ELEMENTARY TREATISE oir PLANE TRIGONOMETRY, WITH NUMEROUS EXAMPLES AND ArrLICATIONS. ^cetgncb for the net ot ^iglt ^rhoolg mib CoUcQee. BY J.'^^ORMSON, M.D., M.A., rniNOlPAIi OF THB WAIiKKUTON HIGH BCHOOTi; BX-PBINCIPAL OP THIS NRWMAUKl.; uioH scHoor,; late mbmbkb of the medical council ajtd examiner ON TTTKOKKTICAL AND PRACTICAL CHBMISTBT IN THE CULLKGE Oi' i*HySIOLANS AND BUaaHONS OW ONTAiUO. TORONTO: CANADA PUBLISHING CO. (LIMITED). 1880, ''.H., ^^-1 ..*?» Entered according to Act of Parliament of Canada, in the year one thousand eight hundred and eighty, by The Canada Publishing Company (Limited), in the office of the Minister of Agriculture. C. B. ROBINSON, PRINTER, JORDAN STREET, TORONTO. if- * ■ PREFACE. The following course of Trigonometry has been prepared with the view of meeting the wants not only of High School but also of Collegiate classes, and in pursuance of this end, the more elementary portions — such as are required for the matricu- lation examination in the University of Toronto — are printed ift a larger type, and form a connected treatise independently of the portions printed in the smaller type, which are intended for candidates reading for University honors. ;f, ; In discussing the trigonometrical functions I have adopted the method of ratios or the Cambridge method ; but on the recommendation of several eminent teachers in our colleges, I have given a full account of the older method or that of the "'.M.n.9 definitions," and have employed the latter method in solving right-angled triangles and in tracing the variations in sign and magnitude of the several functions in the various quadrants. Of course, teachers who prefer to use the " ratio definitions " only, can omit the portions treating of the " lino definitions " without breaking the continuity of the ciourse. The solution of right-angled triangles by the natural func- tions, and the application of Trigonometry to the determina- tion of heights and distances, have been introduced at a very iv. PREFACE. early stage, with the view of leading the sMident to take a deeper interest in the study of the science at the outset. Numerous examples and solutions, chiefly of a practical character, are given, at frequent intervals throughout the work, for the purpose of fixing in the mind of the student the principles of the text, and of illustrating the practical bearing and utility of Trigonometry in Surveying, Navigation and Astronomy. ' ' The more advanced portions of the work have been dis- cussed from an elementary point of view, and will be found to include all that is re(juired on this subject from University students generally. - As my aim has been to produce a work of utility, I have not hesitated to consult, in the preparation of this volume, the various English and foreign works on the subject ; but those to which I am most indebted for suggestions, are the works of Hind, Todhunter, Snowball, Wiegand and Chauvenet. i • The examples have been obtained from various sources ; some are either entirely original or original in the form in which they are presented ; some have been selected from other trigonometries, and some from examination papers set during the last twenty years in the Universities of Cambridge, London, and Toronto, and in McGill College, Montreal, and Harvard College, U.S. The answers have been given wherever neces- sary, and great care has been taken to secure their accuracy; but it is quite possible that some errors have crept in, and I shall be very thankful for any corrections anyone may send to myself or the publishers. I shall also feel very grateful to my PREFACE. V. professional confreres for any remarks regarding difficulties or omissions in the text or examples. Great care has been taken to render the text clear and explicit, and I am sure my readers will unite with me in an expression of thanks to the publishers for the almost faultless style in which they have executed the typographical portion of the work. A second part, on Spherical Trigonometry, with numerous examples and applications to Nautical and Spherical Astronomy and Geodesy, is in course of preparation, and will appear if the present work meet with a favorable reception. J. MORRISON. High School, Walkbrton, June lift, m^o. •r J ... CONTENTS CHAPTER I. PAOB Definitions • • 1 Sexagesimal Division of Angles 3 Centesimal Division of Angles 4 Circular Measure 4 ''^ CHAPTER II. •,,.'•' ■ ■:.,-■:..:;• Definitions of the Trigonometrical Functions 8 General FormulsB 10 Line Definitions of the Trigonometrical Functions 12 Functions of 30", 60% 45% 18% 54°, &o 18 Funtions of half an Angle in terms of the whole Angle 21 Examples...... 24 ...... .•"'7'X;i!, 'Ai:: ;■' ;.i; .■.:'.;•.,.■,, ;;,;„ii;,v, CHAPTER III. -r Solution of Right-angled Triangles 28 Heights and Distances 32 CHAPTER rV. V ... Extension of the Definitions of the Trigonometrical Functions, use of the signs + and — 37 Variations in Sign and Magnitude of the Trigonometrical Functions 38 Line Definitions 43 Functions of 180° -A 45 Functions of 90° + ^ 46 Functions of Negative Angles 47 Groups of Angles having the same sine, dco. 48 Examples 50 viii. CONTENTS. CHAPTKll V. FAOB Functions of th« sum and difference of two Angles 54 Functions of Coniiiound Angles CO General Fonuulit; 61 Functions of the Multiple and Subniultiple of an Anglo 64 Functions of mA 66 Functionsof (vl+i?+C), ( -A+B+C), &c 67 General FormuLe 69 Forniuhe of Verification 71 Examples 72 CHAPTER VI. ' " . Circulax Measure of an Angle > sine, but < tangent 80 - - sin ^ , tan Limiting values of and ~ 81 ^ 6 03 Smd>e-- ..,........: 82 4 Calculation of the Numerical Values of the Trigonometrical :•: J Functions 82 Trigonometrical Tables 86 Increments of the Trigonometrical Functions corresponding to a given Increment of the Angle 89 Examples 92 Dip of the Horizon 94 Examples 96 CHAPTER VII. Properties and Usea of Logarithms . » c 99 Modulus of a System of Logarithms 103 Relation between the Bases of two Systems 104 Common Logarithms 105 Determination of the Charactoristic 1 06 Logarithmic Tables 108 Arithmetical operations 112 Arithmetical Complement of a Logarithm 114 Logarithms of the Trigonometrical Functions 115 Logarithmic Functions of Angles near the limits of the Quadrant 120 Examples 123 CONTKNTS. ix. CHAPTER VIII. PAGE Formuloe for solving Right-angled Triangles 127 FormuliB for solving Oblique-angled Triangles 129 Area of a Triangle 139 Perpendicular of a Triangle 140 Exaniplea 140 CHAPTER IX. Solution of Right-angled Triangles 151 Solution of Oblique-angled Triangles 154 Examples 109 CHAPTER X. Application to Surveying, Navigation and Astronomy 176 Mariner's Compass 184 Length of a Degree of Longitude 185 Moon's Distance from the Earth 186 Parallax 187 Examples 190 '' CHAPTER XI. •■ _;•••':" -'.^ n!.f i\ .. ''i ■ Radius of the Inscribed • "' cle of a Triangle 201 Radii of the Escribed Circxorf 203 Radius of the Circumscribed Circle 207 Distance between the Centres of the Inscribed and Circum- scribed Cir'os 209 Distance between the Centres of the Escribed and Circum- scribed Circles 210 Perimeter and Area of Inscribed and Circumscribed Polygons. 212 Circumference and Area of a Circle 213 Area and Angles of the Inscribed Quadrilateral 214 Examples 215 CHAPTER XII. - Inverse Tiigonometrical Functions .*; . V 225 Examples 227 X, CONTENTS. CHAPTER XIII. PAOH Division of Angles 231 Solution of Equations 236 Auxiliary Angles 238 Examples 241 Quadratic Equations 242 Cubic Equations 244 Eliminati jn of Trigonometrical Functions 246 Examples 248 CHAPTER XIV. The Exponential Theorem . . 250 The Logarithmic Series 252 The Napierian Base 255 Calculation of Napierian Logarithms 257 Calculation of Common Logarithms 258 Jheory of Proportional Parts 258 Examples 262 CHAPTER XV. De Moivre's Theorem , 263 Trigonometrical Expansions 266 The sine and cosine of an Angle expressed in terms of the Circular Measure 268 Expansion of the Integral Powers of the cosine 269 Expansion of the Integral Powers of the sine 270 Expansion of cos tid in terms of Descending Powers of cos ft . . 272 Expansion of cos n6 in terms of Ascending Powers of cos 6 . . 274 Expansion of sin x and cos x in terms of the Circular Measure, independently of De Moivre's Theorem 275 Sines and tangents of Small Angles 277 Examples 278 CHAPTER XVI. Exponential Values of the sin, cos and tan 280 Circular Measure expressed in terms of the tangent — Gregory's Series , . . 281 CONTENTS. Xi. PAOB Euler's Series for computing tt 282 Machin's Series for computing tt , . . 283 Solution of Equations by Series 284 Summation of Trigonometrical Series 290 Examples 295 CHAPTER XVII. 1 Resolution of x^n - 2 cos . ac« +1=0 into Simple and Quadratic Factors 299 Resolution of cc" - 1=0, n being odd 302 Resolution of x" - 1=0, n being even 303 Resolution of x'^ + 1=0, 7i being odd . . . . 304 Resolution of x^ +1=0, n being even 304 Resolution of sin x and cos x into Quadratic Factors 307 Computation of Logarithmic sine and cosine 308 I Miscellaneous Examples qiq [ [ Appendix ^ ; ; ;; ggg Numbers often used in Calculations 332 ^. a- •^■■ryem; THE GREEK ALPHABET. The Greek Alphabet is here inserted to aid those unac- ({uainced with Greek in reading the parts of the text in which its letters are used. A a Alpha. B /3 Beta, r y Gamma. A 8 Delta. E € Epsilon. Z ^ Zeta. Rr) Eta. e Theta. I t loto. K K Kappa A A Lambda. Mfji Mu. NV Nil. : H I Xi. ■ O o Omicron. Hit Pi. P p Rho. S cr Sigma, T T Tau. Y V Upsilon. $ <^ Phi. X X Chi. * i/^ Psi, Q 0) Omega. u. PLANE TRIGONOMETRY. CHAPTER I. ON THE MEASUREMENT OF ANGLES. ' Article i. — Trigonometry (from rpLywvoV) a triangle, and fieTpiio, I measure) is that branch of Mathematics which treats of the methods of determining the unknown parts of a triangle when a sufficient number of them is given ; but in its more extended signification, it embi'aces the investigation of the various relations of angles in general, such investigations being carried on by means of certain quantities called trigonometrical ratios or functions. It is divided into two branches, — Plane Trigonometry and Spherical Trigonometry ^ — the former treat- ing of angles and triangles drawn on a plane, and the latter of those on the surface of a spliere* 2. — Definition of an Angle. In Plane Geometry an angle is the inclination of two straight lines to each other, and must therefore be always less than two right angles; bu^- in Trigonometry an angle may be of any magnitude whatever, and is conceived to be described by a straight line revolving about a given point from one position to another. Thus, the lines AC, JSC, would, in Geometry, bound only one angle ACB; but in accordance with the more extended definition of an angle, B^ they may also be regarded as containing an angle subtended by the arc DE, and described by a line revolving about O^ In the PLANE TllIGCNOMETRY. direction of the arrow-head from the position AO to the posi- tion CB. Again, in order to effect this, tlie revolving line may be supposed to make any number of revolutions. Thus, the minute hand of a watch, at half-past one o'clock, will, since twelve o'clock, have described an angle whose magnitude is six right angles. 3. Let AC and BD be drawn at right angles to each other, then the whole angular space about is divided into four right angles, AOB, BOO, COD, BOA, which are respectively called the first, second, third and fourth quad- rants. Let OPi, OP^, OP3 and OP4 be different positions of the revolv- ing line OP, by the revolution of which about from its primitive position AO, in the direction from A to B, the various angles at are described. The extremity of the revolving line will evidently describe the circumference of a circle. When the revolving line coincides with AO, it is said to make with it an angle zero ; when it reaches the position OP^, it will have described an angle AOP;, which is called an angle in the first quadrant ; when it coincides with BO, it will have described the right angle AOB; when it takes the position OPj, it will have described the angle AOP2, greater than one right angle, but less than two, which is called an angle in the second quadrant, ; when it coincides with CO, it will have described the trigotioinetrical angle AOG, or two tight angles. In like manner, when it takes the position OP3, it will have described the trigonometrical angle AOP^, sub- tended by the arc ABCP3, which is an angle in the third quad- rant^ when it coincides with OP^, it will have described the trigonometrical angle AOP^, subtended by the arc ABC DP i^ DEFINITIONS. S which is an angle in ih^ fourth quadrant, and when it coincides with AO, it will have described four right angles. If the revolution be still continued until the position OP^ is again reached, an angle greater +han four and less than five right angles will be described, and 33 on. The angles described by the progressive revolution of OP in the direction from A to B, are regarded as +, or positive; while those described by OP revolving in the opposite direction from A to D, are regarded as - , or negative, the signs + and - here indicating merely contrariety of direction. Positive and negative angles are always understood in this sense, unless the contrary be stated. 4. — Sexagesimal Division of the Right Angle. A degree is the ninetieth part of a right angle ; a minute is the sixtieth part of a degree ; and a second the sixtieth part of a minute, and so on according to a sexagesimal subdivision, but in practice all subdivisions beyond seconds are expressed as decimal parts of a second. The characters used to denote degrees, minutes and seconds are °, ', " ; thus 17° 5' 24". 4, repre- sent seventeen degrees, five minutes and twenty-four and four- tenths seconds. The comjylement of an angle containing A degrees is the remainder obtained by subtracting it from 90°, and is written 90° --4°. llius the complement of 40° 10' is 49° 50'. If the angle is greater than 90°, its complement will be negative ; thus, the complement of 130° is 90° - 130° = - 40°. In every right-angled triangle, each of the acute angles is the complement of the other. ■ ' The supplement of an angle containing A degrees is the remainder obtained by subtracting it from 180°, and is written 180°-^°; thus the supplement of 70° is 180° -70 = 110"; and of 204°, - 24". Since the sum of the angles of a triangle is equal to two 4 PLANE TRIGONOMETRY. right angles, each of the angles is the supplement of the sum of the other two. 5. — Centesimal Division of the Right Angle. By some of the Frenci: mathematicians the right angle was divided into 100 equal angles, called grades; each grade into 100 minutes, each minute into 100 seconds, and so on according to a centesimal subdivision. K D denote the number of degrees in an angle, and G the number of grades in the same, then, since 90 degrees = 100 grades, we have P : Gf : : 90 : 100 ; or, a = \^ D, and D = ^qG. (1) The centesimal division is now abandoned oven in Franco, as its general adoption would involve a change in our tables, and in the graduation of astronomical and other instruments. 6. From Article 4 it will be seen that the unit of angular measure there adopted, viz : 1°, cannot be directly compared with any lineal unit, such as an inch, a foot, &c., since they are magnitudes of different kinds. In the following articles it will be shown how a lineal unit may be applied to determine the magnitude of an angle. 7. — Circular Measure. The circumference of a circle varies directly as its radius. ,_. In the two concentric circles in- scribe regular polygons ABC... abc... having the same number of sides ; then from the similarity of the tri- angles ABO, ahO, we have ; ?: ah Oa :^ CIRCULAR MEASURE. 5 and since the polygons are equilateral, perimeter of the polygon ABC ... _0A perimeter of the polygon abc... Oa If the number of the sides of the polygons bo increased, their perimeters will approach more nearly to those of the circles in which they are inscri})cd. Now, let the number of the sides be inoieased indefinitely, then the perimeters of the polygons will ultimately coincide with the perimeters of the circles, and, since the above proportion will still hold true, we shall have circumference of the circle ABG ... _ radius OA circumference of the circle abc... radius Oa Therefore the circumference of a circle varies directly as its radius, and therefore, also, as the diameter; hence it follows 1.U i xi- i.' circumference , i • • • 1 1 i that the ratio, ; , has a certain invariable value. diameter It will be shewn hereafter that this ratio is, to five places of decimals, the number 3-14159 which is usually denoted by the Greek letter tt. If, then, r denote the radius of a circle, its circumference will be 27rr. 8. In the circle BCD, take any angle BAG = A°: let CB the arc which subtends it = a, and the radius AB = r. The semi-circumference BCD will be Trt' which subtends the angle 180°; p.n J since " the angles at the centre of a circle are proportional to the arcs which subtend them " {Euc. VI. 33), we have 180° : ^° : : arc ^(7i) : arc ^(7 9 PLANE TRIGONOMETRY. Whence A" — 180° a IT r r- 1= 180' 3-14159 &c. a X — r = 57''-2957795... x.^. (2) r Therefore if a and r be given in lineal measure, the number of degrees in the corresponding angle A can be obtained. The invariable angle 57°*2957795... is usually denoted by w", therefon (2) may be written ^" = a)°x-. . (3) r ■ ' • ■ 9. If a = r, then ^° = a)°, that is, the angle which in any circle is subtended by an arc equal to the radius, contains a)° or 57'''2957795... , which is called the unit of circular measure, and the fraction is called the circular measure of the radius angle. From (3) we have 0) r = — a (4) A' ^ ' A" and a = —j- r. (5) w arc 10. If the circular measure be denoted by Oy we have radius ii° = 57*'-2957795... x^^ : - u, ^'=3437'-74677... x^V (6) X" = 206264"-806... x^) •' - If r denote the radius of a circle, its circumference is 27rr, EXAMPLES. ( and iherefore the circular measure of four right angles is _. or 277. Hence it follows that the circular measure of two right angles is TT, and of one right angle --. Examples. 1. Find the number of degrees in an angle which ia sub- tended by an arc of 5 feet, the radius being 2 yards. .. . Ans. 47M4'47"'34 " 2. Find the circular measure of 42°. Ans. -73303 3. How many degrees in an angle whose circular measure ^^^^' . ' Ans. 30° 0' 43''-45. 4. Find the radius when an arc of 1° measures 10 feet. Ans. 572-957795 feet. 5. The arc of a railway curve is 250 yards, and subtends an angle of 6J° at the centre. Find the radius. Ans. 1-302 miles. 6. How many degrees are in an angle subtended by an arc of one inch to a radius of one foot ? Ans. 4" 46' 29". 7. What angle will an arc of one inch subtend to a radius of one mile 1 Ans. 3"-254. 1 [8] CHAPTER II. TRIGONOMETRICAL RA.TIOS — FUNDAMENTAL PORMULiE. II. Let QAP be any angle, and in AP take any points B, Bif J?2, ' = _. a cosec il = sec i5 = — . a a^ a^ o» cosec- ^ = cot^ -4 + 1. (16) 5 r 12 PLANE TRIGONOMETRY. Line Definitions. 17. The Trigonometrical ratios defined in Art. 11 may be represented geometrically by means of a circle, as follows : Let DHr be a circle whose radius = 1, and let DAB be any angle in the first quadrant DAII. From B, the extremity of the radius AB^ draw BG perpen- dicular to AD, and from D draw DE perpendicular to AD, to meet AB produced in E, Also draw BG and HF perpendicular to AH, the latter meeting AB pro- duced in F. Representing the angle DAB by ^, we have sin A BG AB BG = BG. (1) That is, to a radius of unity, the sine of an angle is the perpendicular drawn from one extremity of the arc subtending it, to the diameter passing through the other extremity. cos A ==^:=^=. AG = GB, AB 1 (2) or, the cosine of an angle is the distance from the centre to the foot of the sine, or the sine of the complement. , ED ED „^ (3) or, the tangent of an angle is the line touching one ex- tremity of the arc subtending it, and terminated by the diameter produced through the other extremity. THE LINE DEFINITIONS. 13 cot A = t2in BAH, (Art. 12) ~AH 1 (4) or, the cotangent of an angle is the tangent of the com- plement. (5) or, the secant of an angle is the produced radius drawn through one extremity of the arc subtending it, and terminated by the tangent drawn from the other extremity. cosec ^ = sec j5il//, (Art. 12) AF AF . ■■■ -AH=-T==^^^ ' '- ■:-^: (6) or, the cosecant of an angle is the secant of the com- plement. l8. From the similar triangles ABG, AFH, we have AG:AH::AB::AF, or, since AG^BC^sinA, and AF= cosec i, sin ^ : 1 : : 1 : cosec A. ^■"■'- .■■■/■-■''■I' "■' . '■'.-" . Hence :, .^ sm^^^^^^^. From the similar triangles ABC, AED, we have '^ ':\^f'''''"' AC.AD-.'.AB.AE, ,"' ' or, cos il : 1 : : 1 : sec -4 H«ncp cos ^ sec A The angle AFH=i\iQ angle BAD, therefore the triangles ADE, IHF, are similar, and we have 14 PLANE TRIGONOMETRY. DE:AH::AD: IIF, or, tan ^ : 1 : : 1 : cot A. Henco tan A = cot -4 The results of this article agree with (8). 19. From the similar triangles ABC, AED, we hav« AC '.AD .'.BG '.DE, or, cos A : \ : : sin -4 : tan A. • A Hence tan A = -, which agrees with (13). cos^ ^ ^ ^ Again, from the similar triangles AGB^ AIIFj cos A cot A sin A From the right-angled triangles ABG^ AED, AHF, we obtain, by Euc. I. 4:7, sin^ -4 + cos^ -4 = 1 sec^ ^ = 1 + tan^ A ' cosec^ A = l + cot^ A. 20. Besides the ratios already defined, GD, the portion of the diameter intercepted between the foot of the sine and the extremity of the arc, is called the versed-sine, abbreviated "versin;" and HG the versed-sine of the complement is the coversed-sine (coversin). The versin and coversin can have no existence according to "ratio definitions" given in Art. 11, since the ratios there defined are independent of the radius ; nevertheless the terms yersin and coversin are used as convenient abbreviations for 1 - cosine and 1 - sine, respectively. EXAMPLES. 15 21. By means of the formuloe (8) (16) any ratio of an angle may be expressed in terms of any other ; or if any ratio is given, all the others may be found. ^o;.— Express all the trigonometrical ratios in terms of the tangent. , . ; 1 1 sec ^ = ± J{1 + tanM) cos A = ^^^ - ±-j^i+ tan^ a)' tan A sin A=±J{1- cos^ ^) = ± -jQ-^^A) , 1 ^ ^(l+tan^^) cosecJ. = -: — 7=—' — +r^"l ' sm A tan A cot A tan A The significance of the double sign will be explained here- after. '...,- Examples. '' ;■■:^■ '.•■■• ^'^^ ,*■-; 1. Given tan A = -r^j find the sine, cosine and secant. ,5 ,12 .13 Am. sm A = -rj, cos A = -y^, sec A = -j^- 2 2. If tan A = ^> find the vcrsin and cosec. 5 ^Tis. (1±— )and ij29. 3. Given sec ^ = 4, find sin ^ and cot 6. •4* / 1 Fi" T ^ris. sin ^ = — -. , cot ^ = ± 4 ' ^15 4. Given cot 6 = 2, find sec and cosec $, Ans. sec ^= ±~-, cosec 6=± ^. 16 PLANE TllIGONOMETRY. 5. Given versin = —, find tan 0. ■ o Atis. tan^=±-^-. 6. If cos 6 = tan ^, find sin 6. Ans. sin 6 = — 7. If tan ^ + 3 cot ^ = 4, find cos e. Alls, cos y = ± -—• or ± 2 VlO 8. Prove cot^ B - cos^ ^ = cot^ Q cos^ 0. cot - sec 1 ^ , . ^ 9. If —;, = -r-r-> find sm B. cot w 16 3 Ans. sin ^= ± — . 5 , - _, sec ^ cot ^ - cosec B tan B ^ - 10. Prove - - - . — ;, = sec Q cosec B. cos ^ - sm 11. If tan ^ + cot ^ = 2, find the value of sin B + cos B. 12. Determine sin a from the equation ;. 9 sin^ a - 4 tan^ a = — . ' 2 ^ws. -iin a = ± — or ± ——. o 2i 18. If sin 03 = m sin 2/> a-nd tan x = n tan y, shew that m^- 1 cos' 03 »: - ?l2-l 1 i T» .1 . (cosec 03 + sec oj)'^ ,. ,„ 14. Prove that — , ~ = (sm x + cos a;)'. sec^ 03 + cosec'* x ^ ' EXAMPLES. 17 15. If 2 tan a = COS a, find ain a. Ans. sin o= J^— 1. 16. If sin^-cose=-^r^, findtan^. Ans. tan 6^= J^. 17. Given sec ^- tan 6=—, find sin Oi A'ns. sin 6=—. 5 22. To radius unity, it will be seen that the resirlta of Articles 18 and 19, deduced from the "line definitions," agree exactly with those previously obtained from the " ratio defini- tions." If, however, the radius is any other number than unity, the sine, tangent, &c., of any angle have not a fixed value as they have in the^ "ratio definitions" or in the "line defini- tions" when the radius is unity, but vary with the radiuS employed. Thus, if AB in the figure of Art. 17 be represented by r, then JBO, ED, AE, &c., become the sine, tangent, secant, / sin 30°- BB Tb^Ybb^I'^'''^^^' ^"*^2. v/5 cos 30° = VI - sin^ 30" - -^ = sin 60°. tan 30° = sin 30° J. 2 cos 30° ^3 ^3 = =cot 60°, cot 30°=. sec 30° = tan 30 1 ^3 = tan 60°. q^o = -7= = cosec 60°. cos 30 ^3 1 cosec 30° = - "-57r, = 2 = sec 60°. sm 30 TRIGONOMETRICAL FUNCTIONS OF 18°, 3G°, ETC. 19 24. To find the sine, cosiiie, , the angle ^Ci^= 54°. Since AB. BC = AC\ we have ■(••:■■ ;;^ 20 PLANE TRTfJONOMETRY. AB{AB-AG) = AG\ AC AC °''' -A^^'Air^' By solving this quadratic, wo oLtain AC -1±./^ AC Since -~ is a positive quantity, the upper sign must be taken j therefore, AB 2 ■is fi. AB AC ^ J6-1 AB 2 ' i ^ . ^., BE 2 BE AC Now, sin 18=^ = 2-1:^=2-18 = -^^ = cos72°. cos 18°= 7(l-sinM8°) ^^('0^^^)= sin 72'. Again, cos 36 = AC 2 AG 2 AC ^^ 1 J^ + l ^ V^+1 x/5-1 (n/5-1)(n/5 + 1) 4 • = sin 54°. sin 36°= 7(l-cos2 36°) < ' ' '; ' ;, , =J 7(10-2^5) = cos 54°. ;■•,;- The other trigonometrical functions of these angles can now be found. The student should verify the following : 1. cot 18°= x/5 J2 + Jq. =3-07768. = tan 72°. 2. sec 72°= ^5 + 1. =3-23607. =coseo 18°. THE FUNCTIONS OF HALF AN ANGLE. 3. tan 54° 4. cosec 36' 5. tan 36° 6. tan 18° v/(2 + ^ JW). = 1-37638. = 1-70130. = -72654. =^ -32492. = cot 36°. = sec 54°. = cot 54°. = cot 72°. 26. To express the functions of half an angle in terms of those of the whole angle, and vice versa. Let BAC be any angle in the first quadrant, represent it by A, and in AG take any point P, and with yl as a centre and AP as radius, describe a circle ; pro- duce BA to meet the circumfer- ence in D, join DP and draw PE perpendicular to AB, By Euc. III. 20, the angle BDP is half of the angle A ; then tan - = tan i?i>P 2i PE DE PE PE AD + AE AP + AE PE ~AP 71 l^lp sin A 1 + cos A (17) Or thus, by the "line definitions." Take JP as unity and with Z) as a centre and ^i> as radius, describe the arc AH and 22 PLANE TRIGONOMETRY. draw AF pcrpciiclicular to AB^ thon ^^=tan — -, FE — ^xn A, and yl^=cos A. From the similar triangles BAF^ DEP^ we have DE'.AD-.'.EP: AF, or 1 + cos -4 : 1 : : sin A : tan — , Z whence tan — = :; r, as before. 2 1 + cos A 27. The formula just proved enables us to find the trigono- metrical functions of IS'' and 75°, from having given those of 30". . Thus, let A = 30", then we have tan 15" = 1_ sin 30° 2 1 + cos 30" Jo = -— i-p = 2- ^3 = cot 75". 2+ n/3 Hence, by Art. 21, we easily find cot 15° = 2 + J3 = tan 75°. sec 15"= J2{ J^- 1) =, cosec 75°. cosec 15" = J2{ 73 + 1) = sec 75°. sin 15" = -^^5^^ = cos 75°. 2 V2 cos 15" = 4^ = sin 75°. 28. Squaring (17) we have THE FUNCTIONS OF H^F AN ANGLE. 23 tan- A 2 sin' A ~ (1 + cos A )'^ 1 - cos" A ~{l + cosAy 1 - cos A ~ 1 + cos A' Bee'^ A 2 1 - cos -4 1 + cos -4 2 1 + cos A 1 + cos ^ = 1 - 1 - cos A 2 From (18) we easily find (18) .Al + cosA ,- -X therefore cos^ — = x •. (ly; sin'^ -^ = 1 - cos' -T (20) l-tan4 cos^= " ■ (21) 1+tan-— . » ' . ^^ VI' / ! I ■ • ■• ■ ; , ■ ' ' , •■ Multiplying (19) by (20), and extracting the square root, we find J . . .. ,. sin -4 = 2 sin -cos —. . ,; (22) 2'^- 2 Subtracting (20) from (19) we have A A . cos ,4 = cos' — - sin'—. (23) 2 2 ^ * 24) PLANE TRIGONOMETRY.' If wo write 2 A for A, \^iich we are evidently at liberty to do, then (22) and (23) become siu 2A = 2 bin A cos A. (24) and cob 2A = cos'^ A - sim'^ A = 1 - 2 sin"'* ^ = 2cob^^l-l. (25) Also, tan 2A ^ ^ ,- cos 2 A 2 sin A cos A y^ ^ cos'^ A - sin'^ A 2 sin A cos A siii^ A cos'^ A 2 tan A 1-tanM* (26) 29. The formiilsB of the last three Articles will be found extremely useful hereafter, and the student is requested to pay particular attention to them. They will be again deduced by a more general process iii a subsequent chapter. Examples. • 1. Find A from the equation tan'^ ^ - 4 tan A + 1 = 0. Adding 3 to both members we have tan" J: - 4 tan ^ + 4 = 3 hence tan A -2= ± J^ tan ^ = 2+^ or 2-^ = tan 75° or tun 15'^ therefore A - 75° or 15°. EXAMPLES. 25 2. Given cosec ^ - sin d =^ ^""q » ^^^ ^^^ ^' By (8) and (17), the given equation becomes 1 . ^ sin - sin = am ^ 1 + cos sin'^ ^ 1 + cos ^ 1 - sin^ 2 /}_ or 1 - cos- ^ ^ COS^ ^ = , ;,- = 1 - cos ^ 1 + COS » COS- ^ + cos ^ = 1 COS^ ^ + COS ^ + i = -r * 4 cos^ + J=+-^ + A-1 hence cos = 3. Given tan + cot = 2 sec 0, find ^. By (8), (13) and (14), the given equation becomes sin cos 2 -;; + cos sin cos sin' 6 + cos- ^ = 2 sin d or 1 = 2 sin 0, hence sin ^ = J = sin 30", therefore ^ = 30'. 0' 4. Prove that cosec 6 = ^ (cot -- + tan — ). 1 cos2-+sin2-- cosec = -r—^ = 7^., ^y (10) and (22) sm ^ ^ . ' -^ ^ ' ^ ' 2 sin- cos— , * ■ s=J(cot-+tan--). 26 PLANE TrvIGONOMETRY. ^ _ , ^ 1 - cos 6 sec ^ - 1 0. Prove that tan -- = — : — —- = — ; -r—. 2 sin tan (^ 9 6 6. Prove that cot 0=^ (cot — - tan — ). 2, A 7. Given sec x tan x = 2 ,^3, find x. Ans. 03 = 60°. 8. Given 6 cof^ 03-4 cos^ a;= 1, find x. Ans. X = 60°. 9. If sin A Hec B= /— , and cos A cosec 5 = ~-, find '^ ^""^ ^' Ans. A = 60°, B = 45°. 10. If cos (2^+5)=-^, and sin {2>A-B)=]-, find ^^^^^' Ans. ^ = 12°, j5 = 6°. 