5 Rewaiy - eee ene ae 7 - rs es mons, Leeann wre eee EE) 2 re eens ene tear NE aan a Ren ne wt renee CORNELL UNIVERSITY LIBRARY MATHEMATICS Mn PLANE TRIGONOMETRY. oe Cambringe: LPRINEHD' BY, O..J. LAY, sta, Jag gue Dluyegsrey reps. PREFACE. THE present work contains all the propositions which are usually included in treatises on Plane Trigonometry, together with about a thousand examples for exercise. The desiyn has been to render the subject easily intel- ligible, and at the same time to afford the student the opportunity of obtaining all the information which he will require on this branch of Mathematics. The work is di- vided into a large number of chapters, each of which is in a great measure complete in itself. Thus it will be easy for teachers to select for pupils such portions as will be suitable for them in their first reading of the book. Each chapter is followed by a set of examples; those which are entitled Miscellaneous Examples, together with a few in some of the other sets, may be advantageously reserved by the student for exercise after he has made some progress in the subject. As the text and the examples of the work have been tested by considerable experience in teaching, the hope may be entertained that they will be suitable for imparting a sound and comprehensive knowledge of Plane Trigonometry, vi PREFACE. together with readiness in the application of this knowledge to the solution of problems. Any suggestions or correc- tions from students and teachers will be most thankfully received. The Miscellaneous Examples at the end are arranged in sets, each set containing ten examples: the first hundred relate to the first eight Chapters of the book; the second hundred extend to the end of the sixteenth Chapter; and the last hundred relate to the whole book. I. TODHUNTER, CAMBRIDGE, August, 1874, CHAP, L Til. IV. VI. VIL. XIV. XV. XVI. XVI. XVIII. XIX. XXII. XXTIT. XXIV. CONTENTS. Measurement of Angles by Degrees or Grades . ‘ so Circular Measure of an Angle. : ‘ if ‘ 7 Trigonometrical Ratios . j A és : 14 Application of Algebraical Signs 7 ‘ ‘ s 23 Angles with given Trigonometrical Ratios - 42 Trigonometrical Ratios of Two Angles oa 50 Formule for the Division of Angles 4 e - 63 Miscellaneous Propositions a A F : "2 Construction of Trigonometrical Tables . . 82 Logarithms and Logarithmic Series . z ‘ F 2 93 Use of Logarithmic and Trigonometrical Tables . 3 «207 Theory of Proportional Parts . : é a RS Relations between the Sides of a Triangle and t the rBeisoind: metrical Functions of the Angles. ‘i 3 % - 148 Solution of Triangles ‘ : ‘ 4 - 158 Measurement of Heights and iietstias . ‘i : - . 172 Properties of Triangles 3 : ‘ > 7 _ - 184 Use of Subsidiary Angles in solving Equations and in adapting Formule to Logarithmic Computation . # 2 . 202 Inverse Trigonomeirical Functions. é : + 207 De Moivre’s Theorem é 2 : : ci “ + 212 Expansions of some Trigonometrical Functions . , 224 Exponential Values of the Cosine and Sine . . . B35 Summation of Trigonometrical Series a ee ee Resolution of Trigonometrical Expressions into Factors . » 252 Miscellaneous Propositions : : . : . 270 ANSWERS . é - ‘ 3 és é é ; ; » 325 Students reading this work for the first time may omit Chapters XII. and XIX.... XXIV. PLANE TRIGONOMETRY. I. MEASUREMENT OF ANGLES BY “DEGREES, OR GRADES. 1. THE word Trigonometry is derived from two Greek words, one signifying a triangle and the other signifying J measwre, and originally denoted the science in which the relations subsisting ‘between the sides and the angles of a triangle were investigated; the science was called plane trigonometry, or spherical trigonometry, according as the triangle was formed on a plane surface or on a spherical surface. Plane Trigonometry has now a wider meaning, and comprises all algebraical investigations with respect to plane angles, whether forming a triangle or not. 2. We have first to explain how angles are measured. Some angle is selected as the wnt, and the measure of any other angle is the number of units which it contains. Any angle might be taken for the wnit, as for example a right angle; but a smaller angle than a right angle is found more convenient. Accordingly a right angle is divided into 90 equal parts called degrees; and any angle may be estimated by ascertaining the number of degrees which it contains. If the angle does not contain an exact number of de- grees we can express it in degrees and a fraction of a degree. A degree is divided into 60 equal parts called minutes, and a minute into 60 equal parts called seconds, and thus a fraction of a degree may if we. please be converted into minutes and seconds. 3. Thus, for example, half a right angle contains 45 degrees ; a quarter of a right angle contains 224 degrees, which we may write in the decimal notation 22-5 degrees, or we may express it as 7. T, 1 2 MEASUREMENT OF ANGLES BY DEGREES OR GRADES, 22 degrees, 30 minutes. Similarly, if a right angle be divided into 16 equal parts, each part contains 52 degrees, that is, 5 degrees, 37 minutes, 30 seconds. 4, Symbols are used as abbreviations of the words degrees, minutes, seconds. Thus 5° 37’ 30” is used to denote 5 degrees, 37 minutes, 30 seconds. 5. The method of estimating angles by degrees, minutes, and seconds, is almost universally adopted in practical calculations. Another method was proposed in France in connexion with a uniform system of decimal tables of weights and measures. In this method a right angle is divided into 100 equal parts called grades, a grade is divided into 100 equal parts called minutes, and a minute is divided into 100 equal parts called seconds. On account of the occurrence of the number one hundred in forming the subdivisions of a right angle, this method of estimating angles is called the centesimal method ; and the common method is called the sexagesimal method on account of the occurrence of the num- ber stxty in forming the subdivisions of a degree. The centesimal method is also called the /rench method, and the common method is called the Znglish method. 6. Symbols are used as abbreviations of the words grades, minutes, and seconds, in the centesimal method. Thus 5% 37° 30" is used to denote 5 grades, 37 minutes, 30 seconds in the centesimal method. A centesimal minute and second are not the same as a sexagesimal minute and second, and the accents which are used to denote centesimal minutes and seconds differ from those which are used to denote sexagesimal minutes and seconds, 7. In the centesimal method any whole number of minutes and seconds may be expressed immediately as a decimal fraction of a grade. Thus 37 minutes is a of a grade, that is -37 of a grade ; and 30 seconds is ——-, of a grade, that is (003 of a grade, 30 (1007? Hence 5* 37° 30“ may be written 5%373; and since a grade is MEASUREMENT OF ANGLES BY DEGREES OR GRADES. 3 one-hundredth of a right angle, 5%:373 may be written as 05373 of a right angle. Notwithstanding this great advantage of the centesimal method, the sexagesimal method has been retained in. practical calculations, because the latter had become thoroughly established by long use in mathematical works, and especially in mathematical tables, before the former was proposed; and such works and tables would have been rendered almost useless by the change in the method of estimating angles, The centesimal method is not practically used even in France. Neverthelesg it is cus- tomary to retain in works on Trigonometry the matter which we shall give in the next two Articles. 8. Zo compare the number of degrees in any angle with the number of grades in the same angle. Let D be the number of degrees contained in any angle, G the number of grades contained in the same angle, Then since there are 90 degrees in a right angle, expresses the ratio of the given angle to a right angle ; and since there are 100 grades in a right angle, —-~ also expresses the ratio of the given angle to a right angle, 100 Hence mies therefore p= g- 2 0-6-4, anit = p= D=D+5D. « The formula D=G— = @ gives the following rule: From the number of grades contained in any angle subtract one-tenth of that number, the remainder is the number of degrees contained in the angle. The formula G' =D +5 D gives the following rule: To the num- ber of degrees contained in any angle add one-ninth of that number, the sum is the number of grades contained in the angle, 1—2 4 MEASUREMENT OF ANGLES BY DEGREES OR GRADES, 9. Again, let m be the number of English minutes contained in any angle, » the number of French minutes contained in the same angle, Then since there are 90x 60 English minutes in a right angle, Ose expresses the ratio of the given angle to a right angle ;‘and since there are 100 x 100 French minutes in a right angle, Too e100 also expresses the ratio of the given angle to a right angle. Hence SE a ae 90x 60 100x100’ _ 9x6 27 therefore m= ioxto"=5o" 50 and B= 7m Similarly, if s be the number of English seconds contained in any angle, and « the number of French seconds contained in the same angle, 4 8 7 co ; 90x 60x60 100x100 x 100’ 81 therefore = 555% 25 and ; c= ae 8. 10. The angles considered in Geometry are in general less than two right angles. We say in general, because angles greater than two right angles are not altogether excluded. For we may refer to the proposition that in equal circles, angles, whether at the centres or at the circumferences, have the same ratio as the arcs on which they stand have to one another ; here there is-no limit to the magnitude of the arcs, and consequently no limit to the magnitude of the angles; and in the course of the de- monstration given by Euclid, an angle occurs which may be any multiple whatever of a given angle, and so may be as great as we please, - MEASUREMENT OF ANGLES BY DEGREES OR GRADES, 5 ll. It is however usual in works on Trigonometry expressly to state that there is to be no restriction with respect to the mag- nitude of the angles considered. Let BAD be any straight line, wa CAL a straight line at right angles to the former. Suppose a straight line AP to revolve round one end A, starting from the fixed position 42. When AP coincides with AC, the angle which has been described is a right angle; when AP coincides with AD, the angle which has been described is two right angles; when AP coincides with AH, the angle which has been described is three right angles; when AP coincides with AB, the angle which has been described is four right angles. Then as AP proceeds through a second revolution, the angle described will be greater than four right angles. Thus if AP be situated midway between AB and AC, the angle between AB and AP will be half a right angle if AP be supposed in its first revolution; the angle will be four right angles and a half if 4 P be supposed in its second revolution ; the angle will be eight right angles and a half if 4P be supposed in its ¢hird revolution ; and so on. 12. The straight lines CAH and BAD form by their intersec- tion four right angles; these are called quadrants. BAC is called 6 EXAMPLES, CHAPTER I. the first quadrant, CAD the second quadrant, DAE the third quad- rant, and EAB the fourth quadrant. Now suppose any angle formed by the fixed straight line AB and the moveable straight line AP; if AP is situated in the first quadrant, the angle BAP is said to be in the first quadrant; if AP is situated in the second quadrant, the angle is said to be in the second quadrant ; and so on. EXAMPLES. 1. The difference of two angles is 10 grades and their sum is 45 degrees ; find each angle. 2. Divide two-thirds of a right angle into two parts, such that the number of degrees in one part may be to the number of grades in the other part as 3 is to 10. 8. Divide half a right angle into two parts, such that the number of degrees in one part may be to the number of grades in the other part as 9 is to 5. 4, ind the measure of 1‘ 5“ in decimals of a degree. 5. Divide an angle of n degrees into two parts, one of which contains as many English minutes as the other does French. 6. If one-third of a right angle be assumed as the unit of angular measure, what number will represent 75°? 7. Determine the number of degrees in the unit of angular measure when an angle of 662 grades is represented by 20. 8. The number of the sides of one equiangular polygon is two-thirds of the number of the sides of another; and the number of grades in an angle of the first equals the number of degrees in an angle of the second: find the angles. 9. Shew that an angle expressed in centesimal seconds will be transformed to sexagesimal by multiplying by the factor -324. 10. Compare the angles which contain the same number of English seconds as of French minutes. 11. Express in the French method 35° 10’ 3”. 12. Express in the English method 69¢ 22° 50°. 8 Il CIRCULAR MEASURE OF AN ANGLE. 13. We have explained two methods of estimating angles, namely, that by means of degrees and subdivisions, and that by means of grades and subdivisions, and we have stated that the for- mer method is that which is most commonly used in practical cal- culations. There is, however, another method of estimating angles which is of great importance in the theory of mathematics, which we shall now explain. The object of the present Chapter is to es- tablish and apply the following proposition: If with the point of intersection of any two straight lines as centre a circle bedescribed. with any radius, then the angle contained by the straight lines may be measured by the ratio of the length of the arc of the circle imter- ' cepted between the straight lines to the length of the radius. We shall require some preliminary propositions ; the proposition in Art. 14 is sometimes assumed, and the beginner may adopt this course and return to the point hereafter. 14. The circumferences of circles vary as their radit. Let R denote the radins and C’ the circumference of one circle ; let r denote the radius and ¢ the circumference of another circle. In each circle let an equilateral polygon of n sides be inscribed, and in each circle draw two straight lines from the centre to the extremities of one of the sides of the inscribed polygon; thus we obtain two similar triangles. Tet P denote the perimeter of the polygon inscribed in the first circle, and p the perimeter of the polygon inscribed in the second circle. By similar triangles a side of the first polygon is to a side of the second polygon as the ' radius of the first circle is to the radius of the second circle; therefore also Now let P=C-—X and p=e-«; thus r(C-X)=R (c-2) ; therefore rC — Re=rX — Rx, 8 CIRCULAR MEASURE’ OF AN ANGLE. Now we assume that by making 2 as large as we please, the perimeter of each polygon can be made to differ as little as we please from the circumference of the corresponding circle ; thus -Y and « can each be made as small as we please, and therefore r.X —Rxcan.be made as small as we please. Hence r('— Re must be ees for if it-had any value a then rX — Ra could not be made less’than a, which is inconsistent with the fact that *Y— Re can be made as small as we please. Thus rU—Re=0 Coe therefore Re 15. Thus the ratio of the circumference of a circle to its radius is constant whatever be the magnitude of the circle; therefore of ‘course the ratio of the circumference to the diameter is also constant. The numerical value of the ratio of the circumference of a circle to its diameter cannot be stated ewactly; but, as we shall shew here- after, this ratio may be calculated to any degree of approximation : 2. that is required ; the value is approximately equal to se , and still 355 more nearly equal to = the value correct to eight places of 113’ decimals is 3-14159265...The. symbol z is invariably used to denote the ratio of the circumference of a circle to its diameter ; hence, if v denote the radius of a circle, its circumference is 2rr: and m= 314159... 16. The angle subtended at the centre of a cirele by an are which is equal in length to the radius is an invariable angle. B CIRCULAR MEASURE OF AN ANGLE. 9 With centre O and any radius OA describe a circle; let AB be an are of this circle equal in length to the radius. Then, since angles at the centre of a circle are proportional to the ares on which they stand, angle AOB are AB rn ; A =5 a four okt angles — circumference of the circle la 7 ea four right angles = F Thus the angle AOB is a certain fraction of four right angles which is constant, whatever may be the radius of the circle. therefore angle A082 = 17. Since the angle subtended at the centre of a circle by an arc which is equal to the radius is an invariable angle, it may be taken as the unit of angular measurement, and then any angle will be estimated by the ratio which it bears to this unit. Let 400 be any angle; with O as centre and any radius Od 0 A deseribe a circle; let AB be an arc of this circle equal in length to the radius; let denote the radius, and / the length ef the arc AC. Then, since angles at the centre of a circle are proportional to the arcs on which they stand, angle AOC _ AC Ls angle AOB AR therefore angle AOC =! x angle AOB ; this result is true whatever the unit of angular measurement may 10 CIRCULAR MEASURE OF AN ANGLE, be, the same unit of course being used for the two angles. If we take the angle AOB itself for the unit, then this angle must be denoted by unity ; thus angle AOC =! 2 18. We have thus proved that any angle may be estimated by a fraction which has for its numerator the are subtended by that angle at the centre of any circle, and for its denominator the radius of that circle. And in this mode of estimating angles the unit, that is the angle denoted by 1, is the angle in which the are subtended is equal to the radius, We have shewn that this angle four right angles Qa angle is a , that is 420 . If we use the approximate value of 2a given in Art. 15, we shall find that 180 _ 57-29577951...; this T ; hence the number of degrees contained in this therefore is the number of degrees contained in the angle which is subtended at the centre of a circle by an arc equal to the radius. 19. Thus there are two methods of forming an idea of the magnitude of an angle which is estimated by the fraction are divided by radius. Suppose, for example, we speak of the angle 3; we may refer to the unit of angular measurement, which is an angle containing about 57 degrees, and imagine two-thirds of this unit to be taken; or without thinking about the unit at all, we may suppose an angle is taken such that the arc subtending it is two-thirds of the corresponding radius, 20. The fraction are divided by radius is called the circular measure of an angle. Since, as we have already stated, this method of measuring angles is very much used in theoretical investigations, it is sometimes called the theoretical method. 21. If, denote the radius of a circle, the circumference is 2ar; 3 a ie . Ian hence the circular measure of four right angles is ——, that is 2z, T CIRCULAR MEASURE OF AN ANGLE. 11 The circular measure of two right angles is 7; the circular measure of one right angle is 3 and the circular measure of 1 right angles is > , where m may be either integral or fractional. 22. We will now shew how to connect the circular measure of any angle with the measure of the same angle in degrees. Let « denote the number of degrees in any given angle, @ the circular measure of the same angle. Since there are 180 degrees in two right angles, [ap expresses the ratio of the given angle to two right angles, And since z is the circular measure of two right angles, é also expresses the ratio of the given angle to two right angles, Hence 2.8. 180°’ thus ; gut T we and O=755° 23. Forexample, the circular measure of an angle of one degree 0 is Tgp a 63 the circular measure of an angle of ten degrees is i isd ; the circular measure of an angle of half a degree is — i 5 G * x35 the cir- cular measure of an angle. of one minute is 180 x60 Oe 63 the circular measure of an angle of one second is TsO x GO TED 3 and so on. Again ; if the circular measure of an angle is = the number of degrees contained in the angle is — , that is — Z a 57-29577951.. if the circular measure of an o is 10, the number of as 12 CIRCULAR MEASURE OF AN ANGLE, contained in the angle is 10. 1) , that is 10 x 57-29577951...; and = : so on. The student is. recommended to pay particular attention to these points; especially he should accustom himself to express readily in circular measure an angle which is given in degrees. 24. Stmilarly we may connect the cirewlar measure of any angle with the measure of the same angle in grades. Let y denote the number of grades in any given angle, 6 the circular measure of the same angle; then the ratio of the given 6 angle to two right angles is expressed by am and also by : Hence Dat. 200-0 er? 2000 thus y=; and = ai E The number of grades in the angle which is the unit of circular measure is = , that is, 63:661977... 25, In Art. 17 we proved that angle 40C =! xangle A0B; where nothing is assumed respecting the- unit of angular measure- ment, except that the swme unit is to be employed for both angles. Since 4O0B is an invariable‘angle, we see that the magnitude of any angle AOC' varies as the subtending are directly, and as the radius ¢nversely. Thus we may say that kx are | ; radius ” when & is some quantity which does not change with AOC, and the value of which depends upon the unit of angular measurement angle AOC.= EXAMPLES, CHAPTER II. 13 which we please to employ, Suppose, for example, that we- wish to take the half of a right angle as our unit; then we require that AOC should be equal te 1 when the arc is the eighth part of the circumference ; thus Qrr hx i 3 therefore k=—. r 7 ls Thus the formula angle AOC' = 3 x Tv are radius gives the correct estimate of the magnitude of an angle when the unit is half a.right angle. EXAMPLES, I. If D, G, C be respectively the number of degrees, grades, and units of circular measure in an angle, shew that D_ @ 2 2. Find the number of degrees in the angle subtended at the centre of a circle whose radius is 10 feet by an are 9 inches long. 3. Find the circular measure of 5° 37’ 30”. 4, Find the circular measure of 18 1’. 5. There are three angles; the circular measure of the first exceeds that of the second by a , the sum of the second and the third is 30 grades, and the sum of the first and the second is 36 degrees. Determine the three angles. 6. Express five-sixteenths of a right angle in circular measure, in degrees and decimals of a degree, and in grades and decimals of a grade. : 7. The angles of a triangle are in arithmetical progression, and the greatest is double the least: express the angles in degrees, in grades, and in circular measure. 14 EXAMPLES, CHAPTER II, 8. The angles of a triangle are in arithmetical progression, and the number of degrees in the least is to the circular measure of the greatest as 60 is to 7; find the angles, 9, Find the circular measure of an angle of an equiangular polygon of 7 sides, 10. Express in each system of angular measurement the angle between the long hand and the short hand of a watch ata quarter past twelve, II, TRIGONOMETRICAL RATIOS, 26. Let BAC be any angle; take any point in either of the containing straight lines, and from it draw a perpendicular to the C A a B other straight line; let P be the point in the straight line AC and P&M perpendicular to AB. We shall use the letter A to denote the angle BAC. Then PM 3 mdicular ieee eee , is called the sine of the angle A; ae hypotenuse AM base aP? that is inotenaae? is called the cosine of the angle 4 ; TRIGONOMETRICAL RATIOS, 15 PM . perpendicular AM’ ee base AM : base pay es perpendicular AP ._ hypotenuse : A that is base a that is ee , is called the cosecané of the angle A. If the cosine of A be subtracted from unity, the remainder is called the versed sine of A. If the sine of A be subtracted from unity, the remainder is called the coversed sine of A; the latter term however is rarely used in practice. » is called the éangent of the angle 4 ; , 18 called the cotangent of the angle A ; , is called the secant of the angle A ; 27. The words sine, cosine, tangent, cotangent, secant, cosecant, versed sine, and coversed sine are usually abbreviated in writing’ and printing; thus the above definitions may be expressed as follows ; smd ait, cos 4-4, ton d= 2, cob = seod ==, cosee d= Se, vers A =1-cos 4, covers A =] —sin A. ‘ 28. The sine, cosine, tangent, cotangent, secant, cosecant, versed sine, and coversed sine are called trigonometrical ratios or trigo- nometrical functions ; sometimes they have been called gontometrical 16 TRIGONOMETRICAL RATIOS, Junctions. A large part of Trigonometry consists in the investiga- tion of the properties and the relations of these functions of an angle. These functions are, it will be observed, not lengths, but ratios of one length to another; that is, they are arithmetical whole numbers or fractions. 29. The defect of any angle from a right angle is called the complement of that angle ; thus if A denote the number of degrees contained in any angle, 90 — A is the number of degrees contained in the complement of that angle. This affords another method of defining some of the Trigonometrical ratios; after defining, as in Art. 26, the sine, tangent, and secant of an angle we may say the cosine of an angle is the sine of the complement of that angle ; the cotangent of an angle is the tangent of the complement of that angle ; the cosecant of an angle is the secant of the complement of that angle. For in the triangle PAJL the angle APJ is the complement of the angle 4; and perpendicular AM Ars = hypotenuse = se as yy perpendicular AM _ ; tan APM = 0 an arp = °t4 } sec APJ = Tepprtense = ie cosec A. These results may also be expressed thus : the sine of an angle is the cosine of the complement of that angle ; the tangent of an angle is the cotangent of the complement of that angle ; the secant of an angle is the cosecant of the complement of that angle. TRIGONOMETRICAL RATIOS. 17 30. The Trigonometrical Ratios remain unchanged so long as the angle remains unchanged, Let BAC be any angle; in AC take any point P and draw Pf perpendicular to 4B; also take any other point P’ and draw P’M’ PM PM’ perpendicular to 4B. Then by similar triangles TPP that is, the sine of the angle A is the same whether it be formed from ZC Mae a“ Pe LP, A a Mt P? B the triangle APJ or from the triangle AP’M’. The same result holds for the other Trigonometrical Ratios. Or we may suppose a point P” taken in AB, and P’M” drawn perpendicular to AC’; then the triangles APJ and AP’” are PM P’M a ae ae * We now proceed to establish certain relations which hold among the Trigonometrical Ratios. similar, and ——; 31. We have immediately. from the definitions 1 1 tan A x cot A=1; therefore tan A eral cot A = ; 1 1 sec A x cos 4 =1; therefore sec d =, cos A = 5 1 ; 1 sin A’ md ~cosec A” PM PM All _ sin A AM AP” AP cosd’ AM AM PM cosd etd = py ap AP sad’ cosec A x sin A =1; therefore cosec 4 = Also tan 4d = 18 TRIGONOMETRICAL RATIOS. 32. To prove that (sin A)* + (cos A)? = In the right-angled triangle AP. we have PIP + AIP = AP’; al? M? therefore ee A = : PM\* /(AM\?* therefore Gra) ze (Gp) =1; a weiN that is: "2. ins (sin 4)’ + (cos 4)’ = 1. “BBr 4 PWith, respect to the preceding proof it should be re- markéd that it is shewn in Euclid, 1. 47, that the square described on the hypotenuse of a right-angled triangle is equal to the sum of the squares described on the sides; and it is known that the geometrical square described on any straight line is measured by the arithmetical square of the number which measures the length of the straight line. From combining these two results we obtain the arithmetical equality PMP + AM? = AP*, It must be observed that (sin A)’ is often written for shortness thus, sin? 4 ; similarly (sin A)* is written thus, sin? A. The~same mode of abbreviation is used for the powers of the other Tri- gonometrical Ratios, and so the result obtained in Art. 32 is usually written thus, sin’? A + cos? A =1, 34. To prove that (sec A)’=1 + (tan A)’, and (cosec A)? = 1+ (cot A)*. In the right-angled triangle API we have AP’ =AM?+ PM; AP? PM’ thereft —= 1+. erefore im Tet se therefore (@)- 14+ PM\* AL, AM that is (sec A)? = 1 4 (tan A)’, TRIGONOMETRICAL RATIOS. 19 Again, since AP? =PM’*+ AM, APS! 4, (AU (eur) =1* (Fiz) » that is (cosec A)? = 1 + (cot A)®. The results here obtained are usually written thus, sec’ A =1+4 tan? A, cosec? A =1+ cot? A. : 4 a 35. By means of the relations established i Kyte, 31...34 we are able to express all the other Trigonometrical Rytiss in terms. of any one of them; thus, for example, we WN exprgspy al pelle: rest in terms of the sine ; stanaineene cos A = ,/(1—sin* 4) (Art. 32); sin A sin A oan aaa = (1 - sin? 4) (Arts. 31, 32) ; _cosd_ ,/(1—sin’ A) ; cot A Sl (Arts. 31, 32) ; 1 1 ; sec A =A ~ (0 — sin? 4) (Arts. 31, 32) > 1 cosec A. Sor (Art. 31); vers A = 1-—cos 4 =1—,/(1-sin’ A) (Art. 32). Again, we will express all the rest in terms of the tangent ; A 1 1 1 7 tan A ~eosec A AJ(1 + cot? A) we ie ) 8 + tan” A) +— tan? A (Arts. 31, 34) ; 1 3 I = SS i 34); ass ~secA ,/(1 + tan? A) [pate ay I ‘ . cot A Fant (Art. 31); sec A =/(1 + tan’ A) (Art. 34) 3 1 J(1+tan?4) | 1 cosee d = = fends meosd laa ant) - 2—2 tan A” 20 TRIGONOMETRICAL RATIOS. We shall now proceed to determine the values of the Trigono- metrical Ratios for some specific angles. 36. To determine the values of the Trigonometrical Ratios fur an angle of 45°. Let BAC be an angle of 45°; take any point P in AC and Cc Wt B draw Pi perpendicular to AB. Since PAM is half a right angle 4 PJ is also half a right angle; therefore Pif= AU. Now PM + AIP =AP’; thus 2PM =AP*; 2 therefore (ep) ae PM | therefore re V2 ¥ PM 1 AM 1 0 , r Thus sin 45° = Ap= 2 3 cos 45° = ap* 7 ; » LM » AM tan 45° = ae eee — 45° — 0 sec =/2; cosec 45°= ae = 4/2 % ] vers 45°= 1 —cog 459 = 1 —— 2° Bo TRIGONOMETRICAL RATIOS. 21 37. To determine the values of the Trigonometrical Ratios for an angle of 60° and for an angle of 30°. Let APB be an equilateral triangle, so that the angle PAB A MM B contains 60 degrees; draw Pi perpendicular to AB, then AM=MB; therefore AM=}4AB=4AP. AM 1 Oe ae Thus cos 60°= p=} sin gor= (1 eost60")=, /(1-7)=,/ (3) a, 5 S60" fT ae, 7. ae _el. tan 60° = ego = og Fg N85 Cobb = Fa = 733 ot ; 1 2 O_ —9- o _ eigtoe! «oF sec 60) rey B08 = 23 cosec 60 sin 60° ~ 33 vers 60° = 1 —cos 60°= 5. And sin 30°= os 60°=35 008 30°=sin 60°=S; tan 30° = cot 60° = cot 30° = tan 60°= /3 ; sec 30° = cosec 60° = ee cosec 30° = sec 60° = 2 ; yers 30° = 1 —cos 30°= ae, Lo lo EXAMPLES. CHAPTER III. 38. The student should render himself perfectly familiar with the values of the Trigonometrical Ratios for an angle of 30°, 45°, or 60°; as they will be perpetually used in the subject. Thus, for example, if an angle of 60° occurs it may be necessary to have the cosine of this angle, which has been found to be 4. And con- versely, if the cosine of an angle is known to be 3, and the angle is less than a right angle, the student will immediately infer that the angle contains 60°. Should there be any difficulty in this in- ference it will be removed by the remarks made hereafter, in which it will appear why we introduce the restriction that the angle is less than a right angle. See Art. 44. It may be observed that if an angle be less than 45° the cosine of the angle is greater than the sine, and if the angle be greater than 45° and less than 90° the cosine is dess than the sine ; these results follow immediately from the triangle PAM (see figure in Art, 26) since the greater side in a triangle is opposite to the greater angle. EXAMPLES. 1. The sine of a certain angle is 2 ; find the other Trigono- metrical Ratios of the angle. 2. The tangent of a certain angle is . ; find the other Tri- gonometrical Ratios of the angle. ; 5 : 2 3. The cosine of a certain angle is af 33 find the other Tri- gonometrical Ratios of the angle. 4, Shew that sec?0 cosec’6 = tan?6 + cot?6 + 2. 5. Shew that sin’@ tan 6+ cos’ cot 6 +2 sin 0 cos 0 = tan 6 + cot 6. 6. Shew that 2 (sin’@ + cos’@) — 3 (sin’@ + cos*6) + 1=0. EXAMPLES. CHAPTER IIL. 23 Obtain solutions of the following equations : a 2 sin’@ = 3 cos 0. 8. siné+cos 6=1. 9. cot @=2cos 6. 10. sin'@—2 cos 9 + 5=0. Il. 3sec*6+8 = 10 sec?6, 12. tané+cotd=2. 13. Given sin (A — B) = ; , and cos (4 + B)= : » find 4 and 2B. 14. Given tan (4 + B)=,/3, and tan (4 —&)=1, find A and B. Iv. APPLICATION OF ALGEBRAICAL SIGNS. 39. In the preceding Chapter we defined the Trigonometrical Ratios, and established certain relations between them; we con- fined ourselves to angles not exceeding a right angle. We shall now extend the definitions so as to render them applicable to an- gles of any magnitude; the relations which were established will then also be found to be true for angles of any magnitude. 40. Let O be a fixed point in a fixed straight line, and sup- pose we have to determine the positions of other points in this MM’ O M straight line with respect to 0. The position of any point in the straight line will be known if we know the distance of the point from O, and also know on which side of O the point lies. Now it. is found convenient to adopt the following convention: distances. measured in one direction from O along the fixed straight line will be denoted by positive numbers, and distances measured in the opposite direction from O will be denoted by negative numbers. Thus, for example, suppose that distances measured from O towards the right hand are denoted by positive numbers, and let M be a point the distance of which from O is denoted by 2 or + 2; then if M’ be as far from O as Mf is, and on the other side of O, the dis- tance of Jf’ from O will be denoted by — 2. \ 24 APPLICATION OF ALGEBRAICAL SIGNS. 41. We have called this method of determining position by means of numbers affected with algebraical signs a convention; we mean by this word to indicate that it is not absolutely necessary to adopt this method, but merely convenient. The symbols + and — are defined in the beginning of elementary works on Algebra as indicative of the operations of addition and subtraction respectively. As the student advances in Algebra he finds that the symbols + and — are also used as indicative of the qualities of quantities ; and that no contradiction or confusion ultimately arises from this double mode of considering the symbols, but that Algebra gains thereby considerably in power. (See Algebra, Chaps. V. and XIV.) It may be remarked, that we are at liberty to take either of the two directions from O as that which will be indicated by positive numbers ; but when the selection has been made, we must adhere to it throughout the investigations on which we may be engaged. 42. Let OB, OC be two straight lines which meet at right Ne P B a a B f ¢ angles ; produce BO to any point B’ and C/O to any point C’. Let P be any point in the plane containing the two straight lines, The position of ? will be known if we know the distance of P from each of the straight lincs BB’ and CC’, and also know on which APPLICATION OF ALGEBRAICAL SIGNS. 25 side of each of these straight lines it is situated. Draw PIM and PN perpendicular to the straight lines BB’ and CC" respectively. We shall adopt the following conventions: the distance ON or PIL will be denoted by a positive number when P is above the straight line BB’, and by a negative number when P is below the straight line BB’; the distance OJf or PN will be denoted by a positive number when P is to the right of CC’, and by a negative number when P is to the left of CC’. 43. 18 positive or Ms 2s . negative ; and also whether cos > is positive or negative ; thus we know which sign is to be taken with each radical quantity. Or if Ae we merely know in which quadrant the angle 5 lies, we can ig Aa determine the proper signs ; for example, if 3 isan angle between 180° and 270°, both its sine and cosine must be negative quantities. 98. By Art. 82 sind =2 sin $ oe 9 ? eg also 1 =sin’ at cos? 3? 2 thus (sing + cos) = =1l+sin A, 2 2 . A A\? ; and (sing — cos 5) =l-sin4; «A A : therefore sin 5 + cos 5 = aif CL APSA) ss asaish waliaal ati (1), _ A A 3 and 7 — C085 = LABIA) ois gaeetrece ern (2); therefore 2 sin 4 5 =A(1+sin A) +,/(1 —sin A), and 2 a = ,/(1+ sin 4) —,/(1 —sin A). 99. Since we may suppose either the positive or negative sign to be placed before each of the radical quantities in equations (1) and (2) of the preceding Article, we seo that corresponding to one ‘ A value of sin A there are four values for cos 5 and four values for FORMULE FOR THE DIVISION OF ANGLES.’ 65 “A sings which has a certain sine, then the formula mm + (— 1)" a includes all the angles which have the same sine; therefore any expression a 2 to give the value of the sine of every angle included in the formula d{nm + (—1)" a}. and the reason of this may be assigned. For if a bean angle which gives the value of sin= in terms of sin a may be expected First suppose 2 even and equal to 2m; then a 5 + cos mz sin a 2 sin 4 {nw + (— 1)" a}=sin (mn + 5) = sin m7 Cos = cos mm sin ¢ =+sin > - 2° 2° Next suppose 7 odd and equal to 2m+1 ; then 7a 7a sin} {n+ (—1)"a}=sin (mx + 75" )ssinmmeos "7% cosmmsin — -- Wis - 7T-Q - 7T-a = cos Mr soar =+ sin 3 ==+ cos Thus four values occur for the sine of half an angle when the sine of the angle is given. Similarly any expression which gives the value of cos 5 in terms of sina, may be expected to give the value of the cosine of every angle included in the formula 4 {nw + (— 1)" a}. First suppose 7 even and equal to 2m; then a a . - @ cos 1 {nm + (— 1)" a} = cos (ma + 5) = cos mm cos 5; — sin mm sin 5 COS mur COS | = + COS = = COS Mr COS = = = COS. 2 2 Next suppose 7 odd and equal to 2m+1; then —-a Ta i 1 in T . a Fe cos 3{nr+(—1)"a}=cos ( mx + — — )=cos mmcos—5——sinmrsin —— ad ol Ta T-a ss) = Cos mm cos —>— = + cos —5—- = + SIN 5 al om 66 FORMULE FOR THE DIVISION OF ANGLES, Thus four values occur for the cosine of half an angle when the sine of the angle is given, 100. If sin A only be given and nothing more be known respecting A, then the ambiguities of sign which occur in Art, 98 cannot be removed, If however A itself be given, or if we merely know in which quadrant the angle A lies, we can determine the proper signs; for in any particular case we may proceed thus, We have _ A A : sin 5 + cos 5 =+ pO PSA), aise a. see ccee «Seen (1), sin? — cos = f= si A) ses sarves corse sacees (2). Now suppose, for example, that A lies between 0 and 90°, then hes between 0 and 45°; thercfore cos : and sing are both positive and cos : is greater than sing 3; hence the left-hand member of (1) is a positive quantity, and we must therefore take the positive sign in (1), and the left-hand member of (2) is a negative quantity, and we must therefore take the negative sign in (2). Therefore if A lies between 0 and 90°, we have —/(1-sin 4); sinS + cos =+(1-+sin 4), sin 4 ~ cos 4 = 2 2A ‘ : therefore 2sin 5=+ J(1 + sin A)- ,/(1 — sin A), 2 cos $= +/(14+sin 4) + /(1 —sin A). For another example, suppose that A lies between 270° and 360°, A, ‘then 5 lies between 135° and 180°; therefore cos is negative, oad oe A, : : and sin zis positive, and cos 3 8 numerically greater than ere hence the left-hand member of (1) isa negative quantity, and wy must therefore take the negative sign in (1), and the left-hand FORMULE FOR THE DIVISION OF ANGLES. 67) member of (2) is a positive quantity, and we must therefore take the positive sign in (2), Therefore if A lies between 270° and 360°, we have sin 4 Ace Sill +sin A), sin - cos = + J(1-sin 4); therefore 2 sin = =—A(1 +sin A) +,/(1-sin 4), Qeos 4 =~ y(l +sin A) — (1 -sin A). 