the circle selected for measuring a radian is a unit circle, we have an alternate definition of a radian. That is, a radian is an angle which intercepts a unit arc on a unit circle. \ arc =: radius m o radius F IG . =r A 3.43. Another system of measuring angles is the sexagesimal system, or degree system, which is commonly used in ordinary calculations involving angles. The fundamental unit of this system is the degree. In Section 3-7 we shall study various relations between radians and degrees, and shall develop rules which allow us to convert from one system to the other. In discussing angles, we frequently use the term angle, in place of measure of an angle, and we rely on the context to make the meaning clear. Thus, when we say "0 = 2," we mean, "0 is an angle whose measure is 2 radians." The word radian is usually omitted when an angle is expressed in terms of radians. 3-7. THE RELATION BETWEEN RADIANS AND DEGREES Since an arc that is equal in length to the radius of a circle subtends an angle of one radian at the center, it follows that the whold circumference, which is 2rr times the radius, subtends an angle of 2?r radians. Furthermore, the whole circumference subtends a central angle of 360. Therefore, 2?r radians = 360, and TT radians = 180. 78 The Trigonometric Functions If the approximate value 3.1416 1 radian 180 = = 7T is used for Sec. 3-7 TT, 180 (approximately), . Q 1 1A U.141D or 1 radian = 57.29578 1 radian = 5717'45" (approximately). (approximately), or Also, 1 = make In order to = T^T radians 0.01745329 radians (approximately). the conversion to radians easier we is expressed in degrees, minutes, and seconds, values : 1' = when the angle give the following 0.00029089 radians, and 1" = 0.00000485 radians. Therefore, one of the following rules can be used to convert from degrees to radians or from radians to degrees To convert from degrees to radians, multiply the number of : degrees by ~ , or 0.0174533. ioU To convert from radians radians by , to degrees, multiply the number of or 57.29578. 7T Note, However, that certain angles are commonly expressed in terms of TT radians, in order to avoid approximate values. For example, 180 90 3-8. = = TT 45 30 radians, 7T/2 radians, = = 7T/4 radians, 7T/6 radians. ARC LENGTH AND AREA OF A SECTOR In Fig. 3-14 is shown a circle of radius r. In such a circle an angle at the center equal to one radian subtends an arc on the circumference equal to r. Similarly, by the definition of a radian, the number of units in the arc s intercepted by a central angle equal to radians is given by the relationship Thus, $ (3-19) = re, or FIG. 3-14. arc = (radius) (central angle expressed in radians) . Sec. 3-8 Now 79 The Trigonometric Functions A denote the area of the sector bounded by two radii s. If is the number of radians in the central the ratio of the area A of the sector to the of the then angle sector, to area of the whole circle, or Trr2 equals the ratio of the angle the angle in the whole circle, or 27r. That is, let and an arc of length , A=A Trr 2 ' 2?r or A =$'* 9 (3-20) it ' If the central angle of an arc or a sector must be re-expressed in radians before expressed in degrees, (3-19) or (3-20) can is be applied. Example 3-10. Express 210 = Solution: Since 1 Thus, 210 = - ^ in terms of radians, 210 lo loU TT radians. ~ =^ = 210 radians. O loU radians. Example 3-11. Express 1215'20" in radians. Solution: Multiply the decimal parts of a radian given in Section 3-7 for 1, I" by 12, 15, and 20, respectively. The results are as follows: 1', and 0' 0" 12 15' 0" 20" 15' 20" 12 Example 3-12. Express Solution: Since TT Example 3-13. Express .20943948 radians .00436335 radians .00009700 radians .21389983 radians. radians in degrees. -^- radians = = = = = 180, ^6 radians = |o (180) 3.5 radians in degrees, minutes, = 150. and seconds. Solution: First, convert the radians to degrees, as follows: 3.5 radians To by To by find the number find thfe 60. The number result is (60) (0.5352) = (57.29578) minutes 200.5352. decimal part of a degree, or 0.5352, = 32.112'. of seconds, multiply the decimal part of 0.112' Hence, 3.5 radians (3.5) of minutes, multiply the = 60. Thus, 0.5352 = = (60) (0.112) seconds = 20032'6.72". = 6.72*. a minute, or 0.112, 80 The Trigonomefric Functions Sec. 3-8. Example 3-14. The radius of a circle is 5 inches. Find the length by a central angle of 30. of the arc of the circle subtended Since 30 Solution: by = > -^ the central angle is ~- Also, r = 5. Therefore, (3-19), =r-0=5--=j| o o Example 3-15. In a central angle Solution: is By circle of radius (3.1416) = 2.618 inches, 6 inches, what is the area of a sector whose 60? (3-20), the area of the sector A = 0(36)7,- = 6?r O t is ^r & 2 0. Since 6 = 60 =~ , 3 square inches. EXERCISE 3-4 In each problem from 60. 2. 45. 1 to 25, express the given angle in radians. 1. 6. 11. 16. 150. 72. 283. 13. 30. 90. 215. 14. 10. 240. 196. 18. 3010'. 19. 4621 / 3. 7. 12. 8. 12. 20. 63. 17. 4. 9. 120. 330. 321. 5. 10. 15. . 20. 23637'. 21. 8216'. 22. 6321'17". 23. 18357'43". 24. 39244'27". 25. 9331'38*. In each problem from 26 to 40, express the given angle in degrees. 26. 7T/6. 31. T/12. 27. 7T/4. 28. Tr/8. 29. 37T/2. 32. Sir/18. 33. 7<jr/2. 34. 5ir/3. 39. 0.763 rad. 36. 3.7 rad. 30. 47T/5. 35. 3?r/20. 37. 8.21 rad. 38. 0.34 rad. 40. 0.8136 rad. In each problem from 41 to 56, draw the given angle in standard position and indicate its terminal side. 41. 30. o 45. 22^- - 42. 7T/4. 43. ir/3. 44. 46. 27T/3. 47. 170. 48. 17T/18. 50. 630. 51. 53. 77T/3. 54. 1000. 55. 97r/4. 57. In a circle of radius 4 49. T/2. 360. 90. 52. 56. - 47T. - llTT/6. feet, find the length of the arc intercepted by an angle of radians. Find the 7ir/6 angle in radians that intercepts a 5-foot arc. A 58. central angle in a circle of radius 15 inches intercepts an arc of 5 inches. Find the number of radians in the central angle. Express this angle in degrees and minutes, rounding off the result to the nearest minute. 59. A central angle of 6214' intercepts an arc of 16 inches on the circumference of Find the radius of the circle. 60. Find the area of a circular sector whose radius is 7 inches and whose central angle is o) 4 radians; 6) 75; c) 3 radians. 61. The area of a circular sector is 72 square inches. Find the angle if the radius is a) 6 inches; 6) 9 inches; c) 5 feet. a 62. circle. The area is a) of 128; a circular sector is 126 square inches. 6) 1.6 radians; c) 30. Find the radius if the anglfe Sec. 3-9 3-9. The Trigonometric Functions 81 TRIGONOMETRIC FUNCTIONS OF ANGLES be an angle in standard position, as shown in Fig. 3-15. associate a real number t, which is the measure of the angle in radians. This concept is equivalent to our previous concept of t, when t was interpreted as the length of an arc laid off on the unit circle by starting at the point (1, 0) and terminating at Let With 6 we can P(*,y) FIG. 3-16. FIG. 3-15. the point P(t). Such an association of the angle 6 with the directed length t of an arc of a unit circle allows us to define the similar cosine and sine of as cos = cos t and sin = sin t. procedure may be followed for the other functions of 0. A Now consider Fig. 3-16, where we show an angle in standard position and a unit circle. By the definition of 0, the terminal side of intersects the unit circle at the point (cos 0, sin 0). This is, of We now extend course, the point designated previously as P(t). with coordinates to an arbitrary point the terminal side of P (x, y). If The length we drop (x,y) to the Therefore, of the radius vector OP is r = V#2 + 2 2/ * perpendiculars from the points (cos 0, sin 0) and the right triangles thus constructed are similar. re-axis, x - = r cos , and : y - = sin r 1 T1 Hence, the coordinates of the point P(x,y) on the terminal side are x = r and cos Using these results with the y = r sin 0. definitions of the functions from Section 3-2, we caii express the values of the six functions in terms of x, y, and r. Thus, sin (3-21) cos tan = = = 2//r, esc x/r, sec y/Xj cot = = = r/y, r/x, x/y. The Trigonometric Functions 82 3-10. Sec. 3-10 TABLES OF NATURAL TRIGONOMETRIC FUNCTIONS OF ANGLES Tables of natural trigonometric functions are so labeled to distinguish them from tables of the logarithms of these functions. Angles in Radians* In Section 3-4 Table I was used to find values of trigonometric functions of the type cos 2 or sin 27T/3. On the basis of the definitions of the functions of an angle given in Section 3-9, Table I may also be used to find the functions of angles meas- ured in radians. Example 3-16. Find the cosine Solution: From Table of cos 1.43 I, an angle of 1.43 radians. = 0.1403. Angles in Degrees. Table II at the end of this text contains the approximate values of the six functions of acute angles expressed in degrees and minutes. It is a four-place table of the functions of angles at intervals of 10 minutes. To find the value of a function of an angle between and 45, first locate the angle in one of the columns at the left, and then same line in the column headed by the name of the desired function. For an angle between 45 p and 90, locate the angle in a column at the right, and then look for the value on the same line in the column with the name of the desired look for the value on the function at Table II its foot. should be referred to in working through these illustra- tive examples. Example 3-17. Find sin 3240'. and 45. Look in the left-hand column to find Solution: This angle is between 3240', and then go to the right to the column headed sin. There find 0.5398. Hence, sin 3240' = 0.5398. Example 3-18. Find cos 5620'. Solution: This angle is between 45 and 90. So look in the right-hand column 5620' is above 5600', and then go to the left to the to find 5620', noting that column with cos at its foot. Thus, cos 5620' The following examples = 0.5544. procedure for finding an angle corresponding to a given value of a function. illustrate the 83 The Trigonometric Functions Sec. 3f-10 Example 3-19. Given tan Solution: Since tan is = 4.511, 6 greater than find is 1, through the columns marked tan at the foot ^ 90. assuming that 6, greater than 45. Therefore, search for the given number 4.511. The corresponding angle in the right-hand column 7730'. or is 7730'. So 4.511 = tan = Example 3-20. Given cos Solution: 6 = By looking through 0.8660, find 6, ^ assuming that 7730', ^ 90. the columns with cos at either the head or the foot, column headed cos, use the left-hand column for find 0,8660. Since this value is in a the corresponding angle, which is 30. Hence, 6 = 30. When either the given angle or the given value of not printed in the table, we can find the desired value Interpolation. a function is or angle by using a method of approximation known as interpolation. We assume that the change in the value of the function is directly proportional to the change in the angle. Although this assumption is not strictly valid, it gives values that are accurate enough for many practical purposes if we limit its use to small changes in the angle. The process of direct interpolation is used if the angle is given and we need to find the value of either an increasing function of the angle, such as the sine, or a decreasing function, such as the cosine. Inverse interpolation is used when the value of a trigonometric function is known and the angle is to be found. Example 3-21. Find Solution: 1820'. sin 1812'. This angle From the table is we not listed in the table, but find that sin 1810' = sin 1820' = 0.3145. it lies between 189 10' and 0.3118, and The desired value of sin 1812' The tabular difference, that will is, then lie between 0.3118 and 0.3145. the difference between the two values listed in the table, is 0.0027. Also, the difference between the angles 1810' and 1820' is 10', while the angle 1812' differs from 1810' by 2'. Since the change in the angle from 1810' to 1812' is 2/10 of the change from 1810' to 1820', we assume that = the corresponding change in the value of the sine will be (0,2) (0.0027) 0.0005, and the amount to be added to 0.3118 is 0.0005. Hence, sin 1812' =0.3123. The accompanying diagrammatic arrangement tabular form: o sin 18 10 ' presents this = 0.3118 \ f 2 10 [ same operation sin 1812' sin 1820' = 0.3118 + x = 0.3145 1* J 0.0027 in The Trigonometric Functions 84 Sec. 3-10 7 Since the angle 1812' is 2/10 of the way from 1810' to 1820 the corresponding functional value will be 2/10 of the way from 0.3118 to 0.3145. Therefore, , 10 0.0027 r x This amount and sin to be is 1812' = (0.2) (0.0027) = 0.0005. added to 0.3118. Hence, the value of the function is 0.3123, = 0.3123. Example 3-22. Find cos 7348'. Solution: The process is similar to that in Example 3-21. However, since the cosine decreases as the angle increases, we subtract 8/10 of the tabular difference from cos 7340'. We find the values of cos 7340' and cos 7350' in a column of the The work may be table labeled cos at the bottom. cos 10 x 7340' = 0.2812 cos 7348' = 0.2812 - cos 7350' = 0.2784 = , or x = x x 0.0028 (0.8) (0.0028) 10 0.0028 indicated as follows: Hence, the amount to be subtracted from 0.2812 is = 0.0022. 0.0022, and cos 7348' = 0.2790, The inverse process of finding the angle when the given value of a function not printed in the table is performed in a similar Here, since we know the value of the function, we find the two values in the table nearest the given value, one less than is fashion. and one greater. Again making the assumption that small changes in the value of the function are proportional to small changes in the angle, we proceed as indicated in the following example. it Example 3-23. Find Solution: / . cot 6 = 0.8780. This value of the cotangent entries 0.8796 and 4850 if We have, 10 = 0.8796 { cot 48 (40 cot 4850' * = 48*43'. lies To 1U + xY = 0.8780 = 0.8744 = 16 W 5J , ' and x =: between the respectively, the angles therefore, the following tabulation: cot 4840' x Hence, 9 not in the table but is and 0.8744. To these correspond, \ 0.0016 0.0052 4840' Sec. 3-10 The Trigonometric Functions 85 EXERCISE 3-5 In each of the problems from 1 to 30, use Table II to find the value of the given function. Interpolate whenever necessary. 2. cot 12840'. 1. sin 3620'. 3. sec 2340'. 4. cos 9650'. 5. sin 13210'. 7. esc 22330'. 8. sec 3930 10. sin 9840'. . 11. cos 75CO'. tan cot 28350'. / (- 13330'). tan 62340'. (- 14. esc 17. cot 5543'. 18. esc 19. sin 5732'. 20. cot 31G'. 21. sin 22. cot 2801'. 25. cos 28. sin ((- 23. tan 15. ). / 271G 31237 2S10'). 12. cot (- 41620 13. sec 39210'). 16. tan 29852'. (- 6. 9. / 24. esc . / 7258'). 26. sin 1647'). 29. esc 289OG'. (- 4451'). (- 28033'). (- 24529'). 27. tan 63602'. . * 30. cos 12619'. In each of the problems from 31 to 60, use Table II to f.nd the values of between and 360 which satisfy the given equation. Express the results to the nearest minute, interpolating whenever necessary. 31. tan 34. cos 37. sin 40. cos 43. tan 46. sin 49. ccs 52. cot 55. cot 58. cot = - 0.11C8. = 0.7951. = 0.5783. = - 0.4147. = 8.345. = O.G702 = 0.9503. = - 1.381. = 7.COO. = 0.1340. 32. sin 35. tan 38. cot 41. shT0 44. cot 47. tan 50. cos 53. cot 56. tan 59. tan = 0.3062. = 0.0553. = - O.G494. = - 0.9959. = - 0.3121. = 0.9043. = - 0.5090. = 0.4230. = - 0.1191. = - LS.CO. 33. cot 36. sin 39. tan 42. cot 45. tan 48. cot 51. cos 54. sin 57. cos 60. tan = 1.091. = 0.2419. = 1.511. = 0.0437. = 1.446. = 2.398. = 0.8519. = 0.2491. = - 0.1323. = 3.235. In each of the problems irom Gl to 72, find the value of the given function. Interpolate whenever necessary. Take 61. sin 0.93. C?. cot 2.46. 63. sec 64. tan 8.71. C5. esc 9.43. 66. cot 0.678. 67. tan 0.333. 68. cot 70. sin - -?} (\ 6 / TT as 3.14. ( 1) 71. esc 0.968. (- 69. cos ^ 72. cot (- 1.24). 0.643). In each of the problems from 73 to 84, find the values of 0, in radians, between 2ir \\hich satisfy the given equation. Use Table I and express the results to and three decimal places, interpolating whenever necessary. * 0.9759. 74. sin 0.9967. 73. cos 76. sec 79. cos 82. cot = = 2.563. = 0.4010. = 0.39. 77. tan 80. tan 83. cos = = 0.9413. = 1.6. = 0.84. 75. tan 78. sin 81. sin 84. cot = 2.066. = - 0.736$ = 0.91. = 1.031. 4 4-1. The Laws of Exponents POSITIVE INTEGRAL EXPONENTS When studying the progress of algebra up to the sixteenth one cannot help but be perplexed by either the total absence of symbolism or, when present, the lack of uniformity in its use. At first, unknown quantities were often represented by words. Later, symbols made from abbreviations and initial letters of these words were used to indicate mathematical concepts, such as number, power, and square. Descartes (1637) is generally credited with our present system of exponents. He introduced the Hindu-Arabic numerals as expo4 3 nents, using the notations a, aa [sic] a a etc. The writing of a repeated letter for the second power of the unknown continued for century, , , , many years. Laws for positive integral exponents were introduced in Section shall now establish these laws and extend 1-11, without proofs. them to apply also to zero, negative, and fractional exponents. recall that if n is any positive integer, a n means the product of n factors each equal to a. In this notation, a is the base and n is the exponent or power. shall proceed to establish the following laws for positive integral exponents. We We We Law of Multiplication. If a is a real number, and positive integers, a m a n = a m+ n (4-1) if ra and n are . Proof. Proof of this relationship follows from the definition of an and the associative law for multiplication. Thus, am = a a a m (to factors), and an = a a a (to n factors). Hence, am a n = = = [a a m factors)] [a m + n factors) cr(to a a a m+n a(to a a(to . For example, x 3 x 5 = a; 8 , and y k yk+3 86 = y 2k + 3 . n factors)] Sec. 87 The Laws of Exponents 4*1 Law of Division. If a is a non-zero real number, and > n, then are positive integers such that = m~ n if m and n m (4-2)f If a an \ a T^ 0, and if . n > m, then nm (4-3) a 1 n a n ~~ m Proof. Proofs of these relationships follow If n is positive. By (4-1) > n, then ~ a m n a n = 0(w n)+n = o, m m : m an , . we have am mn =a n Hence, dividing both sides by , . a For example, 37 ~5 If m < n, then n m Divide both sides by a 35 an ~ = a w+(n = - ( an . an a n-m sides by a n , we have n n = ( an- m \) // a n = a n n1~m / / \a a a Now, dividing both an (4-1), ~ m) to obtain am am By is positive. a n~m - > - I = an ~ For example, 37 Law n are for a Power of a Power. a If _ 32 ' a real number, and is if m and positive integers, then (4-4) (a m = n ) amn . Proof. This relationship can be easily proved as follows By the associative laws for multiplication and addition, the law of multiplication expressed by (4-1) can be extended to three or : more factors. Thus, 0/n A . gn QP . m rr (a =. a m+n an) Q? ap similar relationship can be written for any That is, (a m - ) (of) (a r ) = number of factors. 88 The Laws of Exponents We may now take m = p = (a m (a ) m = r to (a ) m ) (to Sec. get n factors) = amn = am+m+ + = a "*. 1 Hence, (a w w ) . For example, (z Law and 2 3 ) = x 6 and (2 2 ** 1 ) 5 , Power of a Product. for a = 2 10 *+ 5 . a and 6 are real numbers, If m and n are positive integers, then if n (4-5) (ab) = anb n . Proof. In proving this relationship, we make use of the associative and commutative laws of multiplication. Thus, n (ab) = = = (ab) [a (ab) ..... (ab) (to n a ..... a(to n factors)] factors) b ..... b(to [6 n factors)] n n a b . For example, Law 6 T^ 0, for a and Proof. if Power of a Quotient. If a and n is a positive integer, then By applying (a\ (b) n a 6 are real numbers, if the law for multiplying fractions, an a ,, a f N =b'b ..... we have , ^ (ton factors) =^- For example, 2 /3xy^\ \ 22 / _ _3^y^ " 4 9o; 2 "" 2 2 2fe i/ 4 An exponent affects only that quantity to which it is attached. 2 2 2 Q Thus, -5x(y*) = -5xy , whereas (~5xy*) = 25x y*. So far we have defined a n only when n is a positive integer. We shall now introduce zero, negative integer* and rational powers in way that they will obey the same laws which were proved such a for positive integral exponents. 4-2. MEANING OF o We shall define the zero exponent by the equation o (4-7) A few illustrations are = 1 (a 7* 0). : = 5, (o - to)' a 1, Sec. 4-2 If The Laws of Exponents a in (4-2) is 89 m = n, we get not zero and an In this case, the quotient on the left equals 1, while the value of the term on the right is a. Since a 1, by definition, the law of division holds for n = m, as well as for m > n and n > m. The student should note that (4-7) gives the only possible definition of a if the law expressed by (4-2) and (4-3) is to hold for the zero exponent, as can be seen from the foregoing discussion. We shall show that the definition a = 1 is consistent with the five laws of exponents in Section 4-1 that is, we shall show that ; when any exponent these laws also hold is zero. In the following explanations, where a quantity occurs in a denominator, we assume that it is not zero. Also, the exponents are assumed to be non- negative integers. Let us, for sake of discussion, suppose that n a m~ ' Then we have . OP a . a m+n = an am = = m+0 = am =a - . ao ~*~ = in (4-1), that n . a. . -_ x am = . ^ Hence, the law of multiplication holds when n = 0. procedure will verify the law if Now let us suppose that n = in (4-2), that is, in A 0. similar m 1 = a = fr = am ~n am and . = ~ am = aw . 1 Hence, (4-2) holds when n = in (4-3) we have If m 0. , = "" an =: ~~ an an and , ~~ an m = ~~ an ~ = "~ an m It is clear that in (4-2) cannot be zero, and in (4-3) n cannot be zero. Therefore, the law of division holds. Suppose that n = in (4-4) that is, in , (a m n ) = a mn . Then (a If m= m n ) = (a n ) = m ) = 1, we have n n (a) = l = in (4-4) w (a and a mn = a mt = a = 1. , 1, and amn = a' n = Hence, the law for a power of a power holds. a = 1. The Laws of Exponents 90 Now consider (4-5), which n n= 4-2 is (ab) If Sec. = a nb n . 0, (a6) n we Finally, = (a&) let n= = anb n and 1, in (4-6), that /a\ n __ a^ - W Then in " a" , and 1> is, 1. 6 " t = a6 = 1-1 = a = 1 = = L t The demonstrations just given prove that the five laws of exponents, originally stated for positive integral powers, are true for all non-negative integral powers, and that the law of division is true even 4-3. when the exponents are equal. NEGATIVE EXPONENTS In order to extend the meaning of exponents to negative integers, define crn by the following relationship we : or* (4-8) = 1 (a* 0), where n is a positive integer. Several illustrations are: ( v a K-2 5 ~ fa.)-2 ' = 52 ~ - 6z) (a 10- 3 25 - in 1U 3 - innn 1UUU > 2 V in Section 4-2, we shall show that our definition is consistent with the five laws of exponents. Let us first note that (4-8) is true even if n = or if n is a negative integer. If n = 0, then As a _ a - - I-!-!. -_()-!a0 1 j If n = p, where p is fln a positive integer, then cr n = ap = 1 / a an cr p We shall use this result in the proofs that follow. In order to extend (4-1), n be a negative integer, that a 0. Then say, let m be a non-negative integer, and let n= p. In this case = #m # n = dm cpp = am 1 a* = am a* it is also assumed Sec. 4-3 If The Laws of Exponents m S p, we have, by am a (4-2) for non-negative exponents, W = - = am /y n 91 ~~ = p am+(-p) _. am+n^ a" Ifm<p, we have, by (4-3) for non-negative exponents, ~~~ ~~ /tW A similar demonstration establishes m < 0, or in case m < 0, and n< in case (4-1) n^ and 0. Proof of the extension of (4-2) rests on the validity of the law of multiplication just established. If a =Q, and if and n are integers (positive, negative, or zero), then am = a m 1n = am or n = am~ n a an m . Although (4-3) is now an immediate consequence, it is not really needed, in view of the general validity of (4-8). The demonstration just given allows the law of division to be stated as a single relationship as follows : ^= fjTn (4-9) aw ~n (a 7* 0). Thus, a single law applies, regardless of whether m > n, n > m, or m = n, where m and n are arbitrary integers. Now consider the law for a power of a power. ^ 0. Then 0, while p, where p In (4-4) let m n Also, a m-p a mn Hence, (4-4) holds in this case. If p, where p i^ 0, while m (a m n (ar ) p Y ( -. a-mp nS = = a mp we have = a ~ p)n = and a mn 0, ( J a~ pn = holds. If both m and n are negative, a similar proceand the extension of (4-4) holds. To extend the law for a power of a product, let n = p in (4-5), where p ^ 0. Then Again (4-4) dure is used, n- - So (4-5) _ _ * is verified nn--?-/)- * 11 - 1 for negative integral values of n, of a quotient, assume To verify the exltended law for a power that n = -p in (4-6), where p ^ 0. Then n h) (a\ = /a\~~ p (h) = 1 P = /a p */hP ~ bp ~P' , anci an fr ^ arp IT* = bp " "5 The Laws of Exponents 92 Sec. 4-3 Hence, (4-6) holds for negative integral values of n. Thus, the laws of exponents hold for positive integral exponents, zero exponents, and negative integral exponents. In Section 4-5 we shall From consider the case of fractional exponents. the general validity of (4-8), it follows immediately that a factor of the numerator or the denominator of a fraction can be moved from the numerator to the denominator, or vice versa, provided only that we change the sign of its exponent. For example, a2 x3 x^z 2 a 2 b~ 3 = TQ and o = SCIENTIFIC NOTATION We are now in a position to introduce certain simplifications when operating with very large or very small numbers, as are customarily used in scientific writing. Any positive number that is greater than 10 or less than 1 may be written compactly by expressing it in standard form, that is, by writing it as a number that lies between 1 and 10 multiplied by a suitable positive or nega4 tive integral power of 10. Thus, 27,000 would be written 2.7 10 4 Similarly, 0.00031 would be 3.1 10~ . . Example 4-1. The speed number in scientific notation. Solution: of light is 186,000 miles per second. The given number 186,000 may be 10 5 written as 1.86 Express this . mass of an electron is 9.11 10~ 28 grams. How many zeros would he required between the decimal point and the first non-zero digit, 9, if Example 4-2. The the rest number were written Solution: in decimal notation? 28 means that we would have to move the decimal its present position. We would thus have to place The exponent point 28 places to the left from 27 zeros to the left of the 9. Example 4-3. If the sun light to reach the earth Solution: As given 10 7 miles from the earth, 9.3 is in Example 4-1, the speed of light 9 3 . iQ7 4-5. long does it take is 1.86* 10 5 miles per 500 seconds = meaning of exponents from integers to the required time 8 minutes 20 seconds. second. how from the sun? Therefore, is i = "b fi . i?^ 5 10 2 = RATIONAL EXPONENTS We shall now extend rational numbers. the Here again we shall make the extension in such Sec. 45 93 The Laws of Exponents a way that the laws for positive integral exponents will be preserved. Suppose that a is a real number and that n is a positive integer. Let us assume that a 1/n has meaning and that (4-4) applies. Then it would be true that 1 (4-10) (a /")" = a' 1 /^ = = a1 a. This says that the nth power of a l/n would have to be a, or in other words that a l/n would be what is called an nth root of a. For example, (4-10) would yield (a 2 1 /2 ) = a, and (a 1 / 3 3 ) = a. l/n Real nth Roots of a. Before defining a let us examine the situation with respect to the existence of nth roots of a given number a. The following results may be proved with the help of the theory of equations. Case I. If n is an even integer and a is a positive real number, there are two real numbers that satisfy the equation r n = a. One , is the positive nth root of a, which is denoted by tya. The the negative nth root of a, which is denoted by ^/a. We may also denote these two numbers together by ^/a. Case II. If n is an even integer and a is a negative real number, of these other is no real nth roots exist, since no even power of a real number can be negative. Case III. If n is an odd integer and a is a positive real number, there is one real (positive) value of r such that r n = a. In other words, if n is odd, there is a real positive nth root, which is denoted by^a. Case IV. If n is an odd integer and a is a negative real number, a. That is, there exists one real (negative) value of r such that r n if n is odd, there is a real negative nth root, which is denoted by y*. Case V. If n is any positive integer and a is zero, there is only one real nth root, and this root is zero. Thus, the definition of an nth root of a is valid under all conditions except when a is negative and n is even. In this situation, no real nth roots exist. (However, the introduction of complex numbers in Chapter 11 will allow us to eliminate this exception.) We are now ready for the following definition. If a is a non-negative real number and n is a positive lfn a designates the non-negative nth root of a, or ^/a. If a integer, is negative and n is an odd positive integer, then a 1/n designates the real nth root of a, or ^/cT. When a is negative and n is even, a 1/w is undefined: Definition. The Laws of Exponents 94 Sec. 4-5 Meaning of am/n. Let a be a given real number, n a positive m/n has meaning, and if (4-4) holds integer, and m an integer. If a m/n = a (1/n)TO = (a 1/n ) m Under these for fractional powers, then a 1/n m/n would be the mth It is power of a assumptions, then, a . . natural to state the following definition. is any integer such that Definition. If n is a positive integer, if a in and if is a real number which the fraction m/n is lowest terms, m/n n is even, then a is assumed to be non-negative when designates 1/n the mth power of a , that is, the mth power of ^/a. Hence, 1/n m amln = m (a (4-11) . ) If the fraction m/n is not in its lowest terms, it is first reduced to lowest terms, and (4-11) is then applied. When a is given, the value of am/n depends only on the value of and n. the fractional exponent, not on the particular values of m Thus, 2 4/2 = 22 = 26 4, ' 8 = 23 '4 and , (- 2) 2/6 = (- 2) 1 /3 = ^^2. In the last example it would be incorrect to apply (4-11) directly, 1/6 has no meaning. ( 2) It may be shown that, if a is positive, since amfn (4-12) The proof We is = <\/a. omitted. omit the details of the procedure for showing that the five laws of exponents hold for rational exponents and nonnegative bases. The reader is cautioned against using the laws for negative bases, since some fail under certain conditions. To summarize the results now established, we restate the laws of exponents here for easy reference. It is assumed that a and b are and n are rational numbers. non-negative real numbers, and that Furthermore, if either a or 6 appears in a denominator or raised to a negative or zero power, it is assumed to be different from zero. shall also m Law Law m of multiplication: a of division: Law for a power Law for a power am an = an ~ am n = a m+n . . = awn n n n (ab) = a b m n of a power: (a ) of a product: . n -J (a\ Law for reciprocal: Zero power: a = ar n 1. = an 5= an 7- . Sec. 4-5 95 The tows of Exponents Note that the radical notation can be replaced by the simpler and much more convenient exponential form. Everything that can be done with the radical notation in the simplification of roots of numbers and in operations involving roots can be done much more naturally by means of the exponential notation. A few illustrations of the meanings and uses of exponential forms follow : x 1'2 32 2 / 5 = (\X32) = Vx, (32) 2 = 1 /5 22 if ^ * = = (- 0; X/32 27) = - 1/3 2; = \/^~W = - -tf/5 5 = - a 2 3; ; - 4; 4 4x 2 The following examples illustrate the applications of the laws of exponents to the solution of problems involving radicals. (16s Example 4-4. Compute the value of \/2 \/2, 1 '2 yi6x* ) and write the result in expo- nential form. Solution: Using exponential notation and the laws of exponents, we have 2i/3 . Example 4-5. Remove Solution: = 21/4 all 2 3/12 2 4 /* = 2 4/12+3/12 = 2 7/12 possible factors from the radical We may proceed as follows: = (2 3 4 z 4 2 2/ = Example 4-6. Use the laws 1/3 ) = 2 1/3 2 1/3 3 1/3 x 1/3 2/ 2/3 3x = . $ 3 4/3 3 of exponents to express ^/y by using only ^/x one radical. Solution: Changing to we have fractional exponents, Example 4-7. Rationalize the denominator in the fraction -J7=f Solution: To write an equivalent fraction in which no radical appears in the denominator, we proceed as follows: __ Example 4-8. Rationalize the denominator Solution: radical We use the relationship (a + b) -= of the fraction 5 (a b) = a2 v3 b 2 to from the denominator. Thus, 1 5 - V3 = " 5 1 ' 5,- V3 5 + \/3 = " 5 + V3 = 5 + V3 22 25 - 3 + V3 remove the 96 The Laws of Exponenfs . 4-5 ^2 - /a? Sec. x2 + 2 __ Example 4-9. Change - to a simple fraction. ^ Solution: Write the expression in exponential form, as follows: a'2 - x2 Then, multiplying the main numerator and the main denominator have _ ^ (a Example 4-10. Express (x 2 + <r2 - 2 / 2 (a 2 - .r 2 - x 2 ) 112 , 3 '2 2 ) + Zx 2 (x + a 3 /2 2 (a ^2 z2) 3 +a by 2 ) 1 '2 2 ) in a factored form, Solution: Rewrite the expression as (x Removing +a 2 common the (x 2 +a 112 2 (x ) factor (x 2 112 2 ) [x 2 + + a2 + 2 ) + a2 a2) 1 -j- / 2 , 3^ 2 ] Zx 2 (x 2 + a 2 ) 112 . we obtain = (x 2 +a 1 /2 2 ) (4z 2 + a 2 ). EXERCISE 4-1 In each of the problems from eliminate all zero 1 to 20, perform the indicated operations and and negative exponents. 3 1. 3x*y. 2. ari". 3. (jV 5- Trb' 6- UJ* 7. (9)". - 4. 10- 2 . 8. 9. 13. (a;V)*(a:-V 16. x-i 4- y-1. Lz^l!!!. - 10 4 2 ) . 14. (x 1 '^- 8 ) 4 + I/)- 17. (x ^- 1 - 1 '8 ) . 15. (x*' 2 -hi/ 1/2 ) 2 . 18. (a; + 2/)- 1/3 (x + j/)i/. ; 20 Write each of the following expressions in exponential form. Remove all possible from the radical and, wherever necessary, rationalize the denominator. factors 21. V80. s- V* 2 - Sec. 4-7 1)1/2. 4-^4. - 49. (2 4-6. 97 The Laws of Exponents x2) 5 / 2 2)2 + s 2 (2 - z2) 3 / - 50. Or 2 2 . 3) I/2 - x*(x 2 - 3)- 1 '2 . THE FACTORIAL SYMBOL The product of positive integers from 1 to n inclusive is called "n factorial" or "factorial n" and is represented by either of the symbols n or /.n. Thus, if n is a positive integer, all ! n\ = 1 2 - l)-n. 1 2 3 r! = [(r -2-3 ..... 1 (n For example, 3! = 6! 4-7. = - = 3 5!. 6; j = 5! 6; = 71; 4 - 5 = 120; l)!]r. THE BINOMIAL THEOREM The statement known as the binomial theorem enables us to express any power of a binomial as a sum of terms without performing the multiplications. By actually performing the indicated multiplications, we + 6) = 3 (a + b) = 4 = (a + 6) 2 (a These formulas may + 2ab + 6 2 a + 3a + 3ab + 63 3 4 a + 4a 6 + 6a 2 6 2 + 4a& 3 + 6 4 a2 + = a2 (a + 6) = a3 (a + *)4 2 &) 3 , 3 2 2 fe , : + ?a6 + ~fc 2 , 3 2 2 +ja 6 + j^|a6 + |ff^& = a4 + + if Applying this suggested rule to (a (a + 6)5 = a* + . be rewritten in the following manner, so as to suggest a general rule 1 (a find that a4 6 + a3 6 2 ^ + 6) + S-4-8-2 1 2 3-4 we 5 , , obtain f a 4 5>4>3>2>1 ^1.2-3-4-5 1 The justification for writing the expressions on the right in this form will be found in Chapter 17, where the binomial coefficients are "given in terms of the combination formulas. The Laws of Exponents 98 3 10a b we simplification of coefficients, Upon + 2 2 10a 6 3 -I- 5ab* + ft which 3 , obtained by multiplying (a + &) 4 Sec. get (a by (a + 6) b) same the is + 5 = a5 + result 47 + 5a4 b that as . Each of the expressions on the left is of the form (a + b) n in which the exponents 2, 3, and 4 of (a + 6) are special values of n. If we let n denote the exponent of (a + 6) in each of the expres, sions on the n + we note that the expansion of (a terms with the following properties 1 In any term the 1. the left, first term sum an and the is 4- b) term is bn contains : of the exponents of a and b last n is n. Also, . The exponent of a decreases by 1, and the exponent of 6 increases by 1, from term to term. 3. The denominator of the coefficient in each term is the fac2. exponent of b in that term. 4. The numerator of the coefficient in each term has the same number of factors as the denominator. Specifically, wherever 1 torial of the appears in the denominator, write n directly above it in the 1 numerator wherever 2 appears in the denominator, write n ; directly above the it in the number above 1 is numerator; and so on. Thus, in (a 5, and the number above 2 is 4. Assuming that these properties hold for values of n, all + 6) 5 , positive integral we have (4-13) (a + b) n = a* n(n + ^i a^b + "^ a - b 2 2 z i 1 * ^ * O This result is the binomial formula. So far we have verified this formula only for n = 2, 3, 4, and 5. In Chapter 16, we shall prove the binomial theorem, which states that the formula is true for all positive integral values of n. The following example shows the procedure for the expansion of Example 4-11. Expand Solution: and n (x - 6 2?/) by the binomial theorem. -- -- - required expansion will be obtained by letting a = x, b begin by setting up the following pattern of n + 1 terms = The = 6. We x6 + ( + - 2y) x*( + x*( 4 2t/) + - 2y) x( 2 + 2y) x*( + ( 2). 2y) : 2y, Sec. 4-8 The exponents numerators and denominators of the simplifying, (x - and of b in the second, third, fourth, fifth Hence, remembering that the last term respectively. By 99 The Laws of Exponents 22/) we obtain the terms are is 6n , 1, 2, 3, 4, we may fill and 5, in the coefficients as follows: following result: = z6 6 GENERAL TERM IN THE BINOMIAL EXPANSION 4-8. we wish any particular term of the expansion of without considering any of the other terms, a study of the binomial formula in (4-13) Section 4-7 will reveal the' following If (a + b) facts to write n : In every term the exponent of 6 is one less than the number of the term. Thus, in the (r-f l)th term, the exponent of 6 is r. (The expression for a particular term is simplified slightly if the number of that term is called r + rather than 1, The sum of the exponents of a and the (r + l)th term the exponent of a The denominator of the since it is n coefficient + l). obtain, then, for the (r 1) (n - n(n r (n 2) - 1) (n n in each term. + (n 2) For r. coefficient in the (r tors as the denominator. In the (r We is is the factorial of the exponent of The numerator of the n(n 6 r.) + l)th term is rl, b. has the same number of fac- + l)th term, it is the product l)th term of the expansion -r+ 1) alMftr r! Example 4-12. Find he Solution: 6 is r = 5. = Here a sixth term of (3x = - 3z, 6 y 2 , and n Hence, the exponent of a 8<7 z ' 6 Q ' 5 . ' 4 is oxr(v . \ox) = n - ) ; 8 2 ) . Since r 8. 5 if y = = 3. + 1 = 6, the exponent of Therefore, the sixth term is 1 00 The laws of Exponents Sec. 4-8 EXERCISE 4-2 In each of the problems from 1 to 16, reduce the given fraction to lowest terms. 9. In each of the problems from 18 to 32, expand the given expression by the (Hint: In problems 29 thru 32, first consider the first two terms binomial theorem. in parentheses as a single quantity). + y)*. 18, (x 19. (x - I) 7. 20. (a - 21. (2a* 26). /ll/3\6 ^r+V (7/2 iC 30. + 2y + ( 2 ) . 31. (x* +x + I)*. 32. (o - 362)3. 2. (*+ ) ' 2r - a - I) 3 . In each of the problems from 33 to 42, find the indicated term. 33. (1 - 35. (a + 26) 12 (x5 t s) 8 , 2\ X/) 8th term. , 5th term. 34. 36. (a 8 > 4th term. + n)", (m 38. (x - 6), - 2/) l , 10th term. 3rd term. term involving y 4 v*\ n -- ~ ] (xy 41. (V^ - vV) 12 , middle term. 42. (2x - term involving y 7 ?/) 1 , * . , middle terms. 5 Logarithms A LOGARITHM assume here that the laws of exponents stated DEFINITION OF 5-1. We shall in Chapter 4 for rational exponents ark valid also for irrational exponents. The definition of a base raised to an irrational power is beyond the scope of this book. However, let us make the assumption that, if b and x are real numbers, with b positive, a corresponding number designated by" b exists. Without giving an x let us assume that all laws of expoexplicit rule for computing b nents established in Chapter ~4 are valid generally for real powers. Finally, let us assume that, corresponding to any two positive real numbers & and n, where b 1, there exists a unique real number x, such that n = b x We can then give the following definition. 33 , = . Definition. from x = 1, then x log n. & If n= b x, where 6 is a positive real number different called the logarithm of n to the base b. write table shows both of forms several following equiva- We is The lent statements. We shall restrict n to positive numbers, since negative numbers do not have real logarithms. For any positive base &, we have 6 = 1 and 6 1 = 6. Hence, it follows from the definition of a logarithm that ' log b 1 = and log& b = l. f Another valuable, relationship results from combining the two equations n = b* and x = log & n. Replacing x in the first equation by its value from the second equation, we have For example, 2 10** 8 = 8, and 10 10*i<> * 101 = x. 1 02 5-1. Find n, Example , Sec. 5-1 Logarithms if = 2. n logs Solution: Write the given equation in exponential form, as follows: Hence, n Find the base 5-2. Example = 32 n = 9. = 2/3. 4 6, if log* . Solution: In exponential form, the given equation to the 3/2 power and recall that 6 > 0. Then = (52/3)3/2 Therefore, is 6 2/3 = 4. Raise both sides = 43/2, & 6=8. Example 5-3. Find = x. 32 x, if logi/ 8 we have Solution: Writing the equation in exponential form, Express 1/8 and 32 as powers of (2-3)* = we have - 3x From this, 5-2. LAWS OF LOGARITHMS = 2 5 2~ or , = - and x 5, = and get 1/8 2, 1/2 3J! = = 3 2 2~ 3 and 32 , = 2 5 Hence, . 5 . 5/3. Therefore, logus 32 = 5/3. Since a logarithm is an exponent with respect to a given base, the rules for operating with logarithms are the same as the laws of exponents. These laws, expressed in terms of logarithms, have the following form. Law I. rithms of The logarithm of a product equals the sum The logarithmic form is of the loga- its factors. (5-1) log& (m n) = log& Proof: To prove this equation, x logb m = m+ Iog6 n. let and y = log& n. n = Then m= Multiplying, Hence, bx we have and bv . mn =b*+v. . (mn) = x +y= log& m+ log& n. Law II. The logarithm of a quotient equals the logarithm of the dividend minus the logarithm of the divisor. The logarithmic form (6-2) is Iog6 = Iog6 m- Iog6 n. Sec. 5-2 103 Logarithms Proof: The proof follows Let x = logb m and : Then m= Dividing, we have ,, o* _ Hence, , iog& ( \ Law III. m^ = ) 71 x - / The logarithm = , and ?i = b<-. y = n = y Iog6 w. , o^. m logb logb n. power of a number equals the expo- of a nent times the logarithm of the number that is, ; (5-3) = logo (n*) Proof: The first step in * logb n. the proof is to x = logb n. Then n = let . b*. Raise both sides to the kth power and obtain n k = (b x) k = b kx . This relationship, when written in logarithmic form, becomes Iog6 (n Replacing x by its value, k ) we have k log& (n ) = The student should note and (log&n) = kx. k Iog6 n. k carefully the difference between log& (n ) fc . Law IV. The logarithm of a root of a number equals the logarithm of the number divided by the index of the root that is, ; \fn = ^ Iog 6 (5-4) log b n. Proof: This equation follows as a corollary of law definition of a fractional exponent, law we have = n l/k <tyn III. . tyn = \ Example A/51 5-4. Express loga Solution: Iog 2 = Iog 2 = Iog Iog6 (n 2 -~^- as a VSl S1^ + 1/fc ) Iog6 n. linear combination of logarithms. = Iog 4 Iog 2 3 lo g2 = 17 1 / 2 - 2 (3 4 Iog2 3 . 17V' - the Hence, by III, Iog6 By Iog 2 3* 104 Sec. Logarithms Example 5-5. Express 2 Solution: 2 logio 3 - = J logio 3 logio x - ~ logio $ + logio y as a single logarithm. + logio y = logio 3 2 - logio 2 + logio = logio (32 y) - logio x"* a: Example 5-6. Transform the equation logo 5-2 + y = Iog x 1 ' sin x into 2/ an equation free of logarithms. By Solution: transposing, y = we get , loga sin * - = , logo Change to the following exponential form x sin x , logo --- : sin x EXERCISE 5-1 In each of the problems from 1. 5. 9. 2 = 8. 10 3 = 1000. 100- 6 = 10. 3 = 2. 2 6. 10- 3 10. y 1 to 12, write the equation in logarithmic form. 64. 3. 3 4 = 7. 0.001. = e*. = 81. 256 1/8 11. 10" = 2. = x. = 4. 10 8. 216 l/3 12. 10 1. = 6. = x. 10 * * In each of the problems from 13 to 21, write the equation in exponential form. 13. logs 16. log* = 2. 64 = ggg 19. logio 10,000 4. = 4. 14. logs 125 = 3. 15. Iog 2 ^= 17. Iog 7 343 = 3. 18. Iog 9 729 20. logio 0.0001 In each of the problems from 22 to 22. Iog 9 3 25. log* = Z. = x. x = - 29. log, 100 = x. = = 6. 3. 3/2. 33, find the indicated value of x. 26. logo.s = 4. 31. logo 243 21. Iog 4 8 4. 24. Iog 4 x 23. Iog 2 64 4=2. 28. log* 81 = - - 32. Iog 64 x = - = - 27. Iog 3 x 1. 30. log, 2. ~= 7 33. log* x ^ = 0. = 1. = 5. 2. In each of the problems from 34 to 39, use the laws of logarithms to write the expression as a single logarithm. 34. 191 ODD logfr 2-3 log& 5 + log& 7. 36. \ log, 7 + 1 Iog 38. 3 logs 2 + logb 13-2 4 4 + 1 log* 3. log& 5. 35. log& 4 -f log& 37. - 5 log 23 TT + - log& 3+3 log& 2^ 12 log* =f & r. Sec. 5-4 39. z 105 Logarithms - Vu 2 - logb (u - a2 ) & logb (w + Vu 2 a2 ) + log& a. 40. Find the logarithm to the base b of the area of a circle in terms of the logarithms of and the TT The time T 41. where g log?, T radius. for a pendulum The area Find Iog6 a) (s is T = IT \/ - > where s is and c is given by the formula the semi-perimeter ^ (a mean G of n positive numbers x\,x*, b, and . +6 + c). r. xn is defined relationship log* 5-3. c), 6, K in terms of the logarithms of combinations of a, , Show oscillation g. b) (s positive geometric by the make one of a triangle with sides of length a, Vs( s The to a constant representing the acceleration due to gravity, a) Find terms of the logarithms of IT, I, and g. b) Find log& I in terms of the K= 43. / is in logarithms of w, T, and 42. of length that G = n = G \/XiX2 - - xn . SYSTEMS OF LOGARITHMS As we mentioned in Section 5-1, any positive number 6 different be used as a base in a system of logarithms. However, only two bases are widely used in practice. The common, or Briggs, system of logarithms, named for Henry Briggs (1556-1631), employs the base 10 and is used for ordinary from 1 may computations. The natural, or Napierian, system of logarithms, named for John Napier (1550-1617), is generally used in calculus and theoretical work, and employs the more convenient irrational base = 2.71828 In this book, when the base is not indicated, it is understood to be 10. Thus, log n means logic n, and the word logarithm will mean common logarithm unless otherwise stated. e 5-4. In COMMON LOGARITHMS Table 5-1, we begin with a list of powers of 10, give equivalent from these determine the form of the logalogarithmic forms, and rithm of a number that is not an exact power of 10. It should be mentioned that the logarithm is an increasing function; that is, as n increases, log n increases. Another way of stating the conditions is to say that if a > b then log a > log 6. 106 See. Logarithms 5-4 TABLE 5-1 From true Table 5-1 it can be seen that the following statements are : The logarithm of an integral power of 10 is an integer. The logarithm of a number which is not an integral power of 10 consists of two terms or parts an integral part, called the charac: teristic; and a positive or zero decimal which determined from a table of mantissas. is part, called the mantissa, Thus, since log 10 = 1 and log 100 = 2, we may expect the logarithm of any number between 10 and 100, that is, a number between IO 1 and IO 2 to be 1 plus a positive decimal part. For example, we shall find that the logarithm of 35.4, which number lies between 10 , and 100, is equal to 1.5490, to four decimal places. In this case, the characteristic is 1 5-5. A and the mantissa RULES FOR CHARACTERISTIC is .5490. AND MANTISSA study of Table 5-1 reveals that the characteristic changes as the position of the decimal point changes in the sequence of digits 0035400. The first entry in the column headed "Logarithm of the number" is log 354 = 2. + decimal. Sec. 5-5 Logarithms 1 07 In the number 354, or 354.0, the decimal point is two places to the right of the first non-zero digit, 3 (reading from left to right) ; the corresponding characteristic is 2. The second entry is log 35.4 = 1. + decimal. In this number, 35.4, the decimal point first non-zero digit (reading from the is one place to the right of left to right) ; the corre- sponding characteristic is 1. Similarly, we note that the zero characteristic corresponds to the position of the decimal point immediately following the first nonzero digit. This position of the decimal point is called the standard We may now formulate the following rule for position. characteristics : Rule for Characteristics. tion, the characteristic If the decimal point is in standard posiFor every other position of the zero. is decimal point, the characteristic is equal to the number of places the decimal point has been shifted from the standard position. The characteristic is positive if the shift is to the right, and is negative if the shift is to the left. We shall now same for see that the mantissa remains the all numbers having the same sequence of digits. Let us again consider the sequence of digits 0035400. Any number containing this n sequence can be written 3.54 10 where n is a positive or negative integer or zero and depends on the position of the decimal point. Suppose that we consider the form log 3.54 = 0.5490. Then the , logarithm of any number containing this sequence log (3.54 10 n ) = log 3.54 + log = n + log 3.54 = n + 0.5490. 10 is n Thus, a shift of the decimal place in the number affects only the characteristic n, and the mantissa remains the same for the same sequence of digits. EXERCISE 5-2 In each of the problems from 1 to 16, find the characteristic of the logarithm of** the given number. 