11. If tan (3i + 2^) = 2 + J%, and tan (5^ - ^) = 2 - ^'3^ find A and 5. Ans. A = %i^\ 5 = 25xV- 12. Find the tangent of 1\\ Ans. J6- J3+ j2-% 13. Find the tangent of 37^°. Ans. JQ+ J'S- V2-2. 14. Given 2 cosec ^ - cot ^= ^, find 9. 15. Prove that sin A = Ans. ^=60°. 1 cot — ■ - cot A 40 10. If tan (9= ^2"- 1, find cos 29. Ans. cos 29 \/2 EXAMPLES. 27 17 If cos ^ = —^-^, shew that tan -= /^t tan -■ 1,. itcos(; ^_^cos' 2 n/ a-b 2. 18. Prove that tan ^ - tan - = tan - sec e/. 19. Prove that 1 + tan ^ tan - = sec A. 20. If sec + cosec B^m, and sec 6 - cosec ^ = ti, find tan 6. m + n m-n Atis. tan 6 21. Given sin ^ cos ^ = -—-=, find ^. Am. ^ = 22|°. 22. In the fig. of Art. 26, join PK, and prove geometrically that A 1 - cos A tan — = — r — J-, •• '^^ 2 sm il sin ^ = 2 sm — COS — , ^ ^ ' ,, • 1 +COS ^ = 2 cos^— , - '■ A ' • ■■'■■ 1 -cos ^ = 2 sin'^ ^. 23. Prove that -■■• ;. . :■ -■;^''i':'n.:--',- , ^, sin ^ (1 + tan A) + cos ^ (1 + cot A) = cosec -4 + sec A. 24. Prove that tan ^ + sec ^ - 1 = tan 6 + sec 0, tan t; - sec (; + i +.n.Ti fl - sin B and tan ^ - sec ^ + 1 tan ^ - sin ^ sec sin^ ^ 1+cos^' 28 PLANE TltlGONOMETKY. CHAPTER III. APPLICATION OF THE PRECEDING PORMULiE. 30. The formula) of Article 12 are directly available for the solution of right-angled triangles. Thus, if ABC he a. right- angled triangle, G being the right angle, and a, 6, c the sides opposite the angles A, B, C respec- ^B tively, we have according to the definitions oLthe sine, &c., ' A ^ A ^ 4. A ^ sm A = —: cos A—-; tan A = —. c c b A ~- ^ Each of these equations expresses a relation between three parts — an angle and two sides — hence, when any two of these parts are given, the other can be found. In order, however, to solve a triangle trigonometrically, it will be necessary to know the values of the sine, cosine, &c., of any given angle. The trigonometrical tables supply these values for every minute^ and sometimes for every ten seconds, from 0° to 90°. At present, however, we will use only those trigono- metrical ratios which we have already computed in the preced- ing chapter. The construction and use of the tables will be explained in a subsequent part of this work. ,. 31. The solution of right-angled triangles presents the fol- lowing cases : Case I. — Given the hypotlienuso and one ant/le, as c and A. To find a, we have sin A = ~'f whence a = c sin A. . (27) c SOLUTION OF IIIGHT-ANGLED TEIANGLES. 29 To find J, wo liave . . if h I cos^=— , whence 6 = c cos -4. (28) c To find B, we have B = 90°- A. Examjde. — Given c = 24, and the angle A = 60°, to find the other parts. By (27) we have a = 24xsin 60° = 24x-^ = 12V5. By (28) we have • 6 = 24cos60° = 24x^ = 12. - ' " and 5 = 90° -60° = 30°. If the angle B had been given instead of ^, we might first find A, thus, A = 90° - B, and then proceed as above. Case II. — Given the hypothenuse and one aide; c and a. To find A, we have sin A=—. (29) To find B, we have B = 90'' -A. To find b, we have cos A = — , whence 6 = c cos A, (30) or b= Jc'^ -a?= J{c + a) (c- a). (31 ^ Example. — Given c = 10 and a = 5 J% find other parts. By (29) sin^=^=JL = sin45°, • -^y V2 hence ^ = 45° and i? = 45°. By(30) 6 = 10cos45° = 10x-V-5 ^2. . . \/2 30 PLANE TRIGONOMETRY. Case III. — Given an angle and its adjacent side ; A and h. To find a, we have tan -4 = -r-j whence a = h tan A. (32) To find c, we have h h COS A = —. whence c = = h sec A. (33) c cos A Example. — Given -4 = 75° and 6 = 15, find a and c. By (32) o= 15 tan 75° = 15(2+ V3) = 55-98. By (33) * c = 15sec75' n/3 = 15.— ^-=15(76+ ^) = 57-648. *./ 51 — 1 Caie IV. — Given.an angle and its opposite side ; A and a. We may first find B^ which reduces this to Case III., or ^« may proceed thus : To find 6, we have tan -4 = — , whence h = : = a cot A. (34) 6 tan A ^ ' To find c, we have sin ^ = — , whence c = -: — - = a cosec A, (35) c sin A -^ ' Or, using h just found, we have , cos A = — , whence c = h sec A. (3G) C Case Y. — Given the two sides ; a and b. To find A and ^, we have tan ^ = cot 5 = ^. (37) 6 ^ SOLUTION OF RIGHT-ANGLED TRIANGLES. To find c, we have 31 sin A =i -, whence c = -: — r = acosec.4. (38) 32. The formula of the last Article may be illustrated geometrically by the " line definitions," as follows : With A as a, centre and a radius unity, describe an arc DG, draw DU and GF perpen- dicular to AG ; then, Art. 17, i>.^ = sin A, AF = cos A, GF^ tan A, and AF = ^qg A. From the similar right-angled trian- ^- gles, AED, AC 23, we have EG that is, whence AD :AB :: DE : BG 1 : c : : sin J[ : a, a = c sin A, which is (27), and a sin A = —t c which is (29). Again that is, whence and AD:AB::AE:AG 1 : c : : cos J. : 6, >, 6 = c cos ^, which is (28), coSil = — , which is (36). Again, from the similar right-angled triangles, AGF, AGB, we have AG:AC :: GF : GB that is, 1:6 :: tan^ : a, . • . whence ' ^^ "I a = 6 tan A, which is (32), and tan A = a which is (37> 82 PLANE TRIGONOJIETRY. Examples. 1. The radius of a circle is 10 feet ; find the side of the inscribed regular pentagon. Ans. 5( J5 - 1) ft. 2. The base of an isosceles triangle is 20 yards, and the vertical angle is 108° ; find the sides. Atis. 10(^5-1) yds. 3. The side of a regular octagon is 14 feet ; find the radius of the inscribed circle. Ans. 7(^+1) ft. 4. If the earth be a sphere whose diameter is 7912 miles, find the radius of the 45th parallel of latitude. Ans. 1978 ^2 miles. 5. The angular elevation of a tower at a place A due south of it is 45°, and at a place B due west of A, and at a distance a yards from it, the elevation is 30° ; find the height of the tower and the distance from £ to the foot of the tower. Ans. Height =^-^ ^/^ yds. ; distance from B = ^j6 yds. 6. A person observes the angle of elevation of a mountain to be 27°, and approaching one mile nearer, the elevation is 54°; find its height and distance from the first station. Ans. Height = J( ^ + 1) miles. Distance from first station = J(4 + ^/lO - 2 ^5^) miles. 7. A tower 60 feet high casts a shadow 20 ^^feet in length ; find the altitude of the sun above the horizon. Ans. 60°. 5 • ' , EXAMPLES. ' * 38 ^' 8. In a triangle ABC, sin A =* —5--, sin B = -— and the isi sides opposite to these angles are ^X5 ^^^ ^i > ^^^ the other side. 9. A church spire subtends an angle of 60° at a certain dis- tance from its base, and 80 feet farther off it subtends an angle of 45°; find its height, allowing 5 feet for the observer's height. Ans. 125 + 40 J^ feet. 10. The angular elevation of a steeple standing on a hori- zontal plane is observed, and at a point 10 yards nearer to it the angular elevation is found to be the complement of the former. On advancing 3 yards nearer, the elevation is double of the first. Find the height of the steeple, the distance of the first point from its foot, and the tangent of the angle of eleva- tion at the third point. ' Atis. Height = 12 yds. ; distance = 18 yds. ; tan of the angle of elevation at third point = 2 '4. 11. Wishing to know the distance from a given point A, to an inaccessible point B, on the opposite bank of a river, a base lijie AG, 84 rods, is measured along the bank of the river at right-angles to ABj and from its extrem- ity C, equal distances CD, ' CE, 6 rods, are measured ^^ on the lines GB, GA. Finally, DE is measured and found to be 4 rods. Find the distance between A and B, and the tangent of the angle AGB. 3 '34, PLANE TRIGONOMETRY. Bisect DE in F, join OF; then, in the right-angled tri- angle CDF, sin DCF^ DF_2^_\_. CD~ Q~V henco 2 /o~ COB BCF^ Jl-i^ 1±» tan J)CF= sin DCF _ 1 3 V? cosi>Ci^~3'2^2"*^ * The angle DGB is double of the angle DCF, therefore, by (26), 2 tan DCF 2 4 ^5 taniC5 = l-tan2 2>(7i^ -^ Hence, by (32), ii5 = ^(7 tan AGB = 84xi^=48^rods. 12. If, in the figure of the la^t problem, CB = 25 yards, and CD = 0F=5 yards, and DE=i yards, find AB and the tangent of i5. — 17 — Atis. AB = 4: 721 yards ; tan 5 = -- ^21. 13. From the summit of a tower whose height is 108 feet, the angles of depression of the top and bottom of another tower standing on the same horizontal plane, are observed to be 30° and 60° respectively ; find the height of the second tower, and the distance between their summits. Ans. Height = 72 ft. ; distance between summits = 72 ft. 14. One side of a right-angled triangle is 60 rods, and the EXAMPLES. * V ^^ 35 cosine of the adjacent angle equals the cotangent of the opposite angle ; find the hypothenuse and the other side. . . Ans. Hypothenuse = 30 J2{ Jl+^ I); side = 30 j'^ij^^i). 15. The base of a right-angled triangle is 60 yards, and the tangent of the opposite angle equals three times the cosine of the adjacent angle j find the perpendicular. Am. 15 ,^2 yds. 16. In a right-angled triangle whose right angle is C, shew by (17) that tan — = ^— — = ; ' •^ "^ ^ 2 b +c a A Ic-h " ' 'v and by (20) that sin — = / -^-« , , , 17. In any right-angled triangle, given A and c, shew that the perpendicular from the right-angle on the hypothenuse is --sm 2A, .,. ., „:;, _ . ,: .-, ,, , ,, , j 18. The sides of a right-angled triangle are 3, 4 and 5, find the length of the perpendicular from the right angle on the hypothenuse. '"' ■' '■'• ;■-'■-,' '::'- ''-"■'■ ' ' ' Ans. 2.4. 19. A person in a line svith two towers, and at distances of 100 and 150 yards from them, observes that their apparent altitudes are the same. He then walks towards them a distance d of j^O yards, and finds that the angle of elevation of the nearer is just double of that of the former. Find the heights of the towers. ^*:.vJ>^.-;.; 20 _ Ana. The nearer, -— ^7 yds.; the other, 10 7^ yds. • 20. A tower of unknown height stands on a horizontal plane, and at a distance ot 80 yards from the foot of the tower 36 . PLANE TRIGONOMETRY. • a mark which is known to be 50 feet high, has an angular elevation just half of that of the tower. Find the height of tower, the observer's eye being on a level with the plane. Ans. 104 '537 feet. 21. The sides of a right-angled triangle are in arithmetical progression, and the area is 108 ; find the sides. Ans. 9, 12 and 15. 22. Standing straight in front of a house, opposite one corner, I find that its length subtends an angle whose sine is 2 3 — J'5, while its height subtends an angle whose tangent is -^i the height of the house is 51 feet ; find its length. Am. 170 ft. 23. When the altitude of the sun is 54°, the shadow cast by a church spire is 150 feet. Find the height of the spire. Ans. 68.819 yds. 24. The base of a right-angled triangle is one-third of the hypothenuse, and the perpendicular from the right angle on the hypothenuse is 6 ^2^; find the sides. Am. 9, 18 ^2" and 27. 25. A May-pole being broken off by the wind, its top struck the ground at an angle of 36°, and at a distance of 30 feet from the foot of the pole ; find its whole height. Am. 58-878 ft. EXTENSION OF DEFINITIONS. 37 CHAPTER IV. ^ EXTENSION OP THE DEFINITIONS OF THE TRIGONOMETRICAL FUNCTIONS — CHANGES IN THEIR SIGN AND MAGNITUDE. 33. In chapter II., the sine, tangent, &c., have been defined for angles in the first quadrant only, or for those less than 90°. We will now extend these definitions so as to include all angles. Draw the lines PQ, MN, intersecting at right angles in A ; let AB \i<& different positions of the revolving line in the first, second, third and fourth quadrants, and from any point oi AB as B draw BG perpendicular to PQ. In the first and second quadrants BC lies above, and lyr K B B C C -P V in the third and fourth quadrants, below PQ ; while in the first and fourth quadrants, AG lies to the right, and in second and third quadrants, to the left of Jfif. These opposite directions are indicated by the signs + and - ; thus, lines which lie above PQ are regarded as positive, consequently those which lie below it must be regarded as negative. Again, lines which lie to the right of MJ^ are regarded as positive, consequently those which lie to the left must be regarded as negative. Hence BG is posi- tive in the first and second qiiadrants, and negative in the third 38 PLANE TRIOONOMKTRY. and fourth, and AC in ponltive in tho first and fourth quadrants, and negative in the second and tliird. Tho revolving line AB, in always regarded as positive, being measured olF in tho direction from A towards B; if, however, AB be produced backwards, tho produced part must be con- sidered iieyativc. The student will not fail to perceive tliat tho signs + and - , as thus employed, indicate, as in Art. 3, merely contrariety of direction. So it is in geography and astronomy — if north latitudes and east longitudes be considered positive, south latitudes and west longitudes must be considered nega- tive. The student must be careful, however, to observe that no absolutely fixed direction is here meant. Any line what- ever may be regarded as the initial one, ■ "'e object of the conven- tion being merely to indicate that lines measured in opposite directions are affected with ojjposite signs. Thus, in the obtuse -angled triangle ABC, the obtuse angle C is measured from BC as the initial line, and is an angle in the second quadrant ; while the angle B is measured from AB as the initial line, and being less than a right angle, is an angle in the first quadrant. 34. To trace the Variations in Sign and Magni- tude of the sine, cosine and tangent of an Angle, as it varies from 0° to 360°. First Quadrant. — Let a, 6, c represent the sides of the right-angled triangle ABC, opposite to the angles A, B, C respectively. VARIATIONS OF TRIOONOMETRICAL FUNCTIONS. 39 gin A = — , which is therefore positive, c When ^ = 0^a = 0, hence sin 0° = 0. As the angle increases, a in- creases, and therefore sin A in- creases in magnitude. When A = dO%a^c, hence sin 90° =1. cos A = , which is therefore positive. c cos 0° = 1. When hence As A increases, 6 decreases, and therefore cos A decreases in magnitude. When ^ = 90°, 6 = 0, hence cos 90° = 0. % + a tan A= — 7, which is therefore positive, + h When hence ^ = 0°, rt = and 6 = c, tan 0° = 0. As A increases, a increases and h decreases, and therefore tan A increases in maijnitude. When hence A = 90°, a = c and & = 0. tan 90° = X . Second Quadrant. sin A = + a which is therefore jjositiv^. 40 PLANE TRIGONOMETHY. As A increases, a decreases, and therefore sin A decreases in magnitude. When A = 180\a^0, hence sin 1 80^ = 0. COS A = , which is thore- B •*«: \ ~c^-T Xi foro negative. As A increases, h increases, and therefore cos A increases in magnitude. When ^ = 180°, 6 = c, hence aos 180° = = -1. c tan A + a J, which is therefore negative. As A increases, a decreases and b increases, therefore tan A decreases in magnitude When A = l80%a = and b = c, hence tan 180° = 0. Third Quadrant. sin A , which is tbere- c fore negative. As A increases, a increases, and therefore sin A increases in magnitude. rss / V When hence A = 270° (3 right angles), a = c, - c sin270^ 1. VARIATIONS OF TRIGONOMETRICAL FUNCTIONS. 41 COS A = , which is therefore negative. As A increases, b decreases, and therefore cos A decreases in magnitude. When hence il = 270^6 =0, cos 270° = 0. tan A = — - =+-=-, which is therefore positive. -00 -* As A increases, a increases and b decreases, and therefore tan A increases in magnitude il = 270°, a = cand6 = 0, tan270':z:oc. When hence Fourth Quadrant. bin ^ = , which is therefore negative, c As A increases, a decreases, and therefore sin A decreases in magnitude. When ^ = 360", « = 0, hence sin 360" = 0. a "^+B ( n N \ ■ ■ c\ -a \ B cos A , which is there- c fore positive. As A increases, b increases, and therefore cos A increases in magnitude. When . ;. < ^ = 360°, 6 = c, hence cos 360° = 1. 42 PLANE TRIGONOMETRY. -a tan A = —7-, which is therefore negative. As A increases, a decreases and h increases, therefore tan 4 decreases in magnitude. When A = 3G0°, a = and 6 = c, hence tan 360° = 0. 35. As the cosecant, secant and cotangent are the reciprocals of the sine, cosine and tangent respectively, they require no special examination. The variations of all the trigonometrical functions, both in sign and magnitude, are given in the following tabular form : 1st Quadrant. 2nd Quadrant. 3rd Quadrant. 4th Quadrant. sine 0° to 90° 90° to 180° 180° to 270° 270° to 360° positive. to 1 positive. 1 to negative. to -1 negative. -1 to cosine positive. 1 to negative. to -1 negative. -1 to positive. to 1 tangent positive. to oc negative, oc to positi\^e. to cc negative, oc to cosecant . . . positive, oc to 1 positive. 1 to oc negative, oc to -• 1 negative. -1 to oc secant positive. 1 to oc negative. oc to -1 negative. -1 to oc positive, oc to 1 cotangent. . positive, oc to negative. to oc positive. X to negative. to oc 36. From the above table it is seen that the values of the sine and cosine always lie between and ± 1, and those of the secant and cosecant between + 1 and + oc , and between - 1 and - oc ; while those of the tangent and cotangent lie between VARIATIONS OF TRIGONOMETRICAL FUNCTIONS. 43 and ± oc , and therefore may be of any magnitude whatever, positive or negative. It is also seen that each of the functions changes its sign when its value passes through zero or infinity ; hence we may write cos 90° = + or - 0, and tan 90° = + a or - oc , &c. 37. The preceding results of this chapter are easily obtained by means of the line definitions. In a circle of radius unity, draw two diameters, PQ and MN, at right angles to each other, and let PAB be an angle taken in each of the four quadrants. Draw the various lines according to the definitions of Art. 1 7, and let, as usual, the positions of the lines in the first quadrant be taken as the positive positions. Fic. 3. p M , : ■ N^-J \ o(- ^ "^ Tk ' V ! A \y P Fie. 2. \^ y ^ c ' \f. N „/^<; ^ ^ Fic.4.\ V^ B7> N p The sine BG is positive in the first and second quadrants, and negative in the third and fourth. The cosine ^C is positive in the first and fourth quadrants, and negative in the second and third. • 44 ' PLANE TRIGONOMETRY. The tangent PD is 'positive in the first and third quadrants, .and negative in the second and fourth. The cotangent MB is positive in the first and third quad- rants, and negative in the second and fourth. The secant AD, in the first and fourth quadrants, is obtained by producing the revolving radius AB forwards ; while in the second and third, it is found by producing AB backwards. It is therefore positive in the first and fourth quadrants, and Tiega- tive in the second and third. The cosecant AE, in the first and second quadrants, is obtained by producing AB forwards ; while in the third and fourth it is formed by producing AB backwards. It is there- fore positive in the first and second quadrants, and negative in the third and fourth. 38. With regard to magnitude, it is seen from the figures of the last Article, that as the angle PAB increases from 0° to 90°, the sine BC increases from to AM or 1 ; the cosine AC decreases from 1 to ; the tangent PD increases continually from 0, and when the secant AD coincides nearly with AM both the tangent and secant become exceedingly great, and infinitely great when AD actually coincides with AM, or when the angle PAB = 90°, AD being then parallel to PD. Hence the tangent varies from to « , and the secant from 1 to oc . Again, when the angle PAB begins at 0°, AJE is parallel to MB, or the cotangent and cosecant begin by being infinitely great, and then decrease continually as the angle increases, until at 90° the cotangent MB vanishes, and the cosecant becomes AM or I. In the same way we may proceed to find the magnitude of the various iunctions in the other quadrants ; but this we leave to the student. The results will be found to agree exactly with those obtained in Art. 35, ^ .. ; / i , - > . ■ » FUNCTIONS OF 180^— A. 45 From the figures of the last Article it is seen that the versed- sine GP, which equals 1 - ^C, or 1 - cosine, is always positive, and varies from to PQ or 2, in the first two quadrants, and from 2 to in the other two quidrants. The coversed-sine, which equals 1 - sine, is also always posi- tive, and varies from 1 at 0° to at 90°, then from at 90° to 1 at 180°. In the third and fourth quadrants it varies^ from 1 at 180° to 2 at 270°, and then from 2 to 1 at 360" or 0°. 39. To find the Trigonometrical Functions of i8o°-A. • Let PAB be any angle A ; produce PA to Q, and make the angle B'AQ = to the angle P^5; then Pil-B' is the sup- b' • B ,, plement of PAB or PAF .-180°-^. Take AB' = ^a AB, and draw B'C and BG perpendicular to PQ ', then B'G' = BG,&.ndiAG'^AG. C - 6 A + * C sin P^i?' + a -h = ain PAB. cos PAB' = — = - cos PAB. tan PAB' = ^ = - tan PAB. or, we may write these results thus : sin (180°-^)= Bin A. cos (180°-^)= - cosyl. tan (180° -i)= -tan^. The cosecant, secant and cotangent being the reciprocals of the sine, cosine and tangent respectively, we have VsfSl/- ■).■ PLANE TRIGONOMETRY. cosec (180°-^) = • sec (180°- J) = cot (180"-^) = 1 sin (180" -^1) sin^ 1 1 = cosec A. cos (180° -^1) -cos A 1 1 = - sec A, — - cot A, tan (180° - yl) -tan^ The same results are readily obtained from Figs. 1 and 2 of Art. 37, where it is seen that BG is the sine of PAB, and also of BAQ, which are supplementary angles. Also, cos PAB = AC = cos BAQ, but with the contrary sign. Hence, sin 70° = sin 110°; cos 160°= -cos 2(f j tan 170° = - tan 10°, &c. 40. To find the Trigonometrical Functions of 90° + A. In the figure, let the angle MAB = A, then the angle PAB = 90° + A. . Now, the radius being unity, sin cos tan cosec sec cot B D V / \ \ / \ ^ V ! A H 7 1 dO° + A)=+BO = +^i> = cos^. = -BD= -sin A. 90 +A) = )-— — -;= , — = -cot^. cos (90 +A) - sin A 90° + y1) = 90° + J) = 90° + ^) = 1 sin (90° + J) cos^ 1 1 cos(90° + yl) -sin^l 1 1 tan (90° + A) - cot A = sec A. - - cosec A. = - tan A. FUNCTIONS OF NEGATIVE ANGLES. 47 These results, as well as those of the preceding Article, will be deduced by a more general process in the next chapter. 41. To find the Trigonometrical Functions of a Negative Angle. In the last figure, let the negative angle PAK be denoted by -A. Draw KH perpendicular to PQ ; then, radius being unity, we have sin ( - ^) = HK, which is negative (Art. 33.) = - sin A. m cos {- A) = AH^ which is positive = cos A. . . sin (-^) -sin ^ tan {- A) = ) — -f = -— = - tan A. cos ( - ^) cos A COSeC (- A)^ — ; jr- = ; — - = - coscc A, sin ( - ^) - sm j1 sec(-^) = — — - = j = sec^. COS ( - ^) cos A 11 cot(-^)=-- — - — rr = — I — T-=-eot^. tan {-A) - tan A -1- . In the same way, let the student find the various functions of negative angles in the other quadrants. „ ;. 42. It is evident that any given angle has only one set of trigono- metrical functions ; but from the preceding Articles it is seen that, for any given function, there are more than one corresponding angle. We will now proceed to determine the groups of angles correspond- ing to any assigned value of each of the trigonometrical functions. To find a general expression for all Angles which have the same sine. In a circle of radius unity, let PAB be an angle whose sine BC is given. Make the angle jB'^g=to the angle PAB, and draw 48 PLANE TRIGONOMETRY. DC perpendicular to PQ ; then B'C'-BC, and therefore the angles PAB, PAB' have the same sine. Moreover, if to each of these angles wo add 3G0°, 720°, &c. , the positions of the revolving lines AB, AB' will be the same. Hence, if we represent the circular measure of the angle PAB by a, and that of two right angles or 180°, by tt, the positive angles which have the same sine are, a, tt - a, and these increased by 2M7r where w is any integer posi- tive or negative, including zero. For negative angles, the revolving line, in passing from the posi- tion AP to AB', describes an angle whose magnitude is - (Tr + a), the negative sign being prefixed to shew the (direction of revolution. When it attains the position AB, the angle described will be ~{2T-a). Hence the negative angles which have the same sine are, -(tt + o), -(27r-a), and these increased by 2n7r. Therefore the positive angles are 2mr + a and 2mr + tt - a, or, 2mr + a and (2n + l)7r - a. The negative angles are -2n7r-(7r+a) and - 2n7r - (27r - a), or -(2n + l)ir-a and -2(n + l)Tr+a. Now, observe that when a is positive the multiple of tt is even, and either positive or negative ; and when a is negative, the mul- tiple of IT is odd, and either positive or negative. Hence all these angles are included in the general expression, mr + {-lf'a, since. ( -1)"^=+! or -1, according as n is even or odd. Therefore sin a = sin (mtt +(-!)** a). (39) 43. To find a general expression for all Angles which have the same cosine. v Let PAB be the angle whose cosine AC h equal to the given one, and let a be its circular measure as before. Make the angle PAB^ ANGLES IIAVlxXG THE SAME TANGENT. 49 equal to PAB; then, since the cosine is positive in the first and fourth quadrants, AC, the given cosine, is the cosine of all angles corresponding to the positions AB, AB' of the revolving line. Therefore the positive angles which have the same cosine are, a, 27r~a, and these increased by 2mr. The negative angles which have the same cosine are, -a, - (27r - a), and these increased by 2nir. The posit *;^'C angles are therefore • 27i7r + a and 2/t7r + 27r-a, or 2mr + a and 2{n + l)ir-a. The negative angles are -2mf-a and -2mr-{2Tr-a), or -2mr-a and -2(;i + l)7r + a, which are all included in the general expression, 2mr zfc a. Therefore cos a=co3 (2w7r± a). (40) 44. To find a general expression for all Angles which have the same tangent. Let PAB be the angle whose tangent PD is equal to the given tangent. The positive angles which have the same tangent are evidently PAB and PAB' , or a and ir + a, and these increased by 2mr. The negative angles are manifestly -{tt-o) and - (27r - a), and these increased by 2mr. Therefore we have for positive angles, 2mr + a and 2n7r + Tr + a, or 2mr+a and {2n + l)7r + a, . And for negative angles, - 2/i7r - (tt - a) and -2)^- (2Tr- a), or -(2/i + l)7r + a and i . -2{n + l)rr + a, 50 i'lanp: tuigonometuy. which are all inclutled in the gonorul expression, mr + a. Therefore tan a=tan (nTr + a). (41) 45. From the last three results we evidently have cosec a = cosec (mr + (-!/' a). (42) sec a= sec (2w7r±a). (43) cot a= cot (n-TT+a.) (44) In deducing these expressions, the least positive angle which has the given value forthe trigonometrical functions, has been employed ; but they are equally true for any angle which has the given value for its sine, cosine, etc. , as the case may be. Examples. 1. Shew that sin A, when determined from cotj Ay has two values, equal in magnitude but opposite in sign. From (10) we have sin ^ = ± s/l - cos'^ A. This is also seen from the figure of Art. 43, where, ii AC be the given cosine, BC = + sin it, and B'C= - sin A. 2. Given tan 0= J3 , shew that sin $ will have two values. From Art. 21 we have sin 6 = tan 6 J^ ± J{l+tan'&) - 2 = sin 60° or sin 240°, since the tangent is positive in the Jirst and third quadrants. 3 3. The cosine of an angle =» - -^, and the tangent 1 J, in what quadrant is the angle ] '• -4 ns. The third. EXAMPLES. 51 4. Given sec 0= - 2, find the general value of 0. 2 Here ^= 120° = .ttt ; hence we have from (43) o $=(2n7r±^-) or (Gn±2)~. 5. Given tan 56= V^ , fiiid the general value of 0. Here we find 5^ = 60° or - , and the general expression for o all angles which have the same tangent is mr + a, where a is the circular measure. • 1 TT IT Hence, the general value = -- (nrr + -) = (3w + l)^-^- 6. Find the sines and cosines of 300° and 162°. • /J A71S. sin 300" = - ----, cos 300° = J, sin 162°== ^ , cos 162°= -Jn/10 + 2^5-. 7. Find the trigonometrical functions of 1098°. ',c^' Ans. The same as those of 18°. • A- 8. The sine and tangent of an angle are both negative, and the tangent = 2 sine; find a general expression for all angles having this property. '■' tt Ans. {6n - 1)—. o 9. Express in terms of 6 all the angles whose sine is - sin 0. Ans. nTT + { - ly+'^e. 10. Find the values of which satisfy each of the following equations: - . - -.- . (1) cos 6= - 1. . , . J .,j..^,, Am. {2n+ l)-n: : (2) ta.nO== -1. Ana. mr- — 4- 52 PLANE TlUaONOMKTltY. (3) H[n'e = l. A7i3. (Gn.±])Z-. (4) soc ^ = 2 tan (9. Ans. (Qn + ( - lf)''r. (5) 2 cosec - cot ^=- J3. Ans. (Gn ± 1)-. O (C) tan ^ + cot ^= 2 soc ^. yfvis. (Gn + ( - 1)")'^-. 11. Trace the variations in tlie sign and magnitude of the tangent and secant, as the angle varies from 90° to 270°, by means of the line definitions. 12. Prove that (sec - + tan %) (cosec J + t^n '^ ) = 5. o o 13. Find the general value of in sin + cosec ^=2. Ans. ^=:(4?i + 1)--. 14. Shew that tan 52|° = ( ^3" + >J~i) ( v/2 - 1). 15. Shew that sec 2^ - tan 2^ = -?-"**"^ 1 + tan 0' 9 16. Shew that tan Q ■■=^ - cot ,- - tan - - 2 2 17. Shmv that tan ^. = ''" ^^^ ""^ ^ 18. Shew that sin ^ = 2 I + cos 2yl 1 + cos A 2 1 , e~ tan ^ + cot — tan -- +cot 6 19. Shew that cos ^ = sin ^ cot 1. 20. Find all the values of $ which satisfy 1 - cos ^ = 2 sin^ $. Am. e=^{{2n+\)±\}~. o EXAMPLES. 53 21. Trace the variations in sign and magiiitudo of the secant, coflocant and cotangent, in the third and fourth quaih'ants, by the lino definitions. 22. Shew that cos 1 1 J^ = J ^2+ V2T75' 23. Shew that sin — and cos -v-, when determined in terms of COS yl, will each have two values, but in terms of sin A, four values. 