101. It is easy to give general formule for determining the : afi A - A signs of sin 3 + cos > and sin > — Be ‘i A A 1 Ay A +3); For sin 5+ cos y= 2 (Sy8img + 30085 3)- Jasin (5 + Z now sin G + 7) is positive if 4, +7 = lies between Qn and (2n + 1), and negative if = + a lies between (2n + 1) 7 and (2n + 2) a, where : ; is ‘ aa alk wm is zero or any integer positive or negative. Thus sin x F008 5 is positive if = lies between 2na — i and 2nm + Sr, and negative if P or Ur = lies between 2am + a and Qn +7 A sis . A A ‘ 7 Similarly sin 5 — cos 5 = /2 sin G - z) ; and hence we can infer that sin — cos 4 is positive if . lies between 2m + 5 and 9 2m + > , and negative if . lies between 2m + = and 2nur + = ‘ where m is zero or any integer positive or negative. We will apply this to an example: required the limits between which = must lie in order that asin =~ J/(l aaa maid 68 FORMULE FOR THE DIVISION OF ANGLES. To obtain this result the lower sign must be taken in (1) and in (2) of Art. 100; thus from (1) we infer that = must lie between A dae + and 2am + = 3 and from (2) we infer that — y must lie between 2nvr + and 2mm + or hence, combining these results, we see that 4 must lie between 2na + a and 2am+ i, where n is zero or any integer positive or negative. 2 tan 9 102. By Art, 85, tan A 7) 1— tan’ = 2 ‘ ~A A putcfor tan d;thus ec tan gt tan yZ -¢=0; 2 therefore tan a = oa . 103. The reason why two values occur in finding the tangent of half an angle when the tangent of the angle is given, may be assigned as before. For if a be an angle which has a certain tan- gent, then the formula m7 +a includes all the angles which have the same tangent; therefore any expression which gives the value of tan = in terms of tan a may be expected to give the value of the tangent of every angle included in the formula 4 (nm + a). First suppose even and equal to 2m; then 3° Next suppose 7 odd and equal to 2m +1, then tan 3 (nw + a) = tan (mn + 5) = tan 5 2 2 2 Thus two values occur for the tangent of half an angle when the sone of the angle is given. tan 3(n7+ a) = tan (ms + a) a tig ete G + ae cobs. FORMULE FOR THE DIVISION OF ANGLES. 69 104. If tan A only be given and nothing more be known respecting A, then the ambiguity of sign which occurs in Art. 102 cannot be removed. If however A itself be given, or if we merely know in which quadrant e lies, we know whether tan 4 is positive or negative, and thus we know which sign we must take, 105. By Art. 91, cos A = 4 cos® : -3 cos i Thus if cos A be given we have a cubic equation for determining cos ‘ ; and the reason for this may be assigned as before. For if a be an angle which has a certain cosine, then the formula 2n7 +a includes all the angles which have the same cosine ; therefore any a : 3 expected to give the value of the cosine of every angle included in the formula }(2n7+a). Now n is of one of the forms 3m, 3m+ 1, 3m—1. First suppose n= 9m ; then expression which gives the value of cos = in terms of cos a may be a a cos 4 (2n7 +a) =cos (2mm 3) = C08 3. Next suppose n= 3m+1; then Qr+a Qn =a, cos 4 (2n +a) = cos (2s + ) 8 ; 3 => Lastly suppose x = 3m—1; then — is. cos 4 (2n7 +a) =cos (2mn— 27% *) _, Iara 3 3 9 Thus three values occur, namely cos 3 » COs or =, > Se 106, By Art. 91, sin d= Ssin4—4sin? Thus if sin A be given, we have a cubic equation for determining sing ; and the reason for this may be assigned as before. 70 EXAMPLES. CHAPTER VII. EXAMPLES, 1. Shew that 2 ein 4 =—,/(1+sin 4)~,/(1-sin 4), when A lies between 450° and 630° 2. Obtain cos a in terms of sin A when : lies between 405° 9 and 495°. a 3. Obtain sin $ in terms of sin A when 2 lies between ~— 45° and — 135° 4. Determine the limits between which A must lie in order that 2 sin lf =— /(1+sin 24) + ,/(1 -sin 24). 5. Determine the limits between which A must lie in order that 2 cos d=— ,/(1+sin 24) + /(1-sin 2.4). 6. Determine the limits between which A imust lie in order that 2 sind = ,/(1+sin 24) —,/(1 —sin 2.4). 7. Divide a given angle into two parts whose sines shall be in a given ratio. 8. Divide a given angle into two parts whose cosines shall be in a given ratio. 9. Divide a given angle into two parts whose tangents shall be in a given ratio. 10. Given tan = = 2-,/3, find sin A. 11. Given sin 210° =~ a find cos 105° 12. Given tan 24 =-=, find sin A and cos A. 13. Find tan 165° from the known value of tan 330°. _ EXAMPLES. CHAPTER VIL 71 A 2sin A —sin 24 2 2sin 4 +sin 21° 0 0 15, vers (180° - A) =2 vers A vers ie = S a 14. Shew that tan’? 16. (cos «l +cos B)’ + (sin A +sin B)’ = 4 cos? aa li. ced Sos epee amb) 4 sin? = 18. Shew that sin 2 ay — VW Vv?) os ony = YC +A?) and tan 224°= 2-1. gz 19. (tan A + cot A) 2tan$(1 - tan? 5) = ( + tan?) : A\ secd+tan A 9 of™ ,4\_ Sec 4+ tan. oe G i 3) “sec d — tan A° 21. sin 3) +208 G- 3) = _sind 7 ‘Ge 2 4 ~ (vers 6)" 22, Shew that 4sint £ (1 sin $) ={1—y(l +sin 0)}. 2 a ae aot — +.os ee 23, os s* = + cos" 7+ g =: 24. tan 7}=/6—-/3+/2-2. 25. tan 1423°=24+,/2—,/3—,/6. 26. If tanw=(2+ /3) tan 5 , find the value of tan x. Lo ~~ If a= (» + : sk i) zw, where n is any integer, find the value of tana+cota. cosa cos 13a 8. Ifa=s,, find the value o haa ae oO L 29. Ifsec(p+a)+sec(p—a)=2 see $, then cos ¢ =,/2 cos 5 : Ll+e\$ _ cosp-c 30. If ton 5 =(7*" *tan ©, shew that cos O= 5 (-tcae 72 MISCELLANEOUS PROPOSITIONS. VIII, MISCELLANEOUS PROPOSITIONS. 107. To find the sine and the cosine of an angle of 18°. Let A denote an angle which contains 18°, then 2A contains 36° and 34 contains 54°; hence, by Art. 29, sin 24= cos 34, therefore 2 sin A cos A = 4 cos® A — 3cos 4; divide by cos A, thus 2 sin A=4 cos’ dA — 3=1—4 sin’ 4, therefore 4 sin? 4+2sin 4-~1=0; by solving this quadratic equation we obtain —l+ Jo z Since the sine of an angle of 18° is a positive quantity we must take the upper sign, therefore sin 4 = fs 1802 wee and cos 18° = ,/(1 —sin? 18°) = ey 108. To find the sine and the cosine of an angle of 36°. 2 - cos 36°=1~2sin’ 18’=1-2( =") =1- So 2? sin 36° = ,/(1— cos? 36°) = _ 109. Hence the values of the Trigonometrical Ratios for angles of 54° and 72° are known; for sin 54° = cos 36°, cos 54° = sin 36°, sin 72° = cos 18°, cos 72° = sin 18°, _ 110. The reason why more than one result was obtained in Art. 107, is that the equation sin 24=cos3A is true for some other values of A besides 18°, This equation may be written cos (90° — 24) = cos 34, ‘MISCELLANEOUS PROPOSITIONS. 73 Hence we conclude that 90°—24 must either be equal to 34 or to one of the angles which have the same cosine as 34; thus every admissible value of A will be found from the equation —2A =n. 360° 34; where 2 is zero or. any integer positive or negative ; 90°”. 360° 243 For example, if n= 0 and we take the lower sign in the de- nominator, we obtain 4 =—90°; this value of A makes cos 4 =0, and thus we see a reason for the appearance of the factor cos A which was removed by division in Art. 107. Again, if we put 2 =1 and take the upper sign in the denominator, we obtain A= 8 54; and sin (— 54°) = —sin 54°=—cos 36° = ie thus A= and thus we see a reason for the appearance of the other root in the quadratic equation of Art. 107, besides the root which we used. 111. To find the sine and the cosine of an angle of 9°, and of an angle of 81°. By Art. 100, sin 9°+ cos 9°= ,/(1 + sin 18°) = V3 + V5). 9 > sin 9° — cos 9°=— ,/(1 — sin 18°) --MG—/) MB-+4/5)~yf(5~ a5) 4 , YOO 8) + M5— J), therefore sin 9° = cos 9° = And sin 81°=cos 9°, cos a sin 9%, We have now found. expressions for the sines and the cosines of the following angles, 9°, 15°, 18°, 30°, 36°, 45°, 54°, 60°, 72°, 75°, 81°. (See Arts, 36, 37, 92, 107, 108, 111.) Since 3°=18°—15°, we can obtain the sine and the cosine of 3° from those of 18° and 15° by Art. 77; and then by means of Art. 76 combined with results already obtained, we can easily find 74 MISCELLANEOUS PROPOSITIONS. the sine and the cosine of any angle comprised in the series 3, O08 TOF as 112. In Arts, 82and 91 we have given expressions for sin 24, cos 24, sin 34, and cos 34 in terms of sin. A and cos 4; we may also express the sines and cosines of 44, 5.4,,.. in a similar way. For sin (x + 1)A + sin (n—1)A = 2s8in nA cos A; therefore sin (n+ 1)4=2 sinnA cos d — sin (n—1)4; 2=3; thussin44 =2sin 34 cos A—sin 24; n=4; thussin5d=2 sin 44 cos 4 —sin34 ; 1 so on; thus we can find in succession sin 44, sin 54,..., in ms of the sine and cosine of A. Similarly, the formula cos (x +1) 4+ cos(n—1)A=2cosnJd cos A, may be used to find in succession cos 4.4, cos 54,... This subject will be considered again hereafter, and we shall then give general formule for the sine and the cosine of mA in terms of the sine and cosine of A for any integral value of n. 113. It is easy to find expressions for the Trigonometrical Ratios of any compound angle in terms of the Ratios of the com- ponent angles. For example, sin (4+ B+C)=sin (A +B) cosC + cos (4 +B) sin C =sin A cos B cos C+s8in B cos C' cos A +sin (cos A cos B—sin A sin Bsin C. Cos (4 + B+ C) =cos (A + B) cos C—sin(A + B) sin € =cos A cos Bcos C—cos A sin Bsin 7 —cos B sin A sin C —cos C' sin A sin B. sin(d + B+C) cos(4+B+C) sin A cos Beos(' +sin Bos C'cosA +sin C'cosA cos B—sinA sin BsinC : cos.A cos Bcos C—cosA sin BsinC@ —cos B sin A sin C—cos C'sinA sinB’ divide both numerator and denominator of the last expression by Tan (4+B+C)= MISCELLANEOUS PROPOSITIONS. 5° cos A cos B cos C'; thus we obtain tan A +tan B + tan C —tan A tan B tan C sa ene mF bad tan C—tan C tan A — tan A tan B’ Suppose B and C' each equal to 4; thus we have, asin Art. 91, 3 tan A — tan? @—3tan’A 114. When three or more angles are connected by some relation, we may often find that some simple relation exists among some of their Trigonometrical Ratios. : For example, if 4+B+C=180°, then will sin 24+sin 2B +sin 20 =4 sin A sin B sin C. For sin 24+ sin 2B = 2 sin (A + B) cos (A + B) =2 sin C cos (A-B) and sin 20 =2 sin'C cos =—2sinCcos(4+B), (Art. 48); therefore sin 24 +sin 2B + sin 2C' =2 sin C {cos (A — B) - cos (A + B)} =4sin (sin A sin B. Again, if 4+ B+C=180", then will cos A + 008 B +0080 =1+4sin4 sin 2 sin A+B A-B C A-B For cos.4 + cos B = 2 cos —;— cos —5— = 2 sin 5 co8 ——; 2 2 3 2 tan 3.4 = ' cos'C = 1 — 2 sin’ c therefore > cos A + cos B + cos-C' = Lede cos 458 - sin ) 4 ¢ee,0/ado2 Hee = 2 sin (cos 9 — cos 2 =1 Pees ee eee. 2 2 2 Again, if A+ B+C = 180", then will tan A +tan B+ tan C= tan A tan B tan C. For tan 180°=0, therefore tan(d4+B+C)=0; and therefore by Art, 113, tan A + tan B+tan C'—tan A tan B tan C= 0. and 76 MISCELLANEOUS PROPOSITIONS. Again, by Art. 118, 1—tan B tan ( —tan C tan A —tan A tan B cot (4 +B +0) = tan A+tan B+ tan C—tan A tan Btan@ ? now cot 90°=0; hence if 4+ B+C = 90°, then will 1=tan Btan C + tan Cytan A + tan A tan B, The relations which have been obtained in the present: Article hold whether A, B, and C are positive or negative provided that A+ B+C=180°; thus they hold if 4, B, and C’ are the angles of atriangle: but this is of course a particular case, as the angles of a triangle are all positive quantities. Any relation which has been found on the supposition that A+B+C=180° will also hold when we change A, B, and C’ re- spectively into 90° 4 , 90°— a and 90° $ : for on the suppo- sition adopted the sum of the last three angles is equal to 180°, 115. For another example, suppose we have to investigate what relation must exist among the angles 4, B, C, in order that cos’?A + cos’ B + cos’C' + 2 cos A cos B cos C — 1 may be zero. cos?.A + cos’ B + cos’C’ +2 cos A cos B cos C’—1 = (cos A + cos B cos C)* + cos’ B + cos’C — 1 — cos’ B cos’ = (cos A + cos B cos C’)* — (1 — cos’ B) (1 — cos’) = (cos A + cos B cos C’)’ — sin’ B sin’? = (cos A + cos B cos (+ sin B sin C’) (cos A+ cos B cosC — sin B sin) = {cos A + cos (B—C)} {cos A + cos (B + C)} A+B-C A-B+C A+B+0 B+O-A =4 cos 3 cos 5 co; 3 cos oor Hence in order that the proposed expression may be zero, one of the four cosines last written must be zero, and thus one of the four compound angles must be some odd multiple of a right angle. EXAMPLES. CHAPTER VIII. 77 MISCELLANEOUS EXAMPLES. Prove the following formule ; cos (a + B+) cos a cos B cos y sin (a +B +) cos a cos B cos y 3. sin (a—) +sin (8 pee ae oF si tale “3 sin F a 4, 4 sin (9—a) sin i: —a) cos (6 — mé) = 1+ cos (20 - 2m) ~ cos (20 — 2c.) — cos (2m — 2a). sin (a +8) cos B —sin(a + y) cos y = sin(B—-) cos(a+B+y). cos (a+ 6 + y) + cos (a+ B—y) + cos (a+y—f) + cos (6 + y—a) = 4 cosa cos B cos y. =1 -tan ® tan y— tan y tana — tana tan f. Lo = tana + tan B+ tan y — tana tan B tan y. * 4d gine 8 7. cos 2a + cos 28 + cos 2y + cos 2 (a +B +) = 4 cos (a +B) cos (8 + y) cos (y +a). 8 sina si sin 8 ~" sin(a—£)sin(a—y) sin(@—y) sin (8—a) sin y "sin (Y= «) sin (yA) ~ 9. cos (a+ 8) sin R—cos(a+y) siny = sin (a + £) cos B — sin (a + y) cosy. 10. sin (a + — 2y) cos 8 —sin (a + y— 28) cosy =sin (8 — y) {cos (8 +y— a) + cos (a+ -y—) +c0s (a+ 8 —y)}. ll. sin(a+@+y) sin B=sin (2 + @) sin (B+ y) —sinasiny. 12. sinasin @ sin (8 —a) + sin sin y sin (y—8) +sin y sina sin (a—y) + sin (@—a) sin (y—) sin (a-y) =0. 13. cos (a + f) sin (a — 8) + cos (8 + y) sin (B- y) + cos (y + 6) sin (y—8) + cos (8 + a) sin (6—a)=0.. r 14, EXAMPLES. CHAPTER VIII. sin (3 — f) sin (a — y) + sin (8-7) sin (a— 8) + sin (y—8) sin (a— B) =0. If 4+ B+C =z, prove the following formulz contained in the examples from 15 to 35 inclusive. 15. 16. 17. 18. 19. 29. 30. Cc A B C A B cot gee ee ga Obs cot 7 Cot 5: 2 sin A+ sin B+sin C= 40e0s 4 cos cos &. : ; eh . A Bes sin A —sin B+sin (= 4sin 5 cos 5 sin 5 - cos 2.4 + cos 2B + cos 20 +4 cos A cosBeosC( +1=0. cos 44 + cos 4B + cos 40+ 1=4c0s 2A cos 2B cos 2C. A B é aw—A a—B ar—-C cos— + cos = + cos = =4 cos 5 3 5 a Og cos —[—. ee See ee eS aE 2 2 ee 4 4 4 oe ee oe ge ee ee 2 2 4 4 4 sin? A + sin?B + sin’C — 2 cos A cos B cos C = 2. sin? 2.4 + sin? 2B +sin?2C + 2 cos 24 cos 2B cos 20 = 2. tee ee eta fi’ Stig tanta, gS Ass 29 sinA+sinB-snd_, 4, 2B mises 6 1 +cos A cos B cos C =cos A sin B sin C + cos B sin A sin? + cos C sin A sin B. cot A + cot B+ cot C= cot A cot B cot C +cosec A cosec B cosec C. ee _ (sin B + sin C ~sin 4) (sin C + sin A — sin B) 4sin A sin B The expression cot 4 + — Binal will retain the same sin B sin C 8 value if any two of the quantities A, B, C, be interchanged. EXAMPLES. CHAPTER VIII 79 A B tan — tan 5 tan a s 3]. tan A +tan B+ tan C = 2 ; (sin 4+sin B+sin C)? 2 cos A cos B cosC 32. sinnd +sinnB+sinnOl =4 sin cos ee cos - cos - ee if m be an integer of the form 4m +1 or 4m +3. 33. snnA+sinnB+sinnO = + si 7 sin = sin — = ‘if ~ be an integer of the form 4m or 4m+ 2. 34. de aii iy oA goo At Ae : g + C08 3 + cos 5 = 4 cos 4 — cos —] or tan A tan B tan@ tand tan B tan? tan B* tan 0 *tand ‘tanC *tand ‘tnd =sec A sec B sec C — 2. 36. If the sum of four angles be two right angles, the sum of their tangents is equal to the sum of the products of the tangents taken three and three. tan (A — B) 3 tan A a 35. 37. If eo prove that tan 4 tan B = tan’C. tan?a _ cos B (cos z — cos a), = anne an* 8 ~ cos a (cos @ — cos B)’ ot @ Gh 2B shew that tan’ 35> = tan 5 tan 3° 29 _ co8a ‘egy cosa’ tan§ tana pee aE oe a? ae = Ge ae tan 6’ tana’ , shew that tan’ 5 tan? 5 = tan® 5° 40. If cosa=cos B cos f= cos f’ cos ¢’, and , / sina=2sin Ssin ©, shew that tan? tan? 8 tant : sin(a—f) _ sin(a+6) ‘tia 41, If ae ae , shew tha’ cot B — cot @ = cot (a + 6) + cot (a—f). 80 EXAMPLES. CHAPTER VIII: tana tan B\* tan B 2 Se ) — tan?a — tan? ek 42. If (Ga a ee 5) tan?a —tan’8, then cos 6 iene 43, If tandé=cos@ tana, and tana’ = tan 6 sin ¢, ata a—a ¥ ae ig Risers then one value of tan 5 is tan 5 tan 3 44, Find the relation between the angles a, 8, y, when the cosines are connected by the relation 1 — cos’a — cos’ B — cos’y + 2 cosa cos B cos y = 0. = +a) _ tan me é =e ars then will 45. If aay (a B) + USE sin’ (B— 7) +222 sint (y— a) = 0. 46, if oe + Sooo), and S20 _ at shew that tin = a. nb gee then cola B= rea 6 y = ne 6 shew that one of the d. 6 € 5): 48. Having given tang = values of tan 5 ig tan 5 tan a9 49. Given cos @=cosacosf, cos 6’ = cos a’ cos B, yo tang tan == = tan os shew that sin’@ = (sec a—1) (sec a’— 1). 50. Having given that sin(B+C — A), sin(C +A —B), and sin(4 +B-C) are in arithmetical progression, shew that tan A, tan BZ and tan C, are in arithmetical progression. 51. If the sines of the angles of a triangle be in arithmetical progression, the cotangents of the half angles are also in arith- metical progression, EXAMPLES. CHAPTER VIII. 81 52. If the sum of the squares of the cosines of the angles of a triangle = 1, the difference between the greatest and least angle is equal to the mean angle. 53. If A+ B+C=180, and sin (4+9)=nsin Y, A Bo n-l ey that tan 5 tan > =o 4° 54. Pai 4 Cries see x y z then (« -y) cot © + (y—2) cot ++ (e~ a) cot 2 =0, 55. IfA+B+C =m where m is any integer, then tan 4 + tan B+ tan C = tan 4 tan Bian C. 56. Shew that ifa, 8, y, and x are any angles sin (2a +a) +sin (28 + x) +sin (2y +a) —sin (2a + 28 + 2y + 32) =4sin(a+ 6+) sin(B+y +2) sin (y+a+ 2). 57. Fyrom the preceding result deduce two special cases by supposing respectively that «= 0 and that x = - ; and from these cases obtain the first two relations of Art. 114. 58. Ifa, B, y be any angles, shew that Qa sina +sin 8 +sin y—4 cos 5 cos 5 cos 5 _ a in S*B Y= {05 Bas B YA ogg P= Byte 38y—a-B+a a+B+y—7) + cos 4 + cos 4 . 59. Express cos 56 in terms of cos 0. 60. Shew that sin 60 =2 sin @ (16 cos’ @— 16 cos’ 4+ 3 cos 6). T. T. ( 82 ) IX. CONSTRUCTION OF TRIGONOMETRICAL TABLES. 116. If 6 be the circular measure of a positive angle less than a right angle, 0 is greater:than sin @ and less than tan 0. Let AOB be an angle less than a right angle and let OB=04; from B draw BM perpendicular to OA and produce it to C so that MC=MB; draw BT at right angles to OB meeting OA produced at 7’, and jom CT and OC. Then the triangles MOC and MOB are equal in all respects, so that the angle 7OC'=the angle ZOB; therefore the triangles TOC and TOB are equal in all respects, so that 7CO is a right angle, and 70 = TB. With centre O and radius OB describe an arc of a circle BAC; this will touch BZ at Band CZ at C. Now we assume as an axiom that the straight line BC is less than the arc BAC’; thus BM the half of BC is less than BA the BAL BA half of the arc BAC’; therefore OB is less than ——= OB} sine of AOB is less than the circular measure of AOL. B that is, the Cc Again, we assume as an axiom that the arc BAC is less than the sum of the two exterior lines BZ and TC ; thus BA is less than ’ CONSTRUCTION OF TRIGONOMETRICAL TABLES, 83 BT; fierefone o> is less than 2: that is, the circular measure of OB OB? AOB is less than the tangent of AOB. Hence sin @, 6, and tan @ are in ascending order of magnitude if 0 be less than 3 ‘ 117. We have assumed two axioms in the preceding Article ; the first is so obvious that it will be readily admitted; but the second is more dificult. The student is recommended to postpone this point for future consideration. It is however easy to shew that the assumption may be made to depend upon another almost identical with that which we have already been compelled to make in Art, 14. For divide the are BAC into any number of arcs and draw tangents at the points of division; then from the fact that two sides of a triangle are greater than the third, it follows that the perimeter of the portion of a polygon thus formed, is less than the sum of BY and TJC by a jinite difference. Moreover this perimeter diminishes as the number of points of division is increased. Now assume as in Art. 14 that the perimeter of the polygon can be made to differ as litile as we please from the are BAC by sufliciently increasing the number of sides and diminishing the length of each side; thus it follows that the arc BAC is less than the sum of BZ and TC. 118. The limit of = unity. For sin 6, 6, and tan 6 are in ae order of magnitude ; when 0 ts indefinitely diminished is divide by sin 6; therefore My ks sae? and —— a 4 are in ascending order 1 of magnitude. Thus a 5 lies in value between 1 and ee? but when 6 is zero, cos @is unity; hence as @ diminishes indefinitely z approaches a approaches the limit unity. Therefore also sue the limit unity. 6—2 84 CONSTRUCTION OF TRIGONOMETRICAL TABLES, fan nD is Tale ot mae eee a de 0 ~“ cos6’ indefinitely diminished is also unity. And as 119. From the preceding Article we see that the limit of : a . . . . msin — when m increases indefinitely is a. am . a _ a @ bela : For msin —=asin — +— ; and when m is indefinitely great m mm ao 2 Bes sin — +— is unity. mm Similarly the dimit of m tan ~ when m increases tudefinitely isa, m é must be carefully remembered that in the important propo- sition of the preceding Article, 6 is the circular measure of the angle considered. If any other unit of angular measurement be adopted instead of the unit of circular measure, the limit under consideration will not be unity. For example, let us find the limit sin n” 2 of when 1 is indefinitely diminished. Let 6 be the circular nr 180? thus measure of an angle of n degrees, then 6 = sinn® sind 7 sinf n 18U 180 0 Now when 2 diminishes indefinitely, 6 does so also, and the limit. sin 4 6 indefinitely is a , which is the circular measure of an angle of of : ; dod sin n° hha ene is unity; hence the limit of a when 7 is diminished when one degree. Similarly we may prove that the limit of — . de n is indefinitely diminished is the circular measure of an angle of one minute; and soon. Thus we shall find that, whatever be the unit of angular measurement, the limit of the ratio.of the sine of CONSTRUCTION OF TRIGONOMETRICAL TABLES; 85 an angle to the angle, when the angle is indefinitely diminished, is the circular measure of the unit. 120. If @ be the circular measure of a positive angle less than 1s a right angle, sin 0 ts greater than 6 = ‘ For sin @ = 2 sin e cos a and tan 5 a is greater than 5 e , therefore 3 3? 6 0 , : sing is greater than = cos=; therefore sind is greater than 20° 2? 2 él cos” . that is greater than 6 cos? : , that is greater than 2 3 a(1 — sin’ 5) . And ain? ? is less than () , therefore a fortiori 3 ig 6 sin @ is greater than 6 ( — ts that is, sin 0 is greater than 6 — i: 121, Thus we see that if @ lie between zero and a right 3 angle sin 6 is less than @ and greater than 6—- 7 ; and therefore 6 6 sing is less than : and greater than eG 2 Now cosd=1-—2 sin? 9 Thus cos @ is greater than 1 —2 (3) ; 3 2 that is greater than 1— a Also cos 6 is less than 1— 2 (5- 5): e, 6+ 6 g that is less than 1 - = +46 -2 (5: ay therefore a fortiori cos 6 is ae oo less than 1 — +i 6" 122. To calculate approximately the sine of 10”. 107 The circular measure of 10” 18 739 x60 x60 , that is ——_ GanD0 ‘ therefore the sine of 10” is less than ET and greater than 86 CONSTRUCTION OF TRIGONOMETRICAL TABLES. T 1 7 \8 ; 64800 - 7 (sama) . If we take for 7 the approximate value ” 3°141592653589793...we find ——— caso = = 000048481368110...; the sine of 10” is therefore less than this decimal fraction. And———~ $4800 a 00 is less than ‘00005, therefore @ fortiori, sin 10” is greater than -000048481368110...... — 7 00005)"; that is, sin 10” is greater than *000048481368078...... We have thus found two decimal fractions between which sin 10” must lie, and these decimal fractions agree in their first twelve figures; therefore we may say that sin 10” = 000048481368...... 1 ] 10% . The value of cos 10” may then be found approximately since it is ,/(1—sin? 10”); or we may make use of the results established in Art. 121. Thus it will be found that as far as thirteen places of decimals we have cos 10” = -9999999988248...... 123. It appears from the preceding Article that as far as twelve places of decimals we have sin 10”=the circular measure of 10”; and in the same way we may shew that sin 1” =the cir- cular measure of 1” very approximately. And if n be any small number of seconds, we shall have approximately sin n” = the circular measure of »”=m times the circular measure of 1”=nx sin 1”. and we are certain that the error is less than —, the circular measure of n” Thus n= 77 sin 1 number of seconds in any small angle is found approximately by dividing the circular measure of that angle by the sine of one second, approximately; that is the 124, We shall now shew how to calculate the sines of angles which form an arithmetical progression having 10” for the common difference. CONSTRUCTION OF: TRIGONOMETRICAL TABLES. 87 Let a denote any angle, then sin (n+1)a+sin (n—1)a=2 sin ne cosa; suppose 2 cosa = 2-4, then sin (n+ 1)a+sin (n—1)a=(2~—4) sin na, therefore sin (n +.1) a—sin na = sin na — sin (n—1)a—ksin na. Now suppose a= 10”, then sin a is known and cos a is known, and therefore & is known; we put »=1, and thus we obtain the value of sin 20”—sin 10”, and thence the value of sin 20”;, next we put n= 2, and thus we obtain the value of sin 30” —sin 20”, and thence the value of sin 30”; next we put = 3, and so on. It will be seen that the only laborious part of this operation consists in the multiplication by & of the sines as they are suc- cessively found; but from the value of cos10” it follows that &=-0000000023504,..and the smallness of / facilitates the process. 125. ‘When the sines of angles up to 45° have been calculated, those for the remainder of the quadrant might be deduced by the theorem sin (45° + A) —sin (45° — A) = 2 cos 45° sin A =,/2. sin A; this would require the multiplication of the sines already found by the approximate value of ,/2. If however we calculate the sines of angles up to 60°, those for the remainder of the quadrant may be very easily found from the theorem sin (60° + A) — sin (60° — A) = 2 cos 60° sin A = sin A. 126. When the values of the sines of all the proposed angles in the first quadrant are known the values of the cosines are also known, for the cosine of any angle is equal to the sine of the com- plement of the angle. The values of the tangents can be found by dividing the sine of every angle by the cosine of that angle. The tangents of angles greater than 45° may be easily inferred from those of angles less than 45° by the theorem tan (45° + 4) — tan (45° — 4) = 2 tan 2.4, which gives tan (45° + A) = tan (45°—- A) + 2 tan 24, 88 CONSTRUCTION OF TRIGONOMETRICAL TABLES, The cotangents are known since the cotangent of any angle is equal to the tangent of the complement of the angle. The cosecants may be obtained by calculating the reciprocals of the sines; they may however be obtained more simply from the tables of tangents by the theorem 1 A A cosee d= 5 { tan 5+ cot 5}: The secants are known since the secant of any angle is equal to the cosecant of the complement of the angle. 127. In the method adopted for calculating the sines of angles, the sine of 10” was first obtained to twelve places of decimals, and then the values of sin 20”, sin 30”, ... were deduced in succession. It will not however follow that the values of the sines of all the angles are correct to twelve places of decimals, and it is therefore useful to be able to test the extent to which the results are correct; and moreover it is essential to be able to test the correctness with which the calculations are performed. We may for this purpose compare the value of the sine of any angle obtained in the manner which has been explained with its value obtained independently. J5-1 4 of 18° may easily be calculated to any degree of approximation, and by comparison with the value obtained in the tables we can judge how far we can rely upon the tables. There are however two formule which are usually called formule of verification from the fact that they can be easily used to verify any part of the calculated tables. These formule are : sin 4 + sin (72° + A) —sin (72°—A) =sin (36°+ A) —sin (36° — A), cos A + cos (72° + 4) + cos (72° — A) =cos (36° + A) + cos (36°— A); they may be readily demonstrated; for Thus, for example, we know that sin 18°= 3 hence the sine sin (72° + A) —sin (72° — A) =2 cos 72° sin d= VOT sin 4, gin (36° +) —sin (36? A) =2 cos 36%sin 4 )+ eae (sedens (2). Again in (1) put m=n+1, thus we obtain the value of 1 log, or ; therefore for x, and therefore - for , log, (n + 1) —loga 1 1 1 =) geri * sGariF totes sequins } cuxivas (3). As an example of these formule suppose we put a= 1 in the expansion of log, (1 + 2): thus we obtain Pied Sab ol ¢,2=l-=4+5-S+2-5+... log, 1 g+3 +s gt 1 1 1 7 oo a Again, put 1 for m in (3): thus | 1 Bo aie ea ee | log, 2 gtap typ } 147. The series (2) of the preceding Article will enable us to find log, 2; put m= 2, then by calculation we shall find log,2 =°6931471...... From the series (3) we can calculate the logarithm of either of two consecutive numbers when we know that of the other. Put LOGARITHMS AND LOGARITHMIC SERIES. 101 n=, and by making use of the known value of log, 2, we shall obtain log, 3 = 1:0986122...... Put ~=9 in (3); then log, 9 = log, 3°= 2 log, 3 and is therefore known ; hence we shall find log, 10 = 2°3025850...... Logarithms to the base e are called Napierian logarithms, from Napier the inventor of logarithms; they are also called natural logarithms, being those which occur first in our investi- gation of a method of calculating logarithms. We have said that the base 10 is the only base used in the practical application of logarithms, but logarithms to the base e occur frequently in theoretical investigations. 148, From Art. 138 we see that the logarithm of a number to the base 10 can be found by multiplying the Napierian loga- 1 1 rithm by —— log 10 ; by 3-30958509...” or by 43429448... 5 this wiateplerd is called the modulus of the common system. that is, The base e, the modulus of the common system, and the logarithms to the base ¢ of 2, 3, and 5 have all been calculated to upwards of 200 places of decimals. See the Abstracts of the Papers communicated to the Royal Society, Vol. xx. page 27. The series in Art. 146 may be so adjusted as to give common logarithms ; for example, take the series (3), multiply throughout by the modulus which we shall denote by »; thus 1 1 1 plog, (m+ 1) ~ log. = 24 f 5 *3@n+1)* stat} j that is, 1 1 1 logyy (7% + 1) - logy) % = 2p fy ats (Gn ip +5 (n+l)! touch : Similarly from Art. 145 we have at 2 at maton oe eee 102 LOGARITHMS AND LOGARITHMIC SERIES. In future we shall in general use the symbol log without: the suffix e or 10; the student will be able always to infer from the context which base we are employing: we almost invariably employ the base ¢ in theoretical investigations, and the base 10 in practical applications. 149. The quantity e is incommensurable. For suppose if possible e= — , where m and n are integers; thus Le + 1 a gt ist at Multiply both sides by |; then A 1 1 an annie me ea * Gel) +2) “ (n+1) (+2) (n+) Slee 1 1 1 Bey Gal Genes GF Gt Nes is a fraction, for it is greater than and less than the geome- trical progression ot pa ee n+1 (tly *@+ip~ that is, less than 7 Thus the difference of two integers is equal to a fraction, which is absurd. Therefore e is incommensurable. 150. We will conclude this Chapter by investigating two limits which will be useful hereafter. To find the limit of (cos “) when n is increased indefinitely. Let Uu= (cos =" = Qa — sin’ “\A ; then n n LOGARITHMS AND LOGARITHMIC SERIES. 103 - 9 O\F 2 : log w= log (1—sin “) = 5 log (1 sin’ =) tfheet Vago « g@ =—— | Bn" — > « BI — + ge BU" He ai ees . 2 n 2 nr’ n This series is numerically less than Ml. oh 1. 4h . 6G —_ elt — +a fein ob... 7; n n n a «9% sin? — : . n n nN, 4a that is numerically less than — =; ————— or — 5 tan’—. 2 «9 2 n 1—sin*— n But tan = =a when 7 is increased indefinitely, by Art. 119 ; therefore » tan’ < =a tan - =0 ultimately. Therefore log w= 0; therefore w=1. Thus the required limit is unity. - @ sin — \" To find the limit of | —> when n is increased indefinitely. n - a2 sin — We know by Art. 116 that —-— is less than 1 and greater than nm - @ ~ a a sm — sin — — , that is, greater than cos . 3 hence tan— me n nr is less than 1" or a N, 1 and greater than (cos =)" ; and by the preceding Article the « @ n a limit of (cos ) is unity, therefore the limit of ( —— } is unity. n n 104 EXAMPLES. CHAPTER X. MISCELLANEOUS EXAMPLES, 1, Find the logarithm of 128 to the base 3/4. 2. Find the logarithm of 243 ,/9 to the base ,/3. 3. Find the following logarithms, log, 2187, log,, 0001, log, cos 45°. _ 4, Find approximately the value of « from the equation 5°. 27%, having given log 2= 301030. 5. Given log -224 =a and log 125 =8, find log 2 and log 7. 6. Required the characteristics of log, 725, and of log, ,/(:0725). %. ~ Given log 2 = 301030, log 405 = 2607455, find log ‘003. 8. Given log 2 = °301030, log 7 = -845098, find log 98 and 4\3 log (sis) . 9. Given log 2= ‘30103, log 3= -47712, find log (-0020736)), 10. Determine the sum of the series 3B [5 [7 11. Shew that e_ 1 142 14243 1494344 2-e* Bt EG E Find x from the following six equations : ad inf. +... ad inf. 12. 4singwsin (1—a)=2cosa-1. 13. cos B J/(w@-«’)+asina=a sin p. 14. sina+sin (#—a)+sin (2e+a)=sin (w +a) + sin (2a—a). 15. cos 3 + cos +5) =sin : (e+5)e c (« 5) a=sina. 16. 2’ cosacos (a -§) +1005 (a) = 20088 EXAMPLES. CHAPTER X. 105 17. cot 2871. a — cot 27a = cosec 3a, 18. Solve the equation m vers 6 =n vers (a — 6). 19. Solve the equation cos 26 + cos (n— 2) 6 =cos 6, 20. Solve the following equation, and shew that there are seven positive values of 6 greater than 0 and less than 2z, sin 9 +sin 36 = sin 26 + sin 40. 21. Find tan « from the equation tanx=tan f tan (a +a)? and shew that in order that tana may be real, tan B must-not lie between (seca — tan a)* and (seca + tan a)’ 22, Find the least value of @ which satisfies 7 wT _f 8/23 23. Given sin®(z + 1) =sin’n@ +sin®(»—1)6 where (n + 1)6, n§, and (x—1)6 are the angles of a triangle, find an integral value of 7. 24, Reduce to its simplest form and solve the equation cos — cos*a = 2 cos*@ (cos 6 — cos a) — 2 sin®6 (sin 6 — sin a). 25. Shew that all the angles which have the same sine as a are included in the formula (2n + 5) T+ & 7 «) : 26. Shew that all the angles which have the same cosine as a are included in the formula (n $f 5) m+(-1)" («-5) 2 27, In the-formula cos 4 ~sin =+,/(1—sin A) the ambi- guity + may be replaced by (— 1)", where m is the greatest integer 270+A4 — 360? the angle A being expressed in degrees. contained in 106 EXAMPLES. CHAPTER X. *A)-1 28. In the formula tan A = + /(1+tan?A)—-1 2 tan A +may be replaced by (—1)", where m is the greatest integer contained in ae , the angle A being expressed in degrees. 29. If tan (cot«)=cot(tanz), shew that the real values of the ambiguity x are given by sin 2a = , where 7 is any integer except — 1. (Qn+1)7 30. Shew how to express cos. in terms of cos A, where is any positive integer. . 2 31. From the equation cosw=+ “ sea deduce the formula for sing in terms of sin 2x, and shew how the proper signs for the radicals may be determined. A cos (6+ a)+B sin (6 +f) A’sin (6 + a) + B’cos (6 + B) same value for all values of 6, then will AA’— BB = (A'B- AB’) sin (a— B). 32. If the expression retain the 33. If the sum of two angles is given, shew that the sum of their sines is numerically greatest when the angles are equal. If the cosine of the given sum is positive shew that the sum of the tangents is numerically least when the angles are equal. 34. If 4+B+C=90°, shew that unity is the least value of tan’ A + tan’ B+ tan’ C. 35. If 4+B+C=180°, shew that unity is the least value of cot? A + cot? B+ cot? C. 36. If A, B, C are the angles of a triangle shew that 2cot A+2 cot B+2 cot Cis never less than cosec A +cosec B+cosec C. 37. Shew that the sum of the three acute angles which satisfy the equation cos’ A + cos* B +cos* C = 1 is less than 180°. EXAMPLES. CHAPTER X. 107 38. If each of the angles A, B, C be less than 90°, then sin (4+ B+(C) is less than sin 4+sin B+sin 0. 39, Find the limit of (cos =)" when n is increased indefinitely. nn 40. Find the limit of (cos =) when 7 is increased indefinitely. 8 41. Shew that sin @ is greater than tan 6 ee if 6 is posi- tive and less than 5 —1\* : : é 42. Shew that >) continually increases as # increases from unity to infinity ; and find the limit of the expression when x is increased indefinitely. XI. USE OF LOGARITHMIC AND TRIGONOMETRICAL TABLES. 151. In the preceding two Chapters we have shewn how tables of the values of the Trigonometrical Ratios may be cal- culated and how tables of logarithms may be calculated, and we shall now shew how to use such tables; we begin with tables of logarithms. It is obvious that tables of logarithms may be cal- culated to various degrees of approximation ; they may be calcu- lated to 5, 6, 7 or a higher number of decimal places. For a list of logarithmic and trigonometrical tables the student may consult the article Zables in the English Cyclopedia, and the Report of the British Association for 1873. Different tables present some variety in their mode of arrangement, and are usually accom- panied with full explanation of their peculiarities and the methods of using the tables ; we shall not enter into any minute account of the way in which tables may be used with the greatest advantage, but shall give such gencral illustrations as will enable the student to avail himself of any set of tables for the purpose of occasional calculation. The logarithms will always be supposed taken to the base ten. 108 USE OF LOGARITHMIC 152. We may observe that throughout all approximate cal- culations it is usual to take for the last figure which we retain, the figure which gives the nearest approach to the true value. Thus for example, suppose we have the decimal fraction :3726; if we wish to retain only three places of decimals we should write ‘373 and not ‘372; the former is too large and the latter too small, but the excess in the former case is 0004, and the defect in the latter case is 0006, so that there is a smaller error in the former case than in the latter case. Thus we have this general rule, when only a certain number of decimal places is to be retained: strike off the rest of the figures, and increase the last figure retained by 1 if the first figure struck of be 5 or greater than 5. We now proceed to explain the use of tables of common log- arithms ; and we shall use tables of seven places of decimals. 153. To find the logarithm of a given number. If the number be contained in the Table we have merely to take the decimal part of the logarithm immediately from the Table and prefix the characteristic (Arts. 141,142). For example, required the logarithm of 534. The table gives ‘7275413 as the decimal part, and the characteristic is 2; therefore log 534 = 2°7275413. Similarly, log 53400 = 4-7275413, log 0534 = 2-7275413. In the last example the characteristic is —2, and this is denoted by the bar placed over the 2: see Art. 142. Suppose, however, that the given number is not contained in the Table; the Table for instance may give the logarithms of numbers from 1 up to 100000 and we may require the logarithm of 5340234. Here we can take from the Table the logarithm of 5340200, and the logarithm of 5340300; we have log 5340300 = 6°7275657 log 5340200 = 6°7275575 difference = ‘0000082 AND TRIGONOMETRICAL TABLES. 109 The required logarithm of course lies between the two logarithms which we have taken from the Table. Now we see that cor- responding to the increase 100 in the number there is an increase 0000082 in the logarithm; and we asswme that corresponding to an increase 34 in the number there will be a proportional increase in the logarithm. Let «w denote the quantity which we must add to the logarithm of 5340200 in order to obtain the logarithm of 5340234; then we have, from the assumption which we have made, the following proportion : 100 : 34 :: 0000082 : a; therefore c= ax 0000082 = -0000028 (Art. 152) ; therefore log 5340234 = 6:7275575 + 0000028 = 6-72756083. 154, We assumed in the preceding Article that the increase in a logarithm is proportional to the increase in the number ; this is a case of what is called the principle of proportional parts, and although it is not strictly true, yet it is in most cases sufficient for practical purposes. We shall in the next Chapter investigate the subject, and shew to what degree of approximation we can rely upon the principle of proportional parts. 