1. 34.63. 5. 0.1340. 9. sin 6341'. f 34630. 2. 3.463. 3. 6. 2637. 7. 0.00346. 10. 0.000001. 11. 378364. 14. sin 8453'. 15. cos 6143'. 4. 268.1. 8. 12. tan 428'. 7 821 ~~' lUjUUU ' 13, cot 8113'. 16. sec b 24 8'. 1 08 Sec. Logarithms 5-5 In each of the problems from 17 to 24, place the decimal point in the sequence of 7314 corresponding to the given characteristic. digits 17. 3. 18. 21. 6. 22. 5-6. HOW - 2. 19. 0. 5. 23. - 20. 1. 3. 24-1. TO WRITE LOGARITHMS As stated in Section 5-4, the mantissa of a logarithm is always positive or zero, whereas the characteristic may be a positive or negative integer or zero. A positive characteristic or a zero characteristic can readily be combined with a given mantissa. For example, the logarithm of 354 is written 2.5490. But when the characteristic is negative, say fc, where 1 ^ k ^ 10, it is more convenient to write it in the form (10 k) logarithm of 0.00354. The characteristic Let us consider the but the mantissa is 34- 0.5490. regarded as positive. We could write log 0.00354 = For convenience in computation, however, we write log 0.00354 in the form and so (10 - 3) + 0.5490 - 10 = 10. is 7.5490 3, - 10, or 17.5490 - 20, on. Note. It would be incorrect to write log 0.00354 = 3.5490, for this notation means 3 0.5490 and would imply that the mantissa is negative. To perform certain computations, it is convenient to write the logarithm 7.5490 - 10 in the form -2.4510, which equals 2 0.4510. It is important to note that the decimal part of the number since 5-7. it is HOW 2.4510 is not the mantissa of the logarithm of 0.00354, not positive. TO USE A TABLE OF MANTISSAS The following examples will illustrate the procedure in finding the logarithm of a number with the aid of a table of mantissas. The student should work through each example, determining the characteristic from the position of the decimal point in the number and determining the mantissa by referring to Table III at the end of this book. Example 5-7. Find Solution: The log 46.7. characteristic is -f 1. To find the mantissa, locate 46 in the column in the table headed N, and then go to the right to the column headed 7. Here we find the mantissa .6693. So the complete result is log 46.7 = 1.6693. Interpolation. If the number consists of more than three digits, the mantissa is found from Table III by means of interpolation. Since the method of interpolation is the same as that described in Section 3-10 for the table of trigonometric functions, there will be no further discussion of it here. Sec. 5-7 Find log 0.03426. 5-8. Example The Solution: characteristic is The mantissa 2. number 3426 has more than since the 109 Logarithms three digits. It found by interpolation, is of the lies way between the mantissas of 3420 and 3430, as shown in the accompanying tabulation: Number Mantissa 10 13 Since the difference between the mantissas of the two numbers in the table we have x This is rounded and the amount off to 8, Hence, the mantissa is = is 13, A to be added = .5348 and log 0.03426 to 0.5340 8.5348 - is given by x =8. 10. Finding Antilogarithms. The number which corresponds to a given logarithm is called the antilogarithm. That is, if log n = x, then n is the antilogarithm of x and is written antilog x. 5-9. Find n, Example if log n = 1.8710. Search through the body of Table III to locate the mantissa .8710. the columns headed N and 3, is 743. Since the Solution: The corresponding number, from characteristic is 1, n = Example 5-10. Find To We may antilog 7.5349-10. The mantissa Solution: .5353. 74.3. .5349 is not in Table III but these correspond, respectively, indicate the work in tabular numbers whose lies between .5340 and 3420 and 3430. digits are form as follows: Number Mantissa 3420 .5340 9 3420 10 +x .5349 3430 From this, we see that = 13 .5353 * . -JL ~ 10 13 ' Therefore, x 6.9, or 7 after rounding off. Hence, the sequence of digits in the desired number is 3427. Since the characteristic is - 3, the untilogarithm of 7.5349-10 is 0.003427. 110 Sec. Logarithms 5-7 EXERCISE 5-3 In each of the problems from number. 26. sin 1018'. 1 to 30, find 27. tan 4133'. the 28. sec 6416'. In each of the problems from 31 to common logarithm of the given 29. cos 8214'. 30. cob 3116'. 50, find the antilogarithm of the given number. 31. 1.6665. 32. 4.4857. 33. 9.4183-10. 34. 0.0645. 35. 2.7024. 36. 7.7388-10. 37. 9.4409-20. 38. 6.3404-10. 39. 1.8401. 40. 3.9552-10. 41. 2.4658. 42. 1.9501. 43. 9.7367. 44. 4.9960-10. 45. 8.7863-10. 46. 9.8821-20. 47. 0.6584. 48. 3.0150. 49. 5.0300-10. 50. 0.1504. Solve for x in each of the following equations: = 4. 51. 10* 54. 10 57. 60. - 1 = x. 55. ^/10 - a;. 61. 10 1 -* 53. 10 2 * 2.019. = x. 58. 10 1 314 ^10*= 10-*' 2 = 0.0123. 5-8. = 52. 10" 56. =x. 10^ = 7.132. = x. = 0.003146. 10 *- 3 = 0.6735. 59. 10-* = 0.2346. 62. 2 LOGARITHMIC COMPUTATION The fundamental laws of logarithms given in Section 5-2 are applied in the following examples to illustrate the application of logarithms to computation. Example 5-11. Find the product Solution: Let x = (0.0246) (1360). Then log x log 0.0246 (1360). (0.0246) = log + log 0.0246= 8.3909-10 log 1360 logo; = = = Hence, by interpolation, we have x 3.1335 11.5244-10 1.5244. = 33.45. 1360. Sec. 5-8 Logarithms Example 5-12. Evaluate Solution: Let x = (0.506)- = -' (0.506) / 1 ' 3 . 05( L logs =logl 1 log 1/3 log 0.506 log x Therefore, x = 1.255 by 1 1 1 . 1/3 - Then log (0.506) 1 ' 3 = log 1 - (1/3) log 0.506 = log 1 - (1/3) (29.7042-30) = log 1 - (9.9014-10). = 10.0000-1Q = 9.9Q14-1Q = 0.0986. interpolation. Alternate Solution: Let Z = (0.506) -1 /3 Then . - (1/3) log 0.506 = - (1/3) (29.7042-30) = - (9.9014-10) = - - 0.0986) log x s= ( = Therefore, x = 1.255 by 0.0986. interpolation. n i r 10 ^ * Example 5-13. Evaluate i =P (0-352) (1.74)2 -^0.00526 Solution: Let x denote the desired value. Then = log 0.352 + 2 log 1.74 - (1/3) log 0.00526. We find that log 0.352 = 9.5465-10, log 1.74 = 0.2405, and log 0.00526 = 7.7210-10. log x log 0.352 2 log 1.74 log numerator = = = 9.5465-10 (1/3) log 0.00526 0.4810 10.0275-10 = 10.0275-10 log denominator = 9.2403-10 log x = 0.7872. Interpolating, we have x = 6.126. log numerator Example 5-14. Evaluate Solution: Let x = ( 7=7 Then . ) \174/ log x = 1.14 [log 253 log 253 log 174 - Then log x Therefore, x = 1.532. = - log 174]. = 2.4031 = 2.2405 1626 1.14 (0.1626) ' = 0.1854. = (1/3) (27.7210-30) = 9.2403-10. 112 Sec. Logarithms 5-8 EXERCISE 5-4 In each of the problems from 1 to 30, perform the indicated computation using logarithms. 1. d* 2. (13.25) (26.80). (3.142)(2.718). 29.34 4. (0.8134) 1/3 . 68^5* 6. 7. \/(0.003468) (16.83) 836 ft fi . - 3 '*. 1 " 42,860 o 9. J/ (8,321,000) 4 r 10. (4.313)(3,068)(0.000642). (36,250) 11. (63.84)2(0.0134). 12. (8.364)(321.5) 13. 14. 15. V(168.3) (14.21). -K- 42.63). 2 V(213.6) (43.98) 2 . V(23,310) - (20,180)2. (Hint: Factor the radicand.) 2 is *O 567.87 3 1 * 5 7 19 8' 16 4 no> 4 (1,083) (0.0813) 31 1 19. 2 v/ (21.36) 3 3,642^ "/i 4/3188" K 0.8103 20 32* 64' 3 38.63 f 4 21 A/ <8L68 V (8.013) (0.034) (0.0 > 22. (1.08) 10 . 24. 3,648 (1.03) 3 6. 23. (1 25. - v 27. (0.8123)- 3 '4 26. 0.083 . 0.08614. 29, [(3.864)-3.i3 31. If e = 30. 2.718, find log Problem 412 . 28. log 16.84 + 32. Find the geometric ' e, log <\A, log mean c i e', and - (83.14)-* ^483.6. - (0.8134) 2 '* IT*. of 564.3, 8634, 0.1349, 8.316, and (Hint: See 42.61. 43, Exercise 5-1.) 33. Find the area of a circle of radius 6,381 feet. 34. The volume 3621 of a sphere is V = 35. Find the radius of a sphere 36. 5 irr*. Find the volume of a sphere of radius feet. whose volume is 8423 cubic feet. Find the length of a pendulum which makes one oscillation in = 980 centimeters/sec 2 (Hint: See Problem 41, Exercise 5-1.) g . 1 second, if Sec. 5-9 113 Logarithms 37. Find the area of a triangle with sides 6,384 (Hint: See Problem 42, Exercise 5-1.) 38. The stretch s of a wire of length = relationship s > |r where g I is 5,680 feet, and 2,164 feet long. feet, and radius r by a weight m the gravitational constant is given by the and k (Young's modulus) is a constant for a given material. Find how much a copper wire of length 120 centimeters and of radius 0.040 centimeters will be stretched by a 12 weight of 6,346 grams, if g = 980 and k is 1.2 10 for copper wire. 39. The current henrys t i flowing in a series circuit with a resistance of seconds after the source of electromotive force RilL given by the relationship i = Ie~ circuit before the short circuit. If i seconds, and 40. If n is L = , where / = is is R ohms the current flowing in the R = 0.1 ohms, t = 0.25 = 2.718.) 1*2 n. However, Stirling's a positive integer, n\ has been defined as the product it is difficult to L 10 amperes, 0.05 henrys, find /. (Take e When n is very large, and short-circuited is compute this product. formula gives (approximately) n\ = n n e~n \S2irn. Use this formula to estimate 91, and compare the result with the true value which you should calculate exactly. Do - - log 2 + the same with 30!. log TT + - (Hint: log (n!) = n log n - n log e + log n). 41. If the rate of depreciation r per year is constant, the scrap value S after n years n Find the of a machine with first cost C is given by the formula S r) C(l of value 10 a machine after which cost scrap $10,000, if 20 years originally = per cent per year 5-9. is . written off as depreciation. CHANGE OF BASE sometimes desirable to change from one logarithmic base to another. Suppose there is available a table of logarithms to some known base b (say 10, for example), and we wish to find the logarithm of a number n to some other base a. We then let x = log& n w whence, by definition, n = b Similarly, if we let y = Iog n, then we have n = ay It follows that a v = b, and our problem reduces to solving this equation for y. Taking the logarithm of both sides to base 6, we have y log& a = x log* b. = 6 1. But Iog6 Therefore, It is ; . . t y J = x (: Vlogfe /e e\ (5-5) a/) = log& n (| ) \logb a/ 1 1 4 Example 5-15. Find , Sec. Logarithms Solution: By log* 125, where e i It is usually easier to multiply 10K 125 result can 0.4343. Thus, by using a table to the base 10. lSio = -p 125 logic e e is a fairly = n xuo 2.3026, than to divide. Since division by logio frequent operation in practical work, and the 2.7183, - (5-5), loge 6 = 5-9 it should be noted that be obtained by multiplying by 2.3026 instead of dividing by log. 125 = (2.0969) (2.3026) = 4.828. EXERCISE 5-5 Find each of the following logarithms by using a table of common logarithms: 1. log, 10. 2. log fl 100. 3. Iog2 e. 4. log e 5. log, e. 6. log* 10. 7. Iog 2 64. 8. Iog 20 1000. 9. Iog2o 100. 13. Iogi 25 1000. TT. 10. logioo 64. 11. log, 8. 12. logo.i 50. 14. log* 20. 15. logo. 02 0.04. 16. logiooo 100. O 6-1. and Vectors Right Triangles ROUNDING OFF NUMBERS Numbers that arise in the applications of trigonometry are usually not exact, but are sufficiently accurate for a given purpose. Numbers of this kind are called approximate numbers, and the degree of accuracy of such a number significant figures it contains. indicated by is Reading from left to how many right, the number are the digits starting with the first non-zero digit and ending with the last non-zero digit, unless it is definitely specified that the zeros on the right are significant. Thus, in the numbers 2.405, 0.002405, and 240500, the digits 2, 4, 0, and 5 are significant figures. The zeros after the 5 in 240500 may or may significant figures in a not be significant figures. When it is desired to indicate whether final zeros are significant or not, scientific notation is often used. Thus, in 2.405 10 5 the 5 the final two zeros are last significant figure is 5 in 2.40500 10 , , ; regarded as significant. To round off a number in which the last desired significant figure the units place or in any decimal place, drop all digits that lie to the right of the last significant figure. It is sometimes necessary also to increase the last digit in the retained part by 1. is in dropped part is less than 5, the last digit unchanged. If the first digit in the 5 is than or if that digit is 5 and it is followed greater dropped part by digits other than 0, the last digit in the retained part is increased by 1. Whten the dropped part consists of the digit 5 alone or the digit 5 followed only by one or more zeros, we shall use the following procedure as an arbitrary rule in this book: If If the first digit in the in the retained part is left the last digit retained even, it is left is odd, this digit unchanged. This rule, 115 is increased by 1; although popular, is if it is inferior 1 1 6 Right Triangles and Vectors Sec. 61 common-sense rules in many cases. For example, if .245, .165, and .725 are to be rounded off to two decimal places and then added, it would be more sensible to round off two of the numbers in one direction and two in the other direction. To round off a number in which the last significant figure will lie to the left of the units place, first drop all digits to the right of the place occupied by the last significant figure, and replace each dropped digit to the left of the decimal point by a zero. Also, either leave the last digit of the retained part unchanged or increase that digit by one, in accordance with the directions just given for dropping only a decimal part. For example, if 2533.62 is to be rounded off to three significant figures, the result is 2530 and if 487,569 is to be rounded off to three significant figures, the result to .485, ; is 488,000. There are working two rules that are generally adopted by computers in with approximate numbers in order to guard against retaining figures that may indicate a false degree of accuracy 1. In adding or subtracting approximate numbers, round off the answer in the first place at the right in which any one of the : 2. given numbers ends. In multiplying or dividing approximate numbers, round off the answer to the fewest significant figures found in any of the given numbers. The numbers entering a problem involving multiplication or division may be rounded off before the computation is begun. If these numbers are rounded off, they should have one more significant figure than the answer is to have. While these rules point in the right direction, it should be mentioned that rounding off computed quantities to as many significant figures as there are 'in the given numbers does not necessarily produce the degree of accuracy implied by the results. The subject of accuracy of computation with approximate numbers is somewhat complicated and beyond the scope of a book at this level. 6-2. TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES One of the simplest, yet important, applications of trigonometry is in the solution of right triangles. right triangle has, in addition to the 90 angle, five other parts. These are two acute angles A and three sides. If we know the length of any two sides, or either acute angle and any one side, the triangle can be solved; that is, the unknown parts can be found. 63 Sec. Right Triangles and Vectors 1 1 To solve problems involving parts of triwe shall find it helpful to be able to express the trigonometric functions of an 7 ~ A angles, acute angle A of a right triangle ABC in terms of the sides of that right triangle. To derive suitable relationships, let us place the acute angle in standard position, as shown in Fig. 6-1. In the right triangle ABC the side AC, which FIG is - ^ the abscissa of becomes the side adjacent to the angle A the side CB, which is the ordinate of B, becomes the side opposite to angle A and the side AB, which is the radius vector to B, becomes the the point J5, ; ; hypotenuse. If we the lengths of the side adjacent, the side let opposite, and the hypotenuse be represented by the symbols 6, a, and c, respectively, we may express the six functions of the acute angle A in terms of a, b, and c as follows : , n A sm A = + . (6-1)' ^. , cos (6-2) v ' /c ox , tan (6-3) /a .. esc (6-4)J sec (6-5; ff > A AA = = c Side adjacent b ~ = r hypotenuse -r side opposite -7-5 ? = A hypotenuse .^ ,. = ^ , cot A = A side adjacent -r-; . side opposite ' c a T > c a > c T 6 side adjacent (6-6) > o side adjacent hypotenuse = r AA = -~- A = a - hypotenuse side opposite r , A = side opposite -r r^ = b a using (6-1) to (6-6), we can express the trigonometric funcan angle of a right triangle without reference to any coordinate system, since the ratios of the sides remain the same regardless of the position of the triangle. By tions of 6-3. PROCEDURES FOR SOLVING RIGHT TRIANGLES When is solving a right triangle in which two parts are known, it advisable to arrange the work systematically and to follow A definite 1. procedure consisting of the following steps a figure reasonably close to scale, and indicate the known 2. : Draw parts. Write an expression containing a trigonometric function which involves the two known parts and one unknown part. 1 1 8 Right Triangles and Vectors 3. Find the selected unknown part from 4. Find all unknown parts other Sec. 6-3 this equation. of the triangle by a similar procedure. Check 5. all results. Whenever possible, select a trigonometric function that gives a by means of a multiplication rather than a division. solution In the following illustrative examples, the acute angles are represented by the letters A and 5, and the right angle is denoted by C, while the small letters a, 6, and c, respectively, represent the sides a -658 opposite them. FlG> 6 _ 2< Example 6-1. Solve the The triangle Solution: is triangle ABC, if A = drawn approximately 2820' and a = 658. to scale in Fig. 6-2. The unknown parts are the angle B and the sides b and c. 6140'. 90 - 2S20' Since A 90, we have B To find the side b, we may apply either (6-3) or (6-6), since both equations +B = involve the because it = unknown b = A and the known parts and We a. shall use cot A= - means of a multiplication rather we have enables us to proceed to the solution by than a division. Since A = 2S20' and a = 658, Then b In this example, To we take find the side c, we = = 658 cot 2820 (658) (1.855) c we may use the = = 1220.59. A = shall use (6-4), or esc Hence, check, = This result b equal to 1221. CSC 28020' To / is rounded off to four digits. ** - We have, therefore, rrJjg. 658 esc 2820' (658) (2.107) relation cos = 1386. 1221 A = r^^ = loot) 0.8802. Hence, A = 2820'. 2 2 a2 Checking by means of the Pythagorean theorem yields the result b = c 2 = 2 = = = 6 whereas 1490841. (c a) (c + a) (1221) (728) (2044) 1488232, These values of b 2 agree when they are rounded off to three significant figures, = Note. In most situations where we must solve triangles, we are dealing with measured quantities, which are necessarily approximate. Therefore, our answers can be no more accurate than the See. 6-3 and Vectors Right Triangles 1 we begin with. If the original data are approximate, our answers must be rounded off to the degree of accuracy indicated by the data. In example 6-1, for instance, the answers may be given as b = 1220 and c = 1390, both rounded 19 data off to , three significant figures. FIG. 6-3. Example 6-2. Solve the Solution: The triangle ABC, shown conditions are if b = 250 and c in Fig. 6-3. Since - = 371. = C OKf| = j o/ = 0.6738. Therefore, A = 4738 / and B = 90 cos A, - 4738' = we have cos A 4222'. I To hence, find a, we have a we may use either unknown a with choice of combining the -r = tan A or - o = A sin c . We either 6 or c; shall illustrate, in order, the computation with each of these equations, thus providing a check on our work. Using Hence, a we have (6-3), = 250 tan 4738' = ^= tan 47=38'. (250) (1.096) = 274.0, or 274 when rounded off to when rounded off to three figures. Using (6-1), we have = sin 47 38 '' 371 Hence, a = 371 sin 4738' = (371) (0.7388) = 274.09 or 274 three figures. EXERCISE 6-1 In each of the problems from 1. a 3. b 5. o 7. a 9. a 11. b 13. a 15. 6 17. = 12, A - 33. = 62.4, B = 7110'. = 3.187, 6 = 6.249. = 4.318, B = 6716'. = 9.863, A = 3621'. = 78.21, A = 4317'. = 123.6, b = 783.1. = 2.312, B = 4057'. 1 to 16, solve the right triangle. 2. b 4. a 6. 6 8. b 10. 6 12. a 14. a 16. a = 168, A = 3816'. = 42, c = 76. = 63.21, B = 8336'. = 827.6, c = 963.4. = 16.32, B = 8710 = 43.21, c = 63.75. = 36.83, A = 5744'. = 389.3, 6 = 62.34. / . ? A wire stretches frorn. a point on level ground to the top of a vertical pole. It touches the ground at a point 15 feet from the foot of the pole and makes an angle of 63 with the horizontal. Find the height of the pole and the length of the wire. 18. A ladder 40 feet long rests against a vertical wall. If its footjs 5 feet from the base of the wall, what angle does it make with the ground? 120 19. Right Triangles A ladder 65 and Vectors Sec. 6-3 a window 35 feet above the held in the same to the other side of the is moved the and street, it will reach a top position window 28 feet above the ground. How wide is the street from building to feet long is placed so that it will reach ground on one side of a If the foot of the ladder is street. building? 20. The grade of a hill the tangent of the angle the is hill makes with the horizontal. Find the grade of a hill which is 275 feet long and which rises 120 feet. 21. To find the width of a river, a surveyor sights on a line across the river between two points A and B on opposite banks of the river. He then runs a line AC perpendicular to AB. He finds that AC is 250 feet and angle ACB is 4217'. How wide is the river? 22. Find the length of a side of a regular hexagon and the radius of the inscribed circle, if An the radius of the circumscribed circle is 10 feet, 560 feet while flying upward for 2,387 feet along an inclined straight-line path. What is the angle of climb? 24. A pendulum 4.5 inches in length swings through an arc of 28. How high does the bob rise above its lowest position? 25. A man 6 feet tall is walking along a straight horizontal path directly away from a lamp post 10.5 feet high. How far is he from the post at a certain instant 23. airplane rises when his shadow is 5 feet long? Horizontal B """*,, Horizontal FIG. 6-4. 6-4. ANGLES OF ELEVATION AND DEPRESSION Let the line an observer AB in Fig. 6-4 be a level or horizontal line, at the point see an object at the point C. A and let If the C is above the horizontal line AB, then the angle BAG measured up from the horizontal to the line of sight AC is called the angle of elevation to C from A. If the object C is below the horizontal line AB, then the angle BAG measured down from the' object horizontal to the line of sight to C from A. AC is called the angle of depression Example 6-3. From a point on the ground 300 the angle of elevation to Solution: In Fig. = tan 2210'. Hence, ing is 122 feet high. its top is 22 10'. How feet high is from the base of a building, the building? a 6-5, we have b = 300 and A = 2210'. By (6-3), 300 a = 300 tan 2210' = (300) (0.4074) = 122.22. So the build- Sec. 6-5 Right Triangles and Vectors 121 30 W- B \22IQ' S FIG. 6-7. FIG. 6-5. 6-5. BEARING IN NAVIGATION AND SURVEYING In marine and air navigation and in surveying, the direction in which an object is seen is expressed by the bearing or azimuth of the line of sight from the observer. The bearing of a line is the acute angle which its direction makes with a meridian or northsouth line. Such angles are sometimes called quadrant angles, or quadrant bearings. we first To describe the bearing write the letter the letter E or W. The of a given direction, or S, then the acute angle, and finally letters depend on the quadrant in which N the given direction falls. Thus, the bearings of the lines OA, OB, 45 W, respectively. and OC in Fig. 6-6 are N 60 E, S 37 W, and N from its bearing only in that the at north in a clockwise azimuth is the angle measured from and 360. direction. An azimuth may have any value between Thus, in Fig. 6-6 the azimuths of the lines OA, OB, and OC are 60, 217, and 315, respectively, measured clockwise from the north. This method of measuring directions is coming into more The azimuth of a line differs frequent use than that of quadrant bearings. We note also that the is often used instead of azimuth. Thus, we may speak term bearing of the bearing of an object regardless of whether or bearing as here defined. we mean azimuth Example 6-4. A ship heads due east from a dock at a speed of 18 miles per hour. After traveling 30 miles it turns due south and continues at the same speed. Find * its distance and bearing from the dock after 4 hours. Solution: In Fig. 6-7, let A be the point at which the dock is located, let B be the point where the ship turns south, and let C be the position of the ship after 4 hours. orj Since the number of hours required to travel of hours spent in travel from B to C is 4 from - 5^ = 7 ^ A to B is ^5 lo = Hence, a =' 18 f* ~ > the number o ^ o = 42. 1 From the figure, tan0 = 42 B = and b = 30 sec 5428' Hence, the distance from the dock 14428' or S 3532' E. is Sec. = ~~ (30) (1.721) = and 30 Therefore, 6-6. and Vectors Riaht Triangles Right Trianales 22 sec 6 6-5 5428', = 51.6. 52 miles and the bearing of the line AC is PROJECTIONS desirable to consider direction along a line segment. underThus, if Pi and P2 are the end points of a segment, we shall P directhe stand PiP 2 to mean the directed segment from PI to 2 Often it is , by the order in which the end points are named. The non-negative length of the segment PiP2 is denoted by |PiP2 |. as a Frequently a directed segment PiP2 may lie on a line, such tion being specified coordinate axis, on which a positive direction has been specified. Then the positive direction on the line may agree with the direction from PI to P2 or the two directions may be opposite to each other. The directed length of the segment PiP 2 is equal to |PiP 2 when the directions agree or when PI and P 2 coincide and is equal to -|PiP 2 when they disagree. Since the context will make the meaning clear, , | | we shall designate the directed length of the by PiP2 segment PiP2 also . M B FIG. 6-9. FIG. 6-8. We recall that the projection of a point on a given line is the foot in of the perpendicular dropped from the point to the line. If B the if is and line the on of the projec6-8 is I, PI Fig. projection tion of P 2 on I, then the directed segment from A to B is the A projection on I of the directed segment PjP 2 We draw PiM parallel between I and or perpendicular to P2 , to show the angle I, . to Sec. 6-6 We on the Right Triangles shall now assume line L and Vectors 123 that a positive direction has been specified P\M = AB, it follows immediately from Then, since trigonometry that AB = PiP2 | where | cos 0, the acute angle between the positive end of I and the positive half -line determined by the directed segment P\P* In Fig. 6-8 it is considered that I is positively directed toward the right. The is result just given can be applied, as seen in Fig. 6-9, in find- ing the projections of PiP2 upon the coordinate axes. The directed lengths of the projections upon the cc-axis and the 2/-axis are, respectively, (6-7) AB = PiP2 (6-8) CD = PiP2 | | where | cos 0, sin 6, OX and PiP2 as shown in Fig. 6-9. the angle between the coordinates of the end points of the segment PiP2 are 6 is When known, the projections of these coordinates. AB From AB = (6-9) x2 , CD are readily expressed in terms the definitions of horizontal and vertical and distances given in Section 2-2, C | it follows that and xi CD = y* - yi. P,(3,2) (0,2) B(7,Q) A (3,0) \67 P2 (7,-5) 0(0, -5) FIG. 6-11. FIG. 6-10. Example 6-5. What jare the projections of the segment PiPa on the Pi = (3, 2) and P 2 = (7, - 5)? Solution: = -5-2=-7. Also, since AB = x 2 - %i = 7 - 3 = Since AB is + 4, we know that AB In Fig. 6-10, CD = - 7, we know that CD is directed 4, is and CD = t/ axes, 2 - if 2/i directed to the right. downward. 1 24 and Vectors Right Triangles Sec. 6-6 Example 6-6. A ladder 12 feet long leans against the side of a house and makes an angle of 67 with ground. Find its projections on the ground and on the side of the house. The conditions are represented in Fig. 6-11. Let of the ladder. The required projections are found as follows. The projection on the ground is given by Solution: x The = I = cos 6 12 cos 67' projection on the side of the house I sin 12 sin 67 y = = = 12 (0.3907) given by = 12 (0.9205) I = 12 be the length = 4.7. is = 11.0. EXERCISE 6-2 1. Two points A and B are 5,000 feet apart and at the same elevation. An airplane 10,000 feet directly above point A. Find the angle of depression from a and the airplane's distance horizontal line through the airplane to point is B from point B. 2. The Washington monument is approximately 555 feet high. Find the angle of elevation to the top of the monument from a point that is 621 feet from the base of the monument and at the same elevation as the base. a kite is 130 feet above the ground and 150 feet of string is out, find the angle of elevation to the kite, assuming the string to He on a straight line. 3. If 4. 5. Find the angle of elevation to the sun 63 feet long on horizontal ground. if a flagpole 95 feet high casts a shadow A boat leaves its dock and heads N 52 W for 4 hours at 14 knots (1 knot = 1 nautical mile per hour = 6,080.4 feet per hour). It then turns and heads N 38 E for 3 hours at 16 knots. Find the boat's from the dock. 6. The grade rise 7. 8. of a certain railroad bod is 0.1095. final bearing and distance How many feet does a locomotive while traveling 175 feet along the track? An approach must be built up to the end of a bridge which is 40 feet above ground. If the approach is to have a 10% grade and the original ground is assumed to be level, how far from the end of the bridge must the approach start? A surveyor wishes to find the distance between two points A and B separated by a lake. He finds a point C on the shore of the lake such that angle ACB is 90. He measures AC and BC and finds that AC is 640 feet and BC is 285 feet. How far apart are A and B? 9. A smokestack is 175 feet from a building. angle of elevation to the top of the stack base from the same window From a window of the building the 2810 The angle of depression to / is . 2430'. Assuming that the ground is level, find a) the height of the window above the ground and b) the height of the smokestack. its 10. is N Two ships leave the same port at the same time. One travels 42 E at 25 knots. The other travels S 48 E at 33 knots. How far apart are the two ships after 4 hours? Sec. 6-7. 6-7 Right Triangles SCALAR We AND VECTOR and Vectors 125 QUANTITIES shall at this point find it necessary to distinguish carefully between two kinds of quantities, namely scalar quantities and vector quantities. A scalar quantity is a quantity whose measure can be fully described by a number. It is a quantity which can be measured on a real number scale. For example, temperature is a scalar quantity, measured on the scale of a thermometer. Also we shall define the scalar components of the segment PiP2 to be the projections on the coordinate axes, or the directed lengths x 2 Xi and y 2 3/1 given by (6-9) in Section 6-6. The student should note, however, that the scalar components of P2 Pi are not equal to those of P\P2 The components of PJP\ are x\ x 2 and yi y 2 and are the negatives of the respective components of PiP 2 . . A vector quantity, or simply a vector, is a quantity possessing both magnitude and direction. A vector may be represented by any one of a set of equal and parallel line segments. Algebraically, a vector described by the scalar components of any segment all such segments have the same components. We shall, in fact, call these the scalar components of the vector and shall is fully representing it ; ./ o FIG. 6-13. FIG. 6-12. enclose vector them in brackets. In Fig. 6-12, three representations of the The arrowhead indicates the order in [4, 3] are shown. which the end points of the line segment are named. We may denote a vector by a single letter in boldface type, say v, or may represent it geometrically by any one of the segments, such as AB, OP, or CD. Here A, O, and C represent the initial points of the three segments, while the terminal points at the arrowheads are B, P, and D. Actually, any other segment with the same magnitude and direction could have been selected to represent the vector v. Instead of using a single letter in boldface type to denote a vector represented by a segment PiP2 we may also use the notation PiP2. , 1 26 Right Triangles and Vectors Sec. 6-7 We have seen that a vector is unambiguone of a set of equal and parallel line ously represented by any segments. Hence, any point may be taken as the initial point of a segment which represents a vector. If the origin is so chosen, the coordinates of the end^point P(x,y) are actually the scalar components of the vector OP. We can therefore give the following simple definition of the magnitude of a vector and its expression Properties of Vectors. in terms of scalar components : The magnitude, or length, of a vector v of one the segments representing v. any Definition. Let v where Since is the length of = v\ [yi, v2 ] be the vector represented by OP in Fig. 6-13, and v 2 are scalar components. Then we have |v| = \OP\. OP is the hypotenuse of a right triangle, = v (6-10) | | VV + v2 2 . In case the scalar components are given by the coordinates (x, y) of the end point P of the segment, the magnitude of the vector is From it follows that two vectors are their and if equal only respective scalar components are equal. For example, u = [u lt u2 ] equals v = |>i, v2 ] if and only if HI = Vi and u2 = v 2 the definition of a vector, if . = [0, 0] to be the zero vector. also define a special vector It corresponds to the exceptional case in which 2 coincides with We P PI and may be considered as represented geometrically by a segment of length zero, that is, by a point. The zero vector may be regarded as having any direction whatsoever. v = [ Vi, v 2 ] is If a vector v. = [v lf v 2 ] is given, the vector defined to be the negative of v. Thus, ifj^ift = v denotes a vector represented by the segment PiP2 then P2 Pi = v denotes a vector having the same length as PiP2 but oppositely directed, namely, from P2 to PI. We note that , - (- A If is unit vector u= [fa, u^ is is Y) = v. defined as a vector any non-zero whose magnitude vector, then -, is unity. = \T^I r^]\ > a unit vector. Multiplication of a vector by a scalar is performed by multiplying the magnitude of the vector by the absolute value of the scalar, maintaining the direction of the vector if the scalar is non-negative and reversing it otherwise. Thus, by k[ui, u^] we mean the vector Sec. 6-7 Ri ghf Triangles [kui, ku2], Using the equation [MI, and Vectors u2 = ] any vector v = Oi,^2 ] as proportional = V^i 2 + ^2 2 Changing v to choose A: - normalizing Sums and k 1 27 [~ yl we can express ? i to a unit vector v/fc if a unit vector v/fc we is called v. Of the many questions which Differences of Vectors. arise in the study of vectors, the one of greatest importance for us at present concerns the addition of vectors. The sum, or resultant, of two vectors is defined to be the vector which has for its scalar components the sums of the scalar components of the two vectors. Thus, the sum of the vectors u and v is given by the relationship u (6-11) It is + = v [MI + vi, u2 + v 2 ]. important to note that some of the laws of the algebra of also hold for vectors. Thus, the commutative law is numbers u + v = v Also, the associative law of addition (u And, for every vector + v) +w= u + - v) + u. is + (v + w). v, v ( = 0. That these laws are satisfied for vectors can be shown geometrically or can be seen from (6-11) by virtue of the known laws of addition of real numbers. In terms of components, the rule for multiplication by a scalar is given by In consequence of this definition, the following algebraic laws are satisfied : + v) = = (k + m)u fc(u fc(mu) = fcu fcu + &v; + mu; (fcw)u. To find the sum of two vectors geometrically, we proceed as indicated in Fig. 6-14. This graphical representation of the sum of two vectors by means of the triangle construction was probably suggested by the behavior of physical quantities represented by vectors, such as forces, displacements, velocities, and accelerations. Their addition is effected by the triangle law or the parallelogram law. 128 Right Triangles and Vectors Sec. 6-7 1C o FIG. 6-15. FIG. 6-14. The following 2. steps indicate the actual procedure. segment representing one of the vectors, say u, with its initial point at the origin of the coordinate system. Place the initial point of a segment representing the second 3. vector, say v, at the terminal point of the first segment. Draw the segment from the origin to the terminal point of the 1. Select a segment for u 4- v. The v. This segment represents the resultant vector, v of two vectors difference u u If u = [ui, u%~\ and v [y\, v%] U =u+ v V = , manner defined in a is analogous to that used in defining the difference of numbers. Thus, v). ( then [U\ Vi, U2 V2\. To find the difference u v of two vectors geometrically, we proceed as indicated in Fig. 6-15. The steps are as follows 1. Place segments representing u and v with their initial points at the origin. v at the 2. Place the initial point of a segment representing terminal point of the segment representing u. Note that the segment representing v is parallel and equal in length to that for v, but has the opposite direction. 3. Draw the segment from the origin to the terminal point of This segment represents the vector v. the segment for : u v. Example 6-7. Find the sum and Solution: The difference of the vectors required vectors are u+ v = [5 - 2, - 3 + 1] = [3, - 2], and u-? = |5-(-2), -3-l] = [7, -4]. u = [5, - 3] and Sec. 6-7 Right Triangles Example 6-8. Express u Solution: The = [4, 3] and Vectors 129 as proportional to a unit vector. expression representing the vector is Q~1 - > [A note that the magnitude of -r > From Section 6-7, . To check, is -r To add vectors analytically, of the vectors to be added. that - we first find the scalar components we know to*) segments representing u and v make angles a and /?, respectively, with the #-axis, the scalar components are given by the projections upon the coordinate axes. Thus, as shown if FIG. 6-16. in Fig. 6-16, ux = | u | cos a cos ]8 ; = t I u I sin a, and vx = | v | vy = | v Then the components of the resultant algebraic sums X= Ux +V Y = x, Thus, the magnitude of the resultant |R| Uv + R is | sin will ]8. be given by the Vy. given by = Also, the direction angle 6 satisfies the relationship tan 6 = Y To determine the quadrant of in mind and use the correct signs - =1 1 j= important to keep and Y. For example, tan correctly, it is of X might lead one to an incorrect value 45 for 6 instead of the correct third-quadrant angle 225. The student should note that this (analytic) procedure and the geometric procedure (parallelogram law) give the same results. This can be shown by appropriately combining the scalar components of the vectors u, v, u + v, and u - v in Figs. 6-14 and 6-15, where illustrated the geometric procedure. we 130 Right Triangles and Vectors Sec. 6-7 o W- -*-E o ^* C \25 D \ s FIG. 6-17. FIG. 6-18. A block weighing 500 pounds rests on a smooth plane making an with the horizontal, as indicated in Fig. 6-17. What force, parallel to the plane, is necessary to hold the block in position? Example 6-9. angle of 25 The force due to the weight mayjbe represented downward through A, such as vector AC. The component of AC which is parallel to the inclined plane and which must be overcome is represented by the projection AB of AC on the inclined plane. Since angle Solution: Let the block be at A. by a vector acting vertically BAG = 90 - 25 = 65, AB = 500 cos 65 = (500) (0.4226) = 211.3. Hence, a force of 211 pounds must be applied parallel to the inclined plane to keep the block from sliding. (It is assumed that three Example 6-10. Find the magnitude and direction of the resultant of a force of E 110 pounds acting in the direction S 4627' the direction 2914' W. significant figures are appropriate.) and a force of 100 pounds acting in N Solution: The given forces are represented by vectors in Fig. 6-18. Let F denote the magnitude of the resultant force and 6 the angle it makes with OE. Since 90 - 4627' 4333' and 90 - 2914' 6046', the components Fx and = Fy = of the resultant are found as follows: Fx = 110 cos 4333' = (110) = 79.73 Fv = = = - (0.7248) - 100 cos 6046' (100) (0.4884) -48.84=30.89; + 100 sin 6046' + (100) (0.8726) + 87.26 = 11.47. 110 sin 4333' (110) (0.6890) 75.79 Now tan*=^ Therefore B = 2022', the magnitude of = and F is 33 11.47 = ~~ 30.89 F= v '"' *"' 0.3713, 11.47 esc 2022' pounds, and its ^ and """ = direction = F 6 ~~ esc v 11.47 (11.47) (2.873) is N 6938' E. = 32.95. Hence, Sec. 6-8 Right Triangles and Vectors 131 EXERCISE 6-3 1. Draw a diagram showing three different segments representing each of the given vectors. Find the magnitude of each vector. b. [12, 5]. a. [3, 4]. - c. [-2, d. 4]. h. [- 3, - - 5]. e. [1, 2. Express each of the vectors in Problem 1 in terms of a unit vector. In each of the following cases, add the given vectors. Find the magnitude and 3. f. 1]. [5, 3]. g. [0, 4]. [0, 1]. direction of the resultant. and [2, 3]. b. [4, - 2] and [1, - 10]. and d. 3]. [- 6, [2, 0] [1, 2], [4, 3], and [0, 7]. In Problem 3, parts (a), (6), and (c), subtract the first vector from the second, and find the magnitude and direction of the resultant. In Problem 3(d), subtract twice the third vector from the sum of four times the first vector and twice the second vector, and find the magnitude and direction a. [1, 1] c. 4. 5. of the resultant. 6. A force pounds acts at an angle of 63 with the horizontal. What are the and horizontal components of the force? a ship sails N 48 at 30 knots, what are its westward and northward of 40 vertical 7. If W components? 8. One force of 28 pounds acts vertically upward on a 43 pounds acts horizontally on the particle. resultant force, 9. 10. and what is its What particle. is Another force of the magnitude of the direction? A barrel weighing 160 pounds rests on a smooth plane which makes an angle of 22 with the horizontal. Find the force parallel to the plane necessary to keep the barrel from rolling down the plane. Three forces act on a particle. One of 50 pounds makes an angle of 25 with the horizontal; a second of 60 pounds makes an angle of 50 with the horizontal; of 75 pounds makes an angle of 230 with the horizontal. Find the magnitude and direction of the resultant force. 11. Four forces act on a body. The forces are 30, 45, 50, and 65 pounds, and they and the third make angles with the horizontal of 25, 160, 240, and 330, respectively. Assuming that all the forces lie in the same vertical plane, find the magnitude and direction required force of the force necessary to hold the body in equilibrium. The equal in magnitude and opposite to the resultant of the given is forces. 12. A an angle of elevation of 37. Its initial velocity is 2,500 feet second. the Find horizontal and vertical components of its initial velocity. per 6-8. shell is fired at LOGARITHMS OP TRIGONOMETRIC FUNCTIONS in this chapter, we have considered the use of a table Qf natural trigonometric functions and have solved various problems involving right triangles. In many problems, however, the computation is greatly facilitated by the use of logarithms to perform the numerical operations. For this purpose the values of the logarithms of the trigonometric functions are required. Table III might be used to obtain the logarithms of the functions found in Table II, So far 132 Right Triangles and Vectors Sec. 6-8 but the work is considerably lessened by the use of Table IV at the end of this book, which gives the logarithms of the trigonometric functions at once. Table IV a four-place table giving the logarithms of functions to 90, For the sine and cosine from is at intervals of 10 minutes of and 90, the tangent of any angle between any angle between and 45, and the cotangent of any angle between 45 and 90, than 1; hence, the logarithms of 10 after the these functions are negative, and we must write tabulated entry. For the sake of uniformity, 10 has been added to each of the other entries in the table. In using the table, therefore, 10 must be subtracted from every entry. The method of using Table IV is similar to that described for Table II and will be illustrated by the following examples. the value of the function Example 6-11. Find log sin 2310'. Since this angle Solution: is less we given in Table IV, is find that log sin 23 10' = 9.5948-10. Example 6-12. Find From Table IV we Solution: 10' The x log cot 5127'. tabular difference = 0.7(26) = 18.2, obtain the values for the following tabulation: log cot 5120' log cot 5127 log cot 5130' is 26. Since ; - 9.9032 - 10 = 9.9032 - 10 = 9.9006 - 10 5127' 6 if Hence, to 5130', log tan 6 .0018 = = = 9.7816 - + x) = 9.7827 log tan 3120' = 9.7845 log tan 10' x way from 5120' 9.7827-10. The positive part 9.7827 lies between the entries 9.7816 and 9.7845 The procedure for finding 6 may be indicated as follows: log tan 3110' TT of the = (9.9032-10) = 9.9014-10. Example 6-13. Find the acute angle Solution: -^ 26 x and we have log cot 5127' in Table IV. is - II 31(10 f . > and 8 = 3110' + 55(10') 10 11 10 10 29 Sec. 6-9 Right triangles and Vectors 1 33 EXERCISE 6-4 In each of the problems from 1 to 15, find the value of the given logarithm. 2, log sin 2120'. 3. log cos 8620'. log sin 4820'. 4. log tan 8830'. 5. log cot 1020'. 6. log sec 4350'. 7. log sin 1326'. 8. log sec 4857'. 9. log tan 4114'. 10. log esc 7832'. 11. log cot 6843'. 12. log cos 1818'. / 13. log esc 8316'. 14. log cos 18'. 15. log tan 5134 In each of the problems from 16 to 35, find the angle (or angles) 6 between 1. . and 360. = 8.8059-10. = 9.9959-10. = 0.4625. log tan 9 = 9.8483-10. log esc 6 = 0.4081. log sin 9 = 9.9567-10. log cos 9 = 9.9755-10. log sec 9 = 0.1967. log cos 9 = 9.1860-10. log sin 9 = 9.9974-10. 16. log sin 6 18. log cos 6 20. log sec 21. 22. 23. 24. 26. 28. 30. 32. 34. 6-9. = 8.3661MO, = 9.1697-10. = 0.4882. log cot 6 log tan 9 = 0.1430. log sec 6 = 0.3586. log tan 9 = 9.7648-10. log cot 9 = 9.8666-10. log esc 6 = 0.3370. log cot 9 = 1.5976. log sec 9 = 0.3870. 17. log tan 6 19. log sin 6 25. 27. 29. 31. 33. 35. LOGARITHMIC SOLUTION OF RIGHT TRIANGLES The solution of a right triangle by means of logarithms the same as by natural functions, except that is exactly for the actual numerical computation a table of logarithms of the natural functions is used in conjunction with a table of logarithms of The following example numbers. will illus- trate the procedure. Example 6-14. Solve the The values Solution: B = 90 - 2140' = ABC, right triangle which in A = known parts are indicated in To find the side a, we have of the 6820'. tan 2140' = ^ 2140' and b Fig. 6-19. = 8.43. We see that > 8.4o Hence, a = 8.43 tan 2140', or log a = log 8.43 Arrange the work as follows: log 8.43 . Therefore, a To = find side log tan 2140' log a -f log = = = tan 2140'. 0.9258 (Table III) 9.5991-10 (Table IV) 10.5249-10 3.35. c, (Table III) we have = c cos 2140'. 1 34 Right Triangles Hence, and Vectors Sec. 6-9 8 43 . C and log c ' cos 2140' = log 8.43 - Jog cos 2140'. Arrange the work as follows: log 8.43 cos 2140' log log c Therefore, c = = = 10.9258-10 (Table III) 9.9682-10 (Table IV) 0.9576 = 9.07. (Table III) EXERCISE 6-5 In each of the problems from I. 4. 7. 1 = 100, A = 31. A = 8817', c = 108.1. J3 = 279', a = 36.13. 8. 6 a railroad track 30 6 9. If rises 2. c 5. c to 8, solve the given right triangle. = 3.45, a = 1.76. = 6.275, B = 1845'. = 98.34, B = 1848'. feet in 3. A = 6. a = = 63.4. = 396.3. 2520', a 645.3, b a horizontal distance of one mile, find the angle of inclination of the track. 10. A II. A force of 628 pounds acts at force of 341 pounds and another force of 427 pounds act at right angles to each other. Find the magnitude of the resultant force and the angle it makes with each of the forces. 180, and a force of 237 pounds acts at 270. Find the direction and magnitude of the resultant. 12. An airplane is flying due east at a speed of 485 miles per hour, and the wind is blowing due south at 33.6 miles per hour. Find the direction and speed of the phane. 13. The westward and northward components of the velocity of an airplane are 363 and 487 miles, respectively. Find the direction and speed of the airplane. 14. The eastward and southward components of the velocity of a ship are 10.4 and 16.8 knots, respectively. Find the speed of the ship and the direction in which it is 15. A moving. Find 16. 17. pounds is acting at an angle of 4713' with the horizontal. horizontal and vertical components. force of 2673 its A force of 162.4 pounds is just sufficient to keep a block at rest on a smooth inclined plane. If the block weighs* 783.1 pounds, find the angle at which the plane is inclined to the horizontal. Two tangents are drawn from a point P to a circle whose radius is 14,32 inches. the angle between the tangents is 3228', how long is each tangent segment? 18. Two buildings of the same height are 11,640 feet apart. When an airplane is If 19. 8,000 feet above one of them, what is the angle of depression to the other one? cable which can withstand a pull of 10,000 pounds is used to pull loaded trucks up a ramp. If the angle of inclination of the ramp is 3616', find the A weight of the heaviest truck which can be safely pulled up the the cable. 