1- v/T 24. Shew that cos 105" = ^~-. 2^/2 25. Find tan 1G5° and sin 1G5°. 26. If tan 6= J^ + 1, find cos 26, and the general value of e. Ana. cos 26— r=:, 6 = mr± . J2 8 27. Find the general values of the limits between which 6 lies when sin^ 6 is greater than cos'' 6. IT 3 Am. Between 2mr + -- and 2mr + -' tt : 4 4 5 7 and between 2mr + —ir and Inir + — tt. 4 4 28. Prove that sec* 6 + tan* ^ = 1 + 2 sec" 6 tan^ 6. 54 PLANE TKIGONOMETRY, CHAPTER Y. TRIGONOMETRICAL FUNCTIONS OP THE SUM AND DIFFERENCE OF TWO ANGLES, AND OF MULTIPLES AND SUBMULTIPLES OP ANGLES. 45. To find sin (A + B) in terms of the sines and cosines of A and B. Let the angle BAG = A and the angle QAD = B^ then the s^rvglQ BAD =^ A + B. In AD^ one of the bound- ing lines of the compound angle {A + B), take any point .P, and draw FM, PQ perpen- dicular to AB and AC respec- tively, and from Q draw QK perpendicular to PM; then the angle QPK = the angle KQA = A. 8in(yl + ^) = — MK+PK M V AP ^QN PK "AP^AP jig' AP ^ P(i'AP = sin A cos J5 + cos -4 sin B. (45) FUNCTIONS OF {A-\-B). 55 46. To find cos (A + B) in terms of the sines and cosines of A and B. Employing the figure of the last Article, we have AM cos {A ■{■ B) = AP _ AN-NM * AP AN QK "AP'AP ANAQ QK QP ~ AQ'TP'qP'AP — cos A cos J5 - sin -4 sin B. (46) 47. To find tan (A + B) in terms of the tangents of A and B. sin {A + B) * • tan (A-\-B)^ -)--. — ^f ^ ^ cos {A + B) sin A cos B + cos -4 sin ^ cos A cos B - sin A sin B' Dividing the numerator and denominator of this fractiqji by cos A cos By we have sin A sin B + , . „. cos A cos B tan {A + ^) = - sin A sin B COS ^ COS B tan ^ + tan B 1 - tan ^ tan B' (47) 56 PLANE TRIGONOMETRY. 48. To find sin (A - B) in terms of the sines and cosines of A and B. Let the angle DAC = A and CAD = B, then BAD = A -J3. In AB, one of the bounding lines of tho compound an!^'l(> (A - B), take any point P, and draw PM perpendicular to AB, PQ perpendicular to AC, (^iV perpendicular to AB, and PK perpendicular to QN; then the angle PQK being the complement of AQN^, is equal to BAG or A. Bm{A-B) = ~ QN-QK AP ^AQAP Pq AP = sin ^ cos i? - cos A sin J5. (48) 49. To find cos (A - B) in terms of the sines and cosines of A and B. From the last figure we have cos (^-.5) = ^'^ AP _ AN^ NM ~AP AN PK ^AP'^AP ^AQAP '^ PQ'AP = cos A cos B + sin A sin B. (49) FUNCTIONS OF {A-{-B) AND (A—B). 57 50. To find tan (A-B) in terms of the tangents of A and B. tan {A-B) = )- — j/ cos (^l - JJ) sin A cos B - sin B cos A cos A cos li + sin A sin B' Dividing numerator and denominator by cos A cos B, we have 4. /A n\ tan ^ - tan B tan IA-B) = - — -— j. (50; ^ "^ 1 + tan A tan B ^ ' 51. Formul.'B (46) (48) and (49) can be easily deduced from (45), as follows : In (45) write - B for B and we have sin {A - B') = sin A cos ( - J5) + cos ^ sin ( - B) = sin A cos 5 -cos A sin ^5. (Art. 41.) The complement of {A + B) is 90" - (^ + i5), and since the cosine of an angle is the sine of the complement (Arts. 1 2 and 19), we have cos (^ + i?) = sin {90" - (^ + ^)} = sin {(90"'-^)-i?} - sin (90° - ^ ) cos ^ - cos (90° - ^) sin i?, ■■ cos A cos B — sin A sin B, If we write - B for B in the last formula, we have cos {A-B)^ cos A cos ( - ^) - sin ^ sin {- B) = cos ^ cos B + sin vl sin B. In a similar manner, from any one of the four formulas (45), (46), (48) and (49), the others may be deduced. 52 . The preceding f ormulte of this chapter have been established only when A-\-B is less than 90^. We will now proceed to shew that 58 PLANE TRIGONOMETRY. they are true for all angles, whether positive or negative ; and from the results of the last Article it is evident that it will be sufficient to shew this in the case of (45) alone. I. Let A-\-B be > 90°, and let ^=90°- a; and JB=90°-2/, and hence a;=90''-^ and 2/=90°-B. Then ^+J5=180°-(x+2/), and therefore (ac+y) < 90'-. Hence the formula (45) is true for (x-\-y). Now, sin {A + B)=am { 180° - (x+y) } =sin (x+y) (Art. 39.) =sin X cos y-\-coB x sin y =sin (90° - A) cos (90° -B]+ cos (90° - A) Bin(90° - B) =co3 A sin B+sin A cos B. Hence the formulae are true when {A-\~B) > 90°. n. Agam, let a;=90°+^, and therefore ^= - (90° -cc). Then x+B=90°-^A+B and 8in(a;+-B)=8in {90°+(^+i?)} =co8 {A-{-B) (Art. 40.) =co8 A cos JB— sin -4 sin B =cos { - (90°- a) } cos A'- sin { - (90° - «) } sin B =cos (90°-a;) cos B+sin (90° -x) sin B sin X cos jB+cos x sin B. Hence the formulae are true for 90°+ -^4 and B In the same manner it may be shewn that they are true for A and 90°4--B ; hence they are true if 90°, or any multiple of 90°, be added to either or to both, that is, they are true for all angles, since by adding multiples of 90° we may increase the angles A and B to any magnitude whatever. In a similar manner we miy show that the formuliu are true for all negative angles. * 53. The preceding formulae furnish us at once with the trigonometrical functions of all compound angles. * For a gooiaetrical proof iu auy aasiguod case, see Appendij. FUNCTIONS OF (90°H B) AND {180<^-B). 59 In (45) and (46) let A = 90°, then sin (90° + B) = sin 90° cos B + cos 90° sin B = 1 X cos j5 + X sin i? = cos B. cos (90" + B) = cos 90° cos B - sin 90° sin B = X cos i? - 1 X sin ^ = - sin B. /^^o T.^ sin (90° + B) cos J5 ^ _ ton (90° + B) = y^-jr, — jf. = ^-p = - cot B. ^ ' cos (90 + B) - sm B cot(90° + i5) = - — ,,.^o , „. = — „= -tani?. ^ ' tan (JO + ^) - cot i? 1 1 sec (90° + 5) = — .^r^r^ — pT = — ; — jj = - cosec ij. ^ ' cos (90 ■\-B) - sm B 1 1 cosec (90° + 5) = . ,„, - — 5r = ^ = sec B. ^ ' sm (90 + B) cos ^ These results agree with those of Art. 40. 54. In (48) and (49) let A = 180°, then sin (180° -B) = sin 180° cos B - cos 180° sin B = X cos 5 - ( - 1) X sin B = sin 5. cos (180° -B) = cos 180° cos 5 + sin 180° sin B = - 1 X cos J5 + X sin B = -COS B. Hence tan (1 80° - 5) = ^ tan B, cot ( 1 80° - 5) = - cot B, . sec (180' -B)= - sec B, cosec (180° -B) = cosec B. These results agree with those of Art. 39. Here we may observe that if B bo less than 90', its supple- ment 180° - ^ is greater than 90° or obtuse, and tliat the sine and cosecant of an obtuse angle are positive, while the cosine, tangent, cotangent and secant are negative. GO PLANE TRIGONOMETRY. 55. In (45) and (4G) let A = 180°, then ft sin (180° + /?)=- sin B, cos (180° + i?) = ~ cos B, tan (180° + B) = tan B, cot (180° + /?) - cot B, nee (180° + B) = - sec B, cosec (180° + //) = - cosoc /;. 56. In (48) and (49) let A = 270°, then sin (270° -B)= - cos 7?, cos (270° -/?)-- sin B, tan (270° -B) = cot 7?, cot (270° - B) = tan B, sec (270° -B)= - cosec i?, cosec (270° -B)= - sec /?. 57. In (45) and (46) let A = 270°, then sin (270° + B)= - cos B, cos (270° + i?) = sin B, tan (270° + B)= - cot i?, cot (270° + i?) = - tan B, sec (270° + 7^) = cosec B, cosec (270° + 7?) = - sec 7?. 58. In (48) and (49) let A = 360°, then sin (360° - 7?) = - sin B, " cos (360° - 7?) = cos B, tan (360° - B)= - tan B, cot (360° -B)= - cot 5, sec (360° - 7?) = sec B, cosec (360° - 7?) = - cosec 7?, or, the functions of 360° - B are the same as those of - B. 59. In (45) and (46) let A = 360°, then sin (360° + B) = sin B, cos (360° + B) = cos B, or, the functions of an angle greater than 360° are the same as those of the excess above 360°. 60. General Formulae involving the Functions of two Angles. From (45), (46), (48) and (49) we have Hm{A + B) + 8m(A-B)= 2 sin ^ cos 5. (51) sin (A + B)- sin {A-B)= 2 cos A sin B. (52) cos (A + B) + cos (A- B)= 2 cos A cos B. (53) cos (A + B)-coa(A-^^ - 2 sin ^4 sin B. (54) GENERAL FORMULAE. Gl If Ave put A+Ji:= A' and A~B = B', we have 2A=A'^B' and 2B = A'-B', whence A = l{A' + B') and 5-i(^'-J5'). Making these substitutions in the above group, and omitting the accents, since A' and B' may be any angles whatever, we have sin ^ + sin ^ = 2 sin }, (A + B) cos I {A- B). (55) sin tI - sin 5 = 2 cos | (yl + B) sin | (^1 - B). (56) cos ^ + cos i? = 2 cos I (^ + J5) cos ^ (yl - i?). (57) cos ^ - cos i? = - 2 sin \{A^B) sin ^ (vl - ^). (58) These formulse are of very great importance, and the student sliould make himself quite familiar with them under this form. They are especially valuable in trigonometrical computations in transforming a sum or difference into a product, and vice versa. Each of them may be expressed as a theorem ; thus, " The sum of the sines of any two angles is equal to twice the sine of half the sum of the angles multiplied hy the cosine of half their dif- ference,^^ and similarly for the others. 6i. Divide (55) by (56), then we have by (13) and (14) ^n ^ + sin 5 2 sin }^ {A + B) cos \ (A-B) sin ^ - sin ^ ~ 2 cos | (-4 + B) sin ^ {A-B) = tan \{A + B) cot \ {A - B) • _ tan|(^ + ^) "tan^^-^)' ^ ^ In the same manner let the student verify the following : cos ^ - cos 5 , , , ^, , , , T,. • ,»r.s -. 5- = - tan i (^ + B) tan -h (A - B). (60) cos ^ + cos ^ ^ ^ ' -^ / \ / sin ^ + sin ^ , ., „, • .^,, ;; ^ = tanH^ + ^)- (61) COS -4 + cos B ^ ^ ' ^ ' n C2 PLANE TRIGONOMETIIV. sin -4 - sin J9 , , ^ -, ,-- = - cot ^ (.1 + B). (G2) cos ^ - cos i> '' ^ ' ^ ' sin ^ - sin i? cos ^ + cos i? sin vl + sin 5 = tan^(^-Z?). (63) „ = -cot H^ + i?). (64) cos -4 - cos ^ ^ ^ ' ^ ' 62. Divide (45), (46), (48) and (49) by cos A cos B, then wc have by (13) sin {A + B) sin ^ cos -B + cos A sin B cos ^ cos ^ ~ cos ^ cos i^ = tan ^ + tan B. (65) cos (A + B\ 1—77 = 1 - tan A tan B. (66) cos A cos B ^ ' sin {A - B) cos -4 cosi? = tan ^ - tan B, (67) f?liilL^_ = 1 + tan ^ tan 5. (68) cos -4 COS i} Tn a similar manner we find sin (A + B) „ , ,^^, -■ ■ \ . ',- = cot i? + cot A. (69) Bin il sin /j ^ ' COS (^ + i?) , „ , ,^^, ■ . ^, . ,; = cot il cot j5 ^ 1. (70) sinil HUiB ^ ' sin {A - B) . , . „=coti?-cotil. (71) sin il sin ^ ^ ' cos (A - B) , . T, , . \ . ; =cot il cot i? + 1. sin^ sm-B (72) . sin (^ + -B) , , _ ' . ,^_. -r— ^- ^ = 1 + cot ^ tan B, (73) sin A cos B f?li^ = cot^-ta„a (74) sin -4 cos JO GENEEAL FOUMULiE. 63 ^(±'^?1 = 1 - cot i tan B. ♦ (75) sin A cos i^ _^_^y_:U?I = cotJ + tan5. (76) sin A cos j6' 63. Divide (45) by (48) and (4G) by (49), then we Lave sin (A + B) sin A cos B + cos A sin i? sin {A - i^^ ~ sin ^ cos 7^ - cos A sin i? tan A + tan ^ tan A - tan i!^' cot B + cot vl cot A cot ^ - 1 ~ cot A cot i^ + r 64. In (47) and (50) let A = 45°, then „ ^, tan 45" ± tan i? tan (45° ± iJ) = , — r—Tf^ « ^ 1 + t^ii "^5 tan B 1 ± tan i? 1 + tan B' _cot7?±2 ~ corZTqpl* In (50) let i? = 45°, then tan ^ + 1* . 1 - cot A i + cot J.' 65. From (47) and (50) we have by (8) (77) (78) cot B - cot A' cos (A + -B) _ 1 - ta n A tan B .^^. ~coii{A-B)~l+isinAtiiiiB' ;: ; (80) (81) (82) , ^-. tan ^ — 1 ' /oo\ (84) / . T-.N cot B cot A=f-1 yQp,v cot {A±B)= ^^^ ^ . m G4 PLANE TRIGONOMETRY. The secant and cosecant of (A^ Ti) or (A — /?) are easily cleriveil from those of the cosine and sine respectively, by means of (8); thus we have sec (^4-J5)= . . , „v ^ ' cos {A-\-IJ) 1 cos A cos ii — sin A sin B sec A sec B 1-tan A tan B' (80) by multiplying numerator and denominator by sec A sec B. In a similar manner the others may bo derived. 66. The product of (45) and (48), and of (4G) and (49), are sin (A + B) sin {A - B) = sin" A cos- B - cos'' A sin' /?, cos (A + B) cos {A — B) - cos^ A cos'' B - sin'^ A sin^ B. But cos' ^ = 1 - sin' J, and cos" 7i = 1 - sin''' i>, therefore sin (^ + J5) sin (yl - B) = sin' yl - sin' B. (87) = cos' 7? -cos' J. (88) and cos (i + B) cos {A'B) = cos' yl - sin' JS. (89) = cos'^-sin'^. (90) 67. General Formulae involving the Functions of the Multiples and Submultiples of an Angle. In (45), (40) and (47), let jB = J, then sin 2 A = sin A cos A + cos A sin A = 2 sin ^ cos .4. (91) cos 2 A = cos A cos A - sin A sin A = cos'^-8in'^. (92) tan A + tan A tan 2 A = 1 - tan A tan J^ 2 tan .4 1 - tan' A' (93) GENERAL FORMULAE. 65 If, in the last three equations, we write A for 2 A, and there- fore U for A, which we are at liberty to do since A is any angle whatever, we have ■:■■■'■ Bin -4 = 2 sin ^ cos -. (9*) cosil = cos2 .^ ..sin^-. (95) A 2 tan - tan^ = ^. W 1 - ton' ^- From (10) and (92) we have .• : , cos- A + mii^ ^ = 1 -• cos- -4 - sin'^ ^ = cos 2il, the sum and difference of which are •• 2 cos^ yl = 1 + cos 2.4 .->. (97) 2 sin*'' ^ = 1 - cos 2i. (98) A ^' Writing A for 2 A and — for ^, these become ^cor'-^I+cobA, (99) 2 8in'''4 = l-cos^, (100) the quotient of which is ^ fe tan'i = ^^:^. -""^ (101) 2 1 + cos il Multiplying the numerator and denominator of the second member of (101) by 1 + cos A, we get ^A (1 -cos' .4) '^"^ 2"(l+cos^)» sin" A (1 + cos -4 ) ,2> 06 PLAXE TRIGONOMETUV. therefore tan. sin A (102) 2 1 +008.1' Multiplying tlio numerator and denominator of the second member of (102) by 1 - cos A, we get ( 1 - cos A ) sin A 1 - COS'* .4 (1 - cos A) sin A sin'^ A 1 - cos A sin A (102 bis) The formulae of this Article have already been deduced for angles less than 90°, in Art. 28, by a less general process, and the student will now see that they are true for angles of any magnitude. 68. To find expressions for sin mA and cos mA. From (51) and (53) we have sin mA-\-ain (m-2)A=2 sin (m— 1)A cos A and cos inA-{-cos {m—2)A=2 cos (m — 1)/1 cos A, hence sin mA=2 sin (?n. -1)A cos A - sin (m — 2)A and cos mA=2 cos {m — l)A cos A - cos (w- .-2)A I ,-2) A. I (103) If we make m successively 1, 2, 3, &c. , these give sin il=8in A sin 2A=2 sin A cos A sin 34=4 sin A cos^ 4 — sin A, &c., &c. and cos A=cos A cos 2il=co8'* A -sin^ A COB 3i4=4 cos^ A-3 cos A, &c., &c. : , i : u^J Other general formulae for sin mA and cos A will be given in a subsequent chapter. j , » ' (104) GENERAL FORMULA. 67 69. General Formulae involving the Functions of three Angles. Lot A, B,C bo any three anglos, then by (45) and (46) Hin(/l+5+C)=8in {(J-^B)-\-C\ =Hin {A ]B) cos C'+co8 {A^ /?) sin =8in A cos B cos C+cos A sin B cos C +C08 A cos J5 sin C-sin ^ sin B sin C. (105) cos{A+B+C)=coti {{Ai-B)+C\ =cos (^ + -6) cos (7 -sin (^ + B) sin (7 =co8 A cos B cos C7— sin A sin fi cos C - sin ^ cos B sin C - cos A sin -B sin C. (106) In the same way we may develop the sine and cosine of (A—B-^C), &c. If we divide (105) by (106) we get, after dividing the numerator and denominator by cos A cos B cos C, tan ^ +tan B^t an C - tan ^ tan ^tang^ tan U^-B+C)=^^^^ ^ ^.^^ B-tB.nAt&n (J- tan JS tan C ^ '^ If Ai-B+G=mr, then tan (A+B-^G)=0, and therefore from (107) wo obtain , . . . tan ^+tan B+tan C=tan A tan B tan C. (108) If A-\-B-\-C=(2nirl)-- , that is an odd multiple of --, then tan (^ f B+0)=oc , and therefore the denominator of (107) must be zero, hence tan A tan B+tan A tan C+tan B tan C=l. (109) If (109) be divided by tan A tan B tan C, we get cot A + cot JB+cot C=cot A cot B cot C. , (110) 70. Again, let A+B~\-0=mr=2n-, an wen multiple of -, then . " sin ( A-{-B+C)=Bm htt = ^ Bin (-^+B+C)=8in (mr- 2A)= - ( - 1)« sin 2A. _ ,^^^^ sin( ^-B+C)=sin (n7r-2B)=-(-l)" sin25. ; * sin ( ^+B - C)=8in (wtt - 20)= - ( - 1)" sin 20. 68 PLANE TRIGONOMETRY. The sum of this group is, by (55) and (58), 4 sin A sin B sin C=— (— 1)« (sin 2^+sm 2B+8in 20). (112) Again, we have cos( A-\-B-{-C)=coa nv =(-1)". \ cos {- A+B-\-C)=co& (nir-2^)=(-l)"co8 2J[. I cos ( A—B-{- C) =cos (nTT -2B}={- 1)" cos 2B. 1 co8( ^4-B-C)=co8 (nTT -2a)=(-l)" cos 20./ The sum of this group is, by (57), 4 cos A cos B cos C={ - 1)»* (cos 2^+co8 25+008 2C+1). (114) (113) TT ir sin( A-{-B+C)=Bm (2n+l)- ' (1116w) 71. If ^+J5+C=(2»+1) 2", an odd multiple of ^ , we have 8in(-^+iJ+C?)=8in{(2n+l)^-2^}=(-l)»»cos2J[. sin ( A-B+O^Hin {{2n-\--')'^-2B} ={-!)*' cos 2J8. sin ( Ai-B-G)=ain{{2n-\-l)~-2C] =(-1)" cos 2 J. ^ Subtracting the difference of the first two of this group from the sum of the last two, we find 4 sin ^ sin B sin = ( - 1)" (cos 2A-{-cos 25 +co8 20 - 1). (112 bis) Again, co8( .4+B+0)=cos (2n4-l)~ = 0. co8(-^+5+C)=cos ((2»+l)~-2^}=(-l)»» sin 2A. cos( ^-J5+0)=co8{(2/i + l)-|-2jB}=(-l)" sin 2B. co8( J[ + B-0)=co8{(2n+l)|-20}=(-ir sin 20. ^ The sum of which is 4 cos A cos B cos 0=( - 1)»» (sin 2^+ sin 2J9+sin 20). (114 bis) ■ (113 6t») OSNERAL FORMUUE. ¥» Bv combining the eauations of group. (lU) and (113), m»ny „te irp"rt"rre.uUs a.0 obUined. Thus, in (113), takmg ha difflnce between the sum of the first two and the sum of the second two, we find 4 cos A sin B sin C=( - !)« ( " cos 2^+cos 25+cob 2(7-1). (115) In a similar manner, from (lU) we obtain 4 sin A cos B cos C= - ( - 1)« ( " b^ 2^+sin 2B+sin 2C). (116) Again, from groups (111 Us) and (113 his) the following are easily obtained, which will serve as exercises for the student. 4sin^cosBcosC=(-l)M-co8 2^+cos2B+cos2C+l).^ 4co8^sin5cosC = (-l)«( cos 2^ -cos 25+ cos 2C+1) 4co8Acos5sinC=(-l)M cos2^+cos2B-cos2C-fl). ^ ^^^^^ 4cos^sin5sinC = (-l)M-8in2^+sin2B+sin2a) 4 sin ^ cos 5 sin C= ( - 1)« ( Bin 2^ - sin 254-8in 2C). 4 sin ^ sin 5 cos C= ( - 1)'» ( sin 2^+Bin 2B - sm 2(7). 72 The following six formxilffi are of great utility in com- puting the values of the trigonometrical functions, and in trans- forming a sum or a difference nto a product. cos A t'n A cot^+tan^= ^v^ + ^^^ CDs'* ^ + s in'' A ~ sin A cOii ^ 2 \_ sin 2^1 2 sin A COS -4 = 2 cosec 2 -4. (118) COS A sin il cot ^ - tan A = -: — r 7 sin A cos ^ 2 ( cos' A - sin'' ^) ~ 2 sin A cos -4 2 cos 1A ~ sin '2 A « 2 cot 2il. (119) 70 PLANE TK1G0N0METK\ '-,f .,i'» , . , 1 cos A •■'■'!■ ' '' COSec ^ + cot ^ = -: + -: . . , , . sm A am A - , .im 1 +C08 A ' , sin A 2 cos^* |- , by (99) and (94) I'H.' .,. .v.- )., 2 Sin — COS — = cot-. i . , (120) . , . 1 cos A cosec ^ - cot A = -7 sin A sin ^ ;. ,, ,.■ ,■ ,,;,. ■ _ 1 -COS A ■ -::■ . sin ^ A >W!V" • 2 sin2 - = — ^1 J, by (100) and (94) 2 sin — COS 2 2 From (81) we have = tan|-. (121) , /.Ro .V l+tan^ tan (45" + A) = ^ ' 1 - tan ^ tan (45 -A)=- , ^ ^ 1+tani' hence - tan (45- + ^) . taii (45" -A) = l±^^ ^ Ll^^ 1 - tan A 1 + tan A _ 2 (1+tan^^) 1 - tan^ A 2(cos^^+sin'J) cos^il-sinM ' ^y(^^^ 2 cos 2A = 2 sec 2^. ^ (122) F011MULJ5 OF VERIFICATION. 7l 4 tan A tan (45* + A)- tan (45° -A) = Yltan^A cot ^ - tan ^ i__, by (119) .., "2 cot 2^' ^^ ' - =2tan2il. G^S) Formulae of Verification. The following four formula are useful for testing the accu- racy of the trigonometrical tables : sin(36V^)-sin(36»-^) = 2cos36°sin^ = |(^4-l)sinX sin(72°-fl)-sin(72°-^) = 2cos72'sinl = KN/5-l)«i^^- The difference of which is sin (36» ^ A) - sin (72° + A) + sin (72° - A) - sin (3G° - A) = sin^. (Euler's formula.) \^^^) Substituting (90° - A) for A in (124) we obtain sin (54° + ^) - sin (18° 4- ^) + sin (54° - ^) - sin (18° - i) = cos^. (Legendre's formula.) \}^^) By (55) we find sin (30°-^) + sin (30° + ^) = cos A (126) Similarly, by (58) we find cos (30° -A)- cos (30° + ^) = sin A. (1 27) ' • Examples. 1 + sin A , ,, . A. 1. Prove that y^^^^ = K^ + *^" 2 ^ ' 72 PLANE TRIGONOMETRY. From (91) and (99) we liave . ^A A A A ... sin^-- + 2 sm— cos— + cos'^ — 1 + sin ^ 2 2 2 • 2 1 + cos ^ ^ ^A 2 cos^ sin-+cos- =M ^1 — cos- = Hl+tan|y. 2, Prove that tan — = 2 1 + sec 61* From (102) we have sin sin cos 6 tan ^ tan 2 1 + cos (9 1 , 1 + sec ^' cos 6' 6. Trove that tan — - tan — - tan — = tan- tan —tan -. •s o fa 2 3 6 'an~-tan- •tan- = ^ ^ ^ , by (13) and (G5) cos— cosmos — . e sin — ^ ^ ^ cos —cos --cos — J o fa . sin— ^ 5 ^ { Kcos g - cos-) } , by (63) cos-cos -cos ^ 0\ tan ^ tan- tan-, by (58) EXAMPLES. •{ *^^ , Tf ..« /} - ££i^±£^, prove that tan ^ = tan |tan ^. From (101) we have ^ >" **-^-^' ^^ ^ ^ ' cos g + cos P _ ^ 1 - cos (9 _ l+cosaco3^ 1 + cos a cos /3 /■*' 1 + cos g cos j8 - cos g - co s g - ^ 1 + cos g cos /3 + cos g + cos ^ ' " 1 - cos g 1 -cos^ , ' ~ 1 + cos g 1 + cos ^ = tan^|tan^|, by (101), therefore , tan - = tan- tan-^- *,.> e ^ ^ 5. Prove that cos (30° - ^ ) " ^os (30" + -^) = sin -. 6. Prove that cot (30' + |-) - tan(30- + | ) = 2 tan (30" - 0) o ^\ 2 cos ^ - 1 7. Prove that tan (30° + j) tan (30° - ^) = gcos^+l 1 - sin 20 8. Prove that tan'^ (45° -6)^ fTsin '2d' , sin 20 + sin d 3d -^ '■ 9. Prove tliat ^^^jf:^-^J = ^^^ T' , tan a + sec g , g . „„ / i eo . ^\ 10. Prove that ^ = tan - tan (45 +-^). XV. J.XV/ cotg + cosecg ^ -^ ■ sin d + sin 3d + sin 5d ^^ 11. Prove that ^^^^^^^^-^-^^,^ = ta- 3d. . ■ ; sin' g - sin^ /? 12. Prove that tan (g + (S) - -^-^^ ^ ^ sin )8 cos )3* ' = cos d- cos 2d ^ d 3d 13. Prove that ^^^^"^^^ = *- 2 """ T' t^l 74 PLANE TRIGONOMETRY. , . ^ , sin 30 + sin 4^ . 14. Prove that ^— 2^ = t.ana ' cos ^0 + coa |6' 15. Express by means of a sum or difference cos 26 cos B ; sin 2d cos 30 ; sin 50 sin 9. Ana. |(cos 30 + cos 0) ; |(sin 50 - sin 0) ; ^(cos 40 - cos 60). 16. Reduce sin^ 30 -sin'' 20 and 6 50 cos'' (45° + -^) - sin'' (45° — -) to products. Ans. sin sin 50 ; sin 20 cos 30. 17. Prove that (1+cot + co8ec 0) (1 + cot 0-cosec 0) = cot — — tan — . 18. Express by a product each of the following : sin + sin (0 - 2^) ; sin" - sin'' (0 - 2) sin (^ - ) sec" sec^ . 27. Sliew that cos- (^ + ) - sin* ^ = cos <)!) cos {26 + ). 28. Shew that sin (6* + <^) cos d - cos (^ + <^) sin = sin . 29. Prove that sin a + sin /3 + sin y - sin (a + /3 + y) = 4 sin |(a + ^) sin ^(a + y) sin ^{(3 + y). 30. Prove that cos a + cos /? + cos y + cos (a + ^ + y) = 4 cos ^(a + /3) cos |(a + y) COS |(^ + y). 31. Prove that tan a + tan ^ + tan y - tan a tan ji tan y sin (a + y8 + y) cos a cos /3 cos y 32. Prove that cot a + cot )8 + cot y - cot a cot j8 cot y cos (a + )8 + y) sin a sin ft sin y' 33. Prove that cos (a + /3 + y) + cos(a + ^3- y) + cos (a - ^ + y) + cos ( - a + |8 + y) = 4 cos a cos yS cos y. 34. If a + /? + y = 1 80°, prove that sin (a + /8) sin (/? + y) = sin a sin y ; sin 2a + sin 2/3 + sin 2y = 4 sin a sin /3 sin y. 35. If a + ^ + y = 90°, prove that . 1 + tan — cos a + sm y - sin /» 2 cos )8 + sin y - sin a /8' 1 + tan -^ 36. If a, j8, y be in arithmetical progression, prove that sin a - sin y = 2 sin (a - /?) cos /? ; cos a - cos 7 sin a - sin y tan (a -3) = : — ^ = ^. V sin a + sm y cos a + cos y 37. If sin ) = tan 6. 76 PLANE TRIGONOMETRY. 38. If tan J(a + /?) tan ^{a - )9) = tan'' ^, prove that cos a = cos )S cos y. or> Ti. ^ 1 - tan''rf> - oy. It tan - = , n— ., prove that 2 1 + tan-* (f) '^ 2 cot 2 = \, prove that sin {2$ + ^) = -~. 42. If cot 29= - tan <^, prove that tan (6 - = — ; — -^, then tan 0, tan and tan il/ are in harmonical progression. 44. If cos (6 - \j/) cos (f> = cos (^ - <^ + \p), then tan 6, tan and tan ^ are in harmonical progression. 45. If tan a tan 6 + sec a sec 6 = sec ^, shew that sin /? :; sin a tan ^ = ± —~ —. cos a cos p 46. Prove that sin" 30° = sin 18" sin 54°. 47. Prove that tan 50° + cot 50° = 2 sec 10°. 48. Prove that tan 52|° = ( V 3 -+ J 2) {s/2 -\). 49. liA+B + C= 180°, prove that sin A + {^in B -sin C A B = tan — tan — : sin .4 + sin -S H- sin C 2 2 B C A sin (il + — ) + sin (5 + — ) + sin (C + — ) + 1 2i ^ ^ = 4 cos \{A - B) cos \{B - C) cos \{C - A). 60. If sin ^ = sin a sin (/? + 6), prove that tan (^ + ^) = tan | tan" (45° + ^). ii ^ ^ EXAMPLES. 77 51. If sin (2C + y) cos « = 2 cos (y - 2) sin x, prove that cot £c - cot y = 2 tan 2. 52. Prove that 2 cosec 4« + 2 cot ix = cot x - tan aj. 53. Prove that 2 sin ^(.4 + B - 90°) cos ^(^ - ^ + 90°) = sin A - cos B. 54. Prove that (cot'^ _ - tan^ — ) tan ^ = 4 cosec 0. 55. Prove that sin (A - B) + Bin (B - C) + sin (C-A) = 4 sin }^{li - A) sin ^(C - i5) sin J(^ - C). 56. Prove that sin A CO8OC (^ - B) cosec (i - C) + sin J5 cosec (B - (7) cosec (B-A) + sinCcofiec(O-A)cosec(G-B) = 0. 57. If cos (A + B) sin (C + i>) = cos (^-5) dn (C - J)), shew that tan i) = tan A tan 5 tan C. 58. If sin (a -7) cos /3 + sin (/8-y) cos a=0, shew that tan a + tan (3—2 tan y. 59. If cot a + cos 13= JJ cot e sin /i^, ' ••- and cot y + cos ^= ^2^ cot (3 sin ^, prove that sin = sin a sin /3 cosec y. ' ' ' cos A cot ^ cot J - cos A 60. Prove that -—r = -. — TT'- cos A + cot A cos A cot ^ 61. Prove that cosec 4:6 + cot 4:6 = ^(cot - tan 6). . t 62. If cot 6=2 tan <^, then cos (0 -) = 3 cos (^ + <}>), and sec (0 + when ^ = 0. B From these results it follows that when B is very small, sin B = B = tan ^, very nearly. 75. To shew that sin B>B-~, where B is the 4 Circular Measure of any Angle betv/een 0° and 90". By Art. 73 we have ,, . - . , .,. , ■- . r :n. :■ B B ''""2 B :■„ '.■ «OS 2 ^-,^ ' •■ • >' ^. ■>••■• hence 2 sin — >^ cos — • 2 2 . .-- ^ 82 . PLANE TRIGONOMETRY. Multiplying both members of this inequality by cos — , we have by (94) sin 0>6 cos'* — ' >0{l-~), since --> sin 76. To find the Numerical Values of the sines and cosines of all Angles from 0° to 90° at inter- vals of 10". „ r Let 6 be the circular measure of 10", then since the circular measure of 180° is tt, or 3.141592653589793 ... as will be shewn in a subsequent chapter, we hav« 180° : 10" ..TTiO, IOtt or 6 = 180 X 60 X 60 3.141592653589793 ~ 64800 • = .000048481368. J ■ But sin e<^ and >^ - -r-, (Arts. 72 and 75.) 4 hence sin ^>.000048481368 - J(. 000048481368)' '• ; ' ^ >.000048481368-i(.00005)3 ■ ^:. ' >.000048481368 - .00000000000003, from which it appears that the quantity to be subtracted from does not affect the first twelve places of decimals. Hence, the sine of 10" coincides with the circular measure of 10" to twelve places of decimals, and therefore - COMPUTATION OF TRIGONOI^rETRICAL FUNCTIONS. 83 sin 10" = circular measure of 10" = . 000048481368 ; also sin 1" = circular measure of 1" = . 0000048481368. If a be the circular measure of any angle -4°, then it is evident that A"=- " — - =-T-^ = ax 206264".806... (128) cir. meaa. 1 sin 1 which agrees with (6). . The cosine of 10" is obtained from the formula cos 10"= v/l-sinU0"= v/(l-sinlO") (1^+ sin 10") = .999999998824. • 77. From (51) and (53), we have sin (^ + ^) = 2 sin A cos B - sin (^ - .5) cos (^ 4- .^) = 2 cos ^ cos ^ - cos ( J[ - ^, in which let B be constantly equal to 10"; ' * - thus, sin {A + 10") - 2 cos 10" sin ^ - sin (i - 1 0") cos {A + 10") = 2 cos 10" cos ^ - cos (^ - 1 0"). Now, if ^ =»= 10", 20", 30", and for the cosines * ' '•; ; cos 20" = 2 cos 10" cos 10" -cos 0", ■ <\ •-■: cos 30" = 2 cos 10" cos 20" - cos 10", -f •>•' cos 40" = 2 cos 10" cos 30" - cos 20", .84 PLANE TRIGONOMETRY. These results may be further simplified as follows : 2 cos 10"= 1.999999997648, = 2 - .000000002352, = 2 - m, suppose. Making this substitution, we have sin 20" = 2 sin 10" - sin 0" - m sin 10" = .0000969627, sin 30" = 2 sin 20" - sin 10" - m sin 20" = .0001454441, sin 40" = 2 sin 30" - sin 20" - m sin 30" = .0001939254, cos 20" = 2 cos 10" - cos 0" - m cos 10" = .9999999953, cos 30" = 2 cos 20" - cos 10" - m cos 20" = .9999999894, cos 40" = 2 cos 30" - cos 20" - m cos 30" = .9999999812, &c., . &c. 78. Having computed the sines and cosines by this process up to 30°, those for angles greater than 30° may be easily found by (126) and (127), thus, • sin (30" + ^) = cos ^ - sin (30' - A) cos (30° + A) = cos (30° -A)- sin A, in which let A = 10", 20", 30", &c., in succession, and we have sin 30° 0' 10" = cos 10" - sin 29° 59' 50", sin 30° 0' 20" = cos 20" - sin 29° 59' 40", &c., (1 + cot 6 ) ^ sin A6' Ad = - sin d sin A^ (1 + cot ^ tan — -) Ad = - sin d Ad ( 1 + --- cot 6). (131) Ad When d is near 90°, -— - cot d may be neglected, and we have ap- proximately Acos d= -sin dAd. * ' a32) A tan d = tan (d + Ad) - tan d _sin (d + Ad) sin d cos (d + Ad) cos d sin Ad cos'^ d cos Ad (1 - tan d tan Ad) sec^ d tan Ad 1 - tan d tan Ad ' > sec'^^ Ad , ~1-Adtand" , ^^^^^ When d is not nearly equal to 90°, Ad fan, d may be neglected, so that we have approximately A tan d = sec2 dAd. (134) A cot.d = cot (d + Ad) - cot d cos (d + Ad) cos d "sin (d + A"d) ~ siiTd - sin Ad sin2 d cos Ad (1 + cot d tan Ad) - cosec=* d tan Ad 1 + cot d tan Ad -cosec'^d Ad 1 + Adcot d* v :;- vyui \I36) INCltKMENTS Ob' TRIGONUM ETHICAL FUNCTIONS. 91 When is not nearly c(iuul to 0", A^ cot $ may be neglected, and we have approximately A cot 0= - cosec2 Aa. (130) A see ^ = sec (0 + A^) - sec 1 1 cos (^ + A^) cos cos ^- cos (6 + A6) cos $ cos (6 + A^) sec 6 (tan + tan ) tan AO 1 — tan B tan A6/ A0 sec B (tan ^ + ) A^ 1 - A6^ tan ^ (137) When B is neither very small nor very nearly equal to 90°, A^" — — sec B and A^ tan ^ may be neglected, and we have approxi- mately A sec ^==sin B sec^ '^ A^. ■ (138) A cosec B = cosec (^ + A^) - cosec Q '_\ 1 ■'.>,-[■ ^■" -i {B + A^)~8in B '■'. .;'"'•.' sin ^ - &in (B + ^B) amBam(B + M) A6' cosec B (tan — _ — cot B) tan Ad 1 - cot ^ tan Ad Ad cosec B (—r — cot B) Ad (139) 1 -Adept d When d in neither very small nor very nearly equal to 90°, 92 PLANE TllIGONOMETRY. cosec 6 and A^ cot 6 may be neglected, and we have ap- Ji proximately A cosec 6 = - COB $ cosec^ AO. (140 From these forinulse we observe, (1) that a small increment of the angle causes a small increment of the sine, tangent and secant, and a small decrement of the cosine, cotangent and cosecant ; (2) that for all the trigonometrical functions, a small increment of the angle produces a small proportional increment or decrement of the function, with the exception of the particular cases noticed above. Hence, if an angle is near 90°, it cannot" be found from the ordi- nary tables with great accuracy from its sine, tangent or secant ; nor if near 0° or 180°, from its cosine, cotangent or cosecant. In the next chapter, however, we will explain the construction of a special table by which the trigonometrical functions of an angle near the limits of the quadrant, can be found with great accuracy 86. The results of this chapter enable us to solve many interesting and useful problems in surveying and astronomy, without using the trigonometrical tables. Examples. 