155. The process given in Art. 153 is facilitated in large Tables in the following manner. Required the logarithm of 23453487, log 23454000 = 7:3702169 log 23453000 = 7°3701984 difference= ‘0000185 Here by the process of Art. 153 we have to multiply 487 4 8 7 * 5 ee es . et es eee be 0000185 by T000? that is, by to * T00 t 1000° Now the mul tiplication is effected for us, and the results given, in a small Table headed Proportional parts, which is printed on the same page as the two logarithms which we have taken from the Table ; 110 USE OF LOGARITHMIC the small Table shews that 4 x -0000185 ea eral ='0000740, that 8 x-0000185 =-0001480, 2] 370 and that 7 x 0000185 = 0001295 ; and ee from these results, by dividing by 10, 100 5 ie and 1000 respectively, we obtain the 7 | 1295 three parts which we require. The pro- tare cess may be arranged thus: log 23453000 = 7°3701984 add for 4 740 8 1480 7 1295 73702074095 ‘therefore, retaining 7 places of decimals, log 23453487 = 7-3702074. 156. We have taken as our example a whole number; if a decimal fraction, or a mixed quantity formed of a whole number and decimal fraction, be given, we may throw aside the decimal point, and find the decimal part of the logarithm of the whole number thus obtained ; then by prefixing the proper characteristic we have the required logarithm. Thus, for example, required the logarithm of :23453487 and of 23453487. The decimal part of the logarithm is -3702074 ; therefore log +23453487 =1-3702074 log 234:53487 = 2-3702074. 157. To find the number which corresponds to a given logarithm. If the decimal part of the logarithm be found in the Table, we have merely to take the number which corresponds to it, and put the decimal point in the number in the place indicated by the characteristic. For example, required the number which has for its logarithm 27275413. Corresponding to the decimal part *7275413 we find in the Table the number 534, and as the charac- teristic is 2, there must be one cypher before the first significant figure (Art. 141); therefore the number which has the given logarithm is 0534, AND TRIGONOMETRICAL TABLES. lil Suppose, however, that the decimal part of the given logarithm is not contained exactly in the Table; for example, let the given logarithm be 1:3702074, we shall find that the decimal part of this logarithm is not in the Table; we have, however, corresponding to the number 23454 the decimal part of the logarithm -3702169, and corresponding to the number 23453 the decimal part of the logarithm ‘3701984 ; thus log 23454 = 43702169 log 23453 = 4:3701984 difference = °0000185 The excess of the given decimal part of the logarithm above 3701984 is 3702074 — 3701984, that is 0000090. The required number of course lies between -23454 and -23453; let d denote its excess above ‘23453, then assuming that the increase of the number is proportional to the increase of the logarithm, we have 0000185 : 0000090 :: 1: d; 90 therefore d= i857 486. Therefore log 23453°486 = 43702074, and log +23453486 = 1-3702074 ; thus the required number is -23453486. 185.) 90-0( 486 740 158. We may save the labour of dividing 1600 90 by 185 in the preceding example by means 1480 of the Table of proportional parts given in 1 200 Art. 155; the process of division, if per- 1110 formed, will stand thus: 90 Now the products 740, 1480, 1110, are i 740 furnished ready in the Table referred to, so T60 that we need only perform the subtractions 8 1480 and put down the following steps : 1200 6 1110 112 USE OF LOGARITHMIC 159. We will now give some examples of the use of logarithms: we take of course all our logarithms from Tables. Required the product of 3670:257 and 12°61158. Log 3670'2 = 3°5646897 5 60 7 8 Log 3670°257 = 35646965 Log 12°611 = 1:1007495 5 172 8 28 Log 12°61158 = 1:1007695 35646965 by adding the logs 4:6654660 Decimal part of log 46287 6654590 70 7 66 4 “40 4628774 Thus the required number is 46287-74, the position of the decimal point being determined by the characteristic 4. 160. Required the quotient of -1234567 by 54-87645. Log 12345 =1-0914911 6 211 7 25 Log 1234567 = 1-0915147 Log 54876 =1-7393824 4 32 5 4 Log 54'87645 = 1-7393860 10915147 1-7393860 by subtracting 3-3521287 AND TRIGONOMETRICAL TABLES. 113 3-3521287 Decimal part of log 22497 = -3521246 4l 2 39 1 20 2249721 Thus the required number is :002249721; there are two cyphers before the first significant figure, because the character- istic of the logarithm is 3. 161. Required the cube of 3180236. Log ‘31802 = 1°5024544 3 41 6 om) Log 3180236 = 1:5024593 3 25073779 Decimal part of log 32164 = ‘5073701 78 5 67 8 110 3216458 Thus the required number is 03216458. 162. Required the cube root of ‘3663265. Log ‘36632 = 1-5638606 6 71 5 6 Log ‘3663265 = 15638683 We have now to divide 15638683 by 3; that is, we have to divide — 1 + 5638683 by 3. It is convenient to write the num- ber to be divided thus, -- 3+ 25638683 ; then by dividing by 3 we obtain — 1 + ‘8546228, that is, 1-8546228. T. T. 8 114 USE OF LOGARITHMIC 18546228 Decimal part of log 71552 = -8546218 10 1 6 6 40 7155216 Thus the required number is ‘7155216. We shall now explain the use of Trigonometrical Tables. 163. To find the sine of a given angle. If the given angle be one which is contained in the Table of the sines of angles the required sine is furnished immediately by the Table; we proceed then to the case when the given angle lies between two which are contained in the Table. For example, re- quired the sine of 44° 35’ 25”, having given from the Table sin 44° 36’ = -7021531 sin 44° 35’ = ‘7019459 difference = 0002072 The required sine of course lies between the two sines which we have taken from the Table; let x denote its excess above the sine of 44° 35’, and assume that the increase of the sine is propor- tional to the increase of the angle, therefore 60” : 25” :: 0002072 : a, therefore x= = x 0002072 = 0000863. Therefore sin 44° 35’ 25” = -7019459 + -0000863 = 7020322. We have thus again assumed the principle of proportional parts, and we shall assume it throughout the present Chapter, reserving the investigation of it for the following Chapter. 164. To find the angle which corresponds to a given sine. If the given sine be found in the Table the required angle is furnished immediately. by the Table; we proceed then to the case AND TRIGONOMETRICAL TABLES. 115 when the given sine lies between two which are contained in the Table. For example, required the angle which has for its sine “6970886, having given from the Table sin 44° 12’ = -6971651 sin 44° 11’ = -6969565 difference = 0002086 The excess of the given sine above the sine of 44°11" is 6970886 — -6969565, that is, 0001321. The required angle of course lies between the two angles which we have taken from the Table; let x be the number of seconds in its excess above 44°11’, then 70002086 : 0001321 :: 60 : 0001321 60x 1321 therefore n= 60 x *-op09086 ~~ Boge = 38. Therefore the required angle is 44° 11’ 38”. 165. Lo find the cosine of a given angle. If the given angle be one which is contained in the Table of the cosines of angles, the required cosine is furnished immediately by the Table; we proceed then to the case when the given angle lies between two which are contained in the Table. For example, required the cosine of 44°35’ 25”, having given from the Table cos 44° 35’ = -7122303 cos 44° 36’ = -7120260 difference = ‘0002043 Since in the first quadrant the cosine decreases as the angle in- creases, the required cosine will be less than the cosine of 44° 35’, and the required cosine of course lies between the two cosines which we have taken from the Table; let x denote its defect below the cosine of 44° 35’, then 60 : 25 :: 0002048 : a, therefore -2 x 0002043 = -0000851. Therefore cos 44° 35’ 25” = -7122303 — 0000851 = 7121452. 8—2 116 USE OF LOGARITHMIC 166. To find the angle which corresponds to a given cosine. If the given cosine be found in the Table the required angle is furnished immediately by the Table; we proceed then to the case when the given cosine lies between two which are contained in the Table. For example, required the angle which has for its cosine ‘7169848, having given from the Table cos 44°11’ =-7171134 cos 44° 12’= -7169106 difference = ‘0002028 The given cosine falls short of the cosine of 44°11’ by 7171134 — ‘7169848, that is, by 0001286, The required angle of course lies between the two angles which we have taken from the Table ; let 2 be the number of seconds in its excess above 44°11’, then 0002028 : 0001286 :: 60: n, 0001286 60x 1286 therefore n= 60 x Q000088 e028 = 38. Therefore the required angle is 44°11’ 38”. 167. It will not be necessary to give examples for the other Trigonometrical Functions ; the important fact to be remembered is that in the first quadrant the tangent and secant increase as the angle increases, and the cotangent and cosecant decrease as the angle increases ; thus the tangent and secant are treated in the same way as the sine, and the cotangent and cosecant in the same way as the cosine. 168. The Tables of Trigonometrical Functions which we have hitherto considered are called Tables of the Natural Functions to distinguish them from other Tables which we now proceed to con- sider. The Table of sines of angles for example is called a Table of natural sines; if we take the logarithms of the sines of all the angles which have been calculated we form a new Table which is called a Table of Logarithmic sines. Similarly, we can form a Table of the logarithms of the cosines of angles, and a Table of the logarithms AND TRIGONOMETRICAL TABLES. 117 of the tangents of angles, and so on ; these Tables are called respect- ively Tables of Logarithmic cosines, Tables of Logarithmic tangents, and so on. 169. The great advantage which we obtain from these Loga- rithmic Tables is that calculations are much abbreviated with their assistance ; this is especially the case, as we shall see hereafter, in what is called the solution of Triangles. We have stated as sufii- ciently obvious that these Logarithmic Tables may be calculated by taking the logarithms of the values of the Trigonometrical Functions which have been already tabulated ; it will be shewn however in the higher parts of the subject that the Logarithmic Tables can be calculated independently, that is, without the use of the Tables of the Natural Functions. We proceed now to exemplify the use of the Tables of Logarithmic Functions, 170. Since the sine of an angle is never greater than unity the logarithm of the sine will never be a positive quantity ; also the same remark is true for the cosine. The logarithm of the tangent of an angle will be negative if the angle be less than 45°, and the logarithm of the cotangent of an angle will be negative if the angle be greater than 45°. In order to avoid the occurrence of negative quantities in the Tables it is found convenient to add 10 to the logarithm of every Trigonometrical Function before registering it in the Tables; the logarithm so increased is called the Tabular logarithm and is usually denoted by the letter Z. Thus Zsin A means the Tabular logarithm of the sine of A, and it is equal to the real logarithm of the sine of A increased by ten. Of course in calculations we shall have to remember and to allow for this increase of the real log- arithms ; this will be seen when we come to the solution of Tri- angles. In what follows we shall exemplify the use of the Tables of Logarithmic Functions. 171. To find the tabular logarithmic sine of a given angle. If the given angle be one which is contained in the Table of the Logarithmic sines the required result is furnished imme- 118 USE OF LOGARITHMIC diately by the Table ; we proceed then to the case when the given angle lies between two which are contained in the Table. For example, required the tabular logarithmic sine of 44° 35’ 25”-7, having given from the Table L sin 44° 35’ 30” = 98463678 L sin 44° 35’ 20” = 9:8463464 difference = ‘0000214 The required tabular logarithmic sine lies of course between the two which we have taken from the Table; let x denote its excess above the tabular logarithmic sine of 44°35’ 20"; then by the principle of proportional parts 10 : 5-7 :: 0000214 : a, therefore «= at x 0000214 = 0000122. Therefore Z sin 44° 35’ 25" -7= 9-8463464 + -0000122 = 98463586. 172. To find ‘the angle which corresponds to a given tabular logarithmic sine. If the given tabular logarithmic sine be found in the Table the required angle is furnished immediately by the Table; we proceed then to the case when the given tabular logarithmic sine lies between two which are contained in the Table. For example, required the angle which has for its tabular logarithmic sine 9°8432894, having given from the Table LI sin 44° 11’ 40” = 98432923 Lsin 44° 11’ 80” = 98432707 difference = ‘0000216 The excess of the given tabular logarithmic sine above that of 44° 11' 30” is 9°8432894 — 9:8432707, that is, 0000187. The re- quired angle of course lies between the two angles which we have AND TRIGONOMETRICAL TABLES. 119 taken from the Table; let » be the number of seconds in its excess above 44°11’ 30”, then 0000216 : -0000187 :: 10: x, 0000187 10 x 187 therefore n= 10x >oo9916 = aig = 8 , Therefore the required angle is 44° 11’ 38"°7. 173. To find the tabular logarithmic cosine of a given angle. If the given angle be one which is contained in the Table of the logarithmic cosines the required result is furnished immediately by the Table ; we proceed then to the case when the given angle lies between two which are contained in the Table. For example, re- quired the tabular logarithmic cosine of 44° 35’ 25”-7, having given from the Table L cos 44° 35’ 20” = 9°8525789 L cos 44° 35’ 30” = 9°8525582 difference = :0000207 The required tabular logarithmic cosine lies of course between the two which we have taken from the Table, and is Zess than the tabular logarithmic cosine of 44° 35’ 20”; let « denote its defect below the latter ; then 10 : 5-7 :: -0000207 : a, therefore nant x 70000207 = -0000118. Therefore Z cos 44° 35’ 25”-7 = 9:8525789 — :0000118 = 9-8525671. 174. To find the angle which corresponds to a given tabular logarithmic cosine. If the given tabular logarithmic cosine be found in the Table the required angle is furnished immediately by the Table; we proceed then to the case when the given tabular logarithmic cosine lies between two which are contained in the Table. For example, required the angle which has for its tabular logarithmic cosine 9:8555086; having given from the Table 120 EXAMPLES. CHAPTER XI. L cos 44° 11! 30” = 98555264 L cos 44° 11’ 40” = 9-8555060 difference = -0000204 The given tabular logarithmic cosine falls short of that of 44° 11’30” by 9-8555264 — 9-8555086, that is, :0000178 The required angle of course lies between the two angles which we have taken from the Table; let x be the number of seconds in its excess above 44°11’ 30”; then 0000204 : 0000178 :: 10: », 0000178 _ 1780 _ 4, * "0000204 204° Therefore the required angle is 44° 11’ 387. therefore n=10 175. It will not be necessary to give examples for the other Trigonometrical Functions ; the important fact to be remembered is that in the first quadrant the tabular logarithms of the tangent and secant increase as the angle increases, and the tabular logarithms of the cotangent and cosecant decrease as the angle increases ; thus the tangent and secant are treated in the same way as the sine, and the cotangent and cosecant in the same way as the cosine. EXAMPLES, 1. Given log 12440 = 4:0948204, log 12441 = 4:0948553, find log 12440°35, 2. Given Jog 1-:0686 = 0288152, log 1-0687 = 0288558, find the number of which the logarithm is -0288355. 3. Given log 23456 = 4:3702540, log 23457 = 4:3702725, form a table of proportional parts for the intermediate numbers, and find log :2345638. EXAMPLES, OHAPTER XI. 121 4. Find the number whose logarithm is — (1-8753145), having given log 1:3325 = -1246672, log 1-3326 = -1246998, 5. Given log3-855 =+5860244, log 3-8551 = -5860356, find log (00385504)4. 6. Given log 24 = 1-3802112, log 4:8989 = -6900986, log 4:8990= 6901074, find (24)3 to six places of decimals. 7. Given — log 14271 =4-1544544, log 20313 = 43077741, log 20314 = 4:3077954, find (142-71)*. 8. Given log7= -8450980, log 58751 = 4:7690153, log 58752 = 4:7690227, L find (:07)° to seven significant figures. 9. Given log2= ‘3010300, log 5-743491 =-7591760, find the fifth root of 0625. 10. Given log 2-7 =-4313638, log 5-172818 = -7137272, find the value of 277%. ll. Given log 71968 = 4°8571394, diff for 1 = -0000060, find the value of £/(-0719686). 12. Given log 103 =2-0128372, log 7440942 = 6-871628, find (1:03)". 13, Find the value of 64 {1 —(1:05)-*}, having given log 105 = 2:0211893, log 37689 = 45762140. 122 EXAMPLES. CHAPTER XI. 14. Find approximately 5¥*, having given log 2 = 3801030, log 1:562944 = 193943, log 349485=5°543428, log3-655 =-562887, log3°656 =-563006. 15. Having given log 12 = 10791812, log 1:257915 = -0996519, log 1:121568= -0498256, find the value of (1-44)7~% — (1:44)7. 16. Having given log 105 = 20211893, log 5303214 = 6:7245391, log 3768894 = 6°576214, find the value of 17. Given sin 47° = -7313537, sin 48° = -7431448, find sin 47° 1’, 18, Given sin 7° 17’ = +1267761, sin 7° 18’ = +1270646, find sin 7° 17’ 25”, 19, Given Lsin 17° 1’ = 94663483, Lsin 17 = 9-4659353, find Lsin 17° 0’ 19”. 20. Given Lsin 26° 24’ = 9-6480038, Lsin 26° 25’ = 9-6482582, find Lsin 26° 24’ 12”, 21. Given Lot 72° 15’ = 9-5052891, L cot 72° 16’ = 9°5048538, énd L cot 72° 15' 35”. 22. Given L cot 81°46'=9-1604569, diff. for 10” = 0001486 find the angle whose L cot is 9:1603493. ’ THEORY OF PROPORTIONAL PARTS. 123 23. Given Z cos 20°35’20"= 9-9713351, diff. for 10”=-0000079, find the angle whose Z cos is 9:9713383. 24. Given Zcos 34° 24’ = 9-9165137, diff. for 1’= 0000865, find Z cos 34° 24’ 26”, and also the angle whose L cos is 9:9165646. 25. Given Z sin 37° 19’= 9-7826301, diff-for 1’ = 0001657, L£ cos 37° 19’ = 9-9005294, diff. for 1’ = -0000963, find L sec 37° 19’ 47”, and Z tan 37° 19’ 47”, 26. Given Z sin 32°18’ = 9-7278277, diff. for 1’ = 0001998, L cos 32° 18’ = 9-9269913, diff. for 1’ = -0000799, fnd JZsine, L cosine, and Z tangent of 32° 18’ 24-6. XII. THEORY OF PROPORTIONAL PARTS. 176. We shall now investigate the principle of proportional parts, the truth of which was assumed throughout the preceding Chapter. The logarithms in the present Chapter are supposed to be logarithms to the base 10; and we will suppose that the Table of logarithms is calculated to seven places of decimals, and that it con- tains the logarithm of every whole number from 1 to 100000. 177. To shew that the change of the logarithm is approximately proportional to the change of the number. m7 44 — tog (1 + “) , nr 2, d ad @& a and by Art. 148, log (1 n “yee (: Sather), where p is the modulus, so that p= -43429448...... Suppose that n is an integer containing five figures so that n is not less than 10000, and suppose that d is not greater than unity. We know that log (n+ d) —logn =log 124 THEORY OF PROPORTIONAL PARTS. , 1/7 1‘ ara Then ea islessthan i (roa00) ,and a fortiori less than 000000003; 8 ae is less than one ten-thousandth part of this, and so on. Hence at least as far as seven places of decimals we have log (7 + d) — logn = 4, This equation establishes the required result; for it shews that if the number be changed from n ton +d the corresponding change in the logarithm is approximately = , that is, the change of the logarithm is approximately proportional to the change of the number. 178. The principle of proportional parts is thus shewn to hold in the case of the logarithms of numbers to a sufficient degree of accuracy for practical use. For when we wish to find the loga- rithm of a given number we can suppose the decimal point in the number placed after the fifth figure, so that the number is thus made to lie between two which differ by unity and which are both contained in the Table; and we have shewn that as far as seven places of decimals the change of the logarithm is proportional to the change of the number. Then we can if necessary change the position of the decimal point and make the corresponding change in the characteristic of the logarithm ; and thus we finally obtain the logarithm of the original given number. Similarly we may proceed if we want to find the number which corresponds to a given logarithm lying between two in the Table. 179. We will now shew how the result of Art. 177 is applied in practice, We have ake log (n+ d)—logn= os =6 suppose, also log (m +1) ~logn = thus log (n + d) = log n+ 38, THEORY OF PROPORTIONAL PARTS. 125 Now 5 being the difference of two known logarithms is furnished immediately by the Table; and to obtain the logarithm of (n+) we multiply this known quantity § by the given fraction d and add the product to the logarithm of m. This is the rule which was used in the preceding Chapter, Art. 153, in order to find the logarithm of a given number. Again, suppose we require the number which corresponds to a given logarithm. Let 2 and n+1 be integers between which the required number lies, and denote the required number by n+ d. Then log (x +d) —logn is known; call it 2, and let 5 denote the x known quantity log(n+1)—logn; thus d=; therefore d= 5 This is the rule which was used in the preceding Chapter, Art. 157. 180. We shall now proceed to examine how far the principle of proportional parts holds in the case of the Natural Trigonometrical Functions ; this we shall do by considering these Functions sepa- rately. “We shall suppose throughout this Chapter that the angles which oceur are positive angles not exceeding a right angle; this is sufficient because it has been shewn that any Trigonometrical Function of any angle is numerically equal to the same Function of some positive angle not exceeding a right angle ; see Art. 55. 181. Zo prove that in general the change of the sine of an angle is approximately proportional to the change of the angle. We have sin (0 +h) - sin 6 = sinh cos 6 — sin 6 (1 — cosh) +o) sinh =sinh cos 6 (1-tang = sin h cos 0 (1 —tan 6 tan). Let us now suppose that / is the circular measure of a very small angle so that sinh = approximately ; thus, approximately, sin (0+) -sin 0 = cos 6 (1 tan 6 tan5) ; 126 THEORY OF PROPORTIONAL PARTS. let us also suppose that 6 is not very nearly equal to 7 so that tan § is not very large, and thus tan@ tan may be neglected. We have then, approximately, sin (6+ h) —sin 6 =h cos 8, and this establishes the proposition. Similarly sin (@—h) — sin @ =— fh cos 6 approximately. 182. We may however require to know more exactly the amount of error to which we are liable in using the result of the preceding Article; this point we will now examine. The approx- imate value of sin (6+) - sin 6, is hcos 6, while the exact value is sinh cos6 —(1— cos #) sin @ ; thus to obtain the approximate value we change -sin/, into / in the first term of the exact value, and we neglect the.second term of the exact value. First then consider the error produced by writing 4 for sink. The circular measure of an angle of half a degree is 355 3 and by Art. 130 sin A cannot differ from h by so much as , So that it may be shewn that for an angle of half a degree the sine cannot differ from the circular measure by so much as ‘00000012. Hence if our calculations extend to only seven places of decimals an error will hardly be introduced by changing sinh into A even for an angle of half a degree, and a fortiori no error will be introduced by the change if we restrict h to be not greater than the circular measure of an angle of one minute. Next consider the error produced by neglecting the term sin 6 (1—cosh), that is, 2 sin 0 sin*4 . Since sin @ is never greater than unity and sin is less than the value of the term neglected is less than a 2 - 3 and if / be the circular measure of an angle of one minute is less than 0000001. Hence if our calculations extend to only THEORY OF PROPORTIONAL PARTS. 127 seven places of decimals, no error will be introduced by neglecting the term sin 6 (1 —cos h) if we restrict 2 to be not greater than the circular measure of an angle of one minute. Therefore if we have a Table of natural sines calculated for every minute to seven places of decimals, no error will be intro- duced by our calculating to seven places of decimals the sine of an angle which lies between two in the Table from the formula sin (6 + h) —sin 6=h cos 0. 183. We will now shew how this result is applied in practice. Suppose that we have a Table of natural sines calculated for every minute, and that we require the sine of an angle which lies be- tween two in the Table. Let & be the circular measure of an angle of one minute; let 6 and 6 +4 be the circular measures of the angles in the Table between which the given angle lies, and let 6+A be the circular measure of the given angle. Then sin (6 + &) —sin 0 =k cos 6=8 suppose, sin (0+) sin 0 =heos 0="8; thus gama Doe ha Ieee i 60° where s is the number of seconds in the angle of which 4 is the circular measure. Now 8 is the difference between two consecutive sines in the Table, and is therefore furnished immediately by the Table, and we must multiply this known quantity by es and add the result to sin 6 in order to obtain sin(9+h). This is the rule which was used in the preceding Chapter, Art. 163. Again suppose that we require the angle which corresponds to a given natural sine. Let & be the circular measure of an angle of one minute; 6 and 6+4 the circular measures of angles in the * Table between which the required angle must lie, and let 6 +4 be 128 THEORY OF PROPORTIONAL PARTS. the circular measure of the required angle. Then sin (6 +h) —sin 6 is known; call it x, and let 6 denote a oe quantity sin (6 +k) —sin 6; therafore “= =a, therefore ” E75 z 5 3 let s be the number of seconds in the angle of which the teats measure is h, then a Sa! 3 therefore = = . This is the rule which was used in the preceding Chapter, Art. 164. 184. When 0 is nearly 5, since cos is then very small, the term cos @ will be very small if A be the circular measure of a small angle, Thus the difference between the natural sines of two angles, each of which is nearly equal to aright angle, is very small ; this is expressed by saying that the differences in the sines of con- secutive angles are nearly insensible when the angles are nearly equal to a right angle. There is also another point to be noticed in this case ; we have sin (9 +h) — sin 6 = sin hcos 0 — (1 — cosh) sin 6 ; the ratio of the second term to the first is numerically sin 6 (1 —cos h) cos sin h that is, tan 6 tan a and when @ is nearly equal to a this ratio will be a sensible quantity unless be extremely small. Thus the second term ought not to be rejected in comparison with the first. term unless é be extremely small. This is expressed by saying that the differences in the sines of consecutive angles are irregular when the angles are nearly equal to a right angle. In the present. case this irregularity is not of much importance on account of the accompanying insensibility, THEORY OF PROPORTIONAL PARTS. 129 185. We have shewn that, approximately, sin (0 + h) -sin 0=/h cos 6; change @ into - 6’, thus , wT , : T ’ Tv ? sin (5-6 +h)—sin (5-6) =hoos (5-0), that is cos (0 —h) — cos 6’ =Asin & ; and by changing the sign of h cos (6' +h) — cos 6’ =—h sin 6. It is convenient to deduce this formula from that already proved, because we thus know, without a new investigation, the amount of error to which we are liable in using it; it may how- ever be proved independently, as we will now shew. 186. Zo prove that in general the change of the cosine of an angle is approximately proportional to the change of the angle. We have cos (9 — h) — cos § =sin h sin 6 — cos 6 (1 — cos h) 1 — cos ) =sink cin 9 (1 -cot 6 sinh sin fisin 0 (1—cot@ tan 5). Let us now suppose that is the circular measure of a very small angle, so that sink =h approximately ; thus, approximately, cos (6 —h) — cos 6 =h sin 6 (i — cot @ tan 5) 3 let us also suppose that @ is not very small, so that cot 0 is not very large, and thus cot@ tan : may be neglected. We have then, approximaiely, cos (6 —h) —cos 6=h sin 6, and by changing the sign of A, cos (0+ h)—cos0=—hsin 6; and this establishes the proposition, 7, 9 130 THEORY OF PROPORTIONAL PARTS. 187. From the result of the preceding Article, we can deduce the rule used in Arts. 165 and 166 of the preceding Chapter; the method is the same as that which we have already given in Art. 183. The only peculiarity to notice is that the cosine diminishes as. the angle increases. And by proceeding as in Art. 184 we see that the differences in the cosines of consecutive angles are nearly insensible and are also irregular when the angles are very small. me 188; 7 ‘To prove that in general the change of the tangent of an angle as cypproximatel; 'y proportional to the change of the angle. _sin(@+h) sind We hive tan (6+ h) — me oR 2a _sin (64 h) cos6—cos(O+h)sin6 sin(@+h—6) _ sinh cos (6 +h) cos 6 ~ cos (0+h) cos@ cos (8+h) cos 0 sin h tanh ~ cos®@ (cos h—sin i tan 6) cos’ @ (1 — tan 0 tan h) * Let us now suppose that / is so small that we may put A for tan h, and also that 6 is not nearly equal to 5 so that tan 6 tanh may be neglected. We have then, approximately, =hsec’ 0 h 3° 6 ° A also by changing the sign of h tan (6 —h) -tan 6 =—A sec’ 6; this establishes the proposition. 189. From the result of the preceding Article we obtain the same rule for the tangent as we obtained in Art. 183 for the sine. We will now proceed to examine the amount of error to which we are liable in using the approximate formula of the preceding Article. tanh cos’ 6 (1 — tan 6 tan h) = tan h sec? 6 (1+ tan 6 tan h+ tan’ 6 tan’h+...); We have = tan h sec’ 6 (1 — tan @ tan h)™* THEORY OF PROPORTIONAL PARTS. 131 thus if'we take only the first term tan’ sec?@ we neglect a series of terms beginning with tan’ sec’ @tan 6, that is approximately h* (1 +tan* 6) tan 6. Now if we have a table of natural tangents calculated for every minute and we wish to find the natural tangents of intermediate angles the greatest value of h is the cir- cular measure of one minute, that is. , or ‘0003 approxi- Tv ? 180 x 60 mately. Hence the numerical value of the greatest error is not - less th 0003)° (1+ tan*@) tan @, and therefore e ess than (° a ) tan 8, an refor ven ERT greater than q we are liable to an error in the seventh place, ae IG of decimals. If, however, we have a table esloulais for every, ten seconds the greatest value of h is the circular. measure, of ied oa ‘Pie * os. ten seconds, that is , or 00005 approximately ; 180 x ae 6 this case we shall be free from error in the seventh bie of decimals until tan 6 is nearly as great as 31: the table shews that tan 73° 18’ is rather less-than 34, 190. Since tan (6+h)—tan 6=hsec’4 approximately, and sec 9 is never less than unity, the differences of consecutive tan- gents are never insensible ; but as we have shewn in the preceding Article, the differences are irregular when the angles are nearly right angles. 191. We have shewn that approximately tan (0 +h) — tan 6 =h sec" 6; change 6 into : — 6, thus tan (5-0+2) te (5-*) heed (§- e’), that is cot (0 —h) — cot 6’ = h cosec? 6’, and by changing the sign of A cot (0’ +h) — cot 6’ =—h cosec’ 6’. This may be proved independently, as we will now shew. 9—2 132 THEORY OF PROPORTIONAL PARTS, 192. To prove that in general the change of the cotangent of an angle is approximately proportional to the change of the angle. cos (8—h) cos 8 We have cot (6 —h) — cot @= an (00) aaa _ cos (9—h) sin 9 —cos Osin (9-h) sin (9-0 +h) = sin (9 —h) sin 6 ~ sin (6 —A) sin 0 sink sin h ~ sin(@—A) sin @~ sin’ 6 (cosh —sin h cot 6) i tanh ~ sin? @(1 —tan h cot 6)° Let us now suppose that A is so small, that we may put h for tan A, and also that 6 is not very small, so that cot @ tanh may be neglected, We have then approximately cot (6 — h) —cot 6 = =h cosec’ 6, h sin* 6 also by changing the sign of h cot (0 +k) —cot 0 =— h cosec’ 6 ; this establishes the proposition. 193. Zo prove that in general the change of the secant of an angle is proportional to the change of the angle. 1 1 We have Redes) = 800 Oa) eeed _ cos @—cos(O+h) sinhsin @ + (1 —cosh) cos 6 ~ cos @cos (6+h) ~ cos’ (cos h— sin / tan 6) tan hsin 6 (1 +tan : cot 6) = cos’ 0 (1 — tan 6 tan fh) . Let us now suppose that % is so small that we may put A for tanh, and also that @ is neither very small nor very nearly equal THEORY OF PROPORTIONAL PARTS, ‘133 to 3 so that tan @ tanh and cot 0 tan may be neglected. We have then approximately sec(9-+h) — see 0 =” a? cos’ @ =hsin 6 sec’ 0, also by changing the sign of h sec (0 —h) —sec 6=—h sin 6 sec’ 6; this establishes the proposition. 194. We have shewn that approximately sec (0 +h) — sec 0 =h sin 6 sec’ 6 ; change 6 into z 6’, thus sec (§-0 +2) -sec (5-0) =hsin (§-¢) soc'(5-6’), that is cosec (6’ — h) — cosec 0’ = h cos 6 cosec’ 4’, and by changing the sign of 4 cosec (6’ + h) — cosec 6’ =— h cos 6 cosec’ 6”. This may also be proved independently. 195. The amount of error to which we are liable in using the approximate formule of the preceding two Articles may be in- vestigated as in Art, 189. . It will be seen that the differences of consecutive secants are insensible and irregular when the angles are very small, and they are irregular when the angles are nearly right angles ; the differences of consecutive cosecants are zrregular when the angles are small, and insensible and irregular when the angles are nearly right angles. We will now proceed to examine how far the principle of pro- portional parts holds in the case of the Logarithmic Trigonometrical Functions, 196. To prove that in general the change of the tabular loga- rithmic sine of an angle ts approximately proportional to the change of the angle. 134 THEORY OF PROPORTIONAL PARTS. We have approximately sin (0 +) = sin 6 + h cos 4, therefore re at nd =1+hcotd; : ; sin (+A therefore log sin (9 +k) — log sin 6 = log ane and log (1+ cot 6)=phcot 6 approximately (Art. 148), where ps is the modulus ; thus approximately = log (1+ cot 8), log sin (8 + h) — log sin 0 = ph cot 6, also by changing the sign of h log sin (6 —h) — log sin 0 = — ph cot 6. If Z stand for tabular logarithm, we have L sin (0 +h) =10 + log sin (6 + A), L sin 6=10 + log sin 6 ; therefore Lsin (9+h)—-Lsin§ == ph cord. This establishes the proposition. 197. We will now shew that in general the principle of pro- portional parts holds approximately in the case of the other tabular logarithmic functions, and then we will consider the amount of error to which we are liable in using the approximate formule. 198. We have shewn that approximately Lsin (0+h)-L sin 6= ph cot 0, change 6 into 37 0’, thus Lsin 5-6 +h) ~Lsin (5-6) = uh cot 5-8), that is Lcos (6'~h) — L cos & = ph tan 6, and by changing the sign of L cos (6 +h) —L cos &’ =~ ph tan 6’. This proves the principle in the case of the tabular logarithmi cosines. THEORY OF PROPORTIONAL PARTS: 135 199. We have shewn that approximately log sin (9 + 2) — log sin 6 = ph cot 6, and log cos (6 + h) — log cos @= — ph tan 0; then by subtraction log sin (9+h) — log cos (6+) —{log sin 9—log cos 6}= ph (cot 6 + tan 6), i Quh that is log tan (6 + %) — log tan 6 = ">, Quh sin 20? therefore Z tan (9+ h) — L tan 6= and by changing the sign of h Quh Lian (—h) ~L tan 9 =—-— 5G. This proves the principle in the case of the tabular logarithmic tangents. By changing 6 into 37 0’ we obtain 2 Leot (6' wh) — Loot =+ this proves the principle in the case of the tabular logarithmic cotangents. 200. We have shewn that approximately log sin (6 + h) — log sin 0 = ph cot 0, 1 1 therefore log am +8) —log aaeo ph cot 6, that is log cosec (6 + h) — log cosec 6 =— ph cot 0, therefore L cosec (6 + h) — LE cosec 6 = — ph cot 6, also by changing the sign of h L cosec (9 —h) — L cosec 9 = jh cot 0 ; 136 THEORY OF PROPORTIONAL PARTS. this proves the principle in the case of the tabular logarithmic cosecants. By changing @ into 3 0’, we obtain L sec (6 =h)—L sec 6’ == ph tan 0 ; this proves the principle in the case of the tabular logarithmic secants. 201. From the results of Arts. 196...200 we obtain the rules which were exemplified in Arts, 171...174. It will be observed that we have deduced the approximate formule for all the other logarithmic functions from that of the logarithmic sine ; thus if we investigate the amount of error to which we are liable in the case of the logarithmic sine, we shall know the amount of error for all the other logarithmic functions. The approximate formule how- ever for the other logarithmic functions may be obtained inde- pendently, and we will for example give the investigations for the logarithmic cosine and the logarithmic tangent. 202. To prove that in general the change of the tabular loga- rithmic cosine of an angle is approximately proportional to the change of the angle. We have approximately cos (9 —h) =cos 6 + Asin 0, cos (9 — h) cos 0 cos (0 —h) cos 0 therefore =l+htan@, therefore log cos (6 — h) — log cos 6 = log = log (1+ tan6), and log (1 +/ tan 6) = ph tan 6 approximately (Art. 148), therefore log cos (0 —h) — log cos 6 = ph tan 6 approximately, therefore Los (@—h)— Lcos6 = ph tan 6, and by changing the sign of h L cos (6 +h) — L cos 0 =— ph tan 6. 203. Yo prove that in general the change of the tabular loga- rithmic tangent of an angle is approximately proportional to the change of the angle. THEORY OF PROPORTIONAL PARTS, 137 We have approximately tan (6 +h) = tan 6 + h sec’, meh). 1 hse 1+ 2h cosec 26, therefore log tan (9 + h) — log tan @ = log (1 + 2h cosec 26) = 2uh cosec 29 approximately, therefore tan (6+h)—L tan 6 = 2Quh cosec 26, and by changing the sign of h L tan (0—h) — L tan 6 =— Quh cosec 26. 204, We will now proceed to consider the amount of error to which we are liable in using the approximate formula L sin (6+h) —Lsin 6 = ph cot 6. In obtaining this formula log (1 + 2 cot 6) was taken equal to ph cot 6, so that the square and higher powers of hcot@ were neglected. But when 6 is very small cot 6 is very large, and thus h? cot?@ may be too large to be neglected; this case then will require further examination. ‘We have shewn in Art. 181 that sin (0+ l) ~sin 0 = sin cos 6 (1—tan 8 tan 5) therefore let us suppose 4 so small that we may write for sin # and 3 for tan ; ; thus approximately sin (0 +4) —sin 0=/ cos 6 — . sin 6, therefore mn 8) 1+Ahcotd-— sin 6 sin (6 +h) : h? therefore log :—_. = = log (1 t+hooté— 12) Ae =n(t cot 6-5) -E (ih cot 6-5) 4... (Art. 148) g = ph cot + cot? 6) +... ; thus if we omit powers of / higher than h? we have g log sin (6 +h) — log sin 6 = ph cot 6 — es cosec’ 6, 138 THEORY OF PROPORTIONAL PARTS. If our Table is calculated to every ten seconds, then the greatest value of 4 is the circular measure of ten seconds, that is about :00005; and = approximately. Thus the greatest error 6 2 : : — . This error will become sensible in calculations to seven places of decimals if 6 cosec’@ is as large as 10°, that is if sin’@ is as small as -006: the tables shew that the sine of 44° is rather greater than ,/-006. to which we are liable is about Thus we see that the differences of consecutive logarithmic sines are irregular when the angles are very small. When 6 is very nearly a right angle, cot 6 is very small while cosec’@ is not very small; thus the above formula for log sin (9 +h) — log sin 6 shews that the differences of consecutive logarithmic sines are nearly insensible when the angles are nearly equal to a right angle, and that these differences are at the same time irregular. From these results we can immediately infer the corresponding results for the logarithms of the other Trigonometrical functions; they will be found enunciated in Art. 206. 205. It appears from the preceding Article, that when an angle is small it cannot be accurately determined from its loga- rithmic sine nor the logarithmic sine from the angle by means of the common tables, because although the differences of consecutive logarithmic sines are then sensible, yet they are irregular. To obviate this difficulty three methods have been proposed. First Method. We may have a Table of Logarithmic sines calculated for every second for the first few degrees of the quadrant; in this case the greatest value of i is the circular measure of one 2 second, and thus = cosec’@ becomes small enough to be neglected, provided sin @ is as small as /:00006: the tables shew that the sine of half a degree is rather greater than ,/-00006, THEORY OF PROPORTIONAL PARTS. 