20. The angle is 4528'. How high of elevation from one point From a is on level ground to the top of a flagpole point on the ground 25 feet farther the pole? ramp with away the angle is 3956'. Trigonometric Functions of Sums and Differences 7-1. DERIVATION OF THE ADDITION FORMULAS Heretofore, we were concerned with relationships between trigonometric functions of a single angle. We shall now establish certain fundamental identities involving two angles, in terms of the functions of the single angles. The following identities express functions of the sum and difference of two angles in terms* of the functions of the separate angles. (7-1) sin (a j8) = sin a (7-2) cos /3) = cos a cos (7-3) sin (a /3) = sin a (7-4) cos (a /3) = cos a cos (7-5) tan (a p) = - ^ p) K/ = + (a + , + , tan (a v (7-6) We . ^ cos cos tan tan 1 + + oj tan cos a sin a sin /3, cos a sin sin a sin j8, /3 /3 ft + tan tan 1 |8 a a /3, |8, ] tan tan sin p j8 tan /3 now prove the formulas for the sine and cosine by using the derivation developed by E. J. McShane. 1 shall o FIG. 7-2. FIG. 7-1. 1 E. "The Addition Formulas for tho Sine and Cosine," J. McShane. American Mathematical- Monthly, Vol. 48 (1941), pp. 688-89. 135 Trigonometric Functions of 136 Sums & Differences and a be any two angles with the same Let shown On in Fig. 7-1. their terminal sides we Sec. initial side 7-1 OW, choose points P as and Q, respectively, each at unit distance from 0. d represent Let distance the from P to Q. We shall now make two computations for d 2 using first OW, and then OP, as the 0[cos(-0). sin (a -0)] , OW is used as the x-axis When of a coordinate system, as shown in Fig. 7-2, we find that the coordinates of P and Q are (cos 0, sin /?) and (cos a, sin a), respectively. Hence, by the distance formula, d2 = = Since a cos a cos 2 a + + cos 0) 2 (sin 2 cos a cos (cos a sin 0) 2 + 2 sin 2 a = cos = d2 cos + 2 + 2 sin 2 (cos 2 FIG. 7-3. a a sin 2 = 2 1 a sin + sin 2 0. , + cos 2 sin sin a sin 0). Let us now use OP as the #-axis, as shown in Fig. 7-3. Then the coordinates of P are (1,0) and those of Q are [cos (a /3), sin (a 0) d 2 ] Hence, . = = = - [cos (a cos 2 2 - - 0) (a j8) 2 cos (a - + sin 2 2 I] (a '2 cos (a - j8) 0) +1+ sin 2 (a 0) 0). 2 Equating the two expressions for d yields 2 2(cos + a cos a sin j8) =2 sin 2 cos (a |8). Therefore, cos (a (7-4) 0) = This establishes (7-4). Setting a = 90 in (7-4), cos we cos (90 (7-7) If in (7-7) we let /3 = 90 - sin (90 (7-8) From (7-7), (7-8), sin (a + 0) = = = a + cos sin a sin ]8. find that - 0) = sin 0. we have - 7) = cos 7. y, and (7-4), we obtain cos [90 cos [(90 cos (90 - (a + 0)] - a) - 0] - a) cos + sin (90 - a) sin 0, Sec. & Trigonometric Functions of Sums 7-1 137 Differences or sin (a (7-1) + /3) = + a cos /3 sin a sin cos /3. This establishes (7-1). = Since cos (-)8) cos /3 and sin (-/?) = -sin , we have, as a con- sequence of (7-4), cos (a + /3) = = cos [a cos a cos = cos ( |8)] ( j8) + a sin sin ( j8) ( ]8) or cos (a (7-2) sin To prove a cos TT ]8) each /3, a a. cos sin a cos j8) ( + cos |8 We may obtain cos a ]8. a sin a cos B 5 cos a cos p sin numerator the of sin /3. + = sm COS fl a sin ]8 r ~# sin a sm p cos and denominator by a sin )8 a cosjg _ tan a + tan ff 1 tan a tan /3 sin a sin ff __ cos a cos 8 a -+ tan pQ /o\ = tan + ^|8) (a v y ^5 1 tan a tan p cos ^ ' / , . manner from (7-3) and (7-5). Example 7-1. Find the exact value Solution: Substituting 45 sin 75 for = sin = sin of sin 75. a and 30 for (45 + 30) 45 cos 30 + cos 45 V2 V3 in (7-1), _\/2 "2*2^2*2 .-= Therefore, 75 ' '- cos (7-6) in a similar sin and (7-1) C7 _, , tan sin use the relationship tan 6 term we have sin a cos j8 cos a cos j8 cos a cos j8 cos a cos B (7-5)y v from sin j8 fi) ^ or sin , Hence, t* cos follows that it = = we (7-5), , tan Dividing (a: a We then obtain m sin (a + (a + p) = 7 r^\ ~ cos (a + p) and (7-2). cos ]8) from (7-1) Similarly, (7-3) + = V (V 4 3 + D- 1 we obtain sin 30 (7-4), Trigonometric Functions of Sums 138 Example 7-2, Find the exact value angle such that sin = a and , O of tan + (a. Differences /3) a if a third-quadrant angle tan is = It follows that tan 4. = - a ~ x m = + 6) / 4 + ' Using 12 (7-5), = - 7-^-7 , tan (a 3 j 7-1 Sec. a second-quadrant is ft 5 = \JL Let a be a second-quadrant angle for which y Solution: Hence, x 3 = & = 3 and = r 5. we have 16 ^ EXERCISE 7-1 In each of the problems from to 8, find the exact value of the given function. 1 135. 15. 105. 1. cos 75. 2. tan 105. 3. sin 5. tan 195. 6. cos 195. 7. tan 15. is a second-quadrant angle, and a 9. If sin a a third-quadrant angle and is = - 3 and cos -r O = ft /3 5 rz find > lo sin (a -f = r and a - B = 45, find tan 0. = 180, find tan 0. If tan a = 3 and a + 3 = - find sin (a + If tan a = j and tan 4 10. If j(S), 4. sin 8. cos cos (a -f 0), and tan (a -f |8). a tan ,4 11. 1 12. > j8 jft o j8) and cos (a + )8), where a and are first-quadrant angles. 13. If cos a. = - and = cos J > o find sin (a j9) and cos (a j8), where a and are acute angles. 14. If sin a =4 -= and cos (a 15. If sin a 17. ^r LZ > cos j8 = 24 ^rr ~ > + ft) > cos sin j8 |3 = -1 and a and is obtuse, 7^ A/2 = -jp > a Id and tan (a 45) and a and - (sin is obtuse, 22. cos are both obtuse, find sin (a + j3 is acute, find sin (a /3) 0) and ]S are obtuse, find sin (a -f j3) and and is in the third quadrant, find 0). identities: a 19. cos (a cos a). Sin(a sin /3 and > 5 = Prove each of the following 20. a > ^O 2 O 18. sin (a > j9). If cos a = cos (a + |8). 3 If sin a = - tan (a 5 = tan # -f 0). = ^3 cos (a 16. > O 7?=cot|8-cota. a sin p 2 a = cos 2 a - sin 2 a = 2 21. tan cos 2 a - 1 = 1 - (a \ 2 TT) = - cos a. =i-*^. +-) 4/ 1 ~ tan a sin 2 a. 7-2 Sec. o no 23. sin 2 sin 25. ( - Trigonometric Functions of Sums & Differences a a cos (a s 97 fo ( a sin (a 9Q 29 cos ( a =2n sin a cos a. -- sin (a "" 0) rin (a f + a a tan + 0) ---~ = tan a + tan p. - cos a; _ cos^a jS tan 0) ~~ P) 4- + + tan a tan I ' tan a a cot ft $ tan " 1 ' _ 2 p cos (A C cos 2 + 35. sin 2 7-2. 1 cot (3 /?) ft cos (a ^ ^ sec a. ct sin (a cos l^TTtSJ cos 2 cos + B + C) = sin A cos cos (7 -f cos A sin 5 A cos B sin (7 sin A sin 5 sin C. sin 3 a = sin 5 a cos 2 a cos 5 a sin 2 a. = cos 2 ft cos (a - 0) cos (a 4- 0) = cos a - sin 32. sin 34. + 1 g) sin (a cos 2 33. " a - tan 39 cos 2 : 26. ft tan ft) ' tan a, . -ft) - a ----= a n 2a sin - 0) +-_ /3) + sin (a ~_ = + p) + cos (a - p) * an 0= + tan + _ si <n 24. 1 ~~ a cos2 sin 2 a. 2 ~ + a sin 2 ' THE DOUBLE-ANGLE FORMULAS we =a in (7-1), (7-2), and (7-5), we obtain functions of twice a given angle in terms of the functions of the angle itself. If Thus, let ft we have the following identities sin (7-9) (7-10) 2a = a 2 sin cos 2 a = n tan 2 a = cos 2 : cos a, sin 2 a, a and /- * * \ (7-11) We may 2 tan Q! tan 2 a obtain two other useful forms for cos 2 2 using in turn cos by forms are given by the a (7-12) sin 2 1 a and sin 2 a= 1 a from (7-10) cos 2 a. These identities - cos 2 a = 1 cos 2 a = 2 cos 2 2 sin 2 a, and (7-13) The following applications of 1. an indication of the possible illustrations give (7-9), a - (7-10), and The student should (7-11). study them carefully. sin 4 a = sm a = oj cos - = sin 2 (2 a) . sin 2 rt cos 2 /a\ ( ^ ) /a;\ ^ j = 2 sin 2 = 2 = cos 2 . sm 9 a cos a c <> s a - 2 a, a; o . ' 9 sin 2 a g 1 Trigonometric Functions of Sums & Differences 40 Sec. 7-2 Example 7-3. Find the exact value of sin 120 by means of a double-angle formula. We Solution: use (7-9) to obtain sin 2(60) sin 120 = = /1\ (2) \ V Example 7-4. 2 sin 60 cos 60 _ - Derive a formula for cos 3 a in terms of cos a. Solution: Applying the identity for cos (a -f and the double-angle formulas, /3), we have cos 3 a = cos (a + 2 a) = cos a cos 2 a sin a sin 2 a = cos a (2 cos a 1) sin a (2 sin a cos a) = 2 cos a cos a 2 sin a cos a = 2 cos a (cos sin a) cos a = 2 cos a (2 cos a 1) cos a = 4 cos a 3 cos a. 2 3 2 2 2 a; 2 3 _. - Example r 7-5. ,1 -r* i , sin 3 , cos rz Prove the identity sm 30 _ -r- 2. cos Solution: First combine the fractions on the left side to the right side. Thus, sin 30 sin __ __ sin 3 cos_3_0 ~" cos - sin 20 _ ~~ sin _ ~~ - sin (3 __ "~~ result 0) cos sin 2_smJ!0 2 sin 2 cos sin cos 3 cos sin ~~ 7-3. - cos and then reduce the THE HALF-ANGLE FORMULAS Functions of an angle in terms of the functions of twice that angle can be obtained directly from (7-12) and (7-13) If cos 2 a = 1 2 sin 2 a is solved for sin a, we obtain . sin a = /I A d= 4/ Also, = by solving cos 2 a cos 2 cos 2 a = a a 1 for cos a, we have +cos2. ., , ' cos 2 ^ r 2 Since these formulas may be equally well regarded as expressing functions of half an angle in terms of the functions of the given angle itself, the same relationship is retained if the identities are written ,~ * A \ (7-14) sm OL ^ = d /I cos a A / * T- cos a A and 1K x (7-15) v* / cos - = . db Sec. Trigonometric Functions of Sums & Differences 7-3 141 These are the so-called half-angle formulas for the sine and cosine. From (7-14) and (7-15) we obtain, by division, , ^x tan (7-16) ^ ' a sin __ ~~ 2 cos The algebraic signs mined by the quadrant If we in A The student / y 2 (7-15), and (7-16) are deter- 1 - (1 + , numerator and the denominator of we cos 2 obtain a __ - coga)2 (7-17) dropped the same A sin 2 / y (1 + a coga)2 will note that, in deriv- (7-18), we have sign in each formula. The validity of this step should be verified by consideration of the signs cos a. Thus, of tan a/2, sin a, and 1 tan a/2 and sin a necessarily have the ing a I of a/2. the right hand side of (7-16) a _ - cos V + cos a (7-14), rationalize, in turn, the tan */l a/2 __ ~ a/2 and sign, while 1 . cos a is r-24 non- negative. Example Find the exact value 7-6. of tan 22.5. Solution: Since the exact values of the functions of 45 (7-16), (7-17) or (7-18). Selecting (7-18), x rtrteo tan 22.50 _ we obtain _ = xtan 45 = 1~ cos 45 V2 are known, we may use 1 Trigonometric Functions of Sums & Differences 42 Example 7-7. Given tan 2 Find sin a and cos a. Solution: Since 2 a is a. r 576 = -=- where 2 a > = -=- is = and y 7 a second-quadrant angle. is 24 an angle whose tangent the terminal side of the angle with x = V49 + 24 = - 7-3 Sec. * 24, as we may shown find a point in Fig. 7-4. on Thus, 25. Therefore, B ina 4 = ,4/1+7 '25 >|/ 2 =5- ~7/25 = -3 and j cos a. - /I l\/ ' Example 7-8. Given that tan - = O w, find sin __ 1 ~~ 2 1 2 Substituting u for / cos a, we find that i , i i tan - we get u > ^ 2 = If ^ J- cos ; 1 If we substitute this value of cos COS i 1 - = a a cos a we solve this equation for in terms of w. we have cos a -f cos a Solution: Squaring both sides of (7-16), ex a and -f- QJ u 2- u2 in the relationship sin 2 a + cos 2 = a. 1, we obtain Note that we can drop the i sign, just as we did a always have the same sign. in (7-17) and (7-18), since tan a/2 and sin EXERCISE 7-2 In each of the problems from 1 to 8, find the exact functional value by using an appropriate double-angle or half-angle formula. I. sin 22.5. 5. sin 9. It is known cos 15. 2. 67.5. 6. cos that cos Q = 3. sin 67.5. 12 ? and 7. 6 is lo a. sin 26. b. cos 0/2. e. sin 6/2. f. 10. It is known that sin 6 40 and a. sin 26. b. cos 6/2. e. sin 6/2. f. II. It is known that tan 6 c. 6 is 3 and a. sin 26. b. cos 0/2. e. sin 0/2. f. cos 20. tan 26. g. sin 36. 6 cos tan 60. is Find : d. cot 26. h. tan 46. positive in the third quadrant. c. cos 26. = - tan 26. g. sin 30. 90. 4. 8. positive in the second quadrant. cos 26. = - 120. tan 67.5. Find : d. cot 26. h. tan 40. positive in the fourth quadrant. Find: c. tan 20. d. cot 20. g. sin 30. h. tan 40. Sec. 7-4 Trigonometric Functions of Sums & Differences 1 43 In each of the problems from 12 to 28, write the given expression in terms of a a multiple of 6. Make use of appropriate formulas to reduce the answer to as few terms as possible. single function of Prove each of the following 29. sin 40-4 31. sin 20 ^ 1- = tan 2 (1 cos 0(sin + 0/2 cos 7-4. identities: 0-2 sin cos 20) tan 3 = 30. cos 40 0), 32. 1 0. = 2(1- 2cos 0). 4 + sin 8 cos 4 0-8 cos 9+1. = + cos 0) 2 20 (sin = sin + sin 30. AS SUMS, AND SUMS sin PRODUCTS OF TWO FUNCTIONS EXPRESSED . 2 cos sin 2 EXPRESSED AS PRODUCTS By adding and subtracting corresponding members of (7-1), (7-2), (7-3), and (7-4), we obtain (7-19) sin (a + ]8) (7-20) sin (a + /8) (7-21) cos (a + ]8) (7-22) cos (a + j8) + sin (a sin (a + - cos - - (a. cos (a - j8) = 2 sin a cos j8, |8) = 2 cos a sin |8, j8) = 2 cos a cos ]8, ]8) = - 2 sin a sin /3. we reverse these identities, they become the following product formulas, which express given products of sines and cosines as sums or differences If : (7-23) sin a cos ft = (7-24) cos a sin /3 = (7-25) cos a cos (7-26) sin = ^ z [sin (a + 0) + sin (a I [sin (a + 0) - sin (a \ [cos (a 4 a sin /3 = - [cos - - + 0) + cos (a (a ft], 0)], ft)], + 0) - cos (a - j8)]. 1 44 Sums & Differences Trigonometric Functions of Sec. 7-4 To obtain the sum formulas, which express given sums or differences of sines and cosines as products, we first let a+ Then, solving for a and = = ]8 a and x j8 = y. we have /?, *-* and /3=^' Substituting these values of a and /3 in (7-19), (7-20), (7-21), and (7-22), we obtain the sum formulas. These are sin y = 2 sin sin y = 2 cos + cosy =2 cos - cos = - (7-27) sin x + (7-28) sin a - (7-29) cos x (7-30) cos x a cos -y sin (^-y^) -^ cos 2 sin cos 5 a sin as a sum of sines. Example 7-9. Express sin 3 Solution: Using (7-24) and replacing a by 5 a and cos 5a 3a sin = + 3a) - ~[sin (5a 2 Solution: Using (7-29) + n cos 20 11 T^ = 2 cos xi 7-11. Prove the Example r n r Z A . Solution: sin 7x cos 7x -I- 8a - sin 2a], cos by 20, we have = 2 cos 30 cos = tan 0. 2i sin sin = cos 7x /7a? ( + ; 5z\ 2 5x =cos 5z . Bin /) , x. /7 ( \ ft COS 2 cos _ 6x sin x 2 cos 6# cos x __ ; ?/ OA r cos \ sin _ ^fl -^ +-- x-x i 2 COS , o/j I identity rt rt = ^[sin we obtain cos 26 as a product of cosines. and replacing x by 40 and A 4.6 -f 3a)] a, -^ Example 7-10. Express cos 40 cos - sin (5a by 3 _ tan x EXERCISE 7-3 In each of the problems from 1 to two sines or two cosines. 10, write the given expression as a sum or difference of 2 sin 60 cos 40. 4. sin 28 sin 20. 8. cos 1. sin 30 cos 40. 2. 2 sin 40 cos 20. 3. 5. cos 40 cos 20. 6. 2 sin 65 cos 15. 7. sin 9. sin 50 sin 0. 10. sin 110 sin 30. cos 40. 21 cos 31. 7-4 Sec. Trigonometric Functions of Sums & 145 Differences In each of the problems from 11 to 20, write the given expression as a product of and cosines. Hint: In problems 18, 19, and 20, note that cos 6 = sin (90 - 6). sines 11. sin 30 13. sin 60 17. sin 80 40 19. sin 40 15. cos + sin 20. + sin 30. - + + + cos 16. sin - sin 18. + cos 38. 20. 25. cos 44. cos sin cos . A* -r esc 40 sec 40 cos 20 -f sin 27. cos 28. sin 29. sin 30. cos sin 31. a a + cos40~ sin -f cos 33. = __ tan tan v<*" a 9A &* a s*n L , -f _L e\/\ cot a ft = COS + -| ) - sin (0 + 0) - cos (-| cos + = 1 rTTJ sin + 7; - = V2 = -|) 0) (-| sin 0. \/3 cos = - V3 0. sin 0. -^ 2 - ft) = - /3) = (a sin (a cos -|) sin 2 cos 2 a a - sin 2 sin = cos = cos 2 j8 2 j8 ft 2 ft - ( a tan 0. cos 2 a. sin 2 a. ; = cos -f tan 80. cos 33. cos cos i sin 29 26. - (0 ft) j. 2 ft (0 ft) oo 22. tan ' T) cos 30 + -j) + sin + cos sin (a + cos (a + (^ p ^' (0 sin 32. ft ~ cos - sin identities: = COB 30. sin 20 4- sin 40 25. 30 sin 64 20. sin 65 = V2 , 23. cos 40. + sin 20. 40 14. sin Prove each of the following 21. sin - 12. cos n ~~ ft) gin p * Graphs of Trigonometric O 8-1. Functions; Inverse Functions and Their Graphs VARIATION OF THE TRIGONOMETRIC FUNCTIONS In Section 3-2, the trigonometric functions were defined in terms of the coordinates (x, y) of the point P(t), w here the number t represents the directed length of the arc of a unit circle measured r from the point (1,0). Later, in Section 3-9, an equivalent definition was given in terms of an (-1,0) angle 9 in standard position. We shall now consider the variation of the trigonometric functions in the four quadrants as f, and with it 9, increases from to 2n. From Fig. 8-1, we can read off the variations shown in Table 8-1. FIG. 8-1. For example, by noticing the changes tinuously from to 27T, we find that sin t in y as varies t increases con- from to 1 in the in the second, from 1 in the third, to quadrant, from 1 to and from -1 to in the fourth. Similar considerations lead to the results for the cosine and tangent. first Recalling that esc t - sin > we know that if either of these f unc- t tions increases the other decreases. Hence, the variation in esc t can be determined from the variation in sin t. Similarly, we may learn about the variation of sec t from that of cos t, and about the from that of tan t. Example 3-2 that tan variation of cot We found that tan t in t has no value when t = ?r/2 is undefined, 7r/2. 146 which means For the sake of easier tabu- See. 8-2 147 Trigonometric Functions; Inverse Functions TABLE 8-1 Variation of Trigonometric Functions lation of this result in Table 8-1 we have employed the much used symbols oo (infinity) and -co. These symbols merely signify that in the neighborhood of Tr/2 or one of its odd multiples the value of tan t is very large numerically. They are not to be used as numbers. 8-2. THE GRAPH OF THE SINE FUNCTION To construct a graph representing the variation of the sine, we x denote a real number or the value of an angle measured either let in radians or in degrees, and we let y denote the corresponding value of the function. Corresponding values of x and y are plotted FIG. 8-2. 148 Trigonometric Functions; Inverse Functions Sec. 8-2 as points on a rectangular coordinate system. We can infer the general appearance of the curve from the results summarized in the preceding section, but an exact representation is more readily obtained by using a table of sines. Let us now construct the graph ofy = sin x from x = 7r/2 to x = 2?r. The following values, found by the methods of Section 3-2, are used to obtain the curve in Fig. 8-2. While the choice of a scale is arbitrary, a better proportioned graph results if the same unit of length is used on both axes. The unit so selected will represent the number 1 on the y-axis and one radian on the #-axis. In terms of this unit a suitable length can then be marked off on the x-axis to represent 2-rr or 360. 8-3. THE GRAPHS OF THE COSINE AND TANGENT FUNCTIONS Using the table of trigonometric functions, the student should table of corresponding values for y = cos x and one for make a FIG. S-3. Sec. 84 149 Trigonometric Functions; Inverse Functions 7T/2 27T 37T/2 **tzn x FIG. 8-4. y = tan #, similar to that used for y = sin # in Section 8-2. Study the graphs in Fig. 8-3 and Fig. 8-4 on the basis of the tables you have made. If we compare Fig. 8-3 with Fig. 8-2, we see that the graph of = cos x may be obtained from the graph of y = sin x by moving y the graph ofy = sin x to the left a distance of rr/2 units. This fact can be checked by using the relationship cos x = sin (# + ir/2), from which it follows that cos = sin 77/2, cos vr/G = sin 27T/3, and so on. 8-4. PERIODICITY, AMPLITUDE, AND PHASE The trigonometric functions are among the simplest of a large which are periodic. As a preliminary to defining periodic functions, we shall call attention to some examples of phenomena which recur periodically, such as the rotation of the class of functions its axis, sound and water waves, the vibration of a and many other vibratory and wavelike phenomena. The spring, behavior of the object involved in a phenomenon of periodic nature earth about determines the type of function that is required to represent it properly. We note particularly that, because of the recurrence characteristic of such a phenomenon, the values which the function assumes in any interval of given length are also taken on in any other interval of the same length. This statement apparently indicates that a function of x is periodic with period p if, for every value of x, the function returns to the same value when x is increased by p. More specifically, we state the following. 1 50 Trigonometric Functions; Inverse Functions A function f(x) Definition. zero number p is Sec. said to be periodic if there is 84 a non- for which f(x + p) =/(x) for all numbers x in the domain of f(x). Any such number p is called a period; the smallest positive number p satisfying the requirement the period. is called Evidently, if p the period of f(x), then is np is a period, for every integer n. Periodicity of Trigonometric Functions. No matter which of the is considered in the definition of the trigonometric two viewpoints functions, we shall see that, if any trigonometric function g t ^ 2ir, then of (t + 2?r) = same function of t. According to the definitions given in Sections 3-1 and 3-2, P(t) and P(t + 27r) represent the same point on the circumference of the unit circle. To locate P(t) we start at (1,0) and proceed around the circle in the proper direction a distance of \t\ units. To locate P(t + 27r) we continue another 2?r units from the point P(t) This merely adds another complete revolution, and we arrive at the same point P(t). If we consider the definitions of the functions in terms of angles, + 2n is coterminal as given in Section 3-9, we note that the angle with and that any trigonometric function has the same value for . coterminal angles. Period of Sine, Cosine, Cosecant, and Secant. From a study of the 1 it is apparent that sin x assumes all values between In 2ir units. and +1 as x, starting from any value, varies through other words, the graph of the function repeats itself during each interval of length 27r, for positive and negative values of x. Or sine curve, stated more concisely, sin (x This is + 2?r) = sin x. equivalent to saying that sin x is a periodic function of x, remains now to show that 2ir or 360 is the smallest positive number p for which sin (x + p) = sin x and for which cos (x + p) = cos x. and that 2?r is a period. It Since, by definition of a period p, sin (x + p) = sin x for any value x 9 we shall select for the purpose of our proof the particular value x = ir/2. We then have . sin But since sin f -5- + pj /7T (- = * \ + P) = cos p, it . sm 7T = L 1 2" follows that cos p = 1. Sec. 8-4 Trigonometric Functions; Inverse Functions 151 Hence, p must be an even multiple of TT. The smallest even multiple TT is 27r, which must also be the smallest positive period of the of sine function. In a similar manner, we find that 2rr is also the period of the Because of the reciprocal relationships existing between the sine and cosecant and between the cosine and secant, 27r is also the period of the cosecant and the secant. cosine function. Period of Tangent and Cotangent. tangent, we write + p) = tan (x and we let x = 0. A of the tangent. of the cotangent. To tan x find the period of the } Then tan p = similar 0, and we find that TT is the period argument shows that TT is also the period We have just seen that the period of sin x is 27T. We shall now determine the period of sin bx, where 6 is a positive constant. That is, we want to know the smallest positive change in x which will produce a change of 27r in bx. If p represents this change in x, we can find p from the relationship Period of sin bx. b(x Solving for p, + p) we immediately P = bx + 27T. find that = 27T T - Thus, the period of sin bx is equal to the period of sin x divided by b, that is, 2;r/&. Similarly, it can be shown that 27T/6 is also the period of cos bx, esc bx, and sec bx. It is also true that the period of tan bx and cot bx is 7r/b. FIG. 8-5. 1 52 Trigonometric Functions; Inverse Functions Sec. 8-4 Let us consider the graph ofy = sin 2x shown in Fig, 8-5. Since the period of sin 2x is 2ir/b = 2rr/2 TT, the function will assume to TT that sin x takes the same range of values in the interval from in the interval from to 27r. Note that the graph of y = 2 sin x is similar in form to that of = sin x, which is shown in Fig. 8-2. The period is 2ir for both y curves. However, for any given value of x, the corresponding value of y in y = 2 sin x is twice as large as is the corresponding value of y in y = sin x for the same value of x. In the graph ofy = a sin x, where a > 0, the greatest value of y is a, and the smallest value of y is a. The constant a is called the amplitude of the function or of the graph. Thus, the amplitude of the graph of y = sin x is 1, while that of the graph of y = 2 sin x is 2. a sin x, where a is a real number, In general, for the function y the amplitude is equal to |a| and the period is equal to 27r. Also, in general, for the graph of y = a sin bx, where a and b are real, the amplitude is |a| and the period is Phase Angle. Since the graph of y = cos x may be obtained from the graph ofy = sin x by shifting it to the left a distance equal to 7T/2 units, we say that the graph ofy = cos x differs in phase by sin x. The amount of horizontal dis77/2 from the graph ofy placement of two congruent graphs, amounting to Tr/2 radians in this case, is called the phase difference, or the phase angle. The amplitude and the period are the same for y = cos x as for y = sin x. y cos 2x - sin (2x + 7T/2) = sin 2 (x + 7T/4) FIG. 8-6. Now 8-6. consider the graph of y This curve in Fig. 8-5 may - sin 2(x + ~) in Fig. ofy = sin 2x evidently be obtained from that 7r/4 units to the left in the x direction. = sin (2x + ~) 4U sin 2x by a shift of Hence, the graph of y ofy = = cos 2x by 7r/4 radians. differs in phase from that Sec. 84 A 153 Tr/gonomefric Functions; Inverse Functions simple method for finding the phase displacement is the point near the origin for which the function sin equals zero that ; is, to locate (2x+^) to find the smallest numerical value of x that makes the quantity x + j zero. We have then sin2(#-f ~) =0 when x + - = or when x = 7r/4. Hence, the phase difference is and the 7T/4 radians, sign of x is shift of the graph toward the is left since the negative. can be shown that, in general, the phase displacement of any trigonometric function of (bx + c), where b > 0, is to the right or It left by The radians (or degrees). \c/b\ ment depends on whether c/b direction of the displace- negative or positive. conclusion that for the graph of y = we arrive at the Finally, a sin (bx + c) the amplitude is \a\ and the period is 27T/6. Also, its phase differs from that ofy = sin x by c/b. Because of usefulness in its is many we applications, shall illustrate by means of an example a procedure for reducing an expression + B cos & to the form a sin (9 + a) of the form A sin . Example Reduce 8-1. A +B sin + a). to the form a sin (0 cos Solution: Write A +B sin = \/A 2 + B 2 cos (~ _ 2 \\/A The absolute values of the coefficients An ' -f B* B2 = + sin + VA + B 2 = and \/A 2 -f B2 2 cos 0^ / cannot be greater 1, and the sum of their squares is 1. Hence, they may be taken as the cosine and sine, respectively, of some angle a. Therefore, the expression A sin -f B cos B becomes than + B2 where A = a cos (cos a and a sin 6 = J5 Example 8-2. Express 4y and sketch the graph. Solution; Since A = sin 26 If 1 -f- and + sin a cos 0) = \M 2 cos angle and a may = ^ and = sin 26 B = a + a), sin (6 - V3 cos 26 in the form y = \/3, \/3 cos 20 sin B2 a sin a. = we have \/^ 2 2^ sin 20 we identify this result with the expression \/A* we have 4- = be taken equal to ~~ ir/3. -f - ^ B2 -h cos (cos It follows that #2 = 2. + a), Therefore, $ 20Y a sin 20 a a sin (60 is + sin a cos 20), a fourth quadrant 154 Trigonometric Functions; Inverse Functions We have, finally, y The amplitude - - _ = i(sin 26 - V3 of this function is r 4 = cos 29) its > | period Sec. 8-4 . - sin is TT, ^20 and yJ its phase angle is ir/6. FIG. 8-7. Since the interval from (TT/G, 0) to (77T/6, 0) is one period in length, the part of the curve obtained for this interval may be repeated indefinitely in both directions to give the complete curve. we have employed The curve is shown different units of length in Fig. 8-7. (Here for convenience on the two axes.) EXERCISE 8-1 In each of the problems from and phase angle to 24, find the period, amplitude, 1 of the trigonometric function. 1. 3 sin 0. 4. cos = 7. 4 tan \ 8 o 10. sin 3. 2.sin|. 13. cot 60. 6. 2 cot 8. si 9. cos0- 19. esc (ir6 22. 6 - + 7). cos 40. | 4 11. 3 sin 50. 12. 5 tan 7T0. 14. sec 90. 15. + 4). 17. tan (TT i 0. = sin 7 6 4 3 37r 16. 3 esc cos 5. o 7T0. ^ 20. cot (2?r 23. 5 +3 sin \ 40. 18. cos (30 TT). (26 5 cot - - o , - 21. 2 + sin 0. 24. 4 +2 2). cos (20 \ + -| o , In each of the problems from 25 to 30, sketch the graph of the given function by constructing a 25. 28. y = cos y = 5 cos \ x x 6 table of values. 26. y = tan 29. y = ^ 2 tan z | y = 2 sin 30. y = 3 sin 2x. 27. 3.r. 85 Sec. Trigonometric Functions; Inverse Functions 1 55 Sketch each of the following graphs without constructing a table of values. 1 2 33. # = tan x. 31. y = sin ^ a:. 32. y = cos 4z. & 6 = 34. y 2 cos 35. 3.r. ?/ = 3 cos 36. irx. = */ r tan 3x. 4U 37. y = 40. !/ = cos 43. ?/ = sin (2x 46. ?/ = sin 48. y = \/3 sin 20 50. y = V2 8-5. INVERSE FUNCTIONS I sin 3. 1 fa? + + cos 2). + 1). cos 38. y = 41. y = sin 44. y = cos (2wx 47. 0. + V5 30-3 | ?/ cos - (x sin 6 * ^) - = 5 cos 42 * V = cos ( 45. y = sin f 39. x. | - cos TT). ?/ sin 30. ~ 3x j = 13 51. = 2 sin y cos ^~7 - 2 )- + 7V x 49. y 0. cos 20. (x + ~ sin ^ cos 20. we defined a function by setting up a rule of between twa sets of numbers, X and 7, called the correspondence domain of definition of the function and the range set of the funcIn Section 2-3, respectively. The function was called single-valued if just one number y of the set Y is assigned to each number x of the set X. If more than one number of Y corresponds to some value of x, the function is multiple-valued. tion, pose a reverse problem. We corresponding values of x. Naturally, y is limited to lie in the range of the given function, since otherwise no x exists. The function which makes correspond to each such y all values of x for which y = f(x) is called the inverse function corresponding to the given function. We shall begin our discussion of inverse functions with an is the set of all real numbers, Y is the set of example in which If we know assume y that y to be given f(x), we may and ask for all X all non-negative real numbers, and the correspondence is deter- mined by the relationship y = x2 . Ordinarily, we assign values to x in order to calculate values erf x 2 In this case, we have a rule of correspondence that assigns just one number y to ea$h chosen number x. Hence, y is a single-valued function of x. The graph ofy = x 2 is shown in Fig. 8-8. Assume now that y is given and that we wish to determine corr^sponding values of x. To do this we solve the given equation x 2 = y for x, and obtain two numbers x = \/y and* x = ^/y cor. s responding to every, non-negative number y. Since the rule of 156 Trigonometric Functions; Inverse Functions Sec. 8-5 correspondence assigns two values of x to each chosen number y, see that # is a double-valued function of y. In this case, the admissible values of y are restricted to zero and the positive real numbers, while those of x comprise all real numbers, as was indi- we cated in the specification for the sets X and Y. O FIG. 8-9. FIG. 8-8. In the study of mathematics, we generally prefer to use the sym- bol x to represent the independent variable and y to represent the dependent variable. To be consistent with this preference, we shall x 2 and y = inverse of the other. call y = .^/x inverse functions, each being called the The graph ofy = \/Hc is obtained by plotting that of x = y 2 as shown in Fig. 8-9. We note that the roles of x and y are interchanged in the two equations y = x 2 and x = y 2 Thus, we see that the , . curve in Fig. 8-9 is actually the curve of Fig. 8-8 with the axes interchanged and one of them reversed in direction. 8-6. INVERSES OF THE TRIGONOMETRIC FUNCTIONS The Inverse Sine. Let us consider the function y = sin x and attempt to apply a discussion similar to that in Section 8-5. Here, as has been noted, the domain is the set of all real numbers, while is the interval 1^7/^1. Referring to the graph of x in and recalling the periodic properties of this y Fig. 8-2, graph, we see that, for every given number y such that 1 :f y 2S 1, there are infinitely many values of x such that y = sin x. To designate the totality of all values of x such that y = sin x, we write the range = sin x which is read x fully that the is sin sin" 1 or esc y. t/ sin"" 1 y, the inverse sine of symbol which equals - = y. The student should note care- l y must be distinguished from (sin y)~ 9 Sec. 8*6 Trigonometric Functions; Inverse Functions Another notation that function is 57 frequently used to represent this inverse is = x which 1 arc sin y, read x is the arc sine of y. In order to conform to the preferred practice of considering y as a function of x, we may designate the inverse of the sine func- tion is by writing = y sin" 1 x or y = arc sin x. We note the following properties of this inverse function. # is a number such that \x\ > 1, then y does not exist. This property follows from the fact that the sine function takes on only the values from -1 to 1. Hence, the inverse sine function sin* 1 x is defined only when 1 ^ x =i 1. For example, sin' 1 2 is not defined, If since there If # is a is no number or angle whose sine is 2. number such that \x\ ^ 1, then y certainly exists. More- over, because of the periodicity of the sine function, there are * x corresponding to every such infinitely many values of y = sin value of x. For example, it x = 1/2, then y = sin' 1 1/2 means that 1/2. Then y may be y is any number or angle such that sin y from these by integral taken as ?r/6, 5-77/6, or any value that differs 1 multiples of 27r. The totality of these values of sin- 1/2 may be represented as + 2n?r + and 2mr, where n = 0, 2, 1, . convenient to plot the graph of y = sin' 1 x for a further study of the inverse function. Since y = sin' 1 x and x = sin y express exactly the same relation between x and y, the graph of Fig. 8-2 may be used as a graph of the inverse function. We obtain the graph ofy = sin* 1 x simply by interchanging the axes in Fig. 8-2 and reversing one of them in direction. The result is shown in We shall find it Fig. 8-10. The question now arises into intervals within each of ing to each x such that ing a first interval \x\ whether the t/-axis can be subdivided which y has just one value correspond- ^ One way 1. TT ^ ^ of doing this is by select- TT -2 ***2" Other intervals, such as -^ z selected. With the ^ y ^ - z and ~ w O z g y entire y-axis thus subdivided, the graph ofy = sin" x as consisting of many single-valued functions or branches. 1 ~ ^ z we may > are then think of the graphs of infinitely Sec. Trigonometric Functions; Inverse Functions 158 8-6 3TT/2 37T/2 7T/2 O -i i O -1 /I -7T/2 "-7T/2 FIG. 8-11. FIG. 8-10. To avoid any ambiguity in later applications as a result of this multiple-valued property of the inverse sine, we shall often restrict y so as to make the function single-valued. There will then be but one value of y corresponding to each value of x such that sin y shall determine this value from the branch for which ^ y x. ^ We - called the principal branch. The values of y chosen this branch are called the principal values of the inverse-sine This branch from -- = is function and are represented by the equation y Note that = Sin" 1 x or y = Arc sin x. in this case the initial letter of the name is capitalized. This restriction to the two quadrants containing the smallest numerical values of y results in a single-valued function y = Sin- 1 x. f or x in the interval The values of y are such that ^ y ^ 1 -1 g x ^ 1. For example, we have Sin- (-1) = -7r/2, Sin- 1 (-1/2) = 1 = 0, and Sin- 1 1 = ir/2. -7T/6, Sin- The Inverse Cosine. We shall next consider the function y = cos x. With the help of Fig. 8-3, we see that the range of y = cos x is the 1 S y ^ 1, and that for every y in this interval there are interval values of x such that y = cos x. We are thus led to the inverse cosine function, which makes correspond to y all the values of x such that y = cos x. If again we interchange the symbols x and y, we may write the inverse cosine function as infinitely many y = cos~ x x or y = arc cos x. Sec. 8-6 159 Tr/0onomefr/c Function^; Inverse Functions This inverse function has the following properties. If # is a number such that \x\ > 1, then y does not exist, because there is no value of y for which |cos y\> 1. If x is a number such that \x\ ^ 1, then there are infinitely many values of y designated by cos' 1 The graph ofy = cos* 1 x is x. shown in Fig. 8-11. The method for = similar to that used for graphing y sin' 1 x. We first write x cos y. We then plot a cosine curve by proceeding as for Fig. 8-3, except that values of the independent variable y are laid off on the ?/-axis. plotting it is The principal branch of the curve in Fig. 8-11 is the portion of the curve for which g y ^ TT. It is represented by the principal value of the function, which is denoted by = y Cos" 1 # or y = Arc cos x. The Inverse Tangent. The inverse tangent function = y 1 tan" x or y = is arc tan x. 7T/2 -2 -3 o -1 -7T/2 -7T FIG. 8-12. represented by the graph in Fig. 8-12. We note that there are infinitely many valu.es of y for every value of x. The principal branch of the graph of y = tan- 1 x is the portion It is for which ~ < y y < ~ = Tan" This 1 x is or represented by the equation y = Arc tan 160 Trigonometric Functions; Inverse Functions Sec. 8-6 The Inverse Cotangent, Secant, and Cosecant. The other inverse trigonometric functions are 1 or y = cot" x y = arc cot x, : y ?/ = = sec" 1 a: or esc" 1 a: or y y = = arc sec x, arc esc #. \Y FIG. 8-13. show only the graph ofy = cot- 1 x. See Fig. 8-13. The < y < TT. principal branch of this curve is given by Principal Values of the Inverse Cosecant and Secant. The selection of principal values of ?/ = esc- 1 x and y = sec- 1 x is by no means uniform among all authors. Some writers adopt the range between and 7T/2 for both functions when x is positive, and between TT and 7T/2 when x is negative. The authors prefer, however, to use the We shall definitions Csc- 1 x = Sin- 1 Sec" 1 x = Cos" 1 and (-) We have, therefore, the following ranges of principal values of the inverse trigonometric functions : 7T -o- SI 1 Sin" x ^ 7T > ^ Cos" 1 # < Cor - ^ Csc- 1 * ^ -J 1 (Csc^ x * 0), ^ 1 a; Sec" 1 x ^ TT, < TT, ^ TT 1 (Sec" x ^ y) Sec. 8-6 161 Trigonometric Functions; Inverse Functions 34 (Ox Sin- 1 = o Solution: Let Sin- 1 3 = -= Then 6. 5 is is used, o 3\ 1 Hence, cos (SinJ / = = Let Tan' 1 Solution: and ( - x) = and cos = Since only the 5 a first-quadrant angle and cos cannot equal 4 - Example 8-4. Find the value of sin [Tan- 1 x being a positive number. = > 5 1 principal value of sin- = 4 =^ sin = ( - x)], Then tan 0. 6 between w/2 and 0. If angle constructed in standard position, as shown x, 6 is lies / in Fig. then sin 6 8-14, 1 Hence, sin [Tan" ( - x)] is found to be = FIG. 8-14. Example 8-5. Find Arc cos Solution: Let Arc cos (cot (cot 60). 60) = Then 6. = cos 6 cot 60 = .5774, by Table II. Therefore, the value of 6 is Arc cos .5774. Since 6 is restricted to the interval from to 180, 6 must, in this case, be a first-quadrant angle. Hence, Arc cos .5774 5444'. 5444', and Arc cos (cot 60) = = = EXERCISE 8-2 In each - < of the problems ^_ 5 K O/y. from 1 to O-r y = 2x 8. Tan- 1 12. Sin- 1 0. ( 16. 19. 9. - J) &/ Cos- 1 Cos- 1 ( 1 +6 3 4 2' 22, find the value of the given expression. 1. V Tan-' 3 In each of the problems from 7 to 15. y=rnx+b. 3 6 5. Cot- 1 find the inverse of the given function. 1 1 3 11. 6, _/*-! - 13. 17. Arc cos Cot- 1 Tan- -^ ^ ( - V). Tan- Cos- 1 (- 1). 14. Csc-i 1. o 1 20. 10. 1 [sin 18. Arc tan (- (- ir/2)]. 1). 21. Sin22. Tan- (-1.414). ^-0.414). In each of the problems from 23 to 37, evaluate the given expression. 1 1 23. tan 26. cot 1 (sin- j|) Sin- 1 24. sin 27. cos 1 (sin- 1) 1 (Cos' 1) 25. sin 281 tan (Sin- 1 .6450). 162 Sec. Trigonometric Functions; Inverse Functions 30. tan (Cot- 1 2). 29. sin (Cos- 1 .9200). 32. sin 1 33. cot (Cot7 V 4 1 1 34. sin (Cot- 1) 37. Sin- 1 36. cos f Sin- 1 5 \ o 35. cc 1 Sin- i) ( 5/ \ 31. sec (Cos- 1) 8-6 (tan ^ In each of the problems from 38 to 56 simplify the given expression, taking u as a positive number. 38. Cos- 1 (sin u). In 38 to 40, 54, 56, u O/2. In 44, 46, 48, 55, u <1. 39. Sin- 1 (- sin u). 40. Sin- 1 (cos u). 41. esc fsin- 1 \ 42. sin (Sin- 1 u). 43. tan (Tan- i u 1 u). _ VI + u 45. cot ("Tan- 1 - V U 47. tan ( Sin- 1 \ 49. tan 48. esc (Sin50. cos (Tan- 1 54. Sin" 56. Cot- v u). 53. sec (Sin- 1 55. (sin u). u 1 +u 51. sec (Cos- 1 u). 52. cot (Sin- 1 u). 1 l (Cos~ V Cos- 1 14). (cos \ (tan In each of the problems from 57 to 65, draw the graph of the given function by changing from the inverse function to the direct function and using the period, amplitude, and phase angle of the function to assist in plotting. 57. y = --sin- 59. y = 1 x. 58. y = a:. 60. y = ^ sin- Q cos- 1 2 tan- 1 x. 1 - sc 1. Zi 61. y = - cos- 63. y = 65. y = 4 tan- x 1 o (Cos- + 2. + 1 62. 64. y 1). 69. 70. 71. < u g 1. 1. <u Prove that cos (Tan- u) > 0, if u ^ 0. Prove that tan (Cos- it) < u 5 1. 0, if Prove that Sin- u + Sin- (- ti) = 0, if - 1 Prove that Tan" w + Tan- ( - u) = for all 72. Prove that = tan- 1 z 4- 3. 3 sin- 1 (2x + ^ 0, if 0, if 1 1 u 1 1 1 1 Tan- 1 u 73. Prove that Cos- 74. Prove that =4 1 66. Prove that sin (Cos- 1 u) 67. Prove that cos (Sin- 1 it) 68. ?/ 1 Tan- 1 u tt + Tan4- - 1 (l/i*) = y Cos- (-u)=ir, 1 Tan~* v = Tan- > if 1. values of u. w > 0. if-lgul. " 1 j** J 'if ti > and v > 0. 1) - 2. V linear Equations and Graphs 9-1. SOLUTIONS OF SIMULTANEOUS EQUATIONS Often problems arise that involve two or more unknowns and as many equations. The solution of such a problem requires the determination of numbers which simultaneously satisfy the given equations. Equations for which we seek common solutions are referred to as a system of simultaneous equations. If the system has at least one solution, it is said to be consistent; otherwise, it is called inconsistent. Let us investigate the following pair of simultaneous linear equations : aix + biy = f a, ( a2x + b2 y = c2 (9-1) . x and y which satisfy both and equations, excluding from consideration the cases a\ = &i = It is desired to find all pairs of values of a2 = &2 = 0. We proceed by multiplying both sides of the and both sides of the second by bi, obtaining b 2 biy = first equation by & 2 c 2 bi. Subtracting the second equation of this pair from the obtain (9-2) If (aib 2 (aib 2 02&i) ^ 0> i first, we a2 bi we ^ nd that _ we see that we have which may exist. Thus, exactly one value for x in any solution 163 1 64 and Graphs Linear Equations Sec. 91 multiplying the original equations by a? and a lf respectively, the first equation thus obtained from the second subtracting ^nd one, we obtain By - (ai& 2 (9-3) - If (ai&2 a2 6i) T* 0, we = a 2 bi)y a2 a. find that 4i CL 2C\ d\C2 ~~ _____________ a\b 2 Again, we have - aic 2 CL 2 bl exactly one value for y in any solution which may exist. Hence, if (di& 2 &2&i) is not zero, we have at most one solution of the pair of given equations, namely, ' x (9-4) = ~ l - l y > Substitution of these values for x and y in (9-1) shows that we really have a solution. The reader should verify this solution by making the substitutions. Consistent and Independent Equations. If (eii&2 e^&i) ^ 0, we have exactly one solution of (9-1), which is expressed by (9-4), and the given equations are called consistent and independent. We may consider, as an example, the pair of equations = 24, 5x-2y= 22. 2x + 3y I \ Here a^b 2 - 0361 = (2) (-2) and y = 4 give a solution. In (5) (3) = -19, and the values x = 6 other words, when these values are substituted in the two given equations, both equations are satisfied. Thus, 2(6) + 3(4) = 24, - 22. I I 5(6) - 0261 = If di ^ 0, division Cases with ai& 2 di&2 &2&1 = 0. 2(4) 0. , 02 = Let us now consider cases in which = by 2 di gives , 01. 01 We then define k = > and we have ai a2 (9-5) = Now let ai tion, and since k = -2. , we 0. = Then, since a^bi = 0, fan, 61 62 = fc&i. cannot be zero by our initial assump0. In this case, if we define we have a2 = see that (9-5) again follows. Thus, ai& 2 - a2 bi = has Sec. 9-1 shown been and Graphs Linear Equations to mean proportional. note also that We that if ai& 2 the 0261 members of left = 1 (9-1) 65 are equations (9-2) and (9-3) 0, become f X = Ci&2 ( y = aic 2 (9-6) C2&1, - a2 ci. It is clear that these equations cannot be satisfied by any pair of values of x and y, unless both right members are also zero. Accordingly, if we wish to determine whether or not the given system has a solution when ai& 2 &2&i = 0, it becomes necessary to take into account the two cases of the right sides of (9-6) being zero or not zero. Consistent and Dependent Equations. In the first case referred to in the preceding paragraph, a,ib 2 o^bi = 0, and Ci& 2 ^2&i and a2 Ci are also equal to zero. Hence, substituting the values a,iC 2 of & 2 and 02 from (9-5), we have 2)61 (ci/b = c 2 )ai (ci/fc = 0. Since a lf bi are not both zero, it follows that c 2 = kci. This means that one of the original equations is a constant multiple of the other, and any pair of values of x and y that satisfy one equation Under will also satisfy the other. this condition, the equations are and dependent. as an illustrative example the equations said to be consistent We have f ( 2x + 3y 4z + Qy = 24, = 48. (3) = 0. Here ai& 2 - a2 b l = (2) (6) - (4) Also, the coefficients of the unknowns and the constant term in the second equation are multiples of the coefficients of the unknowns and the constant term in the first, and the multiplier is 2 for all three terms. Infinitely many solutions exist. Some of them are x = 0, y = 8 x = 12, y = : and x = t,y 04 - 2t o , where t is ; any number. ; % Inconsistent Equations. Finally, let us suppose that in the origand at least one of the numbers a^bi = equations aj)* inal Ci& 2 ~ c 2 6i, aiC2 a 2 Ci is different from zero. Hence, no solution of (9-1) exists and the equations are inconsistent. This case is characterized by the condition that one of the numbers (cik <%) 61, (cik 02)0,1 is not zero, in view of (9-5). It follows that c2 ^ kci* 1 66 and Graphs Linear Equations Hence, the multiplier for the constant terms. jto the members left Sec. 91 of (9-1) does not apply Consider, for example, the equations 2x + 3y = 24, 4x + Qy = 7. I 1 coefficients of the unknowns in the second equation are of those in the first equation, and the multiplier is 2 for multiples both terms; however, the multiplier for the constants is not the Here the same as that for the other terms. Hence, these equations are inconsistent. 9-2. To ALGEBRAIC SOLUTION OF LINEAR EQUATIONS IN TWO UNKNOWNS and independent system of two linear equatwo unknowns, we reduce the system to one equation in one unknown by eliminating one of the unknowns. The following example will illustrate two commonly used methods for eliminating the unknowns. solve a consistent tions in Example 9-1. Solve the equations 2x+3y= 24, 5x Solution: Since a\b 2 a 2 &i = - 2y = 22. (-2) - (2) (5) (3) = - 19, the equations are and independent, and there is but one solution. If we use the method of elimination by addition or subtraction, the procedure is the same as that indicated in obtaining the solution (9-4) from equations (9-1). To eliminate y multiply the first equation by 2 and the second by 3, in order to consistent t make the coefficients of y numerically equal in both equations. 4z + Qy = 48, 5x - 6y = 66. Adding, we get Solving for Now, x, 19$ = 114. x = 6. we have substitute 6 for x in the 3y first = 24 - of the original equations. 2x = 24 - 12 = We thus obtain Then 12, or y=4. Alternate Solution: If by solving the first we use the method of elimination equation for y in terms of 24 x. -2x by substitution, We thus get we begin Sec. We 9-3 Linear Equations Solving for x, - 2x 24 then substitute ~ for and Graphs 1 67 y in the second equation and obtain we have Ux - - 2 (24 2s) = 66, or - 15x + 4s = 48 Hence, 19z = 66. 114, and Substituting 6 for x, we as before, x =6. find that y = 4. A Example 9-2. grocer has some coffee selling at 80 cents per pound and some at 90 cents per pound. How much of each must he use to get a mixture of 100 pounds worth 86 cents per pound? Solution: Let x = number of pounds of 80-cent coffee, and y = number of pounds of 90-cent coffee. Then x +y = 100, and + 0.90y = O.SOz Simplifying, x + y z + 9y These equations have the single solution x LINEAR EQUATIONS IN THREE 9-3. 0.86 (100). we have = 100, = 860. = 40, y = 60. UNKNOWNS In the solution of a system of three equations in three unknowns, one method is to employ the following steps, which we do not justify here 1. Eliminate one of the unknowns from a pair of the equations; then eliminate this same unknown from another pair of the : 2. original equations. Solve the resulting two equations for the two remaining unknowns. 3. Substitute the values found in step 2 in any one of the original equations to find the third Example 9-3. Solve the system 2x x 3x unknown. of equations + 3y - 2=5, - 5y + 2z = 1, + y - 42 = - 1. 168 and Graphs Linear Equations Solution: Eliminate z from the first 5x Now Sec. and second equations by addition to obtain + y = 11. eliminate z from the second and third given equations - 5z 9-3 90 = by addition to obtain 1. We then consider the equations 5x + y = 11, Solving these equations for x and y by subtraction, we have x Substitution of these values in the first given equation gives z 2. solution of the given system is z 2, y 1, 2 = = =2 = 2. and y = 1. Hence, the = EXERCISE 9-1 In each of the problems from Check 1. (3x I 1 to 30, solve the given system of equations. all solutions. x - 2y 30 = 6, = 4. 