1. An object standing on a horizontal plane subtends an angle of 3' 20" at the distance of two miles, find its height. Let BC be the object whose • height is required. The angle ^ = 3' 20"= e, and AC == 2 ^ miles = a. Then BG = AC tAiiO, by (32) = aO, by Art. 74, — ax circular measure of $ ti" .°'""206264".8' ^y^^^®> 2 X 200 206264.8 10| feet, very nearly. EXAMPLES. 93 2. The equatorial radius of the earth is 3962.8 miles, which is found by astronomical observations and calculations to sub- tend at the sun's centre an angle of 8". 95. Find the sun's distance from the earth. , . Let S and E represent the centres of the sun and earth. The angle ASE=: 8".95 = 6, AE = 3962.8 miles = r, and SE = X. Then and X sin = r x = sin 0' r cir. meas. ' 206264".8 8". 95'" = 91328000 miles. — rx by (35) by Art. 74 t . by (128) r.»/, 3. The apparent semi-diameter of the sun is 16' 1".82 at the earth's mean distance, as given in the last problem. Find the sun's radius. ., , ) In the figure of the last problem, let S be the earth's centre and E the sun's. The angle ASE = 16' 1".82 = ^, SE= 91328000 miles = d and AE the sun's radius -= r. Then r = d sin 0, by (27) "■' ' = dx cir. meas. of 0, by Art. 74 = dx = d X t ' .r. 206264".8 ' 961".82 by (128) 206264".8 425860 miles. f)4 PLANE TRIGONOMETRY. E. r'' Hence the sun's radius = 107.4 times the earth's radius and as the volumes of spheres are as the cubes of their radii, we have ■'''' I- "' ■■ ■'•. •' -- ■*■■ ■ -'■ volume of sun : volume of earth :: (107.4)' : 1 :: 1238833:1, ■ so that the sun is more than a million times as large as the earth. Dip of the Horizon. 87 . The dip of tlie horizon is the angle of depression of the visible horizon below the true horizon, arising from the eleva- tion of the observer's eye above the level of the sea. Let BBG be a section of the eartli re- garded as a sphere, and A the position of an observer above the surface. Join A C and pro- duce it to 6r, also draw AE at right angles to the vertical line AG, and A F touching the sur- face at J), then neglecting the eflfect of atmospheric refrac- tion, D is the most distant point visible from A. ceive AE and AF to revolve about AG as an axis, AE will describe the plane of the true or celestial horizon and AF will describe the surface of a cone toucliing the earth in the small circle called the apparent or visible horizon, and the angle FAD is called the dip of the horizon. Let 7t = ^-B, the height of the observer's eye, r = DC, the radius of the earth, /) = the dip of the horizon. The angle FAD = ihe angle ACD = D. PIP OF THE HORIZON. 95 AD JAG xAB DG~ D J{2r + h)h Then ,.^ tani) = — = ^~-^^ , {Euc. III., 36) [271 /h^^ As h is always very small compared with r, the square of tlie fraction — is quite inappreciable, and therefore may be r neglected, so that we have . ^ •,'-.■ •■ •• : . tan 2) = ^—. ' (141) Again, as the angle D is always very small for all accessible heights above the earth's surface, we may write the circular measure of the dip for its tangent (Art. 74), and expressing it in seconds by (128) we obtain i?" = 206264".8. The distance AD oi the visible horizon = the arc BD very nearly ; let -4 i> = i)5 = c?, then .:,.;, , . ^ . . d = DC X the circular measure of D •';' ■ , =r^^=j2^,. . ' •■ (143) The mean vp' e of r is 20888628 feet. Substituting this in (142) and reducing the constant coefficient of vA we have D" = 63".S2^fh, • (142 his) where h is expressed in feet. Owing to the effect of refraction, this coefficient is dimin- ished by about ^ part. Deducting ^^ of 63.82 we have finally i)" = 58".02>/A. (144) 06'", PLANE TRIGONOMETRY. By this formula, the dip is computed and tabulated for heights varying from 1 to 100 feet, in works on navigation and iihitronomy. Expressing d and r in miles and h in feet we have, by com- puting the constant coefficient in (143), d=l.2Ujh. (145) which gives the distance, unaftected by refraction. The amount of refraction however varies very much with the temperature of the atmosphere, in ordinary states of which, this distance is increased by about -^^ part. Hence, increasing the coefficient 1.224 by -jJj of itself, we have c^= 1.317 s/A = M V^j very nearly. (146) In (145) let d= 1 mile, then we have A = .6675 feet = 8 inches, very nearly. Hence, if the tangent DA in the figure, be one mile, AB va^ inches, which is expressed by saying that on the surface of still water the depression. of the horizon is 8 inches for a horizontal distance oi one mile ; therefore in levelling for canals, aqua- ducts, . ... Examples. 1. Given cos A = .6400566, to find A, From the tables, we hare ' ' '- v , . ? . EXAMPLES. 97 COS 50" 13' = .6398862 cos 50° 12' = .6401097 Difference for 1' or 60" = .0002235 cos -4 = .6400566, the given cosine, cos 50° 13' = .6398862, next less in the table. Difference = .0001 704 Hence we have .0002235 : .0001704 :: 60" : 45".7, Therefore ^ = 50° 12' 14". 3. 2. Find in a similar manner the angle whose cotangent is 2.1453675. Ans. 24° 59' 28".2. 3. Given tan 45° 1' = 1.0005819, find tan 45° 0' 40". Am. 1.0003879. 4. Find the angle whose sine is .7126666. Ans. 45" 27' 8".2. 5. Find the angle whose tangent is .8952524. , Am. 41° 50' 11".6. 6. Given sin 45° = -'^^--, show that tan 22° 30' -.41421 36, and sec 22° 30' = 1.0823922. 7 The circular measure of an angle is .1047683; find the number of degrees in it. Ans. 6°0' 10". 8. Given cosec A + cor A = J^ , find A. Ans. ^ = 60°. 9. Find the dip of the horizon for a height of 25 feet. , ' Am. 4' 50". 98 PLANE TRIGONOMETRY. 10. How far can the top of a lighthouse 200 feet high be seen at sea 1 ' Arts 18| miles. 11. From what height will the horizon be 44 miles distant 1 Ans. 1111^ feet. 12. From the top of a ship's mast whose height is 81 feet, the top of a lighthouse 100 feet above the level of the sea, was just visible ; required their distance, taking the effect of refrac- tion into account. Ans. 25^ miles. 13. What is the difference between the true and the appar- ent level, at the distance of 1000 feet? Ans. .2873 inch. 14. A ship whose mast is 90 feet high, is sailing directly towards an observer at the rate of 10 miles per hour, and the time from its first appearance till its arrival at the observer, is 1 hour 9§ minutes ; find approximately the earth's radius, sup- posing the observer's eye to be on a level with the surface of the sea and no allowance made for refraction. Ans. 3958 miles. 15. When the moon's distance from the earth is 238824 miles, her radius subtends at the earth an angle of 15' 33".5 ; find her diameter. Ans. 2161.6 miles. 16. Find the distance and the dip of the horizon for a height of 6 feet, taking the effect of refraction into account. Am. 3.226 miles; 2' 22". LOGARITHMS AND LOGARITHMIC TABLES. 99 CHAPTER VII. ON LOQARITHMS AND LOGARITHMIC TABLES OF THE TBIGONO- METRIOAL FUNCTIONS. 88. The logarithm of a number is the index of the power to which a fixed number, called the base, must be raised, in order to produce the given number. Thus, if then X is the logarithm of the number R, to the base a, and is expressed thus, a; = logaiV; or, if there be no occasion for mentioning the base, simply by aj = log N. Again, since 10" = 1 ; 10^ = 10 ; 10^ = 100, &c., logio 1=0; logio 10 = 1; logi„ 100 = 2, &c. It is evident that any number except unity may be taken as the base; and if for any base as 10, the logarithms of all * numbers be computed and registered in a table, the table thus formed constitutes what is called "A table of a system of logarithms to base 10." There may therefore be an infinite number of systems of logarithms. .. 89. Although in theory any number except unity may be used as the base, yet in actual practice it has been found most convenient to use only two systems, viz. : I. — Logarithms to base 2.71 8 J8 . . . (denoted by e) which are called Napierian logarithms, from the name of the inventor, 100 PLANP] TRIGONOMETRY. Lord Napier, Baron of Merchislon in Scotland, who published the first table of logarithms in the year 1614, and* II. — Logarithms to base 10, the radix of the ordinary scale of arithmetical notation, which are called common logarithms. These possess some peculiar advantages, and all the tables in common use are calculated to this base. Hence when we speak of logarithms, we mean logarithms to base 10, unless the con- trary is stated. go. Since a" = l, and a} = a^ we have loga 1 = 0, and log^ a= 1. (147) whatever a may bew If a be greater than unity, then a-a = ; a°= 1 and «+• = oc , therefore loga = - x ; log^ 1=0; loga oc = oc . (148) Hence, the logarithm of any number will be between and — oc , or between and + oc , that is, will be negative or posi- tive according as the number is less or greater than unity. Again, if a be less than unity, it may be shewn in a similar manner that the logarithm of any number will be positive or negative according as the number is less or greater than one. Since a* and a—* are always positive quantities, negative numbers have no real logarithms. gi. In any System of Logarithms, the Logarithm of the product of two numbers is equal to the sum of the Logarithms of the numbers. Let a be the base, m and n any two numbers, and x and y their logarithms respectively. Then by the definition of a logarithm, we have V PROPERTIES OF LOGARITHMS. 101 ,2W. = a* or jc = loga wi and n = ay or y = loga ^ • and therefore mn ^a"^ .aV but by the definition, x + y is the logarithm of mn to base a, therefore log^i mn = x + y = loga m + loga n. 92. In any System, the Logarithm of the quo- tient of two numbers is equal to the difference of their Logarithms. Using the same notation as in the last Article, we have ma* . — = — = a«-l/: n av ' but x-y is the logarithm of — to base a, n lliorefore loga— =*-y = loga W- loga W. Hence it follows that ' ^ • loga— = loga 1- loga m • ' • • =-logaW. by Art. 90. 93. In any System, the Logarithm of the zith power of any number is tz times the Logarithm of the number. Let m = a* as before, that is, £C = logrt w». -> • • ^ . Then -' = . ' m» = (««)♦* ^; ^^ vj . ^a^^- ' 102 PLANE TRIGONOMETRY. )mt nx iH the logarithm of 7/1" to base a, therefore loga 7/t" = nx -= n loga m, which is true whether n be an integer or a fraction. 94. We thus see from the last three Articles that, by means of logarithms, the processes of multiplication and division are reduced to those of addition and subtraction ; and those of involution and evolution to multiplication and division. The arithmetical operations of addition and subtraction cannot be performed by logarithms. •The following examples shew the application of the pre- ceding principles : . 189 , 3^x7 , -3 ^i . log— = log— --= log 3^x7* = 3 log 3 + J log 7. log — 7- — = n log a^ — log c — p log h. logQ^=^logi=i{log3-logl7}. log sld'-h' = \\og{a'-h'') = |log{(a + 6)(a-6)}' . = i log (a + 6) + J log (a - h). 95. Having given the logarithm of any number to a given base as e (where e denotes a certain number 2.71828... of which more hereafter), to find the logarithm of the same number to. any other base as a. ;, r Let X and y denote the logarithms of any number N to the bases a and e respectively, that is, ^. PROPERTIES OF LOOARITHMS. 103 let a5 = loga y, and y = loge iV, hence a* = iV^ and eV = -A^, therefore a* = eV. Expressing this equation in logarithms to base e, we have a; log<, a = y l<^ge « = y. by (147) Substituting the values of x and y, we have loga J^ loge a = loge JV, therefore loga -^=q • log- JV. (149) loge « If a= 10, the base of the common system, then logio I^ - j^-iy^ loga ^. (150) From (150) it appears that every common logarithm may be resolved into two factors, one of which is constant and depends upon the base of the system employed, the other is variable and depends upon the number itself. The constant factor or multiplier which thus connects two systems of loga- rithms, is called the modulus, and is denoted by M. Thus, the constant , -~ is the modulus of the common system taken loge 10 ^ relatively to the system whose base is e. The modulus of the common system may, therefore, be defined as the constant factor or multiplier, by which it is neces- sary to multiply the Napierian logarithms in order to convert them into common logarithms, and is equal to the reciprocal of the Napierian logarithm of the Hase of the common system. It will be shewn hereafter, in the chapter on the computation of logarithms, that the Napierian logarithm of 10 is 2.302585... , therefore the modulus of the common system • ^ I^iTlO " 2:302585:: " -4342944... , . 104 PLANE TiunoxoMKruv. DcMotiiii; tlio coniinoa loyariihin ol a number by log, (I'^'O) may bo written log iV^=. 4342944 x log,, N. (151) Hence we also find = 2.302585 X log i\r, (152) which enables us to finil the Napierian logaritlim when th«! common logarithm of any number is given. 96. Since log^ N= Mloge N, and logb N=^ M' loge iV, wlu4-o M and M' are the moduli, we have by division therefore the logarithms of the same number in different systems are proportional to the moduli of those systems. • ' 97. Relation between the bases of two Systems. Let X and y denote the logarithms of any number N to the bases a and b respectively, that is, let a;=log^ N, or a" = JV, and y=zlogj^ N, or b'^=N, therefore , a/>^—.jfV... Expressing this equation in logarithms to base a, we have » and expressing it in logarithms to base 6, we have - xlogj,a=y. ' Eliminating x and y from the last two equations, we obtain logfc a log„ 6=1, (153) M^hich expresses the relation between the bases of any two systems. COMMON LOGARITHMS. 105 g8. Properties of Common Logarithms. In the common system, the logarithms of all nuniborn which are integral powers of 10, are immediately known. Thua 10'>=1, therefore log 1 =0, 10^ = 10, " log 10 =1, 10^=100, « log 100 -2, lO^'^lOOO, « log 1000 = 3, • «kc., (fee. Hence it appears that the logarithms of all numbers which are not integral powers of 10 consist of either a fraction or an integer and a fraction. Thus, the logarithm of every number between 1 and 10, lies between and 1, or is a proper fraction; the logarithm of every number between 10 and 100, lies between 1 and 2, or is 1 plus a fraction ; the logarithm of every number between 100 and 1000, lies between 2 and 3, or is 2 plus a fraction, and so on. The integral part of a logarithm is called the characteristicj and the decimal or fractional part the mantissa. Hence, if a number is between 1 and 10, the characteristic of its log =0, " =1, " =3, ; • ••• « =n-\. Therefore the characteristic of the logarithm of a number which has n integral places, is 7i - 1, or is less by unity than the number of integral places in the number. 10 « 100, (( n (( 100 " 1000, (( H I( 1000 " 10000, (( K (( • ••• .... IQn-l a 10", <( (( u 106 PLANE TRIGONOMETRY. Again, since *. 10" = 1 = 1, therefore log 1 = 0, 10-^= t\7 =.1, " log.l =-1, • 10-*'^=tAu =-01, " log .01 = -2, 10-^ = TT5W--001, " log .001 =-3, &c., . ....■...• .... 10-» lO-Ci+i) « " « u ^ui^\ Therefore generally if a number has n ciphers -after the decimal point, the characteristic of its logarithm is (tmT). The negative sign is written over the characteristic to shew that it alone is negative, while the mantissa, upon this supposi- tion, is Always positive. Hence, we have the following general rules for finding the characteristic of the logarithm of any number. "" . Rule I. — The characteristic of the logarithm of any number greater than unity, is positive and less by unity than the number of integral places in the given number. -■# COWMON LOGARITHMS. 107 Rule II. — The characteristic of the logarithm of a decimal fraction, is negative, and numerically greater by anity than the number of ciphers after the decimal point. Thus, for the following; numbers, 14067. 521.64, 6.721, .364, 04271, .0027, .0001672, the characteristics are respec- tively 4, 2, 0, T, 2, "3, 1. And conversely, if logarithms be given, having character- istics 0, 1, 2, 3.... n, there are in the numbers to which these respectively belong 1, 2, 3, 4....(w + l) digits; and if the characteristics are 1, 2, 3....7^, there are in the corresponding numbers 0, 1, 2 .... (»- 1) ciphers respectively after the decimal point. , :• : 99. When the logarithm of any number is known, we can at once write down the logarithm of any other number having the same significanf digits, but differing only from the given number in the position of the decimal point. Let N be any number whose logarithm is known, then 10" X A^ is a number which has the same significant digits as N, but with the decimal point moved n places to the Hght ; N again, -— or 10-" x iV is a number which has the same signifi- cant digits as N, but with the decimal point moved n places to the left. • . • ^ - Now, log (10» X N) = log 10» + log iV, • = H log 10 + log iV, = n + log N' ; ". • and log (10-« x N) = log 10-» + log N, , . ., = - n log 10 + log Ny — -n + \og N". Hence, as n is an integer, the logarithms of 10" x iV and 10-" x N differ from the logarithm of iV in the characteristic 108 PLANE TRIGONOMETRY. only ; therefore by giving the logarithm of N its proper char- acteristic in accordance with the rules investigated in the last Article, we can at once deduce those of 10" x. N and 10~»* x N. Thus, log 2436000 = 6.386677, log 24360 =4.386677, log 24.36 =1.386677, log. 2436 =1.386677, "log .002436 =¥.386677. IOC. In this system it is only necessary to register the mantissa3 in the tables, for the characteristics can be determined by the preceding rules. These properties render the common tables less bulky and more comprehensive than those computed for any other base, and g've them an advantage over all others. lOl. The arrangement and use of the common tables are easily understood from the following portion of a table taken from Chambers's Logarithms : No. 1 2 3 4 5 9022 9921 0820 1719 6 7 8 9 D 90 4829 4830 4831 4832 6838672 9471 6840370 1269 8662 9561 0460 1359 9 8752 9651 0550 1449 18 8842 974 L 0640 1539 27 8932 9831 0730 1629 36 9112 .011 0910 1808 9202 .101 1000 1898 63 9291 .191 1089 1988 72 9.381 .280 1179 2078 81 Difif. 90 45 54 In this table we have the natural numbers between 48290 and 48329, and the mantissse are between .6838572 and .6842078. The four left hand figures of the natural numbers are placed in the column marked No., and the fifth figure at the top of one of the other columns. As the first three figures of the mantissse are the same for several consecutive numbers LOGARITHMIC TABLES. 109 in the table, they are not repeated for each logarithm, but regis- tered once only. Thus, the mantisssB of the logarithms of all numbers from 48290 to 48305 inclusive, contain the same three initial figures, viz. : 683, which are registered opposite the num- ber 4829 and in the column under 0, while the last four figures of each mantissa, are registered in the columns 1, 2, 3, &c., to which they belong. Thus, the mantissa of log 48296 = .6839112 «* log 48303 = .6839741 &c., ao. In the second horizontal line and in the column under 6, a dot is found in place of a figure. This indicates that the value of the three initial figures has changed, and therefore both here and all along the remainder of the same line, the three initial figures taken from the horizontal line next below, are to be used. The places of the dots are to be supplied with ciphers. Thus the mantissa of log 48308 = .6840191. In this manner we obtain from the tables the mantissas of the logarithms of all numbers which contain, three, four or five digits. The logarithms of all numbers between 1 and 100 are generally tabulated separately. At the bottom of each page there is a " table of proportional parts" by which the mantiss86 of the logarithms of all numbers containing six or seven digits may be calculated approximately ; and conversely, the number corresponding to a given logarithm whose mantissa cannot bo exactly found in the tables, may be found approximately to six or seven places of figures. . > In the construction of the "table of proportional parts," it is assumed that the increase in a logarithm is proportional to the increase in the number. The proof of this principle will be given in the chapter on the computation of logarithms. The 110 PLANE TRIGONOMETRY. ■ .v. results given by it are only approximate, but the error is gener- ally so small that it may be safely neglected, especially if the numbers are large. ' • • The last column marked D contains the difference between two consecutive mantissse ; thus in the above table, the mantissa of 48304 = .6839831 " 48305 = .6839921 Difference corresponding to 1 = .0000090 The significant digits only of this difierence, are registered in column D. Now, suppose the logarithms of 48304.1, 48304.2, 48304.3 (fee, are required, we have by the principle just stated 1 :.l 1 :.2 1 :.3 .0000090 : .0000009 .0000090 : .0000018 .0000090 : .0000027 &c., (fee, &c. '4» HtMice. the mantissa of log 48304.1 = .6839840 log 48304.2 = .6839849 log 48304.3 = .6839858 &C., &C., •~' --TBC. ' The significant digits, 9, 18, 27, 36, i^; • . ; Ans. 7|. EXAMPLES. 125 2 6. If the log of 9 is -, what is the base? Ans. 27. o 7. Given log 98 = 1.991226, and log 112 = 2.049218, find the logarithms of 50, .7 and 1750. Am. 1.698970, T.845098, 3.243038. 8. Given log 1^ = . 096910, and log i =1.045757, find the 9 •logarithms of 2| and 2^. Ans. 0.397940, 0.352183. 9. Shew that 5 log ^ + 3 log — - + log — = log 2. 10. Find the 50th root of 10. Am. 1.047126. 11. Find the value of (-)^*. Ans. .612295. 12. Find the 17th root of .071852. Am. .8565. 13. If a, 6, c be in geometrical progression, prove that loga **> logb w> logo '* a,re in harmonical progression. 14. Shew that log . . oosec^ = log ^ sec ^. ©sin A Ocos/4 15. Find x from the equation 8* = 1000, having given log 2 = . 301030. • ^71*. a; = 3.3219. 16. Given log 1.4 = .146128, log 144 = 2.158362, and log 441 = 2.644438, find the logarithms of the nine digits. 17. Given log 2 = .301030, and log 3 = .477121, find the logarithms of 135, 405, 3.24 and — , 9 Am. 2.130334, 2.607455. 0.510545, T045757 18. Find the value of ^-^IlL^^il^lL . Am. .340653. (19)* X (.061)2 19. Findlog2 6. '. Am. 2.584. 12G PLANE TRIGONOMETRY. 1 20. If 32«.5'«-i =23«+i shew that « = -— .^ -, s- log 45 - log 8 21. Given xV =y^, and o? — y\ find a; and y. 9 27 22. Given 28«'.52a;-i =45«.3a'+i, find x Ans. -.991. 23. Given Log cot 68° 21' 10" = 9. 598661, Log cot 68° 21' 20" = 9.598600, find Log cot 68° 21' 16".4. Ans. 9.598622. 24. Given Log sin 37° 4' 40" = 9.780244, Log sin 37° 4' 50" = 9.780272, find Log cosec 37° 4' 47". Am. 10.219736. 25. Given Log cos 75° 13' = 9.406820, Log cos 75° 14' = 9.406341, find Log cos 75° 13' 22".4. Ans. 9.406641. 26. Given Log tan ^ = 10.572676, find A, having given Log tan 75° 1' 20" = 10.572622, Log tan 75° 1' 30" = 10.572706, Ans. A =-75° 1' 26".2. RIGHT-ANGLED TRIANGLES. 127 CHAPTER VIII. FORMULA FOR THE SOLUTION OP TRIANGLES. RIGHT-ANGLED TRIANGLES. 114. The formulse for right-angled triangles have been already given in Chapter III. They are immediately derived from the definition of the sine, cosine and tangent. Thus, from the right-angled triangle ABC, we have sin A = -, cos ^ = --, tan ^ = -r. c c " (158) Special Formulae for Right-angled Triangles. 115. When A is nearly 90°, the first of this group cannot be accurately computed from the tables, since the sines then^iffer very little from each other; and for a similar reason, the second .of the group cannot be accurately found when A is very small, or when 6 sTearly equal to c. Hence, an angle near 90° should be deter- mined by its cosine, and a small angle by its sine or tangent. Special TormX are therefore frequently necessary, especially in surveying and astronomy, where great accuracy is required. From (100) we have 2sin2 2-=l-cos4. Writing 90°-^ for A, this becomes . . ,;^ ■ A 2Bm2 (45°_-)=l-8inil =1 - - , by the first of (158) c a 128 PLANE TRIGONOMETRY. • whence . «"^ (^^°- 2~^=s]^17^' ^1^^) ■which may be used with advantage when A is near 90°. Il6. From the second of (158), we have sin" ^=1 - cos'^ A=l - —r. , whence sm ^= I — l/(c+6) (c-b) = ^ , -, (160) b which may be used instead of cos A=—, when A is small. . From (100) we have A 2 sin^ 2"=l""Cos -4 whence c — b— 2c sin^ -^ , (161) by which c — b may be accurately found when A is small. .117. From (83) we have s ;; tan A — 1 =— — . , by the third of (158) . a-b '■ ' a which may be used instead of tan A=j-, when A is near 90°. 118. From (101) we have , v . ^ |l-cos A ; ' • tan — =^ 2 Njl+cos^ • ;, OBLIQUE-ANGLED TRIANGLES. 129 1-A by the second of (158) ■J 1+- 6 c-b whence c+6 A — &=a tan -x- , a (163) by which c - 6 may be accurately found when A is near 90^. Since o» - b^—a^, we have from (163) c^-b^ Q-b a' a tan 2 A (164) or c-\-b—acoi-^i by which c-|-6 may be found when A is near 90*. OBLIQUE-ANGLED TRIANGLES. i,%; ^ iig. The sides of a Triangle are proportional to the sines of the opposite Angles. In the triangle ABG^ let the sides opposite to tlie angles A, B and G be denoted by a, h, c respectively. Draw CD perpen- dicular to AB, produced if necessary. Then, in the right-angled triangles ACD, BCD, we have by the first of (158) (Fig. 1) CD = b sin A, and CD = a sin B, whence 6 sin il = a sin 5/ 10-'' 130 PLANE TRIGONOMETRY. and (Fig. 2) CD = h sin A, and GD = a sin CBD = a sin B, since CBD and ABC are supplementary angles, whence 6 sin il = a sin 5. Therefore, in both cases we have h sin ^=a sin 5, which may be converted into a proportion, thus, a '.h '.'. sin A : sin B. In the same way we may shew that a : c : : sin ul : sin (7 and 6 : c : : sin J8 : sin C These three proportions may be written as one, thus, a ; 6 ;c ::siD il : sin ji5 : sin C, or better thus, ^ 'L.^JL^. (165) sin -4 sin ^ sin C 120. The Sum of any two sides of a Triangle is to their Difference as the tangent of half the Sum of the opposite Angles is to the tangent of half their Difference. , From (165) we have a sin-il 6""siniS' whence, by composition and division, * . a + 6 sin il + sin 5 a - 6 ~ sin il - sin 5 ^toi(4+|) (59) (156) tan \{A - B) OBLIQUE-ANGLED TRIANGLES. 131 Since ^+5 + C=180",i(^ + 5) = 90°-2- * G and tan ^( J + J5) = cot - . Hence (166) may be written tani(^-£) = ^cot|. (167) Similar relations may at once be inferred between 6, c, B, C and a, c, A^ G. The last equation may be written thuB »4 c tani(i-5) = — ^cotg. 1+ — a Let — = tan ^ an assumption always possible, since a tangent may have any value between and oc ; then we have ,.. «v 1-tan^ G *^^i(^-^)=rTt^-^^'2 " ^ - = tan(45°-e)cot|. (168) In the last three formulae a > 6, therefore we may put b a 1 - cos (h G then . ^h(^-^)-iT^^''°^2 '■'ir^- = cos <6t . + COS tan^cot^. ' (168 6i5) 132 PLANE TRiaONOMETRY. 121. Geometrical proof of (i66). Let ABC be any triangle of which the side BC is greater than AC. With C as a centre and ACj the shorter of the two sides, as radius, describe a circle cutting AB in F^ BC in E, and BC pro- duced in D. Join AD, CF, AF, and draw FH at right-angles to AB. The angle FAD ia & right angle ; the angle AFC is equal to the angle CAF; the exterior angle AC^ is equal to the angles CAB, ABC, that is, A + B. But the angle AFC is half of the angle ACD (Fuc. III., 20), therefore the angle AFC is equal to ^(il + B). Again, the angle BCF is equal to the diflference between the angles CFA and CBF {Fuc. I., 32), that is, the angle BCF is equal to A - B ; but the angle FAF is half of the angle ECF, therefore the angle FAF is equal to \(A- B). Now, AD = AF tan AED = AF tan \{A + B) and FH=AF tan FAF= AF tan ^{A - B), and since ^5" is parallel to AD, we have BD '. BE '.: AD '. EH that is, a + 6 :a-6 :: il-^tan J(^ + ^) : ^^ tan \{A - B) j: tan J(^ + B) : tan J(J - B) therefore tan \{A -B) = ^5—-. tan i(A + B). OBLIQUE-ANGLED TllUNGLES. 133 122. To express the cosine of an Angle of a Triangle in terms of its sides. From the angle C of the triangle ABC, draw CD perpen- dicular to AB^ produced if necessary. From Fig. (1) we have by Euc. II., 13, BC'^ = A(P^AB''-2AB.AD. But AD = ACQO^Ay therefore BC r.AG' + AB'- 2AB.AC cos A, that is, a'^ = 6^ + c' - 26c cos A. Again, from Fig. (2) we have by Euc. II., 12, BC^ = AC^-\'AB'-\-2AB.AD. ■ . But ^i) = i4C co8(7^i> = ilC 008(180"-^)= -ilC cos J, therefore BG^ ^^AC + AE"- 2AB.AG cos A, that is, a^^b^ + c^- 2bc cos A. Hence, in both cases, we have a2 = 62 + c»-26ccos^. , (169) In the same manner, we find ' i' . : ^^' b^=:a'^ + c'-2accosB. :•, ; ; ./r(170) c2 = a2 + 62_2a6cosC. . (171) From the last three formulae, we have b^ + c^ -a^ cos A = cos B = 2bc ' a' + c'- 6" 2ac (172) (173) 134 PLANE TRIGONOMETRY. 123. The same results may, be obtained independently of Euclid, as follows : In Fig. (1) we have c = BD + AD, = a cos -B + 6 cos A ; and from Fig. (2) c = BD-AD, = a COS B-b cos (ISO" 'A)t = a cos B + b cos A ; and similarly for the othei sides. Hence, in every triangle, we have a = b cos C + e cos B. ' b = c cos A + a cos C. c = o cos B + b cos A. (175) Multiplying the first by a, the second by 6, and the third by c, then subtracting the first result from the sum of the other two, we have a* = 6' + c'- 26c cos -4, , 6« + c' - a» , , whence coSil = rr » as before. 12/,. To express the Functions of half of an Angle of a Triangle in terms of its sides. From (100) we have A 2 sin'^ ^ = 1 - cos i4, . 