139 Second Method. This is called Delambre’s Method. A. Table is constructed which gives the value of log 5 g + Lsin 1” for every second for the first few degrees of the quadrant. Let 6 be the circular measure of an angle of m seconds, then 6=nsin 1” approximately (Art. 123), therefore log ms Ae log eee log sin 2” — log n — log sin 1”, =Zsin n”—logn—Lsin 1”, therefore logn=Lsinn”— (log a 9 a0 + Zsin 1) . If the angle is known, then the Table gives the value of log Bing +ZLsin 1”, and log n can be found from a Table of the log- 6 arithms of numbers; thus the formula enables us to find Z sin n”. _ If the value of Zsin n” is given and we have to find n, we pro- ceed thus: since Z sin n” is known we can find approimately the value of the angle, and then from the Table we get the value of log + Lsin 1”; then the formula gives us log 2, and we can find » by an ordinary table of logarithms of numbers. In this operation we are liable to an error by using an approximate value of me instead of the real value. But it may be inferred from 6 Chap. 1x. and will be more fully shewn hereafter, that when 6 is jz small ss is very nearly equal to 1 — o and thus a small error in 6 0 ie not produce any sensible error in our calculations, since lo og 3 ne will vary far less rapidly than 6. Third Method. This is called Maskelyne’s Method. It may be used if Tables such as those described in the other methods are not accessible, 140 THEORY OF PROPORTIONAL PARTS. It may be inferred from Chap. 1x. and will be more fully shewn hereafter, that when @ is very small we have approximately 2 sin 6 = jo cosO=1-8; : 2 24 therefore ms =l- os (2 - 5) ‘approximately, = (cos 6) approximately, therefore log sin 6 = log 6 + 4 log cos 6 approximately. This formula gives log sin 6 at once if 6 be given. If log sin 6 be given, we must find an approximate value of @, and then find log cos @ approximately ; then we have log 6 = log sin 6 — 3 log cos 6. Here since cos 6 varies far less rapidly than 6, we are free from sensible error by using an approximate value of log cos @ instead of the real value. A similar formula may be found for the tangent of a small angle ; for sin 0 & 6 ; tan 6 aad = (9 - 3) (i - x) approximately, tan 0 6? 6? therefore = ( -5) Q + 5) 6? 6? “4 : =1l+ a (1 -5) approximately, therefore log tan 6 = log 6 — 2 log cos 0 approximately. 206. We will now sum up the results of the investigations of the present Chapter. The principle of proportional parts is applicable to all the trigonometrical functions natural and logarithmic with certain exceptions, which occur when the angles are small or nearly equal to a right angle. In the exceptional cases the differences of consecutive functions are sometimes irregular only; sometimes they are nearly insensible, and then they are also irregular. THEORY OF PROPORTIONAL PARTS. 141 For the natural functions we have the following exceptional cases. For the sine the differences are insensible when the angles are nearly right angles; for the cosine they are insensible when the angles are small. For the tangent the differences are ir- regular when the angles are nearly right angles; for the cotangent they are irregular when the angles are small. For the secant the differences are insensible when the angles are small, and irregular when they are nearly right angles; for the cosecant the differences are irregular when the angles are small, and in- sensible when they are nearly right angles. For every logarithmic function the principle of proportional parts fails both when the angles are small and when they are nearly right angles. For the log sine and the log cosecant the differences are irregular when the angles are small, and insensible when they are nearly right angles, For the log cosine and the log secant the differences are insensible when the angles are small, and irregular when they are nearly right angles. For the log tangent and the log cotangent the differences are irregular when the angles are small and when they are nearly right angles. 207. In using Trigonometrical Tables it is necessary to avoid as much as possible the cases in which the principle of pro- portional parts does not hold. In other words, we must endeavour to use a Table such that the differences of the function corre- sponding to given small differences of the angle are both sensible and regular. If the differences of the function are insensible for a certain number of decimal places we cannot by any method determine the value of the function for any intermediate angle, or perform the converse operation, so long as we are restricted to the certain number of decimal places. If the differences of the function are irregular we cannot determine the value of. the function for an intermediate angle, or perform the :converse operation, by the principle of proportional parts, though we may by retaining the terms which were neglected in the first approxi- mation, 142 THEORY OF PROPORTIONAL PARTS. 208. If we have to determine an angle from its natural sine or cosine it will be advisable to employ the natural sine if the angle be less than 45°, and the natural cosine if the angle be greater than 45°, For the differences of consecutive sines vary approximately as the cosine of the angle, and the differences of consecutive cosines vary approximately as the sine of the angle; thus the differences of consecutive sines are greater or less than the differences of consecutive cosines according as the angle is less or greater than 45°. A similar remark holds for the logarithmic sine and cosine. 209. The student who is acquainted with the elements of the Differential Calculus will see that all the results of the present Chapter may be obtained from Taylor's Theorem; and thus these results may be easily retained in the memory, or at least readily recovered when required. For example, consider the natural sine; we have by Taylor’s Theorem sin (0+) =sin 0 +h eos 6—%sin (0 +2), where A is some proper fraction. This formula shews that if we put sin (9 +/)=sin 0 +h cos0 2 the error is less than : . Moreover we see that when 6 is small the principle of proportional parts is especially applicable, for then 2 the term * sin (0+Ah) is extremely small in comparison with hcos@; and on the other hand, when @ is nearly 5 the principle 2 is not so appropriate, because then sin (9+Ah) may be sensible in comparison with h cos 6. Again, by Taylor’s Theorem, we have log sin (6 + h) = log sin 6 + ph cot 6— ue cosec’ (6 + Ah), EXAMPLES. CHAPTER XII. 143 where ». is the modulus and » some proper fraction. This equa- tion shews that the principle of proportional parts is in general applicable for the logarithmic sine, but that the differences of consecutive logarithmic sines are irregular when the angles are small, and insensible and irregular when the angles are nearly right angles, 210. The following application of Taylor’s Theorem will give a good mode of estimating the amount of error involved in the principle of proportional parts, Take the logarithmic sine for example ; we have log sin (6 + h) = log sin 6 + ph cot (9+ Ah), where A is some proper fraction. Thus the approximation uses cot @ instead of cot(@+Ah), The true value in fact of log sin(@ + h) —log sin 6 must lie between ph cot 6 and ph cot (6 +h), so that the error is less than ph {cot 0 — cot (0 + )}. MISCELLANEOUS EXAMPLES. 1. From one of the angles of a rectangle a perpendicular is drawn to its diagonal, and from the point of their intersection straight lines are drawn perpendicular to the sides which contain the opposite angle: shew that if p and g be the lengths of the perpendiculars last drawn, and ¢ the diagonal of the rectangle, pi+giac. 2. If two circles whose radii are a and 6 touch each other externally, and if @ be the angle contained by the two common tangents to these circles, shew that _4(a—d) J(ab) (a+ by? 5 3. Given seca sec 0 + tana tan 6 = sec , find tan 0. 144 EXAMPLES. CHAPTER XII. 4. Find the limit when 6=0 of ; 7) ; sin 5 cos 20 ie tant vers 6 cot 0’ rai sec 20—1° 5. Shew that ee” is never less than 1+cot 6 if @ lies be 2 tween (0 and =. 6 tanOd+c-—1 6 7. Find the condition necessary that the same value of 6 may satisfy both the equations asec?9—bcos@=2a, bcos’?O- asec 6 = 2b, 8. Eliminate a and @ from the equations a=sin a cos f sin 6 + cosa cos 6, 6 =sin a cos B cos 6 — cosa sin 0, c=sina sin B sin 0. 9. Eliminate a and f from the equations b+ccosa=ucos(a—6), b+ccosB=ucos(B— 6), a—B=28; and shew that w?— 2uc cos 0 + ¢* = 0 sec’ 6. 10. Eliminate x from the equations atan’*O-—a _ 2a tan 6 7 tan 2a tan 2a’ tan 2a+tan2a’ and shew that tan (2a + 2a’) = tan 20. 11. Eliminate 6 and ¢ from the equations sinO+sin d=a, cos6+cosd=b, cos(@—-¢) =e. — x; 12, Eliminate 6 and ¢ from the equations zcos6+ysind=a, xcos(6+ 2p) — y sin (6+ 24) =a, 6b sin (0+ $) =asin ¢. 13. Eliminate x and y from the equations tanat+tany=a, cota+coty=b, x+y=e EXAMPLES, CHAPTER XII. 145 14. Eliminate 6 from the equations a _ soc? 4 ~ cos’ 6 2D _ sec" 0 +008" 6 a@ sec’6+cos?6? y 15. Eliminate 6 from the equations (a +b) tan (6-¢) = (a—6b) tan(6+¢4), acos 26+ bcos 20=c. 2 a 22 s x , 16, Given 730080 =%; cos 6 +5, cos 6, x 7 y _ 2 me sin (0+0) sin(O—0) sin 20” shew that asa sin’ a 17. Eliminate ¢ from the equations ycoosP—axsing=acos2¢, ysingd+xcosd=2asin2¢; and shew that (x + y) +(e- y)s = 2a. 18. Eliminate @ and ¢ from the equations cos = SBE cos p = =~, cos (6—¢) =sin B sin y; and shew that tan’? a= tan’ 8 + tan’ y. 19. Eliminate @ from the equations m= cosec 0—sin 6, m=sec 6 —cos 6. 20, Eliminate 6 from the equations ‘ cos’@ sin’ @ 1 x sin 6—y cos 6=,/(a*+y*), aot errs 21. Eliminate 6 and @’ from the equations asin*6+a'cos’*0=0b, a’sin® 6’+a cos’ 0’ =0',. atan @=a' tan 6’; and shew that 3 + i A 6 6b a T. T. 10 1 ==+ Qa 146 EXAMPLES. CHAPTER XII. 2 “9 : : : cos’O sin? 1 22. Given a? +y’?=a?+0", ay=ab sina, ——+—,-=-, x y a a® shew that = cot 24 = cot 2a + pe cosec 2a. cos@ cos2a cos 3x 23, If i =——., shew that a, a, 3 a2 2a,-a-a aoe 2 1 3 sn’ = = ——__——.. 2 4a, sinv sin 3x sin da 94, = = : a, a, a, @,—24a,+a, a,—-3a shew that ee No eT as Oy 3 ; cos% cos(a4+6) cos(x+20) cos(a+30 25, Given = ( ) = (x+ 28) = ( ) ; a, a, a, a, a,+a Ay + G4, shew that eng a, a, tan agers iret a : cos 2a cos 2a’ A 4 26, If sintp= SS 7288 | then tant? = é cos” (a + a’) 2 tan (7 aa 4 sin (9-8) cosa cos (a +6) sin B sin (— a) cos B * cos(@—f) sina : tan 6 tana cos (a—f) _ 4 tangtanB cos(a+f) ”’ shew that tand=}(tanf+cota), tang=4 (tana —cot f), 2 _sinfsin6 tan (0-2) l+a cos(B—0) cotp ’ 27. If and 28. If a shew that a= (cot > — 2 cot p) (‘an gt 2 cot p) a j Re raude 7 : a. 29. Given sinésing¢=sinasinf#, tang@cosB=cot;, prove = that one of the values of sin § is sin 5 sin f. 30. Given sind=nsind, tand=2tan@, find the limiting values of n that these equations may coexist. EXAMPLES, CHAPTER XII. 147 31, Shew by means of a Trigonometrical formula that if utYyt+e= xyz, 2 2 ae 9 Dy 2 Fs ad Z 3 = z . 2y 3° law l-y l-# l-a& 1-y* t-32 32. Find the values of v, x, y, « from the equations sine siny sing C= Soe SS sing sind sine 33. Find the limit of (eos ace Ses when x is zero. 3 w@tytes2r. 34. From a table of natural tangents which goes to 7 places of decimals, shew that an angle may be determined within about goth part of a second when the angle is nearly 60°. 35. When an angle is very nearly equal to 64°36’, shew that the angle can be determined from its Z sine within about 7yth of a second ; having given log, 10. tan 64°36’ = 48492, and the tables going to 7 places of decimals, 36. Shew that a o & a es ¢ — tan? 5) @ — tan? 5) Q — tan? 5) oh ad inf = —* na” 37. If 4, B, C be positive angles which satisfy the equation suv d +sin?B+sin’C = 1, prove that 4 + B+ C is greater than 90°. 38. AB B Cc therefore b=ecosd, a=ccosB. Since cos A=sin B and cos B=sin A, we may also enunciate the proposition thus: in a right-angled triangle each side is equal to the product of the hypotenuse into the sine of the opposite angle. 213. In any right-angled triangle each side is equal to the pro- duct of the tangent of the opposite angle into the other side. From the figure of the preceding Article we have BC AC ag Ree therefore a=btan'A, b=atan B. tan A = SIDES OF A TRIANGLE AND FUNCTIONS OF THE ANGLES. 149 Since tan A = cot B and tan B=cotA, we may also enunciate the proposition thus: iz any right-angled triangle each side is equal to the product of the cotangent of the adjacent angle into the other side. 214. In any triangle the sides are proportional to the sines of the opposite angles. A A B D Cc B Cc D Let {BC be any triangle, and from 4 draw AD perpendicular to the opposite side meeting that side, or that side produced, at D. If B and C are acuée angles we have from the left-hand figure, AD=AB sin B, and AD=AC sinC; therefore AB sin B= AC sin, e sin€ therefore 5 snk If the angle C be obtuse we have from the right-hand figure, AD=AB sinB, and .1D=AC sin ACD=AC sin (180°-C)=AC sinC; therefore ABsnB=AC sin C, e sin therefore aR If the angle C be a right angle, we have from the figure of Are 212, AC = AB sin B, e 1 sin 7 ee 6 sn Bm 8" e sin@ Thus it is proved that in every case a eae 150 RELATIONS BETWEEN THE SIDES OF A TRIANGLE AND , Similarly Cee coy ee db sneb ¢ sin} The results may be written symmetrically thus, sind sinB sin('. « 6 ¢6? ‘ 1 and we shall shew hereafter that each of these is equal to ap mdb where & is the radius of the circle described round the triangle. 215. To express the cosine of an angle of a triangle in terms of the sides. Let ABC be a triangle, and suppose C' an acute angle. (See the left-hand ‘figure of the preceding Article.) Then by Euclid IT. 13, AB = BC? + AC’ -2BC. CD, and CD=AC cos€ ; therefore ce =a’ +b? — 2ab cos-C. Next suppose C an obtuse.angle. (See the right-hand figure of the preceding Article.) Then by Euclid IT. 12, AB = BC? + AC? + 2BC.CD, and CD = AC cos (180° — C)=— AC cos C, therefore C=a +0? —2ab cos C. 2 = 2 Thus in both cases we have cos C' = tac Moreover when C' is a right angle, a+ 0’=c¢? and cosC is zero ; thus the formula just found for cos C’ is true in every case, V+ —a +a? — 0? Similar] cos A = ————_,_ cos B= —. : y 2be ’ Lea 216. In ewery triangle each side is equal to the sum of the product of each of the other sides anto the cosine of the angle which wt makes with the first side. From the left-hand figure in Art. 214, we have LC = BD + DC =AB cos B+ AC cos C,, that is a=ccosB+b cos, THE TRIGONOMETRICAL FUNCTIONS OF THE ANGLES. 151 From the right-hand figure in Art. 214, we have BC = BD - DC = AB cos B - AC cos (180° ~ C) = AB cos B+ AC cos C, that is a=ccosB+ bcos. Similarly we shall have b=acosC'+ccosA, and e=bcosA-+acosB. 217. To express the sine, the cosine, and the tangent of half an angle of a triangle in terms of the sides. We have by Art. 215, B+c—al cos 4 = eo ee: _, #+e-a? a?-(b-c) therefore ices —— Step therefore sine = (a+ b—c) (ate-b) . 2 Abe Let 2s=a+5+¢e, so that s is half the sum of the sides of the triangle ; then ‘ a+b-c=a+b+e-2c=2(s—e), a+c—b=at+b+c—2b=2(s-6). snr4 0-9 6-9 2 be , _ A s — b)(s—e) and sing = /E-9E-9., P+ e—a? (b+0)?—-a?, Also jiegecrs Therefore 2be 7 2b6e 2 therefore ge = (+b +e) (b+c-a) a8 (s—a@) z 2 4be bc and os A = s a) From the values of sing and cos 4 we deduce A (s — 6) (s —¢) tn4 = ,/C=e—9 . 1 152 RELATIONS BETWEEN THE SIDES OF A TRIANGLE AND The positive sign must be given to the radicals which occur in this Article, because “ is less than a right angle, and therefore its sine, cosine, and tangent are all positive. Similar expressions hold for the sine, the cosine, and the tan- gent of half of each of the other angles. 218. To express the sine of an angle of a triangle in terms of the sides, A A . Since sin 4 =2sin 3 COS Gy» we obtain sin A = pp ae be ae b =a s(s—a) (s—6) (s—e). Or we may find sin A directly from the known value of cos -1 ; ~ 24 ___ (B+c%—a') 20%? + 2elat + 2a°b* —a*— bt — ct thus sin’ 4A =1 wt = We 5 Dh2p2 228 272 74 _ ft __ ot sin AN 20% + 2c’a? + 2a°b? —at— be : 2be the former expression may be shewn to agree with this by forming the product of the factors s, s—a, s—b, and s—c. therefore Similar expressions hold for sin B and sin C. 219. We have proved the formule in Arts. 214...216 inde- pendently from the figures; we may however observe that it is easy to deduce those in any two of the Articles from those in the third. Thus we may first establish as in Art. 216, that a=bcosC+ccosB, b=ccosA+acosC, c=acosB+bcos 4; multiply the first of these equations by a, the second by 6, and the third by ¢; then add the first two resulting equations and subtract the third; thus we obtain a? +0? -- o = 2ab cos C. Similarly the other two formule of Art. 215 may be deduced. ' THE TRIGONOMETRICAL FUNCTIONS OF THE ANGLES. 153 Then from these results we may proceed as in Arts, 217, 218, snd 2 : and shew that = ah A's (s — a) (s —b) (s — 0), abe sin B in C and that — and = are equal to the same expression. Thus sind _sinB _sin@ a 6 ¢ Or we may begin by establishing the formule of Art. 214 directly from the figure, and then proceed thus : sin f = sin (180°— 4) =sin (B+ C)=sin B cos C+ cos BsinC ; therefore 1=cosC ae +cos B a = e cos 0+ © cos B 3 sin 4 snd a a therefore a=bcosC' +c cos B, Similarly the other two formule of Art. 216 may be deduced ; and then those of Art. 215 will follow in the manner shewn in the beginning of the present Article. 220. The reason why an ambiguity of sign occurs in the formule for sing and cos 4 of Art. 217 may be explained as on former occasions. It will be observed that we have an expression ‘ pA ny for cos, and we proceed to deduce expressions for sin 9 and cos 5 ; and in Art, 96 it has been shewn that in this case we may expect two values differing only in sign for each of the required quantities. 221, Since the formule in Art. 217 have been strictly demon- ‘ _ A strated, they must of course always furnish real values for sin =, cos 4 , and ton 4, if the triangle really exist. That they do so may be easily verified from a known property of a triangle. 154 EXAMPLES. CHAPTER XIII. Take for example the formula ~ 24 (at+b-c)(a+e-b), ree Abe : ‘ aah , that this may give a possible value for sin a the expression on the right hand must be positive and less than unity. It is positive, because from the fact that two sides of a triangle are greater than the third, we have a+b —e positive and a+¢—6 positive. And the numerator is a’— (c—)’, and this is less than the denominator provided a* be less than (c— 6)’ + 4bc, that is provided a’ be less than (b+), which is obviously the case, MISCELLANEOUS EXAMPLES. 1, The sides of a triangle are a7+a%+1, 2a+1, and a°-1; shew that the greatest eal is 120°. nA 2. If cos 8B =—_~ mo? shew that the triangle is isosceles. 3. Ina ee triangle of which C is the right angle, A b+ © oo = 4. Ifin a triangle atan A+) tan B= (a+b) ten = that A=Z. +B shew 2 5. The angles of a plane triangle form a geometrical pro- gression of which the common ratio is }: shew that the greatest side is to the perimeter as 2 sin — is to unity. a 6. If A’, BY, C’ are the external angles of a triangle, shew that 2be vers A’+ 2ca vers B’ + 2ab vers C'= (a+b +c)’. 7. From the angle A of any triangle ABC a perpendicular AD is drawn to the base, and from D perpendiculars DE, DF are drawn to AB, AC respectively : shew that AL. EB.cos*C=AF. FC. cos? B. EXAMPLES. CHAPTER XIII. 155 8. Ifa, b, c be the sides of a triangle and the opposite angles be 26, 86, 40, show thot texte = (=) a +¢ 9. ABC is a triangle of which C is an obtuse angle: shew that tan 4 tan & is less than unity. 10. If the sides a, b,c of a triangle be in arithmetical pro- gression, shew that A-C B Cc A 36 = 9s = 2 2 — ay 2 sin}, and @cos’ 5 + ¢ cos*> 5 11. If D be the middle point of the side BC of a triangle cot BAD —cot B=2 cot A. . 12. If an angle of a triangle be divided into two parts such that the sines are in the ratio of the sides adjacent to them respectively, shew that the difference of their cotangents is equal to the difference of the cotangents of the angles opposite to their sides. 13. If the cotangents of the angles of a triangle be in arith- metical progression, the squares of the sides will also be in arith- cos metical progression. 14. Given the vertical angle and the ratio between the base and altitude of a triangle, find the tangents of the angles into which the vertical angle is divided by the perpendicular drawn from it to the base. 15. If the base of a triangle be divided into three equal parts, and ¢,, ¢,, #, be the tangents of the angles which they subtend at GAG) aGe)) q 16. If the sines of. the angles of a triangle be in arithmetical progression, the product of the tangents of half the greatest and half the least is 3. 17. If the side BC of a triangle be bisected at D and AD Le 2c sin A B-c the vertex drawn, shew that tan ADB = 156 EXAMPLES. CHAPTER XIII. 18. If A, B, C be the angles of a triangle and cot i cot a cot 2 in arithmetical progression, shew that cot 4 cot e = 3, 19. Straight lines are drawn from the angles A and B of a triangle dividing the angles respectively into parts whose sines are in the ratio of 1 to ; these straight lines intersect at D: shew that DC either bisects the angle C or divides it into parts whose sines are in the ratio of 1 to 2%. 20. If J be the length of the straight line which bisects the angle A of a triangle and is terminated by the base, 6 the angle which it makes with the base, 2s the perimeter of the triangle, shew that » (sin 0~ sin *) = los 4. sin 6. 21. If6and ¢ be the greatest and least angles of a triangle, the sides of which are in arithmetical progression, shew that 4 (1 — cos 8) (1 — cos ¢) =cos 6 + cos ¢. 22, From the angular points of a triangle ABC’ straight lines are drawn making each the same angle a towards the same parts with the sides of the triangle taken in order. Shew that these straight lines will form another triangle similar to the former, and that the linear dimensions of the two triangles are in the ratio of cosa —sin a (cot A + cot B + cot C+) to 1. Shew that in any triangle the relations given in the following Examples, from 23 to 42, hold: 23. a (bcos C—e cos B) =? - 0°. 24. a(cos B cos C +cos A) =6 (cos 4 cos C' + cos B) | =c(cos A cos B + cos C), 25. (b+ e~a) tan 4 = (¢+a—) tan 2 = (a4d—0) tan ©, 26. bcos B+ccos C=acos(B-C). 27. (a+) cosC'+(b+¢) cos d+ (c+a)cosB=atbte. EXAMPLES. CHAPTER XIII. 157 28. (a? — b*) cot C + (b? —c*) cot.A + (c?— a’) cot B=0. 29. (a—b) cot 5 + (0—a) cob % + (0-c) cot 4 =0. A B 2 30. 1—tan 5 tan = a" 31. (a+6+¢) (cos A +cos B+ cos C) oA B ; = 2a cos? 5 + 2b cos’ > + 2c cost S i 39 sin’.{ cos A cos B 4, 008 A cos an cos B cos 0 = a’ ab ac be 7 33. acosA+bcos B+ccos C =2asin B sin C. 2asin B sin 34. cos 4+cosB+cos 0 =14+ at+bt+e 35. a? —2ab cos (60° + C) = c? — 2be cos (60° + A). 36. cot a opie : ater :b+e-a: 2a, 4 2 2 2 A iB SC A : B C ga a” s/¥ 4x ame aes 3 a Ss ail 37. cos 3 cos 3 cos’ 5 43 € cos >) ( cos) @ cos) 5 A B where 23 = cos 2 + cos 5 + cos z: : ‘ A+B 38. The perimeter of any triangle is 2¢ cont cos 5 sec - . 39. Ifysim’A+asin°B=ssin?B+ysin’C=a2sin'C +2 sin, then e:y:2%:sin 2d ; sin2B: sin 2C. 40. 8 sin 4 sin % sin C is less than 1, except when A=B=C. 41, asin (BC) cos (B+ C - A) +bsin (C — A) cos (C'+ 4 — B) , +cesin(A—B) cos (4 +B-C)=0. } snd snB sn sind sin B sin C ae moa * cos cosd cos cosd cosB = sin A +sinB+sin C + (cos A + cos B + cos C) tan A tan Btan C. XIV. SOLUTION OF TRIANGLES. 222. Inevery triangle there are six elements, namely, the three sides and the three angles. The solution of triangles is the process Ly which when the values of a sufticient number of these elements are given we calculate the values of the remaining elements. It will appear as we proceed that when three of the elements are given, the remaining three can be found except when the three angles are given, and then we cannot determine the lengths of the sides but only the ratio they bear to each other. We shall have occasion to introduce logarithms into our formule, and we shall as before by the word logarithm or the abbreviation log denote a logarithm to the base 10; and by the letter Z placed before any Trigonometrical Function, we shall denote the tabular logarithm of that function, which is formed by adding 10 to the logarithm to the base 10. We shall begin with a right-angled triangle and shall suppose C the right angle. 223. To solve a right-angled triangle having given the hypo- tenuse and an acute angle. Suppose the hypotenuse and the angle A given; then B=90°- A; <= sin A, therefore a=c sin A, therefore loga=loge+logsin A= loge+Zsin A-10; : =sin B, therefore b=c sin B, therefore log 6 = log c+ logsin B = loge + Z sin B—10. Thus ZL, a, and } are determined. SOLUTION OF TRIANGLES. 159 224. To solve a right-angled triangle having given the hypo- tenruse and a side. Suppose ¢ and a given; then sin 4 = *, log sin .f = log a—loge; therefore Lsin A=10+loga—loge; this determines 4; then B=90°-— 4A. And c’=a'+0*, therefore 6*=c?—a’=(c—a) (c+a), therefore b=A/(c—a) (+a), log b = $ log (c— a) + 4 log (¢ +a). Or we may find } from the formula 6 =c cos. Thus 4, 8, and 6 are determined. 225. To solve a right-angled triangle having given a side and an acute angle, Suppose a and .f given; then B= 90°-A; a. a @ sm A, therefore c= and? log ¢ = log a — log sin A = loga—Lsin.1+10; ; = tan 4, therefore b= awl ‘ log b = loga —log tan A =loga —Ltand +10. Thus BZ, c, b are determined. If a and B are given, then 4=90°—B; thus J is known, and we may find ¢ and 6 as before. 226, To solve a right-angled triangle having given the two sides, Here @ and 6 are given ; then tan. =“, therefore log tan A = log a — log ,, b > therefore Ltan A=10+loga—logh; 160 SOLUTION OF TRIANGLES, B=90°- 4; a. a —=sind, therefore c=——, c sin A therefore logc =loga—Lsin A + 10. Or we may find ¢ from the formula c=,/(a°+b’), but this is not adapted to logarithmic computation. Thus A, B, and c are determined. 227. We may remark here that when an angle of a triangle is determined from its cosine, versed sine, tangent, cotangent or secant, no uncertainty can exist about the angle, because only one angle exists less than 180° for which any of these functions has an assigned value. But when an angle of a triangle is determined from its sine or cosecant uncertainty may exist, since there are two angles less than 180° which have a given sine or a given cosecant. But no uncertainty will exist in the case of a right-angled triangle, because each of the other angles of the triangle must be acute. We now proceed to the solution of oblique-angled triangles. 228. To solve a triangle having given two angles and a side. Suppose A and C’ the given angles, and 6 the given side ; then B=180°-A-C; sin A i ae therefore a= z —. A % sin B 6 sin B’ therefore log a =log b+log sin A—log sin B= log 6+Z sin A-—LsinB; similarly logc=logb+ZsinC—Lsin B. Thus B, a, and ¢ are determined. If A and B are the given angles then C=180°-B-A, and we may proceed as before to find a and c, SOLUTION OF TRIANGLES, 161 229. To solve a triangle having given two sides and the in- cluded angle. Suppose 6 and c the given sides and A the included angle. We have —— 3B ; therefore = oa oe 3 sn@ cc’ sin C c therefore pe Cee eee : ; that is sin AcotC’+cos4 = Z sin C thus cotC’ is determined, and therefore C can be found; and then B. But as this process is not adapted to logarithmic computation another is usually given: we have ae = a therefore anf ae = ae smC oc’ . sn B+sn@ b+e tan 3(B-C) _ therefore tan }(B+C) =o8, (Art. 88), and tan 1 (B+) =tan }(180°— ea therefore tan spate! A 2 b+ 3? therefore log tan 4(B—C) =log (b—c) — log (6 +c) + log cot pa 5 therefore L tan }(B - 0) =log (—¢)—log (b+e) +L cot S ; this formula determines }(B-—C); and 4(B+C) is known since it is 90° eee 3 thus B and C can be immediately found, Also — oe sna , from which a@ can be found. We have supposed 6 and ¢ unequal; if however 6=c then B=C, and all the angles will be known, so that a can be found as in Art. 228, T. T. 11 162 SOLUTION OF TRIANGLES, 230. In finding a from the expression just quoted we should require three logarithms, namely, those of c, sin A, and sin(’; in the following method we shali only require two new logarithms. a 6b c a b+¢ Wolere Goa p ang? aaa “en ee and sin B+sin@ = 2sin}(B+C)cos}(B—C) (Art. 84) : , =2cos}4 cos3(B-C), : 1 one ae (b +c) sind _ (b+0e)sin$A. ~2 cos 4.4 cos 4(B-—C) « cosf(B-—C) ’ as the logarithm of 6 +c’has been used in the former part of the solution, we shall only require two new logarithms, namely those of sin }.4 and cos 4(B -C). We may observe that we have also a b-e : sin A sin B-sinC’’ and from this in the way already shewn we can deduce _ (b-c) cost ~ sind (b-C) * @ Thus we have the two formule, acos 4(B-C) =(b+c) sin} A, asin 3(B-C)=(b-c) cos 3A. 231. We can also from the given quantities in the preceding Article determine the third side without previously determining the other two angles. For we have a’ =0°+ ¢c?—2bcecos A, by Art. 215; and we can transform this formula into another, which is adapted to logarithmic computation as follows : > @=0+e0- 2be( 2 costs 2 y = (6 +c)? — 4be cos? = (b+) {1- oar oF cos 3}. A 2 SOLUTION OF TRIANGLES, 163. Now find an angle 6 such that sin? @ = “on cos® A ‘ (6+ ¢) thus a? = (6 +c)? (1 —sin’ 6) = (6 +c)’ cos? 6, therefore a= (b+c) cos 6, therefore log a= log (b +c) + log cos 6 = log (6 +c) + Dcos 6-10; thus a is determined. It is usual to give the name of subsidiary angle-to an angle introduced into an expression for the purpose of putting it in the form of a product of factors. Thus 6 in the preceding investiga- tion is a subsidiary angle. We are certain that an angle exists which has the square of its sine equal to the given expression ; for that expression is positive, and it is less than unity because 4bc is never greater than (6+c)" and cost S is less than unity. The equation for determining 6 gives by taking logarithms 2 log sin 6 = log 4 + log 6 + log ¢ — 2 log (b+ c) salem, therefore 2L sin O=2log 2 +logb +loge—2log (5 +¢)+ 2L cos. 232. The process of Art. 229 is sometimes facilitated by tho use of a subsidiary angle when the logarithms of 6 and c are known. b-c A L(B-C)= = We have tani (B C) =F OG: b b-c_ tand-—1 ca Now let 7 tan d; therefore Fee steno 7 (9-9); A thus tan }(B-C)=tan (6-7) cot 5. Or thus, suppose ¢ less than 0; let c=bcos¢; b-c_ l-cosd_, oo. os bro" Trap 3! : A thus tan 3 (B -C) = tan’ 5 cot 5 - 11—2 164 SOLUTION OF TRIANGLES, 233, To solve a triangle having given two sides and the angle opposite to one of them. Let a and 6 be the given sides, and A the given angle ; then ane sl therefore dpe ind snd a a therefore L sin B=logb-loga+ Zsin A, —e is less than unity, two different angles may be bsin A if found less than 180° which have for sine, one of thesa angles being less than a right angle, and the other greater. Ifa be greater than 0, then 4 must be greater than B, and therefore B must be an acute angle; thus only the smaller value is ad- missible for B, If a@ be less than 6, then either value may be taken for B. When B is determined, C is known since it is 180°— A — B, and then ¢ can be found from sin C sin A“ c a Thus if two values are admissible for B we obtain two correspond- ._ ing values for C and c, so that two triangles can be found from the given parts. i bsin A a =1, then B is a right angle, so that only one tri- angle can be found from the given parts, If 6 sin A is greater than unity, no triangle exists with the given parts, Thus, when two sides are given and the angle opposite the less we can generally find two triangles from the given parts, and this case in the solution of triangles is therefore called the ambigu- ous case. We say that two triangles can be generally found in order to have regard to the exceptions; for the triangle may be SOLUTION OF TRIANGLES. 165 right angled, and then only one triangle can be found, or the triangle may be impossible. 234. The ambiguous case may be illustrated by figures. Cc aA Bee at “B D C i D B he f B ~. - Witkin Let CAD be the given angle A, and AC the given side 6; sup- pose a circle described from C' as a centre with radius equal to a. The perpendicular from C’ on AD is equal to bsin 4; therefore if a be greater than 6 sin A, the circle will meet the straight line AD at two points, which we will denote by B and BY. If a be less than 0, then B and B’ are on the same side of A, as in the first figure; thus two triangles, namely ABC and AB'C, can be obtained, each having the given parts a, 6, A. Ifa be greater than 2, then B’ and B are on opposite sides of A, as in the second figure ; thus only one triangle, namely C'AB, can be obtained hav- ing the given parts a, b, A; the triangle CAB’ has an angle CAB’ which is 180°— A instead of A. If a be equal to bsin A, the circle touches the straight line AD, and the two points B and B’ in the first figure coincide ; thus one triangle is obtained which has a right angle at B. If a be less than }sin A the circle does not meet the straight line AD, and no triangle exists with the given parts a, b, 4. 166 SOLUTION OF TRIANGLES. 235. In Art. 233 we first found the angle B, and afterwards the side ¢; we may however adopt another mode of solution and begin by finding c, For a =0' +c? — 2bc cos A ; therefore ce’ — 2be cos A+ b?- a =0; by solving this quadratic equation in ¢ we obtain c=bcosA+,/(a’—6' sin’ A), and we shall now discuss the values thus found for e. If a is less than 6 sin A, the values of ¢ are impossible, and no triangle exists with the given parts. If a is equal to bsin A, we obtain c=bcosA. If A be an acute angle, c is positive and one triangle exists with the given parts. If A be an obtuse angle, ¢ is negative, and this indicates that the triangle is impossible ; and in fact a is less than 8, since it is equal to dsin A, and so A cannot be an obtuse angle in a real triangle. If a is greater than b sin A, then two values occur for ¢, and these will both be positive if A be an acute angle and bcos A greater than ,/(a’—6’ sin’ A); the latter leads to the condition b’? cos’A greater than a’?—b’sin’A, that is, 0’ greater than a”. Hence we see as before that there are two triangles if A be an acute angle and a be greater than bsin A and less than 0. 236. To solve a triangle having given the three sides. Let s denote half the sum of the sides; then by Art. 217, sin 4 = (s— ee, cos = ge, A_ [{s=d)(s-9), a> = s(s—a) ? and similar formule are true for the other half angles. SOLUTION OF TRIANGLES. 167 The formule for the tangents of half the angles will be the best to use with logarithms, because then we only require the logarithms of s, s—a, s—6, and s—e, in order to find all the angles ; whereas if we use the formule for the sine or cosine we shall require in addition the logarithms of the sides. 237. When all the sides of a triangle are given, the angles may also be found by dividing the triangle into two right-angled triangles. . Thus, with the left-hand figure of Art. 214, we have AD? = AB*— BD’, and also = AC*?—CD?; therefore AB’ — AC? = BD*-~ CD’, therefore (4B + AC)(4B-AOC)=(BD+CD)(BD-CD); from this we can find BD—CD, and then since BD+CD is known we can find BD and CD; then CD BD cos B = —— cos =F; x AB’ thus B and C are determined, With the right-hand figure of Art. 214 we have as before (AB +AC) (4B - AC) =(BD + CD) (BD-CD); from this we can find BD+CD, and then since BD-CD is known we can find BD and CD; then ‘D BD C cos B= — 4: AB?’ thus B and C are determined. cos (180°—C) = 238. We have seen in Chap. x11. that the Tables of trigo- nometrical functions cannot always be used with: advantage; this circumstance guides us in selecting the method~of solution of a triangle to be adopted when more than one method is theoretically applicable, and leads us to modify the method of. solution. in some cases, For example, suppose we have to find A from the equation 168 EXAMPLES. CHAPTER XIV. sin A =n, where n is nearly equal to unity; this is an inconve- nient equation for determining A, because the difference of conse- cutive sines is nearly insensible when the angles are nearly right angles, We have however ae) a VP) AG) and this formula is free from the objection. Similarly if we have to find A from the equation cos 4 =n, where n is nearly equal to unity, we may advantageously transform the equation thus sin = JES aos) )=./ 3°) *) l-cosA 1-7, l+cosA “148? therefore tan 4 Es - SG GS + * or thus EXAMPLES. 1. Find the values of the angle A having given sin B= -25, a=5, b=2°5. 2. One side of a triangle is half another and the included angle is 60°: find the other angles. 3. The sides of a triangle are in the ratio of 2: /6:14+,/3: determine the angles. 4, If 4=30°, 6=100, a=40, is there any ambiguity ? 5. Having given A=18°, a=4, b=4+,/(80), solve the triangle. EXAMPLES. CHAPTER XIY. 169 6. Having given 4=15°, a=4, b=44,/(48), solve the triangle. 7. Ifa, 6, A be given, and a be less than 8, and if ¢, e’ be the two values found for the third side of the triangle, then e* — 2ce’ cos 24 + ¢? = 4a? cos? A. 8. Find the sum of the areas of the two triangles which satisfy the conditions of the problem in the ambiguous case. See Art. 247. 9. If B,,C,, and B,, C, are the angles of the two triangles in the ambiguous case, then sinC, sin C, + sin B, * sin B, 2—2cos A. 10. In the ambiguous case the area of one of the triangles is n times that of the other; shew that if b be the greater of the given sides and a the less, § is less than net . See Art. 247, 1. If loga+10=1log6+Zsin A, can the triangle be ambi- guous ? 12, If 6 be an angle determined from the equation cos 6= » prove that in any triangle A-B_(a+b)sin0 A+B_ csind ae no + ORE | Baby 13. If tan g=2M) sin , then c=(a—6) sec ¢. 14. In a triangle ABC in which a=18, 6=20, c=22, find Ltn, having given log 2= -3010300, log 3 =-4771213. 15. The sides of a triangle are 32, 40, 66: find the greatest angle, having given log 207 = 2°3159703, log 1073 = 3:0305997, L cot 66° 18’= 9°6424342, diff. for 1’= 0003431, 170 EXAMPLES. CHAPTER XIV. 16. The sides of a triangle are 4, 5, 6: find B, having given log 2 = ‘3010300, Zcos 27°53’ =9-9464040, diff. for 1’=-0000669. 17. Apply the formula cos = ee (Ss; = a to find the greatest angle in a triangle whose sides are e 6, 7 feet respect- ively, having given log 6 =-7781513, L cos 39° 14’ =9-8890644, diff. for 60” = 0001032. 18. Two sides of a triangle are 18 and 2 feet respectively, and the included angle is 55°: find the other angles, having given log 2=:3010300, Z cot 27° 80’ = 10°2835233, L tan 56° 56’=10°1863769, diff. for 1’ = 0002763. 19. Two sides of a triangle are in the ratio of 9 to 7, and the included angle is 64° 12’: find the other angles, having given log 2= 3010300, Z tan 57° 54’ =10-2025255, tan 11°16’ = 9-2993216, £ tan 11°17’=9-2999804. 20. If a=70, 6=35, C=36° 52’ 12”, find the other angles, having given log 3= 4771213, L cot 18° 26’ 6” = 104771213. 21. The ratio of two sides of a triangle is 9 to 7,, and the included angle is 47°25’; find the other angles, having given log 2=-3010300, Z tan 66°17’ 30” = 10:°3573942, L tan 15° 53’ =9°4541479, diff. for 1’= -0004797. 22. Ina triangle a= 30, 6 = 20, and the contained angle = 22°: find the other angles, having given EL cot 11°= 107113477, ZL tan 45°48’ = 100121294, ZL tan 45° 49'= 10-0123821, log 2 = -3010300. 23, Given b=14, c=11, 4 =60°, shew that B=71° 44’ 29”; having given Ltan 11°44’ 29"=9°31774, log 2=-30108, log 3 = 47712. EXAMPLES. CHAPTER XIV. 171 24. The sides of a triangle are 7 : 8, 9: determine all the angles, having given log 2 = -3010300, L tan 24° 5’ 40" =9°6505069, LZ tan 24°5’ 50” = 9-6505634, L tan 29°12’ 20" = 9°7474183, Z tan 29° 12’ 30” = 9-7474677. 25. In a right-angled triangle the hypotenuse c= 6953 and b=38: find B, having given log 3-475 = 5409548, log 6-953 = 8421722, Lsin 44° 59’ 15" = 9:8493902, diff for 1” = -0000021. 