3x 2x 2x 3x x 2x 2x 3x - y = 7, + y = 8. + 30 + 1 = 0, 3. 6. - 2/4-7=0. + 2y = 3, + 30 = 1. + 30 - 1 = 0, + 0+3=0. 9. 12. 15. 17. - 3z = 6, + 20 = 3. 2x + 2y - 3 = 0, 5x + 30 + 4 = 0. 3z + 0+7=0, 4z + 80 + 9 = 0. 2x + 60 - 7 = 0, 3z - Sy + 9 = 0. 2x + 40 - 50 = 3, - 7 = 2, 3z + 4x + 80 - 10z = 1. x + 20 - 2=3, + 40 + 2s = 1, + - 30 = 2. - 3x + 80 + 9s - 3 = 0, 2x + 30 + 2-4=0, Bx - 20 - 22 - 4 = 0. 3z + 52 = 0, - 42 = 2, x 4z 32 + 1 = 0. 20 3x 40 + 22 = 3, 2x + y =1, 5x - 3y 4- 42 + 5 = 0. 2x + 30 - 62 + 2 = 0, - 2 = 0, 50 + 3 s x 3z - 19. 21. 23. 25. a- V-21 f ~23 =0A ' 9-4 Sec. Linear Equations 26. By 2z 2z = 169 27. 4, -20 +3* =3, + 4s = 2. 3s 28. + and Graphs (4 29. 13(3 1 1 _2 1 = x y z ~~3 1 30. (3(3 -f 0) 0) -0) =5, - 0) = 5. -4(3 -0) =5, - 3(3 + 0) = 7. * Z I' - 4(3 3(3 4(3 -f 0) (X (X +0) - For each of the following systems of equations, determine whether it is consistent and independent, consistent and dependent, or inconsistent. 31. 34. 37. - = - 5, 60 = - 3. + 40 = 3, [2x x + 20 = 6. \ - 30 = 1, /23 + 6w = 2. - x + 30 = 2, 2x = 60 + 14. /23 43 1 30 32. \ x j [210 38. f - ay 9-4. + 2x 35. 5 40. - (33 50 70 63 3x + + + 63 9z - 80 - 8=0, -f 10 + 1 - 3 120 40 = = = = 0. = 0, = 0. 3, 1. a6. 33. J33 [63 36. J45z J153 43 39. | [150 - 20 = 8, - 40 = 8. - 270 = 21, - 90 = 7. - 30 = 6, -203 +6 =0. bx -{-ay = a6c. GRAPHS OF LINEAR FUNCTIONS The discussion of rectangular coordinates in Section 2-1 set the stage for the pictorial representation of a function. By this representation of a function f(x), we mean the graph of the equation y = f(x). It consists of all points, and only those points, whose coordinates x and y satisfy the equation. In the same section we considered the graphing of lines parallel to the coordinate axes. The equation x = 3 was shown to represent a vertical line, that is, a line parallel to the y-axis which intersects the #-axis at the point (3, 0). This line thus includes all points 3 units to the right of the t/-axis. This example illustrates the fact that a linear equation in x alone represents a line that is parallel to the ^/-axis. Similarly, y = 2 was shown to be the equation of thfc horizontal line whiph is parallel to and 2 units above the #-axis. And this example illustrates the fact that an equation in y alone represents a line parallel to the z-axis. Furthermore, it is proved in analytic geometry that the graph of every first-degree, or linear, equation in x and y is a straight line; and, conversely, that every straight line is the crraDh of a linear equation. 170 Linear Equations and Graphs Sec. 9-4 We shall proceed by first preparing a table of corresponding values in a given problem and then plotting the corresponding 'points on the coordinate system to obtain the graph of the equation, The following illustrative examples will point the way toward an understanding of the procedure in the graphing of linear equations. Example 9-4. Graph the function 2x + 3. = 2x + 3. Then assign any values for or, substitute them in the and obtain the corresponding values for ?/. The table and the graph are Solution: Let y equation, shown in Fig. 9-1. Since a straight line is definitely determined when two points are known, only two pairs of values of x and y are needed in graphing a linear equation. We can, however, use three points in order to check our work. /U,5) y 3 5 -1 FIG. 9-2. FIG. 9-1. Example 9-5. Graph the equation 2x Solution: The equation may be table of corresponding values of 9-5. 3?/ = 6. solved for y in terms of x. y A + = - Then + 2. | # and ?/ and the graph are shown in Fig. 9-2. INTERCEPTS In general, the points where a curve crosses the coordinate axes are the easiest to obtain. The ^-intercepts are the values of x at the points where the graph crosses the x-axis. Since y = on this axis, the ^-intercepts are the values of x that correspond to y = 0. Similarly, the ^-intercepts are the values of y at which the graph crosses the i/-axis. They are the values of y that correspond to x rule = 0. Hence, we have the following : To find the ^-intercepts, set y = To find the ^-intercepts, set x = and solve for the equation and solve for in the equation in x. y. Sec. 9-6 Linear Equations Example 9-6. Find the and Graphs 171 intercepts of the line = 6. Solution: To find the ^-intercept, let y = 0. Then 2x = 6, and x = 3. To find the ^/-intercept, let x = 0. Then 3y = 6, and y = 2. = 3 and = 2 found in this solution correspond Note that the intercepts 2x a; the points (3, 0) and coordinate axes. (0, 2), + 3y ?/ respectively, where the line in Fig. GRAPHICAL SOLUTION OF LINEAR EQUATIONS IN 9-6. to 9-2 crosses the TWO UNKNOWNS In the graphical solution of two linear equations in two unknowns, the graphs of the two equations are drawn with reference to the same coordinate axes. Since the solution of two equations in x and y is a pair of values of x and y which satisf^ both equations, the solution must represent graphically a point common to both lines represented by the equations. Hence, the values of x and y which satisfy both equations give the coordinates of the point of intersection of the lines. We find, therefore, that the two lines intersect in a single point, are parallel, or are identical, according as the equations are consistent and independent, inconsistent, or consistent and dependent. Example 9-7. Solve graphically 2x + 3y = 24, 5x - 2y = 22. Solution: Tables of corresponding values for the two equations and also their graphs are shown in Fig. 9-3. It is seen from the graphs that the lines intersect at the point (6, 4). That x = 6, y = 4 gives the solution of the given equations Y 2x + 3y = 24 5x - 2y = 22 y -11 12 22 5 FIG. 9-3. may be checked by substitution. 172 I/near Equations and Graphs Sec. 9-6 EXERCISE 9-2 In each of problems from x- and to - 3y = 1. = x -f 5. = 3z + 4. x 1. 4. 7. 1 y 9, graph the given function. In each case give the ^-intercepts. 2. y 5. 8. y = 2x - 8. = 3.r. = x - 3. 3. 9. Solve each of the following systems of equations graphically. 12. 11. 130 - 2x = 0, fa? - 3// = 1, 10. \3x 13. (2*/ I2x 2y = 0. = + 3, = 16 - 3y. - a; x + /3x - I 14. i y 3x =2. = = 4, y 6. (20 \2* ?/ = = - a? 4. # 3x - 5. = * -3, = y + 3. 15. -0 1. =4. - ~r y) 16. \2x - y =4, +2y = 12. 17. + 4 - 4 = 50 - 3, =4x +2. 18. 2* - 30 = 12. 10 Deferminanfs DETERMINANTS OF THE SECOND ORDER 10-1. Let us consider the following system of two linear equations in two unknowns: . , ( aix + b\y [ ClzX + b>2 (10-1) The y nn ci, . solution of these equations by the given by (10 - o\ x -2) --- k-*! , "" t)lC2 p- y > C2. method of Section 9-1 - is - 2 0,201 It is understood that a^b 2 We ~ a<2 bi 0. more convenient way of writing the expression (tib 2 Onbi. The notation which we select for this purpose will enable us to express also the numerators of (10-2) by means of the same symbol with the proper changes in letters. In choosing a symbol we shall select a form which will exhibit the numbers a l9 &i, a2 &n in the same relative positions as in (10-1) Thus, we write shall at this point introduce a . , 02 62- This arrangement of the four numbers in a square array, consisting of two rows and two columns, is then enclosed within vertical bars, as follows : ai hi a2 This symbol represents a determinant of second order. Thus, we start with a square array, or matrix, such as 174 Sec. 10-1 Determinants We &2 &i, called then associate with this array a number ai& 2 determinant, which is denoted in the following manner its : The numbers a u b lt a2 The numbers ai and & 2 , b 2 are called elements of the determinant. on the principal diagonal the numbers on the secondary diagonal. Note. We observe that the "expanded" value of the foregoing determinant is equal to the product of the elements on the principal diagonal minus the product of the elements on the secondary a2 and lie ; &i lie diagonal. It is interesting to note also that this value is the algebraic sum of all possible products obtainable by taking one and row and one element from each column. preceded by a plus sign or a minus sign, according only one element from each Each product is to a rule to be stated in Section 10-2. Using the notation of determinants, we can write the solution (10-2) of (10-1) in the form X (10-3) = y = 0,2 We note that the value of each of the unknowns in (10-3) may be written as a fraction whose denominator is the determinant of the coefficients as they stand in (10-1), and whose numerator is the determinant formed from that of the denominator by replacing the column of coefficients of the unknown in question by the column of constant terms. Note. If a2 = kai and & 2 fc&i, where k is any number, then = 0. In this case, the equations of the system (10-1) are inconsistent unless both numerators of the fractions in (10-2) are also equal to zero, that is, unless = and = 0. Therefore, the equations of the system (10-1) represent distinct, = kc l9 or they represent coincident lines parallel straight lines if c 2 if Co = kc\. Sec. 10-2 175 Determinants Example 10-1. Solve the system of equations -.x \x + 3y = 1, - 2y = 22. we have Solution: Using determinants, 1 3 22 -2 - 2 - 66 - 68 4 3 - 8 - 9 - 17 3 -2 = 4. Also, 10-2. 4 1 3 22 4 3 3 -2 88-3 _ - 8 85 - 9 = -5. 17 DETERMINANTS OF THE THIRD ORDER A determinant of the third order is a number designated by a square array of nine elements arranged in three rows and three columns and enclosed within vertical bars. An example is D= (10-4) a2 62 The value, or expansion, of the determinant (10-4) as the quantity is defined &2 (10-5) 63 63 C3 C3 or as the quantity (10 G) D = ai&3C2 ^ 0162^3 Here the products such as ai& 2 c 3 ai& 3 c 2 the terms of the determinant. , , and are known as Minors and Cofactors. In any determinant the minor of a given element is the determinant of the array which remains after deleting all the elements that lie in the same row and in the same column as the given element. Thus, in (10-4) the minors of a x 61, , Ci are, respectively, C2 63 3 tt2 C2 62 &3 176 Determinants The column Sec. 10-2 cofactor of an element which lies in the z'th row and fcth is equal to the minor of that element if i + k is even, and is equal to the negative of the minor cofactor = if i + 1)*+* ( k is odd. That is, minor. Thus, in the determinant in (10-4), the cof actor of &i equals the minor of ai, since 0,1 lies in the first row and in the first column and i + fc = l + l = 2, which is even. Similarly, the cofactor of bi is the negative of the minor of 61, since i4-fc = l + 2 = 3, which is odd. Often the following procedure may prove more convenient for finding the sign corresponding to a given element. Beginning with + in the upper left-hand corner, change sign from place to place, moving horizontally or vertically, until the position for the element in question is reached. The schematic arrangement of signs corresponding to the elements of a third-order determinant is thus as follows : Note that the sign for any position is independent of the path followed in arriving at that position. We shall designate the value of the cofactor of an element by the corresponding capital letter, and we shall use the subscript that occurs with the element itself. Thus, the cofactors of a t 61, c x are, , respectively, A, (10-7) C2 = a.3 C'3 Hence, (10-5) for the expansion of the determinant written as follows may also be : D = (10-8) aiAi + biBi + aCi. This sum is called the expansion of the determinant according to the elements of the first row. We observe at this point that the right member of (10-6) reprepossible products, here 3 in number, that can be formed from the determinant in (10-4) by taking one and only one element sents all ! from each row and each column. of the determinant is It follows also that the the same, regardless of the value row or column Sec. 10-3 177 Deferm/ncmfs according to which the expansion the determinant as D = (10-9) or as D = (10-10) is made. Thus, we may express + b 2 B2 + 03^.3 + 63^3 + c^Cz. These equations represent the expansion according to the elements of the second column and according to the elements of the third row, respectively. Example 10-2. Expand the determinant 2-53 according to the elements of the third column. Solution: 1 7 4 2 1 1 7 4 first The expansion according 2 6 row and according to the elements to the elements of the first 6 1 - 1 4 + 1) + 3(42 + - 6 2 1 7 row of the is This reduces to 2(8 - 7) + 5(24 Expanding according to the elements () 2 -5 -1 7 2 -I- -1 . 7 2 -5 2) = 259. of the third column, = 3(42 + 2) - (14 - we have 5) + 4(4 + 30) = 259. PROPERTIES OF DETERMINANTS From the definition of the value of a determinant we may deduce the following important properties of determinants. These properties supply us with more convenient methods for evaluating a determinant. Note. For a more complete discussion of these properties, the student is referred to any one of the various treatises on determinants or to texts on the theory of equations or on solid analytic geometry, where he will also find proofs which apply to determinants of any order. The properties listed here will be employed in examples that follow, and their usefulness in simplifying determinants will be illustrated. 178 Determinants Property The value of a determinant 1. is not changed Sec. 10-3 if its rows and columns are interchanged. Property 2. If all the elements of a row, or of a column, are multi- same number, the value of the determinant plied by multiplied by that number. For example, the Property b\ ka,2 62 = k 62 0,2 two rows, or two columns, of a determinant are If 3. ka\ is identical or proportional, the value of the determinant is zero. For example, determinant let the first two columns be identical, as in the CL\ D Cii C\ Q<2 C2 Cl'3 Then, expanding according to the elements of the third column, we have 0,2 D = 02 = 0. (73 Property 4. The value of a determinant is not changed if we add to the elements of any column (row) any arbitrary multiple of the elements of any other given column (row) . For example, 61 C2 (12 + rib i 61 Ci + ribz 62 C2 3 The proof first column, last we Expanding according to the elements of the find that + nbi bi 0,2 + nb2 b2 0,3 + 7163 &3 L The follows. ai 61 c\ 2 0,2 b2 02 ^3 dz 63 C.3 c determinant vanishes, since two columns are identical. Sec. 10-3 179 Defermmcrnfs Example 10-3. Evaluate the determinant 43-1 D= Solution: By 5 1 2 4 adding 2 times the elements of the first row to the elements of the second row, we obtain If now we add 3 times the elements of the first row to the third row, the determinant becomes 3-1 4 13 7 14 13 Expanding according to the elements In this example, we first of the third column, 13 7 14 13 = we have -71. converted the given determinant to one in which one of the elements of the third column are zero. For the final but expansion, the given determinant was thus reduced to a determinant of the second order, and was all its value easily found. = 2 satisfies and third rows are proportional. Hence, the Example 10-4. Without expanding the determinant, show that x the equation Solution: Substituting is satisfied by x x3 3 1 6 -4 -x = 0. 2 for x in the determinant, we have This equals zero, since the equation 3x 2 first = - 2. 12 -8 2 3 1 7 6 -4 1 180 Deferm/ncrnfs 10-3 Sec. EXERCISE 10-1 In each of the problems from to 12, evaluate the given determinant. 1 1. 5. 9. * In each of the problems from 13 to equations by means 19, solve of determinants. Check the given system of simultaneous solutions by substitution in the all equations. x 13. 2x 16. If + 30 = - 4y = 14. 5, 7. |2x \3x y= iji) Find the ^-intercept and the 0-intercopt of the x 2 18. Solve graphically the following f 3x y I 4 1 -3 1 + - D represents the determinant in Problem 10, show that D of the straight line through the points On, 17. 15. 1, +2y =4. and line (j 2 , is 20 30 = = 5, 5. the equation 2/2). whose equation is = 0. system of equations: = 0. 19. Find the coordinates of the vertices of the triangle whose sides are the straight x - y + 2 = 0, 2x + 90 + 15 = 0, 7x + 40 - 30 = 0. and lines 20. Find the vertices of the parallelogram formed by the following x y 1 2 1 1 -3 2 1 x = 0, y I 6-11 1 = 0, x y I 1 3 1 2 4 1 = 0, lines: = 0. Sec. 10-4 10-4. Deferm/nanfe 181 SOLUTION OF THREE SIMULTANEOUS LINEAR EQUATIONS IN THREE UNKNOWNS Let us consider the following system of linear equations ix + biy + (10-11) The determinant ciz = C2Z = di, of the coefficients of the i D = : unknowns is 61 C2 0,2 For the solution of such a system, we employ the following theorem, which is known is Cramer's Rule. If the determinant D of the coefficients of the system not equal to zero, the system has just one solution. In this solution, the value of any unknown is equal to a fraction whose denominator is D and whose numerator is obtained from D by replacing the column of coefficients of the unknown in question by the column of constants di, d2 and c? 3 Proof. Let the numerators of the fractions for x, y, z be denoted by DI, D 2 D 3 respectively. We proceed to show that if equations (10-11) are to be satisfied, then Theorem. is . , , , Dx = (10-12) Dy = Di, Specifically, the equation for (10-13) Dz = D3 . x will have the form C2 C2 (12 #3 Z> 2 , 63 63 C3 Similar equations may be written for y and z. To find x the first equation of the system in (10-11) is multiplted by the cof actor At, the second by A 2 and the third by A 3 9 . , After adding and collecting terms, (10-14) (aiAi we obtain + a2^ 2 + a3A 3 )x + (biAi + b2A 2 + bzAz)y + (ciAi + c2^2 + c^A^z = diAi + ^2^2 + ^3^3. x in (10-14) is the expanded value of D according to the elements of the first column. The coefficient of y is The coefficient of biAi + b 2A 2 + Defermmonfs 182 Sec. 10-4 this is equal to b\ 62 &3 &3 <?3 two columns are identical. Similarly, the Hence, we have shown that the left side of the expanded form of the left side of (10-13) and that which equals zero, since coefficient of z is zero. (10-14) is the two are therefore the same. The right side of is (10-14) cM-i + d2 A 2 + d 3A 3 which , is the expansion of the determinant 61 i i This determinant may be obtained from D by replacing the coefficients of x by the column of constants that is, it is the expansion of the determinant that we have called DI. Hence, the equation Dx = DI in (10-12) is established. By a similar procedure we can show that the equations Dy = D 2 and Dz = D 3 are also valid. ; If D T^ 0, the value of x is given by the equation d\ x (10-15) Similar values may bi = be found for y and The proof of (10-12) is valid whether z. The proof is complete. D= 0. If D^ or D ^ 0, the equations of the system (10-11) are consistent and have only one solution, which is of the form (10-15) that is, ; (10-16) If D = 0, = 5"' v = ^' and any one or more of the other determinants, D\, D2 not zero, the given system of equations has no solution and is DZ, is inconsistent. , Sec. 10-5 If 183 Defermmcrnfs D = 0, and all the other determinants are zero, the equations may be consistent or inconsistent. If they are consistent, there are infinitely many solutions. This case will be of the system (10-11) treated in Section 10-5. The following example Example 10-5. Solve the system D ^ 0. of equations - x x 2x Solution: which will illustrate the case for z = 1, 2z = 3, y + +y - - + 3z = 4. y Here = D= 11; = 2. Hence, x = -~ = y ~ = D ~5 ^ = - If O j_J we check by substitution, we find that these values satisfy the given equations. UNKNOWNS WHEN SYSTEMS OF THREE LINEAR EQUATIONS IN THREE 10-5. D We = note that when D = 0, the system (10-11) will not have a any one of the other determinants DI, from zero. Suppose that a solution is given by solution if x = y r, = z s, = D2 D 3 , is different t. Then the equations (10-12) become r = It follows that Z?i = 0, s Z?i, Z> 2 = 0, = D2 Z> 3 = 0. , * = D3 . The following example will illustrate the case of consistent equations where D = and Z?i = D 2 = #3 = 0. The equations are said to be dependent, and they have infinitely many solutions. The student should construct an example to show that the equations (10-11) may be inconsistent when D = 0, even though Z>i = D2 = D3 = 0. 184 Determinants Example 10-6. Solve the system Solution: 10-5 of equations x + y 2x - y 4z - 2?/ - z=3, + 30 = + 62 = Here Z) Sec. = 1 1 2 -1 4 -2 - 1, 2. 1 This equals zero because the second and third rows are proportional. But, we also find that 13-1 1-1 1-1 3 2-2 6 3 D = l 1 3 42 6 2 1 1 3 2-11 4-22 = 0. In this case the given equations have a solution. In fact, the second and third equations are proportional, and so either of those can be solved together with the first equation for two of the unknowns in terms of the third. For example, x =4 - 5 -f 5z 2z Thus, we have a single value of x and a single value of y for every value of However, there are infinitely many values of solutions of the given equations exist. 10-6. and therefore z, infinitely z. many HOMOGENEOUS EQUATIONS The system (10-11) is homogeneous if di = 0, d 2 = 0, and d 3 = 0. = y = z = 0. When Such a system always has the trivial solution x = d2 = d3 = 0, it is seen that DI = D 2 = D& = 0, for each of these determinants has zero for every element in one column. If D 0, it follows from (10-16) that we can have but one solution, which is given by di = the given system is to have a solution besides the trivial = 0, nonmust equal zero. It may be shown that, if solution, Hence, if D D trivial solutions always exist. Example 10-7. Solve the system x of equations - 2x K/p y 3y _ 2?7 + 2=0, + 4z = 0, " - 9. f) 10-7 See. Solution: Determinants 185 D is The determinant = 0. Therefore, nontrivial solutions exist. To find these solutions, we proceed as follows: Transpose z in each of the first two equations, and solve for x and y in terms of 2. Then we have - 1 D= - z 1 1 = - 2-3 -3 -42 1 -2 2 -42 2, Hence, = I -p = 2, j and i/ = 2 = Substitution shows that those values also satisfy the third equation. The given system therefore has infinitely many solutions, and the values of x, y, and z are related by the equations x = z = and y 2z. EXERCISE 10-2 In each of the problems from to 1 equations by means of determinants. 6, solve the given system of simultaneous all solutions by substitution in the Check equations. 1. + y - 2 = +3y - z = + y -3z = 3x x 3 11. I2x 2x - 6x + 2y 20 10, 2x 5. + 2-2, - 2z = 5. x 3jc - y - For each of the following systems find at that there is no nontrivial solution. + 22 = 0, 7. 2x - 8. +2=0, 10-7. 2x x - 2=0. = -1, = 0, 22 7z - 3y least 4- 2z 3 2.r 32 62 x 9. 0, +3y - 2=0, + 3y + 2x 6. - + 62 = 7, + 3y + 62 = 0, + 2y + 92 = 3. - y = 3, - 32 = - 1, - y ~ 2. ?/ one nontrivial solution, or show = 42 x 3. -2y - 4y + z=3. - y - 32 = 7, + 2y - 2 = 10, - 3# + 22 = - 7. 13, 5x = - 4. x 2. 11, [ J2.c = 0. 3x - + 2y y 2/ + 32 = 0, +42 =0, - s = 0. SUM AND PRODUCT OF DETERMINANTS .Closely related to Property 4 of Section 10-3 is a theorem concerning the sum of determinants. We shall illustrate the theorem for splitting the elements of a given column into two parts by means of the following equality for third-order determinants + 61 + 62 03 + &3 ai ci 0,3 02 0,2 2 Cs as 03 d\ d\ 62 : 186 Deferm/ncmfs Sec. 10-7 This can be shown to be true by expanding the three determinants according to the elements in the first column and noting that in the expansion the minors are the same for all three determinants. Example 10-8. Show that Solution: Expanding each of the determinants according to the elements in the This result is first on the left side of the equation column, we have the expansion of the determinant on the right in the given equation Computing the value of each determi- according to the elements of the first column. nant in the given equation, we see that each side reduces to 47. A similar theorem for splitting the elements of a given row is illustrated by the following example : ttl + Ci 61+6 CL2 (12 62 This can be shown to be valid by expanding according to the elements of the first row. Thus, consider the determinant This may be written as a 2 sum 3 4 5 1 7 in various ways. 2 2 Examples arc 4 5 4 5 -2 2 3 5 and 1 We 7 7 1 product of two determinants of equal to a determinant of like order in which the element of the ith row and kth column is the sum of the products of the elements shall state without proof the rule for the the same order. of the ith The product row of the first is determinant and the corresponding elements of the kth For example, for second-order determinants, column of the second determinant. we may write +2-7 1-6 +2-8 1 2 5 6 1-5 3 4 7 8 3-5 +4*7 3*6 +4*8 Sec. 10-7 187 Deferm/ncrnfs That the product of the values of the two determinants on the left equals the value of the determinant on the right is checked by expanding. Thus, the desired equality becomes (4 - (40 6) - 42) = (950 - 946), or (-2). (-2) To is illustrate the multiplication of = 4. two third-order determinants, we have that equal to 3-1 +2(-3) 1-1 3-7+2-94- 5( - +5-5 +(-!)( -3) +2-5 .4-1+ 6(- 3) +0-5 l-7 4. 4) + (-l)9+2(-4) 7+6-9+ 0(- 4) 3-6+2-3+5-2 1 6 + (- 1)3 +2 2 4-6+6-3 +0-2 This reduces to 22 which 30(- is equal to 21), 630. which equals Also, - 19 34 14-10 7 82 42 14 computing the values of the given factors, we have 630. EXERCISE 10-3 In each of the first three problems, combine the given determinants into a single determinant, and evaluate the result. 1. 3. 188 Determinants Sec. 10-7 In each of the following problems, find the product of the determinants without evaluating the individual factors. 4. 8. Complex Numbers 11 11-1. THE COMPLEX NUMBER SYSTEM There are many problems that cannot be solved by the use of numbers alone. We observe, for example, that the equation x2 + 1 = has no real root, since x 2 can never be negative if # is a real number. In order to provide solutions to such equations, a new system of numbers, called the complex number system, was intro- real duced. Later in this book, we shall find many instances of solutions involving complex numbers. We now define a complex number as an ordered pair of real which we denote by (a, b). If the numbers a and b are numbers, shall regarded as the Cartesian coordinates of a point in a rectangular coordinate system, we have a one-to-one correspondence between the set of complex numbers and the set of points in a plane. The plane is called the complex plane. Two complex numbers (a, b) and (e, d) are equal if and only if they correspond to the same point, that is, if and only if - a c and b - d. Addition, subtraction, and multiplication of complex numbers are defined as follows : (a,6) (11-1) (11-2) (a, 6) (11-3) + - (a, 6) =(a + c,6 + d)i = (o - c, 6 - d), = (ac - 6d, ad + (c,d) (c, d) (c, rf) be). For example, (1, 3) (1, 3) + (5, 2) - (5, (1,3). We 2) (5, 2) = (6, 5), = l- 4, 1), = (-1,17). also define the following special (11-4) 0- (11-5) 1 (11-6) i = (0,0), (1, 0), (0, 1). 189 complex numbers : 1 Complex Numbers 90 Sec. 1 1-1 serves as a zero of the complex number serves as a unit, in Accordance with the following The complex number system, while properties 1 : + - 1 The so-called = = = (a, 6) (a, 6) (a, 6) (a, 6) = + = = (a, 6) 1 (a, 6) i= imaginary unit (a, 6), 0, (a, 6). will be discussed in (0, 1) more detail in Section 11-2. If A; is a real number, * (11-7) Also, we (a, we define = (fc, 0) 6) = (a, 6) (fca, kb). define - (11-8) =(-!) (a, 6) = (- a, - 6). 6) = (0, 0), the complex number (a, 6) Since (a, 6) + (a, called the negative of (a, 6). The Reciprocal (x, of (a,b). ^ 0, If (a, &) then it (a, b) is has a reciprocal y) such that Furthermore, the reciprocal (11-9) = (x, y) (a, 6) 1. given by is (x , y) ^ (11-3) and (11-5), the equation written in the form By (ax Since a + & ^ 0, taneous equations 2 2 - ay by, + bx) (& = - 6x + by = 1, ap = 0. '(#,?/) =1 may be (1, 0). we may determine x and ax &) by solving the simul- T/ j 1 The values of x and y can be foundry any of the methods taken up in Section 9-1. The solution is X a ~a 2 + We have, therefore, verified _ ~ y b*' (11-9) Division of (a, b) by (cy d). If as follows b a2 + 6a <* . (c, d) = 0, division can be defined : (a, 6) where (u,v) (11-10) is -5- = d) (c, (a, 6) the reciprocal of (c,d). (a, 6) -!- (c, d) = (u, v), By (11-9), (a, 6) ac ( + bd ad + bc we have Sec. 1 1 -2 Complex Numbers For example, (5, 13) + (3, 1 91 -, -2) = This result can also be verified by the method in Section 11-3. THE STANDARD NOTATION FOR COMPLEX NUMBERS 11-2. The ing property shall i - (0, 1), defined by (11-6), has the follow- : i* (11-11) We number special = (0, 1) now show (0, 1) that = (- (a, 6) 1, 0) = - (1, 0) = - 1. and the binomial form a 4- bi are equivalent, or that (11-12) in (a, 6) which a Finally, = a + bi, + bi means al + bi. By (11-5), (11-6), and al + bi = a(l, 0) + 6(0, 1) = (a, 0) + (0, 6). (11-7), by (11-1), we have (a, Hence, (11-12) is 0)4- (0,6) = (a, 6). established. Real and Imaginary Parts of Complex Numbers. We call a the and & the imaginary part of the complex number a + bi. If a and 6^0, a + bi reduces to bi, which is called a pure imaginary number. If 6 = 0, the complex number a + bi, or al 4- bi f reduces to the complex number al, which may be identified with the real number a. The complex numbers, then, include both the real part numbers and the pure imaginary numbers as special cases. Illustrations of various classes of numbers follow Some real numbers are 2, 5, and \/3. Some pure imaginary numbers are 3i, and x/Bt Some complex numbers are 2 + 3i, and \/S i. Note that the numbers 2, 5, \/3, 3f, and \/5i may be put into the standard form a 4- bi and written, respectively, as the complex + 3f, and - \/5f. Since numbers -2 + Of, 5 4- Of, \/3 4- Oi, = + Of, which may be written briefly as 0, we shall drop the use real : . of bold-face 0; similarly for bold-face 1. Conjugate Complex Numbers. The conjugate of a complex num+ bi is defined as a bi. Likewise, a + bi is the conjugate of a 2f bi. Some pairs of conjugate complex numbers follow: 2f, 3 + 5f 3 own is A its x 5f real x + number and 2yi 2yi, ber a ; , conjugate. ; . 192 Complex Numbers Powers of t. = ft 6 i = i 2 . = _^ = i2 i3 seen that It is readily Sec. = i i, i* = i2 i 2 11-2 = 1, an(j so on Therefore, successive positive integral powers of i have only four different values, namely, these four values are repeated in regular order. i, i, and 1 1, if n is Hence, any positive integer, we have in general i ft 9 ; These relationships afford a simple method for evaluating powers 3 i = i of i, as shown by the following illustrations i 7 = i* +3 38 = 2 = 103 4 26 1 = = = i &*+* i i -1 and i i. : ' ; - ; 11-3. ON COMPLEX NUMBERS OPERATIONS STANDARD FORM IN From the definitions given in Sections 11-1 and 11-2, it follows that a 4- bi and c + di can be added, subtracted, multiplied, and divided as if they were real binomials, except that, where i2 appears, it is replaced by 1. Addition of two complex numbers is effected by adding their real and imaginary parts separately and subtraction is performed by subtracting their real and imaginary parts separately. Thus, in accordance with (11-1) and Algebraic Addition and Subtraction. ; (11-la) (a + bi) (a + bt) + (c + di) = (a (c + di) = (a + c) + (b c) + (b + d) i, d) i. and (ll-2a) - - - For example, + (3 20 + (4 - 50 = (3 + 4) + (2 - = 5)i 7 - 3t, and (3 + 20 - (4 - 50 = (3 - 4) + + (2 5)i = - 1 + 7i. We note that the sum of conjugate complex numbers is a real number, because (a + bi) + (a bi) = 2a. Also, the difference of two numbers a pure imaginary number, because is conjugate complex = + 2bi. (a bi) (a bi) Algebraic Multiplication. To find the product of two complex numbers, multiply them according to the rules of algebra, and 2 replace i by -1 in the result. Thus (a Replacing (ll-3a) i 2 + 60 (c + di) = ac by -1, we have, (a + bi) (c + di) + adi + in = bci + bdi 2 . agreement with (11-3), (ac - bd) + (ad + bc)i. Sec. 11-3 Complex Numbers 193 respects the notation a + bi is more convenient than In (a, 6) particular, the former notation makes it easier to remember how to multiply two complex numbers. For example, In many . + (3 20 (4 - 5z) = (12 + + 10) + 8)i = 15 (- - 22 7t. The student should note that the product of two conjugate complex numbers is a non-negative real number, because (a 4- bi) (a bi) = a2 + b2 . Algebraic Division. To obtain the quotient of two complex numbers, multiply the numerator and the denominator by the conjugate of the denominator. Thus, if c + di ^ 0, + bi c + di a + bi c + di a __ ~~ c _ + adi di __ ac ~~ di c c (ac + bd) c2 Therefore, + . a ,. bi __ ac ~ 'c~+~di The right member is of the , + (be + d2 , , ad)i , + bd ~^ be ad 2 2 c2 + d2 c + d form A + Bi, and , bdi 2 bci d 2i 2 2 . lm this equation agrees with (11-10). ; Example 11-1. Reduce -i The conjugate Solution: to the form a of i is i. Example 11-2. Find the value of 5 Solution: Represent the division as the denominator by 3 5 + 3 - 3 13i ' 3 2i + 2i. We +2i ~ __ + + bi. Then + 13i divided -~ ^ > by 3 - 2i. and multiply the numerator and get (15 - 26) + (39 + 9+4 2i 10)i "" _ 11 13 + 49 . *' 13 EXERCISE 11-1 In each of the problems from 1 to 12, express the given quantity in the form a* and give its conjugate. In working these problems, note that V + bi a l. v^7^. 2. 4. V- 5. 7. 3 10. ^' 2 . V~ V- * V2 + 3 \A 2 2. 2 - 8. 11. V 17^. - 3. V4 - 3 VV15 V- 64a 3(xi 2 . 6. 3 6. V V- 9. 1 16. 4- - 12. 3 a? V+ V- 32a 6 +2 8. 2 3 . Complex Numbers 194 In each of the problems from 13 to 13. i. 26, compute the value . 15. (-i) 13 -i 70 19. 37 183 . i 23. i 14 14. i 12 17. i". 18. 21. i s i 29 22. . - 25. i 25 i + 50 i Sec. - i 75 (-i) of the given expression. - (-i) 17 16. . 235 . - (- 20. i) 11-3 (-i) 24. i 10 18 . + i 30 i ao + . 602 . . 26. i 10 -f i 100 -f i 1000 -f i 10000 . 100 . i In each of the problems from 27 to 36, find the values of x and y which satisfy the given equation. = = 27. (2z, 3y) 29. (to, 5y) (18, + - 34. 2z - (y 40i 36. x)f = = 6 - - 2yi 3-ct 1 +2y,x + = 5y) (20 + 3f. + 8x - + 3, - 50 = 12 + 2x 7 + + 6 5y + = 1) (x, 1). 19). 35. 3x 2yi (8, 9). to y), - 33. 3z 2st. + + 30. (x 25). 31. (to 32. z = 28. (2*, 3?) (3, 1). - 6 - 20)i = - 30)i. - 2)i. (5 = (4 + (x 0. In each of the problems from 37 to 78, perform the indicated operations and reduce to the form a + bi. 37. (2, 40. (1, - 3) - 43. (4, 46.(0,1) 7) + - 3) 38. (1, (5, 6). - (7, - 41. (2, 3) 3). 44. (0, (4, 3). - + 1) - 39. (1, (3, 2). (1, 1). I) 3 . 4 V) 42. (1, m 45. 0) (1, + ' (0, 1) - ~ (1, 1). 47 . _ - ' 3 ^~) 50. V"17^ - V^2 52. ( - 54. (3 2i) + 2i) + 3i) 56. (6 58. (6 61. (5 1) -f- ( - - (6 -f 2i). 67. (3< + 4) + - 8) (2i (1 71. (3 72. (6 3 76 IU ' (2 + i) - 57. - 59. (2 + 5i) (6 62. (3 - 3i) (4< 3t) (2 3i). +2). - ( 4i). -f- 66. (3 69. (2 68. (2 + i) 2 + (5 - f). 2 - V3i) (4 + 5i) - v/3i) 2 + (3 + V30 2 - (4 - 3i) (i - 2i) (1 -f f) (1 - 3i) + (4 - 5i). i). 70. (3 (6 55. 65. (5 4). - + 3i) + (2 - 3t). + 9i) + (5 + 2i). - 3 + 2i) - (5 - 2i). - 5 V 17!). (2i + 3) + (8 60. 2f (5 + 3i). 63. (1 + i V2) (5 2f). 51. (4 3f). 5i). V^ + V^2 - V9. 53. (8 + 5i). 17!). (3i + V" 4- (3 - + a. - 6 (3 -f 64. (5< 73. 1 - - - + 3i) 2 49. ' 5 -s- -4i "' (3-2i) numbers - 5i) *- 3t) -s- - 5i). (2 - 4<). (5 W) (2i 6). 75. i 1. j^+j)^^^ (6 79. Prove that complex -5- f) . 74. (6 5t). - '' 7) satisfy the associative 4- (4 - (4 - 7) if (a + bi} (c + di) = 0, then a + bi = (4 + Gi) + ' ) and commutative laws of addition and multiplication and the distributive law. 80. Prove that 3i). 5i) (8 (3 -I- or c + di = 0. Sec. 11-4 Complex Numbers 195 GRAPHICAL REPRESENTATION 11-4. As we have P point ~ seen, the complex number a + bi determines a definite in the plane whose rectangular coordinates are x = a and Conversely, to every point P in the plane corresponds a combi for which the values of a and b are the respective rectangular coordinates of P. See Fig. 11-1. In this system, the real numbers a + Oi are represented by points on the #-axis, which is called the axis of reals. Pure imaginary numbers + bi are represented by points on the 2/-axis, which is called the axis of imaginaries. y 6. number a + plex a FIG. 11-1. FIG. 11-2. more convenient at times to represent the complex number by the vector drawn from the origin to the point P. The a~ + 6 2 and length of the vector is given by the relationship r = the direction is given by an angle determined from the equations a - r cos 6 and b r sin 6. In Fig. 11-2 is indicated the graphical addition of the two complex numbers a + bi and c + di, which may^ be represented either by the points P l and P 2 or by the vectors OJ?i and OP^ Withjths completion of the parallelogram OPiP-JP^ the sum of OP l and OP2 can be represented by the diagonal OP 3 Thus, either P 3 (a-f c, It is a+ bi V , . d) or OP represents the sum (a 4- c) + (6 + d)L Hence, the vector which represents the sum of two complex numbers is the sum of the vectors representing the given numbers. 6 + To and l3 subtract c c + di from a + bi graphically, we merely add a + bi di. Example 11-3. Add the complex numbers 2 + 3f + 2t let P 2 (6, and 6 graphically. Pi (2, 3) represent the number 2 + 3f and 2) represent number 6 + 2i. Draw OPi and OP 2 in Fig. 11-3, and complete the parallelogram OPiPaPa. Then Ps^represcnts the sum 8 + 5i of the complex numbers 2 + 3i and 6 + 2i, and OP 3 represents the sum of the vectors OPi and OP 2 Solution: Let the . Complex Numbers 196 Sec. 11-4 /'3 (8,5) o FIG. 11-3. The difference of two complex numbers may be obtained same manner if we apply the relationship (a + - to) (c + di) = (a + bi) + (- c - in the di). P and Q represent the numbers a + bi and c + di, respectively, complex plane, as shown in Fig. 11-4. Then c di is represented by Q', which is the reflection of Q through the origin. Let in the P(a,b) Q fad) O FIG. 11-4. Let us recall how the difference of two vectors was represented graphically OP, OQ,jand OQ' represent a 4- Section 6-7 and the vectors respectively, then representing the 11-5. OR is number was explained in Fig. 6-15. If bi, c + the desired vector and (a + (c bi) + di) di, R we in let and -c -di, is the point . TRIGONOMETRIC REPRESENTATION Let the complex number a + be represented by the radius Then the distance = r is called the modulus, or^the absolute value, of the complex \OP\ number and the angle 0, which OP makes with the positive #-axis, vector drawn from the bi origin to the point P. ; is called an argument, amplitude, or angle of the complex number. Sec. 11-5 From Complex Numbers clear that is Fig. 11-1, it = a - = b = and - cos 6 and 8 r cos = 197 sin 0, or r sin 6. Hence, a + bi = + r cos = (r sin 0)i + i sin 0). r(cos This last expression is known as the polar form or the trigonometric form of the given complex number, as contrasted with the standard or rectangular form a + bi. To reduce a given complex number a + bi to the trigonometric form r(cos 9 + i sin 0), we find r and by means of the relation2 2 a = = b 6 r r and r cos 0. sin 0. We have + ships V^ , a Example + = bi rf - + - i) = Represent the 11-4. + r (cos i sin 0). complex number /5 9 + ^r~ and change the given notation i graphically, V3/2 to the trigonometric form. Solution: /I are [ \J a = The P whose point rectangular coordinates 1 ., \/3\ ^- /) represents the number Z ^ , , > 1/2 and 6 quadrant angle. = \/3/2 are both We from Fig. see = 7T/3 or 60. Hence, 1 J \/3 zr Here a = determined from cos 6 1 and 1 1-5 that = = = 1. ^ and Since * r FIG. 11-5. = j^l- The 1/2 and sin 6 f = + ^1 ^ 6 1/2 i l (i \J So sin r = V2 = = ^ + Example 11-5. Express the complex number Solution: 6<r . . positive, ^ is a first- determined from the equations cos 6 let 9 + , ^ f) i \/2, and y= = C os this case 60 / 1 -- \X/2. In angle 6 is we may + i sin 60. in the trigonometric form. is We a fourth-quadrant angle thus have -\/2 -4= - --Ut = \/2(cos 315 + i sin 315). EXERCISE 11-2 In each of problems from jugate graphically. 1. 3 2. 8 2f. + 5. 3 - 5i. 6. 1 10. 1 to 12, represent the complex + 2i. 3. 3 - 7. i. 1. 11. + . number and 4. its 2 - con- 3t. 8. 1. (1 - 12. 5 + 12. 1 Complex Numbers 98 Sec. 1 1-5 In each of the problems from 13 to 24, perform the indicated operations graphThen check the result algebraically. ically. - 30 + (- 4 + 0- 60 - (4 + 30. + 20 - (5 - 0. 13. (7 15. (3 17. (3 21. 3 23. 7 + + 19. (5 - (1 - (4 20. (3 - 16. (6 50. - 40 - (2 + 0- 20 - (- 2 + i V). (1 18. (3 + 30 - (4 - 5i). - 2i) + (6 + 20+ 40 - - 2 - 40. 14. (2 ( 22. (v/2 24. (4 + 2i) - (5 + + 30 - - 40. + V&) - 7i. + 30 - (3 + 20- (1 (2 In each of problems from 25 to 36, change the complex number to the trigonometric form and represent 25. 1 + 29. |(1 QQ * Let graphically. - 26. i. + V20. , d4 ' MULTIPLICATION TI (cos a+ i - V30- W ~ ,, AND 27. 5. 30. ~(1 1 -i 6 -4i -2TT -3-+!' 11-6. it } ,., 4l) - 31. 5 Q(5 ** 2 ' a + i sin a) = fi^Kcos and r2 (cos sin a) /3 r% (cos /3 a cos /3 + i sin sin we have proved 32. -? Q, 3 1 + i sin /?) 3 \/3i. -2i * 2 DIVISION IN TRIGONOMETRIC +< FORM be any two com- their product is given /3) a sin /3) = Thus, 12t. 6~+' plex numbers in trigonometric form. Then, by the relationship TI (cos + - 28. 3 3t. + i(sin a cos + cos a sin ]8)] ]8)+ isin (a+ j8)J. /3 rir2[cos (a+ that the absolute value of the product of two the product of their absolute values, and an angle of the product is the sum of their angles. The result found for the product of two complex numbers can be complex numbers is extended to the product of three or more complex numbers. The quotient obtained by dividing the complex number 1*1 (cos a + i sin a) by the complex number r2 (cos/J + isin/3) is given by the relationship ri(cosa r2(cos ft + fr'sinoQ _ + i sin ~~ ft) = i sin + i sin a) cos i sin cos i sin + - + i sin (a (a ri(cos a r2(cos ft [cos ft) |8 ft ft ft ft) It follows that the absolute value of the quotient of numbers quotient is is ft)]. two complex the quotient of the absolute values, and an angle of the the difference of their angles. 11-7 Sec. Complex Numbers Example 11-6. Find the product f sin 199 + of 2(cos 30 30) and 3(cos 120 i sin the rule for products in polar form, we have i sin 120) 3(cos 120 By Solution: + i sin 30) 2(cos 30 = 2-3 [COP = 6 [cos + (30 + 150 120) i sin + + i sin = 150] 6 (30 + - ^~ + ( 120)] - 3 - V3 + ( ) 1 i is divided + -^A by Zi From Examples 11-4 and 11-5 Solution: - 1 = i and i - 1 1 + 2 _ i \/3 + i sin 315), + cos 60 + ft (cos 315 cos 60 . i sin -f i sin = - [cos \/2 (cos - 11-7. 1) + i Hb ^ (315 - 60)] + i sin 255] 75 + i sin 75). we [0.2588 +i _ \~ (V3 - i sin 255 Or, using exact values of cos 75 and sin 75 L. 60 - 00) + [cos (315 - V2 - V2 315) l a table of trigonometric functions, r 60. i sin 2 = V2 = V2 From = i f. in Section 11-5, \/2(cos 315 ~ + i /^i i Example 11-7. Find the quotient when find that the result is (0.9659)]. previously found, we obtain -i (VS + 4: 1) _J = - | [(V3 - + i (V3 + 1) Z 1)]. DeMOIVRE'S THEOREM we extend the law n factors, we have If [ri(cos di = If + 120). + of multiplication of the preceding section to i sin 0i)] [r2(cos 02 + 2 rir 2 - - how we put - rn [cos(0i TI = r2 = + + i sin #2)] - - + n) [rw (cos + n + i sin (0i + = = rn = r and 2 = ' 2 = i sin + On - = - n )] - + 0*)]. 0, it fol- lows that [r(cos 6 + i sin 0)] n = r n (cos n0 + i sin n0). known as De Moivre's Theorem. Although we have derived De Moivre's Theorem This result values of n, is it can be shown to hold for erly interpreted. all real only for integral values of n, if prop- Complex Numbers 200 Example 11-8. Find the value Since 1 Solution: \/2 + i sin (cos 315 - (1 = i A/2 i) 7=- ( -p-i The two By De + i sin 1 i is 4 )] - = = +i sin 0) = (cos if 2, 315)] by De Moivre's Theorem. sin 26 we have + cos 2 2 sides are equal only and for cos 20 Moivre's Theorem for n 20 the polar form of > = [\/2 (cos 315 + i sin 315 = (V^) [cos (4 315) + i sin (4 = 4(cos 1260 + i sin 1260) = 4(cos 180 + i sin 180) = - 4. 4 i) Example 11-9. Derive formulas cos 20 ] \\/2 V2 / 315). Hence, by De Moivre's Theorem, 4 Solution: 11-7 by De Moivre's Theorem. 4 of (1 Sec. (2 cos sin 0) t - sin 2 0. the corresponding real and imaginary parts are equal. Hence, - cos 20 = cos sin 20 = 2 sin 2 sin 2 0, and 11-8. cos 0. ROOTS OF COMPLEX NUMBERS Let p(cos + tsin$) be an nth root of the complex number r (cos + i sin 0), where ^ ^ 360. Then < [p(cos By De + i sin <)] n = r(cos 6 + i sin 0). $ Moivre's theorem, this leads to p (11-13) n (cos n<f) + i sin rup) = Our problem now is to find for which (11-13) angles all is <f> + i sin 0). KCOS non-negative numbers /> and all satisfied. Separating real and imaginary parts, we have n p cos (11-14) ncf) Squaring and adding, p 2n (cos = we 2 n<t> r cos 0, sin n< = r 2 (cos 2 6 r sin 0. obtain p (11-15) (11-14) n + sin 2 w$) = 2 2* 2 Therefore, p = r , since cos a is then given by the equation From p + = we then have cos n<j> = cos 0, sin 2 a= 1. + sin 2 6). The absolute value p ^ sin 0. sin n< from these equations that the angles n$ and 6 can a only by multiple of 2ir or 360. More precisely, It is clear (11-16) where 'fc n<f> is = any +k integer. 360, or = ft 7i + If . differ 11-8 Sec. For Complex Numbers 201 = , 0, 1, 2, (n-1) in (11-16), we obtain n distinct values of the angle, all of which are non-negative and less than 2?r or 360. Corresponding to these angles we obtain n distinct roots fc given by the formula /b-360 cos (11-17) For example, for k cipal root, 0, we 1 fc . . . J- n sin 360 n obtain one nth root, called the prina with absolute value r 1/n and angle n for k ; 1. The value so on to would yield the same root as k n.860* = 6 cos \ fl .n + sin (\Tb k k n+ n n D ^ 360)f = sin and 0, - and Similarly, it same root as This means that 1 yields the and so 1, only cos - + 360)/ = (\n ; ' n n / we have 360 n k=n k= n 1, + a second root, with absolute value r 1/n and angle since = on. distinct roots exist. It is interesting to note that the points which represent the roots are equally spaced on a circle whose radius is <\/r and whose center is the origin. This fact is illus- trated in the following example and Fig. 11-6. FIG. 11-6. Example 11-10. Find the three cube + i sin 60). Solution: The 60 three cube roots are found + i sin 60) 60 = 2[cos(20 As just shown, the substitution of Hence, for k = 0, we have 2(cos 20 + 140) and for k = by evaluating (11-17). Thus, +k 360- 3 = t sin roots of 8(cos 60 the root is +k = 0, i sin 120) 1, 20) ; . . + 60 + i sin , + i sin (20 i we have 360> 3 +k 120)]. 2 yields the three required roots. for k 1, the root is 2(cos 140 = + i sin 260). + The roots are repre- 2, 2(cos_260 sented by the equally spaced vectors OPi, OP 2, and OPs, terminating on the circle whose radius is 2 and making angles of 20, 140 J, and 260, respectively, with the ; positive z-axis. Complex Numbers 202 Sec. 11-8 EXERCISE 11-3 In each of the problems from expressing the complex form. 1. 3. (1 + i) (1 - number 1 to 18, perform the indicated operations by first Express the answer in rectangular in polar form. 2. V3i). (-1 + V3i) (\/ + (l + i) (V + i). 1 ~ l 4. 0. fci'-VS. 1 +* V3 + 7. - 3f . V3 + i 11 "' " 2 13.,, (1 +* 1 .... +.-.. (3 + 4i) (2 + 1) (3 (2 + 3t) + 1) 14. ) + *) (V3 .7. i(l - _ -1 100 / 20. V2i)J [1 1 1 _ \70 (-i + iV In each of the problems from 22 to 29, find roots as directed and represent them graphically. + i sin 22. Find two distinct square roots of 9 (cos 50 23. Find four distinct fourth roots of 16(cos 36 + i sin 24. Find three distinct cube roots of 27(cos 165 + i sin - 25. Find 26. Find the three cube roots of 27. Find the four fourth roots of 1. 28. Find the two square roots of i. 29. Find the three cube roots of -^ five distinct fifth roots of 32. 1. (1 + V3i). 50). 36). 165). Sec. 11-8 Complex Numbers 203 In each of the problems from 30 to 34, the complex numbers E, voltage, current, and impedance, E when I = 5 = 4 respectively, and E= 7, and 31. + 4i amperes and Z = 30 - Si ohms. Compute / when E = 110 + 3(K volts and Z = 20 - I5i ohms. 32. Compute Z when 33. When two 30. Compute 7 + 3i amperes and E = impedances Z\ and --=+ A L% 2 115 volts. are connected in parallel, the equation determines an equivalent impedance Z. ju\ = 5 34. If 2 + 4f ohms and and z are #2 =8 6i Z designate IZ. ohms. conjugate complex numbers, prove that Compute Z when Z\ Equations in Quadratic Form 12 12-1. QUADRATIC EQUATIONS IN ONE UNKNOWN This chapter provides an extension of the work on linear equations to second-degree, or quadratic, equations. Consider a quadratic equation in one unknown written in the form ax 2 (12-1) where a, 6, and c + bx +c= (a are given real numbers. and called the general quadratic equation in x, ^ 0), This equation is is said to be in standard form. If 6 T^ 0, (12-1) is called a complete quadratic equation; if b = 0, it is called a pure quadratic equation. Thus, 3x~ a; + 4 = is a 4; and complete quadratic equation in which a = 3, b 1, and c = x2 2 1 and c 2. is a pure quadratic with a In Section 12-4 we shall prove that every quadratic equation has two and only two solutions or roots. The roots may be equal or unequal, and they may be real or complex. Their natures depend on the values of a, 6, and c. We eral use for finding these roots, well to the solution of equations shall consider the in gen- and we shall then apply them as which are not quadratic in x but which can be written as quadratic equations ing the unknown. 