2bc'b''-c'+a? "" 26c * ' OBLIQUE-ANGLED TRIANGLES. 135 Ihc * _ (a-b + c){a + b-c) Wc • ^^'^^ This may be simplified by putting whence -a + b + c = a + b + c-2a — 2(s-a). a-b + c = a + b + C'2b = 2{s-b). a + b-c = a + b + c -2c=2(8- c). Hence (176) becomes, after dividing by 2 and extracting the square root, In the same manner we find from (173) and (174) 2 N oc ' ■• ■! (178) ^'"^nI^^^^T- . "•^(^") From (99) we have . ^i c, ; . \ 2 cos'' — = 1 + cos ^, -i+-^6r-' ^y(^^2) 2bo * jb + cy-a' 26c * {a + b + c) (-a + 5 + c) 26c * 136 PLANE TRIGONOMETRY. _ 25 . 2(s - a) 267~' Therefore, cob~:=^'^^^. " (180) Similarly, <^o^o" = \|~^ ' 0^^) -4=nI'-^^- • <^«^) ah Dividing (177) by (180) we have ^ A j(8-b)(s-c) ,,„„, tan^=^ ^ „,„_,, \ '■ (183) «(s-a) and in a similar manner 125. To express the Sum and DilTerence of any two sides of a Triangle in terms of the other side and the opposite Angles. ' v From (165) we have a+6 sin ^4-8in B ' ", - Jl ~ sin ' ^ V BhiMA±B)coah(A-B) , " c > by (55) and (94) - sin 2" cos -g but ^+B+C=180°, therefore -4+5=180°- C, and ^(^-|-JB)=90°-^; and sin ^(^+5)=cos "2 , cos i(^+5)=8in ^. OBLIQUE-ANGLED TRIANGLES. 137 Hence we have '^ ;r> a+6 cos^cos K^-.B) "7""' : G* co8i(J.+ -B)co8 2" ~Zo3l{A-\-B)* and «+»=''«7|(I+BF- * ^ In a similar manner we find » Bin ^(A-E) ^g,^^ 126. From (165) we have t^-j-R-fc gin ^H-si n B+sin >: • - a sin A 2 ai n ^(A4-B) cos ^(.1 -^)+2 sin ^U4-.B) cos hU-\-B) 2 sm 77 COB -X- ., sin \{A\B) Uos ^(^-■B)+cos \{A^B)\ ^ sin ^ COB g- * ' A B ^ B .- ; 2 COB ^ cos y cos ^ 2 cos -^ cos ^ ^^^^ sin 2" COS g- sm ^ In a similar manner are obtained the following, which may be verified by the student : -j^^_L/. 2sm-o sm -5- -a-\-t>+c _ 2 2 ^jg9j a .A Bm«- . -.,.^ -- 2 ^ B . ^ y,^ 2 COS IT sm ■5- ^"^"^^ , ^ ^ (190) COSTT 138 PLANE TKiaONOMETRY. a+h-o 2 81112 cos 2 *:•■!(' a A ■■"" cos 77 ! '■ 2 : (a+6+c), the last four 7 • S cos 17 cos 77 8 2 2 a . ^ sin 77 2 .5.0 Sm 77 Sm 77 8~a 2 2 a . A Bin 77 2 B . 8-b cos 2 sin 2 ^ a ^ cos 77 2 . B 8-6 sin 2" cos 2 oc ^ cos 2 (191) (192) (193) (194) (196) '^V • 8 — C From these we may at once deduce those for -r, — ^— , &c. Thus, writing 6 for a in (192) and (193) we have C a cos 2 *'««2 h' %-h sin . B am 2 ^ . 2 ''''2 . 5 am 2 (19G) (197) > &C. f &C; The product of (193) and (197) gives (179), «S;o. AREA OF A TRIANGLE. 139 127. To express the sine of an Angle of a Tri- angle in terms of its sides. *i . r . '; From (94) we have sin il = 2 sin — cos — , = 2>J<-^^^^-sl'^.by(177)and(180) = ^ Js(s-a) (s-b)(8-c). (198) 128. To express the Area of a Triangle in terms of its sides and angles. Let A represent the area of a triangle; then, from the Figures of Art. 119, we have £S. = iAB. CD, and CD = AC am A. Therefore A = pcsin^. (199) In a similar manner, we find - A = |ac sin -B = ^a6 sin (7, " > . (200) that is, the area of a triangle is equal to half the product of any two sides into the sine of the included angle. Again, from (199) ..' ' A = ^6csinil 'V ==^ . - Js{s-a){s-h)(s-c), by (198) = >/s(s-a)(«-6)(s-c). (201) c sin _o From (165) we find b = . , which substituted in (199) gives _c' sin ^ sin jB c' sin A sin B 2lL^"^ " 2sin(^ + ^) • (^^^^ 140 PLANE TRIGONOMETRY. 129. To express the Perpendicular of a Triangle in terms of the side on which it falls, and the Angles adjacent to that side. Let p denote the perpendicular from C on the side c, then pc = 2A, and P='—* (203) c sin il sin 5 , , ^ , ° sin (^ -Hi?) ' ''y(202) (2(H) If the triangle is right-angled at C, (204) becomes p = c sin A sin JBf -|-sin2il, ^205) since sin jB = cos A. Examples. . .A , .b r* sin»— sin'_ cos2~ 1. In any triangle, shew that + = .^ * a 6 « From Art. 124 we have ■ 2 _(8-b)(s-c) t sin' — (^ abc dn«- , 2 (8~a)(s-c) aba * -^A . ^B sm' — sin' — therefore --+-_j_=JL^(,.„^.. J) -, since 29^a + h + e. ab EXAMPLES. 1*1 s{s-c) 1^ ab 8 ^ by (182). 2. In any triangle, prove that c" cos (^ - J5) = 2ab + (a^ + 6') coa (A + B), From (186) and (187) we have c cos l(A -B)^ia + h) cos J(il + B) c sin J(^ - 5) = (a - 6) sin J(^ + 5). Squaring and subtracting, we get cMcos«i(i-5)-sinH(^-^)} = (a + 6)'cosH(i + 5) ■ ^ ^^ ^ (a - 6)« sin" i{A + i?) c« cos (il - 5) = 2a6 {cos'* i(i + B) + sin' K^ + ^)} + (a» + 6^) {cos' K^ + ^) - «^"' i^"* ^: ^) i or c'' cos (il - 5) = 2a6 + (a^ + 6') cos (^ + 5). 3. In any triangle, prove that ,t * :, C t (a + 6)' sin'* 2" * ("* " *)' ^^^ "2 cos (il -J?) ' From the result of the last example we have c' 008 {A-'B) = 2ab - (a« + b') cos C, since (i + B) and C are supplementary angles. , G C\ Hence c» cos (il - i?) = 2a6(cos' -^ + sin" g") -(a« + 6»)(co8> 2-sin'2) i: =(a + 6)»sin»|'(a-6)'cos«g, 142 PLANE TRIGONOMETRY. ^ (a + by sin^ I - (a - b)' cos» | therefore c' = , . ... cos [A - B) 4. If the sides a, 6, c of a triangle are in harmonical progres- sion, prove that cos' — B sin A sin G 2 cos ^ + cos C * Since a, 6, c are in harmonical progression , 2ac = . n\ a + c ^ f As the sides of a triangle are proportional to the sines of the opposite angles (Art. 119), we evidently have from the above equation . „ 2 sin il sin (7 sm B = . sm ^ + sm (7 For (1) may be written c a + c 1 /a 2a _ 1 /a c \ '"2\a'^a)* sin C l/sin-4 sin (7\ , . „ 2 sin ^ sin C whence BUiB = —, — . i^ ,.- sm ^ + sin G TT o • -'^ ^ 2 sin -4 sin C Hence 2 sm — cos — = a .' 2 2 sin |(ii + G) cos i(^ " ^) * sin ^ sin (7 COS 2 COS ^(A-G) tx. e 1« 4 -^ sin .4 sin C therefore rnr „ i 2 ^ » 2sin-cos J(^-0) EXAMPLES. 143 sin il sin C ; • i . ■ 2co8^{A + C) COB ^(A-C) sin A sin C . ,-_. i^ The student should notice the following transformations de- pending on (165), which will be found very useful in the solu- tion of problems. Whenever the sides of a triangle and the sines of the opposite angles are homogeneously involved, and occur either in the numerator or denominator of a fraction, or. on opposite sides of an equation, we may substitute the one for the other, as in the following examples : 5. In any triangle, prove that (a" + 6' + c') sin -4 = a' sin -4 + a6 sin 5 + oc sin (7. a^ + l^ + c^ Commencing with 7 , we have -.•(,/.-? a^ + b^ •¥ c^ a a h h c c ab " a b a b a b ' , sin' A sin« B sin» G sin A sin B sin A sin B sin A sin B 1 /a sin il 6 sin J5 c sin (7\ sm whence (a' + 6' + c') sin il = a* sin A + ab Bin B-^ ac sin C, 6. Shew that the area of a triangle ABG ^^^^ "^^ sin'^-sin'^C • From (199) we have A =ibc sin il J ,x,j 2\S^^^ sin J5 sin C ■K^ 144 PLANE TRIGONOMETRY. 7. If the angles of a triangle ABC be bisected by the lines AD, BE, CF, meeting the opposite sides in the points Z>, E and F, shew that the area of the triangle ABG is to the area of the triangle DEF, as (a + 6) (a + c) (6 + c) : 2a6c. Here we have ^ ^ ^ sin ^^i) sin^gjg sin ^2) (7 sin DAG ~ sin ^i)5 * sin DAG ' BD AG ^AB'DG^ ^y<^^^> BD^ AG_ ^DGAB BD AB c whence and or DG~AG~~b BD^DG 6 + c BD ^ G a b + c BD c therefore BD = --^^, and i)C^ —, b+c' b+c* similarly EG —/' AE —, a + c\ a + c* a+0 a+b Let A denote the area of the triangle ABG, and 8 that of the triangle DEF, then A = J6c sin ^, whence sin ^ = — be and fin J5 = — , and sin C= -- . ^ . ad ' EXAMPLES. 145 Area of AEF= \AE . AF sin A, by Art. 128 A fee (a + 6) {a->rcy * Area of BFD = \BD . BF sin 5 ~(a + 6) (6 + c)* Area of CiE^i) = JCi> . C^ sin C Aa6 (a + c) (6 + c)* Therefore the area of the three triangles, AEF, BDF, CED, t he ac ah v " \(a + *) (a + c) "^ (a + 6) (6 + c)^('a + c) (6 + c)j' and therefore (ho etc oft \ ^""(a + 6)(a + c)"(^+6) (6 + c)"(a + c) (6 + c)j A . 2a6c , ^ (a + 6) (6 + c) (« + c)* or A : 8 :: (a + 6) (6 + c) (a + c) ; 2a6c. Examples. In any triangle, right-angled at C, prove that A ITTc \, b'-a' ... 1. cos 7r = *.|-7; — C0S2il = £5 2' 2 > 2c 6* + a2 / «x 2a6 . /i m (a + 6) (a -6) 2. 008(^-5) = -^ tan(.l-^) = ^ ^^ . 3. ton 2A = jr ^7T—\- ^^ 2il-sec 2B = j—' (6 + a) (6 -a) ^-a 4. fLti^ J2 Bin (il + ^S"! c= V2»2» cosec 2^1." c 11 146 PLANE TRIGONOMETRY. 5. A = {c^ sin 2i4 = — tan A = \{a + b + c) (a + b - c). In any triangle ABG^ prove that ^\Ti{A-B) a^-l? tan5_a' + ft'-c» ^in (^ + i^) ~ c» * t^:^^"a'-62+c'* versin .1 «(^-& + c) 7? ,«f /. ^ 7. ; — n = 77 J ^ • COS B + sm Ji cot C ■— — . versin i» 6(-a + o+c) o A ^ B s-c A B 8-b 8. tan — tan — = . tan jr cot — = . 2 2 a 2 2 8-a ^ A ^ B BO *^°2'-*^''2 c "^*2-''°*2 a •7. ■ -4^ B a- b ' A 8 - a' tan — - tan — . cot — 40 ^ J 10. COS ^ + COS ^ = 2 sin'' — . tan — tan — tan 77=0. c 2 2 2 2 s*' \, ,A B C a" ,A B G 11. cot — + cot — + cot — = — — cot — cot 77 cot — . cot- + cot- ^ 12. ^ .^ = - . 6 COS 5 + c cos C = a cos (B - C). cot 77 + cot — 13. a cos ui + 6 cos jB + c COS (7 = 2a sin jB sin C. 14. (6 + c) COS ^ + (a + c) cos -B + (a + 6) COS (7 = a + 6 + c. -4 B G 15. (« - a) tan — = (s - 6) tan — = (s - c) tan — , 6 COS C - c COS 5 6 + 16. ; — ' — = • ■ . ., . f. c 17. cot -4 + cot 5 = — cosec J5. ,. . ■. -^ - \ EXAMPLES. 147 18. c^={a + byHin' ~ + {a-by cob''^, 19. ^ (a' + 6' + c^) " be cos A +ac cos B + ab cos C. i4 J5 C 20. »(co8 il + cos -S + JOS C)=>'a cos' 7^-1-6 cos' ^ + c cos' — . 21. (6' - c') cot A + {c*' a') cot 5 + (a' - 6') cot (7 = 0. 22. cos -4 + cos -B + cos C =« 1 sin B sin C. s ABC 23. (6 - c) cot — + (c - a) cot — + (a - 6) cot -^ = 0. ^, sin il + sin C cos C - cos ^1 ,B cos A + cos G sm il - sin C 2 sin(i4-Jg) a'-6' a sin(J?-C)~6'-c'' c' V .:, ^ .: , -. 26. a' sin 2B + 6' sin 2^ = 2a6 sin C. a cos ^ cos C + cos 5 , 6 cos 5 cos (7 + cos A a' - 6' 6' - c' c^ - c^ 28. — r— sin C + — 5- sin il + ,„ sin ^ + 4 sin \{A-B) c' a'* 6^ "^ BinJ(^-C)smMC-^) = 0. • ^' .'■ ABC cot--+ cot-+ cot- ^^' ~ B abc sec - sec 2- sec 2- ,/';;(_ ^ ^ ^ c M •■■•••-'•■I - '-'^ ■' ": cos»2 cos»^ ^^ ^ 31. cot ^ - cot -4 7 — coseo (7. ^i^Aiv- 148 PLANE TRiaONOMETRT. A ^. ^ b + c 32 . Bin (- + B) cosec g = — A S C 33. In any triangle, if cot — , cot -, cot - are in arithmeti- A C cal progression, shew that cot — cot — *= 3 ; and if cot A^ cot B, cot G are in arithmetical progression, shew that the squares of the sides are also in arithmetical progression. 34. If tlie line GE which bisects the angle (7, of any tri- angle, meet the base in jE', shew that ■ < „ lah G ^ . „^ a + 6 ^ G GE = r- cos TT, tan i!AC/ = r- tan — , . a+6 2 a-0 2 and cos 5 - cos -4 = 2 cos -^ cos AEG 35. In a triangle ABGy if sin Ay sin J?, sin G are in har- monical progression, so also are the versed sines. A H C 36. In any triangle ABG^ if tan — , tan — , tan — are in Z Ji ^ arithmetical progression, so also are cos A, cos J?, cos G, 37. If p, q, r denote the lines dravn from the angles of a triangle bisecting them and terminated by the opposite sides, shew that , ,, A B & cos^ cos^ ^_i 1 i jp q r ~ a b G* 38. In any triangle prove that ,^ j A =«(»-o) tan — . *^*- '• -' v. 5s > > 39. A = OOB TT COS -jr- cos 75" . ' K- $ « A a EXAMPLES. 14f9 sin A sin B sin C s2(cps ^ + COS ^ + COS (7 - 1) ^^' ^" ^^rr+^ni? + sinC ■ ' , , siif ^ sin B 42. A«i(^»-6«)^-(jr-^. ,,.: ,:^.hi ■'•■p-^i CTa&a(cot^+ COt B) ' ■■'■•' ^^' ^ ^ 2ca cGsec^l; ' :- "■•:^-; xvi^ ^ ^'^^i'r' •■<*^' '^' .'.1 r.-i-; ■.' .{.,'J I. ..■/.■>^\ii, ■ • l....i; 'u'JfiJ "••t-i'.'; -ti':' ~i -I: .<•!•.'■ •x' ,. ,-. -.,^1.-., .. .X-. oip,i5.i;j ft ^<> vJ /^\ ,{; ><''i*.-."*{ An A : ■ I. ^",d.j H'.hI>< , •-UV-.uV-'I ^^' ^-A{cotA+cotB+cotCy '" ' 46. A= J ^ ^' 4(cOt -„■ + cot — ■ + cot — ) , n?v,!.,j 5.: -jju: ui ;»''- . ^ "" 2(cosecTTcosec B + cosec (7) ' ,. ... . 49. A = i(«^ «i^ 2^ + *' «i^ 2^)- ^"''^ ^^^^^^^^f^'-=" ''^ " ' ■••■ • -■ :• ABC ''^^'i^^ 50. A =«» tan - tan - tan -. (>) t? v> .■■} 51. In any triangle, if pi be the perpendicular from tlie angle A upon the side a, shew that "^ a + b + c b^ sin + c^ sin B P' = B d= 6 + c . cot - + cot — l'">0 PLANE TRIGONOMETRY. 52. If perpendiculars be drawn from ^he angles of a triangle bo the opposite sides, shew that the sides of the triangle formed by joining the feet of these perpendiculars are a cos -4, 6 cos B and c cos C and that its area = ^ab cos A cos £ sin 2(7. 53. If the sides of a triangle are 3^4 and 5, shew that the cotangents of the semi-angles of the triangles are 3, 2 and 1. 54. If perpendiculars be drawn from the angles of a triangle to the opposite sides, shew that the products of the alternate segments of the sides thus made = abc cos A cos B cos C. 55. If jOj, p^y p, be the perpendiculars drawn from the angles A^ B^ C oi a. triangle ABC to the opposite sides a, b, c respectively, shew that ^ p^ sin A +p^ sin B-^p^ sin C Sin B bin U 56. In any triangle shew that 2«* (sin ^ + sin 5 + sin C)» = -t-(« cos ^ + 6 cos 5 + c cos C). (aainA + b sin J? + c sinC)« -= (a" +b^ + c«) (sinM + sin^'iS + sin^C). 57. In any triangle, if — = . )^ — J^ , shew that a», 6», c» c Bin {B-C) are in arithmetical progression. 58. In any triangle shew that the distance from the middle of the base (c) to the foot of the perpendicular drawn from the 1 y-, . , 1 -J. - ■, ' c tar A - tan B angle C to the opposite side, is — , 2 tan A + tan B * 69. In any triangle shew that ^^"^ ^^ " ^^ = ^Iz^ , sin C c* 60. In any triangle prove that a sin (^-C) + 6 sin (C - ^) + c sin (A - 5) = 0. RIGHT-ANGLED TRIANGLES. 151 CHAPTER IX. ^ ON THE SOLUTION OF TRIANGLES. RIGHT-ANGLED TRIANGLES. 130. The various cases of right-angled triangles are solved by (158)-(164). ' .Vr Ex. ^.— Given ii = 65" 17' 20'' and c = 17.216, to find the other parts. (See Fig. Art, 114.) From the first of (158) we have ' a >: e sin ^, ' or in logarithms log a «= log c -1- Log sin ^ - 10. log c = 1.235932 Log sin il = 9.958290 loga-L194222 therefore a - 1 5. 6395. Again, from the second of (158) we have _- , ., :■ . , , ; . h = c cos Ay or ,,,,,,,, log 6 = log c + Log cos ^ - 10. i, ,i Log = 1.235932 ^ J. .., Log cos -4 = 9.621221 : »> ,.,.',;;^ ; '; -i - ^ log 6 = 0.857153 therefore 6 = 7 197 and ' 5 = 90°-^ = 24M2'40''. 152 PLANE TRIGONOMETRY. Ex. ^.— Given a = .1799 and 6 = .2465, to find the other parts. From the third of (158) we have tan A= — y , 1 , OP Log tan ^ = log a - log J + 10. Loga=L255031 *, ' log 6=1391817 i' - - - Log tan ^ = 9,863214 therefore -4 = 36° 7' 2r'.5 and 5 = 63°52'38".5. From the first of (158) we have a sin^* ''/'; ^^^ '^'^' '•'■' ^''^■• o^ logc = loga-Logsin ^ + 10. ,•■• Examples. 3. Given c = 332.49 and a = 98.399, to find the angles and *lie ^ase. Am. A = \T 12' 51", 6 = 317.6. 4. Given 6 = 374 and B = 52° 40' 18", to find the other parts. An8. a = 285.2, c = 470.34. 5. Given A = 25° 18' 48" and a = 8.5623, to find 6. Am. 6 = 18.1028. 6. Given a = 48 and 6 = 36, to find A, Am, ^ = 53° 7' 48". 4. RIGHT-ANGLED TRIANGLES. 15 7. Given c = 197.01 and a =196.64, to find A. By (159). log (c - a) =1.568^02 log 2c = 2.595518 2)16.972684 Log sin (45° -4) = 8.486342 2 4 therefore \ 45°-^ = T 45' 21".7 1. . . . . ■. and ,, \ =43M4'38".3 , . ; . ,1 = 86"29'16".6. -. • l.!'.l ' f \/ '• D J 8. Given a = 984.1 and 6 = .32 14, to find A and c. By (162). log (a - 6) = 2.992919 log (a + 6) = 2.993203 Log tan {A - 45°) = 9.999716 therefore , . , A- 45° = 44° 58' 52."7 and " ' il = 89°58'52".7. By (163). log a = 2.993039 Log tan 4 = 9-999858 k - V ' '■;^A yr. ■'•h'..!^ ; t^ Log tan ^= 9.999789 log(c-6)= 3.608743 c-b= 4062.027 154 PLANE TRIGONOMETRY. By (164). loga= 3.608954 Log cot ^ = 10.000211 log(c + ^/)= 3.609165 c + 6 = 4065.98 therefore h = 1.9765 and c = 4064.0035. 10. Given b = 100.56 and c = 100.64, to find A. 4ns. .1 = 2°17'5".3. r » 11. Given c = 270 and A = 18' 40", to find b. Arts. 6 = 269.9960197. 12. Given A = 89'' 20' 40" and c = 10, to find a. ., J, , -; ; Ans. a = 9.9993674. OBLIQUE-ANGLED TRIANGLES. 131. The relation investigated in Art. 119, between the sides of any triangle and the sines of the opposite angles, fur- nishes us with two equations involving the three sides and the three angles, viz. : sin -4 sin B sin (7 .it ,' ^ ~ -^ :'~^tJ -■''.. We aho have ^ + ^ + C = 180'. It is evident, then, that if any three of these quantities, except the three angles, be given, the other three m&j be iound ; for on substituting the three given parts we would hf.ve three equations, containing only three unknown quantities, which can therefore be found by solving the equations. *"" ' ' If the three angles only are given, we can find the ratios, but not the magnitudes of the sides, from the equations a sin A a sin A b sin jB ' c . a C * OBLIQUE-ANGLED TRIANGLES. 155 Hence it is evident that one side, at least, or what is equivalent to one side, must be given, in order to solve any triangle. The only cases then which can occur in oblique-angled tri- angles are the following : (1) Given one side and two angles. (2) Given two sides and an angle opposite to one of them. (3) Given two sides and a» included angle. Me,/ (4) Given the three sides. We now proceed to the solution of the different cases. Case 1. Given one aide and two angles^ or a, B and C, to find the other parts. First Solution. We have " ^ = 180'' - (5 + C)- From (165) we have ' a sin B • -n ^ a k u- J _ __ — =aam. B cosec A^ «' »* sin il or in logarithms log & = log a + Log sin B + Log cosec A -20. a sin C . ^ „ J n„f1 c=—. — r- = « sin C cosec ^. ^^J. log c = log a + Log sin C + Log cosec A - 20. Examples. 1. Given a = 512.24, ^ = 54° T 35" and C^7V V 41", to find by c and A. A = lSO''-{B + C) = 180' - 125° 9' 16" = 54° 50' 44" loga= 2.709473 loga= 2.709473 Log sin 5= 9.908652 Log sin C= 9.975744 Log cosec A = 10.087457 Log cosec A = 10.087457 log 6= 2.705582 log c= 2.772674 ' ■ 6= 507.67 c= 592.58 156 PLANE TRIGONOMETRY. 2. Given c = 71.984, ^ = 61", and C = 58° 14', to find a and 6. _ Am. a = 73.8838, 6-74.05. 3. Given ^=35° 42', i5 = 7^" 27' and c = 142, tolind a and 6. ^m. a = 89.47, 6= 149.05. 132. Second Solution. Given c, A and B. to find ihe other parts. =180- (A +JB). By (186) and (187) cos A(^+i^) . G * ..- .. . , ' sm — and . ....sinH^-i?) .in ^^ - 7?) sinA(.l+^) C * cos 77 ,. then a = ^(a + 6) + J(a - 6) and b = ^{a + b)- J(a - 6). Examples. 1. Given c = .4781, A = 70" and £ = 60° 40', to find a and h. i{A-£) = i°iO', ^ = 24° 40' logc= T679518 ■ logc= r67951S Logcos J(^- JS)= 9.998558 Log sin J(^ -i?)= 8.910404 G C Log cosec - = 10.379512 Log sec -=10.041555 log (a + 6)= 0.057588 log (a -6)= 2.631477 a + 6= 1.14179 a -6- .04280 therefore a = .59229 and 6 = .54949. , ^ 2. Given c = 100, ^ = 72" and B = 30°, to find a and ft. ^ws. a = ^7.23, 6 = 51.118. OBLIQUE-ANGLED TRLA.NGLES. 157 ' . 1' 133. Case II. — Given two sides and an angle opposite to one of thenij or a, b and A, From (165) we have sin J5 = — sin ^, a , or in logarithms Log sin B = log b - log a + Log sin A^ C=180°-{A + B), then and (165) . , sin A or log c = log a + Log sin C + Log cosec A - 20 sin (7 . c = a — — r = a sm C/ cosec A^ Since the sine of an angle is equal to the sine of its supple- ment, the equation sin ^ = — sin ^ a U ..^,:>^ cannot without other conditions, determine whether B is less or greater than 90°. Hence this is called the ambiguous case, for it is evident that there may be two triangles which will fulfil the conditions of the problem. The ambiguity, however, does not always exist, as we now proceed to shew. Ut When A < 90\ Let CAE be the given angle A. Take AC equal to 6, diaw CD perpendicular to ^ ' AE and denote it by Fic.l. p. Then from the right-angled triangle ACD we have p — b sin A. 158 PLANE I'KIGONOMETRY Fic.2. With C as a centre and radius equal to a, describe a circle. Now, if a > JO and a < 6, the circle will cut ^^ in two points, B and B\ on the same side of -4, and there will be two solutions, since each of the triangles ABC, AB'C fulfils the required conditions. li a > p and a > 6, there will be but oiie solution ; for, although there are two triangles ABC, AB'C, the latter is excluded by the condition that the given angle A is acute, while CAB' is obtuse, and there remains there- fore but one triangle which satisfies the con- ditions of the problem. lia = p there will be but one solution, and the triangle will be right-angled at B^ as is evi- dent from the figure. li a < p there will evi- dently be no solution^ for the circle will neither intersect the opposite side nor be tan- gent k) it. If a = 6 there will be but one solution, and the triangle ABC will be isosceles. FiQ.3. 2nd. When A > 90°. When the given angle is obtuse the other angles must be acute, hence there will be no ambiguity, and a must be always greater than 6. (^«*c. L, 18.) ' \^ OBLIQUE-ANGLED TRIANGLES. 169 Examples. 1. Given a = 925, 6 = 1256 and ^ = 30° 25', to find 7?, C and c. p = b sin A or logp = log b + Log sin A - 10. log 6 = 3.098990 Log sin ^ = 9.704395 log j9= 2.803385 therefore j9 = 635. 9 Since a > p and a < b, there are two solutions. To find B we have log 6 = 3.098990 " Log sin ^ = 9.704395 ar. comp. log a = 7.033858 (Art 106) Log sin 5 = 9.837243 therefore i?= 43" 25' 43".5 and 5' = 136-34'16".5 Hence, (Fig. 1) ACB = 180° - (^ + i? ) = 106" 9' 16".5 and ACB' = ISO'- {A + B')= 13° 0' 43". 5. . . To find c, we have loga= 2.966142 loga= 2.966142 Logsm AGB= 9.982504 Logsin^C5'= 9.352483 Ijog cosec A = 10.295605 Log cosec A = 10.295605 logc= 3.244251 ' . logc= 2.614530 ,T - c= 1754.9 . ';. a c = 411.65 2. Given a = .5412, 6 = .308 and ^4 = 36° 60' 44", to find £ and c (Fig. 2.) Ane. 5 = 19" 67' 16", C-.76519. I no PLANE TRIGONOMETRY. 3. Given ^ = 107" 40', a = 5.32 and 6 = 3.58, to find J9, C and c. Ans. B - 39' 52' 52", C = 32° 27' 8", c = 2 996. 4. Given A = 27° 44', a = 17 and 6 = 40.25. to find the other parts Ans. Impossible. 6. Given A = 52° 19', a = 325 and h = 333, to find the other parts. Am. i5 = 54° 10' 56" or 125 49' 4", C = 73°30' 4" or 1° 51' 56", c = 393.756 I or 13.3668. 134. Case III. — Given two sides andj>li6 included angle, or a, b and G. First Solution. To find A and B, we have from (167) tan UA - B) = r cot — 2^ ' a + 6 2 or, Log tan \{A-B) = log (a - 6) - log (a + 6) + Log cot — , 2 then ^ = J(^ + ^) + J(^-5) and B ^l{A-^B)-l{A'B). ■ To find c we have from (165) a sin G ' . sm A or log c = log a + Log sin G - Log sin A. Examples 1. Given il = 176, 6 = 133 and C = 73°, to find the remain- ing parts. . Here we have a + 6 = 309, a - 6 = 43 and — = 36' 30' r \ 2 OBLIQUE-ANGLED TRIANULES. IGl By (1G7). log (a -6)= 1.G33468 ar. CO. log (» + &)= 7.510042 Log cot ^ = 10.130791 By (165). loga= 2.245513 Log sin C= 9.980596 12.226109 Logsinil= 9.954216 Log tan ^{A -B)= 9.274301 therefore 1{A - li) ^-Wdd' 2".7 and 4(^ + i;) = 53'30'0" hence ^-64° 9' 2". 7 .5 = 42°50'57".3 logc= 2.271893 tlierefore c= 187.022 2. Given a= 16.86, 6 = 9.60 and 6^=128° 4', to iind the remaining parts. Am. A = 33" 34' 40", 7?=18°21'20", C = 24. When A and B have been found by the above method, c may be found by (186) or (1 87-), from which we have cosJ(^+i?) c=-{a-\-h) co^ ^ (A -JJ) and ^ ' sm ^{A-B) 135. Second Solution. — Given a, b and C, to find the remaining parts. . When the logarithms of a and h are given, which often occurs in the computation of a series of triangles, we may find A and B by (168) or (168 his). • ' ' Example. * ^' Given log a = 2.245513, log 6 = 2.123852 and C = 73', to find the remaining parts. (Same as Ex. 1, Art. 134.) 12 162 PLANE TRIGONOMETRY. By (168). log 6= 2.123852 loga= 2.245513 By (168 his). log b =-- 2. loga= 2.245513 log 6- 2.123852 Log tan ^= 9.878339 ^ = 37' 4'39".7 45°-^= 7''55'20".3 Log tan (45° -$)= 9.143510 Log cot ?= 10.130791 Log COS <^= 9.878339 «^ = 40° 54' 54" ^ = 20° 27' 27" 2 Log tan ^= 9.143510 * 2 Log cot 2" = 10.130791 Log tan i(^- 5)= 9.274301 Log tan J (^ - ^) - 9.274301 ^(A - B) = 10° 39' 2".7 1{A-B) = 10° 39' 2".7 the same as by (167). These methods are quite as short in practice as that of the preceding Article. The latter, Lowever, should not be used when a is nearly equal to 6. (Art. 115.) 136. Third Solution. — Given a, b and C, tojindc directhi without finding the angles A and B. From (171) we have c2 = a2 + 62-2a6cosC. This, however, is not adapted to logarithmic computation. In the following forms it is easily computed by logarithms : From (100) we have then C ' cos (7 = 1 - 2 sin'' — , Q c' = a'^ + 52 _ 2a6 + iah sin'* — (a - hf + iah mxi' ~ OBLIQUE-ANGLED TRIANGLES. 163 4a6 sin' — - ('^ -'')'[' '-rrrw' iab sin' — Let ^^^'^-- (a-b)'^ • or then we have /-f . C' 2 vao sin - tan0 = r— , (206) a- c2==(a-6)2(l + tan2(9) = (a-6)'sec'^ and c = (a -6) sec (9. •' ' ' (207) When a is very nearly equal to 6, the denominator a-b will be very small and 6 will be near 90"; in that case the fol- lowing form will be preferable : ^ c2 = a'-l-6'-2a6 cosC = a' + 6'-2a6(2 ^'|-1) by (99) = (a + 6)' - 4a6 cos' — 2 4a6 cos' — I ' . ■ ■ .^ / 'iao cos- —\ Now since the first member of this equation is positive, th(^ iab cos' 2 second member must be so likewise, therefore ^ ^u must be less than 1. Hence we may assume iab cos' — sin' B - . ,y, {a + by 164 PLANE TRIGONOMETRY. 2jai C cos or then we have and sin ^ = ' (a + 6) ' = (a + by cos" e c = {a + b) cos $. Example. (208) (209) Given a = 176, 6 = 133 and C = 73", to find c. (Same as Ex. 1, Art. 134.) By (208) and (209). log a = 2.245513 log 6 = 2.123852 log (a + 6) = 2. 489958 Log cos ^=9.781934 2)4.309365 log Va6" = 2.184682 log 2 = 0.301030 Q Log COS -^ = 9.905179 2.390891 log (a + 6) = 2.489958 log c = 2.271892 c= 187.022 The remaining angles may now be found by (165). Log sin ^ = 9.900933 . ^ = 52° 45' 12". 137. Case IV. — Given the three sides, or a, b and c, to find A,B,C. . First Solution. By (198) we have ,, in J. = — is/s(s -a) (s- b) (s -c), sm be (210) ■which may be i. ed when A is not near 90°. Similar formulae of course exist for sin B and sin C OBLIQUE-ANGLED TRIANGLES. 165 Second Solution. From Art. 124 we have si.-2=s/ f^ B_ ks-a){s-c ) ''''2-Nl ^c ' 2 \ ab Third Solution. COS COS 2 B C COS — 2 Is (a - a) t Fourth Solution. tan -- = 2 8(8 - a) tan tan B 1 (8 -a)(8 - c) \l 8(8- ,(8-h) in which C^_ Us -a)(s- b) 2"\/~7(s-c) .'i^ (211) (212) (213) When all the angles are required, the last group will be the most convenient, since only four different logarithms are required from the tables. Tlie first group is to be preferred when half the angle is les8 than 45°, and the second when it is greater tlian 45°. The third group is accurate for all angles. 106 PLANE TRIGONOMETRY. • Examples. 1 Th(i sides of a triangle are 13, 14 and 15 ; find the anglea L^t a = 13, 6 = 14 and c = 15 ; then s = 21, 8-a = 8, s-6 = 7, «-c = 6. To find A. From the first of the last group, we have log(«-6)= 0.845098 log(s-c)= 0.778151 ar. CO. log 5= 8.677781 (Art 106) ar. CO. log {8 -a). 9.096910 2)19.397940 A Log tan -= 9.698970 therefore and To find B, 2" = 26° 33' 54'' ^=63' 7' 48". therefore and log(s-a)=: 0.903090 log(«-c)= 0.778151 ar. CO. log 5= 8.677781 ar, CO. log (s - 6) = 9. 154902 . 2)19.513924 Log tan-- 9.756962 - = 29M4'41".6 5 = 59° 29' 23".2. OBLIQUE-ANGLED TRLA.NGLES. 167 Tojmd C. log(s-a)= 0.903090 log(«-6)= 0.845098 ar. CO. log 8= 8.677781 ar. CO. log (« - c) = 9.221849 2)19.647818 Log tan -= 9.823909 therefore and Verification C = 33°41'24".4 C = 67''22'48".8. ^ + ^ + (7 = 180^ The following transformation of (213) will facilitate the computation when all the angles are required. Multiply the numerator and denominator of the second member of the first of (213) bj^s - a, and we have tan 1 Us -a -a) (s - b) {s ~ c) 8 Put #- -a) {s - b) {s - c) = r. then also 8 tan- = 2 8-a ^ B r tan — = 2 8-b . C T tan- = 2 «-c (214) 2. Given a = 1468, 6 = 1359 and c = 1263, to find the angles. Ans. A ==67" 58' 51", ^ = 59° 7' 4", ^=52° 54' 5". 1G8 PLANE 'VRIGONOMETllY. 3. Given a = ^56, 6=1 and c = 7, to find A. Am. ^ = 115° 22' 37". 4. Given a = 5, 6 = 4.037 and c = 3.9575, to find A. Ans. ^ = 77° 25' 12". 5. Given a = 25, 6 = 30 and c = 20, to find i4, 5 and C. Am. A = bb° iC 18", 5 = 82° 49' 8", (7 = 41" 24' 34". 6. Given a = 39, 6 = 35 and c = 27, to find A, B and C. Am. A = 7(>' 45' 21", 5 = 60° 52' 33", C = 42° 22' 6". 138. Given tioo sides and the included angle of a triangle^ to find its area. From Art. 128 we have A = |6c sin A = \ac sin B = |a6 sin C. Ex. i.— Given a = 30, 6 = 40 and (7 = 28° 57', to find the area. loga=^.477121 log 6 = 1.002060 Log sin (7 = 9.684887 ar. CO. log 2 = 9.698970 log A = 2.463038 A = 290.427 JUx. 3. — The sides of a triangle are 103.5 and 90, and the inelided angle is 100°, find the area. Am. 4586.75. 139. Given two angles and the included side of a triangle, to fii.d tlie area. From (202) we have c' sin A sin B 2 ' Bin{A'VB)' EXAMPLES. IGO Ex. 1.— Given A = 80% i?=:60° and c = 32 feet, to find th( . area. Log sin ^= 9.993351 Log8inj5= 9.937531 Log cosec (A + B) = m 191933 2 logc- 3 010300 ar. CO. log 2= 9 698970 logA= 2.832085 A= 679.33 square feet. Examples. 1. In the ambiguous case of triangles, given a, b and A, shew that if Cj, Cg be the two values of the third side, Ci + C.2 = 26 cos Af CjC2 = b^~ a\ and that the area of both triangles = 16^^ sin 2 A. From (169) we have a^ = P ■\-c^ - 2bc cos A *■ or c2-26cosyl.c + 62_^2^0 (1) If Ci and Ca be the two values of c in tliis equation, we have by the theory of quadratics, Ci + C2 = 2b cos A and c^Ci = b^~ a\ Now, AI) = '-l±^' = b cos A, therefore q + Cj = 2b cos A , Again, c,c, =^ABx AB' ^AP'. {Euc IIJ , 36.) ^AC-PC^ ^b'^-dK By solving (1) for c we have c = 6 cos vl ± sjoir - h^ sin^ A. (2) 170 PLANE TRIGONOMETKY. Hence Ci = 6 coH ^ + Jn"^ - P sin'' A and Cj = i cos ^ - \lci^ - h^ sin' A. Area ABC = Ihc^nxnA = U'-'sinylcos^ + ^6sin^ sld^ - b'^ sin''* A. Area AB'C= hbc^nmA = Jji^gi^vlcos^ - ^isin^l Jd^ - b'^ sin*'' A. therefore the area of both triangles = b'^ sin ^ cos -4 From (2) it is seen that when a = & sin ^, c = 6 cos A. whirh gives only one value for c and the triangle is right-angled at B. 2. Given the perpendiculars from the angles of a triangle upon the opposite sides, to find the sides and angles. Let Pa, Pb) Pc denote the perpendiculars on the sides a, b, c respectively ; then by (203) 2A 2A 2A Pa = -^, Pb-J-. Po=--* 2A , 2A 2A or o, — ) = , c = f Pa' Pb' Pc Substituting these values of a, 6, c in (172) we have 4A'' 4A2 4A^ . Pl Pl Pl 2 ^^ 'Pb Po Pl"^ pl ' pi _ PlPc + PaPb - Pb Pc 2 ^pfiPbPe PbPo which determines A, Then & = -^» ^--^' ^^' sin A . sin ^ EXAMPLES. 