26. Two sides are 80 and 100 feet, and the included angle 60°: find the other angles, having given log 3=°47712, Z tan 10°53’ 36” = 9:28432. 27. Two sides of a triangle are 3 and 5 feet, and the included angle is 120°: find the other angles, having given log 4:8 = 6812412, LZ tan 8°12’ = 91586706, diff. for 60” = 0008940. 28. g B BD=roots, 3? therefore r (cot - + cot 5) =4, agin ‘ae +0 B.C oe =asin > sin =, therefore r= . therefore 7 sin 3 3? cos > Or thus: r= AB ton 5 = (s- a) ta. 250. To find the radius of the circle which touches one side of a triangle and the other sides produced, Let ABC be a triangle, and let O be the centre of the circle which touches the side BC’, and the other sides produced at the points D, BE, F. Join OD, OF, and OF. The angles at D, £, and F are right angles by Euclid m1, 18, Let 7, denote the radius of the circle, PROPERTIES OF TRIANGLES. 187 The quadrilateral OBAC may be divided into the two triangles OAB, OAC; therefore the area of this quadrilateral is 5M ne Again, the same quadrilateral may be divided into the triangles OBC and ABC; therefore the area of this quadrilateral is gnt8. Thus +8; c b a gregh =ar 1 therefore (¢ +5 —a) 3 =8S, that is r, (s—a) =S, therefore 7, = =. ‘ It is easy to shew by Geometry that AF=AE=s, BD=BF=s-c, CD=CE=s-8. The centre of the inscribed circle is also on AQ, and the dis- ‘ tance between it and O subtends a right angle at B and at C. Similarly, if r, be the radius of the circle which touches CA and the other sides produced, and 7, the radius of the circle which touches AB and the other sides produced, S S y= 7=——, 3 s—b’ 3 g-c¢ A circle which touches one side of a triangle and the other sides produced is called an escribed circle. 251. We may also obtain an expression for the radius of an escribed circle similar to that in Art. 249 for the radius of the inscribed circle. For, in the figure of Art, 250, the straight line OB bisects the angle which is the supplement of B, and the straight line OC bisects the angle which is the supplement of C Thus BD=r, cot (90" -3) , CD=r, cot (s0"- 5) therefore wh (tan : + tan A) =4; 188 PROPERTIES OF TRIANGLES. Cc B @COS > COS5 a COS Cos ae” a) therefore r= Zuo TT ‘ in cos 5 Or thus: r= AB tan4 =stand. 252. To find the radius of the circle described round a tri- Let ABC be a triangle, and O the centre of the circle described round it. Draw OD perpendicular to BC, then BC is bisected at D by Euclid ur. 3. Let & denote the radius of the circle. The angle BOC is double the angle BAC, by Euclid 11. 20; therefore BOD=A; a . 2sin A? thus & is expressed in terms of a side a the opposite angle. By Art. 218, sin A = F- 28 , therefore rae. and BD=Rhsn A= as therefore & = 253. Many theorems have been demonstrated with respect to the circles which have been noticed in Arts. 248...252; as an example we will find an expression for the distance between the centres of the inscribed and circumscribed circles of a triangle. PROPERTIES OF TRIANGLES, 189 Let ABC be a triangle; let O be the centre of the circum- scribed circle. From 0 draw a perpendicular OD on BC, and produce it to meet the circumference of the circle at H. Then the are BE is equal to the are CH; and therefore the straight line AZ bisects the angle BAC. Thus the centre of the inscribed circle will be on AX’; let the point J denote it. Join OJ and IC. The angle EI0=4(4+C) by Euclid 1 32; and the angle ECI=ECB+BCI=3(4+C): therefore the angle ZIC =the angle HCI: and therefore HI = EC. And #C=2Rsin f 3 therefore ZZ =2Rsin e ‘i _A r Hence EI x IA=2R sin 7a? 2Rr, 2 And (R- OL) (R+ OL) =2Rr, by Euclid ut. 35 ; therefore OF? = R° — 2Rr. If we suppose Z# produced through £ to a point J such that EJ = EI, the point J will be the ‘centre of the escribed circle which is opposite the angle 4 ; and we shall have OF* = R*+2Rr,. 190 PROPERTIES OF TRIANGLES. 254. To find the area of a quadrilateral which can be in- seribed in a circle. Let ABCD be the quadrilateral ; let AB=a, BC=b, CD=c, DA=d, The figure can be divided into the triangles ABC, ADC; its area therefore =4 (ab sin B+ cdsin D) =} (ab + cd) sin B, for the angles B and D are supplemental by Euclid 111. 22, Now from the triangle ABC, AC? =a’ + — 2ab cos B, and from the triangle CDA, AC? = 0° + d?—2cdcos D = c? +d’ + 2cd cos B; therefore ce? + d’?+2cd cos B=a? + B’ — 2ab cos B, @+0—-d 2Q(ab+ed) ? 1 _(@+¥-e-a") 4 (ab+ ed)? therefore cos B= therefore sin’ B= PROPERTIES OF TRIANGLES. 191 _ {2 (ab +ed) +07 + d*— a? —b*} {2 (ab +cd) — c?-d? +.0°+ 0" 4 (ab + cd)? _ e+ d)?-(@-8)"}{(@ + bY — (e-4)"} 4 (ab + ed)° _(¢e+b4+d—a)(at+e+d—b) (at+b+d—c)(at+b+e- d) ~ 4 (ab+ cd)? Now let $(a+5+¢e+d)=8; thus 16 (s —a) (s—8)(s-¢)(s—@) argc aS 4 (ab +cdy Hence the area of the quadrilateral =A{(s— a) (8-8) (s—¢) (8— @)}. If we substitute the value of cos.B in the expression for AC’, 2cd (a? + 6° — c? — d*) 2 (ab + cd) cd (a° + O°? —¢° — d*) ab+cd _ (ac+ bd) (ad + be) ab + cd we obtain AC@P=a+d?+ =c4d?4 Similarly it may be shewn that a+d°-b'-e ee _ (ae+ bd) (ab + cd) | ad + be The radius of the circle described round the quadrilateral may be easily expressed ; for this circle passes round the triangle ABC, hence by Art. 252 its raditis AC. (ab 4- cd) (ac + bd) oie. ~2sin BB =j {ee He-aG da) 192 PROPERTIES OF TRIANGLES. 255. To find the radii of the inscribed and circumscribed circles of aw regular polygon, that is of a polygan which has all its sides equal and all its angles equal. Let AB be the side of a regular polygon of m sides; let O be the centre of the circles, OD the radius of the inscribed circle, OA the radius of the circumscribed circle. Let AB=a, OA=R, OD=r. The angle AOB is the n®™ part of 4 right angles, that is, 40B ==", 40D =", AD=S=Rsin—~=rtan-; n n therefore a 2 sin — 2 tan — 256. The area of a regular polygon may be expressed by means of the radius of the inscribed circle, or of the radius of the circumscribed circle. For with the figure of Art. 255, the area of the triangle AOB =34AB.0D= aa nr a gg oti =z cons PROPERTIES OF TRIANGLES. 193 therefore the area of the polygon Also the area of the polygon = zr’ tan® = cot <= nr tan= ‘ 257. To find the area of a circle. The area of a regular polygon of m sides described about a circle of radius r ° T =nr* tan -. n Now suppose ” to increase without limit, then the area of the polygon approximates continually to the area of the circle as its limit, and therefore the area of the circle will be equal to the limit of the above expression, But when 7 is indefinitely great, n ton == m, (Art. 119); therefore the area of a circle of radius r= ar". 258. To find the area of a sector of a circle. Let 6 be the circular measure of the angle of the sector; then area of sector 6 é aes ee (Euclid vi. 33) ; 6 therefore the area of the sector = a7* x 5 => Since @ is the circular measure of the angle of the sector, the length of the are of the sector is r@; hence the area of a sector is equal to half the product of the length of the are into the radius. The area of a segment of a circle can now be found. For a segment of a circle which is less than a semicircle is equal to the: 13 T.T. 194 EXAMPLES. CHAPTER XVI. difference between a sector and a triangle; so that if @ be the cir- cular measure of the angle of the sector the area of the segment is " (6-sin 6). A segment of a circle which is greater than a semicircle is equal to the difference between the circle, and a segment less than a semicircle. EXAMPLES. 1. The sides of a plane triangle are 24, 30, 18: find the area. 2. Two angles of a triangle are 15° and 45°, and the included side is 10 feet: find the area. 3. The sides of a triangle are equal to 3 and 12 respectively, and the contained angle is 30°: find the hypotenuse of an equal right-angled isosceles triangle. 4. The area of a triangle = } (a’ sin 2B + 6’ sin 2.4). : a’ —b* sin A sin B 5. The area of a triangle = 2 sin (d—By’ ; 2abe A B 6. Th fat le = ——_—__ = aa = e area of a triangle aad ae 8g O87 CHS: Shew that the triangle whose sides are proportional to gh(k +P), kg? +i), (hk +l) (hl-ghk) has its area and the trigonometrical ratios of its angles rational. 8. The sides of a triangle are in arithmetical progression, and its area is to that of an equilateral triangle of the same perimeter as 3 is to 5. Find the ratio of the sides and the value of the largest angle. 9. If the alternate angles of a regular hexagon be joined so as to form another regular hexagon, and again the alternate angles of the latter hexagon be joined, and so on, shew that the sum of the areas of all the figures so formed = a where S is the area of the original figure. EXAMPLES. CHAPTER XVI. 195° 10. If we proceed with a regular figure of m sides, and of area S,as with the hexagon of Example 9, and § denote the sum of the areas so formed, shew that & sin 7 sin = ne” ‘ n nN Explain the cases where 2 = 38 or 4, 11. Ifan equilateral triangle be described with its angular points on the sides of a given right-angled isosceles triangle, and one side parallel to the hypotenuse, its area will be 2a*sin 60° (sin 15°)’, where a is a side of the given triangle. 12, Shew that, with the notation of Arts. 248 and 250, Ag Be? rrr,=r cot 5 cot? 3 cot? 5° 13. The straight lines which bisect the angles 4, C of a triangle ABC meet the circumference of the circumscribing circle at the points 4’, C’: shew that A’C’ is divided by CB, BA into three parts, which are in the proportion i oA : 2 sint ae a : sin? 2 sin > Sa 9 Ss. 3 3° 9 . 14. If A be the difference between the sides containing the right angle of a right-angled triangle, and S its area, the diameter of the circumscribing circle is equal to ,/(A’ + 48). 15. The sides of a plane triangle are 5, 5, 6: compare the radii of the inscribed and circumscribed circles, 16. O is the centre of the circle circumscribed round an acute- angled triangle, and AO is produced to meet BC at D: shew that DO cos (B-—C)=A0 cos A. 17. Accircle is inscribed within a given triangle, and another triangle formed by joining the points of contact; within this latter triangle a circle is inscribed, and another triangle is formed as 13—2 196 EXAMPLES, CHAPTER XVI. before, and so on continually: shew that the triangles thus formed ultimately become equilateral. 18. The sum of the diameters of the inscribed and circum- scribed circles of any plane triangle is equal to acotA+bcot B+ecotC. 19. Perpendiculars are drawn from the angles 4, B, C of an acute-angled triangle on the opposite sides, and produced to meet the circumscribing circle: if those produced parts be a, £, y respectively, shew that a45 #552 (tan A + tan B+ tan C), B 20. In any triangle the area of the inscribed circle is to the C ‘ . B area of the triangle as 7 is to cobs cot 3 cot >. 21. On each side of an acute-angled triangle as base an isos- celes triangle is constructed externally, the sides of each being equal to the radius of the circumscribed circle: if the vertices of these be joined a triangle will be formed equal and similar to the original, 22. If & be the radius of the circumscribed circle of a triangle, acos4+bcosB+ccosC =4h sin A sin Bsin C, 23. O is the centre of the circle circumscribed about a triangle ABC; from O the perpendiculars OD, OF, OF are drawn to the sides: shew that 4 (OD? + OF* + OF’) =a’ cot?A +b? cot B + c* cot? C. 24. If +r be the radius of the circle inscribed in a triangle, and 7, the radius of the circle inscribed between this circle and the sides containing the angle A, shew that ee (cos # — sin z 2 4 4 r,=7 =r : ita eee i 4 3 ( ¢tsiny) EXAMPLES. CHAPTER XVI. 197 25. Ifr be the radius of the circle inscribed in a triangle, and r,, 7, 7, the radii of the circles inscribed between this circle and the sides containing the angles A, B, C' respectively: shew that J (rat's) + xii (yr, + if (ra) =r, 26. Ifa triangle A’B’C’ be formed by joining the feet of the perpendiculars let fall from A, B, C on the opposite sides, shew that B’C’ is numerically equal to R sin 2.4, where # is the radius of the circle circumscribed about ABC, 27. Perpendiculars drawn from the angular points of an acute-angled triangle to the opposite sides meet those sides at the points D, #, F: shew that if # and R, be the radii of the circles described about the triangles ABC and DEF respectively, and r, the radius of the circle inscribed in the latter triangle, #,=4F, and r,=2R cos A cos B cos C. 28. Ifr, 7, 7,, 7, denote the radii of the inscribed and escribed circles of a triangle, shew that Cae 20 rr," 29. If A be the area of the circle inscribed in a triangle, A,, A,, A, the areas of the escribed circles, then af Se 1 i 1 V4 V4 V4. VAS 30. If the sides of a triangle be in arithmetical progression, the perpendicular on the mean side from the opposite angle, and the radius of the circle which touches the mean side and the other two sides produced, are each equal to three times the radius of the inscribed circle. 31. The distances of the centre of the circle inscribed in a triangle from the centres of the three escribed circles are respec- Cc tively equal to wee bseo, eset. 198 EXAMPLES. CHAPTER XVI. 32. Two similar triangles have a common escribed circle touching sides not homologous a,, 0, Shew that a,: a,=sinB+sinC—sind : sind +sinC@ —sin B. 33. If 0,, O,, O, are the centres of the escribed circles of a triangle, then the area of the triangle O,0,0, 3 a b c =area of triangle ABC {1 + Fageaa anead + ao} 34. The centres of the three escribed circles of a triangle are joined: shew that the area of the triangle thus formed is = where r is the radius of the inscribed circle of the original triangle. 35. A’, B’, C’ are the centres of the escribed circles of a tri- angle; A’, B’, C’ are joined so as to form a triangle: if r and r’ be the radii of the circles inscribed in ABC and A’B'C’ respectively, A iB ¢ re cot z cot z cot a u Bee ess Sie 2 2 2 36. If +r be the radius of the circle inscribed in a triangle ABC, 2s the sum of the sides, 7’, 2s’ similar quantities for the triangle which is formed by joining the centres of the escribed circles, shew that 2 ay sin sine ain © rs! 2 2 2° 37. Let a, a, be the distances of the angle A of a triangle from the centres of the inscribed circle, and the circle touching the side a and the other two sides produced; 8, 8, similar quantities for the angle B; y, y, similar quantities for the angle C’: shew that apy apy. = (abo. EXAMPLES. CHAPTER XVI. 199 38, Shew also that m4 eae I BY oy" 39, Shew also that a a 2 3+ B é 2 “+ + G z *) =O a 0; Shey aisoAtay 6 e228 2g, aa bB, cy, 41, There is only one point within a triangle, such that if perpendiculars be drawn from it to the sides, circles can be in- scribed in each of the three resulting quadrilaterals: prove this, and if p,, p2, ps be the radii of these circles, and p that of the inscribed circle of the triangle, then I isl WW sf isl 1 1 Wl l\1 G-DE-D-C-DE-D-C-DE-D- 1 PS \Pa P 2 Pp 3s Pp Ps P/ \Pi PLP 42. A circle is inscribed in a plane triangle ABC. Another circle is inscribed so as to touch the two sides AB, AC, and the last circle; again, a third circle is inscribed so as to touch the same two sides 4B, AC, and the second circle, and so on. Circles are also inscribed in the same way so as to touch BC, BA and CA, CB. Shew that the area of the inscribed circle is to the sum of the areas of all the other circles as 1 is to pee a hee a ee S e eg OG 4 2 gn ae 43. O and O’ are respectively the centres of the circles described about and inscribed in a plane triangle ABC. Join OA, OB, OC, O’A, O'B, O'C, and let R,, B,, B,, 72) 7p) V5 be respect- ively the radii of the circles Geeumede tne the ieee BOC, COA, AOB, BO'C, CO'A, AO’B. If R be the radius of the circle circumscribing thé given triangle ABC, shew that rrr, & at ae c 7 “abe a+b+o? E eae 44, From any point P within or without a triangle ABC, perpendiculars PA’, PB’, PC’ are dropped on the sides BC, CA, 200 EXAMPLES. CHAPTER XVI. AB; and circles are described about the triangles PA’B’, PBC’, PC’ A’. Shew that the area of the triangle formed by joining the centres of these circles is one-fourth of the area of the triangle ABC. 45, Three circles touch each other externally: prove that the square of the area of the triangle formed by joining their centres is equal to the product of the sum and product of their radii, 46. If the sides of a triangle be in geometrical progression, and the perpendiculars from the angles on the opposite sides be taken as the sides of a new triangle, then the angles of this new triangle will be equal to those of the original triangle. 47. Ifa, B, y be the ratios which the sides a, }, c of a triangle bear to the perpendiculars on them from the opposite angles A, B, C, then a? + B'+y?-2(aB+ By+ya)+4=0. 48. On the sides of any triangle equilateral triangles are described externally, and their centres are joined: shew that the triangle thus formed is equilateral. 49. The sides of a triangle are 65 and 25, and the difference of the opposite angles is 60°: find all the angles, having given log 3= 4771213, log 2= :3010300, Z tan 52° 24’ =10-1134508, ZL tan 52°25’ = 10-1137122. 50. If perpendiculars be drawn from the angles of an acute- angled triangle to the opposite sides, shew that the sides of the triangle formed by joining the feet of those perpendiculars are * acos A, bcos B, and ccos (’; and thence shew that a’ cos’ A — b? cos’ B — c? cos? C’ 2be cos B cos C = e089 24. 51. Six circles are inscribed between the three escribed circles of a triangle and the angular points, each touching a side and a side produced: shew that the products of their radii taken alternately are equal. EXAMPLES. CHAPTER XVI. 201 52. Straight lines are drawn from the angles 4, B, C of a triangle through any point P meeting the opposite sides of the triangle at the points A’, B’, C’ respectively: shew that AB’. BC’. CA’ = AC’. BA’. CB. 53. Shew conversely that if the relation just expressed holds then the straight lines 44’, BB’, CC’ meet at a point. 54, Shew that the perpendiculars from the angles of a tri- angle on the opposite sides meet at a point. 55. Shew that the straight lines which bisect the internal angles of a triangle meet at a point. 56. Shew that the straight lines which join the angles of a triangle with the middle points of the opposite sides meet at a point. 57. Shew that the straight lines which join the angles of a triangle with the points where the inscribed circle touches the opposite sides respectively, meet at a point. 58. Let a straight line be drawn from the angle A of a tri- angle to the point where the escribed circle opposite to the angle A touches the side opposite to it; let similar straight lines be drawn from B and C with respect to the other escribed circles: shew that these straight lines meet at a point. 59. In the figure of Art. 250 shew that the straight lines BE, CF, and AD produced meet at a point. 60. A quadrilateral figure is so taken that a circle can be described about it and inscribed in it. If its sides be produced in both directions, and r,, 7,, 7,, 7, be the radii of the circles, in- scribed in the triangles formed on two sides, and escribed on the other two sides, then r,7,7r,7,=7', where r is the radius of the circle inscribed in the quadrilateral. ( 202 ) XVII. USE OF SUBSIDIARY ANGLES IN SOLVING EQUATIONS AND IN ADAPTING FORMULA TO LOGARITHMIC COMPUTATION. 259. We shall now shew how to obtain the numerical values of the roots of a quadratic equation by the aid of Trigonometrical ~ Tables, w Suppose the equation to be a — 2nxn + q=0, where p and g are both positive; from this equation we obtain x=p+,/(p’-9)=p {2 «f(t -4)}- Now: if ¢ is less than p’ assume 3 =sin’@; thus x=p(1+cos6)= ap costs, or 2p sin’ § If q is greater than p* the roots are impossible; we may then assume = sec’@ ; thus e=p{1+,/(-1) tan 6}. (2) Suppose the equation to be x’ — 2nu—q=0, where p and g are both positive; from this equation we obtain: “=p + Jere d=p {i+ /(1 +f) Now assume tan’9 = 7 ; thus cos 6+] cosO +1 =,/ cos 8 sin 6 x=p(1+sec0)=p 6 6 = Jy oot 5 or ~/q tan 5. USE OF SUBSIDIARY ANGLES. 203° (3) If the equation is of the form #*+2px7+q=0, where p and q¢ are positive, we can solve the equation x* — 2pa+q=0, and then change the signs of the roots (Algebra, Art. 340). (4) If the equation be of the form 2*+2p2—q=0, where p and ¢ are positive, we can solve the equation x’ —2px—q=0, and then change the signs of the roots, 260. In like manner we may obtain the numerical value of the roots of a cubic equation by the aid of Trigonometrical Tables; we will exemplify this by considering one case. Let the equation be 2°—ga-r=0, and suppose 27r* less than 4q°. Put w=ny; thus n'y —gny—r=0, 9 as ee therefore ye 0 Now by Art, 91, cos’a — cos a ae =0; 3 r _cos3 assume y=cosa, cai then 73 = 7 =} 4 3 3 § thus n= (2) » cos 8a=4r () ; the last equation determines 3a, and thus a is known, then 4q } y=cosa and n= ncosa= (2) COS a. The value of cos 3a is less than unity, since we have supposed 27r* less than 49°. It appears from Art. 105 that we might also suppose Qa é y =.cos GF #2) ; 204 USE OF SUBSIDIARY ANGLES. consistently with the value of cos 3a given above; thus finally the three roots of the cubic equation are s 4 2 (3) cosa and 2 (3) cos (F + a) 5 3 where cos 3a == (-) ‘ 2\9 261. “If in mathematical researches equations like those that have been given of the second and third degree, presented them- selves to be solved, their solution would be conveniently effected by the preceding methods, and by the aid of the Trigonometrical Tables; but the truth is, in the application of Mathematics to Physics the solution of equations is an operation that very rarely is requisite, and consequently the preceding application of Trigo- nometrical Formule is to be considered as a matter rather of: curiosity than of utility."—(Woodhouse’s Trigonometry.) 262. To the examples which have already occurred of the use of subsidiary angles we will add two more, See Art. 231. (1) Required to adapt a+6 to logarithmic computation. If a and 6 are necessarily positive we may proceed thus: assume Be tan’ 6; then a atd=a(1+2)=a(1 + tan’ 6) = asec’ 6, If a and 6 are not necessarily both positive we may proceed thus: assume * tan 6; then a+b=a(1+2)= a (1+ tan 6) = ot (Se am) V2 “2 .0(-9) ‘EXAMPLES. CHAPTER XVII. 205 (2) Required to adapt acosa+bsina to logarithmic computation, Let oa tan 9; thus . db. ‘ acosa+Dsina=a(cosa+ sina) =a (cosas tan 0 sin a) a cos 6 =, 008 (a-6) or “7 00s (a+ 6). os 6 MISCELLANEOUS EXAMPLES. 1. Solve 2°—-62+4=0. 2. Shew that the roots of the equation 2°-—3a—1=0 are 2 cos 20°, —2sin 10°, — 2 cos 40° 3. Shew that the roots of the equation 2°—p2*+qz—r=0 a 4 2 are 2 eG cosa, 2 (5) cos (F+ «) , and 2 (F) cos (F Ea «) ; 5 where cos 5a = 5 (2) 3 = 5 provided p*= 5g and (5) be not greater than (6) . 4, Find the roots of the equation x — 102° + 202-8 =0. 5. A person wishes to ascertain the side BC of a triangular field ABC, but is only able to make measurement of dines within the boundary of a circle which passes through A and touches BC; shew how after measuring four straight lines he may determine BC. 6. Two men standing at the same point C observe the hori- zontal angle subtended by two objects 4 and B; they then both move away, one in the direction AC, the other in the direction BC, until each observes the horizontal angle to be half what it was before. The distance each walked being given and the horizontal angle at C, determine the distance 4B, ‘ 206 EXAMPLES. CHAPTER XVII. 7. The altitude of a balloon is observed at three places A, B, C simultaneously to be 45°, 45°, and 60° respectively; A and B are respectively West and North of C’: form an equation for determining the height of the balloon. 8. The distances 6 and ¢ of a station A from two other stations B and C are known, and the angle BAC is required. It not being practicable to observe the angle BAC, the angle BOC (a) and the angle AOC (8) are observed at a position O situated in the plane ABC, at a small known distance » from A, such that the triangle ABC is entirely within the triangle OBC. Shew that if 6 be the circular measure of the angle (BAC —BOC) then approximately sin(a—f) sin8 g=n{—G—P art. 9. Ata distance of 50 feet from the foot of a tower the eleva- tion of its top is 45°: if the elevation and the distance be correctly measured within 1’ and 1 inch respectively, find approximately the greatest error in the height. 10, A person standing at a distance a from a tower sur- mounted by a spire, observes the tower and the spire to subtend the same angle: if 6 be the known height of the tower, express the height of the spire (c) in terms of 6 and a. If y be the error in the height of the spire corresponding to a small error f in the height of the tower, shew that 1B hy 4b? a? e 6 }. a‘ — b* 11. The side a of a triangle and the opposite angle A remain constant: shew that the small variations of the other sides y and f are connected by the relation ysec C+ PB sec B=0. 12, The angular altitude and breadth of a cylindrical tower on. a level plane are observed to be a and f respectively; and at a point a feet nearer the tower they are observed to be a’ and pf’: EXAMPLES. CHAPTER XVII. 207 find the height and the radius of the tower. Find also the relation existing between u, a’, B, 6’. 13. In the preceding Example if the observed angular breadth be subject to an error 6, and if p be the greatest consequent error in the calculated radius (7), shew that p will be given by the equation 2p 4 r ee zh - B) {cosec F cosee oF cot ef cot B rt 8. If B=60°, 6’=120°, d=the circular measure of 6’, find approxi- mately the ratio of the greatest error in the calculated radius to the radius. 14. P, Q, & are three known positions in a straight line, and PQ, QF are observed to subtend equal angles at a certain point S: find the error in the calculated distance of S from Q in conse- quence of a small error « in the observed angles. XVII. INVERSE TRIGONOMETRICAL FUNCTIONS. 263. The equation sinz=a asserts that w is an angle of which the sine is @; it is found convenient to have a notation for expressing this relation in which a stands alone. The notation used is this, #=sin™a. Similarly the equation #=cos™’a ex- presses that « is an angle of which the cosine is a; and x=tan™'a expresses that x is an angle of which the tangent is a; and so on. 264. Experience will prove that the notation here given is often convenient; and we may shew that it is not altogether an arbitrary notation, but one that naturally presents itself. or, let any function of a be denoted by f(x); then the same function of J (x), that is, f{f(x)}, may be briefly and conveniently denoted by f°(x). Thus, for example, the logarithm of the logarithm of « may be denoted by logs. Similarly f[,/{/(x)}] may be briefly and conveniently denoted by /*(x); and so on, Thus with this notation we have, when m and n are positive integers, SS (=f (2). 208 INVERSE TRIGONOMETRICAL FUNCTIONS. Now we may examine what meaning it will be necessary to... ascribe to f(a), in order that the relation just given may hold when m or ” is zero. Suppose 2 =0, then the relation becomes LP) =F"), this leads us to settle that /°(x) shall be considered equal to x. Again we may examine what meaning it will be necessary to ascribe to f~'(a) in order that the relation f"f"(a) =/"*"(«) may hold when m or n is—1. Suppose m=1 and n=-1; thus the relation becomes SF "(@) =f"(@) =, s0 that f~'(x) must denote a quantity whose function f is a. Thus sin“'x should denote a quantity whose sine is 2; and this is the meaning which we have already assigned to the symbol, It will be observed that consistently with the remarks here made, sin®# should stand for sin (sin x), and not for sinw x sina. But as sin (sin x) is a function which rarely occurs, it is custom- ary to use sin’ x for what should be denoted by (sin x)’. 265. Any relation which has been established among trigo- nometrical functions may be expressed by means of the inverse notation, Thus, for example, we know that 2tan 0 26= ———_ ; ae 1-tan’ 6’ this may be written 2 tan 6 Dhaene a¢=tan (i — tan? i) g let tan 6=a, so that 0=tan™'a; thus 2a -a 2tan™a=tan™ ia" Similarly the relation sin 36=3sin@—4sin’?@ may be ex- pressed thus, 3 sina =sin™'(3a—- 4a"). EXAMPLES, CHAPTER XVIII. » 209 EXAMPLES. Shew that tan~'? = 2 tan}. 2. Find the value of sin (sin 3 + cos™!4). 1 a -1 s—1 3. Shew that sin 85 = sin 5 +s 7 . 4, Find the value of tan (tan™*x + cot™'z). 5. Shew that tan“? 14 tan“? 3+ tan7$+tan7}= 6. Shew that tana = tan? — B + elas +tan™'e. 1+ad 1+ be 7. Find the tangent of 3 tan™ ve tan” il +tan7! i a2 7 as 26 4 8. Shew that tan™{(,/2 + 1) tan a} — tan7'{(,/2 — 1) tan a} = tan“(sin 2a), 9. If tan (@—a) tan (6 — 8) =tan%0; then 6=ttan™ Se 10. Shew that cos , nc 3) +cosec7? 4 i) =f: 11, Shew that sin“ £4 sin 4 sin 28 ==, 12. Shew that 3 tan Teton =i —tan™' aa 13, Shew that tan Tr +tan™ aie =F 14. Shew that tan (2 tana) = 2 tan (tan~’a+ tan™a’), 15. Shew that tan™' (} tan 2.4) + tan™’ (cot A) + tan™’ (cot?4) = 0. TT 14 210 EXAMPLES, CHAPTER XVIII. 16. Shew that 26 ee ON 7 5) G = tan (F+ eos 5) +tan (F 4 cos a): 17. Shew that ° -1% b° 2 10 9 coseo® (tan sy + 3 sec (gtan 7) = (a +B) (a? +0. Sclve the following seven equations in a: oS 18. sin7'2 + sin7 7 bol 8 -1 2a 1 ‘ ~ -, 2b -1 19. sin Teqet Sim Tae? tan x. 20. tan7' (@ —1)+ tana + tan” (w+ 1) = tan” 3x. 21. sin! 2a —sin™e /3 =sin™z, 22, tan™'44+2tan7 1+ tan" 44+tan™ = =a 23. sin 2 cos”' cot 2 tan™'x=0, 1 ; 94. tan7?—~=tan™ 1 fom = z a—l x w@—x“+1 25. If sec @—cosec 6 = ; , shew that 6= 5 sin7 :. 26. If sin (7 cos 6) = cos (7 sin 6), shew that 0=+ 4 sin™ 3. 27, Shew that if sin"@ +sin’6= 4, then (2n+1)5 is one of the values of y which satisfy the equation oO y=sin™ (sin 6+ sin $) + sin“ (sin 6 —sin ¢). 28. Find x from the following equation, —tan7 J = tan ‘ Tee ea ” @ 3 EXAMPLES. CHAPTER XVIII. O11 29. Shew that one of the expressions . ., 2b+a-¢ : NE sin™' —____ + 2 sin ) ate ate, is an odd multiple of ze 30. Find all the positive integral solutions of tan™ «+ cot! y =tan™ 3. 31. Shew that if ¢ be a positive integer, the equation tan” @+tan"'y=tan™¢ has no positive integral solutions; while the equation cot a+ cot y=cot7'¢ has as many as there are different divisors of 1 + c’. 32. Shew that tan™ ; pg hr oe a geaet fa Cy + % C4, + 1 -1_ 37 &g ‘ aC, aaa re +tan7* —§—4 + 00... +tan™ ——*=1 + tan™'—, cc, +1 Cnn th ¢, where ¢,, ¢,,...... C, are any quantities whatever. .Shew that we can express the sum of any number of angles of the form sin™ wo sin" ee, naan in the form sin aa , where m and 7 are rational functions of a, b,.a’, d’,... m+n -1 (- ) m 77? where no 34. Write down the general value of sin is an integer. . 1 (- 1)" 35. Write down the general value of cos™'+—, where m is an integer. 36. Write down the general value of tan™’(—1)", where is an integer. oe 14—2 ( 212 ) XIX. DE MOIVRE’S THEOREM. 266. The student has already learned from Algebra that although the square root of a negative quantity is the symbol of an impossible operation, yet such roots are of great use in mathematical investigations. It is usual to adopt the convention that /(-a@*)=a,/(—1), and that such expressions as a,/(—1) shall be subject to all the laws of algebraical transformations. In the remainder of the present work it will be found that ,/(- 1) occurs very frequently in our investigations; we shall for the present assume that this expression may be freely used like any real algebraical expression, and hereafter we shall give some re- marks on the question of the validity of demonstrations which are obtained by the use of the symbol ,/(—1). (See also Algebra, Chap. xxv.) It is becoming usual in mathematical works to employ a simpler symbol instead of ,/(—1) in order to save room; the letter « is very convenient for this purpose, and we shall ac- cordingly employ it in some of the subsequent Chapters. 267. De Moivre’s Theorem. Whatever be the value of n posi- tive or negative, integral or fractional, cosnO+,/(—1)sinné is one of the values of {cos 6+,/(—1) sin 0)". Multiply cos a+ /(—1) sina by cos B+,/(—1)sinB; the product is cos a.cos 8 —sinasin B + ,/(—1) {sinacos 8 + cosa sin f}, that is, cos (a + 8) +,/(—1) sin (a+ B); multiply the last expression by cos y+,/(—1)siny; the product is cos (a +B+y)+,/(—1)sin(a+ B+). DE MOIVRE’S THEOREM. 213 By proceeding in this way we obtain the product of any num- ber of factors of the form cosa+,/(—1) sina. Suppose there are n of these factors, each factor being cos # + ,/(—1) sin@; we then have {cos 6+ ,/(— 1) sin 6}" = cos n6 + ,/(— 1) sin 26. This proves De Moivre’s theorem when n is a positive integer. Next, let 2 be a negative integer ; suppose 2 =—m, then {cos 6 + /(— 1) sin 6}" = {cos 6+ ,/(— 1) sin 6}" 1 1 ~ feos 0+ ./(— 1) sin 6}" ~ cos m0 + ,/(— 1)sin mb’ multiply both numerator and denominator by cos m6 — ,/(— 1) sin m6, cos m6 — ,/(—1) sin m0 | thus we obtain cos’mO+sin*md — ” that is cos m9 — ,/(— 1) sin mé ; that is cos (—m@) + ,/(-1) sin (— m8), or cosn6+ ,/(—1) sin né. This proves De Moivre’s theorem when n is a negative integer. Thus, since when 2 is any integer, {cos 0 + ,/( — 1) sin 6}" = cos n@ + ,/( — 1) sin nf, it follows that cos6+,/(—1)sin@ is one of the values of {cos nO + ,/(—1) sin nb when 2 is any integer. Lastly, let x be a fraction ; suppose i , then {cos 6+ ,/(—1) sin 6}" = {eos 8 + ,/( - 1) sin 6}¢ : = {cos pO + ,/(—1) sin p6}?, and, by what has just been shewn, one of the values of the last expression is cos V(-1) sin : Thus De Moivre’s theorem is completely established. 214 DE MOIVRE’S THEOREM. 268. We have shewn in the preceding Article that when n is fractional, cosn6+,/(—1)sinn@ is one of the values of {cos 6 + ,/(— 1) sin 6}"; we shall now shew how all the values of ’ the last expression may be obtained. Suppose nat Now both cos @ and sin 6 remain unchanged when @ is increased by any multiple of 27, while by putting 0+2rm instead of 6, and ascribing to r in succession different integral values the expression cos n6+,/(—1) sinn@ assumes gq different values and no more. For suppose r successively equal to 0, 1, 2,...... q—1; then we obtain the series of angles po p(O+2nr) (6447) p (8+ 2gm— 27) a z po i A and we know that no two of these angles can have the same sine and the same cosine, because no two of these angles are equal or differ by a multiple of 27. (See Art. 93.) Hence we obtain q different values of the expression cosn6+,/(-1)sinn6. We shall not in this way obtain more than q different values, for if r=s+mqg, where m is any integer positive or negative, cos n (6 + 2r7) and sin n (6 + 2rm) are respectively equal to cos 2 (6+ 2sr) and sin 2 (6 + 2sz). We can thus find g different values for the expression - foos 0+ f—1) sin OF; that is, we can find g different expressions, which by being raised to the g™ power, produce cos p6 +,/(—1)sin p6. And it is known from the theory of equations that there must be q values of a, and no more, which satisfy the equation x*=c, where ¢ is either real or of the form a+0,/(-1); thus we infer that we know ail the values of the expression {cos 6+ ,/(- 1) sin 6} P a, DE MOIVRE’S THEOREM. 215 269. De Moivre’s theorem may be usefully employed in ex- tracting any assigned root of an expression of the form a+6 ,/(—1). Suppose for example we require the cube root. Assume a=rcos6, b=rsin 6; so that 6 r=ar+ 6%, and tan 0 ==. Then a+b /(-1)=r {cos 6+ ,/(—1)sin 6}; and therefore {a+b J(-D}= + foo3 8 +/(— 1)sin 6}. One value of {cos 6+ ,/(— 1) sin 6}} is cos f+ J(-1) sin $5 and the other two values are respectively comes +f(- Lain sat i cos BHO (C1) sin 4r +6 7: 270. We proceed to deduce some important results from De Moivre’s theorem. In the equation cos nO + ,/(—1) sin nO = {cos 6 + ,/(— 1) sin 6}", suppose ” a positive integer. Expand the right-hand member by the Binomial Theorem, and equate the possible and impossible parts of the two members ; thus ea cos"~*6 sin?@ cos 78 = cos"@ — 42021) (0-2) 9-3) coptp into cos"~* 6 sin’ 6 sin nO = 7 cos" 6 sin gan en”) % n(n—1) oe) (m= 4) oc"-80 sin? — 271. The preceding formule hold whether nm be odd or even, but the last terms of the expressions on the right-hand side are different in the two cases, and it will be useful to distinguish the Cases. 216 DE MOIVRE’S THEOREM. If n be even, the last term of the expansion of {cos 6 + ,/(— 1) sin 6}" is possible, namely, (—1)?sin"6; and the last term but one n-1 is impossible, namely, m(-—1)* cos@sin""’6, which may be n-2 written ,/(—1)(—1)?* cos @sin""’ 6. Thus when 7 is even the last term of cos x6 is (—1)’sin"@, and the last term of sin is n-2 n(—1) * cos sin" 6. If n be odd, the last term of the expansion of {cos 6 + ,/(- 1) sin 6}" is impossible, namely (—1)'sin” 6, which may be written JET(H 2) sere and the last term but one is possible, namely ok co @sin*” 6, Thus, when n is odd, the last term of cosn@ is AS” ee Osin""’ 6, and the last term of n-1 sin nO is (—1) * sin” 6. 272. From the formule for sin 26 and cos 26 we can deduce an expression for tan n@ in terms of the powers of tan 6. sin nO For tan n6 = cos 20 cos”~* @sin* 6+... mens" §sing-2O— HO) - n n(n— 1) cos” 6 — 2 cos*-*@ sin? 6+... Now divide both numerator and denominator of this expres- sion by cos"@; thus we find for tan the expression n(n—1) (n—2) (n—3) (n—4) [6 12D taneg R=?) (n-2) (n~ 3) t9_ [4 tan’6-. ntand a ea DE MOIVRE’S THEOREM. 217 If ~ be even, the last term of the numerator of tan né@ is n(—1)? tan’6, and the last term of the denominator is (- 1)? tan"6. If m be odd, the last term of the numerator is (- 1)? tan"@, and the last term of the denominator is n(—1)7 tan". These results follow from those established in Art. 271. 273. We may also obtain general formule for the sine, cosine, and tangent of the sum of any number of angles which are not all equal. We have seen (Art. 267) that {cosa +,/( —1)sin a} {cosB +,/(— 1) sin} {cos y+ ,/(- 1) sin y}...... =cos(a+B+yt..... )+/(-1) sin (at B+yt.....) Now cos a+/(-1)sin a=cosa {1 +,/(-1) tan a}, cos B +,/(—1)sin B = cos B {1+,/(- 1) tan B}, ee tee ae nee ene ter eaeeeereenes thus we obtain cosacosBcosy...{1+,/(—1)tana}{1+,/(—1) tan B}{1+,/(-1) tany}... = 008 (a+ Bty tee J+V¥(-J])sin(at+B+yt......) Let s, denote the sum tana+tan@+tanyt+...... 3 let s, denote the sum of the products of the tangents taken two at a time; let s, denote the sum of the products of the tangents taken three at a time; and so on. Then by multiplying together the factors 1+ ,/(-1)tana, 1+,/(—1) tan B, 1+ /(-1)tany, ...... and equating possible and impossible parts we obtain cos (a+ B+y+...)=cosacos B cosy... {1—s,+8,—8,+..}, sin (2+ B+ y+...) =cosacos 8 cos y... {8,— 8, +8,—8,+...}. 218 DE MOIVRE’S THEOREM. By division, § —S,+8.—S +... tan(a+B+y+...)=+—*—_*+_1—, l-s,+s,-a +... -2 If » be even, the last term in the numerator is (—1)? s n=)? and the last term of the denominator is (- 1), ; if n be odd, the n-1 last term in the numerator is (—1)* s,, and the last term in the n-1 denominator is (-1)? s,_,. If the angles a, £,... are all equal, the formula will coincide with that given in Art. 272. 274. “We shall now prove formule for the expansion of sina and cos a in series of powers of a. We have, when n is a positive integer, cos n6 = cos"é — 7 cost" sin’ n(n —1) (n— 2) (n— 8) a Es Let n#=a; and suppose m to increase without limit, and let 6 so change that m may remain a positive integer and 7 be always equal to a; thus 6 must diminish without limit. The preceding equation may be written (2-6) g"26 =} as ni ONE Beg cos a =Ccos U — 1.2 6 a s"-*6 sin‘O —...... 42 (a— 6) a (a — 36) ere 5) - Now when m increases without limit, and, therefore, 0 dimi- nishes without limit, me sin 6 of ee to (a) also cos @ is unity and so is every power of is equal to unity, and so is every power DE MOIVRE’S THEOREM. 219 cos 6 up to cos" (Art. 150). Hence the above formula becomes a? at 6 9 “a 16" Suen sin 20 = 7 cos ~asing SO OM") cose 29sintd + ete cosa=1 Also nig Sin 6 _2(a—8) (a= 28) | ost? sin 6 thus sin a =a cos™' 9 —_— =a aoa ge a 0(=- 5 yt Hence, by supposing to increase without limit, we obtain 3 a a a? ees = *B “fe” seeeee The results of this Article are of the greatest importance; we shall make some remarks upon them in the next three Articles. 