12-2. methods in expressions involv- SOLUTION OF QUADRATIC EQUATIONS BY FACTORING form can be the solution of the equation depends on the following factored, If the left side of a quadratic equation in standard important principle The product of two or more numbers equals zero at least one of the factors is equal to zero. : That is, A B= if and only if A= or B = 0. if and only if Sec. 122 Quadratic Form in Equations 205 In practice, we apply this principle by equating to zero each linear factor of the left side of the given quadratic equation, and solving the resulting linear equations. The following examples will illustrate its application. Example 12-1. Solve 2x* - 7x To Solution: find the values of + = 6 by factoring. x which satisfy the equation 2x 2 7x + 6 = 0, write the left side in the factored form - (x This product equals zero and only if - x 2) (2x = 2 2 or x = 3/2. Moreover, Hence, x 3/2 satisfy 2x* - 7x + 6 = 0. Thus, _ = 0. 3) either if - 2s or = 2(2)2 - = 0. 3 2 and 3/2 are solutions, because both 2 and +6=8- 14 +6=0, 7(2) and 2(3/2) 7(3/2) +6=9/2- 2 a) Solve the equation 2 sin Example 12-2. all - 2 non-negative angles x - x +6=0. 21/2 sin - x = 1 for sin x. b) Find than 360 which satisfy this equation. less Solution: a) Factor the given equation to obtain (sin Since sin x 1 = or 2 sin x x - + 1 1) (2 sin = 0, it + x = 1) 0. follows that sin x 1 or sin # = 1/2. Check: 2(1) 2 -(1) -1=2-1-1=0, and b) x = When sin x = 1, = x 90; when = - sin x 1/2, x = 210 or 330. Therefore, 90 or 210 or 330. Check: 2 sin 2 90 - sin 2 sin 210 - 2 sin 2 330 - 2 - 90 1 210 - sin 330 - sin = 1 2 Note that / = 1 2 - 2^ 0, - ^) 2( = - - 1\ 2 - / ^J ( - 1 = 0, - 1 = 0. ^) ( - 1\ ) which the unknown is a trigonometric thus have only two values of sin x which satisfy the this equation is a quadratic in function of the angle x. equation. 1 - = and 1 - We The determination solution of the quadratic, and of the angle, however, goes it may beyond the algebraic happen that there are more than two values of x which satisfy the equation. For this reason, it is recommended that all solutions be checked by substituting in the original equation. 206 Equations In Quadratic Form Example 12-3. Solve the equation 3 - sec x = write - values of x less than 27r radians. = Solution: Since sec x and of fractions, transpose, - 2 cos 2 x 3/2 or cos x x for which cos x number we can > - 2 cos x 3 1 for all = 2 cos x cos x 12-2 non-negative We then clear 1. factor, to obtain cos x = Hence, cos x real 1 cos x Sec. - 3 = (2 cos j; = - 1. When = 3/2. - z 3) (cos = - cos x + 1) 1, x = 0. = There TT. no is Cfcecfc: 3 sec = TT 2 cos TT - 1, or 3(- = 2(- 1) - 1) 1. should be noted that factoring provides a method of solving any pure 2 c = 0. Thus, the equation is equivalent to quadratic equation ax It + which gives + x /i - =0 x or /i/ -= d r or x = This result agrees with that given by writing x 2 - /!/ r (Z and then simply extracting square x = \/ - > 0. "~~ / zb ^ when - a 0, roots both of Note that the roots are real when - ^ & obtain to sides = and pure imaginary EXERCISE 12-1 6. In each of the problems from 9 than 360 which satisfy the given equation. Solve each of the following equations for x or to 20, find Check all all non-negative angles solutions. 1. x2 3. 2x(x 5. x2 = 0. + 5) = - x = 6. o;3 - 7. + + 9. sin 2 6 7x 27 - (sec 6 cot cot 19. 3 sin 2 1 - 3. +3) =0. = 0. = + 13. sec 17 3x(x sin 11. sin 15. (cot less 2 esc + 6) l).(cot +4 ~ +6 2. z2 - 10x 4. z2 + 6.r 6. 6x 2 8. 16x +21 =0. - 27 = 0. - 4x - 192 = 0. - a 2 + 2a6 - 6 2 = 0. = 16. + 2) = 4. + 3 tan 14. 3 cos 2 - 8 sin 2 esc 0=4 sin* - 2 = 0. = 0. 4 12 1 16. 6 3 2+3 cos 1ft 2 3 cos 20. * A 4 cos 2^ a 2 esc esc 8 * + 6) 0-10 sin 0. 0. esc 0. 12. 2 tan 2 143 (cot = 10. sin 6 +3 "*" +4 -I= 3 cos " Q-^-2 8 cos "^ _ - 2 COMPLETING THE SQUARE The method developed here is based on the fact that we can make 2 any binomial of the form x + kx into a perfect square if we add to 12-3. Sec. it let 1 2-3 in Equations 207 Quadratic Form the square of one-half the coefficient of x. To make this clear, us recall from Section 1-18 the formula for a perfect-square trinomial. The formula is + a) 2 = x + 2ax + a 2 2 (x Since the coefficient of x in x 2 2 ^J (fc\ Thus, the left A* or kx 4- is k, . the square of one-half of 2 Adding -j this to x 2 +^ = + fc#, we have + ~) x2 + kx member is a perfect square, namely, the square of (x Applicability of the procedure to a variety of processes, including solution of quadratic equations, is illustrated in the following examples. Example 12-4. Solve x 2 - 2x = 4 by completing the square. We first transpose the constant term, so that the left side will be of + kx. Hence, the equation becomes = 4. x - 2x = 1 is added to the left side to make it a perfect square. Now the quantity I) Solution: the form x 2 2 2 ( To obtain an equivalent equation, The result is the same quantity - x2 2x + = 1 is added to the right side also. 5, or - (x Taking square roots 2 = 5. we have of both sides, So the desired solutions are x I) x - 1 1 + \/5 and x = = \/5. = 1 V5. Check: j (1 and + V5) 2 - (1 - _ \/5) 2 - 2(1 2(1 + V5) -4 = 1+2 \/5 +5-2-2 _ _ - V5) - 4 = 1 - 2 V5 + 5 - 2 + 2 2 Example 12-5. Solve 2x - +3 = 5x Solution: Transpose the constant 2x 2 Since the coefficient of x 2 is not 1, _ \/5 -4=0, - 4 = 0. by completing the square. term to obtain -5x we make = it 1 5 3. by dividing both we have o >/5 3 sides by 2. Then 208 Quadratic Form In Equations 5\\ 2 / (12 \ both sides, thus making the ~ *2 4.=-* + ^ 16 ~ ~ 2 + 25 . or / X take square roots of both sides, we obtain ~ and 2(1) - &) 2 = 2 -5(1) Reduce x 2 Example 12-6. (y ' 5 1 = j =t T Solving for z, 1 l -: -r' 4 4 or # is, = + is 16 we have x K o = X That 16 5\ 2 _J_ ~ 16 4/ \ When we " Tr ~ , 2* 12-3 25 = / / a perfect square. The result _ 3 25 _ left side 5 Sec. = 1. +3=2-5+3=0. + y* - 4x + Qy +4 = the form to (x - 2 ft) r2 . The solution of this problem requires that we complete the square of the terms containing y as well as the square of those containing x. Hence, for convenience, we write the equation in the form Solution: - (x When we 4z ) + (y + 6y 2 ) = - 4. complete the squares in the parentheses, the equation becomes - 4s + 4) + (0* + 6// + 9) = - 4 + 4 + 9. (x* Thus, the solution is - (x + 2 2) (y + 3) 2 = 9. Example 12-7. Reduce 9z 2 - 4y 2 - l&c - 16# A(x -ft) 2 - B(y -k)* - 43 = =C. - 2x - Solution: Write the equation in the form 9(x 2 Complete the squares in the parentheses to obtain 9(x 2 - 2s + 1) - 2 4(i/ + 4y + 4) ) = 43 + 9 to the form 2 4(?/ - + 4t/ ) = 43. 16. Note that the numbers and 4, which are added within the parentheses to complete the squares, must be multiplied by the coefficients 9 and - 4, respectively, to determine the numbers that are added to the right side. The reduced form is, therefore, 1 9(s - 2 Example 12-8. Reduce \/3z Solution: I) 2 - 4(0 + 4z - + 2) 2 = 36. 4 to the form y/a((x For convenience, work with the quantity 3z 2 radical sign until the final result is obtained. Hence, write 3* 2 + 4* - 4 = - + 4# h)* - - A; 2 ). 4 without the Sec. 12-4 Complete the square Now, of the terms in x 209 Quadratic Form in Equations and simplify to obtain write this result under the radical sign to obtain Comparing choose a = h 3, = - 2/3, = and k vX(# form this with the required h) 2 + / l/3(( /c 2 ), we <r) ) see that we may d= 4/3. EXERCISE 12-2 In each of problems from 1 to 15, solve the given equation by completing the less In each of the problems from 9 to 15, find all non-negative angles square. than 300 which satisfy the given equation. Check 2. x 2 + lOx = 40. 1. x 9 - 8x = 20. 4. X 2 + x + i = o. 5. z 2 + x + 2 = 0. 7. z2 12. 1 14. 3 +z - 5 = 0. 8. 2x 2 = 3z + 9. 20. 22. 24. x2 6x 2 9. tan 2 esc 2 - 15. sec 0. + 2 tan 1. 2. cos = 2. to 25, reduce the equation to the form 1(5 17. x 2 4x + + 32^ - 18?y + 37 = 0. 2 2 - 32 = 0. 9x + 4?y Sy x 2 - 9// 2 - 4x 4- 36;// - 41 = 0. 2 4r 2 + 50# -f 32x + 41 = 0. 5// % = ^-4 01 = 13. 2 2 - 7x = 30. - 5x - 1 = 3. 6. + tan-' 6 = sec 0+3. cot + cot = 3 - 4 cot 0. In each of the problems from A(x -h) 2 +li(y -A-) 2 =C. 16. x 2 - 4?/ 2 - 2x + 1 = 0. 18. all solutions. 2 2 19. x 21. 4.c 2 + 4i/ 2 + 4// - 23. 4x 2 25. 3* 2 2 -f 9^/ - 9// y 2 2 2 6x + 16y + 21 = 0. - 40y -h 109 = 0. - Ifxc - ISy -11=0. lOx + 32o; -h 36y -f 64 = 0. + 2Qx - 2y + 11 = 0. In each of the following problems, express the quantity inside the radical or - h) 2 k 2 ). parentheses in the form a((x 26. V& &r - 27. 40. \/x 2 - V(9^ 2 12-4. Gx + - 48,-c - v/2x 30. (x * 29. 32. ~ - 2 16 J 6.r 4- + 34) 41. 28. 2/3 . 31. (2x 20?/ - + 28.c + 34)~ - 14a: + II) 2 - 2 (4// 76) 3 '2 . 1/2 . 7 + 23) 3 . 33. (9z 2 + 24x + 25)~ 1/3 . 34. (7x* SOLUTION OF QUADRATIC EQUATIONS BY THE QUADRATIC FORMULA applying the method of completing the square to the general quadratic equation (12-1), we can obtain a formula for the roots,, either real or complex, of any quadratic equation whatever. The general equation is By (12-1) ax 2 Transpose the constant term ax 2 + bx + c = 0. c, and obtain + bx = c. 21 in Equations Dividing by a, Quadratic Form Sec. 1 2-4 we have c 6 + ax = --a x *2| 2 6\ 2 - = & ^2 to both ) (1 4a 2 o/ sides to obtain . * o + . b 62 . :C+ = - which becomes 2 ft _ 4ac &2 Extracting square roots of both sides gives Solving for x, Vb 2 - 4ac b __ + 25~ * , 25 we have 6 = * -\/& 2 4ac & Hence, to solve a quadratic equation, put form ax 2 + 60? +c= 0, and substitute the it into the standard coefficients a, &, and c in the formula just derived to obtain the roots ,10 ON (12-2) xi = - & + Vb2 - 4ac % , x2 and - = b - Vb 2 - 4ac ^ That these numbers #1 and #2 are solutions of the given quadratic equation is shown by substituting each of them in (12-1). The details of this substitution for x l follow -: /b 2 - 26 b2 Vb 2 - 4ac + 2 (6 +c. - 4ac) + b Vb 2 - 4ac 2a b2 - 2ac - 6 Vb 2 - - - 4ac 62 + 6 \b 2 - 4ac . A 2 Hence, the number x satisfies the equation ax + bx + c = 0. similar computation shows that #2 is also a solution of ax 2 + bx + c = 0. Consequently, the two expressions given for x in (12-2) are actually roots of (12-1). In Section 12-7 we shall study the expressions in (12-2) further and shall determine when the roots are dis- tinct and when they are real. Sec. 2-4 1 Example 12-9. Solve 5z 2 = 5, Here a formula, we obtain Solution: = fr = = 8 6# c 6, 21 Quadratic Form In Equations 1 by the quadratic formula. Substituting these values in the 8. _ _6 ~ \/36 db 6 160 -f Therefore, x\ j^ These values are seen to = _ *~~ Hence, si = 1, -r - +2 = = 1, c = 2, Vr~^~8 _1 6 1 = g when satisfy the original equation 2 Example 12-10. Solve x Solution: Since a 10 10 = 2 and z 2 = = 6 db 14 \/196 db 10 x by the substituted for x. formula. we have \^J "_ 2 1 db 2 1+V? ~ j and z2 1 = Ctecfc: Similarly, the second root may Example 12-11. Solve 2 all non-negative angles x We make Solution: sin 2 less be checked. x -f 3 cos x - 3 = by the formula, determining than 360. + cos use of the identity sin 2 x 2 = x 1 to transform the given equation into one involving a single trigonometric function of x. 2 cos 2 x we have Replacing sin x by 1 t 2(1 - + cos 2 x) - 3 cos x = 0. 3 we obtain Simplifying, 2 cos 2 x - 3 cos x = 0. -f 1 -- Solving for cos x by the formula, we find cos x Hence, cos x The = solutions =3 V9 - =t A mining =3 1 A 4 = or 1/2. Therefore, x or 60 or 300. may be checked by substitution in the original equation. 1 Example 12-12. Solve cos x tan x all 8 4 -f sin 2 x = sin x 1 by the formula, non-negative values of x less than 360. Solution: Making 6 use of the identity tan x - cos x -sin x cos a: . 2 h sin x , = cos x = > we have . 1 sin x. deter- 212 Equations Quadratic Form in Transposing and simplifying, we get sin 2 x + 2 sin x Sec. 1 2-4 = 0. 1 x by the quadratic formula, we obtain Solving for sin sin z - = 2 2 = - - /o V2. 1 1 one solution. However, sin x = \/2must be excluded, since tho sine of an angle cannot be numerically greater than 1. The value = 3 sin Check: For sin x + V2 1 = cos x \/2 = is we 1, - \/2 V2 find that and 2 = tan 3 ~ + (V2 - V2 - 1 +3 2 From - = 2 I) - (V2 - 1 1). 2 This reduces to or 2 we have Substituting these values in the original equation, V2V2-2- V2 -1 _ V2V2 - _ 2\/2 x/2 = 2 the table of trigonometric functions, = - 1 V _ + 1 - V2. = we have x 2428' or 1 5532'. EXERCISE 12-3 Solve each of the following equations for x or 8 by the quadratic formula. In each of the problems from 14 to 24, find all non-negative angles 6 less than 360 which satisfy the equation. Check all solutions. 1. s* 2. x2 - 3. x2 4. x2 + 5. 2 6. 7x2 7. + 6* - 7 = 0. + 2x - 5 = 0. z + 2z + 1 = 0. 2z 2 4- 3* + 2 = 0. 9. (2x - I) - 2 2(2x 13. 3+2 17. - 8 = 0. 2 21. (sec 23. (esc 12-5. 10. (a 2 2X + 3 3 * x + ; ' 3 "*" cos 0+1) + 2) = 0. = 0. _ 9 = o. - 5 = 0. 20 + I _ gj. - 7x - 6 2 )z 2 - ft 14 - ww cot 2 1 + ^ " - cot * 2 cot (sec 22. (sin 2 24. - - 3 3 - 3 (a + 2 - 62) = 0. 1 4 : tan = - + 3 = " 3 " 18. 16 sec 2 on - ' 3 16. esc 2 cos 2 + 2) = sec + 3. + esc = 1. - 4abx 8 6 = 0~ v +r 3x o-Vr +5 -- = 3. tan cot - 2 sin = sin + 3. sin 2 ^ __ ___ _ - + ^ 3 15. 1) - 5 11 11. - 8. 9.r 2 - x x 2 sin 0. 8 sec + 2 cot 6 _ 3 + 3) 1 - = 0. 1 ~~ o cot (3 - sin 0) = T +| = 3(sin + 3). - EQUATIONS INVOLVING RADICALS Sometimes an equation in which the unknown appears under a radical sign can be reduced to a quadratic by raising both sides to Sec. 1 2-5 Equations Quadratic Form in 213 a power sufficient to remove the radical. The process must be repeated until the unknown no longer occurs under a radical. The operation of raising both sides of an equation to a power may lead to an equation redundant with respect to the original that is, the final equation may possess roots that are not roots of the original equation. Such roots are called extraneous roots. For this reason, every root obtained must be checked by substitution. ; + 4 = 2. - 3x + 4 = 8. = 0. Example 12-13. Solve the equation \/x* - 3x Cube both Solution: sides to obtain x 2 - x2 Factoring and solving for x, we 4 Transpose, and get find that = x - 3x x or 1 = 4. Check: >X(-1) 2 -3(-l) and +4 = ^1 +3 +4=^8=2, _ Hence, both 1 and 4 are +4 = 3(4) ^/g = 1 \/x 2. roots. Example 12-14. Solve the equation \/2x - +3 = 1. Solution: Transpose one radical to obtain When = 1 both sides are squared, the result is 2x like terms, Combining - = 1 + \A~T3. _ + 2>A +3 + x + we obtain - 5 = 2 V-c square both sides to get x 2 - Wx + 25 = - Ux + 13 = 0. x = 13. x Now we 1 1 3. + 3. 4(s + 3). Transposing and combining gives ;r* By factoring and solving, we find that _ x , and Hence, 13 is = 1 V2(l) - 1 V2(13) - 1 a root, but 1 is or _ - VI +3 = 1 - 2 ?* 1, - V13 +3=5-4 = 1. not. EXERCISE 12-4 Solve each of the following equations. In each case check for extraneous roots 3. V% - 2 = 4. Vz + 5 = 1. 5. \/z 1. 2 - 16 = 2. 4. 2x l . 6. V3z +4 = 2. ^3z -1=7. >A 2 - 2 = V2.c + 6. 214 - 7. x 9. \/2x 11. V^T = - 3 + x \/5 3 =4 - 4- 3x. IN = 5. Sec. * 8. 0. \X4oTT~5 EQUATIONS 12-6. Quadratic Form in Equations - 5* 1 10. \/z 12. \/% x - ' 1 2 12-5 +6=0. + Vz - 3 = 2. QUADRATIC FORM Frequently, an equation which unknown may be considered not quadratic in the given is as a quadratic in - 3or 2 + some expression and 2(x 2 - 2x) 2 2 = Thus, or 6 (x 2x) may be treated as quadratic equations in the 2 2 or and expressions (x 2x), respectively. This type of situation was met earlier in Examples 12-11 and 12-12. The following 4 involving the unknown. = 2 examples will further in quadratic form. methods used illustrate 4 Example 12-15. Solve the equation or Solution: Let x~ 2 = y, - 3or 2 in solving equations +2=0. so that the given equation becomes _ 7/2 Factor, to obtain + 3y - = 2 (y - 1) (y y = l or y or 2 = 1 or z~ 2 = 2) Therefore, 0. 0. = 2, or = 2. Hence, the solutions are x Example 12-16. Solve Solution: Let x 2 2 (z = - 2 = y, and 1 2) - 2 y - 2 Factor, to obtain 2) 7y get 10 = 2 + - 1 = - 7(.r we so that a: 7= + = 0. - 2) (y y = 2 or y 2 =2 or .T 2 - 2 x2 =4 or z2 = 7, Hence, = 0. 0. (y 5) 10 = 5, or x2 - = 5. Then, and the roots are & = 2 and = \/7. Sec. 127 Equations 2 Example 12-17. Solve x + - x i'n V^ + z + 2 2 215 Quadratic Form = 0. 3 V# 2 + * + 3 = Then we can write + 3) - 2 -\A + z + 3 - 3 = 0, or y - 2y - 3 = 0. (x + Therefore, y = - 1 or 3, and \/z + z+3=-lor Vz 2 + x + 3 = 3. Solution: Let 2 ?/. 2 2 re 2 definition of the radical, By \A +05+3= 2 1 \/a is a non-negative number. Hence, although consistent with the original equation, there are no values is of x which satisfy this equation. Consider, then, V# 2 + # + 3 = 3. z 2 + x + 3 = 9, This leads to x 2 + a? or Hence, x = 2 or a? = - - 6 = 0. 3. Substitution shows that each of these values of x satisfies the original equation. EXERCISE 12-5 Solve each of the following equations. Check 1. 3. 5. 7. 9. ^+ 12 = 0. - 4 = 0. (x + 2) + 3(z + 2) x 4 - 13x + 36 = 0. (3* - 4) + 6(3x - 4) + 13 = 0. f x + 1V + - 48 = 0. + X2 _ 2 2 2 4jr* 2. 6. x4 2 8. (x 10. (a; -) - - 4. r* 2 2(2; all solutions. 6z 1 - liar* = 0. 3 + 8 = 0. = 0. 2 2 + 3z) 2 - 2 - 2 a:) - 14(z 20(x 2 2 + 3x) + 45 = 0. - x) + 36 = 0. THE DISCRIMINANT 12-7. be recalled that the two roots of the general quadratic are equation ax- + bx + c = It will - , and The expression 62 4ac 2a which appears under the radical sign, is the discriminant of the quadratic polynomial ax 2 + bx + c, or called 4ac, the discriminant of equation (12-1). In what follows shall make we shall assume that a, 6, and the solutions Xi and 1. If 6 , 2a are real, and we use of the discriminant to determine the character of the roots without actually solving the equation. 2 c 4ac = x>2 , 0, we By inspection of reach the following conclusions each of the two roots #1 and x 2 and both roots are thus real. is : equal to 216 in Equations If 6 2 2. real, - 4ac and they are If b 2 3. 4ac V& 2 ~~ 4ac both roots are is real, distinct. is negative, then V& 2 ~~ 4ac be summarized as follows may imaginary, and the is numbers of the form a + roots are distinct complex These results Sec. then is positive, 12-7 Quadratic Form fti and a /3i. : 4ac is a perfect Furthermore, if a, 6, and c are rational, and ftrational square, then the roots are rational; otherwise, they are irrational. The following examples will illustrate character of the roots. Example 12-18. Determine the character + = 7x 49 + 2x* Solution: Here a = 2, b 62 The discriminant discriminant is is - = 7, c 4ac positive, - 2kx -f 4 and 15. to determine the of the roots of 15 = 0. Hence, 120 = 169 = (13) so the roots are real 2. and unequal. Since the a perfect square, the roots arc also rational. Example 12-19. Determine kx 2 = - how = all values of k for which the roots of the equation are equal. Solution: The discriminant must equal zero for the equation to have equal roots. 2 2 or k - 4. When Hence, 6 - 4ac = 4& - 16& = 4k(k - 4) = 0, and so k = = k 0, the equation is not quadratic. Therefore, the roots arc equal only when k = 4. EXERCISE 12-6 Determine the character of the roots of each of the following equations by means of the discriminant. 1. 4. 7. 10. 13. + 3s + 4 = 0. 6z - 7x + 3 = 0. 5z 2 + 7x + 2 = 0. 4* 2 - 8z + 2 = 0. 4x2 _ i 2x - 9 = o. ar a 2 + &c - 9 = 0. x + IQx -f 2 = 0. 4z - 12z + 9 = 0. 2x2 _ x + 3 = Q. x 4 4z - 18 = 0. 2. x2 3. x* 5. 2 2 8. 11. 14. 2 2 6. x 9. x2 12. + + 3x - 2x 5z 2 15. x 2 IQx +z - 4z - +2 +1 +5 +7 = 0. = 0. = 0. = 0. = 0. 6 Sec. 12-8 217 Quadratic Form in Equations SUM AND PRODUCT OF THE ROOTS Adding the two roots of the general quadratic equation 12-8. ax 2 we + - xi+x*= b + - - 2 \/b 4ac XlX2 = 0, 6 - Vb 4ac 26 6 -_=--. = Ya we have + Vb - -ac 2 6 - 2 - Vb 2 - b 4ac 2 b __ "" we = c , 2 - - + Also, multiplying these roots, Hence, + bx by using (12-2), obtain, 2 2 (b 4ac) __ "" 4a 2 4ac c __ "" 4a 2 a sum and product have, for the xi (12-3) x2 = ~ xix a =- + and (12-4) - of the roots, > tZ These formulas are used in various ways, for example, in checking roots of a quadratic equation, and in forming an equation if its roots are known. To find the factored form of the quadratic polynomial ax 2 + bx + c, let us make use of the sum and product formulas just found. for 6, and solving XiX 2 = - for c, we have Solving Xi + x.> = b a(xi Therefore, ax 2 a a = + bx -f + x<2 ) and c ax 2 + +c= ax 2 = = a(x if c = x\ bx and 0, its may a(x - (xi 2 - xi) (x - X2). xi) (x = #2) 0. write the equation equivalently as (X its 2 we have + x<)x + a(x\X2) + x )x + (ziofe)) are the roots of the quadratic equation factored form can be written ^ 0, we may This form or a(x\ x>2 a(x Since a Hence, by substitution, a(x^x^). - Xi) (X = 3fe) expansion, X2 (#1 + X 2 )X 0. + XiX2 = 0, be used in writing an equation whose roots are known. Example 12-21 indicates the procedure. Example 12-20. Without solving the equation, find the 5.c =0. of the roots of the equation 3x 2 sum and the product +2 Solution: In this case, a =ft o , = and the product 3, b - 5, of the roots is and c =^ do - = 2. Then the sum of the roots is 218 Example 12-21. Write a quadratic equation roots are (1 \/3 Xl = 1 X2 = (1 sia* = + + \/3 + (1 12-8 form (12-1), given that the and x 2 = + (1 ~~ (1 - V3 l>) \/3 i) 2 is x 2r have t, ^ 4 -f V3 we =2 f) 1 ' i) = = 0. 1 - 3 i a = 4. Writing the desired equation in factored form, we have Alternate Solution: (* - + VSO) (*-(!- V3 <)) = (1 0. we obtain - (z or ((* - - 1 \/3 - (x i) + V3 1 - V3 i) ((* - 1) + - 1) + 3 = 0, (3 1) Hence, and, i ^ + Therefore a suitable equation Simplifying, in the Sec. i). Solution: Since x\ and Form Quacfrof/c in Equations finally, a; 2 - +4 = 2x = 0, i) x/3 f) = 0. 0. EXERCISE 12-7 In each of the problems from of the given equation. 1. z* 2* - 1 0. = + 6x 4. 2 7. 5x* - 2.c + 3 - Qx - 1 Form an 10. 1, 1/t 14 3 - to 9, 2. 3.r 2 - x 5 sum and find the + = 2 8. 5 2 4- 6.c + 1 6. 5.D - = + 3. a* 0. -ir-4*+i= the product of the roots 9. 0. 2 = 2 - lOOx 2 0. 6z + - 40x = 0. 1 +17-0. equation with each of the following pairs of roots. 3. 11. 0, 1 15. 2 '2'3' 18. = 0. = 0. 1 V3, V5. 19. 0, + 2. 12. iV 16. - 13. 3, 6. 1, 1. 17. i. |(1 V3 - 20. a 6f 21. . V3 GRAPHS OF QUADRATIC FUNCTIONS 12-9. To graph any quadratic function the form ax4- bx + c and + bx + c, we set of ax 2 y construct a table of values of y corresponding to assigned values of x. The graph is of the type shown in and is called a parabola. As found by the quadratic formula, the two solutions of the general quad- Fig. 12-1 ratic equation (12-1) are given (12-2) Xl = X2 = 4ac 2a and FIG. 12-1. 6 \/b 2 4ac by Sec. 1 2-9 Equations Since (12-1) states that y solutions Xi Quadratic Form in = in the equation 219 y = ax 2 + bx and x 2 are ^-intercepts of the curve. let A and C be the ^-intercept points, and In Fig. 12-1, the mid-point of AC. or We note Now B is the point ( ^- = ax- + = ax- bx + + bx k c -f c = Q. be which represents a straight fc, may This line parabola, depending on the value of and y B let V > consider the equation y line parallel to the x-axis. the 7 . Therefore, c, that OB = OA + AB, "" Xl X2 - Xl +X2 - - + or may not intersect the Solving the equations y k. simultaneously, by elimination of The roots of ?/ we = k obtain this resulting equation are and 6 \/6 2 4.a (c k intersects the curve in two If the value of k is such that y distinct points, the discriminant in and the roots will be real with abscissa x section, = - ^ci is /c) 2a 2a and (12-5) distinct. is greater than zero, The point on the line y k then equidistant from the points of inter- whose abscissas are x and x z . = k does not intersect the curve, the discriminant in (12-5) is less than zero, and we have a pair of conjugate complex In this case, If, is on the other hand, the value of k is such that y roots, z>a the real part of these roots. The points with abscissa x = and arbitrary ordinates lie 0. on a vertical line, called of the curve; the curve the axis of symmetry, or simply the axis is said to be symmetric with respect to the axis. The point of intersection of the axis and the parabola is called the vertex of the parabola. If the coefficient a of the second-degree term of y = ax- + bx + c is positive, the vertex is the loivest point, and the curve is said to be concave upward. If a is negative, the 220 Equations vertex is in Quadratic Form the highest point, and the curve Sec. 12-9 said to be concave is doivmvard. To find the coordinates of the vertex, the equation x = b = , and (12-6) y That = ax 2 + bx + c a& 2 - -r-o b2 b2 , -^ 4o in another way. Let us demand = k intersect the parabola in two coincident us insist that the two roots x\ and xs in (12-5) that a horizontal line y coincide. = of the vertex are thus found to be The vertex may be characterized points. solve simultaneously of the axis and the equation y The coordinates of the parabola. we is, let Then the discriminant in (12-5) is equal to 0, and = ^-- The value of y corresponding to this value of x is Za the ordinate of the vertex, as found in (12-6) We say that the line #1 x2 . y is = - - b2 4ac 4a tangent to the parabola at the vertex. - 2 Example 12-22. Graph x Solution: Let y ponding values of The =x - ?/, 2 6x as in the -f 4, assign values to accompanying table. x, and compute the corresis shown in Fig. 12-2. The graph coordinates of the vertex are =!(-i.il) (2, + 4. Gx -4) and b2 2/=c ^-=4 9= -5. (7,11) Hence, the axis is the line x positive, the vertex (3, = 3. Since a is 5) is the lowest point on the curve, and the curve is concave upward. Sec. 12-10 Equations in Quadratic Form 2 Example 12-23. Graph y = x - Qx + 9 and ?/ same coordinate system as was used for the graph The s2 - Qx = x2 of y + 14 relative to the 6# Example 12-22 but applying shown in Fig. 12-3. Solution: Tables similar to that in curves are constructed. = 221 -f 4. to the first two three curves are (C) y - j-2 _ Ox + 14 FIG. 12-3. Curve (A) of .r 2 Gx H- 4 roots of x 2 6.c because x 2 Gx 0. + + 14 = 1 has imaginary roots. reader should relate the discriminants of the quadratics to a study of these 1 The two points, corresponding to the roots 3 =fc \/5 Curve (#) is tangent to the x-axis at (3, 0), because both 9 = an equal to 3. Curve (C) does not intersect the z-axis, crosses the x-axis at graphs. 12-10. QUADRATIC EQUATIONS TWO UNKNOWNS IN The general equation of the second degree ax (12-7) 2 + bxy + cy + 2 dx + ey in x +f = and y is 0, and / are given real numbers. An equation of this form, in which at least one of the coefficients a, 6, and c is different from zero, is called a quadratic equation in x and y. By a solution of such an equation, we mean a pair of real or complex numbers which, when substituted for x and y in (12-7), where a, b, c, d, e, will reduce the left side of the equation to zero. infinitely many pairs of numbers which Usually there are satisfy the equation. 222 Equations in Quadratic Form Sec. 12-10 If c T^ 0, (12-7) may be solved for y in terms of x by means of the quadratic formula. Corresponding to each real value assigned to x f we then obtain, in general, two values of y. We then have pairs of numbers (x, y) which, if real, may be plotted in a rectangular-coordinate system. It is shown in analytic geometry that the graph so obtained will be one of a class of curves called conic sections, inasmuch as they may be obtained as curves of intersection of a plane and a right circular cone. At this time we shall confine ourselves to merely listing the curves which comprise this and indicating briefly the form of the quadratic that corremore adequate discussion of this sponds to each of the graphs. class A subject is given in analytic geometry. However, typical examples of these curves are given here. 1. Parabola. When A ^ 0, the equations y = Ax- + Bx + C and = x Ay- 4- By + C represent parabolas with vertical and horizontal axes of symmetry, respectively. the equation x-+ y 2 = C represents a circle whose center is at the origin and whose radius is \A?3a. Ellipse. When the constants are positive, the equation Ax- + By- = C represents a curve called an ellipse. If A = B, the 2. Circle. When C is positive, ellipse is a circle. Point Ellipse. If A and B are positive and C 0, the equaAx- + By 2 = C is satisfied by only one point, namely, the origin. The graph is then said to be a point ellipse. 3c. Imaginary Ellipse. If A and B are positive and C < 0, there are no (real) points on the graph, and we say that the equation Ax 2 + By 2 C represents an imaginary ellipse. 4a. Hyperbola. When A, B, and C are positive, the equations Ax- By 2 = C and Ay 2 Bx 2 = C represent hyperbolas. 3b. tion the equation xy = C represents a curve called an equilateral hyperbola. 5. Pair of Straight Lines. The equation Ax 2 + Bxy + Cy~ + Dx f Ey + F = represents two straight lines, which may be either distinct or coincident, if the left side can be expressed as the product of two real linear factors. We use the quantity b 2 4ac, which is usually called the characteristic of the equation ax 2 -f- bxy + cy 2 -f dx + ey -f / = 0, to determine the nature of the conic corresponding to a particular form of the general quadratic equation. In analytic geometry the following statements are shown to be true 1. If b 2 4ac = 0, the conic is a parabola or two real or imagi4b. Hyperbola. When C ^ 0, : nary parallel 2. If 2 b' 4ac imaginary < lines. 0, the conic ellipse. is an ellipse or a point ellipse or an Sec. 3. 12-10 If b 2 4ac > 0, the conic 223 Quadratic Form In Equations is a hyperbola or two intersecting lines. In Section 12-9 the graph of the parabola y = ax 2 + bx + c was discussed. In the following illustrative examples, the procedures for graphs of other quadratic equations are considered. 2 Example 12-24. Graph x + y2 = 9. = = to obtain the ^-intercepts, which are Solution: Set y 3; and set x to obtain the ^-intercepts, which are it 3. Solve the equation for y, obtaining y V9 - = x2 . construct a table of other corresponding values pf x and y. To yield real y, the numerical value of x cannot exceed 3. As shown in Fig. 12-4, the resulting graph is a circle with center at the origin and radius 3. Then values of IY -3 FIG. 12-4. 2 Example 12-25. Graph 4z + 9?/ 2 = 36. to obtain the ^-intercepts, which are Solution: Set y = 3; and set x 2. Solve for y and obtain obtain the ^-intercepts, which are y = o V9 - x2 to . Construct a table and draw the curve, as shown in Fig. 12-5. an = This illustrates ellipse. X -3 - 2 1.49 1 1.89 2 1.89 1.49 FIG. 12-5. Note that the numerical value yield real values of y. of x must be equal to or less than 3 in order to 224 2 Example 12-26. Graph 4z - 2 9s/ Quadratic Form in Equations Sec. 12-10 = 36. 3. Setting x = 0, Solution: Setting y = 0, we find that the ^-intercepts are 2 - 4. Hence the curve has no ^/-intercepts. however, results in the equation y = Solving the given equation for we have ?/, = y The accompanying table is V* 2 | 9- constructed. FIG. 12-6. Note that This x must be equal to or greater than 3 in The graph of the given equation is shown in Fig. 12-6. in this case the numerical value of order to give real values of y. illustrates a hyperbola. EXERCISE 12-8 Identify and graph each of the following. 1. x2 4. 4z 2 7. y 10. 12. +y = 2 - 9*/ 2 25. 2. = 5. 36. + - ?/ 2 2 ?y 17. 19. 4*/ = = x 2 - 3x + 2. 8. 2 2 2z + 4y =4. 3z 2 - 4xy + 2^/ - 6x + 3?/ = 7. 14. 4* - 4ry + 2 2x - 2 = 0. xy + y 2x - xy - 28y 2 = 0. 9z 2 - 24xy + 162/ 2 + 3x - 4?/ = 6. 2 15. 9z 2 2/ 2 12-11. GRAPHICAL SOLUTIONS QUADRATICS 2 = 36. = 16. 4. 3. 4,r 2 6. x2 9. 5^ 2 - + = 0. = 0. 9?y 2 = 28?/ ftcy -f- 2 9?/ 2 . = 2x + 13. x + xy - 2y* + 3y - 1 = 0. 2x + 4?y - 12 = 0. 16. z 2 - 3x - 3y* + I8y - 27. 18. 4x + 4xy - 3i/ 2 + 4x + lOy = 3. 20. 4x 2 + 3xy + 4i/ 2 - Sx - 8y = 24. 11. 5ar?y ?/. 2 2 OF SYSTEMS OF EQUATIONS INVOLVING In Chapter 9 we solved systems of two or more linear equations both algebraically and graphically. Frequently, however, simultaneous systems include one or more equations of the second or higher degree. We have, therefore, to consider the problem of finding systems of values of the unknowns x and y that satisfy two equa- Sec. 12-11 Equations one of which tions, is in 225 Quadratic Form quadratic and the other of which is linear or quadratic. We shall begin types of systems. by illustrating some graphical solutions of several The graphical method yields only the real solu- tions of a system, but may prove advantageous in suggesting solutions and interpreting results. In general, this method yields at best only approximate solutions. The graphs should be drawn as it accurately as possible. Example 12-27. Solve graphically the system ' x2 x - 8* + 3y = 0, - Zy + 6 = 0. * Solution: Solving each equation for y in terms of x, y= 8x - x2 , and x we have +6 =-3 Construct tables of values, and draw both graphs, using the same coordinate system, as shown in Fig. 12-7. 7 123456 + 3^-0* +*x 7 8\ (A): J?2 -8.1: (B): *- FIG. 12-7. The line and the parabola are seen to It follows that these points represent intersect at the points (1, 7/3) common roal solutions, possibly and (6, 4). only approxi- mate, of the system. A check by substitution shows that the real solutions are, in fact, x = 1, y = 7/3; and x = 6, y = 4. The Example 12-27 suggests a procedure for finding graphical solutions of quadratic equations in one unknown. solution of 226 Equations in Quadratic Form -x Example 12-28. Solve the equation x* Since x Solution: = 2 x + Thus, the original equation 2, is 2 = Sec. 12-11 graphically. both sides of this equation may be set equal to y. by the system y = x\ replaced f ( y=x+2. y =x +2 2 From d=l 1 2 4 3 9 4 16 -2 the accompanying tables of values of x and ?/, the graphs shown in Fig. The graphs show that the line and the parabola intersect at 12-8 are constructed. the points 1, ( their abscissas 1) must and (2, 4). Since these points are x2 = x + = 2. '(-2,0) FIG. 12-8. Example 12-29. Solve graphically the system t ( Solution: From the first and From and to both graphs, 2. 2 Hence, the required roots of the original equation x and x common satisfy the equation the second equation, 3x 2 - 2?/ given equation, 2 = 6. x 2 = are x = 1 12-12 Sec. Equations The necessary The ellipse (A) in 227 Quadratic Form and the graphs are shown tables are given here, and the hyperbola (B) are seen to in Fig. 12-9. intersect in the following four distinct points: (2, V3), (2, - x/3), ( - 2, V3), ( - 2, - >/3). These values of x and y already appear in the tables used for constructing the and need not be checked by substitution in the original equations. However, such checking is usually desirable. graphs, EXERCISE 12-9 Solve each of the following systems of equations graphically. 1. 4. 7. 10. 13. 12-12. As may ALGEBRAIC SOLUTIONS OF SYSTEMS INVOLVING QUADRATICS in the case with linear equations discussed in Section 9-1, it happen that in a system of equations involving quadratics 228 in Equations Quadratic Form Sec. 12-12 part of the graph of one equation coincides with part of the graph of the other. Such a condition gives rise to infinitely many soluUsually, however, there are only a finite number of points of intersection of the graphs corresponding to the given equations, tions. and the algebraic problem consists of finding the pairs of numbers (x, y) which satisfy both equations. We can say in this case that two simultaneous equations in x and ?/, of degrees m and n, respectively, can have at most mn solutions. Thus, a system of one linear and one quadratic equation can have at most two solutions, and a system of two quadratics can have at most four solutions. When a system consists of two quadratic equations, the algebraic solution usually leads to a fourth-degree equation in one of the unknowns. Since we have not presented a general method of solving a fourth-degree equation, we shall consider here only systems whose solutions can be effected by the theory of quadratic equations. The methods of procedure in some of the more important types are shown : One Linear and One Quadratic Equation. A system of type can always be solved by the method of elimination by Case this in the following three cases 1. substitution. Example 12-30. Solve the system x x2 - 3?/ + 6 = 0, - Sx + Zy = 0. Solution: Solve the linear equation for y in terms of x, obtaining x + 6 Substitution for y in the quadratic equation yields ,,_ 8a;+ 3(?L6)=0. Collecting terms gives x2 = - Ix + 6 = 0. and x = 6. Substituting these values in the linear equation, we obtain y = 7/3 and y = 4. Hence, the solutions are x = 1, y = 7/3; and x = 6, y = 4. These values can readily be verified as solutions of the given system. Note that The roots of this equation are x 1 they correspond to the coordinates of the points of intersection in Fig. 12-7. Example 12-31. Solve the system + 2y + 4 = 0, x* + 4y* - 2x x 3 = 0. Sec. 12-12 229 Quadratic Form In Equations Solution: Solve the linear equation for 2y, to obtain = - + 4)2 + 4). We then have - 2z - 3 = 0, + + 2y (x Substitute in the quadratic and collect terms. * + z2 (x or 2x 2 One solution *= The other 6x = 0. 13 is solution - 3 + i V17 y= ' 2 5 + t \/17 5 - 4 is * = - 3 - tVl7 y= ' 2 i yT7 ' Since these values arc imaginary, the graphs of the two given equations do not intersect. by = c. When the consists of two system equations containing only squared terms in each unknown, it can be solved for x 2 and y 2 by the methods used Case Two Equations 2. Form ax 2 + of the 2 for linear systems in Section 9-2. Example 12-32. Solve the system + x* 2x 2 Solution: To 2y* - 2 2/ = = 17, 14. eliminate y 2 multiply the second equation , by 2 and add the two equations, as follows: + 2y* = 17 - 2y = 28 x2 2 4z 2 5.r 2 Solving for x, i Now = 45. we have substitute 9 for x 2 in the first 2 2?/ = = 17 y = or 3. of the original equations. - 9 = Then 8, 2. Hence, we have the following four solutions: (3,2), These may be written (3, -2), (3, =b 2), ( - (-3,2), (-3, -2). 3, d= 2). 2 2 of the Form ax + bxy + cy = d. If the system is of this type, the solution is effected by elimination of the constant term. The resulting equation is then solved for one Case 3. Two Equations terms of the other. This procedure gives us two linear and y which may be combined with either of the given quadratic equations to form two systems of the type con- unknown in equations in x sidered in case 1. 230 Equations Quadratic Form in Sec. 12-12 Example 12-33. Solve the system _ x* 2x 2 xy + 2y = 1, 2 2xy + Sy = 3. 2 Solution: Multiply the first equation from the new equation, as follows: 3z 2 2x 2 - by 3 and subtract the second given equation xy + 6?/ = 3 + Sy = 3 - 2y = 0. (x - 2 3xy 2xy x2 2 2 Factor, to obtain + y) (x 2y) = 0. Hence, x - 2y = 0. x + y = or combine each of these two equations with the form the following two systems: f x 2 - xy + 2y 2 = 1, x + y = 0; and f x 2 - xy + 2y 2 = 1, - 2y = 0. We may We first given equation to then proceed by the method for case 1. solutions of the given system consist of the solutions of these two systems. The Hence, we have s = - 1/2, y = 1/2; x = 1/2, y = - 1/2; * = l,y = l/2; -1,0 = -1/2. in connection with systems described another method is effective Occasionally, under case 3. The following example illustrates this method. *= Example 12-34. Solve the system f + x2 = 37, 2 9?/ xy=2. I Multiply the second equation by equation, to obtain x 2 + 6xy + 9y 2 Solution: the first Also subtract Qxy = 12 from the 6, to obtain 6xy = 12. Add this to* = 49. equation to obtain x 2 - Qxy + 9?/ 2 = 25. The left side of each of these new equations is a perfect square. We chose the multiplier of xy, which is 6 in this case, so as to obtain perfect squares. We now first have the system / I (x (x + 3y) 2 = 49, - 3y) 2 = 25. i Hence, x + 3y = 7 or x + 3y = - 7. Also, = 5. x x or 3y = 5 3?y We now solve the following four systems of linear equations: 3 + + 3y = 7, + 3y = 7, + 3y = ~ 7, -3y = 5; \a? - 3y = - 5; \a?-3y=5; \x fa; fa? f Sec. 12-13 Equations in Quadratic Form 231 An equation in x and y is said to be symmetric in x and y if the equation is unchanged when x and y are interchanged. When a system consists of two quadratic equations both of which are symmetric in x and y, the substitutions x = u + v, v will give an equivalent system which may in some cases be solved by previous methods. y-u EXERCISE 12-10 In each of the problems from 1 to 21, solve the given system of equations algebraically. 22. Complete the solution of the system in Example 12-34. In each of the problems from 23 to 26, solve the given system by the method of Example 12-34. 23. x2 xy 25. X2 xy + = = 25. + V2 = 12. 7/2 24. 50, xy 26. 25, + y2 = 4; xy + x + y = 10. z /?/ + A = 56, X2 29. 2 2 28. 30. 3+0=2. 12-13. An EXPONENTIAL *2 a?y Solve each of the following symmetric systems. 27. z2 + 42/2 = 13, = 3. + = 144, = 56. 7/2 + 02 - x - = 2, xy + 3x + 30 = 2. x 2 + 02 = i 3> 3x 2 + 2x0 + 30 2 = 42. x2 AND LOGARITHMIC EQUATIONS equation in which the unknown occurs in an exponent is an exponential equation. Such an equation is usually solved by taking the logarithm of each side and solving the resulting equation. When this latter equation is in linear or quadratic form, it may be solved by preceding methods. called 232 Example 12-35. Solve Quadratic Form Equations in for x: 5* +3 = 625. Sec. 12-13 Solution: Write the equation in the form 5*+ 3 5*. This equation and only is satisfied if x if = + 3 = 4, = 3 ** 1 Example 12-36. Solve the equation 2 Solution: that is, = x 3 4 *. of each side to the base 10, Taking the logarithm log (2**+*) = 1. we get 4 log (3 *), or (3z + 1) log _ _ Therefore, - 4 log 3 From Table III, log = 0.3010 2 = 4z log 3. 2 3 log 2 = and log 3 0.4771. Substituting these values, we have _ ~ X __ 0.3010 4 0.4771 - + 1) 0.3010 3 ' or Example 12-37. Solve for x: log (x = 1 - log (3a? + Collecting terms containing logarithms on one side as a single logarithm, we have Solution: member 2). log (x + + log 1) (3x + 2) = and writing that 1, or + log (x + 2) = 1) (3x 1. Writing the members in exponential form, we have (x + 1) (3* + 2) + 5z - This equation reduces to 3z 2 Solving for x, we find that a; = 1 or a; = 10' = 8 = 0. 10. g = ' o Checking, we find that x = 1 satisfies the original equation, whereas x gives rise to logarithms of negative numbers. Since negative real logarithms, this latter value of x is not to be used. Example 12-38. Solve y = log, (x + \/l +# 2 ) for y. Solution: Write the given equation in exponential form, obtaining e*v Solving this equation for x, - we have 2xe -1=0. is o o numbers do not have x in terms of Transpose and square to remove the radical. The result = Sec. 12-14 233 Quadratic Form in Equations EXERCISE 12-11 Solve each of the following equations for the 1. 2* = 64. 7. 5* = 15. 10. (3*) (2*) = 36. 5. 4* = 24. 8. 3* = 11. 5"* 17. 21. 23. e 2 * 3) = 1000. = 2.403. 22. Iog 2 = 1) t= 1 - x 25. log (to + 2). + 3) = 3. 4e*+* = 7. (-!)+ Iog 2 (x 16. 29.,=^. 12-14. = 32. = 3. 15. (1.5)* 18. Iog 2 6*. = 0.4136. 8 12. log* = 4.83. 2 26. e* + 2 *- 2 = 9. 3(2*) 3 z 10*** = 243. 3 3 ** 1 3. = 0.6932. 14. s l4 =0.04681. 17. 3*+ 2 = 5.03 = (3.17)i'<*-i>. 20. log (4* 19. log (3x - 5) = 3 - log 7. 13. log* 2 16. x. = 256. 2. 4** 1 = 0.0001. 4. 10* unknown 31. e 4 * - - e 2* 10 = 0. GRAPHS OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS The graph ofy = log x is shown in Fig. 12-10. By assigning values to x, one finds corresponding values of y from Table III. few pairs of values are shown in the A accompanying tabulation. FIG. 12-10. The graph = be obtained from a table of exponential Here, however, we shall proceed as follows: Take the logarithm of each side to the base 10, to obtain of # e* may functions. log y = x log e. 234 Equations Prepare the accompanying Quadratic Form in table, Sec. 12-14 and construct the graph in Fig. 12-11. EXERCISE 12-12 Graph each 1. y = Iog 7 of the following. 2.x x. = Iog 7 3. y. T/ = 3-. *-"*-? x 4.y = . 7. y =3+4(2-). 10. y = 100e- 06 *. 2 y = y = e"'. 11. v = 7 3 *- 1 2 10-* 5. 8. e2 . 6. 9. . 12. y = 16(5-). y = log c x. = 3.9 2 *- 3 . ?/ 13 13-1. Theory of Equations INTRODUCTORY REMARKS With the ever-increasing importance of mathematics neering and the physical in engi- problems are constantly occurring that involve the solution of equations. Often these equations are of the simple algebraic or trigonometric types which we have already learned to solve. There are many other problems, however, which require the solution of equations of higher degree than the second and of some types of somewhat more complicated transcendental equations. Equations of the third and fourth degree can be solved by methods analogous to those which we used for quadratic equations. Because of their complexity, however, these methods are seldom used. It has been proved that no such procedures exist for equations of degree higher than the fourth. sciences, In this chapter we shall consider various properties of polynomial equations in general. Some of these properties will be of considerable use in later studies of mathematics, while others are considered here merely for the aid they give us in determining roots of equations. 13-2. A SYNTHETIC DIVISION simplification of the ordinary method of long division, called synthetic division, will be presented here. This abbreviated method not only enables us to quickly find the quotient and remainder when a polynomial f(x) = a Q x n + a^ x n 1 H h a n is divided by a binomial of the form x r, but also affords a simple process for substituting values of the variable into a polynomial. We shall divide 2x 3 - 9x- + I3x + 5 by x - 3 to illustrate the pro' cedure in synthetic division as compared with that of long division considered in Section 1-19. 235 236 Sec. Theory of Equations By long division 2z 3 we have - Qx 2 + 13-2 : a- 13x 2x 2 3 - 3x +4 I3x Qx 4z 4a + 5 - 12 + 17 3x + 4, and the remainder is 17. Thus, the quotient is 2x Since like powers of x are written in the same vertical column, 2 the work may be shortened by writing only the coefficients, as in the following schematic arrangement: 1-3 2-3+4 + Next, we note that the first 17 term in the divisor x written, since the divisor is always linear, in it is term and the r need not be coefficient of x always unity. Moreover, it is not necessary to write the first row that is to be subtracted, since its coefficient is in each always the same as that of the term directly above. Also, only the first term of each partial remainder needs to be written down, for the second term first row. is the same as the term directly above it in the Finally, the coefficients in the quotient need not be written, since these are precisely the leading coefficient in the dividend and the remaining partial remainders, excepting the Hence, we may indicate the process in the following way : 2-9+13+5 -6 + 9 - 12 + 17 | -3 last. Sec. 13-2 237 Theory of Equations This scheme can be written compactly as follows - 2 9 13 - 5 6 9 12 2-3 4 17 | : 3 If we replace -3 by +3 in the divisor and add the partial products in the second row instead of subtracting them, we obtain the same The synthetic result. division then takes the following 2-9 13 5 6 -9 12 2-3 4 17 form : 3 | Here the numbers 2, -3, and 4 in the third row are the coefficients of the quotient, and the last number 17 is the remainder. We can now outline the procedure for synthetic division. Note that in every step of the procedure, immediate reference is made to the illustrative example. To divide f(x) = a x n + a^ x n ~* H h a n by x r, first arrange in of for zero the coefficient of f(x) descending powers x, writing any missing power of x. process in three rows, as ( >, its in the : row, write the coefficients in f(x) in order, as !, right, put the constant term of the divisor with sign changed. We have then Step a Then arrange the numbers involved shown in the following steps In the 1. an . first At the Example a Step 2. ai a2 an Bring down the 2 r - | first coefficient place of the third row. Thus, 9 13 5 3 | a of f(x) into the first we have Example i an a2 \_r_ / 2 9 13 5 | 3 i; r, and write the product a r in the second, Bring down the sum of ai and a r into the third row. Thus, we now have Step 3. row under Multiply a by ai. Example a an ai a 2 |_r_ 2 - 9 6 apr + i) 23 13 5 . 238 Sec. Theory of Equations 13-2 In the example, multiply 2 by 3 and write the product 6 in the second row below 9. Then add 9 and 6, writing the sum 3 in the third row. Step 4. Multiply o r row under a2 and , 4- di by r, place the product in the second add. Continue this process until finally a product has been added to a n The complete . example follows solution of the illustrative 2-9 13 6-9 2-3 4 5 | : 3 12 17 9 In the last operations, 3 is multiplied by 3, and the product 9 equals 4. Finally the is written below 13. The sum of 13 and 5 to product of 4 and 3, or 12, is written below 5 and added to a n give 17. Step 5. When the process is completed, the last number in the directly below a n is the remainder. The other numbers in this row, read from left to right, are the coefficients of powers of third row x in the quotient arranged in descending order. The entire synthetic process of dividing a polynomial f(x) by x r, although it is somewhat complex notationally, can be conveniently exhibited as follows : an a n _i a2 ao ai r [ b n ~2r bpr b\r bo b n -\r R b n -i 62 hi bn of the Here the expressions for the coefficients 6 &i> b 2 = = a r 6 + 6 a of are x in the quotient Oo r2 &i 2 Oi, powers -1 2 = b n -i + a! r- + + a n -i. Hence, the quotient + ai r + 02 do r , , , _i () , 11 , may be written q (x) (13-1) = b x*- 1 + Also, the remainder assumes the R = (13-2) a r n The expression for R is inf(x). In other words + air"- 1 Bi xn ~2 + + & n -i- h a n ^r + an form H . precisely the result of substituting r for x B=/(r). (13-3) we r and q(x) from f(x), subtract the product of x Finally, we obtain the remainder R = f(r). Or, if we transpose the product, we have the usual statement found in discussions of division. That is, if (13-4) /(*) = (*-r).j(* Sec. 13-2 239 Theory of Equations The following examples will further illustrate the process of synthetic division. Example 13-1. Divide 3x* Solution: Since x of the missing 4z 2 +x - 3 , we have the + 2. 