171 3. Given the sum of the sides a and h = k, the side c and the angle C, of a triangle, shew that a = k cos'* ^ , h = k siv? ^ , . ^ . n/A^c^ C where sin ^ = ± — sec - . k A From (171) c' = a^ + i^ - 2a6 cos G = a2 + 2a6 + h' - 2ab (1 + cos C) Q = J^ - iah cos^ — , C therefore 4a6 = (J^ - c^) sec'^ — A which is possible since {a + 6)^ > 4a6, then 1 - 7 rrr, = 1 - sin'* 6 ' ' a-h or r = cos a + 6 2a + 6 1 + cos ^, or -?- = 2 cos'* ~r k 2 and a = A; cos^ — - 2 Also r = 1 - cos 6 a + OP 7- = 2 sm^ -^ A; 2 and h = k sin' — . A 172 PLANE TRiaONOMETRY. This prohlem may also be easily solved by (18C) and (187). Thus, from (186) we have cos 1{A -B) = - con i(A + B) G h . Hence the angles are known. Fiom (187) we have a — o = c »» , ■ . cos 2 Ilonce the sides a and b are known. 4. The sides of a triangle are 13, 14 and 15, find tho area. Ans. 84. 5. The sides of a triangle are 13. 20 and 21, find the perpen- d cuLar on the longest side and the area. Ans. Perpendicular =12, area = 1 26. 6. Tho sides of a triangle are 25, 6.l and 52, find the area and the angle opposite the shortest s?de. Ans. Area = 624, angle = 28" 4' 21". 7. The sides of a triangle are 137, 111 and 124, find the area. Ans. 6510. 8. The angles of a triangle are 70°, 60" and 50°. and the perimeter is 150, find the sides. (See formula 192.) Ans. 64.81. 50.51, 44.68. 9. The sides of a triangle are 85 and 75 and the included angle is 75°, find the other angles. Ans. 57° 10' and 47° 50'. EXAMPLES. 173 10. The ratio of two sides of a triangle is 7:3, and the angle they contain is 6° 37' 24", tij^d the other angles. Ans. 168° 27' 25".4 and 4° 55' 10".6. 11. The aides a, 6, c of a triangle are as 8 : 10 : 16, find the angled. Ana. J5 = 55° 46' 16". 12. Given two sides of a triangle and the difference of their opposite angles, to solve the triangle. (See formula 166.) 13. In any triangle prove that tan - = ^ ^ -, whore d is the difference between the sides a and c, and p is the per- pendicular upon the side h, and that tan B = ; — - . c-b cos A 14. If a, y8, y be the perpendiculars from the angles of a triangle upon the opposite sides, prove that a? be 15. In the ambiguous case, if A and 8 be the areas of the two triangles, prove that ; ' ' • , i ; 1 »\ A'^ + 8'''-2A8cos 2^1 = ^, (A + 8)' a, b and A being given. 16. In the ambiguous case, if c^, c^ be the values found for the side c, when a, b and A are given, prove that 'i c* + cl- 2c ^c^ cos 2 A = 4^2 cos^ A and ^! ~ ^8 = 4^ *^^^ ^ "^^ ~ ^^ sii^^ A. , 17. In any triangle, given a, B and the sum of the other two sides (equal to m)j to solvo the triangle. • ' . C m-a B Ans. tan — = cot — . 2 m + a 2 18. If a, )8, y be the distances from the angles A, i/, (7 of a 174 PLANE TRIGONOMETRY. triangle to a point P within it, from which the sides subtend equal angles, find the sides and the area. Ana. The sides are J{jS^ + y^ + ^y) si {o? + -/ + ay) and the area is -— - (a/3 + ySy + ay). 19. In a triangle, given C, c and ab = m\ to find the sides a and b. Ans. a + b — c sec 0, a-b = c cos 0, , . /, 2w C , . ^ 2m . G where tan v = cos — and sin d> = sm — . c 2 c *>. 20. In a triangle, given C, the perpendicular from C=p and a + b = wi, to find c. Ans. c = m tan — , where tan = — tan — . J p 2 ^^ 21. The perimeter of a triangle is 100 rods, ^ = 102° 51' 30", J5 = 25° 42' 45" and (7 = 51° 25' 45"; find the sides. Am. a = 44.51, 6 = 19.8, c = 35.69. 22. The perimeter of a right-angled triangle is 24 rods, and one of the angles = 30° ; find the sides. ■ Ans. 5.072, 8.784, 10.144. 23. Given the base a, the vertical angle A, and the difference of the other two sides = c?, to find the sides. (See formula 187.) 24. Two adjacent sides of a parallelogram are a and b, and the included angle is 6 ; shew that the diagonal drawn from this angle is ' . . : . , ■ sj{a'^-^b'^-\-2abQose), • ', and that the other diagonal is ^ EXAMPLES. 175 25. The sides of a triangle are 17, 25 and 28 ; find the area of the triangle formed by joining the feet of the perpendiculars drawn from the angles to the opposite sides. (See problem 52, Chapter VIII.) Arts. 24:^V4V 26. In a right-angled triangle, the sum of the hypothenuse and base is 2986, and the angle at the base is 52°, find the perpendicular. Ans. 1456.37. 27. Two sides of a triangle are 356 and 294, and the angle opposite the latter side is 51° 27', find the other side. Ans. 316.309 or 127.4079. 28. The perimeter of a triangle is 128, the angle C = 28° 4' 21" and the perpendicular from C on c = 49.92, find the side c. • Ans. 25. 29. Given 6 = 39, c = 51 and B = 115°, to solve the triangle. . An^. Impossible. Why 1 30. In a triangle given (7, c and d^ - b^ = 7ri?, solve the triangle. •. '^^ - n , Ans. sin {A - B) = — ^ sin C. When will there be two solution*? ; ' : ■.^.;Vr^'-:.' M!-., iX 176 PLANE TRIGONOMETIIY. CHAPTER X. APPLICATION OP TRIGONOMETRY TO SURVEYING, NAVIGATION AND ASTRONOMY. 140. In this chapter we will give some examples shQ;wing how the formulae of Trigonometry are applied in the solution of some very interesting and useful problems in Surveying, Navigation and Astronomy. Horizontal and vertical angles are most conveniently measured with a theodolite. This instrument is composed essentially of two graduated circles having their planes perpendicular to each other. When the instrument is in use, one of the cix'cles is placed in a horizontal position by means of spirit levels. On this circle horizontal angles are measured ; on the other are measured vertical angles whether of elevation or depression, that is, whether the object is above or below the horizontal line passing through the centre of the axis of the vertical circle. A telescope is attached to the axis of the vertical circle, by means of which a clearer view of the object may be had. The sextant and reflecting circle are employed to measure angles in any plane whatever. They are the only instruments which can be conveniently used at sea for measuring the altitudes of the heavenly bodies above the horizon. For the descriptions, adjustments and modes of using these in- struments, the student is referred to treatises on surveying, such as Gillespie's Land Surveying, or Sims's Treatise on Mathe- matical Instruments. Their construction and adjustments will, however, be best learned by a careful study of the instruments themselves. We V HEIGHTS AND DISTANCES. 177 shall therefore suppose that the manner of adjusting and apply- ing them to practice is known, and proceed at once to give a collection of problems which they enable us to solve. 141. To find the distance 0/ an inaccessible object upon a horizontal plane. Let it be required to find the distance from A to an object C situated on the opposite bank of a river. Measure a base AB, whose length we will denote by c. Measure also the horizontal angles CAB, CBA; then (Art. 131) AC_ AB sin B whence or in logarithms AC = c sin (7' sin B »• 11 :A'i'- (■■ ' i V sin {A-\rBy ^•'>-- log ^C = log c + Log sin B - Log sin {A + B). This problem may also be solved without the use of any instruments for measuring angles, as follows : Having measured fhe base AB as before, measure any length Ad along AG and an equal one Ag along AB; then measure dg. Similarly mea- sure Bm and Bn, equal to each other, and then measure mn. Bisect dg in h and join Ah, then since Adg is an isosceles triangle Ah bisects the angle A, and we have 13 178 PLANE TRIGONOMETRY. similarly 81112 = . B sin :r- = Id'' mk Bm gd inn '2 Bm Hence the angles A and B are known by the aid of the tables, and the solution may now be completed as before. 142. To find the height and the distance of an inaccessible object standing on a horizontal plane. Let DC be the ob- ject. Measure any base AB, and denote its length by c. Measure the horizontal angles CAB (a) ABC (13) and the angle of elevation CAD (y). Then from the have triangle ABC we AC^c sin B ; = C' sin P sin (7 "sin(a + /3) Ih the right-angled triangle ACD we have AC^ just found, and the angle CAD (y) to find DC. Hence DC = AC i&nCAD sin /8 tan 7 ~ = Or thus : At any convenient point A measure the angle of ele- vation CAD (a); then mea- . sure a base AB (c) directly from the object, and at -5 -g measure the vertical angle sin (a 4- j8) HEIGHTS AND DISTANCES. 179 ABD (/3). Then the angle ^Z) B = a - /?, and from the triangle ^^i) we have by (165) c sin )8 AD=: sin (a - yS) * then in the right-angled triangle ADC DC = AD Bin CAD c sin a sin /3 and AC = AD COS CAD c cos a sin j8 sin (a - /8) •■'-:'», i-;;o 143. To find tlie distance between two incuccessible objects on a horizontal plane. » ... Let G and D be the two objects. Measure the base AB equal to a, and the angles CAB^ DAB, ABC, ^i?A' which put equal to a, )8, y, 8 respectively. Then from the tria^^le ABC we have by (165) *n An sin^5C smy AG = AB —. 77v-9r=a sin ^C-S sin(a + y)* sin ^5i) sin 8 similarly, ii) = ^5 ^^^j^ = « ^1^(^ + 8) ' Then in the triangle ACD, the sides AC, AD and the in- cluded angle CAD = (a - (3), are known, therefore CD can be found by the methods of Arts. 134-136. 180 PLANE TRIGONOMETRY. 144. Three inaccessible objects A, B, C (not in the same straight line), situated on a horizontal plane and at known dis- tances from each other, are visible from another point P on the plane, and the angles A PC, BPC are given; to find the distance from P to each of the objects A, B,C. Let A, B, C be the three inaccessible objects, and P the point of obser- vatioa. Let the observed angles APG, BPC be de- noted by a, )8 respectively PAChye,QXidiPBChy. Since the sides of the triangle ABC are known, J^ the angle C can be found by (179) or (182) ^ + <^ + a + ;8 + C = 360°, , ., , ., therefore K^ + <^) = 180''-i(a + )8 + O). ' From the triangles APC, BPC, we have by (165) 6 sin ' - '^ a sin are supplementary, ^ = 45°, and the solution fails when the point P is without the triangle, in which case the quadrilateral AGBP is inscriptible in a circle. , . 145. To determine the height and distance of an inaccessible objecty by having given its angles of elevation observed at three points at given distances from each other, and in the same straight line. Let DE be the object, A, B, C, the three stations of observation: let AB = af BC =- 6, and DF - x; and let the observed angles of eleva- tion at Aj B, C, be a, ^, y, respectively : Then AD = x cot a,BD = x cot l3,CD = x cot y. Draw i>/ perpendicular to ACy then in the triangles ABB, BCD, we have by Fuc. II., 12, 13, ,. . . x^cof^ a = a'^ + x^cot^ ^ + 2a. Bf a;2 cot*'' y = 52 + a;2 cot^ )3 - 26 . 5/: Eliminating Bf and solving for x, we have i i. >=\|^ ab (a + b) cot^ y-{a + b) cot^ /i + b cot^ a ' And thus AD, BD, and CD become known. 182 PLANE TRIGONOMETRY. 146. A flagstaff of kriown length, standing upon the top of a tower of known height, subtends at the eye of a spectator on the same horizontal plane as the tower, a given angle ; to find tJie distance of tlie tower. Let AB = a, the height of tower ; BC = h, the length of the staff*; P, the position of the ob- server, at which the staff" sub- tends an angle BFC = fi. Let AFB=^e, then APG==e + /3. AF = a cot e and AF=(a + h) cot {6 + ^), whence . a tan (^ + /8) = {a + h) tan 6 X and tan 6 = ^ cot ^ ± ^ A'' cof^ ^ - 4a (a"T X) 2{a + h) ' therefore AP=\ { A cot /8 + Jh^cot^ ^-ia (a+T) } , from which we see that there are two values of AF, unless K^ cot'* ^ = ia {a + h). If a segment of a circle be described on BC containing an angle equal to yS, it will in general cut the horizontal line in two points, F and Q, corresponding to the two values of AF. When A cot ^ = 2 J a (a + h), the circle will touch the hori- zontal line, the points F and Q will coincide, and ^ will then have its maximum value, so that tan $= \ 7 . \a-hA' and then / ' AP= J a {a + h), \Euc. III., 36) Or we may proceed thus : From the centre E draw ED, JS^iJ perpendicular to CB and HEIGHTS AND DISTANCES. 183 AQ; join CJE and EP ; the angle CFD = CPB = CQB = ^. In the right-angled triangle, CDE^ we have h h DE tan )3 = -, or BE = — cot fi^AH, and CE sin ^8 = ^, or 0^ =\ cosec /? = ^P. Pir= JEP'-EH^ = JEP'-AD^ -J — cosec* B - (a+ — ) 4 ' ^ 2 therefore ilP^^fi'- PZr=i{7i cot )8- ^/t^cot^^- 4a (a + A)} and iig = ilfl' + Pir=j{Acot^+ Jh^cot''fi-ia{a + h)} the same as before. -/;,.;. 147. In Surveying, the course or bearing of a line, is the angle which it makes with the meridian passing through one extremity, and is reckoned from the north or south point of the horizon, toward the east or west. N Thus, if I^S represent the meridian and the angle B'PQ be 30", then the bearing of PQ from P is 30'' to the east of north, or as usually read " north thirty degrees east " and written N. 30° E. ^ The reverse bearing of a line is the bearing taken from the other extremity. Thus the bearing S of P^ from ^isS. 30* W. 184) PLANE TRIGONOMETRY. In Navigation, the course of a ship is generally referred to the points of the Mariner's Compass which consists of a circu- lar piece of card board attached to a magnetic needle, and is so balanced that it can move freely in any direction. The circumference is divided into thirty-two equal parts called points, and each point into four equal parts called quarter ooints, The points are read as follows, beginning at the north and going east : " north, north by east, north-north-east, north east by north, north east," and so on as seen in the figure. The angle between two adjacent points is W° or 11° 15'. A quarter point is therefore 2° 48' 45", a half point 5° 37' 30", and a three-quarter point 8° 26' 15". LEaiGTH OF A DEGREE OP LONGITUDE. 185 148. To find the length of a degree of longitude on any 'pa/rallel of latitude. Let V be the pole of the earth, C the centre, EQ an arc of the equator containing one degree, and AB ^rv arc of one degree on any parallel of latitude. Join CE^ CQ and draw AD and BB perpendicular to the raalus PC ; then it is evident that the angle ADB is equal to the angle ECQy and therefore CQ : BB :: EQ : AB. But regarding the radius CQ as unity, DB is the sine of the angle BCF or the cosine of the angle BCQ, that is, the cosine of the latitude : hence 1 : cosine of the latitude : : EQ : AB and therefore AB = EQ cos. lat. From which it follows that similar portions of different parallels of latitude are to each other as the cosines of the latitudes. «. , Ex. — Find the length of a degree c )ngitude at Toronto, latitude 43° 39' 24". The equatorial radius of the earth is 3962.8 miles, but in consequence of the flattening in the direction of the polar diameter it is only 3956.514 miles at Toronto: and consider- ing th5 earth a sphere with a radius of 3956.514 miles, the length of a degree of longitude on the equator would be 69.054 miles =EQ. " • '■vv'.'. ',-■-■■'■■•'.•. ^ -- ■- •'■.:'.■:•-■ ,■ ;-i'^r:- Hence ' ? - log ^^= 1.839190 '"' Log cos lat. = 9.859432 Vr«'i ,..,, .n ■ . . log ^5= 1.698622 : , therefore .4^ = 49.96 or 50 miles very nearly. 186 PLANE TRIGONOMETRY. As a degree in longitude makes a diflTerence of four minutes in time, it follows that fifty miles east or west on the parallel of Toronto is equivalent to four minutes difference in time. 149. To Jmd the MoorHs distance from tlie Earthy a/nd her diameter. ; . Let PF be the earth's axis, JEQ the equator, and A and B the positions of two observatories on the same meridian and whose latitudes ACQ, BGQ are known. Draw the radii, CAs CBy and produce them to meet the celestial sphere in Z and Z' ;' * through A and B draw AH, BH at right angles to CZ and GZ' respectively, then Z and Z' are the zeniths, and AH and BH are the horizons of A and B respectively ; let M be the moon, and when she is on the meridian of the observatories, let her altitudes MAH, MBH be measured with a sextant, theodolite or any other suitable instrument. Join AB, then regarding the earth a sphere in order to render the problem as simple as possible, we have in the isosceles triangle ACB, the sides AG^ GB and the angle AGB, the sum of the latitudes given, hence AB and the angles GAB, GBA can be found. The angle HAB = HBA=\iqM of ACB, and since MAH and MBH are deter- mined by observation, the angles MAB, MBA are known; hence AM and BM can be found. Finally, in the triangle ACMf PARALLAX. 187 the sides AM^ AG and the included angle MAC are now known, therefore MG can be found, which is the distance sought. From A draw AN tangent to the moon and join MN, then ANM is a right angle and the angle MAN the moon's apparent Bemi-did.meter can be determined by observation, therefore MN= AM sin MAN, whence her radius is known and therefore her diameter. ■ Parallax. 150. The parallax' of a celestial body is, in general, the apparent angular displacement which is produced by viewing the body from two different points. The term is used in astronomy to express the difference of altitude or zenith dis- tance of a celestial body when seen from the surface and the ' centre of the earth respectively. Let G be the centre of the earth, P the place of an observer on its surface, M a celestial object seen in the horizon, M' the same seen at the zenith distance M"PM\ and M" the same object seen in the zenith. Now it is evident that when the object is at M" it will ap- pear in the same direction whether viewed from P or G ; there is then no displacement and therefore no parallax. If the object be at M\ its apparent direction is PM' -while its true direction is GM\ and the angular displacement TM'G is the parallax due to the zenith distance M"PM' or to the altitude MPM. It is evident that the angle PM'G increases sis the body 188 PLANE TRIGONOMETRY. approaches the horizon, and when it is in the horizon as at M the parallax PMC has its maximum value, and is then called the horizontal parallax which is, in fact, the largest angle which the earth's radius subtends at the object. Parallax increases the zenith distance and consequently diminishes the altitude. The angle M"FW is the apparent and M"CM'the true zenith distance of the object when at M'. Hence, to obtain the true zenith distance from the apparent, the parallax must be subtracted; and to obtain the true altitude from the apparent, the parallax must be added. It is evident from the figure that the efiect of parallax is wholly in the vertical circle passing through the observer and the body, the azimuth or angular distance from the meridian of the observer is therefore not affected. 151. To Jind the parallax at any altitude when the hori- zontal parallax is given. Let P=(7i/P, the horizontal parallax, p = CM'F, the parallax in altitude, Z' - M'PM", the apparent zenith distance, R = CM= GM\ the distance of the object, r = (7/*, the earth's radius, then in the triangle CPM' we have ^mCM'P ^ CP^ ,;,;.,!. ■ ■;■ .;, ■, ... . sinCPJ/' ~ CM' ,v.-5 . . .., „ sin jo r . • ; or -: — ^= ^=smP, sin Z R ,, , , whence i') ^/_ . sin p ::= sin P sin <^'. ,.^ , ,.. (215) In the case of the sun and planets, the horizontal parallax is so small that we may, without sensible error, use the more convenient formula p"=P' Bin Z'. . ,;.;>!.:. • PARALLAX IN ALTITUDE. 189 Jjf the true zenith distance M'CM" = Z he given, instead of the apparent, we have ' sin CM'P CP I i sin GFM' CM sinp r . _ or -: — 7-^ r = — = sinP; (a) sm {^+p) R hence sin (Z+p) + sin p 1 + sin P sin (Z + p) - sin p~ 1 -sin. P^ tan (-+;?) p or —4 tanM45° + --), tan- Z P Z whence tan (— +^) = tan2 (45° + — )tan--. (216) This equation males known ^ Z+p^ from which we obtain Z p by subtracting — . ■ ' . ^ -, ■ ^- , -) yi'.^^.^//''h t1^ '■■h:;t;;^ ^^'f Another solution of equation (a) will be given in a subse- quent chapter. • . . . .* . ^ ^ . .:■■■.■■■.■' ',;,><;.■■} --;■■.;; ,\.^,^i ^'.kX-- 152. To find the augmentation of the MoorHs semi-diameter on account of her altitude above the horizon. The apparent diameter of the moon is the angle which her «lisk subtends, and is not the sa'- ■ for all points on the earth, on account of their different distances from her. Supposing the moon's distance from the centre of the earth to remain constant, her distance from the observer must dim- inish as her altitude ajaove the horizon increases, and therefore her apparent diameter must increase. This increase of the apparent diameter is called the augmentation of the diameter due to altitude. 190 . PLANE TRIGONOMETRY. In the figure of Article 151, let 8 denote the moon's apparent semi-diameter at i/, and 8' the augmented semi-diameter when at M\ then we shall evidently have 8:8':: PM' : CM' or CM. 8 PM' sin PCM' sin Z' B'~CM sin CPM'~ am Z* sin Z slKZ'* whence ^' = ^^^^ (217) .. Examples. 1. To determine the distance of a ship at anchor at (7, a base line ^5 of 100 rods is measured along the shore, and the angle ABC was observed to be 83*' 18' and the angle BAC 32° 10' ; find the ship's distance from B. Ans. 58.96S rods. 2. From the decks of two ships A and B, 880 yards apart, the angle of elevation of a mountain C due north of both ships was observed at each; at A the angle was 35° and at B 64°; required the height of the mountain above the surface of the sea, the deck of each ship being 21 feet above it. . , . • , . , , , Am. 942.75 yds. 3. A tower subtends an angle of 39° at a distance of 200 feet from its base, what ejiigle will it subtend at a distance of 350 feet from its base 1 Ans. 24° 49' 53".5. 4. There are two monuments whose heights are 100 feet and 50 feet respectively, and the line joining their tops, when produced, makes with the horizontal plane on which they stand an angle of 37°; find their distance apart. Ans. 66.352 feet. 5. To determine the distance between two inaccessible rocks C and Df in the sea, a base line AB of 670 yards was measured EXAMPLES, ^ - - 191 on the shore and the following angles were measured at the extremities of the base: BAD^^O" 16', BAC = Q7° 56' ; ABC = 42° 22', ABD^IW 29'; find CD. Arts. 1174.26 yaxdg. 6. The hypothenuse of a right-angled triangle rests on a horizontal plane, and is 100 feet long; one of the angles is 36° 40' and the inclination of the triangle to the horizontal plane is 60° ; find the height of the right angle above the plane. Arts. 41.48 feet. 7. An object stands on the top of an inaccessible hill, and from a certain point on the horizontal plane the angle of eleva- tion of the top of the hill is 40°, and that of the top of the object 61°. Going back 100 yards in a direct line from the object, the angle of elevation of the top of the object was then 33° 45' ; find the height of the object. Ans. 46.663 yards. 8. The angle of elevation of a tower 100 feet high, due north of an observer, was 50° ; what will be its angle of elevation after walking due west 300 feet? Ana. 17° 47' 50". 5. 9. "Wishing to find the distance between a battery at B and a fort at Fj which cannot be seen from the battery in consequence of the ground between the battery and fort being covered with a forest, distances BA and AC to points A and C where both the fort and battery were visible, were measured, the former being 2000 yards and the latter 3000, and the angle BAF was found to be 34° 10', FAC 74° 42', and FCA W 10'; find the distance between B and F. Ans. 5422.3 yards. 10. The elevation of a balloon was observed to be 20° bear- ing N.E., and by another observer 4000 yards due south of the former its bearing was found to be N.b.E.; required its height. Ans. 511.24 yards. 11. Coming ift from sea, at a certain point P I observed 192 .PLANE TRIGONOMETRY. two headlands ^ and 5, and inland at C a tower which appeared between the headlands : the distance between the headlands was 5.35 miles, and the distance from A to the tower was 2.8 miles, and from B to the tower 3.47 miles. I observed the angles APC, BPC, the former being 12° 15' and the latter 15° 30'; required my distance from each of the three objects. A-,v6. ^P=11.257; CP=12.7523; i?P= 11.034 miles. 12. Bequired the distance of the three objects Aj B, C from the point P situated within the triangle ABC, from the follow- ing data: ^5 = 267 rods, ^C = 346 rods, BC = 209 rods, angle ^P(7=128°40', angle^P^ = 9r 20'. . • Aiis. ^P= 248.854; J?P = 91.134; CP= 130.81 rods. 13. While sailing along a coast a headland G was observed to bear N.E.b.N. ; having run E.b.N. 15 miles to B, the head- land bore W.N.W.; find the distance from the headland at each observation. Ans. 8.499, 10.81 miles. 14. ^ and B are two points lying N. and S., and 50 rods apart ; what must be the distance of a third point C from each, that it may bear N.b.W. from B and W.b.S. from A f Ans. 9.754 and 49.04 rods. 15. The courses of two ships are N. and E., and their rates of sailing are equal, the bearing of the former -from the latter was E.N.E, ; but after each had sailed ten miles the bearing of the latter from the former was S.S.E, ; find the distance be- tween the ships at the first observation. Ans. 7.653 miles. 16. The elevations of two mountains, in the same line with the observer, are 9° 30' and 20°; on approaching two miles they both have an elevation of 38° 15'; find their heights and the distance between them. ' ' ' ' * >^>*-^!' '--'■>■ v>;i. An^. The nearer, 747.77 yds. ; the more remote, 2380 yds. The distance between them, 2070.48 yds. " ' EXAMPLES. • 11)3 17. From a ship sailing N.W., two islands appeared in sight, one bearing W.N.W., the other N, and after sailing eight miles farther the first bore W.b.S., and the other N.E. ; required their bearing and distance from each other. Ans. S. 58° 40' 50" A W.; distance, 12.95 miles. 18. Two ships sail from the same port at the same time, the one due N. at the rate of six miles per hour, the pther on a course N. 60° E. at the rate of ten miles per hour for two hours, she then tacks to cut the other off or to overtake her ; how far must she sail to do it, and on what course ? Ans. 23.75 miles, on a course N. 46" 49' 35" W. 19. From two stations on the deck of a ship 100 feet apart, the bearings of an object on shore were N.E. and N.N.E., and the ship's head was N.b.W. ; find the distances of theobject from each station. Atis. 145.177 and 217.27 feet. 20. A cape C bears from a headland II, W. ^ S. 4.23 miles; how must the cape bear from a ship which runs in towards the headland on a course N.b.W.^ W., until the headland is 2.3 miles distant from the ship? ; Atis. W.H^f.W. : 21. A ship was 2640 yards due south of a lighthouse, and after sailing N. W.b.N. 800 yards, its angle of elevation was 5° 25'; required its height. Ans. 191.94 yards. 22. A tower, 65 feet high, subtends an angle of 60° at the eye of an observer standing on the same horizontal plane as the tower ; find his distance from it, his eye being five feet above the plane. ,. ,. , Atis. ii.S ieet. 23. A lighthouse standing on a rock, is observed from two points in a line with it, and one quarter of a mile apart ; from the nearer point the elevations of the top and bottom are 52° 14' and 48° 38' respectively, and from the more remote point 14 194 PLANE TRIGONOMETRY. the elevation of the top is 16° 28'; find its height and eleva- tion above the sea. Ans. Height, 60.82 feet; height above sea, 445.2 feet. 24. From the top and foot of a lighthouse 58 feet high, standing on a cliff by the sea coast, the angles of depression of a ship's hull measured from the visible horizon are 5° 47' and 5° 8'; find the ship's distance, the diameter of the earth being taken at 7926 miles. Ans. 1688.7 yards. 25. At noon, a column in the direction E.S.E. from an ob- server, cast a shadow the extremity of which lay in the direction N.E. from him ; the elevation of the column was found to be 46°, and the length of the shadow 80 feet ; find the height of the column. Ans. 61.23 feet. 26. From the top of a hill, two telegraph posts standing on the horizontal plane below, were seen in a line with the observer ; their angles of depresyion were 30'^ and 45° respectively and their distance from each other 176 yards; find the height of the hill. . ■ ! • An^. 240.416 yards. 27. From a ship a lighthouse bore N.NvE. ; after sailing E.b.S. seven miles it bore N.W.b.N. ; find its distance from the ship at each station. Ans. 5.953 miles; 8.257 miles. 28. A flagstaff 24 feet long, standing on a cliff by the sea shore, subtends at a ship an angle of 38' ; the elevation of the cliff is 14*; find the height of the cliff and the ship's distance from it. Ans. Height, 508 feet ; distance, 2038 feet. 29. From the top of a hill, a vertical pillar 220 feet high, standing ou the horizontal plane below, subtends an angle of 1° 12', and the depression of its top i& 12° 20'; what is its dis- tance, and the height of the hilH J- ; i. . ^-^ Ans. Distance, 9977 feet; height, 2401 feetw ! EXAMPLES. 