275. It must be observed with respect to the formule esta- blished for the expansion of sina and cosa, that a is the circular measure of the angle considered; for it is only when an angle is estimated in circular measure that = Oe coniey gehen @ stator nitely diminished. It is easy to obtain the requisite modification of the formule when any other unit of angular measurement is adopted. Thus, for example, sin n° =a— 3 +Bo where a is the circular measure of the angle of n°; thus a=Tg5> and we have : nir 1 wy at ne VS a = T80 -3 (Ta i (80 aan ur 4 Similarly cosn°=1— eee (te) idee 220 DE MOIVRE’S THEOREM. 276. The series for sina and cosa are convergent for all values of a. (- 1)" a 2n-1 2n-1 numerical value of the ratio of the (n+ 1)" term to the n®™ is a? 2n (2n+ 1) The n® term in the series for sina is ; hence the 3 and whatever be the value of u we can take n so 2 a large that for such value of ~ and all greater welt ond) shall be less than any assigned quantity ; hence the series is con- vergent (Algebra, Art. 559). Similarly it may be shewn that the series for cosa is always convergent. 277. The proof given in Art, 274 involves one point that may not at first appear quite satisfactory. The (r+ 1) term of cos a is strictly (- yee) eet 1) cos" sin?6 : this we write in the form (-1y ~a(a—6) (a— 26). (a= 278 +8) | os"""6 sin 6\"" ome v0 (OF) - Now it is proved in Art. 150 that the limit of cos*-*"6 is unity, and also that the limit of ay is unity; the only ques- 0 tion is whether the limit of a (a— 6) (a — 26)...(a— 2r6 + 6) 2 a” Be for all values of r. TE is obviously true when r=1; that is, the limit of * = 9): is : ; and we can shew by induction that the DE MOIVRE’S THEOREM. 221 required result is always true. For assume that a (a—6) (a— 26) ... (a — 276 + 6) So a LR where R diminishes without limit when 6 does so, so that the limit ar — of the right-hand member is “— ; introduce a new factor a 3 & — att a Ra. 270 { +Rh; © [2re 1" Qr+ 1 Qe ed Bes? and when @ diminishes without limit all the terms on the right- arti hand side vanish except 5 [r+ left-hand member. Similarly we can shew that when another a-— att? factor a is introduced the limit is 3 and so on, 2r +2 [2r+2 » which is therefore the limit of the 278. The following example will shew how the series for cos @ may be practically useful. Suppose two sides @ and bof a triangle are known, and the included angle C; if C be a very obtwse angle we can give a convenient expression for the third side of the triangle. For suppose +—6 to be the circular measure of the angle C, so that 6 is very small; thus =a’? +b°—2abcos C=a' +b? + 2ab cos 6 =a7+b° + 2ab Q - 5) approximately, =(a+0)'—ab6'= (048) {1-2}. Hence, by extracting the square root, = b)21 ee approximatel c=(at+ ){ e557} PP y: GS to Lo EXAMPLES. CHAPTER XIX. EXAMPLES, 1. Extract the square root of cos 44 + ,/(— 1) sin 44. 2. Find the values.of (—1)}. 3, Obtain the six values of (— 1), 4 Find the three values of {1 +,/(- 1)" 5. Given 2 zIG6" shew that 9 is nearly the circular measure of 3°, 6. Given sin (G+ 0) ='51, find approximately the value of 6, neglecting powers of 6 above the second. ce . ‘= eae shew that fy Gard) Qn ae (2n +1) 2n a 1) (2n - 2) ee +... +(—1)""1 (2041) a,+(-1). 8. If OcotO=a,+a,0+ 4,04 shew that a,.= 54 - > tat at + ne hence find 6 cot 6 to four terms. 9. If sec O=a,+a,6? + a6 + ... + a,,0°" + shew that = Don—e _ Van—a {= lj"a, a, 2 if A aveck Ba 10. If cos 2a + ,/(—1) sin 2a be substituted for @ in the ex- be pression +b a4)’ and similar quantities for 6 and ¢, and the result reduced to the form 4+ B,/(- »; find the values of A and B in terms of a, B, y. : EXAMPLES. CHAPTER XIX. 233 11, Shew that {cos 6 + cos d + ,/(— 1) (sin 0 + sin )}” + {cos 6 + cos @ —,/( — 1) (sin 6 + sin ¢)}" = Qe (cos 2) cos 3 nore) = #): 2 12. Shew that if =e’, and J(1-¢) =ne- 1, c n dl +ecosd= 5 (1 +n2) (1 +2) : 13. Prove the following rule for finding the length of a small circular arc: from eight times the chord of half the are sub- tract the chord of the whole arc, and one-third of the remainder will give the length of the are nearly. 14. Shew that the following rule for finding the length of a small circular arc is more accurate than that in the preceding example: to 256 times the chord of one-fourth of the are add the chord of the arc; subtract 40 times the chord of half the arc, and divide the remainder by 45. 15, From the identical equation (x — b) (2 — ae (% —c) (a— a) (a—a) (@—b) _ (a=0) (a=0) * (=e) (5=a) * (6= a) (0-8) deduce the following by assuming x = cos 26+ ,/(— 1) sin 26, and corresponding assumptions for a, 6, and ¢: =I, sin (6~8)sin(9~¥) - 9 (94 sin (a= B)sin (a=) 7 0-*) gents y) sin (9 — a) ‘sin (B—7) sin (B= a) sin (9 — a) sin (9 — 8) sin (y— a) sin (y— f) sin 2 (0-) one sin 2,(0-y)= 0. XX. EXPANSIONS OF SOME TRIGONOMETRICAL FUNCTIONS. 279. Let x denote cos 6+,/(—1) sin 6; then 1 1 z eens thus + 2=2c0s8, and w—>=2,/(—1)sind; also x" ={cos 6+ ,/(—1) sin 6}" = cosn6@+ ,/(—1) sin nd, 1 1 1 a {cosd+J(—1)sin 6)" cosnO + /(— 1) sin nO =cos n§ —,/(—1) sinn6; 1 1 : thus a" + == 2 cos nb, and a” — = 2,/(— 1) sin nb. We shall find this notation useful in the following investi- gations. 280. To express cos"@ in terms of cosines of multiples of 6 uhen n 18 a positive integer. 2" cos" O=(a +2) =o" +na, a 1,2) 1) or? i+ a mie = = 1 1 die a. 3 tne. gt toa Now rearrange the terms on the right-hand side, putting together the first term and the last, the second and the last Lut one, and so on; thus we obtain 1 spl n(n-1)/ 4. 1 . mtn (at +5) + 1s (# tom) te but at +4, =2coand, a? + A a” 2 =2cos(n—2) 6, and so on; EXPANSIONS OF SOME TRIGONOMETRICAL FUNCTIONS. 225 therefore n(n—1) Led 2""" cos" = cos nO +n cos (n— 2) 6 + cos (1-4) 04+... 4 2(n-}) a 1) The last term of the series on the right-hand side will take différent forms according as m is even or odd. In the expansion cos (n — 2r) A+... of (« + =) by the Binomial Theorem there are n+ 1 terms; thus th when n is even, there will be a middle term, namely the G + 1) ; which is nin). fondant Mg 3 ie tea (yet 1) Hence, when n is even, the last term of 2""' cos" @ is n(n—1)...(4n+1) 2|4n ‘i When x is odd suppose it =2m+1; there are two middle terms in the expansion of (+2) , namely, the (m+ 1)" and (m+ 2); their sum is 2(n—1)...(n—m +1) («+2) ee Hence when n is odd, the last term of 2"~* cos" 6 is n(n—1)... (+8) 3(n—1) 981. We shall find that sin’@ can be expressed in terms of cosines of multiples of 6 if x be an even positive integer, and in terms of sines of multiples of 6 if be an odd positive integer ; this will appear in the following two Articles. cos 0. TT. 1b 226 EXPANSIONS OF SOME TRIGONOMETRICAL FUNCTIONS. 282. To express sin"6 in terms of cosines of multiples of 6, when n is an even positive integer. m (m= 1) ne lee n(n—1) é ai . ( A ¢ (-3) +3 a. (-5 rele s) Fa) + Now rearrange the terms on the right-hand side, putting .. together the first term and the last, the second and the last but one, aud so on; thus we obtain a coe n(n — De ae wr ion(e +oa)+ 7 “+ a ia et poe) “ia (gz+1) “afar en(2ee ne} ts 2" (— 1)*sin -( a eae + tates Therefore n Ss a ont 1): sin" @ = cos 26 —n cos (nm — 2) 0+ cos (n— 4) O— + (— Lye neil D gly apt, ai $n(n—1)...(3n+1) ais 283. To express sin"6 in terms of sines of multiples of 6 when n is an odd positive integer. +(-1) :. 1\" @ do wea TT fn 2 16 2 58) aol = n-2 2" ( 1) sin 6 = ( Js 0" ne casa 13 Berri: n(n —1) 2 1 + a 1 ee ge et oe to Now rearrange the terms on the right-hand side, putting together the first term and the last, the second and the last but one, and so on; thus we obtain EXPANSIONS OF SOME TRIGONOMETRICAL FUNCTIONS. 227 a" -5 =n (e = =) ic c a (« ( 7) - atents n(n—1).. Ele -2); d(n- 1) x, but at —2,=2,J(-1) sin nd, f reek, ft 20 (- 1) sin (2 — 2) 6, and so on; therefore WAY? site Seine n sin (n—2)6+2 1) sin (n—4) 6 _n(n—1)(n-2 *n(n—1).. -3(W+8) is Deus 6)O+...+(- 1)? 1 (m—1) 284. If x be not a positive integer, the expressions for cos" @ and sin" @ in terms of the cosines and sines of multiples of 0 are very complicated. For these we may refer to Peacock’s Algebra, Vol. 11 pp. 435...440. 285. In Art. 270 it is shewn that when n is a positive integer, sh Sel @ a) cos"”*4 sin’ 9 cos 26 = cos" 6 — 9 RO ROH?) coset sin 0 .. since sin*@=1—-cos’@, _ sin*@ = (1 — cos? 6)’, and so on, it is obvious that cos” can be expressed in terms of. powers of cos @; we will now give a direct investigation of this expression. 286. Zo express cosnO in a series of descending powers of cos 6 when n ts a positive integer, Let x =cos6+,/(—1) sin 8, 15—2 228 EXPANSIONS OF SOME TRIGONOMETRICAL FUNCTIONS, . 1 so that a+-=2cos6, and a + = = 2 cos 06 ; pe x 1 2 now (1-22) (1- a ~#(w 45) +a*=1-2(e-2), where c=2 cos 0. Take the logarithms of both members ; thus log (1 — 2x) + log ( -=) = log {1-2z(c-2)}; 2 therefore 2a + }2%n"+ ho’? +...4-4+$ 54454... x 2 a Bla =2(c—2) + h2%(e—2) + 4H (c—2) +... +3 at(e—a)"+ ae In this identity we may equate the coefficients of 2. On the lefthand side the coefficient of 2" is ~(a+5), that is 2 ‘ . : 7008 nO; the coefficient of 2° on the right-hand side must be obtained by picking out the coefficient of 2* from the expansion of ‘ef (c—2)” and of the terms which precede it. a The coefficient of z” in Ly (c—2)" is < 5 gn (c ~z)" : the coefficient of 2” in is — a (n-1)0"?; n—-1 . - @ "(ez)" 7. 1 (n—2) (n- 3) ott the coefficient of 2” in og A eG IG aaa : and generally the coefficient of 2" in a a "(e—2)"" is (-1)' (n-1) (n-1r- 2 .(n— 27 +1) n-Tr n—or c . EXPANSIONS OF SOME TRIGONOMETRICAL FUNCTIONS. 229 17 Thus 2 cos 28 = (2 cos 6)" — n (2 cos 6)""* + - 3 (2 cos 6)"~* — —r- —r-2)...(n- we #(-1Y oii i Pe eclentesy (2 cos 6)" +... Ir The series on the: right hand is to continue so long as the powers of 2 cos @ are not negative. 287. It is obvious either from the above series or from that in Art. 270, that when 7 is an even positive integer cos” can be arranged in a series of powers of sin’, Thus we may assume in this case cosnO=1+A,sin?6+A,sin*6+ A,sin°O+...+A, sin", It is clear that-the first term must be unity, because when 6=0 we have sin@=0 and cosnf=1. Now we shall adopt an indirect method of determining the values of the coefficients Bp Aa, ones Change.@ into +A; thus cosn6 becomes cos 26 cos nh — sin nO sin nh ; now put for cosnf and sinh their values in terms of nh by Art. 274; thus the above expression becomes nh? 2 cos 26 — nh sin né — cos 20 +... Again in the term A, sin” 6 change 6 into 6+; we thus get A,, (sin 6 cos h + cos 6 sin h)”, that is, A,, (sin 8+ heos 0—© sin 0 — ays If this be expanded in powers of 4 the term involving 4° is A,, ee sin”~* 6 cos* @ —7 sin® o} Equate the’ coefficients of 4”; thus 2 - cos nO = A, {cos’d — sin’6} + A, {2. 3sin®@ cos*O — 2 sin‘d} SEL + A,{ Er sine" O cos! 9 sin®™ 0} +. 230 EXPANSIONS OF SOME TRIGONOMETRICAL FUNCTIONS. Now put 1—sin®6 for cos’@ on the right-hand side; then the term containing sin” @ will be 2r (2r — 1) (27 + 2) (2r +1). ~4, fF Maye and this coefficient must be equal to that of sin” 6 in the series 2 = for -5 cos 26, that is, to -5 A,,; thus nr 2 A,, = ar by, Fe Anes (r + 1) (2r ie 1), gz 2 therefore A a (2H) a — Ore) Qr+ 2) 2 By means of this law we may form the coefficients in succes- sion; we may consider 4,=1; then nv n? ag ae A ee oF _w te 2?) “ 3.4 RO 28 and so on. Hence, finally, gt nay lat 2) «an (nt 29 nt 4) cosn6 =1 gg = erg In the above process by equating the coefficients of A we shall obtain Fae —nsinn6 =A,2sin 6 cos0+A,4sin’ 6 cos6+..+4, 2rsin”~' 0 cos6 +... Substitute the values of A,, A,...; thus ‘ 2 2 2 2. r sin nd =n cos {sin @—™ == sinto == N A) sins. [3 [ When n is odd, we may start by assuming sin n@ = A, sin + A, sin? 6+ A,sin’0+...4+A, sin"; EXAMPLES. CHAPTER XX. 231 then, by proceeding as before, we shall find n (n* — 1) ee) is cos nd = 008 {1 5 = sen DE Pha.) sin nf =” sin 0— sin’ 6 + sin’ 6 — 288. In the four formule obtained in the preceding Article change 0 into = 6; thus we have, if m be an even integer, n? (n? — 2°) [4 (-1)" sinnO= =nsin0{cos6-" = B Petes a oe n 2 (= 1)' cos nf = 1— > cos? + cost@—... and if 7 be an odd integer, ue n (n?—1) 3 (-1)? cosn@ =n cos 6 — cos’6+ cos*O—.., n(n?—1) (n?—3*) iB (- nn sin nf = sin 61 a's 9 post 94 =O coat - et MISCELLANEOUS EXAMPLES. 1, Expand (sin 6)*"** in terms of cosines of multiples of 6. 2. Expand (sin 6)*"*? in terms of sines of multiples of 0. 8. Expand (cos 6)" in terms of cosines of multiples of 0. 4, Prove that in any triangle a cos} (B-C) , b! cos $ (C — A) 4, 200s 3 (A - B) cos 3 (B+C) cos 3(C'+ A) cos 4 (4 +B) =2 (ab + be + ca). 5. From the angles of a triangle ABC, perpendiculars 4D, BE, CF are let fall on the opposite sides: prove that asin (BAD - CAD) + bsin (CBE -—ABE)+ sin (ACF — BCF) = 0, 232 EXAMPLES. CHAPTER XX. 6. From A and B two acute angles of a triangle draw AD and BD at right angles respectively to AC’ and BC. If p be the radius of the circle inscribed in ABD, then AB=p (sec A +secB+ tan A + tan B). 7. Threc equal circles of radius a touch each other: shew that the area of the space between them is (vs - ) a’, 8. The area of a regular polygon inscribed in a circle is a geometric mean between the areas of an inscribed and of a circumscribed regular polygon of half the number of sides. 9. The area of a regular polygon circumscribed about a circle is an harmonic mean between the areas of an inscribed regular polygon of the same number of sides, and of a circumscribed regular polygon of half that number. 10, If the side of a pentagon inscribed in a circle be ¢, the radius is ae : 11. Three circles whose radii are a, 6, ¢ touch each other externally: prove that the tangents at the points of contact meet at a point whose distance from any one of them is ( abe \ at+b+e/]~ 12. The sides taken in order of a quadrilateral whose opposite angles are supplementary are 3, 3, 4, 4: find the area and the radii of the inscribed and circumscribed circles. 13. The area of a regular polygon inscribed in a circle is to that of the circumscribed polygon of the same number of sides as 3 is to 4: find the number of sides. 14, If the radii of three circles which touch each other be a, b,c, and u, B, y be the chords of the arcs between the points of contact in each, shew that anf et) ba) ea): EXPONENTIAL VALUES OF THE COSINE AND SINE. 233 15. Shew that the limit of SF diminished, is e. tan =o @2 6 when @ is indefinitely 16. The two diagonals of a quadrilateral figure whose oppo- site angles are supplementary cannot be equal unless some one of the sides be equal to the opposite one. ; 17. Two circles whose radii are a and 6 cut one another at an angle y: shew that the length of the common chord is 2ab sin y J(a* + 2ab cos y + 5°)” 18. The radius of the circle inscribed in a triangle can never be greater than half the radius of the circle described about the triangle. XXI. EXPONENTIAL VALUES OF THE COSINE AND SINE. 289. If we expand ¢* and e* by the exponential theorem we obtain “+ kia 4 oe "EE? Fea? ka 5 ka 7 BEE If it were possible to make %°=—1, so that h*=1, k*=-1, and so on, then the right-hand member of the first equation would be the expansion of cosa, and the right-hand member of the second equation would be the expansion of sin a (see Art. 274). Hence we are led to these results, evi) 2 envi) ° eV) _ eo tVi-b iS ee 5(e +e") = l+T5 a -e*)=a+ 234 EXPONENTIAL VALUES OF THE COSINE AND SINE. The meaning of these equations is simply this: if we expand eV and eV", by the exponential theorem, in the same way as if ,/(—1) were a real quantity, we shall by the above formule obtain the known series for cos # and sin a. These expressions for cosa and sin # are called the exponential values of the cosine and sine. 290. From the exponential values of the cosine and sine we may deduce similar values for the other trigonometrical functions, Thus, for example, eNED _ entyi-D mS TED We shall now use the exponential values in establishing certain results. In the remainder of this Chapter and in the next Chapter we shall employ the letter « instead of the symbol ,/(— 1). 291. Zo expand 6 in powers of tan 6. eft — e- & By Art. 290, ttn = ee l+ctan@ — ee = — =e, »l-ctan6 e% therefore Take the logarithms of both members; thus 26: = log (1 ++ tan 6) — log (1 —« tan 6) =2ftam 6-3 tan’@ + F tan? “ae therefore G=tan 6-5 ton’ +> tan’d—... This is called Gregory's Series. Let tan =a, so that O=tan'2; 1 thus tan ga Z att pat... EXPONENTIAL VALUES OF THE COSINE AND SINE. 235 292, The preceding investigation is unsatisfactory, because it gives no indication of the extent to which the result may be relied upon as arithmetically intelligible and true. The mn term of the n—-1,,on—1 last series is ot a ; hence the numerical value of the ratio of n= the (n+ 1)™ term to the x" is st at ; therefore the series is convergent if « be less than unity (Algebra, Art. 559). The series is also convergent when # is equal to unity (Algebra, Art. 558). For values of x greater than unity the series is not convergent, and is therefore not arithmetically intelligible. 293. Moreover tan™’s has an infinite number of values corre- sponding to the same value of a, so that one member of what appears as an equation admits of more values than the other; this point is left unexplained in the investigation which has been given. The subject of series cannot be adequately treated without using the Differential Calculus. The student must therefore be referred to treatises on that subject for a satisfactory demonstra- tion of Gregory’s Series. It is there shewn that so long as 6 lies between -fandG, the result Q=tan 0 —F tan’ +5 tan’ 9 — .. is absolutely true. (See Diferential Calculus, Chapter VII.) If, however, 6 =na + $, where ¢ lies between — . and 7 then g=ten p—ytan'g + } tants — me that is, O—nr=tan 0 —ytan’O +7 tan'O—... 294, In Gregory’s Series put 6= ; 3 then since tan = a 1, ata4s 9 LA i} — ! oo] + * — cs al 236 EXPONENTIAL VALUES OF THE COSINE AND SINE. This series might be used for calculating the value of 7; but it is very slowly convergent, so that a large number of terms would have to be taken to calculate z to a close approximation. 295. Luler’s Series. Lge 1 R38 7 oe cl SWS -1 as =f eer tan 3 + tan 3 tan ; T tan7* 1 2 “6 ar 1 1 1 1 tus G5 gp te 7g ak 1 1 1 5 8.2 8.8 7.00" 296. Afachin’s Series. We shall first shew that Cie ee qf tan 5 tan 35° 2 1 5 10 5 fae, ed sad edhe eS Pb, 2 tan 5 7 tan se 94 ~ tan 19? 25 10 1 5 12 120 Sle dt ae em -1 = ase 4 tan 5 2 tan eI tan : 5 tan Ti9° ~ 144 1 Hence 4 tan™ 2 is a little greater than “ ; suppose 120 1. i : then Ti9= tan G + tan x) —— from this we find 2==<.;: ‘therefore i =4 tan"! EXPONENTIAL VALUES OF THE COSINE AND SINE. 237 Therefore — pat fe- ; J : \ 5 3.5°°5.5° 7.5 {3 I ge dl ~ 1239 ~ 3(239)" * 5(239)* ~ 7(239)" wo}, 1 1 1 i -1 ad, SS dyes 297. It may be'shewn that tan 7397 tan 70 tan 593 pg Se pee eo thus qf tan 5 ~ tan 7 + tan 99° The series for tan™ he and tan™ a are convenient for pur- poses of numerical calculation. The value of w has been calculated by two computers inde- pendently to 500 places of decimals, and by one of them to 707 places of decimals: see Contributions to Mathematics...by William Shanks, London, 1853; and the Proceedings of the Royal Society, Vol. xxi. page 319, and Vol. xx11. page 45. The value of “ has been calculated to 52 places of decimals: see the Proceedings of the London Mathematical Society, Vol. rv. page 308. 298. Given sinw=nsin(xt+a), required to expand x in powers of n. Here eget =n { eltteye etal}, therefore em — Lan {otto grat, therefore e (1 —ne™) = 1 —ne™, l-—ne* therefore et = ei therefore 2Qar= log (1 seat ne-**) an log (1 = ne*) =n (e*—e* Zen ea) oo (eo — ee fi cee 2 8 therefore. e=nsina+>-sin 2a +3 sin Sat siceee ol 238 EXPONENTIAL VALUES OF THE COSINE AND SINE. As an example, suppose a= 7 — 2a, then n=1; thus : de, 1 ‘oe @=sin 2e—5 sin 4¢ + 3 sin 62—7sin 8x +... 299. Given tanzx=ntany, required to find a series for a, et! — e-*t ev — ev Here fae n Ore" oe ore 2yt em] et — 1 therefore ol ea]? lenje™+1l—n therefore t= ao (l-n)e“+l+n 1+ me l-n ‘ ee SA ae where m= Tan’ therefore 2ar= 2yr + log (1 + me") — log (1 + me”) mn = 2y. — m (6 — €™) fea) Shans 3 2 “sin dy - “sin Gy + ++ therefore w=y-—msin 2y + 5 800. Zo find the coefficient of x" in the expansion of e* cos ba in powers of «. Here = e* cos ba = Le™ (e+ e-™) = J etn 4 5 a, Expand these two exponential expressions by the exponential theorem ; then the coefficient of 2” is In {(a+ but + (a—b)"} mi(s ne) (: 0) Sead (Soe) Ff Ses i 2nl\ro¢r ror \ Now suppose <= cos 6, ° = sin 6, so that r= a+ 6". EXPONENTIAL VALUES OF THE COSINE AND SINE. 239 Thus the coefficient of 2" becomes (a? + BF 2|n (a + 6°) = Bn (cos nO +1 sin nO + cos nO — usin n6) {(cos 6 + usin 6)" + (cos 9 —csin 6)"} 2 22 ot cos 70. Similarly the coefficient of 2” in the expansion of e*sin ba in (a+b? nm powers of a is sin 70. 301. The series in Art. 298 may sometimes be of assistance in the solution of triangles. We have sin B=" sin A ="sin (B+); hence, by the formula, pee nC OF. 20 eB , 30 Gan +g sm +3738 Acs If } be less than a the series is convergent, and if = be a small fraction a few terms of this series may give B to a sufficient degree of approximation ; the series gives the circular measure of B, and - the measure in degrees or minutes or seconds may be deduced by the aid of Art. 22. 302. Given two sides of a triangle and the included angle, to Jind a series for the logarithm of the third side. Suppose a and 6 the given sides and C’ the circular measure of the given angle; suppose 0 less than a, we have ce =a’ +b° — 2ab cos C =a? + 0° — ab (e+e) = (a — be) (a — be) =a'(1 ~o 6) (1c); a a 240 EXAMPLES. CHAPTER XXI. b b thus 2 loge = 2 loga +log (1 = ©) +log (1 = i) =?2] 6 ce Cu b° aCe 2c . e oga—" (¢ +€ )- 5a $e) — os b b? oe therefore log ¢ = log a ~ ao C- Jae 08 20- qa C8 3C-... This series is convergent since b is supposed less than a, and if ° be small a few terms may give loge to a sufficient degree of approximation. EXAMPLES, 1. Apply the exponential values of the sine and cosine to sin 26 1—cos 26 2. If the sides of a right-angled triangle be 49 and 51, shew that the angles opposite to them are 43° 51’ 15” and 46° 8’ 45” nearly. shew that =cot 0. 3. If the angle C of a triangle be given, and the other two adjacent sides a, b be nearly equal, shew that the other angles are nearly equal to C 180°(a-b | C Olya-b ON? 9 —_ ——| —— — pee (eco 57 3(aa3%"3) }. 4. In any triangle, if d—B be small compared with (, shew that the circular measure of d — B is equal to 2 sin B+ (=) sin 2B nearly. 9@ 5. If @ and b be the sides of a plane triangle, A and B the opposite angles, then will log b— log a =cos 24 —cos 2B+ (cos 44 —cos 4B) 2 (cos 64 —cos 6B) +... < Vv EXAMPLES. CHAPTER XXI. 247 7 1 1 1 6. Shew that s=73t5-—7+ p+ 7. IfA+Beslog (m+n), shew that tan B=™, and 24 =log (n! +m’). 8. Reduce cos (9 + x) to the form a+ fu. 9. Reduce sin (6 + dx) to the form a + fu. 10. Ifw=(a+ d.)?*, express log w in the form a + fu. 11. Reduce (a+ :)’*™ to the form a+ Bu 12. Prove that {sin (a — @) + e** sin 6\" = sin" a {sin (a — n6) + e** sin nO}, XXII. SUMMATION OF TRIGONOMETRICAL SERIES. 303. Zo find the sum of the sines of a series of angles which are in arithmetical progression. Let the proposed series consist of the following » terms, sina +sin (a+ 6) + sin (a+ 28) +... +sin {a+ (n—1) Gl. We have ted oe cos ( — 58) ~cos (+56) =2sin5 Asin, ae in 5 Bsin (a+ 8), eee eee eee mee tee teeter samme tenn ae eee TT. 16 242 SUMMATION OF TRIGONOMETRICAL SERIES, Let S denote the proposed series ; then, by addition, cos (2-58) cos (o+*2) = 2Ssin 5 B; 1 2n-1 cos ( 3 ) = 008 (a+ 5 B) 2sin 5 B therefore S= sin (2+"5*)sin“P sin! g 304. Zo find the sum of the cosines of a serves of angles which are in arithmetical progression. Let the proposed series consist of the following n terms, cos a + cos (a + 3) + cos (a + 28) +... +c0s {a + (n—1) fi}. We have sin(a+58)—sin (a—56)=2sin 5 005 sin («+3 8)—sin (a+ 6) =2sing 6 eos (a+ 8), sin (a+3 8) —sin (o+38)=2 sin 5 Beos (a+ 2P), eee ee mee cent eet eee ee ete erecercres sin (0+ Sp) —sin (4-45 pl) = 2 sin 5 Boos fa + (n=l) Bh Let S denote the proposed series ; then, by addition, SUMMATION OF TRIGONOMETRICAL SERIES. 243 sin (a+ 2851) —sin («J 6) =28 sin 5 8; sin(o+ 5) -sin(n-5 8) therefore S= 2 sin} B cos (a+2>* ) sin sin 305. Suppose in Arts. 303 and 304 that p="; then since sin ™B = sin =0, the sum of the sines or the sum of the cosines of the series of angles u, ata att,..a ee is zero. This is a very important result, and the student should pay great attention to it. Moreover we may give this wide extension to our result: let m and n be positive integers, m being less than n, and ° B= 7 , then the following sum is a number independent of angles, sin" a + sin” (a+ 6) + sin” (a + 28) +... +sin™ (a+ — 16). The same theorem is true when sine is changed into cosine. The theorem is established by the aid of Arts. 280...283. Suppose, for example, we take m=4, We have sin*a=1} {cos 4a — 4 cos 2a + 3}, sin‘ (a + B) = 4 {cos (4a + 48) — 4 cos (2a + 28) + 3}, and so on. Thus the proposed series can be replaced by other series ; the sum to 7 terms of cos 4a + cos (4a+ 48) +... is zero by Art. 304; the sum to » terms of cos 2a+cos (22+ 28)+... is zero by the same Article; thus the proposed series reduces to “ ‘ 16—2 244 SUMMATION OF TRIGONOMETRICAL SERIES. The condition that m is less than m ensures that the denomi- nators in the expressions for the sums of the sines and cosines do not vanish. 306. The series in Art. 304 may be deduced from that in T 2 quired so often in the solution of problems, that the student should be able to quote them from memory. As we have just intimated, if the first result be known it is suflicient, since the second can be obtained from the first by changing sine into cosine in the first factor of the numerator. It will be seen that the results are obviously correct when n=1, and when n=2; thus there is a test of the accuracy with which the formule are quoted. The cases in which 8 =a may be specially noticed ; we have then Art, 303 by writing a+~fora; the sums of these series are re- gh US na $ asim — 2 2 sina+sin 2a+sin 3a+...+sinne=—————_ , . a sin — 2 egg sin 2 eo sin 2 2 cos a+cos 2a+ 0S 84+... + COS 20 =e. si a in — 2 307. We may now deduce the sum of the following m terms : sina—sin (a + 8) +sin (a+ 28) —... + (—1)"""sin {a + (x — 1) BI. This series may be written sinat+sin (a+ B+7)+sin (a+ 28 +27) +...4+sin {a+ (n—1) (B+z)}. We have then only to change B into B+7 in the result of Art. 303. infor = E+) gn MB) Hence the required sum is _ Btor ae SUMMATION OF TRIGONOMETRICAL SERIES. 245 Similarly cos a — cos (a+ 8) + cos (a + 28) — ... + (—1)""! cos {a + (2 - 1) B} | cos {a + mot er) sin sii) in® = 308. Zo find the sum of the following n terms: cosec @ + cosec 2% + cosec 4a + cosec 8a +... + cosec 2"! x, x We have cosec # = cot a cot x, cosec 22 = cot # — cot 2a, cosec 27~* a = cot 2"-2a— cot 2"71 a. Let S denote the proposed series ; then, by addition, x 2 S= cot 5 — cot 2" are a 309. Zo find the sum of the following n terms : 1 tats tan’ + 5 tan %, +. sie tea, We have tan «= cot #— 2 cot 2x, Looe x g tang = 5 Coby — cota, 1 x 1 1 a gmt tan ao yi = 9 cot == ae 33 cot Re: Let S denote the ee series ; then, by addition, é S= ari o0 b ae — 2 cot 2x. 246 SUMMATION OF TRIGONOMETRICAL SERIES. The term gen cot =~ * 00 B ee where B= 55 5 aaa; if we ea suppose ” to increase indefinitely, cos 8=1, and ae tae Thus the limit of the proposed series, when n is indefinitely increased, is = 2 cot Qa. 310. To find the sum of the following n terms : sina +csin(a+ 8) +c’ sin(a+28)+...4+¢" "sin {a + (n—1) 6}. Let S denote the proposed series; substitute for the sines their exponential values: thus Qu = et + celetBle + cPelat28)e 4... + cP-lelatmB—B)e — e-*— ce~ (2+ Be Fe (a-+28)e_ |, — oI g-(a+nB-f), We have now two geometrical progressions ; thus no 2S = ee 1 — cherie peat —cMe7 Be — ceBe 1l—ce-F ea a_cfela— B_e- Denne (a+nB)c Apentlfe(uB+a— Blt—eW(nBta— Bley L—c (e+ e-F) + 7 2 therefore Sa sina—c sin (a— fs) —c"sin(a+ mB) + o°** sin {a+ (nm — 1) BY : 1-2¢ cos B+? If ¢ be less than unity, then when m is indefinitely increased c® and c’** diminish without limit; hence if ¢ be less than unity, the limit of the proposed series when n is indefinitely increased is sin a —c sin (a — f) 1—2¢cosB+e?~ Similarly we can shew that cus a + ¢ cos (a+ B) +6’ cos (a+ 28) +...4+ c"' cos{a+ (a —1) Bt _ cosa —¢ cos (a— ) —c* cos (a+) + 0**? cos {a + (n — DB 1-2¢cosB +0? SUMMATION OF TRIGONOMETRICAL SERIES. 247 This result may also be obtained from the preceding by chang- ing a into a +5: If ¢ be less than unity the limit of the proposed series, when 2 is indefinitely increased, is cos a — c cos (a — f) 1—2¢cos B+? 311. To sum the infinite series sin (a+p)+% een and ¢ cos (a+p)+5 cos (0+28) +5 cos (a+ 38) +... Denote the former series by S and the latter by C’; multiply the former by + and add it to the latter: thus a 2 3 C +S = cetlat+8) + ¢ et(e+2) + gla +38) he 2 [3 = e*( ere? _ 1) a e'#(e¢ cosB+tcsin B = 1) = 6600s B gt(a-+esin B) — eu = e038 {cos (a + ¢ sin 8) +esin (a+ ¢ sin B)}— (cosa+csina). Equate the real and imaginary parts: thus C = e¢°38 cos (a + ¢ sin 8) — cosa. S =e¢s8 sin (a + ¢ sin 8) — sin a. The method of this Article might be used in Art. 310; or the method of that Article might be used here. 312. We shall not solve any more examples of the summa- tion of Trigonometrical Series; the student will find more exer- cise of this kind in the sollection of examples for practice, In many cases the summation is effected by the artifice which is employed in Arts, 308 and 309, by which each term of the pro- posed series is resolved into the difference of two terms. Practice alone will give the student readiness in effecting such transforma- tions. If he cannot discover the necessary mode of resolution in any example, he will find no difficulty in recognizing it when 248 EXAMPLES. CHAPTER XXII. he sees the result of the summation given in the collection of answers. Thus, for example, required the sum of the following n terms : sec a sec 2a + sec 2a sec 3a + sec 3a sec 4a +...4+ sec na sec (n+ 1)a. The result is coseca{tan (n+ 1)a—tana}; and by putting n=1 this suggests the necessary transformation, namely, sec a sec 2a = cosec a {tan 2a — tan a} ; then, sec 2a sec 3a = cosec a{tan 3a — tan 2a}, and so on. The student who is acquainted with the Differential and In- tegral Calculus, will be able to deduce numerous series from known series by differentiation or integration ; and when the results are obtained they can frequently be established by more elementary methods. Thus, for example, differentiate both members of the equality established in Art, 309; then 1 1 ate 1 0 sec? w+ 5,800 5 + 55S C0 og t+ + panme SOC ar 1 2 = — yume cosec * sai + 4 cosee 2a. Again in Art. 310 put B=a; thus sin a [roe cosa ng? Sina +e sin 2a + ¢sin da +c’ sin do + see a a Integrate with respect to a; thus 2 iL * Cc c é — 7, log (1 — 2c cos a + ¢*) = cosa +5 cos 2a +3 cos 3a +7 cos dat... No constant is required; for when a is zero both sides are equal. EXAMPLES. 1. Find the sum of » terms of the series sin’ a + sin’(a + 8) + sin? (a+ 28) +... 2. Find the sum of 7 terms of the series sin’a + sin’(a + 8) + sin’ (a+ 28) +... EXAMPLES. CHAPTER XXII. 249 3. Find the sum of xz terms of the series cos* a+ cos* (a + 8) + cos* (a+ 28) +... sin 6+sin 36 +sin 50+... to m terms 2: Shey thus tanas = cos 6 + cos 36 +cos 56+... to » terms” 5. Sum to 2 terms the series cos 6 cos (6 + a) + cos (9 + a) cos (6 + 2a) + cos (8 + 2a) cos (9 + 3a) +... 6. Shew that sin @—sin 20 + sin 30~... tom terms _, oy +8 cos 6—cos 2040s 50—...tonterms "2s ™ ) 7. Sum to ~ terms the series sin (p +1) 0 cos 6+ sin (p + 2) 0 cos 26+... 8. Sum to » terms the series sin asin 2a + sin 2a sin 38a +sin 3a sin 4a+... : 9. Deduce from the result of Example 8 the sum to x terths of the series 1.242.343.44... 10. Sum to 2 terms the series sin 36 sin 6 + sin 66 sin 26 + sin 126 sin 40 +... Sum to infinity the following series contained in the Examples from 11 to 16 inclusive: cos 8 cos? 0 cos? 6 peace 9 ll. cos6+ t cos 26 + Ty 008 30+ B cos 46 +... : sin 26 sin 30 12. sin 6 — T.2 ae cos 26 cos 40 Oo aa iy eos 14. 2 cos 8 + 5 cos 0 + 3 cos 6 +7 cos O+... in 6 c08 0 sin 26 cos’ 0 sin 36 cos? | 15. sin @cos og : oh 16. ona EO” ae fae 804 os 1 1.2 250 17. 18. 19. ZS 20. EXAMPLES. CHAPTER XXII. Shew that cos 9-5 0s 20 +5 cos 30—... =log € 0035). 1 1 1 Shew that cos 20 + 3 cos 60 +5 cos 106 + ... = 5 log (cot 6). Shew that 2 gs a. ng sin 20 4 wsin 36 c= ("+ cot 8) 2° 3 x Shew that sin | log cos 6 + log eos § + log cos 5, + ... = log ar 2 Sum the following series to 2 terms contained in the Examples from 21 to 35 inclusive : 21. 1) lo 26. 27, 28. 29, 30. sin 6 (sin) +2sin5(sin$) +4sin 2 (sin’) hee 6 6 6 6. 6 tan 5 sec @ + tan 7 sec 5 + tan 5 sec 7 + et cot 6 cosec 6 + 2 cot 26 cosec 26 + 2? cot 276 cosec 276 + .., 1 1 1 sin Osin 20 ‘sin 20830 smoddsm40 °°" 1 1 1 sin @cos26 cos 26sin 36 Taman eosae 1 1 1 -1 -1 -1 ten isla noe isoeo =1 ana © -1 i tan™* 2+ tan lal 7a ee ee ee . ion qe Os 8a, a ae. sinasin 3a +sin5sin-> +sin 5, sin + ... 1 1 1 cos +008 30 * cosd +0850 cos @+c0s70~ ae sin 6 z sin 20 i sin 36 % cos 26+cos@ cos40+cos6 cos60+cosd ~~" ho eu _~ EXAMPLES, CHAPTER XXII. sin 6 i 3 sin 36 be 3° sin 3°60 1+2cosd 1+2c0s3d~ 1+2c0s30* 32. cob™ (2a7* + a) + cot? (2a7} + 3a) + cot! (2a! + 6a) +cot™ (2a7'+ 10a) +... 31. 33. eee 20 + 3,800 0 see 26 sec 2°64... 2 34, peg tan 26 +55 ae tan 276 + = 5 log tan 2° + . 0 6 6 - a a 5 + 2cos = cos +2 ee 35. cos 5 + 2 cos 5 cos 55 + 2° coss 5 008 ga COS 53 + 36, An oiftitedaeal polygon is inscribed in a circle and from any point in the circumference chords are drawn to the angular points: find the sum of the squares of the chords and the sum of the fourth powers of the chords. 37. Circles are inscribed in triangles, whose bases are the sides of a regular polygon of m sides, and whose vertices lie in one of the angular points: shew that the sum of the radii of the circles is 2r (1-nsin’ x) » where 7 is the radius of the circle circumscribing the polygon. 38. Circles are inscribed in triangles whose bases are the sides of a regular polygon of » sides and whose vertices lie in one of the angular points ; 7 is the radius of the circle circumscribing the polygon : shew that the sum of the areas of the circles is Gar? sin? 7 sin’ se 39. Shew that if n be a positive integer nsin 6 + (n—1) sin 20+ (n—2) sin 36+... +sin nd _v+l1 (6 sin(n+1)0 See see 4sin's 40. Shew that if 2 be a positive integer (n+1)nsind+n(n— aed +(n—1)(n—2) sin 30+...4+2, lsinnd n(n +3) 1 si 30 reo gre cots — J cosee? 5 4 C08 —p- — COS —p 6}. XXII. RESOLUTION OF TRIGONOMETRICAL EXPRESSIONS INTO FACTORS. 313. It is known from treatises on the Theory of Equations that the expression #”—1, where 2 isa positive integer, can be resolved into factors, each of the form x—a, where a is either areal quantity or an expression of the form a+ ,/(—1), where aand are real: and there is only one such set of factors. We proceed now to resolve the expression x*—1, and some similar expressions, into component factors. The factors of the expression x"— 1 are found by solving the equation «£*—1=0; every root of the equation determines one factor of the expression: thus if a denote a root the corresponding factor is « — a. 314. To resolve x” — 1 into factors. The expression cos oO J(-1) sin where r is any in- teger, isa root of the equation «"=1; for the n™ power of this expression is by De Moivre’s Theorem cos 2r7 + ,/(—1) sin 2rz, that is 1. First, suppose » even. If we put r=0 we obtain a real n 2 we obtain a real root —1, and the corresponding factor is # +1. root 1, and the corresponding factor is «—1; if we put r= If we put for + in succession the values 1, 2, 3, ...... 5-1 we obtain n—2 additional roots, since each value of r gives rise to two roots. These roots are all different, for the angles are less than w and all different, and thus gee cannot have two coincident values. Therefore 2"-—1=(«—1) (# +1) P, where P is the product of n—2 factors obtained by ascribing RESOLUTION OF TRIGONOMETRICAL EXPRESSIONS. 253 to » in succession the values 1, 2, 3, ...... 5 1 in the expression ad 2Qer 2rar = % — Cos —— eeef(=1) sin. Qrar 200 The product of the two factors «-cos ——,/(—1) sin— and «—cos mm J(-}) sin = , is the real quadratic factor Qrn\8 las : (x- cos ==) +sin?=”, that is, 2° — 2a ge 1, 2 2 n Hence when m is even xz"-1=(e-1) (#+1) (2-20-05 = +1) (2*- 220s 41)... o eae? sg far Qu cos ~ og + 1} {a* — 2m c08 — =SiIN—~—7, «se 2n 2n Qr cos s=—= sin 2n 2n 317. Zo resolve a" — 2a" cos 0+ 1 into factors. If cos 6 =1 the expression becomes (a*— 1)’, and if cos9=—1 it becomes (a"+ 1)’; in these cases the resolution into factors is effected by what has already been given in Arts. 314 and 315, and we will therefore suppose these cases excluded from what follows. If we put au" — 24" cos8+1=0, we obtain 2"=cos@=,/(—1)sin@; hence w is an n™ root of cos 6+,/(—1)sin@; the n™ roots are found from the expression ogo +6 2rr +0 — a Te for sits is obvious from De Moivre’ Theorem that the n™ power of the last expression is cos (2rr+ 6) +,/(—1) sin (2r7 +6), and if r be an integer this reduces to cos@+,/(—1)sin 6. If we ascribe to 7 in succession the values 0, 1, 2,...%—1 in the expression Qrr +O Ira +O a EN J )sin the expression. For if r=p and r=g could give the same value to the expression we should have by ascribing integral values to 7, we obtain 2n different values for cos in “e+, op PE*E. H—1)sin a (-1)sin 2pm +O _ gant n nN nt EXPRESSIONS INTO FACTORS. 257 now by Art, 93 we cannot have cos PETE _ og "ae +8 and i SE ss, “gr 6 3 nor can co E+ 8 gg and - 2pr+O qu +8 v7 2 » for that, by Art. 94, would require Qpr +O , Jar +8 n = to be a multiple of 27, so that @ would be a multiple of 7, and this value of @ has been expressly excluded above. Thus we obtain 2 different values of x. 2rar+6 ee -J(-1)sin wna ee! J(-]) sin ara a4 is the real quadratic factor ( trey . .2rr+6 ‘ ovr +0 x— cos +sln° n n The product of the two factors a—cos +1. » that is, x — 2a cos Thus 2-22" cos6+1 27+0 ~(#*- 2 cos + 1) (2*- 2a cos +1) G ies eee 1) 7b Es In - fat ~ 2 cos 2 DE 1h fat — 2urcos CRA =F, 1}. 318. We shall now deduce some important results from the preceding general theorem. Suppose «=1; then 2 (1 — cos 6) = 2" ( - cos) (1 aor ee °) ( = at 6) _ ( pete 4) .. ( 1 —cos ———_—_——_ } . n Let 6 = 2n¢ and a =a; extract the square root; thus # sin nd = 2""' sin sin (2a +4) sin (4a + g)......8in (2na - 2a + 4). 17 T. T, 258 RESOLUTION OF TRIGONOMETRICAL We shall now prove that the upper sign must always be taken on the left-hand side. First, suppose ¢ to lie between 0 and 2a; then every factor on the right-hand side is positive, and so is sinzgd. Next suppose @ to lie between 2a and 4a; then every factor on the right-hand side is positive except the last, and sinng is negative. Next suppose ¢ to lie between 4a and 6a, then every factor on the right-hand side is positive except the last two, and sinn@ is positive. By proceeding in this way we see that for every value of ¢ between 0 and 2na, the upper sign must be taken, so that we have for all values of between 0 and x sin np = 27" sin > sin (2a + P) sin (4a + ¢)...... sin (2na — 2a + ¢). We shall next shew that this formula is true for all values of @; for suppose 6=mr+wW where m is any integer, positive or negative, and y is between 0 and 7; then we know that sin ny = 2”" sin Wsin (2a +) sin (4a+y)...... sin (2na— 20+); but sin mp =sin (né —nmz) = sin nd cos nmm = (— 1)" sin nd, sin y =sin ($— m7) = sin ¢ cos mm = (— 1)"sin ¢, sin (20+) =sin (20+ ¢—mz)= sin (2a+¢) cosma = (—1)”sin (2a+¢), and so on, Substitute these values of sin ny, siny, sin (2a+ y),...... in the formula which expresses sin ny in factors; then divide both sides by (1) and we obtain the required formula for sin nd, whatever may be the value of ¢. In the expression for sinn¢ change $ into $+; then n¢ is T 53 hence ad changed into nd + cos nb = 2" sin (p + a) sin (p + 3a) sin (f + 5a)...sin (2na —a + ¢). In the last result put $=0; thus 1 =2"" sin asin 3asin da...... sin (2na—a), Tw where a=5-. ald EXPRESSIONS INTO: FACTORS. 