2 by x = x + 2, we have r = r power x - Writing zero for the coefficient 2. following result: 30-4 - 1-2 6 12-16 30 3-6 8-15 28 |-2 Note that the first -coefficient 3 is brought down into the first place of the third Next 3 is multiplied by 6, is written in 2, and the product, which is the second row under 0. The sum of and - 6, or - 6, is written in the third row row. directly below 0. Proceeding in this way, is 28. + 8x fix 2 we find that the quotient is 3x 3 15 and the remainder Example 13-2. Given /(x) =z 3 2x 2 + 5x - 4, find By (13-3), /(r) equals the remainder obtained in the division of f(x) Hence, we have the following results Solution: by x r. : a) Therefore, /(- Therefore, /(I) = - 1) 12. = 0. c) Therefore, /(3) = 20. EXERCISE 13-1 In each of the problems from find the quotient 1. 3. 5. 7. 9. 11. 1 to 15, divide the first and the remainder, by using synthetic - Sx + 7, x - 1. - 5x + 6, x - 4. z 4 - 3x 2 - 6x + 6, i + 3. 2x 4 - x 3 - 6z 2 + 4z - 8, x - 2. 3 + 3x 2 - 2,r - 5, x - 2. 4 z 4- z 3 - 59s 2 - 69z + 030, z 2 x2 2. x* x2 3 , .c divide successively by each factor.) a? - 42. 4. z 6. x3 function by the second, to division. - - x2 Sx 2 Sx + + fix - 3. - 3. 6, x - 4. x 24, 6, - 8. 2x 3 3x 2 + Or + Sx 2 + 4, x 10. 2x 5 - x + 2. z + 2. (Hint: Factor, the divisor and 3z 3 + 2z 4- 1, 240 Theory of Equations 12. x* 13. 14. z5 aw - + 15. x* 3x 3 + So; 2 2x 4 - 21x 1. 1, o x y* y. - Sec. 13-2 + 2, x - 3x + 2. + 18, 2 + 2x - 3. 2 3x a; y For each of the following polynomial method of synthetic division functions, find the indicated values by the : = 3x - 7x2 _ 5x + 6 Find /(I) and/(- 4). = - 2x + 2x - 5x + 2. Find/( - 2) and/(0). = x - 4x 3 - 4x + 24x - 9. Find/(3) and/(9). = x* - 3x3 _ 13^2 + 21 X + 18. Find /(I) and/(3). = 7x 4 + 37x3 _ X _ 14a + 4. Find /(I) and/(- 5). 3 16. f(x) . 17. /(x) 2 3 a;* 2 4 18. /(a?) 19. /(x) 2 20. /(x) . THE REMAINDER THEOREM 13-3. In Section 13-2, it was shown that the remainder in the division of a polynomial by a binomial x r can be found without actually performing the division. Thus, in establishing (13-3), we have proved the following theorem. Remainder Theorem. If a polynomial f(x) is divided by x r, the remainder is the value of f(x) for x = r; that is, the remainder from this theorem that f(x) is exactly and only if /(r) = 0. Hence, we have proved It follows x r if divisible by also the fol- lowing theorem. Factor Theorem. If f(r) = nomial f(x), and conversely. Example 13-3. 3 157. Therefore, since /( therefore, that x is a factor of the poly- 2 2 According to the factor theorem, /( of /(x). But /(-3) = (-3) 4 - 2 Example 13-4. r + 3 a factor of x 4 - 2x + 3x - 5? = x 4 - 2x 3 + 3x - 5. Also, x - r = x + 3, and r = - Is x Solution: Here/(x) = then x 0, - 3) 7* 0, x 3) must equal zero if x -f 3 is 3. to be a factor +3 (-3) -5 =81 +54 +27 -5 + 3 cannot be a factor of x 4 - 2x 3 + 3x 2 - 5. 2 (-3) 3 Given /(x) = x 3 - 3x 2 2 is a factor of /(x). + 5x - 6. Show that /(2) = and, = x 3 - 3x 2 + 5x - 6, we have by substitution 3 3 (2) 2 + 5 (2) - 6 = 0. /(2) = (2) From the fact that/(x) equals zero when x = 2, it follows from the factor theorem Solution: Since /(x) that x - division 2 is and a factor of /(x). The student should check this result find that /(x) = x3 - 3x 2 + 5x - 6 = (x - 2) (x 2 - x + 3). by synthetic Sec. n 13-4 241 Theory of Equations Example 13-5. Find under what condition an integer and a ^ 0. (x -f a) is a factor of x* + an , where is =r + Solution: In this case, f(x) equal zero only if ( a) = n - an and/( a) , an that , = ( only when n is, a) + an w This . sum can odd. is EXERCISE 13-2 In each of the problems from 1 to IS, determine of the first. If it is a factor, find another factor. 1. 2z 3 3. z3 5. x3 7. xs 9. x4 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. + x + 6, x - 2. - 3z - 1, x - 1. - z 2 - llx + 15, x - 3. - 256?/ 16 z - 2?y 2 - 4z 3 - x 2 + 16.c - 12, z + - + . , the second function 2. z3 + 4z 2 4. z7 - 6. z5 + 243, 8. 2z 3 Gx 2 2z2 if 1, -f z 3z 2 -f .r - a factor + 6. x + x + 2. 2 bx - is 1. + 3. - 9z 111, z - 1. 1. + 5s 4 + 20z + GO* 2 + 120z + 120, x + 1. + aft (5 - 6r?& )z - 2a 6 2 (5 + 3a6 2 + 12a 5 x - ab. 2x* - 3x* - 3z - 2, x + 2. i - 10.r + IS* - 24.1* + 75, x - 2. 2x 4 - 31x + 21x - I7x + 10, x + 1. - 4(Xr - x 2 + lllj - 90, 2x - 3. 12.x 12z - 22.r 2 - 34z + GO, 3x + 5. 24* - 122x + 159x - Ix - GO, 3z - 4. 24x^ _ 74^.4 __ 85.r 3 + 311x 2 - 74x - 120, 2x + 1. Show that the equation x + 2.r 3 - 9r - 2x + 8 = has the roots 1,2, - 1, - 4. Find all roots of the equation 12.r - 4Qx* - x + lllx - 90 = 0, given that .r 5 3 5z 3 2 2 2 4 )a: 4 , 8 3 4 fc 2 3 3 4 3 2 4 2 2 4 a double that is, that (x a divisor of a n 3/2 root, b is Prove that a 22. Prove that a + b is a divisor of a n 23. Prove that a + 6 is a divisor of an 24. Prove that neither a + b nor a even integer. is 21. 13-4. We 3/2) 2 is a factor of the left side. b n for - bn -f- bn b is if every positive integral value of n. n irf a positive even integer. if n is a positive odd integer. a divisor of a n + 6n if n is a positive THE FUNDAMENTAL THEOREM OF ALGEBRA usually assume that every algebraic equation with real or coefficients has at least one real or complex root. Although complex the existence of such a root is not to be taken for granted, a proof is beyond the scope of this book. Accordingly, we shall accept the following theorem as true. Fundamental Theorem of Algebra. Let f(x) be a polynomial of degree n with complex f(x) =0 has at least one The first complete and was given by Gauss in Since that time, many coefficients. complex Then the algebraic equation root. rigorous proof of the fundamental theorem the beginning of the nineteenth century. proofs have appeared, but most require knowledge of the theory of complex functions. 242 Sec. Theory of Equations 13-4 repeated application of the fundamental theorem, it can be roots of any polynomial equation with real or complex coefficients is equal to the degree of the polynomial. We shall state the following theorem and give its proof. By a shown that the number of Theorem. If f(x) is a polynomial of degree n, the equation f(x) =0 has exactly n roots. Proof. Let f(x) be a polynomial of degree n with real or complex coefficients. By the fundamental theorem, there is a number TI such that f(ri) =0. Then, by the factor theorem, = f(x) where q\(x) - (x n) qi(x), a polynomial of degree n is 1 whose leading coeffi- cient is OQ. has a root by the fundamental theorem. Likewise, q\(x) = root denote this by r2 then QI (r 2 ) = 0. Also, we If , qi(x) = (x - r2 ) 92(3), 2 with leading coefficient a is of degree n tf 2 (#) can continue in this way also has a root. Similarly, # 2 (#) = until we come to a polynomial of the first degree with root rn Then where . We . (13-5) f(x) = a (z - n) (a - r2 ) (a - r n ), where a is the coefficient of x in our final first-degree polynomial. Since f(x) equals zero when we substitute for x any one of the n has at numbers r lf r2 it follows that the equation f(x) = , r, rn least the roots r a r 2 Moreover, there are no other roots. For, suppose that r is some rn Substitution of r for x in (13-5) yields root other than n, , , . , /(r) . , , = a (r - ri) (r - r2) (r - r n ). The right side of this equation cannot equal zero, since no one of the has no factors can equal zero. Therefore, /(r) ^0, and f(x) = more than the n roots found before. It may happen that a certain root appears more than once among the numbers r t r2 r. In that case it will be counted as many , , , times as the corresponding equal factors appear in (13-5). If a certain factor (x - r) appears times in (13-5), then r is said to be a root of multiplicity m. A root is called a simple root, a double root, a triple root, and so on, the proper name depending on how many times the same factor appears. Hence, combining the statements that "f(x) = has at least n roots' 1 and "f(x) = has no more than n roots," we can conclude that a polynomial equation of the nth degree has exactly n roots, a root of multiplicity being m m counted as m roots. Sec. 13-5 243 Theory of Equations It is to be noted that a rigorous proof of this theorem requires the use of induction. (See Chapter 16.) 13-5. OF COMPLEX ROOTS OF AN EQUATION PAIRS We wish to remind the student that while an equation of the nth degree has n complex roots, the number of real roots may be less than the degree of the equation. For example, x 2 + 1 = has no real roots. Determination of the number of real roots may be simplified by use of the following theorem. Theorem. if If all the coefficients of f(x) = are real numbers, and number a + bi is a root of f(x) = 0, then the conju- the complex gate a bi is also a root. It understood that a and b are real and is 6^0. Let x in f(x) Then we have Proof. a + bi. f(a + bi} = a (a = a xn + bi) n + + a x xn~ l ai(a + H h bi)*- 1 + a n be replaced by + . If we expand the powers of a + bi by the binomial theorem and simplify the resulting expression, then all terms which contain even powers of i will be real, while all terms which contain odd powers of i will be pure imaginary. Denote by P the aggregate real part, and by Q the aggregate imaginary part. Then we have = P + Qi = 0. with (11-4) P = Q = 0. /(a Hence, in accordance Now, replace x by a + bi) , 0. In those terms of f(a bi) f(x) raised to an even power, the result will remain the bi in in which bi is same as in f(a + bi) However, all terms in f(a bi) in which is raised to an odd power will have their signs changed. Hence, . /(a But, since we have shown - bi) that = P- P-Qi In other words, a bi is also a root of Example 13-6. Solve x* - x3 - Qi. P=Q= = 2x 2 + 6s bi 0, we conclude that 0. f(x) - 4 = = 0. 0, one root being 1 + i. 1 + i and 1 i are roots of the sum for the and and (12-4) product of two given equation. Using Eqs. (12-3) 2 are roots a: 2x +2 = 0. these numbers of we find that roots, conjugate complex Dividing the original polynomial by this quadratic function, we get the quotient Solution: According to the last theorem, both x2 + x roots, x - 2. = - Solution of the equation x 2 I. 2 and x = + x - 2 = yields the remaining desired ; 244 Theory of Equations Sec. 13-5 - vX EXERCISE 13-3 1. Solve X s 3 2. Solve x 3. Solve z 4 4. Solve x* 5. Solve a; 5 + x 2 .- 2x -f 12 = 0, one root being 1 + V3f. + 3z + I2x - 16, one root being -2 - 2 V~ 3. - 2x - 7x + ISx - 18 = 0, one root being 1 -f 3z -f 7z -f 6z + 4 = 0, one root being - 1 - Vi. - 8z -f 27z - 46z + 38x - 12 = 0, one root being 2 3 2 3 2 4 3 and one root being 6. Solve x* 7. Find a Find a 8. 13-6. f. 2 +x + 1 = 2 by considering the equation to be a quadratic equation in x 2 two whose roots are 2 and 1 -f 2i. equation of lowest degree having the roots i and 1 - real cubic equation, real 2 1. THE GRAPH OF . of A POLYNOMIAL FOR LARGE VALUES OF i. x. In graphing a polynomial function, it is helpful to know the location of points on the curve for numerically large values of x. It can be shown that, when x is numerically sufficiently large, the term a x n of highest degree is numerically larger than the sum of all the other terms combined. Therefore, the sign of this term determines the sign of the entire polynomial. Let us consider the values of the function f(x) = a; 3 + 5x 2 - Ix 13 as x assumes various values from left to right along the #-axis, that is, for increasing values of x. The results may be tabulated conveniently as follows : When x is negative, but numerically sufficiently large, it is seen that f(x) is negative. Thus, when x = - 10, x* = - 1000, while the sum of other terms, or the value of 5x 2 -Ix - 13, is only 557 and ; = -443. For points far enough to the right of the origin, = for x say 10, f(x) is positive. For example, /(10) = 1000 + 417 = 1417. Hence, the graph of y = f(x) would be located below the #-axis on the left but would rise above the #-axis as x gets larger /(-10) and larger. As a second x4 + 7x 2 -8x + = illustration, let us consider the function f(x) 10. Here f(x) is positive for large numerical values of x, regardless of the sign of x. Hence, in this case the graph Sec. 1 3-8 245 Theory of Equations would be above the right and left of the numerical values of x to both re-axis for large origin. The following helpful conclusion can be drawn from the preceding discussion involving large values of x. When x is sufficiently large and positive, f(x) has the same sign as the leading coefficient a. When x is negative and sufficiently large numerically, f(x) has the same sign as a when n is even, and has the opposite sign when n is odd. The following symbols are sometimes found in discussions of the values of functions for numerically large values of x. When the symbolic statement /(-foo)>0is used, what is meant is that, for all sufficiently the statement tive values of large positive values of x, f(x) ig positive. Similarly, means that for all sufficiently large posioo) < means oo) > x, f(x) is negative; the statement /( /(+ which are sufficiently large numerco and means that, for all negative ically, / (x) positive /( ) < of x values which are sufficiently large numerically, f(x) is negative. 2 Thus, if f(x) =x* + 5x -7x- 13,/(- oo) <0 while /(+ oo) > 0. = x* + 7x 2 - Sx + 10, /(- oo)>Oand/(+ co)>0. However, if f(x) that, for all negative values of x is 13-7. ; ROOTS BETWEEN a AND b IF f(o) AND Another helpful theorem relating equation is f(J>) HAVE OPPOSITE SIGNS to the roots of a polynomial the following. Theorem. If the coefficients of a polynomial f(x) are real, and if a and b are real numbers such that f(a) and f(b) have opposite has at least one real root between signs, then the equation f(x) = a and b. We shall not give a proof of this statement here, but shall merely mention the following geometric considerations. The graph of a polynomial is a continuous curve; that is, it has no "breaks." Therefore, if the points (a, /(a)) and (b,f(b)) lie on opposite sides of the #-axis, the graph apparently has to cross the #-axis at least once between these points. 13-8. RATIONAL ROOTS The following theorem is fundamental for the solution of equa- tions having integral coefficients. Theorem. If the equation f(x) with integral - a xn coefficients + aix n~ l H h an = has the rational root -, a > where c and d 246 Theory of Equations Sec. 13-8 are integers having no common factor > 1, then c is a divisor of the constant term an , and d is a divisor of the leading coefficient a . We shall make use of a principle from the theory of numProof. bers: If an integer c divides the product of two integers a and 6 and if 6 and c have no common divisor other than 1, then c is a divisor of a. XI Let -3 a common a Multiplication w Since c divides now ^ c "+ ' a "~ l l + a" = a d gives + aicn~ all and d are integers with no c Then 1. cn ^ + ai d^r + n by d aoc where 0, divisor other than cn term. If = be a root of f(x) ~l + a ndn = n h a n -icd d H terms before the 0. final one, c also divides that c is factored into primes, none of these primes is a divisor of d, and therefore of d n . Thus each prime divides a n , and so c itself divides an In a similar fashion, it may be shown that d . divides a . - rational roots of 3z 3 Example 13-7. Find the I7x 2 + I5x + 7 = 0. st Solution: The possible rational roots are of the form -r > where c is a divisor of ct the constant 7 and d is a divisor of the coefficient 1/3, 7, 1, Using synthetic division to check - 3 3 By the remainder 17 15 7 3 20 -20 35 -35 -28 theorem, /( - = We see, however, that /(O) there must be a root between x 1) = - = 1 hope that a rational root sibility is 1/3. The check by 3 3 - 28. Hence, by following result: 1 - 1 is not a root. the theorem in Section 13-7, and x = 0. Examining our list of possible 1 and 0, we see that a posbetween lies synthetic division follows: 17 15 18 21 7 -16-7 - |- 4-7. Therefore, roots, in the possible roots are 7/3. we obtain the 1, The only 3. |- 1/3 = 1/3 is a root. After dividing the given polynomial by x + 1/3, and 2 18z 4- 21 to 0, we obtain the quadratic x 2 6x + 7 = 0. equating the quotient 3x Hence, x This equation has the roots 3 =h \/2, which are not rational numbers. Therefore, = 1/3. the only rational root is In this example, as is sometimes the case, it has been possible to find all the roots of the given equation. Sec. 13-8 247 Theory of Equations EXERCISE 13-4 1. Show that l&r 3 - 33z 2 + 5 = = 0. + - 1 and 0, between and 1, and between and 2. Find these three roots. Find the rational roots of 2x 3 - 9z 2 + 3x + 4 = 0. 3x - 5 = has at least one positive root. Prove that x 3 + 2x 2 x 3 4- x 2 + x Prove that x* 3 = has at least one positive root and at least one negative root. Prove the following corollary of the theorem in Section 13-8. If f(x) = x n + aix*- 1 + + a n = has integral coefficients and has an integral 2x has real roots between 1 2. 3. 4. 5. root 6. r, then r is a divisor of a n Find the integral roots of x* . - 1 + x + 1 = has no rational roots. = 0. Find all the integral roots of 8 + z a + x + I = 0. 5 3 Show that 3 is a root of x 3 - ^ x - 2x + - = 0. Why does this z & Show 2 that the equation x 1 8. Solve the equation x 3 7. 9. 10. a: 2 diet the not contra- theorem in Section 13-8 or that in Problem 5? In each of the problems from 11 to 20, find all roots of the given equation. 12. 4x 3 - 16z 2 - 9z + 36 = 0. 11. x 4 - Sx 2 + 16 = 0. 13. 3x* 3 15. 2z 17. 5z 3 19. 2x* + x 2 + x - 2 = 0. - x 2 + 2x - 1 = 0. - 13z + 16s - 6 = 0. - x 2 - 4z + 2 = 0. 2 + 3z 2 - 6x - 9 = 0. x llz 2 + 37x - 35 = 0. x* - Sx 3 + 37z 2 - 50z = 0. 3z 3 - 13z 2 + 13z - 3 = 0. 14. 2x* 16. 18. 20. s 14 Inequalities INTRODUCTION 14-1. In previous chapters we have explained some methods of determining the roots of an equation. By applying these methods, one can find the values of an unknown for which a certain function of the unknown equals zero. Often, however, it is necessary to solve an inequality, that is, to discover for what values of the unknown a certain function is less than or greater than another function. The present chapter is concerned primarily with the solution of inequalities, and the following discussion is essentially an extension of the study of the order relation undertaken in Section 1-8. We, therefore, recommend that the student thoroughly review Section 1-8 before starting the study of the present chapter. Since the solution of inequalities often involves the use of absolute values, a thorough mastery of Sections 1-9 and 1-10 is also a requirement. The classification of inequalities corresponds to that of equalities or equations. As in the study of equations, there are two kinds of inequalities involving unknowns, namely, absolute inequalities and conditional inequalities. An absolute inequality is an inequality that is satisfied by all values of the variable or variables for which the functions appearing are defined. A conditional inequality is one that is true only for certain values of the variable or variables. Thus, x 2 + 1 > (where x is real) is an absolute inequality, because it is true for every real value of x but x 1 > is a conditional inequality, because it is valid only ; when x > 1. PROPERTIES OF INEQUALITIES 14-2. The gous have rules for dealing with inequalities are to some extent analoto those for equations. In transforming inequalities, we shall occasion to use the following elementary principles which 248 Sec. 14-3 249 Inequalities follow at once from the fundamental properties proved in Section 1-8. Here are three From From From 12 8 > < 12 < a If 1. Principle then a &, illustrations c < b c. : + 3 > 8 + 3. - 2 < 12 - 2. 12 - 8 > 0. follows that 12 8, it 12, it follows that 8 > 8, it Principle 2. If a follows that < b c> and < then ac 0, be and - < c Two illustrations are given here From 3 < 5, it follows that 3 2 < 5 Q A From 8 < 10, it follows that < ~ c : 2. 1 , Principle Here 14-3. an is From 3. If 3 < a < b and illustration 4, it c < then ac 0, > be and c > c : -3 > follows that -4. SOLUTION OF CONDITIONAL INEQUALITIES The process of solution of a conditional inequality consists in values of the variable which satisfy the inequality. finding solution consists of a set of values of the variable, rather than one or more isolated values as is usual in the case of a conditional equation. A all The discussion in this section is limited to inequalities involving rational functions in only one variable. If this variable is x, the or f(x) < 0. For inequality can be written in the form f(x) > instance, suppose we want to solve the inequality x2 We may - > x 2. then obtain the following equivalent inequality: /(a) = x2 - x - 2 > 0. seen that this transformed inequality has the same solution set as the original inequality. In solving this transformed inequality, we find first the values of x, if there are any, for which f(x) changes sign as x increases in magnitude. If f(x) is a polynomial, such a change of sign occurs It is easily when f(x) =0. In the example under consideration, changes of sign are obtained only at points where 2 /GO = x -x-2=Q. we must find the roots of x 2 - x - 2 = 0. These are 1 and and 2, they determine on the #-axis three intervals throughout each of which f(x) retains the same sign. In other words, to find the Hence, 250 Sec. Inequalities 14-3 > 0, we find the interval or intervals within which the has f(x) sign indicated in the given inequality. In the example under consideration this sign is positive, because f(x) is to be > 0. solution of f(x) The method is applied in Example 14-1. In general, the solution of an inequality is obtained by equating the function f(x) to zero and solving the resulting equation. If the is inequality of the i?i form ^ 0, ps) where p(x) and q(x) are poly- 2 nomials, it may be cleared of fractions by multiplication by [q (x) ] Since the square of any non-zero real number is positive, the sense of the inequality is not changed by multiplication by this factor. This leads to the form f(x) ^ 0, where f(x) is a polynomial. The . values of x for which f(x) changes sign are called critical values. When the critical values are arranged in increasing order, they determine on the #-axis intervals, throughout each of which f(x) cannot change sign. Consequently, the required solution is represented by the set of values of x for which f(x) has the same sign as that indicated in the given inequality. Note. In general, it can be shown that if a factor of f(x) appears an odd power (that is, if f(x) = has roots of odd multiplicity), the function will change sign at values of x for which this factor vanishes. If a factor of f(x) appears to an even power (that is, if to =0 has roots of even multiplicity), the function will not change sign at values of x for which this factor vanishes. Therefore, it is sufficient to test only one value of x in one interval. This test gives the sign off(x) in that interval. The sign of f(x) in each of the other intervals can be quickly and easily determined from the multiplicity of the critical values. Substituting into f(x) a value of x in each interval then provides a check of the solution. f(x) Example 14-1. Solve the inequality x > 2. = x2 - x - x2 Solution: An equivalent inequality f(x) From the preceding discussion it These roots are As shown in 1 and 2 > 0. follows that the critical values are the roots of x2 - is - x - 2 = 0. 2. Fig. 14-1, the points 1 and 2 determine on the z-axis the following three intervals (a) x < - 1; (b) - 1 < x < 2; (c) x > 2. Sec. 14-3 251 Inequalities Throughout each of these intervals, f(x) retains the same sign. This condition 2 x 2 shown in Fig. 14-2. Here may also be seen from the graph of y = x we note that in each of these intervals the curve lies either entirely above the a>axis or entirely below it. The = (- > solution of f(x) - (- can now be found by examining Thus, for a value such as x of these intervals. -2=4. Hence, f(x) is =x x 2 shown in 1. x-axis to the left of x = For the value x = in the interval (6), we have /(O) = - 2. Hence, f(x) is nega2) 2 2) 2 In the graph of y = - the sign of /(a?) in each 2 in the interval (a), /( - 2) positive throughout the interval (a). above the Fig. 14-2, the curve lies throughout this interval, and the graph below the g-axis. For the value x = 3 in the interval (c), tive lies = /(3) tive, (3) 2 - (3) - = 4. 2 and the graph again So/(.r) lies is posi- above the .r-axis. These same may results also be obtained following much shorter procedure: Select a value of x in the interval (6) for by the which /(x) easily evaluated. is = isz=O.Since/(0) -2,/(x) this interval. Therefore, f(x) vals (a) and (c), changes at the Such a value <0 throughout > for inter- because the sign of f(x) values x critical = 1 _ and * A x=2. We see, therefore, that in the intervals (a) and (c), f(x) has the sign indicated by the given inequality. Since f(x) must be greater than zero, the solution set of the - 1 and x > 2. given inequality is described by x < Example 14-2. Determine the values Solution: We shall of x for which \/x 3 2x 2 - 3x is real. solve the equivalent problem f(x) 3 Solving the equation x 2x 2 = z 3 - 2z - 3x ^ 0. 3# = 0, we find that the critical values are 2 1, 0, 3. (d) As shown in Fig. 14-3, these critical values determine on the #-axis the following four intervals: (a) (c) x < - 1; < x < 3; (6) (d) x Throughout each interval f(x) has the same sign. In this example we may select z = 1 in the interval - 3(1) = - 4, f(x) < throughout this interval. 1 > (c). <x < 0; 3. Since /(I) = (I) 3 2(1) 2 252 Sec. Inequalities As we proceed into the interval > Therefore, f(x) and becomes < (6), we 14-3 when x = 0. when x = - 1 find that/(z) changes sign in the interval (6). Again f(x) changes sign from the interval (c) we find that/(#) changes sign when x = 3 and is > in the in the interval (a). Proceeding to the right into the interval (d), interval (d). Thus, we have the following results: in (a), /(a) in (),/(*) < 0; in (&),/(*) <0; in(d),/(x) Hence, the inequality /(x) 2x 2 3x = condition x 3 1, =x 2x 2 3x ^ are which the original expression is satisfied is 1 solution. ^ x ^ V# Example 14-3. What values in intervals (b) and 0; 0. (d). And, since the also allowed in the original problem, the values the 3 are included in 0, 3 > > > 3 2# 2 3x solutions 3x 2 ^- > 0? X t Solution: Clear the given inequality of fractions by multiplying by (x - 2) (& 8 - 3rc 2 ) > 0. Solving the equation z 2 (,r - 2) (x (x we find that the critical values are 0, 0, 2, 3. obtain f(x) = Hence, we have (a) x < 0; the, for f(x) These are the values of x for 3. is real. x3 of x satisfy the Therefore, and x ^ following four intervals, as < x < 2; (c) 2 (b) shown in Fig. 14-4: < 3; x < (d) Z > 2) - 2 3) and = 0, 3. Throughout each of these intervals /(x) has the same sign. z = 1 in the interval (b). Since /(I) = (1 - 2) [(I) 3 = 1) 2) 3(1) 2, it follows that/Or) > throughout this interval. We see that /Or) does not change sign for the critical value x = 0, because x is a double root of f(x) = 0. Hence, f(x) > in the interval (a). Let us - 2 ] initially test f(x) for ( = ( As we proceed to the right from the interval (b) the function f(x) changes sign for in the interval (c), each of the critical values x = 2 and x = 3. Hence, f(x) < and f(x) > in the interval (d). Therefore, f(x) is positive in the intervals (a), (6), and (c?). That is, the solution set of the original inequality is described by x < 2, excluding the value x 0, = and x > 3. Example 14-4. Solve the inequality x 2 = 2 2x +3 > 0. = + x Solution: Let f(x) 2x 3. Since the roots of f(x) are imaginary, there are no critical values. Hence, the graph of y f(x) lies either entirely above the x-axis or entirely below that axis. = Testing for x above the = 0, we find that/(0) = 3. z-axis. This result indicates that the graph lies is" satisfied for all real values of x. Consequently, the inequality 14-3 Sec. 253 Inequalities - 2x 1 1 Example 14-5. Solve the inequality By Solution: Section b-a<x<b+a. r 3~ Hence, be written as follows may the inequality 1-10, < I which 1, < 1 < We may 2x 1 and 5 O 2x - - or 1 i 1 a equivalent to is equivalent to 2z | 1 < 1 3, < 2x < 2 4. 5 O < and determining the common 1 3<2x we have O described is by proceed by solving individually the two inequalities or 1, 1 - 1 < 3, or x < 2. The common solutions < 2. So we again have - 1 < x < 2. 2x x 1 < | that the solution set of the original inequality Alternate Solution: 2x is 6 : l-3<2x<l+3 We thus find - 1 < x < 2. - x | < satisfy the 2x 1 < 1, we have - 1 < x and inequalities For x. For solutions; r o EXERCISE 14-1 In each of the following problems, solve the given conditional inequality or inequalities. 1. x - 3 4. x - 1 7. 6r < 2. x 0. 5. 4x 8. 4x 11. - 14. - + 3 < 0. 10. 3 - 13. < 4x + 0. < 2x + 2-i + i 1. < 1. < 0. - 16 < - 8 > 3x - 3 < 6x < 3. 1 < ^? + ? < 1 0. 10. x 6. 4x - 16 > 0. 9. 4x < - 3x - < 3. 12. 2x 2 22. 4z 2 25. 3ic 20. 32. + 2 5x < - 1. 3# < 4. 27. (x + !)(*+ 2) 29. (x - 31. Vz 2 33. x x 35. 1) (x +2 +1 25 - (x 2) (x is real. > o. 23. ^ ^ X 2 I < 3 < 0. _ 1 X + 3) < 0. 28. (x - 0. 30. (2x 3) > - - 32. \/x* 34. 1) (x - 5z 3 +2) x2 < | 21. x(3x < 4) (x | 144. 24. 6a? 0. > - 26. 3# 2 4- 1 - 2* | . - x | IB. 1. 18. 19. +5 > 3, 2 4. < < -- - ^ X X 7. 1. 1. 9. + 5) (x + 2) + 6 - 6) > (3x 4- 1) is real. 0. ^ 0. 254 Sec. Inequalities 14-4 ABSOLUTE INEQUALITIES To prove the truth of an absolute inequality, one must use the known properties of the order relation. When none of these seems readily applicable to the given inequality, it may be helpful to replace this inequality by an equivalent one which may be more Repeated replacements may have to be made. In carrying out a sequence of replacements, one need not verify equivalence at each stage, provided that the final inequality can be shown to imply the original one. The methods for proving absolute inequalities may be used also to prove theorems involving inequalities, as in Example 14-7. easily treated. + 62 ^ 2 Example 14-6. Prove that a The given Solution: inequality a2 that is 2ab for all real equivalent, - 2ab + (a - 2 b2 by numbers a and Principle 1, b. Section 14-2, to 0, to is, This last inequality 6) 0. because the square of every real number is true, negative. Therefore, the original inequality is is (I Example 14-7. a .a . , prove that -r Solution: o < distinct positive real r < inequality T , , is is by equivalent, is true because Similarly, the inequality , < ad this inequality, by Principle -3 is -f ed 1, is < < ab 1, CL I the inequality a T o / + be. be. < , , is a -f c , +a o . is -5 by true also. equivalent, < < by Principle 2, to be -f ed, < be. equivalent to the given condition T c <3 Hence, /* <^ , < > to a ,... C -5 equivalent to ad This last inequality again inequalities < Principle 2, Section 14-2, to equivalent to the given condition T it is , by equivalent, Principle Principle 2. Therefore, the inequality T and T d, ad This inequality if , . ab -h ad This inequality numbers, and +c <c 3 a + b The and d are If a, 6, c, non- true also. , < is true also. c j are true, a Therefore, it follows that the original 14-4 Sec. 255 /nequcrfiffes EXERCISE 14-2 1. Prove that a 2 2. Prove that +1 2 a > Prove that a 2 4. Prove that a > 5. Prove that T + -a > Prove a2 o a2 7. If that + b2 a if ^^ ^ ^TT + 3. 6. 2a 2t a a if a3 if 2 if a 6 > 1 < a if > and < > 6 and that a 2 ; + Note that a 2 is real. < 1. 0. a < if < a 1. 1. a and b are positive and a + b + c ^ a& + + c 2 ^ 26c, and 2 = 2a only when a = 1 2 fcc 2 2a6, 6 c2 9* b. + ca. (Hint: From Example 14-6, + a 2 ^ 2ca. Add these inequalities.) - a and 6 are two positive real numbers, the quantities and \^ab, > ~-r are called, respectively, the arithmetic mean (A), the geometric mean (Cr), and the harmonic wean (H) of a and b. Prove that < A, except when a 6. = H <G (In this case, we have 8. Prove that a 3 + 63 > 9. Prove that ab + cd ^ C2 10. = -I- rf2 Prove that a 3 z\ %2 13. If 21, z 2 14. If z is , 1 1, if a, 6, c, a and 6 are positive and a and d are ^ as a 3 , positive, and if a2 > 6. + &2 = 1 and = \ | 21 6, if a, 6, and c are positive. 6. a complex number, the modulus or absolute value of z is \/x 2 4- y 2 Show that \z + z 2 \zi\ +\z*\ and that is z | $ + c o according 6 + ab < a 4 + 6 4 if a ^ = x + yi denoted by I 6), if 3ab(a ^T Prove that 7 12. If 2 //.) 1. b 11. A =G= | . | 22 | , for all l \ complex numbers and z 3 are complex numbers, prove that a complex number, prove that 2 | | | | Zi and z\ +z 2 +| \ 22. \z\ +23 <2||. 1 -h |2 -2 3 |. 15 Progressions SEQUENCES AND 15-1. SERIES Sequences. An infinite sequence, called more simply a sequence or sometimes a progression, is a single-valued function whose domain of definition is the set of positive integers. A finite sequence is 1, a single-valued function whose domain consists of the integers for some positive integer m. 2, , m In specifying a sequence, it is necessary to give a definite rule of correspondence which assigns to each integer n a single definite number, or term, of the sequence. This term may be denoted by a n In particular, there is a first term ai corresponding to the integer 1, a second term a2 corresponding to the integer 2, and so on. The sequence may be specified by the array of numbers . (15-1) aij This sequence a,2, as, , an , . denoted briefly by {a n }. (As usual, throughout of dots stands for numbers assumed to be present but not written.) A finite sequence has a "last" term and am or simply by {an }. may be designated by a i9 o-2 The nth term, or general term, of a sequence is denoted by a n From the rule specifying the nth term for each n we obtain the is this discussion, a row , . f first, for second, third, and other terms of the sequence by substituting 1, 2, 3, and so on in turn. For example, if an = 1/n, n the values the sequence is 111 lig'a'i""' We should note that {an } is the symbol for the sequence or function as a whole, whereas an is the symbol for the nth term or value of the function corresponding to the integer n. 256 Sec. 15-1 257 Progressions There are many methods for specifying the function in the Two of these methods follow tion of a sequence. of Explicit Formula. In one method, the nth n itself by means of an explicit formula. Here are a few If T, If an an = n, the Tl T* If an = sequence ~T~ ' L is given in terms . .123-^ ti16 se(l uenc e is ~ > ^ the sequence ' 1 = > 9 O Zi n jr ^71 is 1, 2, 3, ., 2 _j_ term illustrations. n = defini- : .,23 ^ is 1, i . 1U > > . O o Recursion Rule. In another method, one or more of the first several values of a n are given explicitly, and a rule is then given whereby a n can be calculated from some or all of its predecessors. A few illustrations are given here. Let a n+ i = an 2 + I with a2 as a4 Let a n+i = (n + Let a n+2 in this = = = example a n o n with = <Wi + as a4 05 = = = 0. Then + 1 = O2 + 2 a2 + 1 = I 2 + a3 2 + 1 = 2 2 + ai 2 l)a n with a\ a2 as a4 Note that = a\ = = = = 2ai 3a 2 4a 3 = nl ai = + ai + 02 3 4 + a3 1 1 = = = 1, 2, 5, Then 1. = = = fife 1 2 1 3 2 4 6 = = = and 02 = = = + + + 1 1 2 2, 6, 24, = 1 1 1. = = = Then 1, 2, 3, Let {an } be a given sequence of terms ai, 03, , a*, Form a new sequence {s n }, where s n is obtained by adding the first n terms of {an }. The sequence of partial sums is then given by Series. . Si sn = = = Oi, 52 ai + 02 + Oi + 02, + an The sequence {s n } formed in this based on the given sequence {an }. The ' > (n , way is called = 1, 2, 3, ) the (infinite) series series as just defined is usually written in the following abbreviated form: (15-2) ' ai + a2 + 258 Sec. 15-1 Progressions Two illustrations of series follow. Illustration partial Let an 1. = - and = sn n + a^ H ai series may si = ai = 1, 2 = Oi + a2 = 1 +o = 1 , 1 3 9 > 11 1 = Let an 2 4. = an(^ $n + + #2 + &i <&n- Then the sums are given by = The Then the . then be written Illustration 2. partial an sums are given by 1 The I- ai + = a2 112 + - g g 3 series is 2 n2 12 6 + ' n Limit of a Sequence. One of the most important questions relating to sequences is whether or not a given sequence an has a limit as n increases indefinitely. If such a limit exists, the sequence is said to be convergent or to converge to the limit <. Symbolically this statement may be expressed as follows { } : = lim a n w-oo This notation is is . read, "the limit of a n as n increases indefinitely ." In order to help make the concept of limit clear, the sequence given by a n = - The terms we shall consider are 71 dl = 1, d2 = 1 H Z ' a3 1 = 7j O >'> 0n =-1 fl When we examine n is, these terms, we notice that the larger the number the smaller the term becomes. In other words, as n increases, the closer to zero as we please, if is the number - we merely n In fact, - can be n made as small choose n sufficiently large. For example, Sec. 15-1 259 Progressions - < n 1000 for every n larger than 1000. We may conclude that for an arbitrary real positive number d, we can find a value of n, say any - < n - N integer - < n d. n for every 100 The ^ such that for > ^ a We larger than 100. all see also that n > N, integers is it true that and the sequence <-> are therefore related as limit (nl indicated by the following statement. The sequence < - 1 (n) converges to the limit = < to each arbi- 0, if trary positive number d there corresponds an integer that 0-d<-<0 + dfor n In general, the limit Definition. number d > < d {an converges to the 0, there corresponds + d for < an every n > N. : sequence a n converges to the limit <, there exists a positive integer such that \ { : } number d > also be put as follows A be defined as follows A sequence N such that may may of a sequence , a positive integer such > N. every n Definition of Limit of Sequence. if to each arbitrary real limit This definition N> if for each N n | < | Convergence of a Series. d for every n We > N. shall again consider the infinite series ai (15-2) which is + a2 + + an + - , sums the sequence of partial Slj 52, - * ' ' , Sn ' ' ' , of the sequence {a n }. By the following definition the series vergent if the sequence of partial sums is convergent. is con- sequence of partial sums of the infinite series (15-2) converges to a limit, and if lim S n = Sj then the series is Definition. If the n- said to converge to the limit S, and S CD is called the sum of the infinite series. The new use of the word "sum" for the value S of an infinite perhaps unfortunate, for it seems meaningless to talk about adding up the terms of an infinite series. Actually, S is not a sum, but it is rather the limit of a sequence of partial sums of the series. If a series does not converge to a limit as n becomes infinite, we say that it is divergent, or that it diverges. series is 260 Sec. Progressions 151 EXERCISE 15-1 1. 2. Given = a\ 1 and a+i = Given a l =4, a 2 the sequence { a n } 3. Given ai = 1, a2 of the sequence 4. 5. Show Show of 6. five terms of the sequence {a n }. an Find the = 2a n +i and a n + 2 = an } 2, = and a n+ 2 . first six terms of + (n na n Find the l)a n+ i . first six terms . = 3 n satisfies the recursion rule an+ = fln+1 + 6an w = 2) also satisfies the recursion relationship 2 that the sequence a Problem Find the . . that the sequence a n . ( 4. A and B are any real numbers whatsoever, show that the = A3 n + #( 2) n satisfies the relationship of Problem 4. = 2a n + 1 with ai = 2. Let sn = + 2 + + an Find Assuming that sequence a n 7. { 3, = n + an Given an +i ai, 02, , 8. Show 9. Prove that 10. If s n + that =n (2n 15-2. - . i 05 and if s cr n 2 , 1) s lf s 2) ai = sn + a% s. , + sw -i for + n > 1, = 2n show that a n = n2 , then s n+ i sn = a n +i = SL and therefore that given ai 1 1 +3 +5 + . ARITHMETIC PROGRESSIONS An arithmetic progression is a sequence of numbers in which each term after the first is obtained from the preceding one by adding to it a fixed number; this number is called the common difference. Note that the common difference may be found by subtracting any term of the sequence from the one that follows. Thus, 1, 5/2, 4, 11/2, is an arithmetic progression with the common difference 3/2, since 11/2 -4 = 4-5/2^5/2-1 = 3/2. Also, 5, 1, -3, -7 is an arithmetic progression with the common difference 4, since -7 -(-3) =-3-l = l-5 = -4. necessary and sufficient condition that three numbers A, B, and C form an arithmetic progression is It follows, therefore, that a C-B^B-A. 15-3. THE GENERAL TERM OF AN ARITHMETIC PROGRESSION Let a denote the first term of an arithmetic progression, and let d denote the common difference. Then, by definition, an arithmetic progression with n terms may be written as follows : a, Hence (15-3) if ln a + d, a + 2d ; a + 3d, , a + (n represents the value of the nth term, ln = a + (n - l)d. - l)d 15-4 Sec. Also, we may n terms write an arithmetic progression of following manner a, We 261 Progressions in the : a + d, a shall be concerned + - 2d, with , - ln 2d, Zn - five quantities in d, t. connection with an arithmetic progression. These are the first term a, the terms n, the nth term ln , the difference d, and the sum number of of the n Sn terms. SUM OF THE 15-4. TERMS OF FIRST n AN ARITHMETIC PROGRESSION Sn represent the sum of the first n terms of an arithmetic progression. If we write the indicated sum in both direct and Let reverse orders, Sn = a + In + (a we have + d) + + + 2d) + (a (l n - 2d) + - (l n d) + Zn , and S = (l n - Adding the right d) + (Zn sides, - 2d) + + (a + 2d) + we have n terms each (a + d) + a. of which is a -I- ln . Thus, 2S n = (a + + Zn ) (a + Zn) + + (a + + ^). k) + (a + Zn) = n(a + t). Therefore, Sn = (15-4) (a - l)d from (15-3) for (n have another useful form for the sum. This is If we substitute a + Sn = (15-5) Example 15-1. [2a + (n Z in (15-4), we - Determine which of the following sequences are arithmetic progressions: a) 3, 7, 10; 6) 6, 1, - 4; c) 3z - y, 4x + y, 5x + 3y. 10-7 and 7 3 are not equal, the sequence not an arithmetic progression. 5. Since these differences are 6 = b) In the second sequence, - 4 is an arithmetic progression. equal, the sequence 6, 1, Solution: a) Since the differences 3, 7, c) = x 10 is -4-1=1 We find that (5x + 3y) - (4x + y) = x + 2y and (4s + y) - (3& - y) + 2y. Since there is a common difference, the given sequence is an arithmetic progression. Example 15-2. Find the twelfth term, and of the arithmetic progression 4, 7, 10, . also the sum of the first 12 terms, 262 Solution: We have a = 4, n = 12, and d = 3. - l)d = 4 + 12 = a + (n Z Also, Sec. Progressions by Then, by (15-3), - (12 1)3 = 37. (15-4), Si. = + Z.) = y (4 + 37) = 246. (a \ Example 15-3. The sixth term is Solution: third term of an arithmetic progression Find the twenty-second term. 3/2. By (15-3), h = a 2d and -f f a {a By solving these l (15-3), 15-4 n = 1/4 two + 2d + 5d = a -f = 3/4, = 3/2. we linear equations, + 21(1/4) = Z6 5d. is 3/4, and the Thus, we have = find that a 1/4 and d = By 1/4. 11/2. Example 15-4. Find each value of x for which the three quantities 3# 5, x + 4, 2 form an arithmetic progression. 3x C B =B A, we have + 4) = (x + 4) - (3x - 5). Solution: Applying the condition (3x o, - i Solving, 15-5. i. A we obtain ar - 2) 15 =~ - (x m, The -*u . i- arithmetic progression . is 25 -j- 37 31 > -7- > -j- ARITHMETIC MEANS The terms of an arithmetic progression between any two given terms are called arithmetic means between the given terms. If we let the given terms be a and Z n any number, say fc, of means , be inserted between a and ln by using the formula (n l)d with n = k + 2. As soon as we have found d, insert the required means. may Example 15-5. Insert three arithmetic means between Solution: Let a = 0, I* = and n 1, 113 Hence, the three means are j > ^ > = 3 + 2 = 5. Then 1 ln =a+ we can and 1. = + 4d, and d = j If only a single arithmetic mean is to be inserted between two given numbers, then the inserted value is called the arithmetic mean of the given numbers. Thus, if a, x, b form an arithmetic progression, x is called the arithmetic mean of a and 6. Since x = x a, 6 Thus, the single arithmetic half their sum. mean of two numbers is equal to one- Sec. 5-5 1 263 Progressions EXERCISE 15-2 In each of the problems from 1 to 9, determine if the given sequence is an arithmetic progression. Find the next two terms of the extension of each arithmetic progression. - 1. 5, 8, 11, 14. 2. 4.4,12,19,27. 5- -H' 8. a 7. a, 6, - 26 - 36 a, 2a. 1, 7, 13, 12. 22, 18, 14, 14. 2, 4, 6, to 7 terms. - - to 50 terms. to 10 terms. 16. a, 2a, 3a, 9. ~-^ , ^ 3, 4, 11. a, ^^ 13. 1, 2, 3, to 10 terms. 15. 1, 3, 5, to 75 terms. 17. 0.2, 0.5, 0.8, to 35 terms. 18. 1, 8, 15, - 10, ^ and S n for the arithmetic progression. 11. 3, 6, 9, to 26 terms. 19, find l n to 12 terms. - - -|'fi'-i'-Ig- '*' + 6, a - 6, a - 26, a - 36. In each of the problems from 10 to 10. 2, 8, 14, 1 3. 19. 19. - to 20 terms. to 100 terms. 1, 2, 3, In each of the problems from 20 to 27, three quantities relating to an arithmetic progression are given. Find the other two quantities. 20. a 22. = S2 i 5, = h = 653, = 4. = 21, a = 6. n 36, n 21. a 23. n = 10, n = 10, d = 10. = 45, d = L 845 = 63. i 24. 26. Z 2l a = 8, = - n 1 i 28. Find the 29. = 21, , d = sum d =\ 3 = i-- , 27. a | = 7, & = 52, d = = 45, d = = 1, S. ^ 1. of the first 100 even integers. Find the sum of the first 31. Insert ten arithmetic 32. Insert six arithmetic n odd integers. means between 20 and means between 100 and means between mean 33. Find the arithmetic of 10 30. 40. 3 and and 56 and that 2. of 4 and 28. Find the arithmetic mean of 28 and 65 and that of 33 and 78. 35. Insert k arithmetic 36. Z 12*5 30. Insert five arithmetic 34. 25. o means between - and a, where a ^ 0. A display of cans in a grocery store is in the form of a pyramid whose base is an equilateral triangle. If each side of the base contains 20 cans and the number of cans decreases by one for each successive row, how many cans are in the display? 37. A numbered consecutively from 1 to 100. Customers and pay according to the number of the ticket, except that numbered above 50 cost just 50 cents each. How much money isi lottery contains tickets draw tickets tickets collected if all tickets are sold? x 3x be an arithmetic progression. 38. Determine x so that 39. The sum of the first and fourth terms sum of the third and twelfth terms is 40. Find the sum of all x, 2, will of an arithmetic progression is 20. The Find the sum of the first 15 terms. 36. multiples of 5 from 100 to 1,000, inclusive. 264 Sec. Progressions 1 5-6 HARMONIC PROGRESSIONS 15-6. A harmonic progression is a sequence of non-zero numbers whose reciprocals form an arithmetic progression. Thus, harmonic progression a if > r o > are in b, c a, - form an arithmetic progression. c The terms of a harmonic progression between any two given terms are harmonic means between the given terms. To insert a desired number of harmonic means between two numbers, we insert the same number of arithmetic means between the reciprocals of the two given numbers and then invert the resulting terms. The harmonic mean ing manner. If a, x, b two numbers of found is in the follow- form a harmonic progression, then - - T a x u > form an arithmetic progression, and x and r Hence, mean the arithmetic is i > of a o mean Solution of this equation for x yields the harmonic X Clearly the harmonic mean ~ __ _2o6_ + a b' a exists only if + = b Example 15-6. Insert three harmonic means between Solution: Is =- > The corresponding and n = 5. Hence, - arithmetic progression = - - -f 4d, - It follows that the arithmetic progression is Therefore, the three harmonic means are = and d - > 0. 3 is and 2. Here > o & a= $ o > ' ^1 ~ o ' 77; ' 1Z O O TTT ' > o Z ZT: 24 8, 12, -=- EXERCISE 15-3 In each of the problems from progression. Find the next ,111 3'7'U' L *5'i'g- 1 to 6, two terms 9 2 determine if the given sequence of the extension of each 1 1 f 8 '5'l6'l6' .-2 8,f. ^ - 7. Find the tenth term of the harmonic progression 8. Find the seventh term of the harmonic progression Insert four harmonic means between 1 and 2. 9. a harmonic .111 1 -4'8'l6* 5.16 is harmonic progression. Z ^ > = 7 > U 6, 3, 2, > . . f 2,|. Sec. 15-8 265 Progressions harmonic means between ^ and 7 4 3 11. Insert four harmonic means between 6 and 24. 12. Find the harmonic mean of 6 and 9. 13. Find the harmonic mean of 24 and 72. 10. Insert three 14. Insert nine 2 3 harmonic means between - and ~ 15. If a 2 6 2 r 2 form an arithmetic progression, show that 6-f-c, o , , 2t a harmonic progression. 16. If a, 6, c form an arithmetic progression and show that ad = be. 17. If the harmonic is a? 18. If a, 6, 19. If a, a = and b, show that x = 5 r b x _ = a h T * b , = form a harmonic progression, show that C harmonic mean of a and and conversely. 1 a j- ~~" C 6 is equal to their arithmetic mean, show that GEOMETRIC PROGRESSION 15-7. A by of a a+& form d form a harmonic progression, 6, c, d form a harmonic progression, show that c, 6, c 20. If the mean a+c, geometric progression is a sequence of numbers in which each first is obtained from the preceding one by multi- term after the plying it by a fixed number; the multiplier ratio may be found by is called the common ratio. The common immediately preceding it. Thus, 1, - > Zi in which the common ratio is .