195 30. If B denote the earth's radius, h the height of a moun- tain and B the dip of the horizon from its summit, show that 6 6 A = 2j5 sin^ — sec = ^ tan ^ tan — . 31. A tower standing on a horizontal plane is surmounted by a flagstaff; from a certain point on the plane, the tower subtends an angle /8, and the flagstaff an angle a ; from another point c feet nearer to the base of the tower, the flagstaff sub- tends the same angle a ; shew that the height of the tower is c tan (3 l-tan;8tan(;-^^) • ' '^'■^-' -'' 32. From the bottom of a tower 100 feet high, the angular elevation of the summit of a hill was 32°, and on retiring 180 feet from the foot of the tower its top is seen to be in a straight line with the top of the hill; find its height. .J,' Ana. 901.5 feet. 33. A person walking along a straight road observes the greatest angular elevation of a tower to be a, and from another straight road he observes the greatest angular elevation of the tower to be p. The distances of the points of observation from ' the intersection of the two roads, are a and b ; shew that the height of the tower is .^ : ,_ . , ,, . ^ I a'-b^ 'cot'"^ a-cot^ /3* 34. Three observers -4, B, C situated in the same straight line, A and C being each at a distance of 1000 yards from B, find at the same instant the angular elevations of a balloon to be at A 48" 10', at B 54° and at G 60° 30'; find the height of the balloon. Ans. 2169.05 yards. 35. A person itanding on a horizontal plane observes that 196 PLANE TRIGONOMETRY. the top of a telegraph post standing on the same plane, is in a line with the top of a building which stands on a hill at a greater distance than the post ; the distance of his station from the foot of the post is b and the angle subtended by the height of the building is /?. He then moves to a station farther off from the foot of the post by a distance a, and in the same lino with the former station and the foot of the post, and he there observes that the angle subtended by the building is the same as before, and that the top of the post is in a line with the foot of tlie building. Shew that the height of the post is J ab + b'\ and the height of the building a (a+ 2b) tan ^ a - 2 Jab + b^ tan /8 36. The angular altitude and breadth of a cylindrical tower are observed to be a and /8 respectively, and at a point c feet nearer to the tower they are y and 8 ; shew that its height is , and its breadth cot a - cot y . /5 . 8 sin 2" sin - sm---sin- 37. A tower, 51 feet high, has a mark at the height of 25 feet from the ground ; find at what distance the two parts sub- tend equal angles to an eye at the height of 5 feet from the ground* , ., .4 ws. 160 feet. 38. The angular elevation of a tower at a place A due south of it is a, and at a place B due west of A, and at a distance o from it, the elevation is /3; shew that the height of the tower ig c sin a sin ^ . ' V'{sin(a+^)sin(a-/3)} * ' -; ' EXAMPLES. 197 39. A ship sailing on a S.S.W. course bore due south, and the angle subtended by the ship was 20' 15", and her length was known to be 160 feet ; find her distance. Ans. 1.94 miles. 40. From the top of a mountain three miles high the true depression of the horizon was 2° 13' 27"; required the diameter of the earth, supposing it to be a sphere. Ans. 7952 miles. 41 At noon, on the shortest day of the year, the shadow of a perpendicular post was seven times as long as its shadow at noon on the longest day ; find the latitude, the sun's declina- tion being 23° 28'. Ans. 38° 27' 47".5. 42. When the sun's declination was lb" T 12". 5 N. the shadow of a perpendicular post was to the height of the post as 5 to 3 ; find the latitude. Ans. 74° 9' 23" N., or 43° 54' 58" S., according as the shadow fell N. or S, of the post. 43 The length of the shadow of a perpendicular object was 4 feet, arfd its longest shadow when sloping was 5 feet ; find the sun's altitude. Ans. 36° 52' 11".5. 44 If a ship sail from a certain place 174 miles eastward on a parallel of latitude, then due south 5°, and then west- ward on a parallel of latitude 194 miles, and reach the same longitude ; required the latitude of the place arrived at, sup- posing the earth a sphere 7912 miles in diameter. Ans. 48° 43' 22" N. 45. A ship sails westward on a parallel of latitude 125 miles which are found to be equal to 2° 30' of longitude; find her latitude. C / \. •.. .■& Ans. 43° 36' nearly. 46 An object 12 feet high, standing on the top of a tower, subtends an angle of 1° 54' 10" at a point 250 feet from the base ; find the height of the tower. Ans. 160.85 feet. 198 PLANE TRiaONOMETRY. 47. A statue 10 feet high, standing on a column 100 feet liigh, subtends at the eye of an observer in the horizontal plane on which the column stand.s, the same angle as a man () feet high standing at the foot of the column ; find the dis- tance of the observer from the column, his height being 6 feet. Ans. 121.095 feet. 48. A flagstaff 8 feet high, standing on the top of a tower, subtends an angle of 57' 17". 75 at 100 yards from the foot of the tower; find its height. . Ans. 232 feet. 49. A tree leans towards the north, and at two points due south at distances a and 6 respectively from the base, the angular elevations of the tree are a and ft. If 6 be the incli- nation of the tree, and h the perpendicular height, shew that tan = r — — -, h = h cot a- a cot )8' cot ^ — cot a ' 50. Four inaccessible objects A^ By C, D are situated in the same straight line, and visible from only one point E. The distance between A and B is 20 chains, and between C and 1) 12 chains; the angle AEB=20\ AEC = U\ and yl^i) = 50°; find the distance between B and C. Ans. 16.48 chains. 51. If the three segments AB, BC, CD of the straight line in the last problem, be represented by a, x, b respectively, and subtend at E angles a, ^, y respectively, shew that sin ^ sin (a + /8 -l- y) a? + (a + h)x = ah sm a sin y cosec' )8 /cot a + cot )8\ /cot y8 4- cot y\ ah V a + x /\ 6-t-a; /' 52. A vertical pillar standing on a horizontal plane appears of the same breadth all the way up, to a spectator standing at a given point on the plane; shew that at any point of the pillar EXAMPLES. 199 whose angular elevation is ^, the radius is a sec tf, where a is the radius of the base. 53. From the top of a mountain the angles of depression of two stations on the plane at its foot are observed to be a and /?, and the diflference of their bearings is y. If a be the distance between the stations, shew that the height of the mountain is a sin a sin B . . „ . sin 2a sin 2B „ y - — -. ^r 2 > where sin" 6 =-- — r— ,-7 ^r- cos'' j- . sm (a + ^) cos 6 sin- (a + y8) 2 54. The boundaries of a tract of land are described in a deed as follows: " Commencing at the intersection of two roads, thence on a course N. 52° E. 21.28 chains; thence S. 29° 45' E. 8.18 chains; thence S. 31° 45' W. 15.36 chains; thence to the point of beginning." Find the bearing of the last course and the area of the tract of land. Ana. Bearing, N. 61** W.; area= 19 ac. 2 r. 36 p. 55. Calculate the bearings of the last two courses and the area from the following field notes : Commencing at a post, thence N. 45° W. 20 chains; thence K 18° E. 12.25 chains; thence E. 12.80 chains; thence N. 32° E. 6.50 chains; thence S. 45° 30' E. 13.20 chains; thence 14.75 chains; and thence to the point of beginning 16.30 chains. Ans. S. and S. 65° 25' W.; area = 59 acres. 56. The boundaries of a tract of land are: A£, W. 25 chains; BC, N. 32° 15' W. 16.09 chains; CD, K 20° E. 15.50 chains; BU, E. 25 chains; BF, S. 30° E.; and FA, S. 25° W. to the point of beginning. A line is run from A cutting off 70 acres 1 rood 33 perches from the west side; find the second point in which this line cuts the boundary. Ans. The side DF, 18 chains east of D. 57. A ship sailed from latitude 51° 24' N. as follows: S.E. 40 miles; N.E. 28 miles; S.W.b.W. 52 nules; N.W.b.W. 30 200 PLANE TRIGONOMETRY. miles; S.S.E. 36 miles; S.E.b.E. 58 miles; find her bearing and distance from the point of departure. • ' '^« • - Ana. Bearing S. 25° 59' E. ; distance = 95.87 miles. 58. A ship sailed 320 miles on a parallel of latitude from longitude 81° 36' W. to longitude 90° W, ; in what latitude was she? Ans. 56° 30' 47". 3. 59. Twilight ceases when the sun is 18° below the horizon, find the latitude of the places at which twilight lasts all night at the time of the summer solstice, the sun's declination being 23° 27' K ■ ' Ans 48° 33' N" 60. The meridian altitude of a star is 64° 10', and its depression below the horizon at midnight is 28° 30' ; find the latitude of the place and the declination of the star. Am. Latitude, 43° 40' N ; declination, 17° 50' N. 61. At two places on the same meridian, one in latitude 59° 31' 30" N., the other in latitude 33° 56' S , the moon's meridian altitude was observed to be, at the northern station 43° 47' 40", and at the southern station 41° 21' 40"; find the moon's distance from the northern station, the earth being re- garded a sphere whose radius is 3956 miles Am. 238020 miles. 62. When the moon's horizontal parallax is 57' 32", and the apparent altitude 50° 40', what is the parallax in altitude? •^ --^... .... .v: ,,.. :. . Am. 36' 28". 63. Find the augmented semi-diameter of the moon when her true semi-diameter is 15' 42", her horizontal parallax and apparent altitude being the same as in the last question. Am. 12".41.* 64. When the moon's horizontal parallax is 58' 10" and her true altitude 50°, find her parallax in altitude. .-■v- Am. 37'52''.6> INSCBIBED CIBCLE. 201 ■ >\ I' : r i c CHAPTER XI. CIRCLES INSCHIBED IN AND CIRCUMSCRIBED ABOUT A TRIANGLE, POLYGONS, AREA OP A CIRCLE, ETC. 152. To find the radius of the Circle inscribed in a Triangle. . „ , ,. The centre is in the intersection of the three lines bisec- ting the angles of the triangle. Let the in- scribed circle touch the sides of the triangle in the points D, E and F. Join OD, OE, OF; the , angles at D, E and F are right-angles. {Euc. B" III., 18.) Let OD=^OE=^OF= r, the radius of the circle ; then area of triangle £0C = iBG. 0D = -^, br area of triangle COA = \AC.OE^ — , ■;f ■'■'.'■■ area of triangle i 05 = i-4^.<5^= -g i therefore, by addition, ..^ ; . :^ j^ ;;,^ . (a -h 6 + c)^ = area of triangle ABC = A. ; ? Put, as before, a + 6 -1- c = 2», :\ ^V;'v*^» th^n A r== — . 8 (218) 202 PLANE TRIGONOMETRY. Tiiat is, the radius of the inscribed circle is equal to the area of the triangle divided by half the sum of the sides. Different forms can therefore be obtained for the radius by employing the various expressions already given for the area of the triangle. Again, from the triangle BOG we have by (204) BC sin OBD am OCD 0D== Bin (OBD+OCD) or . B , G asm. — sm — 2 2 sin ^{B+G) . B , G ^ a sm -- sm -- 2 2 A "V cos^ From the figure we have AE=AF, BD=BF, GD^GE; AE+BD+GD=s, ■■ AE+a=s, , AB=s-a=::AF. \ £D=s-h = BF. GE=s-c = GD. hence or therefore Similarly or From the right-angled triangle AGE, we have .■ OE=AEtsinOAE A 2' B 2' Similarly r=(s-a) tan r=(a-6) tan 'hi , r={s-c) tan G Multiplying the last three equations together, we have 0» A B r^ = (s- a) {s -b) (s- c) tan -~ tan - - tan A" A B G =— tan — tan — tan — « 2 2 2 ... A ., Bsr^s tan — tan — tan — , 2 2 2 ' , A B G '-. r=s tan - tan — tan — . ■ 8- .1 _, .J G (219) (220) (221) ■?'!";' ;"Mt. (222) ESCRIBED CIRCLES. 203 From the figure we have , . ! . i AO=AE sec OAE ' ^ , . A s-a = (s-a)sec-- = — ^, iv:.'.7 (223) Similarly be s ac s ah A cos ~-= 2 £ cos-= C cos -= cos - 2 c s-a .be. B0-- s-6 . ac. C0= = (-« ah. (224) 153. To find the radii of the Escribed Circles. The escribed circles are the three circles which touch one side of a triangle and the other two produced, and are exterior to the triangle. ''V■<.i■:^ :'.{ i\i >\1K_ ?ii)vi TietABC be a triangle, bisect the exterior angles at B and C by BOi, COi, then Oi, will be the centre of the escribed circle 204 PLANE TRIGONOMETRY. which touches BC and the other two sides AB, AC^ produced. Join AOi which will also bisect the angle A, and draw O^D, OiE, OiF, to the points of contact 2>, U, F; then the angles at 2>, E and F are right angles. Let OiD = O^F=OiF=ri, the radius of that circle which touches the side a, and let ra, r^ denote the radii of those that touch the sides b and c respectively. The quadrilateral ABO^^C = the tria.ngha AO^C, AO^B, \ * = the triangles uijBC, BO^G, therefore, the triangles A 0^0, A O^B = the triangles ABC^ BO^C^ A therefore r, = In like manner we find r, = and '.'i .« r3 = ■ ^vA ''; S - C (225) Bisect the angle ACS by CO, then is the centre of the inscribed circle, and OCO, is a right angle; heuco the angle 0501=90° sLiidBC0^==9QP~~, ;: ^ '- > Ci From the triangle JBOOi we have by (204) ^ _- sin OBOi sin BCO, ^ Bm(CBO^+BGOi) ■'\, . . ■ , ^ ; • asm(90''-|-)8in(90<'-^) - :. , n.. V,^. sin {180°-i(JS+0)^ ESCRIBED CIRCLES. 205 that is ri = Similarly B C ^ a cos — cos — 2 ^ A , cos- COS — COB ~ 2 ^ r- = ,^» r« = 5 cos- ^ B c COS — cos — 2 ^ 008- J226) i '.:(':( .v/ (82^ r-f'-.' 1. . From the figure we have , ^ .;;„ „ j^. AE=AG+CE=AC+OD, AF=^AB+BF=AB+BD. But AE=AF, therefore 2^^=2^J'=^5 + ^.C+BC ■ >' ■•' =a4-&+c, "-■■■■ ,''C!/ ■ ;i . -'--''■■ '•- and AE=AF=^a'\-h-\-c)=s. ^ Also CD=CE=AE-h=s-h, and BD='FB==AF-c=-s-c. . t ^ »'■■ ;•.! <;,v D/;. or ,|vil (227) From the right-angled triangle AO^E we hav« Oi^=^-E^tanj&^Oi or Similarly r, =5 tan — . if Ha •> ' ^ 2 JO r^=s tan — . \- rm 'h~J)^ (228) C r«=» tan — .^ ft:"S^Qi.> 206 PLANE TRIGONOMETRY. find From the right-angled triangles AEOi, CDOu BDOi ^e easily A ' ' ' AOi =AE sec EAOj^ =s sec — . BO^=BD sec DBOy_ = {s'-c) cosec -. COj =DC sec DCO^={s-h) cosec -. (229) Similar expressions may be easily deduced for AO^, BO2 , &c. From (228) we find ^ A ^^ 2 B ^ A ,, or t*! tan —=r2 tan — , tan B that is, BF=AEf and therefore AF^BH^^a. 154. To find the distance between the centre of the Inscribed Circle and that of one of the Escribed Circles. . ■ ■ ^'■, ^ . ■ ' '. ■ , From the figure we have '* l' t ; '; , 00i=AOj,-AO A , A =» sec — - (a _ a) sec — , by (229) and (223) 2 A -a sec Similarly 00^=^ sec — , 00^=0 sec O ■i} (230) CIRCUMSCRIBED CIRCLE. 207 155. To find the radius of the Circle circum- scribed about a Triangle. Let ABC be a triangle and the centre of the circum- scribing circle whifth is found geo- metrically by Euc. IV., 6. Draw the diameter BOA' and join A'C, then the angle BA'C:= angle BAG^ because they are in the same segment, and BGA' being in a semi-circle, is a right angle. Denote the radius BO by R, then OP whence A' B sin BAV = BG, > 2B sin ^ = a, f ■ ^ -■ a i?= 2 sin A h ' 2 sin^ ' "■©■■ ^■ by (165). 2 sin (7* Or we may proceed as follows: Draw OD perpendicular to BCj then , OB Bin BOD ^BD, r.' -'yt.u' a ,'j. or whence ^ sin -4 = — , a , as before. (231) 2 sin A To express R in terms of the sides and area we have R-= a 2 sin ^ abc 2bc sin A abc "IK by (199). (232) • 1 1 208 PLANE TRIGONOMETRY. From (180), (181) and (182) we have ■ \ \ A B C As cos ~ cos — cos — = , which combined with (232) gives 2 2 2 abc JJ=— sec -- sec -— sec — •, (233) 4 2 J ^ , 156. Relations between the radii of the InscribiBd, Escribed and Circumscribed Circles. , The product of (218) and (225) is ''• t'l.. From (225) we have 111 s—a-]-8—h-\-8—o I 1 = - ri ra r^ A 3s-(aH-6+c) 5 1 (235) A A r From (222) and (228) we have , ^ ^ Cf ^ ^^ JB ^ O, ri4-»*2+**3 -*•=«(*»« ^+tan -4-tan --tan - tan - tan ~). Dividing by (233) we obtain ' '^ l^ '"'\ ri+r2+r3-r ^^, A B C ^''''' A . B C ^ *•** —— — =4(8m- cos - cos -+C0S - sm - cos - .. A B . C . A . B . C-' . +COS - cos — sin ^-sin - sm - sin -) ^^. A B-\-0 A . B+C =4 (sm - cos --—fees - sin — -—) A A • id =4(sin'' — •+cos'^— )=4, \"yi m 'A ;^^^t]^i-|-i>3+I>3=4B=+2B(ri-fr2+»'a-r) (241) ==12E% by (236) 159. To shew that the distances between the centres of the Escribed Circles and the centre of the Inscribed Circle are bisected by the cu-cum- ference of the Circumscribed Circle. ,, • ( Let Q be the centre of the circum- scribed circle of the triangle ABC Bisect CB in D and draw DE perpen- dicular to CB, and join AE ; then the angle GAB is bisected by AE, and the centres of the inscribed and escribed circles are in AE ; let them be at O and Oi respectively. Join EB and OB. Since the angle GBE is equal to the »ngle GAE, we have EBO=GBE+GBO=UA+B), EOB=OAB+OBA=^{A+B), EO=EB=DB sec DBE and therefore {Euc. I. , ;-i2j but by (230) therefore _ a A 0E=^— sec — - 2 2 A 00 1=0' sec -, 00i=20E. 212 PLANE TRiaONOMETRY. We may here notice, for the sake of problems, that if OQ be joined and produced both ways to meet the circumference, we have hy Euc. III., 35, {B-{-OQ) {B-OQ)=EO . AO a A .in- ar 2ar ^ . A A 2 sin ^ 2 sin — cos — 2 2 .2Br, whence OQ2=i22-2i2r. i6o. To find the Perimeter and Area of a Regular Polygon of any number of sides, which is inscribed in or described about a Circle of given radius. Let AEB be an arc of a circle whose centre is 0; AB a side of the inscribed regular polygon of n sides ; OE at right angles to AB, and therefore bi- secting it; CD a tangent at E^ A) and meeting OA and OB pro- c duced in C and D; then CD is '* a side of the circumscribed regular polygon of n sides. TT •TT Let ^ = r. The angle AOB = — , and therefore AOF^- n n IT AB = 2AF= 2r sin AOF= 2r sin - ; n IT therefore perimeter of inscribed polygon = 2wr sin — . (242) n Area of the inscribed polygon = 7i x triangle AOB AO.BO = n- sin AOB = inr^ sin . n !VV (243) AREA OF A CIRCLE. 213 Again, CD = 20 E = 2r tan COE = 2v tan - , therefore perimeter of circumscribed polygon = Inr tan - . (244) Area of the circumscribed polygon = n x triangle COD = w. OE. CE^nr" tan -. (245) From (243) and (245) we have area insc ribe d polygon ^ area^triangle_^0^ ^ ^^ area circuiii^ibed polygon area triangle COE OE' ^% = co.'-. (246) OA' n • i6i. Circumference and Area of a Circle. . The circumference of the circle is evidently intermediate in length to the perimeters of the inscribed and circumscribed polygons, hence the circumference of the circle lies between that is, between IT "" 2wr sin — and 2wr tan — , IT "^ sin — tan — 71 ^ 27rr .^— and 27rr ' IT n n n Now, let the number of sides be indefinitely increased, then is very small, and when 7i = oc,^ = 0, therefore (Art. 74) sin — tan — !L = 1 = 1, when- = 0. n n Hence when n is infinite, the perimeters of both polygons be- tween which the circumference of the circle lies, become 27rr, therefore the circumference of the circle = 27rr. 214 PLANE TRIGONOMETKY. Again, the area of the circle lies between ^nr sin and nr''' tan — , 27r , TT sin tan — or between nr^ and irr^ — , Zir IT n n each of which becomes irr^ when n = cc . Therefore the area of the circle which is always intermediate in magnitude to the areas of the two polygons = Tr^. (246 bis) Hence, if 6 be the circular measure of the angle of a sector supH as AOE in the last Figure, then the area of the sector A0M=^6r\ 162. To find the Angles and the Area of a Quadri- lateral inscribed in a Circle. Let AB=^a, BC=b, GD=c, DA=d; join ^ AG: then Area il^Oi)=area ABO-\-ADC =^ab sin B-\-icd sin D =^{ab-\-cd) sin B. (1) We have by (169) AC'^=a^-\-b^-2ab cos B=c^-{-d^-^2cd cos £ since cos .D= — cos B, (Art. 39) therefore cos B= ■ , 2{ab-^cd) ' hence 2 sm^ — =1 — cos 5=^^ 2 2{ab+cd) (2) and 2 cos" — ==l4-co8 5= — ^— ^^ ^^ ^, ^3) 2 ^ 2(abi-cd) * ■ . ^^^ therefore tan^ — — EXAMPLES. B (c+d)2 -(a-b) 215 2 (a+6)2-(c-d) 2 2 (247) " {a-\-h-c-\-d) {a-\-h-\-c-d) {s -a){s- h) '='{s-c){s-d)' where 2s=a-\-h-\-c-\-d. Multiplying (2) and (3) together we easily find 8inB=-T^l/(«-«)(«-^)(«-^)(^-'^) (W-\-cd which substituted in (1) gives Area ABGD=V (s -»)(«- &) (« - c) (« - '(M-)=»- 222 • PLANE TRIGONOMETRY. 32. In any triangle prove that r=- rira + rira+rarg 33. The radii of the escribed circles of a triangle are 16, 48 and 5:^ ; find the sidesL Ana. 25, 51 and 52. ABC 34. In any triangle prove that rirjra =r^ cot^ — cot^ -- cot^ --. ^ ^ 2 35. If r be the radius of the inscribed circle, and ra the radius of the circle inscribed between this circle and the sides containing the angle A, shew that 1 - sin ^ ra = r ^ = r tan^ (45° - -). 1 + sin — 36. The sides of a triangle are in arithmetical progression, and the distance between the centres of the inscribed and circumscribed circles is a geometric mean between the greatest and least; shew that the sides are as |/¥-l : i/6~: 1/5^ + 1. 37. In any triangle prove that (1) ^=2-2(sin2 ^ + 8in2 ~ + sin2 -); (2) ah + ac + bc=rri+rr2+rr3+rir2+rir3+r2ri ; (3) ^j^^ Jr^+r^Ur.+r^U r^+r,) ^ 38. If Pj, Pj, p, be the perpendiculars from the angles of a triangle on the sides a, b, c respectively, prove that 111111 (1) -_+_+_- = _+ — +_j Pi P* P» r, r, r, „ 'Zrr.rtr. m Ji= — ^-^; . P1P2PS 2r,r3 .. 1111 (4) -+ =-; Px P, P» ••« p* pl pl be ac ah ' P^Pb PiPs PiP» a' * c EXAMPLES. • 223 39. If r,, ra, r, be the radii of three circles which touch one another externally, shew that the area of the triangle formed by joining their centres is K' (r, + r, + r^) r^rtr,. 40. If and Q be the centres of the inscribed and circumscribed circles of a triangle, and if r^ , r^ , r^ be the radii of the circles which circumscribe the triangles BOO, AOG, AOB respectively, and Ba , Bb , Be the radii of the circles which circumscribe the triangles BQC, AQC, AQB respectively, prove that a h c _ abc '"'oTb^c _ B ~Bl'^~B'b^~B^~~W' *^ abo ~^'h^c* where B is the radius of the circle which circumscribes the triangle whose sides are a, h and c. 41. Shew that there is only one point within a triangle from «rhich, if perpendiculars be drawn to the sides, circles can be in- scribed in each of the three resulting quadrilaterals ; and if ri , r2, rj be the radii of these circles, and r that of the inscribed circle of the triangle, then (1.1) (i-i)+(l-M (i-MH^-i) (--i) = ^. - 42. Shew that the area of the triangle 0x0^0^ (Fig. of Art. 153) = All+ r + 7 + T \t \ -a + o + c a-o+c a+o-cf where A represents the area of the triangle ABC. 43. ABCD is a quadrilateral inscribed in a circle, shew that : AC am A=BDamB. ,'i 44. Shew that the perimeters of an equilateral triangle, a square and a hexagon, each containing the same area, are as t/27 : t/'16 : */12. . ", v • 45. In the Fig. of Art. 152, prove that . . • ' AO^'-i-BO^ + CO^zzab + ac+bo- ^^° -. a-\-b-\rc 46. Prove that the area of a regular hexagon inscribed in a circle is geometric mean between the areas of the inscribed and circum- scribed equilateral triangles. 224 PLANE TRIGONOMETRY. 47. In a regular polygon of n sides, of which a side is 2a, prove TT that Bi-r = a cot — , where B and r are the radii of the circum- scribed and inscribed circles respectively. 48. If A, Ai, A2, -^j denote the areas of the four circles which touch the sides of a triangle, A being that of the inscribed circle, prove that 1111 r=z= + ■ Va Va[ VA2 VA3 49. The external bisectors of the angles of a triangle are produced to meet the circumference of the circumscribing circle ; shew that the area of the triangle formed by joining the three points thus R obtained is ~ (a + b + c). 4t 50. li r^, r^, r^ denote the radii of the circles inscribed in the triangles BOC, AOC, AOB, in the Fig. of Art. 152, prove that — + — + — = 2 cot — + cot — + cot — ). Ta n r^ 4 4 4' 51. The square of the side of a pentagon inscribed in a circle is equal to the sum of the squares of the sides Qf a tegular hexagon and decagon inscribed in the same circle. 52. Shew that the areas of the circles in Euc. IV., 10, are as 6+/ 5": 2. ■ . 53. If through any point within a triangle three straight lines be drawn from the angles A, B, C, meeting the opposite sides in D, Ej F respectively, prove that OP OE OF AD^BE^GF' ' 54. If a, )8, y be the angles which the sides of a triangle a ibtend at the centre of the inscribed circle, prove that 4 sin a sin )8 sin 7=sin ui +sin f+sin (7. INVERSE TRIGONOMETRICAL FUNCTIONS. 225 CHAPTER XII. INVERSE TRIGONOMETRICAL FUNCTIONS. 163, If sin 0=a, then 't) is an angle whose sine is a; but instead of writing this description of 6 at full length, the following notation Id employed to express this relation : ^=3in~' a. Similarly, if cos (p=b and tan i{)=c, then ^=cos~' h and i/^=tan~' c. The student must be careful to remember that the index - 1 does not here imply an algebraic operation, but merely expresses the relation between the angle and its functions as above enunciated. The functions sin~^ a, cos~^ b, &c., are called inverse trigono- metrical functions, from the nature of the notation which is anala- gous to that employed in Algebra, where x~^ is the inverse of x. Since there are several angles which have the same sine, several which have the same cosine, &c. , it follows that the above equations are more correctly written as follows : ' sin~' a=mr+( - 1)" 0, COS"' b=2mr± and cos =±av/(l-6«)±6v/(l-a«), therefore (9+0=sin-'{ ±av/(l - 6'*)±6n/(1 -a')}, or sin-' a+sin"' 6=sin-' { ±a>/(l - 6*) ±b\/{l - a'^) } . (250) In a similar manner we find Bin-« a - sin-' 6=sin-'{ ±a^{l - b^)l^by/{l - a^) }. (251 ) cos-' a±co8-' fc=cos-i{a6=F v/(l - a*) (1 - b^) }. (25::) 165. Oimn tan~^ a and tan~^ b, to find tan''^ a±tan~'^ 6, Let 0=tan~' a and ^=tan— ' 6, then tan d=a and tan ^=&, tan d±t8i.n tan (0±0): ITtan^ tan^ a±b ~lTab * (a+b \ - ), tan-»a±tan-i 6=:tan-'(— 5=^). (253) Xn a similar manner we find cot-J a ± cot-i 6=cot-' ( ^^^^-^ ) . (254 ) V 64-a / '^ • ■' ' ■ 166. Here we may remark in regard to (253), that since there are several angles whose tangent is a, several whose tangent is 6, and several whose tangent is ~ , it follows that the sum or dif- 1-rao ference of any two of the angles whose tangents are a and b respec- EXAMPLES. 227 tively, is o»|uaJ to flome oiui of thu angles whose tanguut is ^ - a - l> ... or respectively. Si)nilar roniuvks ui)ply, of course, to the other formiilio of this ohapter, and in faot t(» all fonnula; containing the inverse functions. Examples. 1. Prove that tan-'{(v/2'-/ 2. Given versin"' versin-' (1 - 6)=versin-i — , find x. a' a Since the versin=l — cos, the above becomes X hx cos-' (1 ) - cos-' 6=cos-' (1 ), a a or cos-' \ (1 - -)hWa - J>') (1 - (1 y) ] =co8-' (1 - -), by (25i>) Cancelling cos"' and solving the equation for x, we have M"Vrs)- . ^ V a cos 6 a — sin ^ 3. Prove that tan-' ;; :— --tan"' — =f ;, 1 - a sm