259 Again we have Hn 8G _ on" sin (20+ 4) sin (4+) ......8in (Ina —2a-+) 3 sin now let ¢ diminish without limit; then since the limit of Sie is m we obtain n= 2"" sin 2asin 4a sin 6a...... sin (2na.~ 2a), These two formule are sometimes useful; the first includes (3) and (4) of Art. 316, and the second includes (1) and (2) of Art. 316, . If we divide the expression for sinno by that for cosnd we obtain an expression for tannd; when n is odd this takes a simple form which we may obtain more readily thus: in the expression for sin @ change ¢ into +5 3 we obtain aad nr x = 2"* cos p cos (2a + ¢$)...cos (2na —2a+¢). cos np sin Divide the expression for sin n¢ by this; hence when m is odd inns =) aadt +2) tan (p +"=*n) nnb=(—1)* tang an (¢ = ) sree an (¢ aT) 319. The expression for sinn@ in Art. 318 may be put into a different form; for sin (2na — 2a+ ) = sin (+ —2a+¢) =sin (2a— ¢), sin (2na — 40+) =sin (r—4a + ) =sin (4a — 9), and so on. Then by multiplying together the second factor and the last, the third and the last but one, and so on, we have sin ng = 2" sin ¢ (sin? 2a — sin’ ¢) (sin’ 4a — sin’ 4)... It will be necessary to examine separately the cases when x is even and when 7 is odd. 17—2 260 RESOLUTION OF TRIGONOMETRICAL First suppose even; then the factor sin (na + ¢), that is, - cos, will occur without any factor to multiply it: hence if x be even, we have sin nf = 2""’ sin ¢ cos ¢ (sin? 2a — sin’ $) (sin® 4a — sin’ ¢)... ... {sin’ (n — 4) a — sin’ $} {sin® (n — 2) a—sin’ gt. Next suppose 2 odd; then we have sin nd = 2" sin ¢ (sin’ 2a — sin’ ¢) (sin® 4a — sin’ ¢)... ...{sin’? (2 — 3) a — sin” $} {sin® (x — 1) a — sin’ ¢}. Similarly from the formula cos np = 2"! sin (p + a) sin (p+ 3a) sin ($ + 5a) ... sin (2na— a + ¢) we obtain if m be even cos np = 2°"? (sin’ a — sin’ #) (sin? 3a — sin’ ¢)... ... {sin’ (n — 3) a — sin’ $} {sin* (n — 1) a— sin’ ¢} ; and if n be odd f cos np = 2""? cos ¢ (sin’ a — sin’ ) (sin® 3a — sin’ ¢)... ... {sin’ (rn — 4) a— sin’ $} {sin? (n— 2) a — sin’ ¢}. 320. We can now resolve sin @ and cos @ into their factors. Suppose nd = 6 and that 7 is odd; then by the preceding Article sin 6 = 9" sin © (sin® 2a - sin* *) (sin? 4a sin *) a n n n Divide both sides by sin 8, and then diminish 6 indefinitely ; since the limit of sin 6-sin$ is 2 we obtain n= 2" sin? 2a sin? 4a...; therefore by division, med 90 sin? — sin? — in ; 1 n n sin ¢G=nSM— a mma fee ee n sin’ 2a sin? 4a EXPRESSIONS INTO FACTORS. 261 Now suppose ” to increase without limit; then since a= 5 n . 6 ‘ sin a0 sin 9 the limit of —= , the limit of and so on; ais B57 3 thus finally, e° e? Ge sin = o(1 -5) (1-353) (1-gm)-- We shall obtain the same result if we begin by supposing n even. Similarly we may shew that 46° 46° 46° cos 0 = (- =) (1 -F2) (1-5) 3ie 321. In the same way as 2”—2a"cos6+1 was decomposed in Art. 317 we may decompose a” —2z"a" cos 0+a™", and each quadratic factor of the last expression will be of the form “ 2rr +6 A c wv” — 2x0; cos nis a’, where r is an integer ; and all the factors are found by giving to r in succession the values 0, 1, 2,...v—1. 2(n- = a os says (2 Be +O cos = o cos 2h” SER apa a and so on; thus all the factors will be found if we take +@ 4 +@, and use both signs and give to 7 in suc- cession the values 0, 1, 2,... up to a if be odd, and up to 2Qrr 2° — 2aa cos . if 2 be even; in the latter case when r= we must take only +60 4 +a’. ‘ nr one factor x? — 22a cos Now suppose a= 1 +e, and @ = 1-=; thus an 2\" z an (gate gyorg) 262 RESOLUTION OF TRIGONOMETRICAL is the expression to be decomposed into factors; and the general form of the factors is 2° a Few ee (1+) ~2 (1-5) 008 +(1-¢), ad 9. that is, 2( <)-2 (1- = ) cos si 4n 4n n Ira = 6 oa Ira =O : pao) ees 2 A that is, 4sin oa (1 +t cot’ In ) Suppose 2 to increase indefinitely ; then on mi Qn a) a9 (.-5) =e", (Algebra, Art. 552), vice oe ie 2rr+ 0 2 => cot? —.—_ = ~— -—_; 4n® Qn (277+ 6)?’ and by putting z= 0 we obtain o27+O, . ,4r+8 4sint $= 4 sin? 4 sin =O asin thus finally 2 ar. . 2 2 a 2 1 “—2cosd+e =4sin?5 {145 5 1+ aan + aa Let «stand for ,/(—1); then we may put - o- € —2 cos @+e7*=2 (cos w — cos 6) = sin 8 sin PO , 6-« “ ‘i and sin it will be a useful exercise to resolve ae into factors by Art. 320, and to shew that the result agrees with that which has just been obtained. For @ put 7+¢; thus we can obtain a formula for resolving +2 cos +e into factors. EXPRESSIONS INTO FACTORS. 263 322. It is usual in works on Trigonometry to give a brief though unsatisfactory demonstration of the results of Article 320 in the following manner. Since sin @ vanishes when 6=0, or +7, or + 27,... it follows that sin 6 must be divisible by 0, 0+7, 0-7, 04+2n, O-2Qz, . therefore we may assume that sin 0 = AO (0-7) (0 + x) (0 — 27) (0 + 27) (8 —37)( + 37)... where A is some quantity independent of 6; thus we may suppose ats e\(, & 6 sin 0= 00 (1) (1-53) (1-3) ~ where a is also some quantity independent of 6. Divide both sides by @ and then suppose = 0; thus a=1, and consequently sn 0=0(1-4) (1-565) (1-34)... T 2 Again, since cos@ vanishes when 06=+2, or a, wien RE follows that cos@ must be divisible by el Ga, i". 3 = » «+» therefore we may assume that cos d= 4 (0-3)(0+3) (0-2) (0452) (9-52) (992)... where A is some quantity independent of 6; thus we may suppose 46 46° 46° cos 0=a(1- r ) ( - #3) where a is also some quantity independent of 6; and by putting @=0 we find a=1; thus 46° 46° 46° cos = (1- (-s)( - 5p) The portions of the preceding investigations which are printed in italics involve assumptions which cannot be considered legitimate. 6+ 2u4 RESOLUTION OF TRIGONOMETRICAL 523. De Moivre’s property of the Circle. Let O be the centre of a circle, P any point within it or without it; divide the whole circumference into 7 equal arcs BC, CD, DE, ..., beginning at any point B, and join O and P with the points of division B,C, D,... Let POB=6; then will OP” — 20P". OB" cos nb + OB" = PB’, PC*. PD’... to n factors. For PB’ = OP? —-20P. OB cos 6 + OB’, PC?=0P*—20P. 00 cs (8 : =) + OC, PD* = OP*~20P.. OD cos (8 + =) + 0D’, and the radii OL, OC, OD are all equal. Thus, by Arts. 317 and 321, the product of all the terms on the right-hand side of these equations is OP™ — 20P". OB" cos nb + OB" ; this proves the proposition. EXPRESSIONS INTO FACTORS. 265 The particular case when P is on the circumference may be noticed ; then 20B" sin = PB .PC. PD... to n factors. Cotes's properties of the Circle. These are particular cases of De Moivre’s property of the circle. Let OP produced if necessary meet the circle at A, and sup- pose 1B=B0="; then nO=27. Thus we obtain (OP" —- OB") = PB’. PC". PD* ... to n factors ; therefore OP*~OB"=PB.PC. PD... to n factors. Again, let the arcs AB, BC, ... ‘be bisected at a, 6, ...; then by the theorem just proved, OP* ~ OB" =Pa. PB. Pb.PC... to 2n factors ; therefore by division, OP*+OB"*=Pa. Pb. Pc... to n factors. 324. It has been stated in Art. 169, that the tables of the logarithms of Trigonometrical functions can be calculated without the use of the tables of the Natural functions ; we will here briefly indicate how this may be effected. We have sind = o(1- =) (1 - - ga) (1- re) put = - for 6 and take logarithms ; thus Wt -_ m m wT mM Jog sin ™ ¥ = log ™ + log 5 + log (1 -7) n 2 m? m + log (1-grys) tes Q - arp) +... 266 EXAMPLES. CHAPTER XXIII. The terms in the last line may be expanded by Art. 145 in series which will converge with sufficient rapidity ; thus we shall have if » denote the modulus pe sin. 5 =logr+tlogm-+log (2n+m)+ log (2n—m)—3 (log 2+log n) . ap me s Gina m2 ee ie gt) Ts (ptaty m' pte gt) 3s Similarly we may find log cos ~ es (Airy’s Trigonometry. EXAMPLES, 1. Sum the infinite series Ie MM Sy, ad rp +55 +3 +e +o, 2. Sum the infinite series Toth, och, ol yt xt 3 an +e Sum the infinite series Tie hg era Tt 3a +B +B be tea 4. Sum the infinite series Mey | bt | a, 1 yt ta +7 oo a 5. If a=7, shew that sina sin Basin 9a......sin (4n— 3) a=27**4, EXAMPLES. CHAPTER XXIII. 267 6. * to BES 7. The product of all the straight lines that can i WHY from one of the angles of a regular polygon of 2 sides insbri a ine a circle whose radius is @ to all the other angular points is na"™’. 8. If p,, Pase+-Pon-1» Po, be the perpendiculars drawn from any point in the circumference of a circle of radius @ on the sides of a regular circumscribing polygon of 2n sides, shew that a” P,PsP+++ Pony + PoPa+++ Pon = oe 9. A polygon is described about a circle touching it at the angular points of an inscribed polygon ; the product of the perpen- diculars drawn to the several sides of the inscribed polygon from any point in the circumference of the circle is equal to the pro- duct of the perpendiculars drawn from the same point to the several sides of the circumscribed polygon. 10. Shew that 16 cos A cos (72°—A) cos (72°+.A) cos (144°— A) cos (144°+A)=cos5d. 11. From the expression for sin 6 in factors shew that 36 144 324 576 mods sera 323° 575 12. Shew that at de\ /, 4c? Ae? e+e* =2(145 =) (+33) (1+ ges) 13. Shew that 268 EXAMPLES. CHAPTER XXIII. 14. Find the sum of the series formed by multiplying to- ul, ale) Jal) wal gether every two of the terms of the series [> 5 3a gee 15. If be even shew that 7 Qa n—-1 e tan p tan (+2) tam (#+=) ... tan (+7 n)=(- 1)’. 16. Shew that sin 54 — cos 5A = 16 cos(A — 27°) cos (A + 9°) sin(d + 27°) sin(A — 9°) (cos A — sin 4). 17. Shew that mw 2.2.4,.4,.6.6.8.8. 2° 14.3.38.5.5.7.7.9. 18. By aid of the formula cos = sin deduce the expression for cos @ obtained in Art. 320 from that for sin 0. 19. Shew that nat 36: 100.196. 324... “3.35. 99.195. 323...” 20. Shew that J3_ 8.80. 224. 440... 29.81.2265. 441..." 21. Shew that cos + tan % sin « = eGo is. 29, Shew that cos a— cot sin 2= Qa Qa 2Q2 2a Qa 0-2)Q+e5)0- ty) l +a 5) (\-aes5) EXAMPLES. CHAPTER XXIII. 269 23. Shew that S87 SY _ l—cos y a 2 2 2 8 DU Gay Gea eal ae ( PC Cea Bae Geo (te +9)" 24. Shew that “S*+ COSY _ 1+ cosy 2 zs » 3 2 = a s}{1- x ji a Ma - a pe { (7-y) (7+y) (37 -y)? (37+ y)° 25. Shew that Sine tsiny (Ds )0-F) (oma) (-e) 26. In Example 21 by expanding both sides in powers of x and equating the coefficients of x, shew that y 2 2 | 2 2, 2 De 2 a-y w+y 3r-y 3rt+y Sa-y Sar+y 27. Shew in like manner from Example 22 that epee eee 23 20 y Yr-y I+y 4n-y 4aty 28. Shew that ae ate hes 55 22°57 2° 29. Shew that = tintgtoh es Jyh. 20 er I as ae ee 30. Shew that ——-= sin y 1 1 1 1 1 1] I J y ey de-y wty dety” Ba—-y 4a—-y 3ety MISCELLANEOUS PROPOSITIONS. bo =I o XXIV. MISCELLANEOUS PROPOSITIONS. 325, Many demonstrations have been given of the very important formule of Arts. 76 and 77; see, for example, the Afessenger of Mathematics, Vol. 111. pages 100 and 123, The demonstrations which we have adopted have the great advantage of being readily applicable whatever may be the size of the angles. The following process is very simple for the case of angles which are not too large. Let the angle COD be denoted by A, and the angle (DU by B; draw CW perpendicular to OD, and DM perpendicular to OC produced. Then the angle DCJ[=A+ B. Now by similar triangles, or by two expressions for the area of the triangle OC'D, we have O05 DUE OD SM cmadvtemnntieer nae (1). DM _OD.CN (ON+ND)CN Thetiot op "9.00 0.0m _CN ND ON oN “Oo Cn” OC ED that is, sin (4 + B)=sin A cos B + cos A sin B. Again, by similar triangles, or by Euclid, 111. 36, Cor., OC .0OM=ON.OD; MISCELLANEOUS PROPOSITIONS. 271 therefore OC .CM + 0C?=0N.ND+0N°, therefore OC .CM=ON.ND-ON®........0c000. (2), CM _ON ND CN CN CD 00° CD” 06° CD’ that is, cos (A + B)=cos A cos B—sin Asin B, Again, from (1) and (2), DM _OC.DM___OD.CN (ON + WD) Cw CM O0C.CM” ON. ND-—CON* ON.ND-ON CN i‘ CN _ ON“ ND ~ [OW oN’ ON’ ND tan A + tan B that is, tan (4 + B) = 7 therefore 326. Having given sin (4 + B)=sin A cos B + cos Asin B, and cos (4 + B)=cos A cos B—sin A sin B, we can deduce the formule for sin (4 — B) and cos (d — B). For put. 4+B=S; therefore d=S—B. Thus sin S = sin (S —B) cos B + cos (S— B) sin B, cos S = cos (S' — B) cos B— sin (S—B) sin B ; multiply the first by cos B, and the second by sin B, and subtract, and we obtain the formula for sin (S—.£): multiply the first by sin B and the second by cos B, and add, and we obtain the formula for cos (S—B). Similarly from the formula for tan (4 +B) we can deduce the formula for tan (4A — B). We might even deduce all the other formule from that for sin(A +B). For since (sin A cos B+ cos A sin B)’ + (cos A cos B—sin A sin B)’ = 1, io 72 MISCELLANEOUS PROPOSITIONS. it follows that if sin (4 + B)=sin 4 cos B + cos A sin B, then cos’ (4 +B) = (cos A cos B — sin A sin B)’, and therefore cos (A + B)=+(cos 4 cos B—sin A sin B); and a little consideration shews that we must take the upper sign. 327. In Chapter vitt. we have given exact expressions for the sines and the cosines of certain angles; we may add that the sines and the cosines of some other angles can be readily obtained by calculating the roots of certain equations. For example, we know that sin 30°= 3 sin 10°— 4 sin? 10°; put x for sin 10°: thus : = 3a — 4a*. Now by Horner’s method which is explained in the Theory of Equations we can easily calculate the numerical values of the three roots of this equation; the least positive root will be equal to sin 10°, and the greatest positive root be equal to sin 50°, and the negative root to —sin 70°: see Arts. 105 and 106. It is obvious that sin 10° + sin 50°—sin 70° = 0, so that the accuracy of the calculation can be easily tested. Since sin 10° can thus be found, and sin 9° is also known, we can by ordinary arithmetical calculation find the sine and the cosine of 1°; and then the sine and the cosine of any multiple of 1°. 328. The propositions which are given in Chapter 1x, admit of some extensions beyond the enunciations to which, for the sake of simplicity, we have there confined ourselves, It will be suffi- cient if we consider only positive angles. We have shewn in Art, 116 that sin @ is less than 6 so long as 6 is less than e it is obvious then that sin @ is less than 6 for every value of 0. MISCELLANEOUS PROPOSITIONS. 273 Now consider Art. 120, The demonstration there given de- pends on the fact that tan § is greater than a Thus it is really Ky shewn that sin @ is algebraically greater than 9-5 as long as @ is less than 7: on examination we shall find that this holds for ; 3. every value of 6; that is, sin 6— (- <) is always positive. For 3 we find, by calculation, that ot is greater than unity when 6=7, and it increases as @ increases beyond this value: thus 3 6 Rssay ig 7 6+sin 6 is always positive. And sin 6 is arithmetically greater 3 than gf certainly as long as both are positive, that is certainly up to 6= 2, which is beyond 6 =5 ; Next consider Art. 121. From that Article combined with the extension just given to Art. 116, it follows that cos 6 is always Ed algebraically greater than 1 — : . And from Art, 121 combined with the extension just given to Art. 120, it follows that cos @ is algebraically less than (a - )) certainly as long as is less than 2: hence it follows that cos @ is always algebraically less 2 than Q a , for this expression is greater than unity if is not less than 2, But cos@ is not always arithmetically wD 2 greater than 1 a , even if @ is less than ae On the other hand g cos @ is arithmetically less than (1 =f , while @ lies between 0 and some value which is greater than . and less than 7. Now consider Art. 130. In the same manner as we extended Art. 120 we can shew that sin @ is algebraically greater than T.T, 18 ‘OT . MISCELLANEOUS PROPOSITIONS. 3 7 6- - for every value of 0, and that sin 6 is arithmetically greater 2 ‘ than 0- S certainly up to 6=,/6. And cos@ is algebraically e 6 : 6. less than 1 — 3+ 9g: certainly as long as 3 8 less than ,/6: hence it will follow that cos @ is always algebraically less es 6 eo than 1- > +55: And cos @ is arithmetically less than 1 — > gt 5g? 7 ge 6 certainly while 6 lies between 0 and 33 for 1- gta is positive ‘throughout this range, and until 6=,/(6—/12). 329. We may add the following proposition to those of Chapter IXx.: If 6 be the circular measure of a positive angle less than a 3 right angle tan 0 is greater than 6 +4 2 3 For sin@ is greater than 0-— z and cos @ is léss than 6 2 4 1- oe = ag? therefore Pa tan @ is greater than FF g e ‘3 * oa hence, by division, we find that 65 9? e 8 (1 = 3) tan 6 is greater than 6 +o + —a ae 1 : ote eo 24 Now if 6 is less than 5 both the numerator and the denomi- nator of the last fraction are certainly positive; and so tan @ is greater than 6 + . ‘ MISCELLANEOUS PROPOSITIONS. 275 Also, since 26 — 2 sin @ is less then S and tan 6-0 is greater than e ; we have 26 — 2 sin @ less than tan 6 — 0; therefore 6 is less than tan 6 + : sin 0. (Serret’s Trigonometry.) 330. The following is an extension of Art. 149: the quan- tity e cannot be the root of a quadratic equation with rational coefficients. For suppose, if possible, that ae’ + be+c=0, where a, b, ¢ are integers: we may take a positive. Divide by e; thus ae+ce*+b6=0. Ae cst. Cl Replace ¢ by the series 1+1+ 5 a* 73 3 7 3 and e7' by the series 1—1 £5 i3 + [a-- ..» Which we obtain by putting —1 for « in the expansion of e*: then multiply by |r. Hence we obtain a 1+ Lig #f1- 5+ } = an integer n+1 nmt+2 “JS n+l meZo os Bre positive by taking m odd when c We can always make + ari c is positive, and ~ even when c is negative. Then by supposing n large enough we have a fraction on the left-hand side of the equation, which of course cannot be equal to an integer. (Liouville’s Journal de Mathématiques, 1840.) We may add that it has also been demonstrated that no commensurable power of e can be a rational quantity : see Algebra, Art. 803. 18—2 276 MISCELLANEOUS PROPOSITIONS. 331. The following theorem is given for the sake of an important application : EDH is a triangle having I the sides ED and DH equal. Produce DH to any point N; and £Z to a point J, such that EI’?=4DH. DN. Draw DM at right angles to DE, and JM parallel to DE, Then the circle which has the centre J and the OD “Mt radius Lif will touch the cir- cle which has the centre V and the radius VD, Let DH=h, EI =i, DN=n; DEH =6. £ Then LH =2hcos 6; and from the triangle HN IN? = (i—2h cos 6)? + (n—h)? — 2 (n—h) (i— 2h cos 8) cos 6 =0'+(n—h)’—2i(n+h) cos 6 + 4nh cos? 0 =(n +h)? — 2i(n +h) cos 6 +7? cos’ 6 =(n +h—icos 6)’. Thus JV =DN-JM; therefore DN =IN + IM, which de- monstrates the theorem. 332. The application of the preceding theorem which we propose to make is this: the nine points circle of any triangle touches the inscribed circle and the escribed circles of the triangle. For an account of the nine points circle the student is referred to the Appendix to Euclid, pages 317, 318, where the following theorems are demonstrated: ABC is a triangle, and P is the intersection of the perpendiculars from A, B, C’ on the opposite sides; the circle which passes through the middle points of PA, PB, PU passes through the feet of the perpendiculars and through the middle points of the sides of the triangle; the diameter of the nine points circle is equal to the radius of the circumscribed circle of the triangle. MISCELLANEOUS PROPOSITIONS. 277 Let ABC he a triangle, O the centre of the circumscriped circle, D the middle point of BC; let AG be perpendicular to BC, let P be the intersection of the perpendiculars, /’ the niiddle point of PA, Let OD be produced to meet the circumference of the circumscribed circle at #; join OA, AH, and FD. Since the nine points circle passes through D, /, & it follows that DF is a diameter; and therefore DF =OA. Hence since OD is parallel to AF’, we see that DF is parallel to 04. Thus if H be the point of intersection of HA and /'D we have ED = DH. Suppose that in Art. 331 the letters D, #, I indicate the same points as in Art. 253. Let . ek 3A. _R.. t=2ksing, h=2K sin’ 33 then n=5 - thus W is the centre of the nine points circle of the triangle; and therefore the nine points circle touches the inscribed circle. Again, suppose that in Art. 331 the letters D, H indicate the same points as in Art. 253; and let J now denote what was de- noted by J in Art, 253; then we see that the nine points circle touches the escribed circle which is opposite the angle 4, 278 MISCELLANEOUS PROPOSITIONS. Since the nine points circle of the triangle ABC passes through the middle points of AB, BP, and PA, it is also the nine points circle of the triangle APB; and so it touches the inscribed and escribed circles of that triangle. A similar remark holds with regard to the triangles BPC and CPA. 333. For the following investigation of the theorem of Art. 286, and of a corresponding theorem, I am indebted to the late Professor De Morgan. It is easy to see that cosn6 can be developed into a series of powers of cos @; we require the law of the coefficients. We know from Art. 310 that if « be less than unity 1-2? oe 2 ‘A [Se cos Op gi = 1 + 2x eos 6 + 2x cos 20 + 2a* cos 30 +... ; but without using imaginary quantities this equality may be demonstrated by clearing of fractions. The left-hand member may be expanded by the Binomial Theorem into the series 1-3’ (I-22 9 sO+ , (-#*) 2” Tea (+a)? aoe (nape EES and by comparing this with the right-hand member we obtain the following result : the coefficient of (2 cos 6)™ in the development of 2cosn@ is equal to the coefficient of x® in the expansion of (1 —x*) x™ 1-x’ that is to the coefficient of x*-™ in the expansion of Hence n—m must be even, say equal to 2r, so that m=n — 2r, The coefficient required is therefore that of y’ in the expansion of oer and, by the Binomial Theorem, this is a yy {eh eee x (n—2r4+1) earner) ‘MISCELLANEOUS PROPOSITIONS. bo =I © (n— 2r +1) (n— 27+ 2)...(n-r-1) fe thatis (- 1)'n In the same manner, starting from x sin 6 = i 2%. 3.8 To Be cosO pat 2 in 8 +a'sin 20+ asin 39+... , we see that the coefficient of (2 cos 6)™ in the development of innd , ‘ . s =e is equal to the coefficient of x" in the expansion of xmt =———xsa1- Hence we shall find that the coefficient of (1 +x’)™ sin nd . = is sin 6 (n'— Ir) (n—- 2r +1)... (m-—7-1) iF S (2 cos @)"-*r"' in the development of (-ly 334. The following is a very general theorem in the summa- tion of Trigonometrical Series. Having given the value of 2 C,+C,0+ 6,0" +... + 6,2", where ¢,, ¢,, ... ¢, are independent of x ; required the value of ¢, cosa +e, cos (a+ 8) +... +¢, cos (a+ nf), and ¢, sina +c, sin (a+) +...+¢,sin (a+nB), Let f(x) denote the known value of 6, +.0,0 + 0,07 +... 6,0". For a put in succession e'? and e~*; multiply the former result by e'*, and the latter by e~'4, and add: thus c, cosa +c, cos (a +B) +... +¢, cos (a +n) lz 3{es SF (e#) + ete f (-#)} . 280 MISCELLANEOUS PROPOSITIONS. Similarly ¢, Sina+c, sin(a +f) +... +¢,sin (a +) =5 {es f (8) — en F(e-#)} The expressions thus obtained on the right-hand side must be reduced and simplified ; we have ef (e8) = (cosa +esin a) f(cos B + csin B), e-2 f (e~?) = (cos a—esina) f (cos B — sin £) ; and when a definite meaning is assigned to / (x) we can obtain a definite result. 335, The investigations of Arts. 310 and 311 will furnish examples of Art. 334; for another example take the following: let _ n(n—1)(n—-2) c,=1, ¢,=nh, c,= a MW’, ¢, 3 Wey rvaeuse ; thus S (a) =(1+ ha)” Then Sf (e#) = (1+ heos B + hb sin B)" =1" (cos @ + esin p)” =?" (cosnp + isin nd), where rcos@=1+hcosB, andr sind =hsin B, so that r?=1+2hcosB+h’, and tan a : Similarly S (€#) =7" (cos nd —csin ng). Hence finally cos a + nh cos (a+py¢ 2) h? cos (a + 28) +... +h" cos (a + nf) =7" (cos a cos np — sin a sin np) = 7” cos (np + a), and sina +nhsin (a+) +2 =") pein (a+ 28)+...+ 2" sin (a + nf) =r" sin (nd + a). MISCELLANEOUS PROPOSITIONS. 281 Various particular cases are included in these general results ; for instance we may put h=1 or—1; we may put a=0 or B; or we may put a=my.and B=—y or — 2y; and so on. 336. Algebraical identities of more or less interest can be obtained by ascribing special values to the angles which occur in Trigonometrical formule. For instance put 6 = i in the two formule of Art. 270; put 6 =0 in the four formule of Art. 288, observing that the limit of om ; sin. 6 the two results of Art. 333. is n when 6=0; put 0=0 in 337. We will now make a few remarks on the symbol ,/(— 1), which has been used very often throughout the latter portion of this book. We may consider that the symbol has been used in an experimental manner, and many results have been obtained by means of it: the point now to be considered is how far these results can be received as true. In the first place, some of the results obtained by using the symbol ,/(— 1) may be shewn to be true by other methods. Thus for example in Art. 333 we demonstrated without the use of the symbol ,/(—1) a result obtained in Art. 286 with the aid of the symbol. So also the values of sin n@ and cos 7@ obtained in Art. 270 may be verified by induction. Moreover the resolution of an expression into factors in Art. 317 may be effected without the use of ,/(—1): see Cambridge Philosophical Transactions, Vol. x1. Again, the following example will shew how in some cases a strict demonstration may be obtained even with the use of the symbol ,/(—1). Let be a positive integer, and suppose it required to expand cos” in terms of cosines of multiples of 6; we may proceed as we did in Art. 280, supposing a to stand for V(-D, Now we know that (ore trate + afer pe} + a {erp eth 282 MISCELLANEOUS PROPOSITIONS, 2 4 6 n thus get {1 + Etats oh an} 2,2 4,4 6,,6 1. ye nyt ny 2 + c gs (n-2)?y? (n—2)ty* | (n= 2)" {1+ at iE + [6 ve...} Now this is true for all values of y, that is, if all the opera- tions indicated be performed, the two members of the equation are identically equal. We may therefore put — 6? instead of 7’, and the result will still be true. Thus eC 6 " nO nih nm—~) ges et eh? is Lal Tesco pete Reed ah yet 2 {1 ta ae ato (n— 2) (n— 2) | ) +n f1-So + [4 ae if BS Sonate Thus 2"-' cos" 6 = cos nO + n cos (1 — 2) 6 +...... See the Article Hquations in the Encyclopedia Britannica by Ivory, and Airy’s Trigonometry. Finally, the student may be informed that a theory has been constructed which offers a complete explanation of the symbol J(- 1), and thus enables us to obtain rigid demonstrations by the use of this symbol. It is not consistent with the plan of the present work to give any account of this theory; the student, however, is recommended hereafter to read the J'rigonometry and Double Algebra of Professor De Morgan. 338. The ratio of the circumference of a circle to the diameter is denoted by 7; this quantity + cannot be commensurable, nor can 7*: we will give a demonstration of the latter statement, which of course includes the former. The demonstration depends on the theory of Continued Fractions which is: explained in the Algebra, Chapter Lvi1, and will be readily intelligible to a student who has mastered that Chapter. MISCELLANEOUS PROPOSITIONS, 283 Let f(y) stand for 1- = + e - + Ley L.2.y(yt)) 1.2.3. (y+ 1) (y+ 2) th en J(y+)) —S(y) = yt) ay" +2); 2 9 therefore SM) _ Pag fy +2) ‘ LG xy t) FOSh Fiv+)) 1 _. Put « for FG) , then ae where p, = eel)! 2, is what 2 becomes when y is changed into y+ 1. Thus Sfy+l) can be transformed into an infinite continued SM) fraction of the second class; see Algebra, Art. 778. Put i for y, sin 6 and § for 2 then f(y) = cos 4, and f(y + 1)=—5 ; then multiply the seal previously obtained by 6, and gue the fractions. Thus we find that tan @ is transformed into an infinite continued fraction of the second class, in which the first component is 6, 3 the second component is : , and generally the 7** component is 6? Divide this result by 6, then invert and transpose; thus we find that 1—@cot@ is transformed into an infinite continued 2 fraction of the second class in which the first component is > 3? = th c and the 7 component is ral’ ‘ 2 In the last result put 5 for 6; and if (5) is commensurable denote it by where m and 7 are integers. Thus we have unity equal to an infinite continued fraction of the second class, in which 284 MISCELLANEOUS PROPOSITIONS. by simplifying the fractions we find the first component is =, the ze ? second component is —, that is —, and the r* component is bn’ 5 xa that is Sei . So that 1.” Gee mn s mn a Tn... But this result is impossible, for we know that such an infinite continued fraction must be incommensurable, and so cannot be equal to unity: see Algebra, Art. 792. Hence z* cannot be commensurable. 339. There is in the investigation of Art. 320 a point which may require examination, Let ¢ stand for : Be and 8 for =: then we have to find the limit when 7 is indefinitely great of , where 7 is an integer which in’ a lies between 1 and nt inclusive. We suppose that is odd ; the process is similar if n is supposed even, Now if r denote any jixed finite number, then when n is in- Ad Pe fied in making this statement when 1 itself increases indefi- nitely with n, as for instance when — or" - The fact definitely great the required limit is but we are not justi- 2 however is that in such a case both and 4 are indefinitely Tv ig small, and we are not led into error oe our use of the latter in- stead of the former. This we will now shew. “MISCELLANEOUS PROPOSITIONS. 285 Let m be an integer which may be as large as we please pro- vided it remain fixed, so as not to change when n increases ; then by the method of Art. 320 we shew strictly that sto0(1-8)(1-£5)(1-B)a where Q is the limit when n is indefinitely great of'a set of factors of the type 1 — a the integer r taking all values between m+1 and = inclusive, Now we have to shew that instead of 2 Q we may take the product of a set of factors of the type 1 — 2? the integer + taking the same values as before. Let R denote the product which we propose to substitute for Q; then we must shew that A= Q, and this we do by shewing that each of them is equal to unity, Now we may suppose m large enough to ensure that sin ¢ is less than sin rf ; hence every factor in Q is positive, and less than unity; so that the limit of @ cannot be greater than unity, a ‘ Let o=" 5m; then @Q is greater than {1 - ame vp ; for every factor of Q is greater than the first factor. But the limit of the last expression when » is indefinitely great is unity, as we see in the manner of Art. 150. Hence the limit of Q is unity. In a similar manner we may shew that the limit of R is unity. 340. There is a point in Art. 321 of the same nature as that just noticed with respect to Art. 320, and which we may treat in the same manner. It is the statement that when m is indefinitely great we have 2 cot? Ire + 6 m Z ane ° 2 ~~ (2rr+ 6)" 286 . EXAMPLES. CHAPTER XXIV. EXAMPLES, ; 6... 6. .7-8 . or +8 1. Prove that sin 6 cos 5 = 8 sin 5 sin’ —— sin’ ——. 2. Prove that 0 6 -6 0 6 6 26 2 2 — 2 — 2 —_— 2 _ ate (cosec 67 sec 3) tan g= (tan 3 cosec’ 6 sec 5) cot 3° 3. Prove that “*y -tan 39 —tan 20- tan @= tan 36 tan 26 tan 6. 4, Find « from the equation tan® w+ cot? e= m’— 3m. 5. The circumference of a circle is divided into 2n equal parts at the points A, P, Q,.... Tangents are drawn at the points A, P, Q, ... and perpendiculars OA, OB, OC, ... are let fall on them from O the extremity of the diameter OA. Shew that OA? + OB? + OC? + w..0.. = 3n (radius)*. 6. ABC is a quadrant; AP, AQ, AR are three arcs in ascending order of magnitude, each being less than AB, and their sum equal to twice AB; radii CP, CQ, CR are produced to meet the tangent at A at p, g, 7, and a triangle is formed with Ap, Ag, Ar. Find the condition that this may be possible, and the inferior limit of Ag and the superior limit of Ap. Prove also that in all such triangles the radii of the inscribed and circumscribed circles are inversely proportional. 7. ABC is a right-angled triangle, C being the right angle, F is the point at which the inscribed circle touches BC, and J the point at which the circle drawn to touch AB and the sides C'A, CB produced meets C'A ; shew that if HJ be joined the triangle FEC is half the triangle ABC. 8. Through the angular points of a triangle straight lines are drawn bisecting the exterior angles. If S be the area of the original triangle and S’ that of the new triangle, shew that wen tle A B = 5 S’ cosec y cosec 5 cosee >. EXAMPLES. CHAPTER XXIV. 287 9. ABCD is a horizontal straight line. From a point. imme- diately above D the known distances AB and BC are observed to subtend:the same angle a. If A4B=aand BC=8, shew that the height of the observer’s position above D is ; 2ab (a +b) tana (a—6)*+ (@+ 0) tan? a" 10. If in any arc not greater than a quadrant a -point~be taken, and from this point two straight lines be drawn, gne , a the extremity of the arc, the other perpendicular to “its cond bad” terminated by it, prove that the sum of these two straight lines is less than the chord of the are. 11. Suppose u the angle of elevation of a cloud, £ the angle of depression of the image of the cloud seen by reflection from a lake, 2 the height of the observer's eye above the lake, then the height of the cloud is Asin (8 +a) sin (B—a) * 12. At noon a person standing on a cliff h feet above the level of the sea, observes the altitude of a cloud in the plane of the meridian to be a and the angle of depression of its shadow on the surface of the water to be 8; the sun being behind the observer when he is looking at the cloud: shew that, if y be the sun’s altitude at the time of observation, the height of the cloud above the surface of the water will be Asin y sin (a + £) sin B sin (y+a) * 13. Shew that the formula of Art. 280 may be verified by induction. 14. Shew that the formule of Arts, 282 and 283 may be obtained from that of Art, 280 by changing @ intos -6. 15. Express cos 6 (tan™’ a) in terms of a. 16. If a quadrilateral can be inscribed in a circle and can also have a circle described about it, the area of the quadrilateral is equal to the square root of the product of the four sides. 288 EXAMPLES. CHAPTER XXIV. 17. The sides of a quadrilateral figure are a, 8, c, d; and the sum of two opposite angles is 6. If S denote the area of the figure, and s half the sum of the sides, shew that S? = (s— a) (s— 6) (8-0) (sd) — abed cos" 5. 18. Shew that n(n+1) [2 19. Shew that tan’? cos” 6 cos n6 = 1— 9 tant. n(n +1) (n+2) 399 Is s ak ry 6 20. If 6 is a positive angle less than 3 shew that ind cos” @ sin x6 =n tan 0 — + vee continually increases with 0, : “° T 21. If 6 is a positive angle less than 3 shew that a continually decreases as 6 increases. 22. In the diagram of Art. 332 if PO be joined, shew that it bisects DJ’, and is bisected by DF. 23. Shew also that PO divides DA into parts which are in the ratio of 1 to 2. 24, Shew that the following four points connected with any triangle are in a straight line: the centre of the circumscribing circle, the centre of the nine points circle, the point of intersection of the perpendiculars from the angles on the opposite sides, and the point of intersection of the straight lines drawn from the angles to the middle points of the opposite sides. 25. Shew that the length of the perpendicular from the centre of the nine points circle on BC is 5 F cos (C — B). 26. Shew that the length of the perpendicular from the centre of the nine points circle on AG in the diagram of Art, 332 is 5 Resin (C-B). EXAMPLES. CHAPTER XXIV. 289 27. In the diagram of Art. 332 shew that OP? = R* (1—8 cos A cos B cos C). 28. Shew that the distance of the centre of the nine points circle from the angular point A is 2 J(1 + 8 cos A sin B sin C). 29, The centre of the nine points circle cannot coincide with the centre of the circumscribed circle unless the triangle is equilateral. 30. The centre of the nine points circle cannot coincide with the centre of the inscribed circle unless the triangle is equilateral. ( 290 ) MISCELLANEOUS EXAMPLES. 1. Ifan angle of 3° be represented by ‘15 find how many de- grees are contained in the unit of that measure. Find also what number will represent a right angle in the same measure. 2, The difference of two angles is 1°; the circular measure of their sum is 1: find the circular measure of each angle. 3. Find tana from the equation tan «+ ab cot w=a+ b. 4, If sin 30=sin 6 cos 26 then 0= F where nis zero or an integer. 5. Ifan angle be divided into two equal and also into two unequal parts, the product of the sines of the unequal parts together with the square of the sine of the angle between the dividing straight lines is equal to the square of the sine of half the angle. 6. Shew that (sec 6 sec f + tan 6 tan ¢)’ — (tan 6 sec f + sec 6 tan ¢)’ 2 (1 + tan’ 6 tan’ p) — sec’ O sec” _ sec 26sec 2h ~ see? @ sec? ph * 7 IfA+B+C=360°, shew that 1 — cos’ A — cos’ B— cos? C + 2. cos A cos B cos C= 0. ’ OF 25 12 ; 7 8. If sn A=, sin B=7,, and sin C= 55, where 4, B, and C are positive angles less than 90°, find sin(4+B+C). 9% If a= rsin 5 (0- a)and y= rsin5 (0+), shew that w — Ixy cosaty’=r* sin’ a, MISCELLANEOUS EXAMPLES, 291 10. Eliminate 6 from the equations (a— 6) sin (9+ ¢)= (a+ 8) sin (8 — 4), aton$ - btan f=, 11. The number of degrees in one of the acute angles of a right-angled triangle is three-tenths of the number of grades in the other: determine the angles in degrees. 12. Shew that if the circular measure of an angle is F where 7 is any integer, the angle can be expressed by an integer both in degrees and in grades.” 13. If sin (a+ £) cos y=sin (a+ y)cos f, shew that B—y isa Tv multiple of 7, or a an odd multiple of 5 14. Shew that sin 44 tan’ 4 + 4 tan®A 4+ 2 sin 44 tan?A —4 tan 4+sin 44 =0. 15. Shew that sin? 24° — sin’ 6°= eS 1 16. If A+B+C=360°, shew that 2 (cos A sin B sin C + cos B sin C'sin A + cos C'sin Asin B) +sin’A +sin’ B+ sin*C=0. 17. Ifa and B are the two values of 6 in the equation cos 6 cos y 4 Ain dsiny _ 1 a b c? shew that (0° + c? — a”) cos acos 8 + (a? +c? 6°) sinasin B =a7+b?—c*, 18, If sind =5, and sin B= >, where A and B are positive angles less than 90°, find sin 2(4 +B). 19. Solve the equation cos 4a + cos 2a + cos x = 0. 20. If 4+B+C+D=360°, shew that B C+A cos 4 +cos B+ cos ('+ cos D = 4 cos = cos B+ 0s ‘= 19—2 292 MISCELLANEOUS EXAMPLES, 21. With two units of angular measurement differing by 10° the measures of an angle are as 3 is to 2: determine those units. 22. If sinw+sin’«=1, find sina; and shew that cos’a+ cos*w=1, 23. Solve the equation tan’ «+ cot? = 2. ) 24, Ifasiné +b cos @=c=acosec 6 +b sec O, shew that sin 20= as 25. Simplify cos? (A + B) + cos? (A — B)—cos 24 cos 2B. 26. If2tan 4 =3 tan B, then tan B sin 2B tan (4 —B) = 553 tan? B ~ 5 —cos 22" 27. Solve the equation sin®*) 94 sin®> hosing, 28. If tan (2a — 38) = cot (3a — 28), and tan (2a + 38) = cot (3a + 28), shew that both a and f are multiples of aa 29. Solve the equation 2 t _ 1—2cos 2a areola 1+2 cos 2a’ 30. If 4+B+C+D=360°, shew that A+B. B+. G+A 3 $12. 3 sin. _ Zz sin A+sin B+sinC'+sin D=4 sin 31. One angle of a quadrilateral contains 60 degrees, another contains 50 grades, the circular measure of another is > : express : all the four angles in degrees, MISCELLANEOUS EXAMPLES, 293 40 60 32. If cos A = 37 and cos B =3 where A and B are angles A-B_ 2 ~ 41x61" less than a right angle, shew that sin? 33. Solve the equation sin 36 = 8 sin® 6. 34. Eliminate 9 and ¢ from the equations a’ cos’ §—bcos*P=c*, acosO+bcosd=r, atan O=6 tang. 35. If tan A, tan B, tan Care in Arithmetical Progression, and tan A, tan B, tan D in Harmonical Progression, then - tan 0 _, _8sin’ (4 — B) tanD "sin 24 sin 2B" ae 36. Find cosa from the equation cos 2 sin? x= sin a cos*a, having given as one solution cos x= sin a, 37. Shew that (2 cos 6 — 1) (2 cos 20 —1) (2 cos 2° 8-1)... (2 cos 2""" §—1) _ 2cos2"d4+1 ~ 2eos0+1 * ’ 38. If tan (m cot 6) = cot (w tan 6), shew that n+], J(4n? + 4n—15) q 4 : tan 6= where x is a positive or negative integer. 39. Express in factors cos’ A + cos’ B + cos* C — 2 cos A cos Bcos C'— 1. 