= 1, V2 i 2 is is a geometric progression = --f-- = 4 ^ -~-l=i a geometric progression in which the = u 4= V2 4= 2 > 4 o \/2, 4 ^ O --r-- is dividing any term by the one 1 = i Also, Zi common ratio * V2 = -i A/2 \/2 that a necessary and sufficient condition that three nonC = R zero numbers A, B, and C form a geometric progression is -^ -j It follows 1 15-8. THE GENERAL TERM OF A GEOMETRIC PROGRESSION Let a denote the first term of a geometric progression, and let r denote the common ratio. Then the progression may be written as follows : a, ar, ar Hence, (15-6) if ln 2 , ar 3 "1 , , ar 71 . represents the value of the nth term, ln = ar*- 1 266 Sec. Progressions SUM OF THE 15-9. TERMS OF FIRST n represent the sum of the progression. Then ar ar 2 H Sn = a by r, we have S nr = ar + ar 2 Multiplying first n terms ~2 + + 5-9 A GEOMETRIC PROGRESSION Sn Let 1 n h ar of a geometric + ar"- 1 . + arn~ + ar n + ar3 + l . Subtracting the first of these equations from the second, term by term, we have Sn = Snr ar n a. Therefore, n a Sn = (15-7) 1) ^ j" 1 = a( * I (r /* ^ 1). n Submultiply both sides of (15-6) by r, we get rln = ar in This for S is form we obtain useful another H stituting (15-7), we If . . S n = 5L=j (15-8) (r * 1). Note. If r = 1, these formulas do not apply; in this case, however, the geometric progression becomes a + a H haton terms, and Sn = na. Example 15-7. Determine which of the following sequences are geometric progressions: a) 2,6,18; 6) 5,10,30; c) *,^~y y Solution: a) The ratios found by dividing each of the second and third terms by the preceding one are 3 and 3. Hence, the sequence 2, 6, 18 is a geometric progression. 6) In this sequence, the ratios of consecutive terms are 2 and 3. Therefore, the sequence c) is 30 5, 10, is not a geometric progression. is a geometric progression in which the The given sequence common ratio x/y. Example 15-8. Find h and S& The common ratio Solution: is in the geometric progression 6, 2/3, 2/27, (2/3) * 6 = 1/9. Since a = 6 and n = 5, we have \ o(l 6 1 - is - 96. Solution: Find the By - (1/9)*) _ 1-1/9 r Example 15-9. The term 6(1 r*) fifth V 59,049/ 1-1/9 term of a geometric progression ratio and the first term. _ 14,762 2187 is 3, and the tenth common the formula (15-6) for the nth term of a geometric progression, we have ar 4 =3 and ar 9 = - 96. 15-10 Sec. 267 Progressions Dividing each side of the second equation by the corresponding member of the 5 32. Hence, r = - 2. Therefore, a( - 2) 4 = 3, or equation, we obtain r = first Example 15-10. Find each value x + 1 x of which the three numbers for form a geometric progression. C = B -r we we apply the condition Solution: If > A -5MJ we obtain z =4/5. Hence, the geometric have is progression - - = 4/5, - x 4- 1 X ~~~ 2, 2 =x Solving, X & x x, 6/5, 9/5. GEOMETRIC MEANS 15-10. Terms of a geometric progression between any two given numbers are called geometric means between the given numbers. Let a and ln be given numbers. Then k means may be inserted between them by using the formula ln = with n ar"- 1 = Example 15-11. Insert three geometric means between Solution: Let a = the three means are 2 ' Zw 4 , 2 = 1 ' and n 2, 2 , 2 = Then 5. /5 = ar mean of a and 6. 2. 1 and 2. 4 and r = 2 1/4 . b Since # X = a geo- called is x - = a db we note that a geometric mean of two numbers mean proportional between the two numbers. same the is Hence, as a Hence, . 3 '4 form a geometric progression, then x If a, #, 6 metric 1, 1 + k EXERCISE 15-4 In each of the problems from 1 to 9, determine whether the given sequence is a geometric progression. Find the next two terms of the extension of each geometric progression. - 1. 2, 8, 32. 2 .\ 4.2,4,6. 5.27,18,12. > 3. 4, 16, 64. 1, 2. & * V3 V6 V3 7 7' "T ' ' ~6~ 8> In each of the problems from 10 to & a2 . T' a ' ' 2" . Q 9. 3" 15, find ln and S n 12. 3, 9, 27, - - ; n = 10., n = 45. 13. 100, ; 14. log 2, log 4, log 16, 16. 11. 3, 2, 4/3, Find the sum of the ; n first = 10. n terms - -L f J 1 1, - 1 1, 1. 1 in the given geometric progression. 10. 4, 2, 1, , - ; n 10, 1, 15. log 9, log 3, log = ; 15. n = V5, of the geometric progression 1, =r 101. ; n j = 6. . 268 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 15-10 For what values of x do x - 2, x 6, 2x + 3 form a geometric progression? For what values of x do 3x + 4, x 2, 5x -f 1 form a geometric progression? 1? For what values of x is x -f 1 a geometric mean of 2x + 1 and a; Find a geometric mean of 4 and 16. Also, find their arithmetic mean. Find a geometric mean and the arithmetic mean of 3 and 12. Insert four geometric means between 1 and 32. Insert five geometric means between 1 and 1,000,000. Insert ten geometric means between 1 and 2. If A, G, and H denote the arithmetic mean, a geometric mean, and the harmonic 2 mean, respectively, of two numbers a and 6, prove that G = AH. If the arithmetic mean of a and 6, when a, b 7* 0, is A, a geometric mean is G, H. and the harmonic mean is H, find A G, G H, and A 2 4 Show that the sum of the first n terms of the geometric progression 1> o > ' * * ' Q is3(l 28. Sec. Progressions (j 1 Discuss how this sum varies as n increases. Show that the sum of the first n terms of the geometric (sli* 16(1 Discuss how this sum varies as sum of the first n terms of the sequence xS n .) be the sum. Then compute S n 29. Find the Let Sn n increases. 1, 2x, 15-11. sum INFINITE of the terms of the finite sequence 2, = 3x 2 4x 3 , Q f\ 30. Find the is progression 8, 4, 2, > =^ . , 11 > =j (Hint: Q<i > > i O =^ GEOMETRIC PROGRESSION A geometric progression in which the number of terms is infinite an infinite geometric progression. In Section 15-9, we found an expression for the nth partial sum Sn of a geometric progression. Hence, Sn is the nth term of the geometric series based on the given progression. Thus, we have is called Si = a; 82 = a + ar; 83 = a + ar + ar 2 ; ; Sn = a + ar + + Furthermore, Let us consider what happens to the sum of n terms of a geometric progression when the number of terms increases indefinitely. If a = 0, evidently Sn = whether r ^ 1 or r = 1. In this case, the number meets the requirement of a limit of Sn , so that the sum S of the infinite geometric series is equal to 0. Now suppose a Case 1. 0. For this condition, we consider four cases. Assume that \r\ < 1. If r ^ 0, the numerical value of r decreases as w increases. Moreover, by making the number of terms * Sec. 15-12 we can make \rn as small we can make Sn differ from sufficiently large, that if |r please < that ; 269 Progressions 1, S n approaches is, as \ we please. It follows by as T-^_ little as as a limit. This condition ^ we may be stated symbolically in the following manner: S = (15-9) where S lim S n n-oo = r^ 1 - > r sum of the infinite geometric series. Equation (15-9) is true also if r = 0, since in this case Sn has the constant value a. Case 2. If r = 1, then S n na. Since a 0, |S n increases indefinitely as n increases. Here S n diverges. Case S. If |r| > 1, then jr increases indefinitely as n increases. ~^Again {Sn diverges. Hence, so does S n = 7-^ is the = | { } 71 ] | \ | r 1 Case ( 4- l)^ 1 a. If r then the progression becomes, 1, n If is \ | r 1 even, Sn = 0. If n is odd, Sn = a, a, a, a, In this case, a. , we say that S n oscillates between and a. Here also the series diverges. The sum of an infinite geometric progression can therefore be found by (15-9), but only when \r\ < 1. When r = 1, r 1, or > sum. the series no and has that the series we say 1, diverges, r\ For a further discussion of this topic, the student may refer to a treatise on the theory of limits. Example 15-12. The owner of a fleet of trucks finds that refined for re-use, 20 per cent of the oil gallons of refined the total amount Solution: We oil and is lost re-refines this oil in the process. each time it if If used motor oil is he starts with 100 becomes of oil he has used before the entire 100 gallons begin with 100 gallons. After the first dirty, determine is lost. reclaiming operation, we have 80 gallons of good oil. When this becomes used and is re-refined, we have 64 gallons; and so on. Theoretically, we would never use up the entire amount of oil. However, the limit of the sum of the amounts of oil reclaimed is approximately reached after a large number of operations. Hence, we have an infinite geometric progression in which a re-refining the oil as of = it is 100 and r = used, the fleet 4/5. Then S 100 _ = , 500. Thus, by owner has had the equivalent of 500 gallons oil. 15-12. REPEATING DECIMALS JL a decimal contains a fixed sequence of digits which are repeated indefinitely, we call it a repeating decimal. Thus, 0.135135 is a repeating decimal. This decimal is written 0.135, the dots indicating the first and last digits of the sequence which is to be repeated. Also, 0.34516 = 0.34516516516 repeating decimal is wholly or partly an infinite geometric series. For If . A 270 Sec. Progressions 1 5-1 2 0.34516 = 0.34 + .00516 + 0.00000516 + it is , composed of the decimal 0.34 and an infinite geometric series in which a = 0.00516 and r = 0.001. Hence, since |r| < 1, since example, ' 00516 a00516 - ~~ ' 34 172 100 0.999 ' 33300 33300 Note that if we divide 172 by 33300, we obtain the repeating mal 0.00516. Example 15-13. Express the repeating decimal 0.26*3 as deci- an equivalent numerical fraction. We Solution: can write this decimal in the form 0.2 + 0.063 + 0.00063 + . Hence, the required number consists of the decimal 0.2 plus an infinite geometric series in which a = 0.063 and r = 0.01. The sum of the series, since \r\ < 1, is 7 0.063 0.063 - 1 " 110 0.99 90 7 1 ~" ~~ 0.01 0.263=1+^=^. Therefore, EXERCISE 15-5 In each of the problems from on the given 1. - , 1 1 , .,24 7. 1, , g , 13. 8. . 5 o 1 , ~ > . sum of the 1 > > ^ 3 2 g , convergent series based 3. 3, . ... 6. 8, V3, - - - 1, 4, 2, - . -... Qi * *' - 39 " > ' ' 5 .". ' ' 25 11. 0.18, 0.0018, 0.000018, .... + 0.012 + 0.00012 + 0.0000012 + Find the sum of the 8, 1 fl , g , 5. 8, 4, 2, . 10. 0.5, 0.05, 0.005, 12. 0.3 to 12, find the 2. 1 . | 1 , 1 geometric progression. - , 1, | 4. infinite based on the series Also, find the . sum infinite geometric progression 24, of the first 20 terms of this progression and com- S instead of $20What would be the error if S were used instead of S n for the sum of the geometric pute the error introduced by using 14. progression 48, 15. - 36, 27, - to 10 terms? A ball is dropped from a height of 3 feet. On each rebound it bounces back to three-fourths the height from which it last fell. Assuming that this bouncing continues indefinitely, find the distance it travels in coming to rest. How far has 16. it traveled after bouncing ten times? A swinging pendulum will gradually come to rest as a result of friction. If, on each upswing, the pendulum swings through 98 per cent of the arc through which it fell, and if the initial arc for one complete swing was 20 inches, find the distance traveled before the pendulum comes to rest. Sec. 15-13 271 Progressions Convert each of the following repeating decimals to fractional form. 17. 0.1. 18. O.i5. 19. 0.90. 20. 0,243. 21. 0.16. 22. 0.142857. 23. 2.9. 24. 1.1234. 25. 0.11542. 26. 2.123. THE BINOMIAL SERIES 15-13. We shall now number. If we becomes /ie IA\ /t (15-10) (1 consider the binomial expansion when n is any real let a = 1 and b = x in (4-13), the binomial formula 1 + nx + + x) n = 11 \ i + i ...+ n(n - L- n(n ^ 1)- 2T - 1) (n The right member of (15-10) We saw in Section 4-6 that, o x2 - + (n 2) , ~ n(n - 1) (n + r 1) 2) x3 +.... x called a binomial series. is n a positive integer, the series on the right in (15-10) terminates with x n and (15-10) is true. Otherwise, the terms of the series continue indefinitely, giving rise if is , to an infinite series. The question which now n is not a positive integer arises is ; that is, whether (15-10) is valid when whether the series on the right n It converges, and, if so, whether its sum is equal to (1 + x) proved in the study of series that (15-10) is indeed valid if \x\ < is . It follows that, if \x\ accurately as we < 1, we may obtain the value of (1 please by taking sufficiently + many terms n 1. as x) of the series in (15-10). The expansion can a positive integer. In this case, the follows : (15-11) Here x + 6) n when n is not expression may be written as readily be extended to (a (a = a > + &) = fo(l L. \ and the expansion = + -Yr a/ j of (a a" fl L. + -Td-j + 6) n is valid if \ Example 15-14. Find the first five terms of the expansion of Solution: By a (1 < + z)~ 3 it\x\ < (15-10), (-8)(-4)(-6) 3! ja (-8) (-4) (-5) (4! 1. \ 1. 272 Sec. Progressions 15-13 Example 15-15. Find ^L04. j/TM = Solution: pansion (1 + 0.04) "3. 1/3 and x = 0.04, and the ex- 5 + 0.04)3 = 1+| (0.04) + (1 = Hence, n is The approximate value 1 We have of the sum + 0.013333 - + (0.04)2 2! of this series 3 ^ 3 ^ (0.04)3+ .... is + 0.000004 = 0.000177 3^ 1.013160. here a case of an alternating series, that is, a series in which the terms and negative. It is proved in the study of series that, if in each term is numerically less than the preceding term and are alternately positive an alternating lim a n := 0, series by using S n the error introduced as the sum of the series is - numerically less than the value of the first term omitted; that is, \S n S\ < \a n +i\. If in the present example we take for S the value S 3 = 1 + 0.013333 - 0.000177 = 1.013156, the error in so doing 15-16. Find the Example F Solution: First we apply is less first than the value of the fourth term 0.000004. four terms of the expansion of ., - x* (15-11) to convert the given expression to a suitable form, as follows: 1 -^ (8 r2\\-i/3 / / = -.)-. = (8(1-!)) 1 ^T) =K (/v2\-l/3 Hcnco, by (15-10), if x* < _i 1 -!-) 8, a2 \ ~ 2\-l/3 / , (-1/3) (-4/8) / 1-2 V "*" s/"" \ i 1-2-3 \ x*\* (-1/3) (-4/3) (-7/3)7 "*" x* 8/" ""/ 1 V "" 24 288 20,736 EXERCISE 15-6 In each of the problems from 1 to expansion of the given expression. i. 6. ^ 1 (1 ~T" X - z)-i/'. 2 - T1 7. X -4+y x 38. 10, find vrr^. 4. ' x (x _? + v the first vr^. 9. 7 2 (x i/) four terms of a binomial 3 +^-r-- 5. ( l/aj 10. (1 . -r / v +ir) 2/) Find the approximate value of each of the following numbers by means of a binomial expansion, using four terms of the expression. 11. (1.02)io. 16. 49 4 . 12. (1.01) 13 . 13. (1.04) 8 17. (0.99)'. 18 . 51 3. . 14. (l.l) 10 . 15. (0.98) 8 19 . 20. (0.97)~ 2 . (1.03)1/2. . 16 16-1. Mathematical Induction METHOD OF MATHEMATICAL INDUCTION When a certain type of formula or proposition has been verified known to be true in general, the method of mathematical induction is often found extremely valuable in in specific cases but is not determining its validity. Suppose that a statement involving a positive integer n is to be proved true for all values of n greater than or equal to a particular initial value. We begin by showing the result to be true for the first value of n. We then assume that k is some particular integral value of n for which the statement holds. With this assumption as a basis, we establish the validity of the statement in the next suck 4- 1. In other words, we ceeding case, namely, that in which n prove that if the statement is true for any specific integral value of n, say n = k, then it is also true for the next larger value of n, namely, n = k 4- 1. Suppose for example, that 1 is the initial value of n. Then the second step establishes that if the statement is true for n = l, it is also true for n = 1 + 1, or 2 if it is true for ; n= 2, it is also of this, we true for ft = 2 + 1, or 3 ; and so conclude that the statement greater than or equal to the is on. As a consequence true for here all values of n A proof by mathematical induction, therefore, consists of two parts and a initial value, 1. conclusion. Part 1. Verification that the statement is true for some initial, value of n, generally n = 1. (This initial value is the smallest value of n for which the statement is to be proved true.) Part 2. Proof that whenever the statement is true for some parn = k, then it is true for the next larger ticular value of n, say for value of n, that is, for n =k+ 1. 273 274 Sec. 16-1 Mdfhemafi'ca/ Induction Conclusion. If both parts of the proof have been given, then the is true for all positive integral values of n greater than statement was made or equal to the one for which the verification in Part 1. The reasoning process involved here, which consists in taking and then repeatedly taking successors, can be exemplified in terms of climbing a ladder. Part 1 puts us on the bottom rung of a ladder. Part 2 shows us how to get from any rung we have reached to the next higher rung. The conclusion states that if we know how to get on the bottom rung of the ladder, and if we know how to get from any rung to the next higher one, then we can reach all rungs, and hence can climb the ladder. an initial integer The following examples Example 16-L If n is will illustrate the positive integer, prove that any 1+1+1 +. 1-22-33-4 (16-1) method. ^ + n(n + n 1) 1 Solution: Part 1: The formula is = when n true 1 1-2 = since 1, 1 1=1. r 1+1* 2 2* Part 2: Let k represent any particular value of n for which (16-1) - 1 tI I l-2" "2-3" "3-4" "*" tI I + fc(fc is Then true. - ~~ " fc 1) + 1 We now wish to prove that (16-1) is true also for the next larger value of n, = k + 1. The sum on the left in (16-1) when n = k + 1 can be obtained n namely, by adding we have its last term, which r-m iwy + 1) (* + 2) is 7, . (* /LL+JL+JL + VI -2 "^2-3 ^3-4 T . k(k + !)/ > to both sides of (16-2). Hence, ^ (k + k k But, since fc we obtain , lft ox (16-3) + ! j +D 1 +2 (t ) ! + + 2oTo3 + ^ ^ 3oTT4 ^ . TTo 1 2 (t . ^ . 1 (* + 1) (k _fc +2fc+l_fc+l_ ~ ~ ~ + i) + 2) t + 2 2 1 + + +2) l)(k k . (fc (fc 1 ^ (A? + 1) (k + 2) _ " (* k + + 1) (* + 2) k + l + 1) + 1 I l + 1 = k + 1. Hence, The members of (16-3) are the same as those of (16-1) when n we have shown that (16-3) is true if (16-2) is true, in other words, (16-1) is true for n = k + 1 if it is true for n = k. We have shown by Therefore, since (16-1) is true for n Conclusion: for every positive integer n. verification that (16-1) is true = 1, it when n follows from Part 2 that (16-1) is = 1. true 16-2 Sec. 275 Mathematical Induction Example 16-2. Prove that - (x a factor of (xn is y) y n n if ) is any positive integer. Solution: Part l:Itn = Part 2: Let A: We shall - (a*M now y*+i). _ x *+i 1, then x n We have *+i _ yk+i = x By assumption, (x factor of y k (x - y). Therefore, if - By by n =x ?/, which Hence, Part 33,* 4. X2/ - * y 1 (x - is seen to have n = of the proof, (# - true for virtue of Part 2, =x ( 7/ , - y) as ?/*), it is also xk y*). Also, it is y) as a factor. (a; has (x ) _ y *) by y) is a factor of the left n is *+i - y) is a factor of (x* the conclusion Conclusion: Therefore, y n specific value of n for which (x k if is factor of that a (x (x y) prove be a member (x =k + - y n ) true for y) - k+l n n y) is a factor of (x is y). inspection, (x also true for the desired conclusion _ 4. <,*( a factor. a factor of is y** a 1 ). 1. when n any = 1. positive integer n. 16-2. PROOF OF THE BINOMIAL THEOREM FOR POSITIVE INTEGRAL EXPONENTS We shall now prove that the binomial formula, n n n n n~ b T ^ an" + (16-4) (a + b) =a + na ^ l 2fc2 ... - 1)1 n(n-l)...(n-r+l) an_r6r (r r! is + . . . +^ true for every positive integral value of n. Proof. Part 1. Hence, (16-4) is Par #. Let Then we have (a + fc 6)* When n = 1, each true for n = 1. be any specific value of = a* + ka k ~l b + k ^k becomes a side of (16-4) ~ n X) for which (16-4) is + 6. true. a*~ 2 6 2 1> Multiplying each (a + 6)*+i = member of this equation by (a ' a** 1 H---- + ab k +ab k -\ ' + "" r (k &), + 1 we obtain 276 Mathematical Induction Hence, by combining terms, + (o 6)* +1 = a** In this result the follows - k(k + 1 sum (Jfc we + See. 16-2 get l)a6 + . . . of the coefficients of a k ~r+l br obtained as is : - (k 1) r + 2) (k - + r 1) + r! fc(Jfc 1) (fc - r + 2) (r-1)! r fc-r+1 ~L _ (fc r + 1) (fc) (fc - 1) (k - r + 2) r! We + note that the value of (a 1 obtained as the product of (a + b) and (a + 6), is exactly the same as the expansion which would be obtained from (16-4) with n = fc + l. Hence, we have 6)** , k shown that n = fc + l. if the binomial formula holds for Conclusion. n= The binomial formula was seen Therefore, by virtue of Part every positive integer 2, we may k, it must hold for to be true for conclude that it is n= 1. true for n. EXERCISE 16-1 Prove by mathematical induction that each of the statements in Problems 20 is true for all positive integral values of n. . 2. 1 3. 2 + 3 4- 5 +4 +6 H---H---- + (2n - 1) = n + 2n = n(n + 1). 2 . 5.3+6 + 12 H---- 6. 4 7. 1+4+7 + 9. 10. 8 4- .. . + 4n = 2n(n +(3W _2) -f 1). 1 to 16-2 Sec. 12. I 8 + 2s + 38 + 13. I 3 +3 +5 + + 48 + 6 + + 22 + 2 + +2 = + 32 + 33 + + 14. 2 15. 3 2 16. 3 17. 4 18. I 2 277 Mathematical Induction + + 3 3 8 3 + 42 + 43 + - (2n (2n) 8 3 = 2 2 I) (2r* = 2n 2 (n + I) 2 rc 2(2" - & + 4 = | (4 - 1). = | (3 - 3 + 3 2 + 5* + 7* + 1). . 1). -,,1). +frt = n(2n -% - ^ 19 ^' "" " 1 3 3 1 2ft 5 ^2 7 5 2 3 (2n 1 21. Prove that x 2n 1) + (2n "*" 4 ~l 2/ "^ i 3 4 5 2n + i "*" 2n is divisible + y*n-i "" 1) 1 i 3 - 1 i 1 " n(n by x + 1) (n + 2) 1 n(n+3) = 4(n Prove that x 2n 23. By using mathematical induction, prove the formula for the metic progression. 24. By using mathematical induction, progression. 1} (n ^ + 2) + y for every positive integer n. 22. {$ divisible + by x -f y, for every positive integer n. prove the formula for the sum of an arith- sum of a geometric 17 17-1. We Permutations, Combinations, and FUNDAMENTAL Probability PRINCIPLE begin the study of permutations and combinations by con- sidering the following principle, which entire subject. is fundamental for the Fundamental Principle. If one thing can be done in a ways, and for such each if, way, a second thing can be done in b ways, then the two together can be done in a b ways. To understand why the principle is true, note that for each of the a ways of doing the first thing, there are b ways of doing the second hence, both the first and the second things taken together can be done in a b ways. The following examples will illustrate the reasoning upon which the principle is based, as well as an obvious extension of the principle to the case when more than two things are to be done. ; Example 17-1. In how many ways can two officers, a chairman and a secretary, be selected from a committee of five men? By the fundamental principle, the problem is equivalent to determining which the two ways things can be done together. The first position can be filled in 5 ways; that is, there are a = 5 ways of selecting a chairman. For each of these possible selections, there are 6 = 4 ways of filling the position of secretary from the remaining men. Hence, the number of ways of selecting a chairman and secretary is a 6 = 5 4 = 20. Solution: the number of in How many Example 17-2. digits 0, 1, 2, , 9, if three-digit numbers can be formed from the ten a) repetitions of digits are not permitted; 6) repetitions are permitted? Solution: place can be a) Here we have three things to do or places to fill. The hundreds 9 ways, since must be excluded from this place. The tens filled in 278 Sec. 17-2 279 Permufaf/ons, Comb/naf/ons, oncf Probability place can then be filled in 9 ways from any of the remaining 9 digits. Finally, the units place can be filled from any of the remaining 8 digits. There are, therefore, 648 three-digit numbers in which no two digits are alike. 9.9.3 = b) If repetitions are permitted, there are 9 ways to fill the hundreds place and 10 10 900 each of the tens and units places. Hence, there are 9 three-digit numbers. 10 ways to fill = EXERCISE 17-1 1. 2. Nine persons apply for each of two vacant apartments. In how many possible ways can both apartments be rented? A large room has eight doors. In how many ways can a person enter the room by one door and leave by a different door? thrown, in how many ways can they fall? There are eight men and six women in a club. In how many ways can two officers be selected so that one is a man and one is a woman? How many possible four-digit numbers are there in a telephone exchange which 3. If three dice are 4. 5. uses only four-digit numbers and excludes 0000? 6. 7. 8. How many six-digit numbers can be formed from the digits 2, 3, 4, 5, 6, 7, 8, 9? How many of these are larger than 700,000? How many numbers of six different digits can be formed from the digits 2, 3, 4, 5, 6, 7, 8, 9? How many of these are larger than 700,000? A woman has seven guests at a party. If she chooses her seat first, in how many ways can she seat her guests? 17-2. An PERMUTATIONS ordered arrangement of all or any part of a set of things is called a permutation. Specifically, suppose we have n distinct things and wish to select r of these to be arranged in a definite order. Each such ordered arrangement is called a permutation of n things r at a time. The number of all such permutations is denoted by n Pr Thus, 5 P2 is read "the number of permutations of 5 different things 2 at a time." In Example 17-1, the possible number of ways of selecting a chairman and secretary from a committee of five men was found by means of the fundamental principle to be 5 4 = 20. This is pre. cisely equal to 5 P2 since it is the number of ways in which two men can be chosen from among the given five men and arranged in the two offices. The number of permutations of n things r at a time is given by the formula , (17-1) n Pr = n(n The truth of (17-1) is - 1) (n - 2) (n - r + 1). readily shown as follows. The first of the n ways. Then the second can be filled in r places can be filled in (n 1) ways, the third in (n 2) ways, and so on. In general, the 280 and Permutations, Combinations, number of already filled. 17-2 Sec. Probability of filling each place is n minus the number of places when the rth object is chosen, (r 1) ways Therefore, places have already been filled, and the rth place can then be r + 1, ways. in n 1) , or n (r In particular, then have (17-2) if n r = n, Pn = the last factor becomes - n(n 1) (n - = 1 2) n n+ 1 = filled 1. We nl This formula gives the number of permutations of n different things taken all at a time. If both the numerator and the (understood unit) denominator of (17-1) are multiplied by (n r) !, we obtain the following alternate formula : H7 * P _ (17-6) n r r Here r it is - n(n 1) (n - (n 2) -r+ (n 1) - r)l agreed that by definition n\ _ ~ ( n -r)! (n ! = ' - r) ! Hence, (17-3) holds for 1. ~n. For example, by (17-2), 8 P6 = n(n - 1) (n - (n 2) - r + 1) = 8 7 6 5 4 = 6720. Using (17-3), we have also aft 17-3. _ ~ _8!_8-7-6*5-4-3-2.1 " nl (^=7)1 PERMUTATIONS OF T2^ 8! n 672 ' THINGS NOT ALL DIFFERENT required to find the number of indistinguishable n things all at a time, if n\ things are regarded as indistinguishable, n^ other things are regarded as indistinguishable, and so on. Let us consider, for example, the number of permutations of the letters a, a, a, b taken four at a time. For convenience, indistinguishable objects are given the same notation. Denote by P the desired number of distinct permutations. Evidently, P is less than ^P^ which is the number of permutations that could be effected if all the letters were distinguishable. For in any one of the P permutations, say (a b a a) any rearrangement of the a's among themselves would not change the permutation. If, how- Suppose it is distinct permutations of , ever, we assigned subscripts to a in this permutation, (eh 6 02 da), we could permute these three distinct letters as in among ways. This can be done for each of the P permutations of the letters a, a, a, 6. We would then obtain P 3 permutations of the four distinct letters al9 o^, a 3 b taken four at a time. There are therefore 4 permutations altogether. Hence, themselves in 3 ! ! , ! P- 3! =4!, or P= Jj 3! = 4! ~3!1! Sec. all 17-4 Permutations, Combinations, and 281 Probability In general, the number of distinct permutations of n things taken at a time, if HI things are alike, n2 other things are alike, n 3 other things are also alike, and so on, equals n\ P= (17-4) n\\ COMBINATIONS . A combination is a set of all or any part of a collection of objects without regard to the order of the objects in the set. We use the symbol n Cr to represent the total number of all combinations of n different things taken r at a time. The different sets of the four letters a, 6, c, d taken three at a time, without reference to the order in which the letters are arranged, are (abc) (abd), (acd), (bed). From each of these four combinations, we can form 3 !, or 6, different permutations of the four letters taken three at a time. For example, from the combination (abc) we can form the distinct permutations (abc), (acb), (bac), (bca), (cab), (cba). Hence, each of the four combinations 9 P3 = 3 permutations to the total number of permutations. Thus, there are 4 C 3 combinations of the four letters if we disregard order, and there are 3! ordered arrangements or permutations of each combination. Hence, we have J3 = 3 4^3, or contributes 3 ! ! = 4>3 >2 C3 = 5 o number we 4 2 1 ! l ' = This 4 combinations. is the same as the o obtained at the beginning of the paragraph. In general, if we divide the total number of permutations of n things r at a time, or n Pr by the number of permutations, or r!, , contributed by each combination, we obtain the total combinations. Symbolically, we have n4 or we = ? * of rbt> p C = 7T' (17-5) If r number ' note that r esting relationship = rPr I , then we can : .. Cr ~ n* r ~~p~ r r r Replacing n Pr by its equivalent expression from (17-1) or (17-3) we have __ write the following inter- n(n - 1) * (n ~ r + 1) _ ' n\ 282 Permutations, Combinations, If r is replaced in (17-6) r) by (n -r) n C n_ r and Probability we obtain , Sec. 17-4 n\ = (n- r)!r! Hence, nC r (17-7) Afote. Having agreed that ! = nCn-r- 1, we can to permutations or combinations of supply (17-1) or (17-3) at a time. Con- n things zero we have sidering (17-5), = n Co (17-8) 0!(n l = L = o! 0)I This result agrees with the intuitive conclusion that there one such "empty" combination. Similarly, n P = 1. is only BINOMIAL COEFFICIENTS 17-5. referring to the development of the binomial formula in Sec- By tion 4-6, we note that the expansion of (a + b) n involves the product (a + 6) (a + 6) (a + 6) taken without regard to order. This fact suggests the use of combinations in the coefficients of the expansion. We a n-r b r term involving said in Section 4-7 that the coefficient of the " ' - " . is This r! n things r the combination of j g precisely the f or mula for at a time, or Cr obtained in Section 17-4. The binomial formula may therefore be written as follows n- 2 2 n b +' -+ n C n b\ (17-9) (a+b)* = n Coa+nCia -ib+nC 2 a n , : .+#'"&+ For example, the expansion of (a + 6) 4 takes the following form : This reduces to a4 If we let a = 6 = + 4a 3 6 1 in + 6a 26 2 + the expansion for (a b) n in (17-9), we obtain + 1) = n C + nCl + n C 2 + + nCn = 2 Ci + C 2 + - (1 Since n C = 1, + nC n . 1. Thus, the total number o;f combinations of n things taken succesn ft at a time is 2 1. sively 1, 2, , EXERCISE 17-2 1. 2. Evaluate 5^2, 7^3, 12^6, 21^4, Evaluate 10^4, nCa, looC's, 21^5, 100^95* Sec. 3. 17-6 Form a. and Permutations, Combinations, all 283 Probability word possible distinct permutations of the letters of the How many are there? b. How many begin and end with a vowel? theory. c. How begin or end with a vowel? 4. many How many distinct permutations are there of the five letters a, three at a time? Write 5. them 6, c, d, e taken out. how many different ways can a dime, a quarter, and a half-dollar be disamong five boys? How many different sums can be formed with a penny, a nickel, a dime, and In tributed 6. a quarter? 7. 8. 8 using combination symbols. the six digits 1, 2, 3, 4, 5, 6, form all permutations taken five at a time. a. How many are formed? b. How many begin with 4? c. In how many does Expand (a -f 6) , From the digit 3 not appear? 9. How many distinct permutations can be made from the letters of the word probability taken all at a time? 10. How many different combinations are there of 4 identical nickels ; 5 identical dimes, and 6 identical quarters? 11. How many of 12. In different straight lines are determined which are in a straight how many on a different by twelve points, no three line? ways can a student select seven questions out of ten test? 13. How many different weights can be 14. 16, and 32 pounds, respectively? In how many different ways can signals be made with seven different where a signal is a set of one or more flags arranged in a specific order? 15. In how many different formed with ways can the 52 cards six objects weighing 1, 2, 4, 8, of a bridge deck be dealt flags, among four players? MATHEMATICAL PROBABILITY 17-6. If, can in a given trial, an event can happen in happen in / ways, and if all the h fail to likely, h different + / ways ways and are equally then the probability p that the event will happen in this trial is p (17-10) = The probability q that the event happen is g- (17-11) Note that will fail to O^p^ 1, O^g^l, and p + q = 1. Two illustrations of probability follow. If a bag contains 3 green marbles and 4 yellow marbles, all exactly alike except for color, the probability of drawing a single green marble is 3 3 284 Permufaf/ons, Comfamaffons, and Sec. Probability 17-6 A die can fall in six ways. The probability of getting 5 or more 2 1 are two possible ways, th with ith one throw of a die is - or - since there o o either a 5 or a 6, for the event to happen. > > MOST PROBABLE NUMBER AND MATHEMATICAL EXPECTATION 17-7. Let p be the probability of occurrence of some event. Furthermore, suppose that n trials of the event are made, of which h are Then - successful. U is called the trials which occurred. It the relative frequency of success for h not to be expected that - is = p. It is TV shown in more advanced treatments of probability, however, that if n is large, it is very likely that the relative frequency is approximately equal to p. Also, the larger we take n, the more likely it is that - approximates p closely. Moreover, it can be shown that the id most probable or expected number of occurrences of the event for n trials is up. For example, when a coin head is tossed, the probability of getting a In 1,000 trials the expected number of heads is thereThis does not mean, however, that if the first 100 trials is 1/2. fore 500. result in 75 heads tails in and 25 the next 100 trials. we should expect 25 heads and 75 Actually, since one toss of the coin does tails, not affect the next one, we should expect about 50 heads and 50 tails in the next 100 trials. Moreover, we may expect about 450 heads and 450 tails in the next 900 trials. the probability of winning a certain amount of money in case a certain event occurs and is the amount of money to be won, If p is m defined to be pm. For instance, if a person can win $12 provided he throws an ace with a die, his expectation is - ($12) = $2. Hence, $2 is the fair amount he should be the mathematical expectation willing to 17-8. pay to STATISTICAL, make OR the is trial. EMPIRICAL, PROBABILITY frequently impossible to have sufficient knowledge beforeall the conditions that might cause an event to happen or fail to happen. In such a case, however, it may be possible to determine the relative frequency of the occurrence of the event from a It is hand of number of trials. Thus, if an event has been observed to happen h times in n trials, and n is a large number, then until addi- large Sec. 17-$ tional and Permufaf/ons, Combinations, knowledge is we available, Probability define the statistical probability, or empirical probability, to be P (17-10 where - l> the relative frequency. is n = Example 17-3. A molding machine turns out 12 parts per minute. Inspection experience has shown that there are 20 defective parts per hour. What is the probability that a single part, picked at random, will be defective? In a run of 10,000 parts, how many defectives should be expected? Solution: defective. defective The parts are produced at the rate of 720 per hour, and 20 of them are Hence, the probability that a single part selected at random will be is =7^ (ZO = ob In a run of 10,000 parts, we should expect ^ oO (10,000), or approximately 278, defective parts. 17-9. MUTUALLY EXCLUSIVE EVENTS Two of more events are mutually exclusive if not more than one them can happen in a given trial. The following theorem may or be stated. Theorem. The probability that some one of a set of mutually exclusive events will happen in a given trial is the sum of the individual-event probabilities. Proof. Consider, for simplicity, a set of two mutually exclusive events. Suppose that the first can happen in hi ways and the second can happen in h2 ways, and n be the let which the two events can happen or and p2 = fail to number of ways Then happen. p\ = total in are the corresponding probabilities of the two events. it Since the n events are mutually exclusive, the hi ways are different from the h2 ways, and the number of ways that either the first or the second event can happen is therefore hi + h%. Hence, the probability p that either the one event or the other will happen .,.&.,.. p.idt*. (17-12) is For example, suppose that a bag contains 2 green marbles, 3 yellow marbles and 5 brown marbles. If a marble is drawn at random, the probability that 2 it is green is o yellow is Hence if either green or yellow a marble is and the probability that it is drawn, the probability that it is ^ is > 23 51 +^ ^ > 1U lu or > lu or ^ ^ 286 Permutations, Combinations, and Sec. Probability 1710 DEPENDENT AND INDEPENDENT EVENTS 17-10. In case two or more events are not mutually exclusive, they are dependent if the occurrence of any one affects the occurrence of the others, and they are independent if the occurrence of one does not affect the occurrence of the others. if a card is drawn from a deck of 52 cards and the not replaced before a second is drawn, then the second drawing is dependent on the outcome of the first. If, however, the first card is replaced, then the second drawing is independent of the first. In the latter case the two drawings are equivalent to simultaneous drawings from two decks. We shall now state and prove the following theorem relating to For example, card is dependent and independent events. Theorem. The probability that two dependent or independent events will occur (successively if dependent; successively or simultaneously if independent) is the product of their individual probabilities. Suppose that the first event can happen in hi out of a ways, and that the second event can happen in ft out of n^ different ways. Then it follows, by the fundamental principle in Section 17-1, that the two events can happen together in hih^ ways out of a total of n^n2 different ways. Therefore, the probability that both events will happen is Proof. total of ni different P * (17-13)' H2 ni Example 17-4. Two cards are drawn from a deck containing 52 cards. Find the when the first card is not replaced before the probability that both cards are aces second is drawn. Solution: 1) The We shall begin with a listing of the following useful probabilities: probability of drawing an ace from a deck of 52 cards 2) If the first card is another ace is an ace and it is is oZ not replaced, the probability of drawing -^ 01 3) If the first card is not an ace and second being an ace is not replaced, the probability of the is 51 4) If the first card is replaced, the probability of the second being This drawing ~4 an ace 4 is -r^ is entirely independent of the first drawing. the Consider, now, given problem. The probability that the first card is an ace is 41 If the first card drawn is an ace, then the probability that the Pi 16 O/ = = Sec. 1 7-1 second card drawn will be aces 287 Permufat/ons, Comb/naf/ons, cmcf Probability 1 pl p a is is an ace is p2 =-._ = 3 = ^r = T1 oi Hence, the probability that both ii . REPEATED TRIALS 17-11. Theorem. If p is the probability that an event will happen in any = 1 - p is the probability that it will fail, then the probtrial, and q ability that it will happen exactly r times out of n trials is n r nC,tr<r* (17-14) = W- i - Proof. The happening of the event in exactly r trials and its failure in the remaining n r times are independent events. Hence, ~ the probability, by the theorem in Section 17-10, is p rq n r But these r trials can be chosen from the n trials in n Cr ways. Since these ~r r Note ways are mutually exclusive, the total probability is n Crp q n that this expression is the (r + l)th term of the binomial formula . * + for (q (q + p) M pY = , q since n + n Ciqn~ p + n C qn~ 2p 2 + l + nC q - p + n 2 r r r + pn . The successive terms of this expansion give the probabilities that n times in n trials. the event will happen exactly 0, 1, 2, r, An event will happen at least r times in a given number of n r + 1, or r times. Since these events trials if it happens n, n 1, are mutually exclusive, the probability that an event will happen at le&st r times is given by the sum , , Example 17-5. What is the probability of tossing an ace exactly three times in four trials with one die? Solution: Since the probability of tossing PJ ability of failure is formula. The ^ * we may an ace in trial is ^ and the prob- substitute in the term n V<r~ rP r of the binomial result is * (f )'(!)'= '(I) (s> Example 17-6. What trials one is = sir the probability of tossing an ace at least twice in four with one die? Solution: The event Hence, the probability will is happen at given by the least twice following - if it sum happens : '(!)'= 4, 3, or 2 times. 288 and Permufaf/ons, Comfa/naf/ons, Sec. 17-1 Probability 1 EXERCISE 17-3 1. A certain event can happen ways and can in four to fail happen in six ways. the probability that it will happen? If $60 can be won on the event, what is the mathematical expectation? 2. A box contains 5 white balls, 4 red balls, and 13 black balls, a. If one ball is What drawn that 3. a. If is what out, it is the probability that is it is red? b. What is the probability white or red? one die is thrown, what is the probability that a "1" or a "2" will turn up? What is the probability that a "3" or larger number will turn up? When a coin was tossed 100 times, 80 heads and 20 tails turned up. b. 4. If the tossing were continued until 200 tosses had been made, what would be the most probable number of tails in the second 100 tosses? 5. A bag contains drawn, what 6. 7. 8. five $1 bills, ten $5 bills, and twenty $10 one If bills. bill is the mathematical expectation? the probability of throwing a "7" or an "11" on one throw of two dice? is What is An automobile owner carries $1,000 theft insurance on his car. If, during the past year, 237 out of 97,864 automobiles registered in his area were stolen, what is the mathematical value of the policy? In a city of 77,000 families, a careful sample of 800 families showed that 120 sample families owned their own homes, a. What is the probability of the that a family selected at 9. random 23,100 voted for his opponent. random voted for the winner? 10. in the city owned its home? b. What expected number of families in the city who own their own homes? In a certain city 28,600 persons voted for one candidate for an What is is the and office, the probability that a voter chosen at A bag contains 5 red balls and 9 black balls. If two balls are drawn in sucand the first is not replaced, find the probability that the first is red and the second is black. cession, 11. In a baseball tournament the probability that team A will win is = > and the ' probability that team two teams will win. 12. The B I will team probability that win A is will Find the probability that one of these ^ reach the finals of a tournament 2 is = > ' I and the probability that it will win the team A will win the tournament. finals is - Find the probability that 13. Find the probability of throwing three successive fours on a pair of 14. The probability of A winning a game when he plays it is j He is dice. scheduled to play four times, a. Find the probability that he will win exactly three times. b. Find the probability that he will win at least three times. 15. Three dice are tossed, turn up. b. What is a. Find the probability that exactly two threes will the probability that at most two threes will turn up? Solution of the 18 18-1. General Triangle CLASSES OF PROBLEMS There are certain relationships among the lengths of the sides and the trigonometric functions of the angles of every triangle. If one side and any two other parts of a triangle are given, the remaining parts can be determined; that is, the triangle can be solved. The three given parts may comprise any one of the following four combinations : Case I. One side and two angles. Case II. Two sides and the angle opposite one of them. Case III. Two sides and the included angle. Case IV. Three sides. In this chapter, we shall discuss methods for treating these four cases. For convenience, we shall let ABC denote any triangle whose angles are A, B, and (7; and we shall let a, 6, and c represent the lengths of the corresponding opposite sides. 18-2. THE Law tional LAW OF SINES ABC be any triangle lettered in the convenmanner. Then the following relationship between the sides of Sines. Let and the sines of the angles may be written a (18-1) This relationship . sin is A = b . sin commonly B = : c sin C called the law of sines. B A FIG. 18-1. 289 290 Solution of the General Triangle Sec. 1 8-2 Proof. We first note that all angles may be acute, as in Fig. 18-1 (a), or one angle, say B, may be obtuse, as in Fig. 18-1(6). (The case where B = 90 entails no difficulties and will therefore be omitted.) In each diagram, let h denote the altitude from the vertex sin B= C to the side AB, Then, in either case, sin h a Dividing the sin sin equation by the second, first A B a a _ QJ j b sin - A A= - and we have --B b sin In a similar way, by drawing the altitude from the vertex we get A to the side BC, b c B sin ~~ sin C The equations thus obtained may be combined law of to give the sines " sm B sm A sin C Note. The law of sines is well adapted to the use of logarithms because it involves only multiplications and divisions. Since any pair of ratios in the law of sines involves two angles and the sides opposite, it may be used in the solution of problems in Cases I and II. As noted above, the law of sines also applies in the special case where ABC is a right triangle. In this case, one of the angles is 90, and 18-3. the sine of that angle SOLUTION OF CASE TWO ANGLES I is 1. BY THE LAW OF GIVEN ONE SIDE SINES: AND When one side and any two angles of a triangle are known, the third angle can be found from the relation A + B + C = 180, and each of the required sides is uniquely determined. These sides may be found by the law of sines. Example 18-1. In a triangle ABC, A = 3814', B = 6720', c = Solve 329. the triangle. Solution: The values of the given parts are indicated in Fig. 18-2. In this case, C = c To 180 find a, - + (3814' we use the 6720') 329 a sin 3814 Therefore, FIG. 18-2. 