40, If4+B+C+D=360°, shew that sin 4—sin B+ sin 0 ~sin D=4 c0s $7 cos 2* © sin ee 41. Express in each system of angular measurement the - angle described by the minute hand of a watch in 25 minutes. 42. Shew that (cos 6 + sin 9) (cos 26 + sin 20) = eos + cos (30 - 5) 5 204 MISCELLANEOUS EXAMPLES. 43. Shew that cosec A cosec 2.4 + cosec 24 cosec 3.4 = cosec A (cot A — cot 34). 44, Shew that 1 cot? : A- cottS sec? 3 A-sec A z =8, 1+cot? -A 2 45. Shew that {see + coseo A (1 +800 4)} {1 — tan? 54} {i 1 - tan? 4} al “G4. = (s005 A +cosec = 34) see 46. If (a—6)sec6 = Jt a‘ +) , and oe J ( as oon’ then fant 1 $¢)= = aad p' "47, Eliminate 6 and ¢ from the equations a _sin(—6) Cc 6 sind i eae a) eo uae: 48, Find cos 5 from the equation B ; x COs & Cos i - 5) =sin Boos 5. 49, If cos (6 + 3¢) = sin (20+2¢), and sin ($ + 36) = cos (20 + 24), Tv Tv Tv wv shew that 6 =(3m—5n) 5 +7 and $= (38n— 5m) Bt ie? or else 6-6 = 2mm — a where m and n are integers. 50. Shew that (1 +sec 20) ) (1 + sec 46) (1 + sec 84)...... (1 + sec 276) tan 2"6 “tand * MISCELLANEOUS EXAMPLES, 295 51, The circular measure of a certain angle is equal to the ratio of the number of degrees in it to the number of grades: find the magnitude of the angle in degrees. 52. Shew that {sin (A — B) + sin (A + 3B)}sec 2B = (cos 2B — cos 2.4) cosec (A — 8). tanO 14 cos?6 Oh Tt eae LR shew that sin (30 +a) =7 sin (9 —a). 54, Solve the equation cos 30 + cos 56 + ,/2 (cos 6+ sin 6) cos 0 = 0. 55. Eliminate ¢ from the equations nsinO—mcosO=2msind, nsin 20—m cos 26=2. 56. Solve the equation 8 sin (9-3) cos" 9+ 8 cos (9-5) sint 06 sin (20~3) =,/3. 57. Among all values of 6 between 0 and = find that which makes sin 6 cos(B—6) greatest; 8 being a given angle between 0 and 3 58. Shew that cos 55° + cos 65° + cos 175°= 0, cos 55° cos 65° + cos 65° cos 175° + cos 55° cos 175° = + : 2 14+,/3 0 0 Ona oN cos 55° cos 65° cos 175° = 8/2" 59. If axcos(a+f)+cos (a— p) =2 cos (8 + y) + cos (8 — y) = # cos (y + a) + cos (y— a) tana _-—s tanB tan y then = T = T - tan 5 (8+) tan 5 (y+ 4) tan 5 (2 + 8) 60. If 4+B+0+D=360°, shew that A+B. B+ seta cos A — cos B + cos (- cos D = 4 sin gq Sin —g— 008 9 296 MISCELLANEOUS EXAMPLES. 61. The number of degrees in an angle-of one regular polygon is to the number of grades in an angle of another as 3 is to 5: find the number of sides in each polygon, shewing that there are only three solutions, 62. Solve the equation sec? 5 + cosec’ 5 = 16 cot x. 63. Eliminate 6 from the equations msin 26=nsin 6, pcos 20=@ cos 0. 64. Find 6 from the equation cos:6 — sin § = cosa — sin a.’ 65. Shew that if sin (4 +B+C+D)=0, then: sin (4 + C) sin (A + D)=sin (B+) sin(B+ D). 66, Shew that all the values of @ which satisfy the equations sinf+sind=p, cosO+cosp=gq, are contained in the expression nr—a+(—1)" 8, where a and B are angles determined by the equations tona=2, sin P= 5 /(p' +9). 67. Shew that cog ot 3a om ee 5a 6x gis 3) COS 5 ‘0: T5 OP 75 8 5 15 8 5° = . 68. Shew that whatever 6 may be, asin’ 6 +6 sin 6 cos 6 + ¢ cos? 6 lies in value between 1 l Be 1 Lips g(ate)+ 5b +(a—c) and 5 (a+c) - 3 vb +(a—c)*. 69. Shew that ee + COs Be + =0 cos a (Fee) cos (F-a)= A Qa Qa Qa Qa 3 cosacas (=F +a +cosacos (J = ) + 008 +a)eos(“—a =—5 COS a COS oe ) cos Be ao ae (Fre (F-e)= 4 . MISCELLANEOUS EXAMPLES. 297 70. Find an expression for the product (cos § 3 + COS é) (cos 5: eB + cos 5) aeeeee (cos 5 5a + cos 5): 71. An angle is the excess of a’ b’ above p’g': find the ratio of this angle to a right angle. - 72, Solve the equation 2 sin? x + sin® 2x = 2. 73, Shew that tan 4 + 2 tan 24+ 4 tan 44 + 8 cot 84 =cot A. 74, Solve the equation cos 2” — cos 4% = sin x, 75. If the sum of the angles A, B, C, D be four right angles, and their tangents in geometrical: progression, shew that the ratio=—1; or else that tan A tan D=tan Btan C =1. - 76. The angles A, B, C’ ofa triangle are in Arithmetical Progression ; and cosec 24, cosec 2B, cosec 2C are in Arithmetical Progression: shew that the cosine of the common difference of the angles is oe 2 7. Shew that cos 4 + cos 24 + cos 34 = : . 34 cos 2.4 sin el 7 sms al 78. Shew that (2-2 e087) (« 2-2 00857) («2 008 52) = =oe+a°-2e-1. 79. IfdAs B+ C= 180°, shew that 34 3B 3c sin’ 4 + sin®°B+sin®C' =3 cos 4 cos S cos 2 + cos -5- cos —- cos —— 27° 2 2 2 2 80. Investigate the ‘conditions which must hold in order that the equation sin’#+2bsinz+c=0 may give two admissible .values for sin x, when 0 is positive. 298 MISCELLANEOUS EXAMPLES, 81. The number of the sides of one regular polygon is to the number of the sides of another as m is to m; and the number of degrees in an angle of the first is to the number of grades in an angle of the second as p is to g: determine the number of sides in each polygon. 82. Solve the equation cosa + cos 7x” = cos 4x, 83, Eliminate a from the equations xtan(a—B)=ytan (a+), (c—y) cos 2a+ (a+) cos 28 =z. 84. If w and y vary so that their sum is constant, find be- tween what limits sin x sin y ranges, and its greatest value. 85. IfA+B+C=180°, shew that : ; CY. A cin(4+5)+sin(B+$)+sin(¢+5)41 A-B 3B-€ C-A : = ea 86. If cos’ dA +cos’B+cos’C=1, cos*a +cos* 8 + cos" y =1, =4 cos and cos A cosa+cos Bcos 8 + cos C' cos y=0 ; shew that sinasin2a sinfsin28 sinysin2y 2cosacosBcosy_ cos A cos B cos 0 cos A cos B cos C Os 87. Show that sin 7 = 1 (1+,/2—/3) /3- V2 2 88. If tan p= 5+ 7S tan 4, ar o\ l+te fg then either sin 6 = = or else cos 6 = 0. 89. Find cos# from the equation cos 2x+bcosa+c=0, Investigate the conditions which must hold in order that there may be at least one admissible value of cos x, supposing 6 positive, MISCELLANEOUS EXAMPLES, 299 90. Ify is not greater than i shew that sin 6 {1 + sin (y—6)} continually increases as increases from 0 to y. 91. Shew that there are eleven, and only eleven, pairs of regular polygons which are such that the number of degrees in an angle of one of them is equal to the number of grades in an angle of the other ; and that there are only four pairs when these angles are expressed as integers, 92. Ifsecasec8+tanatan @=tan y, shew that cos 2y cannot be positive. 93. Find the general value of an angle such that its cosine is to its tangent as 3 is to 2. 94. If w and y vary so that their sum is constant, find between what limits sin «+ sin y ranges, and its greatest value, 95.. Solve the equation cos 6 — sin 6 =,/2. 96. Express in four factors sin? A + sin? B + sin’ C —2sin A sin Bsin 0-1. 97. Shew that 008 5 = ; (-1 + /24+4/3),/2+,/2. 98. Eliminate 6 between : \ 3 ec asin (9 + 7) +5sin (6-7) =p? : ™\ 7 , 7 and acos (9-5) +8008 (6+7)=csin (2047). 99. If 4, B,C be any quantities, and a, 8, y angles such that A cota+ Bcot B+Cevot y=(4+B+C) cot acot B cot y, and. (B+ C) cot B cot y + (C +A) cot y cota + (A + B) cota cot B=0; shew that Asin 2a + Bsin 28+ Csin 2y=0, 300 MISCELLANEOUS EXAMPLES. 100. If cos 6=cosacos#—sinasin B ,/(1—c’ sin’ 6), and cos # = cosacos 8+sin asin B ,/(1 —c’ sin’ 4) ; shew that 2 2 cos 6+ cos = EOSaeNe and 1+ cos@cos p= ss ene 2 lc’ sin’asin’p’ 1l—c’sin*asin’£° Hence find sin § sin ¢ and ian § tan : * 101. Shew that cos114+3cos94 +3 cos74 +cos5A = 16 cos* A cos (44 + 7) cos (44 es 7) - 102. Find approximately the distance at which a coin an inch in diameter must be held from the eye looking towards the moon so as just to conceal it, the apparent angular diameter of the moon being half a degree. Hl OO 103. Solve the equation cos’ @ sin 3 + sin® 6 cos 36 = 104, Eliminate 6 and ¢ from csind=asin(0+¢), asingd=dsin§, cos 6 — cos d = 2m. 105. In any triangle asin (B-—C)+6sin (0— A)+esin (A-B)=0. 106, Shew that cos? 18° sin? 36° — cos 36° sin 18° = al 107. The perpendiculars from the angles of a triangle on the opposite sides meet at O; and OA =a, OB=y, OC’ =z; shew that a bc abe ae ay & aya” 108. The top of a pole placed against a wall at an angle A with the horizontal plane just touches the coping; and when its foot is moved a yards further from the wall and its angle of incli- MISCELLANEOUS EXAMPLES. 30 uation is B, it rests on the sill of a window: shew that the per- pendicular distance between the coping and the sill is : A+B a 109. AB is the diameter of a circle, C its centre; a straight line AP is drawn dividing the semicircle into two equal parts ; @ is the circular measure of the complement of the angle PCB: shew that cos 0 = 0. a cot 110. The sides of a regular pentagon and of a regular decagon inscribed in a circle of radius R are a and a’, and the radii of the circles inscribed in them are x and 7” respectively: shew that ¥ a @ 22R a —a?= FR, and —+G=-—-. ror lll. IfA+B+C=(2n+1)7z, shew that 4 cost? scout 2 (cot st sont ont cst cot 2) cost 5 + cos* 5 +008" 5 — 2 (cos gon 3 +008 5 008 gt cos 7 008 3 +4 cos? 5 cos" B vost e =0, 112. If tan’6 is less than unity, shew that 1 ds tan? @— tan‘ 9 +5 tan® 6 ee = sin’ °0 +5 sin" 9+ qsin’ 6+ vepeies 113. Solve the equation cos 6 — cos 20 =sin 30. 114. In any triangle a’sin(B—C) , b’sin( (C- 4), e*sin (A — B) = sin A sin B sin C 115. Solve the equation cos 38 + sin 36 =cos 6 +sin 0. 116. If tan’e=tan (a+ a) tan (a— ) then .sin 2a = 2. sin a 117. If tan? A tan A’= tan’ B tan B’= tan’ C tan C’ =tan A tan B tan C, and cosec 24 + cosec 2B + cosec 20 = 0, shew that tan (A — 4’) = tan (B-B’) = tan(C—C’). 302 MISCELLANEOUS EXAMPLES. 118. If cos 60°= sin 36° cos A, cos 36°=sin 60° cos B, and cos C'= cos A cos B, then one value of 4+B+C is 90° 119. Two rows of houses of equal height stand at an angle to each other. Ona sunshiny day the distance of the corner of the shadow from the corner where the rows meet is observed twice and found to be a and } respectively. Shew that if h be the height of the houses and a the difference between the altitudes of the sun on the two occasions 1? +h(b—a) cota+ab=0, 120. P is any point in the are of a semicircle APB; two... circles are described touching the semicircle and also touching AP, BP at their middle points: shew that the rectangle contained by the radii of these circles is one eighth of the square described on the radius of the circle which is inscribed in the triangle APB. 121. Shew that sin a sin (8 —y) cos (8 + y—a) + sin B sin (y — a) cos (y+a— f) +sin ysin (a — £)cos (a+B—-y)=0. 122. Shew that log cot = cos 20 + 5 (cos 26)? + ; (cos 26)*+...... 123. Ifd+B+C=(2n+1) z, shew that cot A + cot B+ cot C—2 (cot 24 + cot 2B + cot 2C) 2 124. Shew that sin A sin B sin (A~B)+sin Bsin C sin (B—C) +sin C sin Asin(C—A) = 1 {sin (24 — 2B) +sin (2B -2C) + sin (20 —24)}. = (cot G+ eotZ + cot 5) (see 4-1) (sec B—1) (sec C — 1). 125. In any triangle Bg oe aeat iat cee Os oy a 2 °3 oe SB tee MISCELLANEOUS EXAMPLES. 303 126, Find « from the equation sec G + x) + sec G - ) = 2/2. 127.. Shew that sin (9—a) , —_sin (6-8) sin (a—f) sin (a—y) © sin (@—a) sin (B— ) sin (8 - y) =0. sin (y— a) sin (y—f) 128. From each of two stations on a horizontal plane, at a distance ¢ from each other, a pillar on a distant hill, in the vertical plane passing through the stations, is seen under the same visual angle, and the angles of elevation of the top of the pillar at the stations area and 8. Shew that the length of the pillar is equal to ecos (B+ coos (6 +a) sin “sin (B — a) a)" 129. Ifin a right-angled triangle twice the distance between the centres of the inscribed and circumscribed circles is a mean proportional between the hypotenuse and the excess of the sum of the sides over the hypotenuse, shew that the radius of the in- scribed circle is equal to one sixth of the hypotenuse. 180, In any triangle shew that 1 1 l r 2 27 2 ee cos 54 + cos 3 Bt cos 5 C= 2+ oR: sin(e+A) = /sin24 sin (w+ B) ~ sin 2B” then tan’? ¢=tan A tan B. 131, If 132, IfA+B+C=(2n+1) a, shew that sin’ 24 + sin? 2B + sin? 20 + 2 cos 2A cos 2B cos 2C' = 2, 133. Sum the infinite series * (log 2)"+ = (log 2)? + (1 +2) log 2+ 1+ 2 304 MISCELLANEOUS EXAMPLES, 134. Shew that sin (4 —B) cos (C'—B) cos (A —C) + sin (B - C) cos (A —C) cos (BA) +sin (C'— A) cos (B— A) cos (C— B) #5 [sin (2B—24) + sin (20-28) + sin (24-20). eh Bie : sin(d-B) a?-B? 185. In any triangle fee $36. ~The diagonals of a quadrilateral figure are in length hand k respectively, and inclined at an angle A: shew that the area of the greatest rectangle which can be drawn with its four sides passing through the four corners of this quadrilateral is 5 hk (1+sin A). 137. A person standing at the door of a house observes that he can just see the top of a church spire over the intervening wall at an angle a; he then ascends to the roof, whence he is just able to get a view of the entire building, and he observes that the elevation of the spire top is 8. Having given the height of the house, that of the observer being neglected, determine the heights of the spire and the wall. 138. In any triangle shew that acos* A + boosts B+ ecos' 5 O=8+'p. 139. 5 g te sin —j—cosec 5 =0. 168. Shew by the aid of Trigonometry that if x+y+2=ayz, 3e-a° 3y-y? 3e-2° (3a— 2°) (3y— 7’) (82—2°) het get 137 1-88 81 ae ey 169. If2, m, m be the perpendiculars from the centre of the circumscribed circle on the sides of a triangle, shew that (5 b “) abe 4(-4+—4+-J=7—. Lo m n/ Inn 170. If A, B, and C are the angles of a triangle, shew that agit a, B a gl sin? 3t sin’ 5 + sin 3 cannot be less than Zt 171. Find the limit when @ is indefinitely diminished of sin ao nd of Vets ad sin 60 vers 60° 172. Shew that 1 1 1 ind ) log J2= 447 i*337 Say: 173. If d4+B+C=180°, shew that A A B C 4 B BC tan 5 + cos 5 set 5 Bet 5 = tan, + COS 5 See 5 RECS i oe a ae = 3 COS 5 Be 9 Sy MISCELLANEOUS EXAMPLES. 309 174. If sin (7 cot 6) =cos(m tan 6) shew that either cosec 26 or cot 26 is of the form m+ 0 , where m is an integer positive or nega- tive. 175. In any triangle om Ate Eten acosA+6cosB+ecosC a+b+e ~ abe , 176. A circle of radius r is inscribed in a sector of a circle of radius a, and 2c is the chord of the sector: shew that 177. If tana+ tan 2a + tan 32+ tan 4u= 0, shew that either 5a =n, or 20% = (2m +1) 7. 178. Shew that sin (0 — 8) sin (8 — Ms sin (8 — y) sin (6 — a) sin (a — 8) sin (a — y) * sin (B—y) sin (6 — a) , sin (8—a) sin (6—f) _ sin (y— a) sin (y— f) 179. ABC is a triangle; straight lines are drawn bisecting the angles A, B, C and meeting the opposite sides ut D, H, F respectively : shew that the area of the triangle DEF Cc : 28 sin4 sin 2 sin $ “fee, 0-2 ioe C=4,., 4B cos 3 OB ag Rg. 180. .From the top of a mountain the angles of depression of two stations in the plane at its foot are observed to be aand B, and the difference of their bearings is observed to bey: -shew that if ¢ be the distance between the two stations the height of the: mountain will be csinasin B sin 2a sin 2B Aut pane? sin’ = “Gen 310 MISCELLANEOUS EXAMPLES. 181. Find approximately the angle subtended by a target two feet wide at the distance of 450 yards, 182. Sum the following ae series : 1 a 4 re +t BoB it 183. If 2 sec 6 =sec (6 + 2a) + sec a — 2a), shew that cos’ 9 = 2 cos’ a, 184. Solve the equations 2 (sin 26 + sin 26) = 1 = 2 sin (6 + ¢). 185, In any triangle (sin 4 + sin B+ sin C/) (tam . +tan Bu tan 4) oA 2 Bs =4+4sin5 sin 5 sin 5 - 186. A, B, C, D, #,... are the angular points of a polygon inscribed in a circle; from the centre of the circle perpendiculars are drawn to the sides, and a second polygon is formed by joining the feet of the perpendiculars: shew that if the area of the first polygon is double that of the second, sin 24 + sin2B+sin2C+sin2D +sin 227+... =0. 187. In“any triangle Y ogee cas nea ¢ asin — —8 =0 3 2 2 2 ag Se 188. In any triangle a’ sin (B—C) , Jésin (C'— A) e* sin (A — B) | smB+snC@ sinC+sndA snA+sinB 189. If abe the side of a regular polygon inscribed in a circle of radius 7, and 6 the side of another regular polygon of twice the number of sides inscribed in the same circle, shew that pale er5)= 9/4) MISCELLANEOUS EXAMPLES. 311. 190. If4+8+C=180", shew that Q —sin 5) Qa —sin 5) cos 4 + (1-sin 5) (i — sin 5) cos 3 ch ON ae SO se tlle + ae D, Sl OB 9 = cos 3 cos 3 cos 3° 191. A person walks in a straight line towards a very distant object, and observés that at three points A, B, C, the elevations of the object are u, 2a, 3a respectively: shew that AB=3. BC nearly. 192. Ifa is less than unity so also is une 1 * log log (1 — x) ° 193. Having given 6sin B 3sin 2B 2sin3B cos (A + B) cos (A+ 22) cos (A + 32)’ shew that it is impossible that any value can be assigned to B other than nz. 2 tog Te RE 8 ang em ie find a general expression for the value of 4. 195. IfA, B, C are the angles of a triangle, and sin A, sin B, sin @ are in Harmonical Progression, then 1—cos 4, 1—cos A, 1—cos C are in Harmonical Progression. 196. If R be the radius of a circle described about a regular pentagon whose side is a, shew tha * = a nearly. d sind om cos Ap. 197. Ina triangle * aa a" and aa G: shew that 312 MISCELLANEOUS EXAMPLES. 198. If O be the centre of the circle inscribed in a triangle ABC, and D, £, F the points of contact with BC, Cd, AB respectively, shew that OA.OB. OC (AF +BD+CE#)=4R,. AF. BD.CE. 199. In the semicircle ABC, whose centre is O, and radius p, the straight line OB is drawn at an angle 2a to OC. Circles are inscribed in the triangles OAB, OCB. Shew that the distance between their centres is pal —sin 2a) »/(1 +sin a) (1 + cos a) 200. A man walking along a straight road observes the directions with respect to the road of two objects when the angle which they subtend is greatest, and then measures the distance from the point of observation to the point whence they appear in the same straight line: find the distance between them. 201. Shew that r,+7r,+7,-r=4R, z Stall i -19_ 7 202. Shew that sin 7 +cot’*3= Z° 203. ADC is a triangle; a second triangle is formed by the external bisectors of the angles of ABC ; then a third triangle is in like manner formed from the second, and so on: determine the angles of the n" triangle. 204. Find in terms of a the value of cos 4 (tan™ a). 205. Find the general term in the expansion of e* cos (bz +c) in powers of a. 206. A circle touches the sides AB and AC of a triangle produced, and touches the side BC at D: shew that a (s° — AD) = 48 (s—b) (s—c). MISCELLANEOUS EXAMPLES. 313 207. If cos*(a+f eau =6+¢,/-1, where the ‘letters denote real quantities, shew that a® By a? B 1 cos’ § sin? Bre ae (eo +e-%)? q (c@—e-8)? 4 208. Shew that log sec 6 3 209. Shew that the sum of n terms of the series sec a sec (a + 8) +sec (a + B) sec (a + 28) + sec (a + 28) sec (a + 38) =2{sin* 6-5 din? 3d a gargs — sin’ 40+... I = cosec 8 {tan (a + 8) — tan a}, 210. In a regular hexagon, one of whose sides is equal to a, a cirele is inscribed; and in this circle another regular hexagon, and so on until there are in all 2 hexagons: shew that the sum of the areas of the hexagons is 6/3 {1 2 ()} a’. 211. Adapt the expression a cos 4+bcosB+ccosC to logarithmic computation, the letters denoting the sides and the angles of a triangle. 212. A, B, C are telegraph posts at equal intervals. by the side of a rail-road; ¢ and ¢’ are the tangents of the angles which AB and BC aubjend at any point P; 7 is the tangent of the angle which the road makes with PB: shew that 2iiil Lf ¢ 213. Shew that T el ek 7 1 ga ftan 572 tan” Zog t 1393: 214, If P denote the point of intersection of the perpen- diculars from the angles of a triangle on the opposite sides, shew that PA°=4R?- a’, ‘S14 MISCELLANEOUS EXAMPLES. 215. Ifa=15°, find the value of (cos a + 4/=1 sin a) (cos 2a +,/—1 sin 2a) | cos 3a—,/ —1 sin 3a 216. ABC is a triangle; a new triangle is formed by the external bisectors of the angles: shew that the sides of the new triangle are 4R cos : A, 48 cos "2, and 42 cos : C re spectively. 217. Shew that the sum of the squares on the sides of the triangle formed as in the preceding Example =8& (42 +r). 218. Reduce to its simplest form cos 66 + 6 cos 404+ 15 cos 264+ 10 cos 56 +5 cos 36 + 10 cos@ 219. If 6 be a positive angle less than 5 shew that ,/cos 6 is less than cos +5 ; 220. If any point be taken within a regular polygon of an even number of sides from which perpendiculars are drawn to the sides taken in order, then the sum of one set of alternate perpendiculars is equal to the sum of the other set. 221. Shew that /rr,r,r,=8. 3 29 ie al = 2 222. Shew that 7 5 tan 7 + 2 tan 79° 223. Assuming the expression for tan 76 in terms of tan 6, shew that if » be an odd integer the following two series are numerically equal, n(n—1) n(n—1)(n—-2) (n—3 _2O=V) ,mr=In= 2) (n= 8)_ 1 nn — 2} (n= 2) 4, (m=) (n—2) (n—3) (n-4) | [3 5 enone eeey and if n be an even integer one of the two series is zero. MISCELLANEOUS EXAMPLES, 315 224. Shew that sin‘ 6 cos’ 6 = 557 (cos 96 + cos 76) — 6a Z (cos 56 + cos 36) + —. = cos 6. 225. Shew that sin? x _3 3°=1 = 1 on : ay 3 hs i yo A etl \ 226. If = ig shew that cos f + cos 3p + cos 96 = and cos 5¢ + cos 7h + cos 11d = be 227. A point is taken inside a regular polygon and per- pendiculars are drawn from it to the sides of the polygon; a new polygon is formed by joining the successive feet of the per- pendiculars : find the sum of the squares on the sides of the new polygon. 228, Shew that if B= 50 then cos 56 + sin 58 =— 2*sin (9-38) sin (0+ i cos (9 + 38) cos (8— 8) (cos @ + sin 6). 229. Given sin 6 {1 +tan*a tan? By + cos 6 {1 —tan’a tan’? B}4 = tan a+ tan £, shew how to determine 6 by a formula suitable to logarithmic computation. 230. If A, B, C are angles of a triangle, shew that sin A + sin B+ sin C is never less than sin 24 + sin 2B +sin 2C, 231, 41. O=nm or nae. 42. 20=(n4})7 or O=2nw a 43. 26=mnm or 6=2nra 2 44, 26=nr+(—- Ire. 45, O=(n+4)m or 44=nr+(-1)"F. 46, 6+ pans. VII 70, 71 2. 2 os l+sind in A . pages 70, 71. ; cos 5 =/( +sin A)—,/(1 —sin A). 3. asin $=—J(1+sin A) —/(1-sin 4), 3 5 4. 2nn + and 2am +. 5. One 42 and one TE ee e 1 » =I 6. 2am —7 and dna +7. 10. 5° ll. - ‘272 ANSWERS. VIII. 327 ; 4 12. sin d=+5, cow dae®; or sind =a, cos A=. 1s Sak 26. «1, a7, 4, 08. 3, VIII. pages 78...80. Example 20 may be deduced from Example 16 by changing A into }(180°— A) and making similar changes for B and C; Example 21 may be deduced from Example 17 in the same way. cost—cosa_ sin’acos8 cos%—cosB sin? £ cosa’ sin” 8 cos’a—sin*acos’B cosa+cosB | sin’ B cosa—sin’acosB ~~ 1+ cosa cos B’ therefore cos 2 = s 2 thon Aad l—cosz tan?@ = tan’a ———.. 39. —5 =z; that is l+cosa tan’ @ tan’a’’ 2 , 2 cos6—cosa cos8—cosa tan*’a B : B 7— = >——; therefore cos a cos a tan® a cos —cosa _ sina cosa’, sin’ a’ cos’ a — sin’ a cos” a’ Sees, as Se ag ee cd = neha oe cos B—cosa’ sin’ a’ cosa os B =— — B sin’ a’ cos a — sin’ a cos a’ cos® a — cos’ a” cos a + cos a’ > ~ cos a — cos a’ — cos a cos’ a’ + cosa’ cos’a = 1 + cosa cos a’ then find Te? : 41. We have to shew that 1+cosf cot B —cot (a+ 6) = cot 6 + cot (a— 8), sin(a+@—f) _ sin(2a—B+8) sinBsin(a+@) sin@sin (a—£)" 43, Find cos from the first equation and substitute in the that is o, costa second; thus we shall get cos*¢ = Epa 48. Put 2 nse for tang; then solve the quadratic; thus l—tan’i¢ — (cos 6 + sin 6’) + (1 + sin 6’ cos 6) sin 0 cos 0’ we shall find tani ¢?= 5 the upper sign gives the required result. The lower sign gives . 0 7 «=O — cot 5 oot (7-5 ‘ 328 ANSWERS, IX. 52. By Example 23, page 78, we get cos 4 cosBcosC =0, so that one of the three angles is a right angle. 59. cos 56=16 cos’ 6— 20 cos? 6+ 5 cos 0. IX. pages 91...93. (n—1) tand n—1 l+ntan'd {/(ntand) —Jcot gd} +2 Vn? the greatest value of this is when the first term of the denomi- 5. tan (6—¢) = nator vanishes. 6. 2Qsin 6 sin? 6 * o , 17607 8, The height in yards = 1760 x tan 1’= “Te 7G0 2 rly. 0 9. Let x inches ‘be the distance, 5 = tam 5 i ; thus - Sas x x 180x4 nearly. 10. We get sin A aoa, 12. 6. 13. 2cos?A=1+cos2A; therefore 4 cos*A=14+2 cos 24+ cos’24; but 2cos?24=1+cos44; therefore, by substitution, we have 8 cos*4=3+4c0s3 24 +cos4A4 ; square again and reduce. Simi- larly proceed with 2 sin’A = 1 —cos 24, 16. 8. 17. 6 T= 2nm. 18. 6+ =Inre =, 19. 5 20=2nm +8. 3 20. 2 =n or 73 = onwa 21. G=nn+ or 20=nm+(-1"%. 22. O=ne +3 or sin 26=2(Y2- 1). 23, O=(2n+1)5 or net 24. O=(2n+1)F. 95. @= 2” or n a 26 es nr or Int@*— . nT io qr se To’ . 57 TT Tt 3° nT T Tv oF. ba 28. 6= nm =F or nme. 29. O=nm+ 7. 380. 0 =m or nr + = 31. sin = 0, or cos6=0, or 00s $ =0. 32. cos 6+sin30=0, that is, cos @=cos (36 +3) : ANSWERS, IX. X. 329 33. 206=nr+ is 34, sinf=—1, or sin $= 0, or tan § =2. 35. 26=(2n+1)5, or 70 = nm +(—1)"'% It should be remarked that answers may be given under apparently different forms; thus, for example, suppose we have to solve the equation sin 20 = cos 6, or 2 sin 6 cos = cos 6, this gives 0 = 2nr +5 and 6=nr+(- Ne ; but we may write the equation cos G = 26) = = cos 6; therefore = oe 260 =2nr + 0. 84 X. pages 104...107. 1. a 2. 243 3/9=(,/8)%. 3. 7;-45-4. 06. ° 4 ee ee 4. 1-06. 6. 8; -1. 10. 2 Bf je =e. 12. Qn—a= Ine ee. 13, «=a cos (a—) or —a cos (a+ f). 7 3a 1d. w= Inw + or nw. 15. 00s (2 +1) a= eos (5 ~ 5 5 16. w= see( -§) or ~ 2 003 8 see a, 17. We can get iG ak _a-9 /m\t. 60 1. . 6 sin 2*7a=sin3a. 18. sin —= (=) sin 5; this gives tan 5. 19. 0=mm+5 or (2-1) 0=2mr+5. . 0 50 T 20. cos 4=0, or sin 5=0, or cosy = 0 22. 6" 23. n= 2 24. sin 2>* sin ue *=0. 31. Write for x successively p-wand S42, 34. By Art, 114, tan’ 4 + tan? B+ tan’C =14+4 (tan A—tan B)*+ } (tan B tan C)* +4 (tan 0 —tan A). 36. cot B+ cot C —cosec A _ sin (B+ C) 1 sin? A-sin Bsin@ & -smBbsnCG smd smAsnZBsn€ ’ 330 ANSWERS. X. XI. XII. 37. If A+B+C=180", we have * a incps’ A + cos? B + cos? C =1—2. cos A cos Bc08 Coeeeeeeeeee (1. Thus, if A, B, C, are all acute, the sum of the squares of the A.dihes ¥s-less than unity. Hence if we require the sum of the ‘sguares.of the cosines to be eyual to unity, one or more of the ' scutesaagles must be diminished, so that their sum will then be less than 180°, 38. From the value of sin(4+B+C), given in Art. 113, it will follow that sin A +sin B+sin C—sin(4+B+C) = sin A (1—cos B cos C) + sin B (1—cos.A cosC’) +sinC (1—cos A cos B) +sin A sin B sin C’; and every term of this expression is positive. a? 39. e 7. 40. Zero, 41. It depends on (1—cos 6)?(1+ 2 cos 6) being greater than zero. 42. Take the logarithm of the given expression. XT. page 122, 14. Denote it by w; then logu=,/5 log 5 2/5 log /5; therefore log (log u) = log 2 + log ,/5 + log (log ,/5) = 301030 +-349485 +1-543428 = 193943; therefore log u = 17562944; &e. Time ieee, ae a ay . cos a cos 5, cot 5 — cob O=— 5. 6. 1 or */ St 7. @=b. 8, atte a1, ll. @48?-20=2. 12. wey'nat(1+4). 13. cote =a — 5. 14, Ba, 15. B’=a?— 2ac cos2p+e?. 17. Find x and y from the given equations ; hence we shall see that #+y=a (sing+cos ¢)*, «—y=a (sind —cos ¢)’, &e. 18. From the last equation we get sin’ 6 sin’ ¢ = (sin B sin y — cos 6 cos ¢)’. ANSWERS, XII. 331 Eliminate 6 and ¢ by the first two equations, and we get sin’ a — sin® B — sin’ y = (sin? a — 2) sin* 8 sin? therefore sin’ a (1 — sin’ 8 sin’ y) = sin? B cos y + cos’ # sin? Hence find cos’ a, and then by division tan’ a. | 2 2 19. (mn)i {mi +n3}=1. + 20, SF ee Le e] 8 gS 22. We have a’ + y+ (y’—x*) cos 20 = 20? sin’? a; therefore a? +B? + cos 20 {(a? + b*)°— 4a°b? sin’ a}? = 26? sin?a; &e. at: saa 29. sin@sin d=sinasin§, therefore 4 sin® = 4 set = mn ein 8 2 sin’ - 40 . 4 sin’ a sin? B A 2 ck Lith Le ee therefore 4sin 3 4 sin xt l=1 aa 2 t cot” 5 and sin’ ¢ =—————$—— ; therefore cot? a + cos’ 8 wg 0 2 sin’ 5 -l=+ ft = tsint § (cot*S + cos" B) sin’ | 4 this reduces to Qsin'S—1 ge (1 ~2sin’$ sin* 8) 30. 7 must lie between — 2 and —1 or between I and 2. 31. By Art. 114 we may suppose # = tan A, y=tan B, z= tan C, where 4+B+C=180%. Therefore 24 + 2B + 2C = 360°; and tan 2A + tan 2B + tan 2C = tan 24 tan 2B tan 2C. This gives the required result. 32. vsine=sinz=—sin x cosy —cosxsin y =-—vsina cosy—v sind cosa, or sinacosy=—sine—sinb cos 2; and sinasiny=sindsin«; square and add, thus sin’ a = sin? 6 +sin’c+2sinbsin ccos«; therefore eede sin’ a ~ sin? b —sin*c 2 sin 6 sin ¢ Similarly cos y and cos z may be found. 332 ANSWERS. XII. XIII. a? 33. 6 28, 37. We have universally sin’ (4 + B)=sin’ A + sin? B+2sin Asin Beos(A+B)...... (1); also in the present case sin? A + sin’ B=cos’(...... (2). If 4+B is greater than 90°, then a fortiori 4 + B+ C is so also, If 4+B is less than 90°, then sin? (4 +8) is greater than sin? 4 + sin? B by (1), that is, greater than cos’C by (2); therefore 4 + B is greater than 90°- C. XIII. Ee 154...157. 5. Let i =a so that the angles of the triangle are 2a, 4a and 8a. Then the ratio of the greatest side to the perimeter sin 8a sin 8a “sin 2a+sin4a+sinSa sin 2a+sin 4a +sin 6a = 2 sin 4a cos 4a sin 4a ~Qsin 3a cosa + 2sin 3a cos 3a cosa +cos 3a 2 sin 2a cos 2a ~ 9 cos 2a cosa =2 sina, 8. ue = “= , therefore 2 cos 6 = sin 36 21. sin 6+sin d=2 sin (6+ ¢); therefore aia! oS 2 cos ee 6 6. od. therefore COS 5 COB 5 = 3sins 3 sin 5 53 ei 0 oP 3 therefore (1 — sin’ 5) (2 — sin’ = 9 sin? : sin prea ane therefore 8 sin’ Z sin gos sin? 77 sin’ = “ 2? therefore 16 sin’ sin’ 35> 2-2 sin’) — 2 sin? 5 0080+ cos. = Or thus, wae ta? wipes, ge 6 ba—3e, 2a°—idk ’ 5c—3a 4c * a+e rt bo a; therefore cos 6 = similarly cos f= ANSWERS. XIII. XIV. XV. 333 37. This will follow from Examples 20 and 21 of Chapter vit. 40. We have to shew that (b+ c—a) (c +a—b) (+b—c) is less than abe except when a=6=c. By squaring, this amounts to shewing that {a?— (c¢—)*}{b°—(a—c)"}{e? =e b)*} is less than a°b°c?; and each factor on the left-hand side is less than the cor- responding factor on the right-hand side except when a=) =e. XIV. pages 168...171. 1. A =30° or 150% 2. 30°, 90°. 45°, 60°, 75°. 4, The triangle is impossible. B=90°, C=72, c=4,/(5 + 2/5). 6. B=45° or 135”. From Art. 235 we have ¢+¢ =2bcos.4 and cc’ = 5? — a’, b’sin A cos A. 11. No; the triangle is right-angled. Palle) Oi TA SY. pe 12. We get sin 0 = sin3C’; and see Art. 230. 13. Sissel he eerie &e. 14. 9°6733937. 15. 132° 34’ 32”, 16. 55° 46’ 16” 17. 78°27’ 47”, 18. 119° 26’51"; 5°33'9”. 19. 69°10’ 10”; 46° 37’50”, 20. 116° 33’ 54”; 26°33’ 54” 21, 82°10'50”; 50°24’ 10”. 22. 124° 48’ 59”; 33°11’ 1”. 24, 48°11’ 23”; 58° 24’ 43”; 73°23’ 54”, 3 2A 3475 ‘i 25, cos A =F , therefore sin’ 37 6953° B=1' 29". 26. 70°53’ 36”; 49° 624”, 27. 38°12’ 47"; 21° 47°13” 28. 26°33'54”, 29. 69°49’ 35”; 50°10'25”. 30. 30° or 150°. XV. pages 177...183. In order to solve some of these ex- amples the student must be acquainted with the Mariner’s Com- pass. In the Mariner’s Compass the circumference of a circle is divided into thirty-two equal parts, so that each part subtends a at the centre of the circle an angle of => gy degrees, that is, an angle of 11}°. The following names are a to the points 334 ANSWERS. XY. of division of the circumference, North, North by East, North North East, North East by North, North East, North East by. East, East North East, East by North, East, East by South, East South East, South East by East, South East, South East by South, South South East, South by East, South, South by West, South South West, South West by South, South West, South West by West, West South West, West by South, West, West by North, West North West, North West by West, North West, North West by North, North North West, North by West. : 100 a. 1. 20,/3; 20. 2. 880(3+,/3). 3. re 5. rcosee 5 sin p. 8. The distance of the eye from the foot of the tower 5 (Ut b— 2h a 10. Let « denote the required height ; then eliminate 6 between ez=btan6, a+«e=btan (6+). ll. 10,/(115) feet; neglecting the height of the observer's eye from the ground. 12. 40,/3 feet. 13. Height 40 /6 feet; distance 40 {,/(14) + /2} feet. 15. Let & be the height of the tower in yards; r the radius of the circular arc which the observer walks over, also in yards: then — = 100, fe tan 15°, 18. 8+4,/2 miles per hour. 4 22, Let h’ be the height of the higher hill, 4 of the lower; then ,—¢t)) sina sin B au h’ kh cota —e ~sin(B-a)’ hk heotB+1 23. 180/38 feet; tangent of the altitude = ae, 25. B= 60° or 120°; approximate error 6”. 2c sina sin B sin (a + B) sin’ a + sin’ 8 — 2 sina sin B cos (a + f) * ANSWERS. XV. XVI. 3395 27. Let AB denote a side of the fort, C the position due South of A; let D be the second position, so that CD =a, and the angle ACD =90°; let # be the third position, so that # is on CD produced, DH = 0, and the angle BED=90°. Let ¢ be the angle between AB produced through B and C# produced through L. Thena+b=ABcos¢; BE=EC tan BCE and = LD tan BDH; and BDE = BAC, for a circle would go round 4, B, D, and C, so that (a +6) tan (90° — a) =6 tan (90° — ¢). 29. /(a*+ 2ab cos B+ 6°), where B=a or 7a, 30. Suppose both straight lines OC and O’C to fall within the angle ACB. Let AC =a, ACO =¢; then from the triangles ACO and BCO we get oc =o. (p+a) _ cos (p— B) | Hence tan¢ is known and then sing and cos¢. Thus we a? cos’ (a, + 8) ‘ 3 shall get OC" = raya B—2sinasin Bam(a+)" A similar expression can be found for OC? in terms of a’ and f’. Then 0’C?= 00?+d?. This finds a and then 4B =a,/2. tan a (1 — tan f) tan 6 (1 — tan a) ° tan a’ (1 — tan f’) tan f’ (1 — tan a’) Then OCO' = ¢’ — ¢, and OC = d cot OCO’ =d cot (¢' - ¢). Thus OC is found ; and then @ can be determined. 1. - 63° 26’ 6”, 32. 104-93 feet, 33. 30° 40’ 37”. 34. 269°40031 feet. Or thus ; find as above tan d= Similarly tan ACO’ = = tan ¢ say. 25(,/3—1) Js 7. The area is proportional to ghkl(hk+gl)(hi—gk). 8. Tto 3; 120°. 49, See Art. 230, 82°24’39"; 22°24'39”; 75°10’ 42”. 54....59 follow from 53, XVI. pages 194...201. 1. 216, 2 3. 6. 336 ANSWERS. XVII. XVII. pages 205...207. 1. 2, ,/3-1, —,/3-1. 4, —2, 2,/2cos 9°, 2,/2cos63°, 2,/2cos81°, 2/2 cos 153°. 7. Let « be the height of the balloon, and a, 6, ¢ the sides of the triangle ABC ; then 4c’x* — 36a°b*x? + 9a7b°e? = 0. 9. About 14 inches. 12. Suppose % the height of the tower, 7 the radius, x the distance of the first place of obser- vation from the centre; then ” = cosec a =~ * — cosec s r 2 r 2 h=atana, h=(x—-a)tana’. From these four equations we may eliminate a, and find & and 7, and also the required relation between a, a’, B, f’. 13. From the preceding question 4 cosee ® — cosco® If we suppose that an error 8 of the same sign is made in 8 and these errors tend to compensate each other ; the greatest possible error in r will be determined by supposing that errors of opposite signs are made in Band fp’. Suppose then that instead of B we ought to have B-8, and instead of B’ we ought to have B’+8. Then by Art. 194 we shall find cos = B cos B (cos e + cos 5) (1 — cos B cos F) ap_8( 253 OER )_a\eea a 2 Blea V1=5 v8 sint F sint & sin? E sine & Divide by the value of - and the required regult is obtained. 14. If PQ=aand QR=S, it may be shewn that 1 (a-6b) (a+b) So" ae age OBS then the change in SQ arising from a small change in B can be calculated, ANSWERS, XVIII. XIX. XXII. 337 1 , 2 XVIII pages 209...211. 2.1. 7. 2087 ° 18. a =75 (5 — 2,/2). 19. onthe 20. w=Oor+} 21. w=0 or +}. 92. a=. 23, e=+1 or +(1+ ,/2). 24, aw=aora@—atl. 28. 2=2. 30. w=l,y=2;e=2,y=7. 34, ma + (-1)"*" @ or (m+n)a+ Clyz . 35. (Qn+m)r+ a : mo 36. nmr+ (- 1) Z . “XIX. page 222. 6. The expression on the left-hand side is 2 sin 7 cos 6 + cos & sin 9 ; then put 6 for sind and 1 € for cos 6. cos (B+y—2a) __ sin(B + y— 2a) 4 cos (a—f) cos(a—y)’ ~~ 4c0s(a—f) cos (a—y)° _ XXITL. pages 248...251. 1. Use sin?a=3(1l—cos 2a). 2. Usesin®a=}(3 sina —sin 3a). 10. A= cos (20 + na) sin na. 5. Foosa+ 10. 3(cos 26 —cos 2**? 6). 2sina 11. 6°" cos (8+ cos Osin 6). 13. 3 (e289 + e8in®) cos (cos 6). 1, 282 t0g(1—cos 6). 15, ore sin (“S**). 16, ¢sin@co88 og (9 + sin’ 6). 20. See Art. 129. geo @ 1s . 6 21. 2 *sin seni — 7 Sin 20. 22. tan 9 —tan 5, - 1 or 23. 24, cosec 6 {cot 6 — cot (m+ 1) 9}. 2sin® $6 sin? 2°76" 25. cosec (9 + 5) {tan (n+ 1) G + 5) —tan (2 + 3} 5 26, tan“ 2 . 27, tan ne 28. 5 008 53 — cos 4a) - 29. Acosec 6 {tan (n + 1) 6 —tan 6}. T. T. 22 338 ANSWERS, XXIII. XXIV. 1. 6 Qn+1 6 1 6 on... 3"6 30. J COsee 5 {scot 6—sec 3} . i Z zfeot ao 3” cot >}: 82. cot™ 57 cot™? sa a. 33. cos @- sin 8 cot 2" 6. ‘ites log 2 sin 2"*? 6 sin nef 6) 34, log 2 sin 26 -—=$—5, —— 2 35, foot seri ~ cot 3s" XXIII toast. a=. 2, 2 . pages. ves . ‘6° ve 90° . 3s: . 96° ‘The four preceding results are obtained by expanding the values of log = and log cos 6, which are given in Arts. 274 and 320, in powers of'0, and equating the coefficients of like powers, on 14, 790° XXIV. pages 286...289. 4, We have -_ +cot x)?—3 (tan «+cot 2)=m- 3m; &e. ‘15. Ts ara {1 — 15a? + 152*— x}. -MIScELLANEOUS EXAMPLES. Pages 290...324. 1. 20°; 42. 25 5 (1 +am) (1 - im): 3. a or b. 8. = 10. a? tan $ = (s+ctan$) (c+dtan$). ll. 222, 673°. a 19. @=(2n+1)% or 5 (tere). V5 -1 21. 205,30. 22. sinw="——. 238, ean iT. 25. 1. 27. O=(2m4+1)m or 6= {mn + (15h. 29, wants. 31, 60°, 45°, 135°, 120°. 33. O=naornm +5. 34. _ 36. 39. 41. 48, 54, 55. 57, ANSWERS, 339 The other values are to be found from cos*x-+- cos «sin a =cos*a. - A+B+O . Bad-A. - = ~4 sin tB +0 + . sin + 4 Fin At Cc. 5 150 degrees, 1662 grades, 47, 2° (b- a) = (2e—a—0). cos (7-§) or ae 51. 2h ke 2 cos = - 4°92 6=(2n+ nF or i [2na (FZ + 0)}. (nsind+mcos6)'=2m(m+x). 56, 2045 =ne+(-1 0 =f + i 61. In the first polygon 5, 4, or 5; and in the second 4, 8, or 20 respectively. 62. 20=nr + (- es 63. 70. 72, 81. 82. 84. p(n? — 2m*) = gmn. 64. 045 = Ines (a+). 1 _cos a~cos B 71, 600+ 100p +g os 6 * 60x90 100x100° cos =, — cos 5, 2" 2? € 7 1 = (2m+1)7 or (2m +1). 74, w=nz7 or sym tly ae 2 (9gn — 10pm) rue fae ee espennely where 2 msg — 1p) a= (2n ‘ay a + epma 83,. A Any Salat 9) seep, Ae The value of sia. esny Fanges betwesr — cos* z and sin? 3 ; where A ‘is the austen sum, 93. mr + (— Be 340 ANSWERS. 94. The value of sinw+sin y ranges between — 2 sin 4 and 2 sin = where A is the constant sum. "95. 6+7=2nn. Tv Tv Tv Tv 96. —4cos (3- 7) cos (3-4 + 4) cos (s-2 + 7) cos (3-¢+ 4) ji where 3=1(44B+0). 98, aus 6, 2 2 cos? a — cos 0 cos a— a B tan tan 2 = anon 100. sin ésin ¢ = + —,—_,—_;,, = + ¢ 1—c’sin’ asin’ B’ 2 2 cosat+cosB 102. 114:6 inches. 103. 0=(4n+1)5. 104, (a7— 6°) (a+ b)= (a—b) c? —4abem. 30 av 0 30 7 7 113. ig em, Or g- ga 2nts 5. 115. 30-7 =2nn+(9-7). 126. 22=2Qn7r+-x. 133. 4. 141. 0 or+,/3 or “5: Ts ae 156. zat 162, ev". 163 ee. 2 cosa ,/cos B 2 cos B Vcos a r/cosa+,/cos 8 ,/cosat,/cosB 181. 5 minutes. 182. e—1. 15 13 152. =, 5. 164. cos(a+y)= » 00S (w—¥) = 171. oe o|8 184. 0+ p=nr+(-1)%, and 0-p=2mr* 3. 1 s th tome amareks om Feb D C9) IMS i ie: a i ea Sen ey ey ew le ser ANSWERS. 341 211. The given expression = 2a sin B sin C. 218. fT. 918, 2cosé. 227. Let m be the number of sides in the polygon, r the radius of the inscribed polygon, c the distance of the assumed point from the centre of this circle, Reet n B 6 241. 3./(p* + 9°—6) =p J(q?- 3) + ¢/(p*- 3). 247. a {sin 116 + sin 99 — 5 (sin 76 + sin 56) + 10 (sin 30 +sin 6)}. then the required sum is 4nr? sin? 5 + xc* sin? B. 256. Nearly 7. 259. If 6=5 we have s=cosz and o=sinz. 260. If m be the number of sides of the polygon, and ¢ the 2 distance between the two points, the required sum is >: n 3"6 6 oi 1 _ 38 | _ 6 270. tan a! 277. log sin 2 log sin 5 - 1 1 1 9 aS ae esas AS 5 576. 280. Fecad {ind eel) i} ; 289. cosec 0 tan? nf. n(n+1) a | 1 290, *™* Game, 298. ttresa Toma}: CAMBBIDGE: PRINTED BY C, J. CLAY, M.A. AT THE UNIVERSITY PRESS- MATHEMATICAL WORKS BY I. TODHUNTER, M.A., F.R.S. Euclid for Colleges and Schools. New Edition. 18mo. cloth. 38. 6d. Mensuration for Beginners. With numerous Examples. Third Edition. 18mo. cloth. 2s. 6d. 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The Conflict of Studies, and other Essays on Subjects connected with Education. 8vo. ros. 6d. Contents. The Conflict of Studies. Competitive Examinations. Private Study of Mathematics. Academical Reform. Elementary Geometry. The Mathematical Tripos. Edited by Mr TopHunter. An Elementary Treatise on Differential Equa- tions. By GEORGE BOOLE, D.C.L., F.R.S. Third Edition. Crown 8vo. cloth. 143, A Supplementary Volume. Crown 8vo. cloth, 8s. 6d. MACMILLAN AND CO, LONDON, paige Neng yore See eens oatunada lta aeeee net oleae ca ares ei ante emeee Sl rhs Sanam ete cmscssiorh fees Saree et ra no See ee arereiee ban ciate ee rey ee ev alae eee 3 Sie - Seta Serene ane ce Sie Reet ees Snail Sere: heh ares euneney er cate EN eed EL AT) poarwe ines a Bash ener . feats peewee Saas sa ae Si eS yt yy Cie esa eee rr by ~ TO el naa bern aaa Ses er Pea eth oat baa