7426'. relationship a c-329 = a _ "" / sin 7426' 329 sin 3814 / ' sin 7426' 18-4 Sec. 291 Solution of the General Triangle and - log sin 7426'. log sin 3814' log a log 329 indicated operations may be performed as follows: 2.5172 log 329 = The + log sin 3S14' = = - 9.7916 12.3088 log sin 7426' = 9.9838 10 10 10 2.3250 log a = a =211. To determine 6, we use the relationship 329 ~~ sin Hence, 6720' _ "" sin 329 sin 67 20' sin and log b follows: The work 7426 / 7426 / ' = log 329 + log sin 6720' log 329 log sin 6720' log sin 7426' = = 2.5172 9.9651 12.4823 log b b As shown log sin 7426'. = 9.9838 = 2.4985 = 315. - 10 10 10 in Fig. 18-1 (a), c = b cos A + cos B. This relationship may be used as a check. Thus, c = 315 cos 3814' + 212 cos 6720' = which agrees 18-4. (315) (0.7855) + (212) (0.3854) satisfactorily with the given value of = 329.1, c. SOLUTION OF CASE II BY THE LAW OF SINES GIVEN THE ANGLE OPPOSITE ONE OF THEM TWO SIDES AND Case II is called the ambiguous case, because the data may be such that two, one, or no triangles are determined. The number of solutions when a, 6, and A are given is indicated by the accompanying table. TABLE OF POSSIBLE SOLUTIONS Solution of the General Triangle 292 We Sec. 18-4 shall use Fig. 18-3 to illustrate in turn the different possiand the sides a and b considered in the table. If the angle A bilities AX are given, we first construct the angle A with the initial ray and the terminal ray AR. Next, we lay off the distance AC = b along the terminal side. Then, with C as the center and the length of the side a as the radius, we describe an arc. We mark the point of the or points in which this arc intersects the initial ray AX angle A. -X X A (<*) Figure 18-3 (a) corresponds to (18-2) in the table. Since BC = a = b sin A, this segment is the altitude of the triangle drawn from the vertex C. Hence, the arc with radius a is tangent to the initial side at B, and the triangle is a right triangle. a Figure 18-3(6) corresponds to (18-3) in the table. Since b sin A, the side a is too short to intersect AX, and there is no < triangle. In Fig. 18-3 (c), a < 6 and 6 sin and the arc will intersect table, A < a, as stated in (18-4) in the AX in two points marked B and B'. Therefore, two solutions exist. The angle B' in the triangle the supplement of the angle B in the triangle ABC. AB'C is Figure 18-3 (d) represents the case in which the side a is longer than the side 6, as stated in (18-5) in the table. Hence, there is only one point B in which the arc with radius a intersects the initial ray AX of the angle A. There is only one solution. Sec. 1 8-4 293 Solution of the General Triangle In Fig. 18-3(e), the angle A is obtuse. Since a < b, as stated in (18-6), the radius a is too short to intersect the initial ray AX, and no triangle exists. A is obtuse and a > 6, as stated in can Here arc the intersect the initial ray AX in only one (18-7). point. There is, in this case, only one triangle. The following examples will illustrate some of the possibilities. Finally, in Fig. 18-3 (/) Example 18-2. In a triangle , ABC, A = = 3615', a = 9.8, b 12.4. Solve the triangle. Solution: Draw Fig. 18-4 approximately to scale, showing the given parts. After A and side 6 have been drawn, an arc is described with C as center and a as angle radius. The arc intersects the side AX in two points, B\ and B 2 and we apparently have two possible triangles, ABiC and AB 2 C. , FIG. 18-4. To find sin B, we use sin B __ sin A ~T~ ~~~T~ Then , 10 12.4 ~ . . . sm 36 9.8 and = The log 12.4 log sin B follows: + log sin logarithmic work log 12.4 log sin 3615' = = 3615' = log 9.8 1.0934 9.7718 10.8652 log 9.8 - - 10 10 0.9912 B = 9.8740 - 10 B = 4826'. A < a < b. The left inequality follows from 6 sin since log sin There are two solutions, the fact that log sin B = log sm < 0, and so sm , < 1. If we let BI = 4826', 294 then Solution of the General Triangle - 4826' = B = 4826', B2 = 180 l 13134' leads to another solution. Thus, = 180 - (3615' + 4826') = C2 = 180 - (3615' + Ci 18-4 Sec. 9519', and # 2 = ISl^', To we find Ci, = 3615' sin 16.5. we have " 2 "" 1211' 3615' 9.8 sin sin As a = / 9519 9.8 sin Similarly, 13134') use the law of sines again. Thus, = 3.5. = b cos A -f a cos B\ may be used. Thus, 3615' + 9.8 cos 4826' = 12.4 (0.8064) + 9.8 (0.6635) = 16.5. the same as the value previously calculated. The same method may partial check, the equation Ci 12.4 cos This result is be applied to check c2 . Example 18-3. Given A = 5630 , = a 13.0, b = ABC. 10.7, solve the triangle Solution: Here we clearly have only one solution, since a > b. This can be seen geometrically if we draw Fig. 18-5 approximately to scale and show the given parts. Since a is greater than b, an arc with C as center intersects on opposite sides of AX A. Obviously there is only one triangle, ABiC, containing the angle A. To find BI, we use the relationship sin a- 13.0 BI sin 5630' 13 10.7 whence = log sin log 10.7 -f log sin 5630' - log 13. This gives B = FIG. 18-5. = Then C To find c, l (5630' + 4320') = 8010'. we use the law of sines and obtain 180 13 c sin This gives c 4320'. - = 8010' sin 5630' 15.4. Example 18-4. C Given A = 6740', a Solution: = 16.0, b The given parts we have = 17.3, solve the triangle. shown are in Fig. 18-6. By the law of sines, sinff 17.3 a m 16.0 B = Hence, B FIG. 18-6. / 16 log 17.3 -f log sin = c 6740 Therefore, log sin 6740' sin __ ~~ The 6740' - (1.2380) -f (9.9661 - 10) log 16 - (1.2041) = 0. B = 90. be completed by applying the theory of right triangles. Only one solution exists. solution may 18-4 Sec. Example 18-5. Given From Solution: work To A = 4723 it Fig. 18-7, ; a , = 230, 6 = 720, appears that no triangle solve the triangle. is possible. The following verifies this fact. find B t use the relationship sin btaining B __ sin A *720sinV23', . Sm The 295 Solution of the General Triangle 230 logarithmic work follows: log 720 log sin 4723' = = 2.8573 9.8668 - 12.7241 log 230 = 10 10 2.3617 B = 10.3624 - 10 = 0.3624. sin B would have to be FlG 18~7 * log sin Since log sin B > 0, Therefore, there is no solution. ' - greater than 1, which is impossible. EXERCISE 18-1 In each of the problems from 1 to 12, solve the given triangle by the law of sines. a = 12.30, A = 3625', B = 4437'. 1. 2. b 3. c 4. b 5. a 6. c 7. o 8. 6 9. a 10. a 11. 6 12. 6 13. = 12.18, A = 4733', B = 6751'. = 461.3, B = 6719', C = 2314'. = 0.6384, B = 3939', C = 8716'. = 6.714, A = 3753', C = 13636'. = 7832, A = Way, B = 4358'. = 21.23, c = 64.21, C = 6231 = 0.8146, c = 31.63, B = 1119'. = 987.4, b = 503.6, A = 5413 = 0.003862, c = 0.0008157, A = 2613'. = 1.386, c = 2.451, 5 = 83 19'. = 4.395, c = 9.806, C = 3746'. / . / . A surveyor wishes to find the distance across a stream from point A He to point B. from A to a point C on the same side of the stream is and angles BAG and BCA are 4953' and 816', respectively. Find finds that the distance 687.4 feet, the distance 14. A surveyor edge of the ran a line AB. was running a swamp A line N 2723' W. How original line produced? 15. due west when he reached a swamp. From the for 2500 feet, and from this point he S 63 far had he gone on this line when he reached his far was it across the swamp? line he ran a How W building 63.7 feet high stands on the top of a hill. From a point at the foot of the hill the angles of elevation to the top and bottom of the building are 4216' and 3831', respectively. Find the height of the hill. 296 16. Solution of the General Triangle Sec. 18-4 From a certain point on the ground the angle of elevation to the top of a building 46 17'. From a point on the ground 83 feet nearer the building the angle of is 17. 6823'. Assuming that the ground elevation is building. One side and a diagonal respectively. is level, find the height of the of a parallelogram are 14.63 inches and 21.4 The angle between the diagonals and opposite the given inches, side is 11623'. Find the length of the other diagonal. measure the distance between two artillery pieces A and B. from an observation point C to gun A is 2447'. Sound travels at the rate of 1140 feet per second, and the sounds from guns A and B reach C in 2.3 and 1.7 seconds, respectively. Find the distance AB, assuming that points A, B, and C lie in the same plane. A body is acted on by two forces, Fi = 2643 pounds and F 2 = 2341 pounds. The resultant F 3 lies on a line making an angle of 4633' with Fi. Find F 3 and 18. It is necessary to The angle 19. of depression the angle between the lines of action of Fi and forces is their vector sum.) 20. A buoy, F2 (The resultant of two . located at a point B, is 6 miles from a point A at one end of an island is C at the other end of the island. If the angle BAG and 10 miles from a point 132 18-5. 16', find THE Law the distance between the points A and C on the island. LAW OF COSINES ABC be any triangle. Then a = b 2 + c 2 - 2bc cos A, b 2 = a 2 + c 2 - 2ac cos 2 2 2 c = a + b - 2ab cos C. of Cosines* Let 2 (18-8) (18-9) , (18-10) These relationships constitute the law of cosines. FIG. 18-8. Proof. We shall establish (18-8) by considering the case when A and the case when A is obtuse as in = 90 involves no difficulty, and it will acute, as in Fig. 18-8 (a) is Fig. 18-8 (&). The case A therefore be omitted. Let h denote the altitude from denote the length AD. Hence, DB c + x in Fig. 18-8(6). C to the side is c - AB. Also, let x x in Fig. 18-8 (a) and is Sec. 1 8-6 Solution of fhe General Tr/ongfe 297 In Fig. 18-8 (a), (c - xY + h2 = + = 62 a 2, and x2 Subtracting, we have c 2 Since x = 6 cos =a = 62 + 2c# a2 A, a2 h2 2 6 c 2 - 2 = 62 + c2 - (c + x) . or , 2cz. 26c cos A. In Fig. 18-8(6), + h2 = 2 a2, and + x2 Subtracting, we find that c 2 + 2cx a Since x = b cos A 2 in this case, a2 = 62 = = + = h2 a2 6 62 . 6 2 or , +c + 2 2 c2 2bc cos A. and proceeding in a obtain (18-9) and (18-10). Note. For a simple algebraic proof of the law of cosines, see problem 22 in Exercise 18-4. The law of cosines applies equally well if ABC is a right triangle. In this case, one of the formulas reduces By drawing similar manner, altitudes to the other sides we = to the pythogorean theorem, since cos 90 18-6. SOLUTION OF CASE III AND CASE Since the law of cosines IV BY THE 0. LAW OF COSINES expressed by formulas involving addition and subtraction, it is not well adapted to logarithmic computation and its use is not recommended unless the given sides are is easily squared. Example 18-6. Given Solution: By b = 9.0, c = 13.0, A = 11510', solve the triangle the law of cosines, a2 52 _|_ C 2 _ 2bc cos A = = 9 + (13) - 2(9) (13) cos 11510' = 81 + 169 - 234( - cos 6450') = 250 + 234 (0.4253) = 349.52. 2 2 Hence, a We = 18.7. employ the law of sines to find angle B. Thus, sin _ "" 9 Therefore, C= sin 11510' _ ~~ sin 18.7 5=25'50'. / 180 (11510' + 2550 ) = 39. 6450 18.7 / ABC. 298 Solution of the General Triangle Example 18-7, Given a Solution: From a 2 = 62 = +c 2 A = 8-6 Find the angles. 7. -+ 52 A = Similarly, = 1 we have the formula C2 _ a2 -T > Hence, We 5, c 26c cos 4, cos Therefore = 3, 6 Sec. B = 3813' and C = 21*47'. 120. can check these by the equation A -f B +C = 180. EXERCISE 18-2 In each of the problems from to 1 solve the given triangle 6, by the law of cosines. l.o= 300, 6 = 250, C - 5840'. 2. a = 50, c = 240, 5 = 11050'. 3. 6 = 65, c = 310, A = 6710 4. o = 130, c = 90, B = 10020'. 5. 6 = 50, c = 110, A = 150. 6. a = 1.63, 6 = 3.45, C = 2610'. = 7. If a = 15, 6 12, c =5 20, find A. 8. If o = 25, b = 30, and c = 35, find B. 9. If a = 100, 6 = 300, c = 500, find C. 10. If a = 15, 6 = 12, and c = 20, find B. 11. If o = 16, 6 = 17, and c = 18, find A, B, C. 12. If o = 260, 6 = 322, c = 481, find A, B, C. 13. The distance between two points A and B cannot be measured directly. Accordingly, a third point C is selected, and it is found that AC = 3000 feet, BC = 4500 feet, and angle ACB = 4620'. Find the distance AB. / . 14. 15. Two sides of a parallelogram are 125 feet and 200 feet, and the included angle is 11030'. Find the length of the longer diagonal of the parallelogram and also the angle between that diagonal and a longer side of the parallelogram. Two 16. Two is feet and 200 feet, and the 6733'. Find the perimeter of the plot. and 420 feet, and one diagonal is Find the length of the other diagonal. In a triangle ABC, a = 25, 6 = 27, and the median from A is 20. Find c, A, B, C. 600 17. ground are 250 sides of a triangular plot of included angle 18-7. sides of a parallelogram are 700 feet feet. THE Law LAW OF TANGENTS ABC of Tangents. Let relationships exist between Ma in (18-11) a a ~ + , b b be any triangle. Then the following sides and the angles opposite them two = : tan i (A 2 - B) = tan > (A + B) Sec. 18-7 299 Solution of the General Triangle tan (B - C) (B + C) (18-12) b +c C + ^ 2 = 5_IL (18-13) tan a 1 x tan s /^Y (C X + A) \ , These relationships constitute the law of tangents. Proof. Let us denote the common ratio of the law of sines by Thus, a = r sin A, 6 = r sin JS, c = r sin C. Then a a b + _ ~~ A A + r sin 6 r sin r sin J? __ sin ~" B r sin sin A A + sin sin B B ^ Substituting from (7-27) and (7-28) of Section 7-4, D sm g . sm AA sin 2 cos . A+ sin B 2 + B) (A y sin 2 (A - r. we have J5)' ~~ _. + JS) cos z- (A - 1 , 2 sin - (A A . , 1 , , . Hence, tan A similar procedure may (A - be followed to prove (18-12) and (18-13). We two shall sides now use the law of tangents to solve a triangle in which and the included angle are given. Note that this law is well adapted to logarithmic computation. Example 18-8. Given Solution: Here c To -6= find 123, c C - 6 = = 249, c A = 5622 372, ; , solve the triangle B, we use the formula + b = 621, C + B = 180 -A = 12338', (C + B) = Therefore, tan i (C - B) = g tan 6149', and 1 log tan i (C - B) = log 123 + log tan 6149' - log 621. ABC. 300 Solution of The logarithmic work Genera/ Triangle f/ia Sec. 1 8-7 follows: = = log 123 log tan 6149' 2.0899 0.2710 - 10 - 10 12.3609 - B) = = - B) = 2017'. log 621 log tan i (C 2.7931 9.5678 i 1 (C Hence, 1 C=i(C+)+i(C-)=826', and B = ~ (C + ) - ~ (C - B) = 4132'. Cfodk: 4 To find a, we 4. 5 +C= 5622' + 826 + 4132 = / 180. use the law of sines. Thus, 249 sin 6622' sin 18-8. 7 4132' Q1Q =313 ' THE HALF-ANGLE FORMULAS The following relationships are very convenient for the logarithmic solution of Case IV, where the three sides a, 6, and c are known : 2 In these formulas, 5 Proof. From = r - (a 2 + b (* + 6) c) c). (7-16) in Section 7-3, we have A ~ l-cosA 1 + cos A 2 Also, from the law of cosines, cos Therefore, _i -l and &2 A = 62 + c2 - 2 + c2 - a 2 _ -a 25^ (6 -a2 - c) 2 _ ~ 2bc (a + b - c) (a - b + c) 2bc Sec. 18-8 we If Solution of the General Triangle let a 301 b + c = 2s, then a b - c = a 6 a - 6 c = a 6 + + + c - 2c = 2(a - c), + + + c - 26 = 2(s - 6), = 6 + c a a + 6 + c - 2a = 2(s - a). .A _ (a + & - c) (a - 6 + c) _ 2 " (g c)2 (6 + c + a) (b + c t( a) a) + Therefore, ( 6) and (18-14) v ' We r 2 s(s a) can derive (18-15) and (18-16) in a similar manner. = 379, b = 227, c - 416, find the angles of the triangle. Here 2s=a+b+c = 1022. Then Example 18-9. Given: a Solution: =511, 8 - a = 132, - 6 = 284, - c = 95. s s s Hence, A . The calculations by logarithms log 284 log 95 log (284) (95) log (511) (132) log tan . ., , Similarly, /(284) (95) follow: = 2.4533 = 1.9777 = 24.4310 = 4.8290 y == A A - log 511 log 132 20 19.6020 - 20 9.8010 - 10 6438 4.8290 ; . . B _ ^7 (132) 2--r w" Hence, we = 2.7084 = 2.1206 find that B = C "" _ (95) (511) (284)' |/(132)(284) y 2 3246' and C V (511) (95) = 8236 / . A + 5 + C = 6438' + 3246' + 8236' = 180. EXERCISE 18-3 In each of the problems from 1 to 10, solve the given triangle by the law of if an angle is given, or by the half-angle formulas if three sides are given / 2. 6 = 17.1, c = 22.3, A = 2116 . a = 50, b = 60, C = 60. / = = = = = = B 6 c A 4. a 9510 . 79.3, 113, 230, c 106, tangents 1. 3. Solution of the General Triangle 302 5. 6 7. a 9. a 11. = 41.82, c = 75.89, A = 7849'. = 625, 6 = 725, c = 825. = 67450, 6 = 84380, c = 98630. 6. a 8. a 10. a Sec. 1 8-8 = 0.1028, 6 = 0.8726, C = 14S13'. = 60.65, 6 = 38.64, c = 23.57. = 0.1146, 6 = 0.3184, c = 0.6379. diagonals of a parallelogram are 6 inches and 10 inches, at an angle of 63. Find the sides of the parallelogram. The and they intersect A and B are separated by an obstacle. In order to find the distance between them, a third point C is selected and it is found that AC = 126 rods and BC = 185 rods. The angle subtended at C by AB is 9614'. Find AB. Two circles whose radii are 14 and 17 inches respectively intersect. The angle 12. Points 13. 14. 15. between the tangents to the circles at either point of intersection is 3846'. Find the distance between the centers of the circles. The sides of a parallelogram are 13.4 inches and 18.5 inches, and one diagonal is 15.6 inches. Find the angles and the other diagonal of the parallelogram. Three circles whose radii are 10, 11, and 12 inches, respectively, are tangent to each other externally. Find the angles of the triangle formed by joining their centers. 16. The sides of a triangular field are in the proportion 4:5:6. The area of the field is 18 acres. If there are 160 square rods in an acre, find the length of each side of the field in rods. 17. In triangle ABC, prove the following: a sin -6 ^-*) n cos-C I a +b -s C 1 sm-C . These formulas are called Mollweide's equations. They may be used in checking the solution of a triangle. 18-9. We AREA OF A TRIANGLE can readily see that the area K (18-17) In either triangle, h = = \ch & K of the triangle = JcbsinA. <u 6 sin A. In like K (18-18) = in Fig. 18-1 is manner, we obtain lac sin B, Zi K (18-19) c By substituting sm sm C = ^ab sin C. for 6 from the law of transform (18-17) to obtain ~o(\\ n (18 20) * A sin B KK -~ c sin 2sinC sines, we may Sec. 1 8-9 303 Solution of the General Triangle cyclic interchanges of letters, we obtain = a2 sin B sin C (18-21) By K K= (18-22) A 2 sin A b 2 sin C sin 2smB To derive a formula for finding the area of a triangle when its three sides are given, we first transform (18-17) in the following manner ' : K = ybc sin A = sin ^bc = 1,/ sm A (2 A . y cos rpc = , A . oc sm -jrZi yJ (2 2 A cos -^- But, from (7-14) and (7-15) in Section 7-3, = A cos /I v A , and 2 cos 2" Using the values from Section 18-8 for have . sm A 2 A /(s ^ be c) (s A/ V A = 6)'- cos 1 ^ A A - , and /I y cos -^ 2 2 and 1 -f . /s(s --r \/ V be Consequently, the formula for area in terms of the sides = \/ s (s The following examples a) (s = 372, A = a)- is c). triangle in Example 18-8, in which b ='249, 5622'. Solution: Since two sides and the included angle are given, (18-17) may be used* Thus, K = ^bc sin A = ^(249) Z 2 The we illustrate the use of the area formulas. Example 18-10. Find the area of the c t) (s cos A, logarithmic work follows (372) sin 5622 / : = 9.6990 log 249 = 2.3962 log 372 = 2.5705 sin 5622' = 9.9205 log log K = 24.5862 = K 38,600. log 0.5 10 10 20 . 304 Solution of the General Triangle Example 18-11. Find the area of the triangle in Sec. 1 8-9 18-1, in which Example A = 3814', B = 6720', c = 329. Solution: Since two angles and a side are given, (18-20) may be used. In sin B _ (329)* sin 3814' sin 6720' _ Q0 inn "" * 2 100 ~ 2 sin 7426' 2 sinC _ c 2 sin A *^ - = 227, c in which a = 379, = 416. Solution: In this solution, (18-21) K= ' ' Example 18-12. Find the area of the triangle in Example 18-9, 6 this case, Vs(s - a) (s - 6) ( - is c) We used. = have V(511) (132) (284) (95) = 42,700. EXERCISE 18-4 1. 3. 5. In each of the problems from 1 to 8, find the area of the given triangle. / a = 12.30, A = 3625', B = 4437'. 2. c = 461.3, B = 6719', C = 2314 = = = = = = 3746'. 4. 6 a 5413'. 9.806, C 4.395, c 503.6, A 987.4, b 6 = 65, c = 310, A = 6710'. 6. a = 300, 6 = 250, C = 5840'. = = 7. a 9. In triangle 15, b Au j = 12, c ABC, r 20. let r j.1. x and, therefore, that r = /4 /-($ /I/ ~ a) r 10. Find the radius of the feet, = a 8. 100, 6 -- be the radius of the inscribed 6) (S ( - = = 300, c 500. Prove that circle. . K = rs C)- circle inscribed in the triangle whose sides are 48.92 63.86 feet, and 72.31 feet. 11. A cylindrical tank is to be built on a triangular lot having sides whose lengths are 200 feet, 186 feet, and 176 feet. Find the radius of the largest such tank which can be built on the lot. 12. In triangle ARC, let R be the radius of the circumscribed a b 2B= sin A 13. In triangle ABC, show scribed circle and that = sin R = -r^ > B = circle. Show that * sin where R C is the radius of the circum- K is the area of the triangle. 15. The sides of a triangle are 23, 29, and 46 feet. Find the areas of the triangle and the inscribed and circumscribed circles. The sides of a triangular plot of grass are 42 feet, 65 feet, and 87 feet. Find the minimum radius of action of an automatic lawn sprinkler which will water all parts of the plot from the same point. 16. An 14. arc of a circle of radius r subtends a central angle bounded by this arc and its chord is jr r2 (6 - 0. Show that the area sin 8). t Find the area of the largest pentagon which can be cut from a circular piece of metal 4 feet in radius. How much metal is wasted? 18. In triangle ABC, prove that the median from any vertex to the side opposite 17. divides the angle at that vertex into two parts whose sines are proportional to the lengths of the parts into which the side opposite is divided by the median. Sec. 1 8-9 19. In triangle ABC, prove A cos a 20. In triangle 21. In triangle 2 -f 6 (6 ABC, prove + c2 = a 2 (cos 2 + B + 6 cos C "" _ a2 -f fr 2 -f c 2 1 2a6c c + c) cos A + (c + a) cos B + (a + b) cos C. 2 2 that C + ABC, show 22. In triangle that cos that ABC, prove +b +c = a a2 305 Solution of the General Triangle sin 2 B) + 6 2 (cos 2 4 2 -f sin C) +c (cos B + sin 2 A). that a 6 c = b cos (7 -f c cos 5, = c cos A -f a cos C, = a cos B + 6 cos A. Multiply the first equation by a, the second by 6, and the third by c, to give a second proof of the law of cosines; that is, prove (18-8) by showing that 62 +c 2 23. Consider a > = a2 any 2bc cos A. Similarly prove (18-9) and (18-10). triangle ABC. If a > b, prove that A > B. If A > B, prove that b. 24. In triangles ABC and A'B'C', let A and A', B and B', C and C" be pairs of corresponding vertices, and let the corresponding sides be a and ; c and c'. If a = a 6 = 6', and C > C', prove that c > c'. If a , and c > c', prove that C > C'. a', b = a', and 6 b', = &', Appendix A Tables Appendix TABLE A I FOUR-PLACE VALUES OF FUNCTIONS OF NUMBERS 30* 310 Appendix TABLE A I (continued) Appendix TABLE A I (continued) 311 Appendix A TABLE I (continued) Appendix A TABLE II FOUR-PLACE VALUES OP FUNCTIONS 313 314 Appendix TABLE A II (continued) Appendix TABLE A II (continued) 315 316 Appendix TABLE A II (continued) Appendix TABLE A II (continued) 317 318 Appendix A TABLE II (continued) Appencf/x TABLE A II (continued) 319 320 Appendix A TABLE II (continued) Appendix A TABLE III FOUR-PLACE LOGARITHMS OF NUMBERS 32V 322 Appendix A TABLE III (continued) Appendix A TABLE III (continued) 323 324 Appendix A TABLE IV FOUR-PLACE LOGARITHMS OF FUNCTIONS Appendix TABLE IV A (continued) 325 326 Appendix TABLE IV A (continued) Appendix A TABLE IV (continued) 327 328 Appendix A TABLE IV (continued) Appendix TABLE IV A (continued) 329 330 Appendix TABLE IV A (continued) Appendix TABLE IV A (continued) 331 332 Appendix A TABLE V SQUARES AND SQUARE ROOTS Appendix TABLE V A (continued) 333 334 Appendix TABLE V A (continued) Appendix A TABLE V (continued) 335 336 Appendix TABLE V A (continued) Answers fo Odd-Numbered Problems PAGE EXERCISE 1-1. 1. (a) Commutative law 11 of addition. (c) Associative law of multiplication. (e) Commutative law of multiplication. 2. (a) 2. (c) 1. -35. 3. (a) -6. (c) 4. (a) -5. (c) 0. 5. (a)l. (g) -5. 1. (e) -10. (e) (c)5/17. (g) 2. (e) 1/1.02. 2a. 1. (a) -3, -2, 0, 4, 5. > 3 PAGE 1/3. 3. (a) False, (c) 0. 4. (a) 0. 5. (a) 6. (a) 7. 1.414. (e) False. (e) 2. (g) ^6. 7 -1. 3. x a. < 7. 5.1,000,000. 57. 59. 69. IT. -2 < x < (e) < V3 < (e) 1 1. 4. 2. PAGE 20 -216. 9.2401. 21. a 7 . ~4 23. a 4 6 4 11. 13. II. 19. a". 27. 33. 35. a 4 2 . 2 2 2 2 2 2 ). ), 2 2 3 3 2 2 2 ). 14 7 a: 3 3. x*y*. + -2a 3^ 2 + 26 + 35. 2x 2 43. 2z?/. .? 3 63. x 3 - a-?/ 5. 3.r - + 3?y 45. -1. 66 2 PAGE 24 13. -4.r//. 21. z2 2. 3 2 ?/ 4.T 47. 2. + z y + xy* + + 3x 2 + 3a: + 3 2 2/ 1 ; 2a;?/ 29. 24a . 37. - . x2 4 - -4a. 7. . 2. 27. -lOx- - - 5a 2 2c. 2 2 ), 2xy. 2 ), ?/ 25. -18x?/. 55. 25. a 4 6. . 37. 5 - 6. 41. 3a - 46. 39. 46. ^a -2a - 36. 45. (a - 2c)x* - (a + 2b)xy + (6 - I)?/ 47. 4a - 56 - 2c. a - (6 - c a + (6 + c), a - (-6 - c). 51. a + (c - 6 z z - (y + z ). 2a + (6 55. x + ((3c 6). 3c), 2a -a 6 + (a& + 6 -a 6 6 (- ab 61. 12. 63. 20. 65. 17. 67. 1. 2.r + (- 3y - 4z), 2x - (3y + 4s). -6. 71. 8. 73. 10. 75. 13. 77. 13. 79. |- 81. ~31. a 4 . EXERCISE 1-4. I. x/3, 3. lu 19. 125a. 17. 32. 29. a 2 6 2 *. 53. 1, 10. 15. 49. 0.1. -2, -3/2, (e) 01 43. - (k) r (i) 0. EXERCISE 1-3. 1.32. x. 15 10. > 22/7 (e) True, -1 < x < 1. (c) -a 4 < 5 < 6. (c) -1 < + a V2 > (c) (c) - (k) 3 x+y. -4, -2, -1, -1/3, (c) -a6. (i) (g)-||- EXERCISE 1-2. 2. (a) 29. (i) - (g) 36 -2/3. (e) (k) -7. -2. (i) 5x 15. /> Sx*y 2 4.r 49. -z?/ -h 2x 57. x 3 - + 2x + .r?/ 1. 2 // + 2xy. 2 ^x 2 ?/ +y 65. 339 59. , 2.r 4 ^ - - + 26 2 x 2 + ~ o^ + y y. . 4jy. - 33. 2xy* -f .r 9a: 2 3 i/ - 41. -2*. + 5. 53. + y. + xy + y 2 61. 2y - xy + 6. 67. Odd values. 51. x 3y. 2 - 3 39. x* . - 3a 2 17. . 31. -Go;* c. - 2 23. -f y*. 4 z2 9. .-c . 340 Appendix B PAGE 30 EXERCISE 1-5. 1. + 2y). 3. 2(2x + - 2?/ + 3z). 11. 3(x 9. a(x 21. a2 2 ?/ - 15. (2x z 2 (z - 25. (7xy 33. (5p 3 2 2 */ 2x - 2 (* ) 2/ w) 4 + .r 4 4- * ). 2 4- 3z) 4- (2y 4- 3z) - (z Z/) w) .r 2/ ]. - -f (z . ) ?^) 2 ]. . (.r?/ . 3. - x 5. 1. - 19- x 2 SOxtyV. 3x 3 ?/. 7. y. f.r - (a) 12. (c) (a) . 4a. (C) 347' x2 - (y (e) l&r// 3a: . 3) 1. (e) 2 , _ (g) i^T5* , ^ 94 3 24" x2 /. (x + +x +4 -I \ / . 1) (x - \ I) o 2 (*+y)(-y) >- J.W. *"' / . + __ (x 21 2(a / 2) (a; - ab 2 + rtx + o\ 3) (x 2 4- 6 ) a n -5.a x (x - 1) (x 7) (* + 2) - 2) ' 1& " (9x 2 3>c _ x 6 7a + 64 M. 1 2 - (2.t 2 ' ^2 * - xy x2 4- -y 2 41 2(x - 5) ' 4 2 , 2j?4-?/4-3 ' "' + 5 - I 4) - 3x .. ' + (to + 4) + ? 1+* ^\ i PAGE 3 x2 ,, ^?y n 9x 4-2 W 2*+8* * 27 (g x). ,, x3-l q ** -6 2 EXERCISE 1-9. 54 /_. " 2 4). ^r+,-2^ +5 12f . _ 3a+b, (s)^ 6 9- 'x^T- + x) (b PAGE 38 _ x' - 3.r 2 - -2 _ 7 --20' 3* 2 t - x(x + 5y) iq 57 _ 5 --2l' - (.r - 7 x4-2* W _ , (q)i. EXERCISE 1-8. ,17 L 15. 24,r?/z. 2 8) 3>c ... ^i' _ . _ (0)^-3,. WT^&rr^r- - (a (i) 1 ().r?y , - 3) (x* PAGE 35 - 3 ?/ 13. 24. 11. 1. - 49) (x .r (g) . 2 ~IT' - 3 1). PAGE 33 9. 2xy*z. 21. (x 2 4. . ) . 101 . a; ). EXERCISE 1-7. *** - 2 ). - ay* 2 (6 ) - 9) 2 - y) (36a: 2 4- fay 4- 2 ). 43. 3(6 - 2 2 45. 2 2 _ 2 _ _ _ x r X 49. (4. 51. -f 1). 3) (2 2?/) (X 3). 5) ( X 3 2 2 2 57. (1 4- 15.r 55. 4- 1) (* 4- 2). 4) (X + 3). (x 63. (2x - 1) (x + 3\ 61. + 1) 6). 2). (x (2x 5x(x 4- 4) (x 2 69. (a 4- 3) 67. (x 4- 2y). 0.6) 6). (6.r 1) (x 2 - 3). 73. (x 4- 3) (2a + y - z). 75. (x 2 - 3) (x + (4a; 5) (2x 1. 2. 2' a) (3y 4- a). ?/ ). EXERCISE 1-6. 1. 2 (a- 59. 17. - 8 4- (x 4- 2/) 4) (x 19. (3y 11). ( 53. 65. 4- + x\ + 4). 2c 6) (0.1 4- 6). r 10 3x 2 (2y 2 41. (te 71. ry 4- + r6 3 *V + [9.r [(x - - (.r + 11) (7* 27. j(0 2 5p 44 (2y 4- 3z)] 4- - fl 13. - -x(a 31. (jy 4- z 2 ) (* 2 ?/ 2 4- 4). (25p g ) 39. (x 4- y 47. - 7. 2). a). 23. (0.1 3a). 4 r6 (x ) - 37. 3[3s 2 - (x* 4- <? 35. (3 4- (5 I2ab) (7xy 4- 12a6). + 2) 29. 2(x + 3a) (x + a(3x + 3?/ - 17. (7x 3) (2* 4- 3). - 5. 7). 5) (4x (fa + + 1) (x 1) (* - 6) ' 5) y 2 341 Appendix B - + 5) 3(4s +-3) 2L (2s 28 V 5) 23 70 37.2i. .i|. 35.1^. x + 5 5z zy 33 2 tK 15. x = 23. y 17. ~ = y ' ,' 4 39. ^jp 5. tft 19- 39.2. PAGE 47 = x y -2. 41. 4. (e) V2. (c) =x 1 ~y~~ 27.7. V34. The area of a circle 3. The area of a trapezoid 5. The volume is (g) 4. II. ~ oZ S = -2, 2 3, 288, Cy , 23. 5/7. 29. 2 49. 40, 58. 35. ~ 4 51. 6, 7. 20, 40, 120. 57. PAGE 52 (i) V^T^. (k) 2. PAGE 56 a function of its a function of is altitude its A = Trr 2 bases, A =-(bi circle, and . height and the radius of life insurance policy is +62). its base, a function of the applicant's age of the type of policy, of the 27. 25. 0. ^/3(57r F2 . 43. all x. 53. 3 -2. 5* 0, V2 - -3, -3, 3/4, 2y + 7. 41. all x. A = HT*, 45. s * C = 2xr, 0, 1. 57. a: 3,2.3, | x | . 3. 0,0.5. .5. 3, 17. | p 2 A = 33. all x. 31. all x. 55. all x. 15. 14. 13. 0. company's rate policy, etc. -1. 1,0,1,0,1. 3,^3' + q* + r 3 , C = 3. 49. \ 59. . 19. 0. x -2 ^ x g PAGE 57 21. 30. 2 37. all x. 35. all x. 47. \x EXERCISE 2-3. 1. 33.-^J formula can be written. -3, -5, - is of a and physical condition, 9. 31.3. irr*h. The annual premium No -3 60; 2 a function of the radius of the of a cylinder 13.10. 11.7. (g) \/29. (e) 7. I. 7. = 47. -- 45. 2. 43. -11. 9. 01 21. y ' 29.1/4. EXERCISE 2-2. V = - 15 55. 170 adults, 330 children. ft. (c) 4. 3. (a) 5. ^2 7. EXERCISE 2-1. 2. (a) 5. 41. 4-^8 -5. 25. ~ 53. 921,600 sq -3/7. ^ f 17 2g 007 37. = -= 4 -23s -. -2. x 3. -5/2. 32/ '5T 1 EXERCISE 1-10. 1. 25 '59* = 0. 2. 39. all a. 51. all s. 342 Appendix B PAGE 59 EXERCISE 2-4. 1. = y y 3. 567. x. 15. 1000/1, 100/1. 7. 5. 5. 327C. 17. -2. 9. 10/3. 1. (a) (1, 0). 3. (a) PAGE 67 sin 7. (e) (1, 0). (c) (0, 1). V5/2. (c) (-0.99,0.14). (c) 1. -1/2. (e) cos t (e) (-0.65, -0.76). -2. (g) tan t V3/2 (a) I/ (i) cot t V3 - V5/2. sec t V% esc t 2/V3 t. 2. (c) 5/13 12/13 5/12 12/5 13/5. (e) A/3/2 - V V V3 2/ V3. (g) 1/V5 1/2 (i) 2/ \/5 1/2 4/5 3/4 V5/2 (c) 0.58. 1. (a) 1. 2. (a) 0.9927. PAGE 74 (e) 0.86. (e) -1.500. (c) 24.52. (g) 0.2571. 1. 3. 7T/6- 7T/3. 45. 27. 7. 5. 27T/3. 270. 29. 15. 31. 33. 35. 0.5925. 5. 0.7412. 3. 1.092. 15. 9.010. 25. 0.2930. 27. -9.462. 45. 5520', 23520'. 47. 427', 2227'. 61. 0.8016. 63. 3.079. 73. 0.220, 6.060. 81. 55. 730', 18730'. 1.143, 1.997. 75. 65. -100.00. 1.120, 4.260. 43. 8310', 26310'. 7.81. 9. ; 51. 3135 32825'. 57. 9736', 26224'. 59. 9311' 273 67. 0.3459. 77. ~ 0.755, 3.895. 7 13.1. !!'. 71. 1.214. 79. 1.158, 5.122. PAGE 96 11. ay". , 69. -0.7073. 1 5.100. 23. 0.5154. 49. 189', 34151'. EXERCISE 4-1. 3.^. 11. 0.2504. 83. 0.574, 5.706. 64 1. 3y. -0.2462. 21. 0.9831. 37. 3520', 14440'. 41. 26450', 27510'. . 39. 4343'. 31. 17310', 35310'. 29. -1.059. 39. 5630', 23630'. r 9. 19. 0.8437. 35. 310', 18310'. 2474 37. 47023'54". 25. 1.6323. PAGE 85 -1.453. 7. 17. 0.6817. 33. 4230', 22230'. 53. 674', 27. 13. 437T/36. 23. 3.2107. (c) 0.04 radian. (b) 16/9 radians, EXERCISE 3-5. 13. 1.181. 11. 27T/5. 21. 1.4358. 630. (k) 1.011. 5.798. 59. 14.74 in. 57. 147T/3, 1.25 radians. 61. (a) 4 radians, 9. 47T/3. 7T/15. 19. 0.8090. 17. 77T/20. (i) PAGE 80 EXERCISE 3-4. 15. 1077T/60. V5. -5/3. 5/4 4/3 EXERCISE 3-3. 1. 21. 0.0324 in. 19. 99.5 Ib, 95.2 Ib. EXERCISE 3-1. 2. (a) (0.54,0.84). 13. 9/4, 27/8. 343 Appendix 8 21. 4(5"*). 33. (5292) ". - 2 49. 2(2 - x2) 3 ' 5. 6,652,800. 15. - - + -f -4+ 7*8 ^--4-+6 27. x 12 29. x 31. 2 z8 X4 ?/2 - +y + z 2 -f 4x 7. 15. 8=3. Iog 256 2 = 10x6 -f 16z -14x 2 80x~ 1/2 + 5 2- 27. 3. =~ I 80ar 3 39. log* (u 41. (a) log* - + 16x 3 35 5 \/u + TT - - 0.5. 10 31. 5/2. 2 1. 3. 4. 19. 7.314. 5. 1.5441. a No 33. - 1 = 9. = y M- 1. -8x7. 41. 924x 3 i/ 3 . 3. Iog 10 x. / 2 = 82 13. 21. 4 3 10,000. = 64. 23. 6. 8. 35. Iog6 25. 2. 37. T fe 7 -f Iog6 g - 2 log* TT. PAGE 107 13. 11. 5. -1. -1. 15. 17. 7314. PAGE 110 7. 4.3636. 17. 7.8452-10. 15. 0.4536. 45. 0.06114. 47. 4.554. 57. 3.728. 9. 9.5490-10. 19. 9.8908-10. 39. 69.2. 31. 46.4. 41. 292.3. 49. 0.00001072. 59. 2.5023. 11. 8.8215-10. 21. 0.4972. 29. 9.1306-10. 27. 9.9476-10. 37. 0.0000000000276. 55. 1.585. - (b) 2 log g. -1. 7.7931-10. 5. 25. 9.9824-10. 53. 0.4266. . 23. 0.007314. 13. 9.9279-10. 43. 5,454,000,000. + 4* + 13x 33 ?/ 8 solution. 1/2 logb -3. 21. 7314000. 35. 504. 10x 2 4- ). 7. 3. 2.0212. -f- 1000 11. 19. 10 4 23. 8.9439-10. 33. 0.262. - PAGE 104 EXERCISE 5-3. 1. 7x 2 1/2 Iog6 -1. -f- 2 / 7 EXERCISE 5-2. 1. l)n. - 3 2x0. 5. logio = 343. 17. 73 29. 1/10. + 11. (n -f 19x 4 = 4. 9. Iog 100 - 1). ^8 39. 2 11 . 81 3. logs Iog 2 - n(n 35a; EXERCISE 5-1. 1. ) 276*. 2?/2: 37, 9. _ 2/ 2 + 35. 7920a 8 6 4 . X 7 + 15z 8 4 - 2 + 2xy + + Qx l y 2 7 2 PAGE 100 7. 5/24. 19. x 612 i/4 a; t ~T~ 21. 8a<> 25. + . 71 23. _^ x 2 (l 2 3. 3/2. 13. s 35. -...- m x EXERCISE 4-2. 1. 54. 23. 61. 1.6297. 51. 0.6021. 344 Appendix B PAGE 112 EXERCISE 5-4. 3. 0.04292. 1. 8.540. 5. 3.183. 15. 11,670. 25.1.708. 27. -1.021. 33. 127,900,000 sq 7. 17. 0.1795. 13. 48.91. 29.1.249. 35. 12.62 ft. ft. 0.0008416. 0.1104. 9. 19. 0.02950. 11. 54.61. 23. 538,100. 21. 20.56. 31.0.4343,0.2171,9.5657-10,23.1,22.46. 37. 6,070,000 sq 39. 16.5 ft. amp. 41. $1,074.00. PAGE 114 EXERCISE 5-5. 1. 2.3026. 13. 1.4307. 7. 6.0001. 5. 0.8735. 3. 1.4429. 11. 2.0794. 9. 1.5373. 15. 0.8228. PAGE 119 EXERCISE 6-1. B = 57, 6 = 18, c = 22. 3. A = 1S50', a = 21, c = 66. 7. A = 2244', 6 = 10.30, c = 11.17. 5. A = 27!', B = 6259', c = 7.012. 11. B = 4643', a = 73.66, c = 107.5. 9. B = 5339', 6 = 13.40, c = 16.64. = = = B A c 13. 15. A = 493', a = 2.663, c = 3.528. 793.0. 858', 812', = 33 ft. 19. 113 ft. 21. 227 ft. 23. 1334'. 25. 4 ft. 17. h = 29 ft, 1. Z PAGE 124 EXERCISE 6-2. 3. 6326', 11,000ft. 80 ft, 173 ft. 1. 9. 60. 5. N1124'W, 74 PAGE EXERCISE 6-3. 1. (a) 5. 2. (a) 3. (a) 5, 5. 12, 2 x/5. 5 [3/5, 4/5]. (g) 4 4. (a) (c) 2 538'. (c) 5, 1438'. 9.8733-10. 18, 198. 3. 8.8059-10. 5. 0.7391. 13. 0.0030. 15. 0.1004. 23. / 5416 23416'. 29. 5340', 23340'. , 25. 5. 9. 020'. 11. 671 Ib, 15. 1815 Ib, 1962 Ib. 117. a = 20040'. 3. B = 5.94. 13. 17. 49.19 in. 7. PAGE Ib, 25045 r , . 9. 9.9427-10. 17. 120', 18120'. 64 2 / 133 7. 9.3661-10. 31. 2724', 15236'. B = 59, 6 = 60, c = A = 7115', b = 2.01, 49 11. Ib. EXERCISE 6-5. 1. 131 V VS, 6326'. (c) V73, 15927'. 0. 7. 22 knots, 20 knots. 9. 60 II. 9.5906-10. 21. 400ft. [0, 1]. EXERCISE 6-4. I. 7. (g) 4. (e) y/2. (e ) nautical miles. 29558'. 19. 830', 17130'. 27. 3011', 33. 127', 181 27'. 210 35. 6547', 29413'. PAGE 134 6440', 6 A = = 133.9, c 6251', b 607 mph, = N3642'W. 19. 16,900 Ib. = 11'. 148.2. 18.53, c = 40.61. 345 Appendix 8 PAGE 138 EXERCISE 7-1. -. -3. 11. 13. 12-VS- . (V5 -2 V2), J g + 2 V). (1 - 2 7. 15. 4/5, -3/5. 17. 33/56, 63/16. PAGE 142 EXERCISE 7-2. 26 169' 239 11. (a) -3/5, -^ (b) 25. cot ? 27. tan* - 17. sin 2 20. tan 60. 15. ^ (g), (0-4/5, , 13. cos 30. ^ 2t 7. 70 - sin C oS 48 - [sin _ 1 ^ [ PAGE 144 cos 8]. + 100 - 9. \ 2i sin 20. [cos 60 5. - ^ cos ^ 2* 0. 3, 8 radians, 15. - 2 sin 50 sin 30. 3. 27r, 1/2, 1, 0. 0. 17. 1 radian, 0. 11. 27T/5, 3, radians. , 19. 0. x 7. 5?r/2, 13. ir/6, 2 radians, 0. , = 3. x = -=- 2 13. 7T/3. T/2. 23. 12/5. 37. -7T/2. 53. 1 , 0. 15. Tr/4, , 0. 7 radians. CD, 21. 2ir, 1, 0. 7T PAGE 5. m x = -12T/ 161 - 22. 7. Tr/6. 2ir 11. |. 2i ^ radian. 3, EXERCISE 8-2. 1. ^& cos 2 sin 3230' cos 730'. 17. 7T TT, 2 sin PAGE 154 5. Sir/3, 1/3, 4 23. 11. t EXERCISE 8-1. 9. cos 40]. 43 cos 3. 19. 2 sin 1. 2?r, [cos 60 -f cos 20]. ^ 0/3 Q/a 13. 2 sin 23. cos 9. - 3. sin 0]. OOy21. tan 20. 19. cos 20. EXERCISE 7-3. 1. (d) 4/3, (c) 3/4, f/50+5x/I6, - 25. 5/13. 39. -w. 17. -7T/3. 15. 7T/6. 27. 3/4. 41. u. 29. 0.3919. 43. w. 45. 19. y 31. 5\/6 21. 9. ir/6. -2427'. 33. 3/4. 47. u. 35. 49. u. -3/5. 51. 1/u. 346 Appendix 8 PAGE 168 EXERCISE 9-1. 1.x- 10/7, = y = -==-, 13. 23. x 27. z _ j. 2L * = 10 / 7 = -21/11, z = -9/2. 25. x = 4, z = 20/3. 29. x = 0, y = 5/7. 10/7, y 35. Consistent 33. Inconsistent. and independent. = 41. Inconsistent unless c 9. (5/3, 0), (0, -5). 11. x = = 5/8. 17. x = - x 15. 1, = y 3.0. 6/5, y j = 25/13, x 15. 19. 7.7. 5.0. y = 17. -5/13. (-3, -1), (2,4), 5. -4). x 5. z = 2, = 2, = 3, = 3, y y z z = = 3. 0. 5. 7. 3 II. 19. + 4t, - 4i. \/2 + 3 V2i, -2. 3. 2. 7. (y, = = x - 3 \/2 \/15 +8|a|Va6i, f. 21. 1. 31. x = 39. + | 1, y ^ 23. 0. = t. -4. 41. o), 1, y = 80, 7. = x 13. (-4/3, y = 0), (0, 4). 22/7. 4.1, = 0.3. y -17). (0, PAGE 185 + 3f. 3i, -3 \/2i. = 33. x + K. 2/3. No nontrivial solution. 9. 3z. PAGE 187 PAGE 193 - 5. 9. 1 + 4 V&, 13. 2, y = 5/2. 43. 25 1 -i. = 35. + Oi. + 6|a|t, -8|a|Va6*'25. 0. 27. & = 3/2, y = = 2 -2, y VT5 -1 1. 9. 2184. 7. 308. 3. = if c = 23/7, x 13. 11.22. EXERCISE 11-1. I. 1. -3). (6, -110. o PAGE 180 EXERCISE 10-3. 1. 0. ~ = 31. Inconsistent. (0, 0). = 4/5. EXERCISE 10-2. 1. 2; |,y=|. -11. 9. 7/5, PAGE T72 EXERCISE 10-1. 1.5. */er 39. Inconsistent. 3. (4, 0), (0, (1/4, 0), (0, -1/3). = y and dependent. EXERCISE 9-2. 1. , Consistent and dependent 1. 1. = 25/7, * = -1/7. = 9/5, y = -1/5, z = */ > 16/11, y 37. Consistent - = o ^= = -2, y -10 IT 17. x T x 15. Inconsistent. = x 5. 15/7. O =J y = 1 -ir J?H 2/ 75 = = = -9/7, y -47 ~ to = x 3. -6/7. -1 18 a? 45. 4 15. 6|fl|i. Vf. -t. 29. z 7/3, y = 4/3. -1 + f. -i. 17. = 1/3. = - 47. 6, 1 y = -5. 37. 7 + Oi. + 3i. 347 Appendix B -3 + 49. - 57. 11 -2 - 75 ' ~ 1+ , , +\ 25 - 2i. 23. 5 + (2 - 29. ~^(('os 5444' + t sin 0). 33. 3 (cos 2 V2(cos 15 + t sin -8 + V + 2 71. 33 t. 55. - 22t. 4t. + 5 <>/) 73. ~ + ^ (2 PAGE 197 - 35. + i sin 45). (cos bl 32012' + + 3f. 21. 4f. + t sin 270). 27. 3(cos 270 6723' 31. 13(cos 5444'). - 19. 6 3f. + f sin f sin 6723'). 32012 / ). PAGE 202 EXERCISE 11-3. a 5, 2i. , ^ A + 7. 1 -1. 9. i. 11. -1/2 - QSO -2o. 17. 19. + ^ i. 0.7951 A +A 13. 21. t). i. ot 31. - - fc /<* amperes. , =-=^ PAGE 206 EXERCISE 12-1. -7. 9. sin 3. = 0, 5. 3, -1/2, -3/4. 1; 13. sec 0=2, 15. cot = 17. cot 0=7, 19. sin = 0, 0, 90, 180. -2. 11. sin 7. 3, 3, = 1, in 10, 9. tan II. sec o 2. -3. -2; 90. -8; 00, 300, 9711', 26249'. -3, 2; 16134', 34r34', 2(>34', 20634 2, | Q in 3. 10, =1 = 1, / . -17, 88', 1888', 17G38', 35638'. ; 0, 180. PAGE 209 EXERCISE 12-2. i I. 1. A- -- 1. 0, 1+ 1 -1. fc 29. 15. + 90) + f sin (9 + k 90)], k = 0, 1, 2, 3. = 0, 1, 2, 3, 4. 2[cos (36 + k 72) + i (36 + k 72)], = 0, 1, 2. cos(SO + k 120) + f sin (80 + 120), 14 + 18t 828 154i + 00 33. ohms. 23. 2[cos (9 25. (0.6065 | i. ' + 17. 9f. 25. V5)*. - 1019. + 3233 * 1233 -1 - 15. lit. 63. 5 IGt. EXERCISE 11-2. 13. 3 + 53. 13 g-~ 69. i. 625 77 ' *' 25 + 24t. 67. ~ 2(K. + Of. 61. 28 + 51. 6 \/3)i. 27 59. 3f. 65. + (\/2 \/2; Q 3. K 5. --1 = 6730', =fc V7i ^ 7. -1 24730', 15730', -3; 0, 10928', 25032'. - =fc TT \/21 27. 1, t, -1, -t. 348 Appendix B Q = == 36 23', 14337', 23730', 30230 4 = 0.4142, -2.414, 6532', 29428'. 17. (x - 3) 2 + 4(y + 2) = 4. cos - 5) 2 + 4(y - 5)2 = 16. 21. 4(s - 2) 2 + 9(i/ - I) = 36. (x 2 2 4(s + 4) - Q(y - 2) = -36. 25. 3(s + 10/3) - (y + I) = 64/3. / 13. esc 15. 19. 23. 27. . ; 2 2 V2[(z-4) 2 +9/2]. 33. [9((o; + 4/3) 2 + 2 * 29. V (3 I)]' 1 / 3 3) 31. [2((x 2 II. - -1, -1 3. 13. 11/8. -1, -1. 5. >/6. = -r o , no values of ; 0. ; , 23. esc 3. -4. 5. 7. 5. 9. PAGE 215 EXERCISE 12-5. 2i. A/3, =fc \/15, 3 3. i, it A/6. 5. 2, db3. Conjugate imaginary. 3. Real, 5. Real, unequal, irrational. 9. Real, equal, rational. 11. 13. Real, unequal, irrational. 17. -2, -1. 2 J7 s2 + 1 5S PAGE 216 unequal, irrational. 7. Real, unequal, rational. Conjugate imaginary. 15. Conjugate imaginary. PAGE 218 EXERCISE 12-7. 1. 7. 2 A/2. EXERCISE 12-6. 1. . ~ 13 4 x/5. 8 / PAGE 213 EXERCISE 12-4. -4 -1/2. ; = 0.414, -2.414; 11428 24532'. = -0.6972, -4.3028; 19326', 34634 21. sec 9. 32)]~" 2 -1, -15/7. 19. cos 1. d= 9. 5/2, ^ 17. sin 6 1. 18. - 3 7. = 1, -5/2, 45, 225, lllW, 29148'. = 3.7913, -0.7913; 23218 30742'. 15. tan 2 PAGE 212 """ -7. 7) . EXERCISE 12-3. I. 1, + 16 3.0,2. 5.3/2,6/5. 7. 6/5, -1/5. _ = 0. 19. z2 - (\^ - \/5)x =0. 21. x* 2 - 2 A/5 z _ + 8 = 0. . 349 Appendix B PAGE 224 EXERCISE 12-8. 3. Intersecting lines. 1. Circle. 5. 13. Intersecting lines. 11. Hyperbola. 15. Intersecting lines. 17. Intersecting lines. (-3/2, 3/4). 7. (3, 2), (-1, -6). No 13. (4, 5. (2, 4), (-4, -3). (2 V, 11. 9. (2, 3). 15. solution. 3. (3, 4), 7. (4, 0), 13. (-5, 9. (4, 6), 3). "^ ( 3. (3, 4), (-3/2, 3/4). (3, 3), 19. (4, 1), (-4, -1), 21. (-4, -2), (V6, (4, 2), (14, + i V2, 2 , 5. . 5. (3, 11. (2) 4) -3), (-2, 6). (4, (3 VI6i, 1) . . (1> 2)> (2> 1} . -2 (- V6, V6,), 2 VG). 23. (5, 5), i -3 - \/7, - i \/7), (2 i V2, 2 27> 7. 1.682. 5. 2.292. 9. 1. Q Q II. Q 15. Q : : - x 7, R : Q x - 3. 0. : + 3z + 4, 72 + 2* - 15, R a;*- + x n ~*y + 2 2*3 x2 : : : 0. 13. -f R : 2. PAGE 239 : : x* 2x* - 2x - 3. 3. Not a r* # 1 : , + 2xV + 4a*V 7. cc' 9. Not a 15. 12a 3 factor. - 22* 2 -f 8j: 11. 5x* - 34o: factor. 4 2/ + + 2a6(5 - + 60. 17. x* 2/ 24s 3x 2 + 6x - 24, R : : 17. 52, 2. 0. -f 2z - 32x^i<> -f - 19. 24, -36. PAGE 241 3c* 2 )x 3 : x2 5. 16x 3 : 5x + 8, R 11. + 3s - 6, R 0. EXERCISE 13-2. 1. Q 5. + x2 Q Q 0. 1 : 1, 9. 3h06M4> ^- EXERCISE 13-1. I. 15. 8.547. 25. -0.44. ^ t log c 13. 2.718. 11. 1.836. 23. 0.7S74. 21. 10, 0.1. 19. 49.3. log(l-ac)-log6-logc 7. * PAGE 233 EXERCISE 12-11. 3. 4/3. + i v^), ' ' 17. 2.944. (-5, -5). (-4, -3). - i V), (-3 + (-3-tV7, -3+iV7). - 3 V93 31+3 Vm\ /31 + 3 x/93 31-3 \/93\ ^' /31 V 31 31 )'\ 31 31 / 1. 6. -10). -4), (-14,4). 25. (3, 4) (4, 3), (-3, -4), 27. (2 9). PAGE 231 (-4, -3). (-3, -1). 3^85) i} (-3, 1). 2). EXERCISE 12-10. 1. 19. Parallel lines. PAGE 227 EXERCISE 12-9. 1. (3, 3), Parabola. 7. Hyperbola. 9. Intersecting lines. - 9(te 5. + 64.ri/ + 12a 3 6 4 2 . 13. 128y". Not a + 39z + 45. factor. 78. Appendix B 350 PAGE 244 EXERCISE 13-3. 1 1. -3. \/3i, - 7. x* 3. 1 + 9x - 4x* 5. 3. =fc i, 2 -\/2 1, 1, 2. *', = 0. 10 PAGE 247 EXERCISE 13-4. "1 1. 15. 1/2, -1. 9. -1/3, 1/2, 5/3. 17. 3/5, 1 t. -2, -2. 11. 2, 2, PAGE 253 EXERCISE 14-1. < 3. > -5. II. -i<z<i. & & 13. 19. -4 25. No x I. 81. a; z 21. 4. values of x. | | 3. 2 33. x 5. 5. x < 4. 7. < a; -|<*<i. O o -1 < x < 1/3. 4fl ~ 23. x , & = 15. /7s = 21. /io 31. ~ o d = S 75 = S 10 = 100, 1, Not an 5. 36 ll. & 149, n = -^5 = . 11 5625. J 26 = /2o 23. a 13 or a = 33. 33j i 6 . 35. 4 = - n 60 3. 1/20, 1/25. IT" ' T 120 ' , II. I,. Ifcfc = |J a; < > x 2, 3. = ho 13. 61. 7. 19. 46 30, 5b = 55. 10, Sio i 100 = 40. Sioo=5050. 100, o 27. 16. ^~ ~T~ 5. 4, 16/5. / = 9, n = 9. 29. n2 . 1 . 37. $37.75. 39. 282. ly PAGE 264 7. 1/47. 9. ^, | , y , |. , "IT ' 3. 256, 1024. , Si, = 9[1 - 5. 8, 16/3. (2/3)'*]. =!,& '"' ={. 16 < =~ Z 45 EXERCISE 15-4. I. 128, 512. 2. fi9 EXERCISE 15-3. 120 1053. S 2Q = 5.9, = -^ -1/2, n > 16 13. PAGE 263 40 550. -13 < x < 17. PAGE 260 = 78, S 26 = 17. 1 arithmetic progression. tt\K 1. 1/15, 1/19. -5/3, s -1. 7. 2, 5, 11, 23, 47; 2, 7, 18, 41, 88. 3.1,2,3,5,11,35. . 9. 25. 5a < < -3, -2 < x < -1. 29. < -2, s > 0. 35. x < -1/2. 27. x 3. 18, 25. 26 < x 9. -i<*<5. & A EXERCISE 15-2. 1. 17, 20. -1/2. 15. EXERCISE 15-1. 1.1,2,4,7,11. ' V- 19. 1/2, t. t& 13. 2/3, 7. PAGE 267 , ^jJL& 13. ljol 17. 3, -14. = ^j" JL& 10-", 9. -1,1. S101 = 19.i^. 21. 6, 15/2. 351 Appendix B 23. 10; 100; 1000; 10,000; 100,000. As n 27. sum approaches increases, the 3 as a limit. + 9 3. 13. 36, 5 ' 16 2 36 - [1 19. 10/11. (1/3) 20 36(l/3) ], 23.3. 21. 1/6. 7' 3 ' 20 . 9 - 5/8. ' 15. 12 1.1* ,o -+-. 2 * 7 .i_4+*!_y!. X2 X* X3 X 9 . 3 3. 216. 5. 9999. C a8 4- + 8 C2a sCitfb C6 8 Ca/>7 X 12 [1 - (3/4) 10 ] 17. 1/9. feet. PAGE 772 ^ - 8- 1 a? 3z 2 5z 3 -2 + -T-lG' , 11.1.2190. 13.1.3684. PAGE 279 7. 20,160; 7560. 6 62 + PAGE 282 3. (a) 720, (b) 48, (c) 480. 20; 210; 95,040; 143,640; 970,200. 7. 8 * 19. 1.0149. EXERCISE 17-2. 1. ft, 6 & EXERCISE 17-1. 1. 72. JL 11.2/11. + 6_i?f. .J__f X X* X X 17. 0.9415. 15. 0.8508. 3 # s A i.n.--_+_. * X) 25. EXERCISE 15-6. -. ~~ PAGE 270 EXERCISE 15-5. 1.64/65. 29. sC 3 a 5 6 3 5. 125. + . f 11.66. 9.9,979,200. 13.63. 15TJ jy EXERCISE 17-3. 1. 2/5, $24.00. 3. 1/3, 2/3. A 15 15. 5. $7.29. 7. $2.42. 5. 9. 9. 0.553. 11. 16/63, 215. , EXERCISE 18-1. 1. PAGE 288 PAGE 295 = 14.55, c = 20.46. 3. A = S927', a = 1169, b = 1079. B = 531', 6 = 1.051, c = 7.513. 7. A = 173', B = 10026', 6 = 71.18. B = 2427 C = 10120 c = 1193. 11. No solution. 13. 615.3 ft. C = 15. 9858', b ; ; , 449 ft. , 17. 12.6 in. 19. 3158 Ib, EXERCISE 18-2. PAGE 298 = 270, B = 5130', A = OO^O 3. a = 290, B = 1150', C = 9. No solution. 5. a = 100, B = 10, C = 20. 7. 4820'. 13. 3257 ft. 15. 700 ft, II. A = 5420 B = 5940', C = 66. 17. c = 20, A = 63, 5 = 73, C = 44. 7 I. c . / , 101. 352 Appendix 8 EXERCISE 18-3. 1. 5. PAGE 301 A = 50, B = 70, c = 56. 3. A = 61, C = 2350', 6 = 262. B = 3111', C = 70, a = 79.25. 7. A = 47, J5 = 58, C = 75. A = 4220', B = 5730 C = 8010'. 11. 4 in., 5 in. 13. 11 in. / 9. , 15. 55 30', 5950', 6440'. EXERCISE 18-4. 1. 88.41. 17. 38 sq 3. 243,900. ft, 12 sq ft. 5. 9285. 7. 90. PAGE 304 11. 54 ft. 15. 45 ft. Index Abscissa, 50 Absolute value, 14, 57 Absolute value function, 57 Addition, 2 of algebraic expressions, 21 of fractions, 36 fundamental laws for, 2 of real numbers, 2 Addition formulas, 135 Algebraic expressions, 17 Binomial theorem, 97 combinations applied to, 282 general term of, 99 proof of, for positive integral exponents, 275 Braces, 18 Brackets, 18 Cartesian coordinates, 50 Characteristic of a logarithm, 106 Characteristic of the general quadratic equation, 222 Circular system, 77 Classification of functions, 61 Coefficient, 17, 18 Combinations, 278, 281 and the binomial coefficients, 282 Common difference, 260 Common logarithms, 105 Common ratio, 265 Commutative law, 2 Completing the square, 206 Complex fractions, 40 Complex numbers, 189 absolute value of, 196 addition and subtraction of, 192 amplitude of, 196 argument of, 196 congugate of, 191 definition of, 189 De Moivre's Theorem, 199 division of, 193, 198 graphical representation of, 195 modulus of, 196 multiplication of, 192, 198 roots of, 200 trigonometric representation of, 196 Composite number, 25 addition and subtraction of, 21 division of, 23 multiplication of, 22 symbols of grouping of, 18 transposing terms of, 43 Amplitude of a trigonometric function, 149, 152 Angle coterminal, 76 definition of, 75 degree measure of, 77 of depression, 120 of elevation, 120 initial side of, 75 measurement of, 76 negative of, 75 positive of, 75 radian measure of, 77 standard position of, 75 terminal side of, 75 trigonometric functions vertex of, 75 Antilogarithms, 109 Arc of a circle, 78 of, 81, 82 Area of a sector of a circle, 78 of a triangle, 302 Arithmetic means, 262 Associative law for addition, 2 for multiplication, 2 Axes, coordinate, 49 Axis of symmetry of a parabola, 219 Composite (reducible) polynomial, 26 Conditional equation, 18 Constant, 53 Constant function, 53 Convergence of series, 259 Coordinate systems Cartesian, 50 one-dimensional, 49 rectangular, 49 Coordinates, 49, 50 Cosecant, definition of, 65 Base change of logarithmic, 113 exponent and, 16, 86 of logarithms, 101, 113 in navigation and surveying, 121 Binomial, 17 Binomial series, 271 Bearing 353 354 Index Cosine, definition of, 64 Cotangent, definition of, 65 Coterminal angles, 76 Counting numbers, 1 Defective equations, 45 Degree of a polynomial, 17 relation to radian, 77 of a term, 17 term of highest, 17 unit of angle measure, 77 De Moivre's Theorem, 199 Dependent events, 286 Descartes, Rene, 50 Determinants, 173 expansion of, 176 minors and cof actors of, 175 principal and secondary diagonal of, 174 properties of, 177 of the second order, 173 solution of linear equations by means of, 181, 183, 184 sum and product of, 185 of the third order, 175 Directed line segment, 122 Discriminant, 215 Distance between points, 50, 51 Distributive law, 3 Division, 5, 16, 23, 25, 39 of a function, 53 Domain Double-angle formulas, 139 Empirical probability, 284 Equality symbol, 12 Equations conditional, 18 consistent, 163, 164, 165 defective, 45 definition of, 18 dependent, 165 equivalent, 43 extraneous roots of, 45, 213 inconsistent, 163, 165 independent, 164 linear, 42, 43, 163, 166, 167 in quadratic form, 204, 214 solution of, 42, 163, 171, 181, 183, 184, 212, 224, 227 Exponential equations, 231 Exponents, 16, 86, 88, 90, 92 Factor theorem, 240 Factorial symbol, 97 Factoring, 25, 27 Fractions, 9 addition and substraction of, 36 Fractions (Cont.) complex, 40 fundamental operations on, 9, 10 multiplication and division of, 39 reduction of, 33 signs associated with, 34 Functions absolute value, 57 algebraic, 61 classification of, 61 constant, 53 definition of, 53 domain of, 53 exponential, 233 graphs of, 146, 147, 148, 169, 218, 233, 244 greatest integer, 57 identity, 53 inverse, 155 irrational, 62 linear, 54, 169 logarithmic, 233 multiple-valued, 53 notation for, 55 periodic, 149 point, 63 polynomial, 61 of, 53 rational, 61 rule of correspondence of, 53 single-valued, 53 transcendental, 61 trigonometric, 63 Fundamental assumptions, 1 Fundamental operations, 1, 9, 10, 20 Fundamental theorem of algebra, 241 range General term in the binomial expansion, 99 Graphs of exponential functions, 233 of inverse trigonometric functions, 158, 159 of linear functions, 169 of logarithmic functions, 233 of polynomials for large values of x, 244 of trigonometric functions, 146, 147, 148, 149, 151, 152 Greater-than symbol, 12 Greatest common divisor, 30 Greatest-integer function, 57 Half-angle formulas, 140, 300 Highest common factor, 31 Highest degree term, 17 Horizontal line, 50, 169 Identity, 18 Independent events, 286 355 Index Inequalities, 248 absolute, 248, 254 conditional, 248, 249 involving absolute values, 14 properties of, 248 solution of conditional, 249 Inequality symbol, 12 75 Intercepts, 170 Most probable number, 284 Multinomial, 17 Multiplication of algebraic expressions, 22 of fractions, 39 fundamental laws for, 2 Mutually exclusive events, 285 Initial side, Interpolation, 83, 108 Inverse functions, 155 Irrational functions, 62 Irrational number, 1 Irreducible polynomial, 26 Law of cosines, 296 Law of sines, 289 Law of tangents, 298 Law(s) of exponents, 86 Least common multiple, 32 Less-than symbol, 12 Like terms, 21 Limit of sequence, 258, 259 Line horizontal, 50, 169 vertical, 51, 169 Linear equation, 42 graphs of, 163, 169, 171 in one unknown, 43 Literal parts, 17 Logarithmic computation, 110 Logarithmic equations, 231 Logarithms, 101 base of, 101 change of base, 113 characteristic of, 106 common, 105 computation by, 110 definition of, 101 laws of, 102 mantissa of, 106, 108 Napierian (natural) ,105 tables of, 108 of trigonometric functions, 131 Mantissa of a logarithm, 106 Mathematical expectation, 284 Mathematical induction, 273 Mean arithmetic, 262 geometric, 267 harmonic, 264 Meaning of a m/n 94 Meaning of a, $8 Measurement of angles, 76 Monic polynomial, 18 , Natural (Napierian) logarithms, 105 Negative exponents, 90 Negative numbers, 7, 8 Number algebraic, 62 complex, 189 irrational, 1 negative, 1, 7, 8 positive, 1, 7, 8 prime, 25 real, 1 transcendental, 62 Number scale, 6, 49 One-to-one correspondence, 49 Operations on fractions, 9, 10 with zero, 5 Order of fundamental operations, 20 Order relations for real numbers, 12 Ordered number pairs, 50 Ordinate, 50 Origin, 6, 49 Parentheses, 18 Period, 150 Periodicity, 149, 150 Permutations, 278, 279 Phase, 149, 152 Point function P (t), 63 Polynomial, 17, 18, 26, 61 Positive integer, 1 Positive integral exponents, 16, 8 Positive numbers, 7, 8 Power, 16, 17, 86, 87, 88 Prime, 25, 26 Principal branch, 158 Principal value, 158, 159, 160 Probability, 278 empirical, 284 mathematical, 283 Product, definition of, 2 Product formulas, 143 Progressions, 256 arithmetic, 260 geometric, 265 harmonic, 264 infinite geometric, 268 356 Index Projections, 122 Tables Quadrant, 49 Quadratic equations methods for solving, 204, 209 in one unknown, 204 in two unknowns, 221, 224, 227 Quotient, 5 Radian, 77 Radius vector, 52 Range of a function, 53 Rational exponents, 92 Rational function, 61 Rational number, 1 Real number, 1 Reciprocal, 4, 5 of logarithms, 108 of trigonometric functions, 71, of, Terms of algebraic expressions, 2, 17 Theory of equations, 235 Transcendental functions, 62 Triangles, 115 solution of general, 289 solution of right, 117, 133 Trigonometric functions, 63 of angles, 81, 116 definitions of, 64, 65 of, 146, 147, 148, 149, 151, 152 of important special numbers, 65 inverse, 146, 156, 158, 159, 160 graphs logarithms Rectangular coordinates, 49, 50 Redundant equation, 45 Remainder theorem, 240 Repeated trials, 287 Repeating decimals, 269 Root of an equation, 42, 45, 213, 217 Scalar quantities, 125 Scientific notation, 92 Secant, definition of, 2 64 Tangent, definition Terminal side, 75 of, 131 of sums and differences, 135 variation of, 146 Trigonometric identities, 68 Variable, 53 dependent, 53 independent, 53 Variation, 57, 58 Vector, 125 65 Second degree equation, 204 Segment, line, 122 Sequences, 256, 258, 259 Series, 256, 257, 259 Sexagesimal system, 77 Simultaneous equations, 163, 171, 181, 224, 227 definition of, 64 Sine, Special products, 22 Standard position of an angle, 75 components of, 125 magnitude of, 126 multiplication of, by a normalization of, scalar, 126 127 projection of, 125 representation of, 125, 128 sums and differences of, 127, 129 Vertex of an angle, 75 Vertical lines, 51, 169 #-axis, 49 as-coordinate, 50 y-axis, 49 ^/-coordinate, 50 Sum, definition of 2 Symbols of grouping, 18 Zero, Synthetic division, 235 Zero polynomial, 61 , 3, 5