a cos a ~ siu c/> or tan-i : tan~> =^. 1 - a sm <> cos ^ .>: a,fr -y , a--o, 4. Prove that tan"' -=tan-' — — -■+tan-' -— ~ 4-tan-' ^-+ . ... tan-' ^-t-tan"' -. a,aj+l ani^-i + l On where a, , a,, . . . . <(» are any quantities whatever. By (253) wo have X 1 , <*i«-y tail"' tan~' — r=tan"' , y «! «i y ■*- » 1 1 a» - a, tan-' tan-' — =tan-' — , 1 1 a, - a, tan-' tan-' — =tan-' , 1 , 1 , ttn - <'f'n— 1 tan—' tan-' — =tan-' a n— 1 On a„a„_i+l by addition we have • tan-' tan-' - =tjvn-' - — ^+tan-' — — ^^+... H-tan-' -5~ ^-1-, y a« ftiy+a ajai+l a„o»_i4.1 therefore by transposition X a,x-v a^-a, . an-On-i . 1 tan-' -=tan-' - — ^+tan-' — — \+... +tan-' ^ ' -f tan-'- , y ftiy+x a,ai+l a,ja»_i+l a,i 6. Prove that cot (fl+tan-i tan^ e)=2 cot 26. .• , , " 6. Prove that cos sin"^ cos sin"^ fi=±d. \l — X |1+X ;,: V ' 7. Provethatco8-ix=2Bin-^ l-^'=2co8-i 1-^, EXAMPLES. — 229 8. Find the general values of cos"^ sin 6 and tan"^ cot 0. Am. 2w7r±(— -0), nir+{—-d). 3 1 9. Prove that tan"^ — =2 tan-^ — . 10. Prove that (1) sm-i-+sin-i-=-. 4 ,-. (2) sec-i 3+tan-i 2V 2 =tan-i ( - - v" 2 ). , 2 1 , , 12 S) tan-^ -=- tan-i -. 4) 2tan-i-+tan-i-=j. (5) sin-i (_v/2+s/"2) + sin-i (- v^2^^V?)=|- 1 1 , 1 TT (6) 2tan-i— +2tan-i— +tan-iy=:— . 11. If sin (tt cos 0)=co8 (tt sin 0), shew that 0= - — sin -. 6 ' I ^ 12. If 2 sin — =co8 d, shew that 0=2 cos-^ . Icos — . 2 >| " f^,. 13. Find the values of tan (tan-^ e+oot-^ 6) and sin (sin-^ — +cos-^ — ). Ans. «, and 1, or 14 If tan {n cot a5)=cot (» tan «), shew that mn ^ ^. 1 . . 4w m and r being any integers ■ 15. If 2 tan-i a;=8in-i ^y, shew that ]/=j--j-^. 1 2 2'^^ _ PLANE TEIGONOMETRY. 16. Find the value of x in the following equations : (1) tan-J2a;+tan-i3ic=:y. Ans. cc=-lor--. 4 6 (3) sin-i 2a;-sin-i>/3' x=sin-i cc ^m. a;=0 or +— 2 ' (4) 8ec-i--sec-i-+sec-ia-sec-i6=0. ^m. x=±a6. (5) sin2cos-icot2tan-ia;=0. Am. a;=±l,or±(l±v/¥). (6) ver8in-i(l+a;)-ver8in-i(l-a;)=tan-i2\/rr^. 1 AtiS,. aj=— or— 1. (7) sin-i a;+tan-i a;=^ -4rw. a:=:J^::l^^. (8) Bin (tan-i a;)+tan (sin-'a;)=m!r. Am, x=—Vm* - 2m'i T 2v/l + 2m2-2. -'-•■. . , '. 17. If sec 6>-cosec 0=2\/2", shew that 3+cosec-* s/?—. 4 DIVISION OF ANGLES. 231 CHAPTER XIII. DIVISION OF ANGLES — SOLUTION OP EQUATIONS — AUXILIARY ANGLES — ELIMINATION OP TRIGONOMETRICAL FUNCTIONS. Division of Angles. 167. We shall here determine, a priori, how many values any assigned trigonometrical function can have when determined from any other function of the angle or of a submultiple of the angle. 168. Cfwen sin A, to find how many values sin — cam, ham when expressed in terms of it, » Let a be the circular measure of the least angle whose sine is equal to sin A; then all the angles whose sines are equal to sin A axe included in the general expression ^ W7r+(-l)"o. f^rt. 42.) " " ' A Hence all the values which sin — can have when expressed in terms of sin A are included in rn7r+(-l)'»a> . / nn+{-ira \ Now n must be of one of the forms 22. or 2X+1, since every number is either divisible by 2 or divisible by 2 with a remainder 1. Let n=2A, then ^^-^j=8in (A7r+--)=±Sin -, according as A is even or odd. 232 PLANE TRIGONOMETRY. . Let n=2A+l, then Sin (—2 )=sm(;i-+^-) (IT a\ a ---^=±C08-, according as A is even or odd. A ■■ Therefore sin —, when expressed in terms of sin -4, has four different values, viz. , + sin — and + cos — . A 169. Given cos A, to f/nd how marvy wdues cos — cam have when o expressed in terms of it. Let a be the circular measure of the least angle whose cosine is ' equal to cos A; then all the angles whose cosines are equal to cos A are included in the expression 2w7r-j-a, and therefore all the different A values which cos — can have when expressed in terms of cos A are o included in * - ' • ■ • 2n7r4-a ' : ^ ,. Now n must be of one of the forms 3p, 3p+l, Sp+2, since every number is either exactly divisible by 3, or divisible by 3 with a remainder 1 or 2. Taking n=Sp, we have 2r(i,7r4-a ,^ a. , o. . a cos — - — =cos (2i)7r±-)=cos (±-);=cos -. Taking n=3p+l, we have 2n7r4-a / 2Tr4-a\ cos — - = cos i 2^n-+ — W~)'=^ Taking 71=3^+2, we have 2n7r -fa / 47r + a\ _ 47r-j-a cos — :;:: — =cos yipir^ — - — |=co8 — - — /„ 27rTa\ 27rTa =C08 I 27r — j=cos — - — , DIVISION OF ANGLES. 233 A Therefore cos —, when expressed in terms of cos A^ has three o diflferent values, viz. : a 27r+a , Qtt — a cos — , cos — - — and cos — - — . 170. Given sin A, to determine how many values sin f J. can have when exp, essed in terms of it. Let a be the circular measure of the least angle whose sine ig equal to sin A ; then all the values which sin ^A can have are in- cluded in sin|(w7r+(-l)«a}, where n is of one of the forms ip, -ijj+l, 4p+2, 4p+3, since every number must be exactly divisible by 4, or divisible by 4 with a remainder 1, 2 or 3. If n=4p, which is even, - . sin I {w7r+(-])'»a}=sin (3p7r+|a)=±sin fa, according as p is even or odd. If w=4j9+l, which is odd, sin |{n7r+(-l)"a}=sin {3p7t+|(7r-a)} =-j-sin |(7r-a), according as^ is even or odd; and so on. Hence we find that sin |-4, when expressed in terms of sin .4, has eight different values, viz. , ' ± sin I a, -t-sin |(7r-a), ±cos |a, ±8in|(7r-3a), 2r7r+e 171 • To find the number of vahies which cos has when n suAxessive integral vcdv^ are assigned to n. Here r, being an integer, must be of the form mn+p where m is or any integer, and j:> is or any integer less than n; that is, r must be exactly divisible by n, or divisible by n with a remainder which is 1, 2, 3 ... or M— 1, # 234 PLANE TRIGONOMETRY. Hence giving r, the values, 0, 1, 2, 3 . .. n- 1, n, n+1, . . . &c., in succession, we have when r=0, 2r7r+^ e cos =cos — , n n r=l, 2rK+e 2Tr + d cos cos , n n ' r~2, cos =cos » n r=3, 2rn+d 67r + cos =cos , n n &c., &c., &c. r=»-3, 2rTv+d Qir-e cos =cos ,« n r=n-2, 2r7r+d 4:Tv-e COS =cos n n r=n— 1, 2rTr+d 2ir-e cos =cos n n r=n, 2r7r+d d cos =cos — , n n r=n+l, 2rTT+e 2ir + e COS ^=cos n n &c., &c. , &c Therefore there are n and ordy n (liferent values of cos -— — corresponding to the values 0, 1, 2, ... n— 1, of r; for the same values of the function recur in the same order when r is succes- sively made equal to n, n+l, &c. ^ 2r-K-\-d In a similar manner we may shew that sin has also n dif- ferent values. n The preceding examples are quite sufficient to shew the mode of prooofding in any assigned case, SOLUTION OF EQUATIONS. 235 Examples. 1. Shew that sin ^, when determined from tan A, has two values. 2. Prove, a priori, that sin mA, when expressed in terms of sin A, will have one or two values, according as m is odd or even; and that cos mA, in terms of cos A, will have only one value, m being in each case a positive integer. \^ 3. Prove that tan — -, when expressed in terms of sin A^ will 4 ,■•..,■,.■.. have four different values. IT 4. If tan ^=sin 2^, find A. Ans. nir or ri7r+(- 1)" — . 4 • 5. If cos 0+cos 20+cos 3^=0. then will, r being any integer, e^{n+{-iY5\ {(7+(-ir'i)^±i}^^. Solution of Equations. 172. An equation in which the unknown quantity is a trigono- metrical function of an angle, is, in general, readily solved by the aid of the ordinary trigonometrical transformations. We shall here illustrate the mode of solving a few easy equations, such as are most frequently met with in Spherical Astronomy. 173. Given dn 6=cos (3 sin {6+ a), to find 6. Developing the second member by (45) we have sin ff=cos a cos ^ sin ^+sin a cos /3 cos 8, or tan 0=cos a cos p tan 0+sin a cos /?, sin a cos /? :' i* . ■ whence tan 6=- -. 1 — cos a cos p ^ Now it is evident that if ^ is not limited to any particular quad- rant by the nature of the problem under consideration, there wilJ be an indefinite number of solutions; for all the angles ", ^+180", ^+360°, 0+540°, &c. , that is, all the angles included by the expression 6+nir, have the same tangent. (Art. 44.) In practice, however, only the first two values of 0, viz., 8 and 0+180", or those less than 3G0^, are considered ; tind the conditions of the problem are generally such as enable us to determiae which of these is to be takl^. 236 PLANE TRIGONOMETRY. It is evident then, that when an angle is determined by a single trigonometrical function, there will be two values less than 360°; but if the values of two functions of the required angle, which have not the same sign — such as the sine and tangent, or the cosine and cotangent — can be found from the problem, the solution is deter- minate under 360°. Suppose, for example, that the required angle is found by its sine and tangent ; if the sine is positive and the tangent negative, the angle will evidently be in the second quadrant or between 90° and 180°; if both functions are negative, then the angle will lie between 270° and 360°, and so on. The solution of the last equation cannot be eflFected by logarithms ; . a formula adapted to logarithms is easily deduced as foflows : Put the given equation in the form sin 6 =cos 13, then sin {d+a) Bin {d+a)+Bin 6 1+cos/? — — • — ■ sin (d+a) -sin d 1 -cos fi tan(e+|-) taai- a a ■ B whence tan (0f— )=tan - cof^ — -. a This determines ^+— , and therefore Q becomes known by de- 2k a a B a ducting — , thus 0=tan~i (tan --- cot" — ) - — . 2 i2 2 ^ , 174. Given tan {6+a)=sin ^ tan d, to find 6, Putting the given equation in the form tan(e + a) . ^ — -=sm B. tane * we have, by composition and division, tan(6/+o)+tan0 1 +sin j3 tan(^+a)-tan0~'i-sin,'3' • f' SOLUTION OF EQUATIONS. 237 whence — : =tan2 (45°+-), sin a 2 therefore sin (20+a)=tan2 (45°+ M sin a and <9=4 sin-' j tan' (45°+ 1) sin a ( - -|. 175* Given tarn, {a+6) tan 6=ian fi, to find 6, Here we have 1- tan (a+e) tan d l~tanj3 1+tan ^a+e) tan 6 ""l+tan )3 * cos (a + 2d) ,,^„ , whence * -^-^^ ^=tan (45°-/3), cos n therefore cos {a + 26) =tan (45° - /?) coi a and 6=^ cos-' { tan (45° - /3) cos a J - -. 176. Criven x sin d=a and x cos 0=b, to find 6 and x. By division we have a tan 0=7" > which gives two values of 6, one less and the other greater than .180°, and also two values of x from the equation x=a cosec 6. Limiting the values of 6 to those less than 360°, the solution is determinate under the following restrictions : 1st. When X is positive. The signs of sin 6 and cos will be the same as those of a and h rsspectively, and therefore the quadrant in which 6 must be taken is determined. 2nd. When x is negative. The signs of sin 6 and 908 are the opposite of those of a and b, and therefore must be taken out accordingly. 3rd. WJien 6 < 180° or > 180°. Under either of these conditions the equation tan 0=t gives only one value of 6 (a and b being unrestricted aus to sign). Under 238 PLANE TRIGONOMETRY. the former condition, x has the sign of a; and under the latter, the opposite sign to that of a. 4th. When 6 is limited to acute values, positive »/ negative. , Under this condition x will always have the same sign as 6, Auxiliary Angles. • 177' Ij^ t^6 solution of equations by logarithms it is necessary to express the sum or difference of two quantities by means of a pro- duct. This can always be effected by introducing the sine, tangent or some other function of an angle chosen for that purpose. An angle which is thus introduced to assist in trigonometrical calculation, is called an auxiliary atigle, and is of great utility and extensive ajjplication, particularly in Spherical Trigonometry and Spherical Astronomy Auxiliary angles have already been employed in a few examples. (See Arts. 120 and 144) Ex. 1. — Adapt jc=v/(a«±62j to logarithms. nI-:-.- (1) x—v'(ci''^+h^)=a Assume tan 6=—, an assumption always possible, since the tau- a gent may be of any magnitude whatever, then a;=av/(l+tan2 6)=a sec 6; hence x will be found by the two equations Log tan 0=log 6-log a+10; log x=log a+Log sec 6-10. (2) x=^{a^ - ¥)=y/(a+h) (a - 6), which is in a form adapted to logarithms; or we may proceed as follows : x=a. 1 • 'i .6.6 Assume sin 0=—, since — must be less than 1, a a •then a=:av/(l - sin* ft)=a cos 6. AUXILIARY ANGLES. 230 Ex. 2. — Adapt x=a±\/a^+¥, to logarithma, Tho equation may be written thus, Assume tan ^=--, a' then «=» (l±sec ^) l+COS ^ 1-C0S6 and — ct ~ cos (ft COS a sin 6 1+cos sin d=sin a, to find ^ in a form adapted to logarithms. Assume x sin 0=cos 6 cos h ) and X cos ^=:sin 6, i ^ ^ that is, tan ^=cot d cos ^ j (2) then the given equation becomes X (cos sin &+sin cos d)=am a, or x sin (0+^)=sin a :;. and £c=sin d sec ^; ., , . , «\ sin a .... therefore sm (0+0)= sm d sec ^ =sin a cosec S cos ff, (3) wrhich gives two values of (0+0). 240 PLANE TRIGONOMETllV. Tn this example, let (I=-30' 22' 47".5; o=20" 10'; h=-lu\ I'm:! (,.. 13y (1) Log COB (T= 9.035857 Log cos /i= 9.984944 . Logwsinfl =9.920801 Log « cos ^ = Log Bin +e) = 9.698947 ( 29° 59' 53^5 ^'*' ""lor 150" C 6". 5, therefore = ] lor - 28° 44' 53". 5 9r 16' 19". 5. If we take the acute value only of 6, we have ff = - 58° 44' 47", which gives ^+ft = -29°59'53".5 or -150° (T 6". 5 and 0= 28° 44' 53".5 or - 91^16' 19''.6 as before. Ex. 5. — Given tan l=:{coit a cos ft- cos 6) cosh- si/n a cos p sin /»., to find h in a form adapted to logarithms. Writing the given equation in the form cos a cos B - cos 6 , . , tan A cosec a sec 3= : cos a - sm h. sm a cos p and assuming cos 0=cos a cos /3, since cos a cos p is always less than 1, we have - ^ cos 6 - cos d , . , tan A cosec a sec p=—. — cos h - sin h sin a cos p 2wnJ(d+e)_BinK^-W) =: ; -r QOB h - SlU h, sm o cos /? • AUXILIARY ANGLES. 241 Again, assuming cot = --,-'— --^— \ wo Imve Bin a COB li * tan A cosec « sec /?=cot cos /i - sin /j, cos cos /i - sin , therefore /i=co8-» (tan A cosec a sec /:J sin f/>) -0. * Examples. 6- If *=^"^ ) shew that x=cos 20, where tan 0= -. 7. If a=v/a+6 + v^a-6, shew that a « a;=2v/a cos (45°--), where cos cos (0+/^), where tan 0=cos /i cot d. 14. Given cos h=cot ^ tan d, -cos (af/i)=tan tan 6, to find J^ and \ • 5 .0. • CO8 ^ . ^ - ' • COS — -Bm« -^ 1 - tan" — 2 2 2 " '-. ■;''r,>' (• ".,.'; ' ..- . - whence tan« ^+2 cot tan ~-l=0. (256) 2 2 . " ■ QUADRATIC EQUATIONS. 243 In (a) let x=y\/h, where the radical is taken with the positive sign, X and y having the same sign. We thus reduce (a) to V which compared with (255) gives a A ' - - 2 cosec ^=— — , y=tan -jr , ■•/■, or ' sin 0=- , and a;=\/6 tan — , a 2 which gives two values of 0, and consequently two values of «, Let B be the smaller of these two values of 0, then all the values of which have the same sine are . — t ,, 0, TT-e, 27r+0, Stt-^, &0., ,: : ,-f' -* ■■. . and all the values of tan — are , ,. * a tan — , tan ^(tt-^), tan ^(27r+5), tan K3^~^)> **'•» ' ' ^ (? <^ . OV tan — , cot — , tan — , cot — , &c. a a a 2 ^ Hence the roots (wj , Xj) of (a) are found by the formulas \ sin 0= , a5,=v/y tan — and x»=v/r cot — , a 2 ' 2 where (9 is always to be taken less than 90°, with the sign of its sine, v/ 6 being regarded as a po^tive, quantity, and a either positive OT negative. When 2v/6 is greater than a, sin is impossible, and both roots are imaginary. ., 179* To solve the equation a«+ax-6=0, (c) where -h is essentially i^ative, a h«mg either positive or negatim. Let x=y\/ h , then (c) becomes . V 244 PLANE TRIGONOMETRY. which compared with (256) gives a 6 2 cot i>=-7^> y=tan -^ , or tan 0= , and x=\/b tan -^, i which gives two values of ^ and also two of x. If ^ be the smaller of the two values of 0, the values of which have the same tangent are e, TT+e, 27r+0, 3n-+0, &c., and all the values of tan — are 2 Q d e e tan—, -cot—, tan—, -cot—, &c. 2' 2' 2' 2' ' Therefore the roots of (c) are found by the formula) tan 0= , «,=v/6 tan— and a5„=-\/6 cot—, ' a ' ^ 2 ' 2' where 6 is to be taken less than 90°, with the sign of its tangent, the radical with the positive sign, and a either positive or negative. tSij Here tan (^ is always possible, therefore both roots are real. ' Ex. —Given x^ - 1. 7246x+. 72681=0, find x. Here we have a=- 1.7246 and 6=. 72681. ' ■ log ( - 2)=0. 3010300/1 log v^= X 9307104 log v/ 6 =1.9307104 ^ e Loff tan o q^aqi *7a ar. CO. loga=9.7633116n n ** 2 Log sin (9=9. 9950520 0=81° 22' 3^ ■^=40° 41' r.5 2 ■ log Xj= = 1.8650282 -..■ ■. Xl= : .732872 Log cot "2" =10.0656822 • log Xj= = 1.9963926 x,= = .991728 Cubic Equations. 180. Let the equation be transformed, if necessary, to another which wants the second term, so that it may be of the form x3-qfx-r=0; (a) CUBIC EQUATIONS. 245 y if x=— , this becomes From (104) we have, by writing for JL, ,5 ^1 3 cos 3^ ^ C083 - —cos . ; 4 * . , , ; i ';,■■ v> ;..■, -.W' '.-''' * which compared with (6) gives cos 6 cos ^=1/ and as= , ,3 lis" 7i«flf=_ or n=— ^ I—, ^ 4 2SJq ^-21^^n^ or cos 30=4n3r=^^ ^ , where the radicals I— and ^ I— are to be considered positive. Sjq Sjq^ If 6 be the circular measure of the least angle whose cosine is equal to cos 3(f>, then by Art. 171 the three values of cos ^ are e 2ir + e ^ 2n-d cos — , cos — r — and cos — - — , 3 00 and therefore the three values of X are ' . 2 ll.cosi, 2 \l.coB?l±l and 2 JI . cos ?I^. \J3 3 \/3 3 Nj3 3 r I27 T* cfl Since cos 3^ < 1, — ^ — < 1, or — < -i . 2 \ 33 4 27 8 Ex. Given aj^- 4a-— =0, find x. o *^^' .444 .. V ' ^^ -4m. — :=. cos 10% - — =r cos 50°, -—-r: cos 70°. -■ v/3 n/3 v/3 The other ^rms of cubic equations can be solved by Trigo- nometry, but the solution is a matter more of curiosity than of utility. We shall therefore pursue the subject no further, but refer the student to the standard treatises on Algebra for a fuller eluci- dation of this subject. 246 PLANE TRIGONOMETRY. Elimination of Trigonometrical Functions. l8l. The elimination of the trigonometrical functions from a given number of equations, is frequently requii-ed in some of the higher branches of mathematics, and is generally effected by the aid of the various trigonometrical transformations given in the preceding chapters. The following examples illustrate the mode of proceeding in most cases: - Ex. 1. — Eliminate d between the equations ■> a sin d 6 cos d v. k « . y I' =0. (2) a sin2 6 b cos'^ Clearing of fractions gives us ,;,!,. \ bj' cos 6+ ay sin d=ab sia 6 cob 6 f . . f' (3) bx oos2 d+ay sin^ 6=0. (4) Adding bx sin^ 6 to both members of (4) we have bx (cos« 0+sin2 6)={bx - ay) sin^ Ol , jt \/bx=i\^bx- ay sin 6 f Vbx whence sin 0= s/{bx~ayy n '<•••>• and :. ' ,,,.vi ■: cosg= ^/^\ ^ , y/{ay-bx) wliich substituted in (3) give after reduction s/bx y/(bx - ay) + s/^y s/{ay - bx)=ah. Ex. 2, — Eliminate between the equations ...v . y cos ^ - X sin ^=a cos 2^ (1) - X cos (ft+y sin 0=2a sin 2^. » ' ) (2) From (1) and (2) we have by division - . ^ i «cos0+yBi n^ • : ., " : .- — =2 tan 2a y cos - X sin ^ 'i' 1 • » : . =F l=(tan ^l)\ Subtracting (3) from (4), and using first the upper and then the lower sign, we get whence 3(^-1) tan* +-^ l=(tan

.: [. Ans. m'+n'=l. 4. Eliminate 6 between the equations , ; . cosec2 6=m tan d sec* 6=n cot 6. , . , Ans. (mny=(s/m+y/n)\ 5. Eliminate B and (7 between the equations a-h cos C-c cos jB=0 6 - c cos A -a cos 0=0 c - a cos B - 6 cos A=0. Ans. a2=62+c«-26cco«X 6. Eliminate 6 and ^ between the equations tan 0+tan ^=a '' tan 6 tan ^ (cosec 25+cosec 2^)=6 cos (^+0)=c cos (6 - (j)). Ans. o,=h +hc. 7. Eliminate 6 between the equations sc=a(cos 6+coB 26) ■ .. '«/=6(sin e+sin 2<^). 8. Eliminate a and ^ between the equations a B . 14. Eliminate ^ between the equations ' ' " '^^^ ■ • '' ^' ••'"'' ' M sin (? -mcos0 =2m sin ^ . n sin 2^ - m cos 20=n, Ans. (n sin d+m cos 0)'=2m(m+»j)- 15. Eliminate 6 and r between the equations < r=2acos20, a;=rcos0, i/=rsinfl. 16. Eliminate 6 and r between the equations . r — =:o-. as r ""'^ ^ ' " ■ ■ v>^-v: ....... 250 ' PLANE TRiaONOMETJiY. ■iU.X *■ 1 CHAPTER Xir. ON THE COMPUTATION OP LOGAEITUMS. We will here prove all the formulf* necessary for the computa- tion of Napierian and Common Logarithms ; but before commencing this chapter the student shpuld read carefully Articles 88-97, of Chapter VII. .> .-- The Exponential Theorem, 182. To expand a" in a aeries of aacending powers ofx, ,au*n!^i M • a'-^=l+A{x+y)+B(x+yy+C(x+yy+kc.f the last series is evidently equal to the product of the two former ; therefore we have l+A{x+y)+B{x+yy +C(x+yy + &c. =(l+Ax+Bx'^+Gx^+&c.)(l+Ay+By^+Cy^+&c.}, THE EXPONENTIAL THEOREM. 251 Expanding both members of thig equation we have 1+Ax+ Bx^+ Cx^ + Dx* +&c.\ Ay+2Bxy+ZCx'^y+4.Dx^y + &c. -Bl/2 +3(7x1/2 +Gi)a2y 2 +&C. Cy^ +U)xy^ +&C, 1+Ax+Bx^ + a»' +• Dx* -f.Stc.\ Ay+A^xy+ABx-y+ACx^y +&c. By"" +ABxy''+B^xhr-t&c. - . i. \>l , » . S-' ,1, Oy3 +ACxy^ +&C. JOy* -ir&C. Cancelling the terms common to both series, we have 2Bxy+3Cx^y+iDx^y + &g.\ (A''xy+ABx^+ACxhj+ &c. +3Cxy^+eDx^y'i+ &c.\ = \ H-ABxy^+B^xY + &c; ' ■i'AChyy^+ &=^(7, OP J3 ^ ^* / ^'*' ' 1.2.3.4* &c. '^ &c. Therefore a»=l+^at+ + + -h&o.^ L2 1.2.3 1.2.3.4 ^ where '' ' ^=(a-l)-i(a-l)!^+|(a-)3-&c. '':.« -J^v Since this result is true for all values of x, take x such that 1 Ax=^, or «=7, then ^ ^ y 1.2 1.2.3 1.2.3.4 =2.718281828459.... =e, henoe J a=e^ and J.=loge a. 252 PLANE TRIGONOMETliy. Therefore we have finally a« =Tl+(log, a) Y + Goge ay —^ + (log, a)' —^ + Ac. (257) which ia called the Exponential Theorem. IS asati, we have e* =1+35+ + + + &o. (258) 1.2 1.2.3 1.2.3.4 ^ ^ The Logarithmic Series. 183* To express loge (1+ar) in a aeries of ascending powers ofx. From the last Article we have log, a= A . : - • = (a-l)-K<»-l)'+K«-l)'-i(a-l)' + &c., in which write l-¥x for a and we get log, (l+a5)=x-^+| -^ +1 - &a (259) In the last seriefl write -x tor x, then we have x* a.3 X* log, (l-a!)=-»- 2 - 3 - 4 - *c- (260) 184. To prove the Logarithmic Series indepen- dently of the Exponential Theorem. A«iume Ax-¥Bx^+Cx^ +Dx* + — =loga (1+a) and Ay+By^+Cy^+Dy*+.... =loga(l+y) By (subtraction we have , . ,, il(x-i/)+B(»a-ya)+C(x»-i/»)+....=loga(l+x)-loga(l+y) -~"'V c,5 =loga(l + ^ We may consider -~ as a simple quantity, and therefore / X ""1/ V loga ( 1+^j — ) may be developed in the same manner as log„(l+x). THE LOGARITHMIC SERIES. 253 Thus, Therefore A{x- y)+B{x^ - 1/») +C{x^ - 1/3) + . . . . Dividing hy x-y we have A+B(x+y)+Cix''+xy+y^)+. ...=a{ ±-)+B^^,+G^~^,+ .... Since this equation is true for all values of x and y, it must be true for x=y ; hence, writing x for y, it becomes A A+2Bx+ZCx^+.... 1+x =ui(l-x+a2-£c='+.... ). Equating the coefficients of like powers of x, we have A=::A, 2B=-A or £=--, V. ■, I. A '■ ■ ' -O - SC=A, or C= -, , 4D=-^, or I>==-T» 4 AQ^ • &C. ; Therefore log„ (1 +x)=:A {x- -+—-—+ ....). fv. 'ijv 2 3 4 Dividing by x we have 1 CT 35 CT "'^ - l0ga(H-x)=^(l--+- --+.... ), X J o 4 but ' ■ - log« (l+x)=loga(l+x)« , (Art. 93.) ,, , 1-x l-3x + 2x« =log„(l+l+ — +-^— + by the Binomial Theorem. ^54 PLANE TRIGONOMETRY. Therefore Since this equation is true for all values of x, it must be true when «=0, hence ^=Iog,.(l+l+-^ + j|^ + -^ + ....) =loga e=- , (Art. 97) log, a =the modulus of th i system whose base is a. (149) Therefore we have generally If axsef we have «' X^ 05* loge (l+a;)=«=x -■^ + ^-~r"^"** *" t)eforo. ^ O 4: 185. To deduce the Exponential Theorem from the Logarithmic Series. From (261) we have, by writing A for loge a* a 3 3 ^=l+x, and raising both members of this equation to the power -— — , we have ■ * Ax . y y' «n/ , y* a ;y'+2xy' ^ 1.2.^» 1.2.ui» L2.3.i» 1.2.3^2 + •» /; ^* '■' md when x=^, this becomes THE NAPIERIAN BASE. 25.^ Restoring the value of A, we have ov = 1 +(log, a) I + (log, ay^ + (log, a)' ^^ + &a'^ i86. The Napierian Base. The sum of the series which we have denoted by e, is the base of the Napierian system of logarithms. This base renders the logarithmic series simpler than any other base would, as is evident from a comparison of (259) and (261). Napierian logarithms are sometimes called natural loga- rithms, because they occur first in the investigation of formulee for their calculation. The Napierian system is used for the most part in the higher Analysis, and, with the exception of the common system which is universally employed in arithmetical and trigonometrical calculations, it is the only one which we shall ever have to use. 137. The Napierian Base is incommensccable* For, if possible, let e=— , where m and n are integers, then" m _L._?_ 1 '1 n" "*" '''1.2 1.2.3'*' '"1.2.3. '..». 1.2.3... (n+1) "^•" Multiply both members by 1.2.3... n, and we have 1.2.3...(»-l)m==2.2.3.4...n+3.4.5...n+... "J (1_ 1_ ) ' ' . ^ ^Wl"*'(n + l)(n + 2)"*'-7' but the former member being integral, the latter must also be so, 1 1 •which is impossible since — - + ; -7- — + ... is. greater than n+1 {n + l){n + 2) t ..11 and less than the sum of the geometrical series — - + -- fi+l n+1 (n+iy^ + — ——+... , that is, less than — : therefore e is incommensurable. 256 PLANE TRIGONOMETRY. i88. Converging Series for the immediate calcu- lation of Napierian Logarithms. From (259) and (260) we have by subtraction jgS jjgS /)p7 log, (1 +X)-I0ge (1 - X)=:2(X+ -^ + -g- + y + •••) /l+x\ „. x'^ x^ x"" '°8'(tIv„-)=^(*+3+-6 + 7 + --'' . . . . 1+x . , , m— 1 in which let =wi, and therefore x= , thu8 we have 1-x m+1 which, however, converges too slowly to be of much utility in the calculation of the logarithms of integral numbers. In (202) let m = , and therefore X m + l~2x + l' • thus we obtain 1+x f 1^ 1 1 I ^^^* X -^|2x + l'^3(2^Tl)^+-[ or log,(l+x)==logeX+2{^ + |-„^+...}, (263) which converges very rapidly, especially when x is large. In the last equation, lot l+x=y'^, and therefore x=i/ — 1 and 2x+l=2i/2 — 1; then we have log. ,/=log. (v.-l)+2(2^^ + 1^-^- + ...} or logc (i/ + 1) =2 loge y - log, (y-1) which also converges very rapidly when x is large. By judicious substitutions, many other series for the calculation of Napierian logarithms, may be deduced, but practically considered CALCULATION OF NAPIERIAN LOGARITHMS. 257 they are now unnecessary, as tables have been already computed to the highest attainable degree of accuracy. We shall therefore pursue this subject no farther, but refer the student to the third example at the end of this chapter for two other converging series, which may be advantageously used for the same purpose. Calculation of Napierian Logarithms. 189. By the last three formulee we are able to compute the Napierian logarithms of all numbers ; but the properties established in Articles 91, 92 and 93, render it necessary to apply them to prime numbers only. Thus, in (262) let m=2, then • l°S'2=2(- + -(-) +-(-)+-(-) +....} =.6931471.... In (268) let tK=2, then f 1 1 /lv» 1 /lx» ) log. 3=log, 2+2 {-+-(-) +-(-) +.... I =.6931471 +.4054650=1.098612. log, 4=log, (2 X 2)=2 loge 2=1.386294. In (264) let t/as4, then log, 5=2 log 4-log.3-2{l+i(^^)V|(|^)V.... } =2.772588 - 1.098612 - .064538 =1.609438. • '*• loge 6=loge 3+log« 2=1.791759. In (264) let i/=6, then log. 7=2 log. 6-loge 5-2|- + -(-) + .... J =1.945910. Iog«8=log«2a=3 log, 2=2.079442. . 'i loge 9=log« 32=2 log* 3=2. 197225. log. 10=log, (5 X 2)=log, 5+log« 2=2. 302585. &c. &c. 258 PLANE TRIGONOMETRY. Hence the modulus of the common system is 1 1 logelO 2,302586 as was shewn in Art. 95. =.4342944819..,, Calculation of Common Logarithms. 190. Having computed the Napierian logarithms by the method of the last Article, we may convert them into common logarithms by (151), thus, Log 2=. 43429448 log, 2=. 3010300 log 3=. 43429448 log, 3=. 4771213 &c. &c. By means of M, the modulus, we may adapt the series of Art. 188 to the calculation of common logarithms ; thus, by (151) log (l+a,)=Iog .+2M{^-i^ + |-jljj-.+ ....}. (265) log (i/+l)=2 log y-\og iy-l)-2M 11 1 (266) [2y^-l 3(2i/2-l)^ Having found the logarithms of prime numbers by the preceding aeries, the logarithms of composite numbers are easily found by the principle of Art. 91. Thus, log 3360=log (2^ x 3 x 5 x 7) =5 log 2+log 3+log 5+log 7 =3.5263393, &c. Theory of Proportional Parts. 191. We shall now investigate how far the principle of propor- tional parts can be depended on in finding the logarithm of a number which is not found exactly in the tables. In the following investi- gation we will assume that the logarithms are calculated to seven decimal places, and that the tabie contains the logarithms of all whole numbers from 1 to 100000. THEORY OF PROPORTIONAL PARTS. 259 To shew that in general the Increment of the Loga- rithm is approximately proportional to the In- crement of the Number. Let N and N+h be two numbers, the former containing five digits and the latter six, the last {h) being after the decimal point. Then we have log {N+h) -log JV=log — ^ =log (l +-) where M is the modulus .43429448 .... Now, if N is not less than 10000, the second term of this series is less than .000,000,002,2, which docs not affect the seventh decimal place, and may therefore be neglected. - Hence, as far as seven places of decimals at least, we have log (N+h) -log N-^h, which shews that the change of the logarithm is approximately pro- portional to the change of the number. We will now proceed to ascertain to what extent the principle of proportional parts can be applied in the case of the logarithmic trigonometrical functions. ig2. To shew that in general the change of the Tabular Logarithmic Function of an Angle is approx- imately proportional to the change of the Angle. Let 6 denote any angle and h any small increment such as 1' or 10", then we have sin (B + li) sin ft cos /i + sin h cos ft sin 6 sin ' =l+h cot ft, approximately J since COS h=l and yin h=^h very nearly. 260 PLANE TRIGONOMETRY. Therefore Log sin (6+h) — Log sin 0=log (1+^ cot (?) h^ cot* . , =M(h cot e + &c.) • 2 =Mh cot 6, approximately. But when 6 is very small cot 6 is very large, and therefore the h^ cot2 e second term may be too large to be neglected. Thus, if h=l' and 0=2', the value of the second term is .0000151, which is far too large to be disregarded. If, however, h=10" and 0=2", the value of the second term is .0000004, which will affect the seventh figure but not generally the sixth; therefore we conclude that when is not very small, Log sin (e+h) - Log sin d=:Mh cot d. (267) that is, with the exception just stated, the change of the logarithmic sine is approximately proportional to the change of the angle. 193. In a similar manner we find Log cos (d+h) -Log cos e=-Mh tan 6 (268) approximately. When 6 is near 90° the second term, omitted in (268), is too large to be neglected. Therefore, with this exception, the change of the logarithmic cosine is approximately proportional to the change of the angle. 194* In vhe case of the tangent we have " '' ' tan 04-tan h tan {C+h): i. - tan h taii tan d+h —tan d+hsec^ 6 1 — h tan 6 approximately, ., ^ tan (e+h) ^ „, therefore =1+2^ cosec 26. tan and Log tan [O+h) ~ Log tan e=2Mh cosec 2fl (269) approximately. THEORY OF PROPORTIONAL PARTS. 261 When 6 is very small cosec 2d is very large, and the second term is too large to be neglected; therefore, with this exception, the change of the logarithmic tangent is approximately proportional to the change in the angle. 195. A similar proof may be employed for each of the other Logarithmic functions. The following are the results which may be verified by the student : Log cot {e+h) - Log cot e= - 2Mh cosec 2d. (270) L 'y sec (d+h) - Log sec d=Mh tan 6. (271) Log cosec {d-h)- Log cosec 6= - Mh cot 6. (272) 196. From the preceding Articles it is seen that the change of the Logarithmic sine is equal to that of the Logarithmic cosecant, but with the opposite sign. Hence a column of " Differences for 1' " is printed in some tables between the columns of Logarithmic sines and cosecants, serving to the former as a column of increments for 1', and to the latter as a column of decrements for 1'. In like manner the columns of cosines and secants have the same differences for 1', and so also have the tangents and cotangents; these columns serving as increments to the secants and tangents, and decrements to the cosines and cotangents. 197. From the preceding investigations, and from an inspection of the tables themselves, the student will see that the principle of proportional parts is not applicable to angles which are very small or nearly equal to a right angle. In the case of the Log sin and Log cosec, the differences are irregular for small angles and insen- sible for angles near 90° ; for the Log cos and Log sec the differences are insensible for small angles and irregular for angles near 90°; for the Iiog tan and Log cot the differences are irregular both for small angles and for angles near 90°. For the methods of computing the Logarithmic functions of angles near the limits of the quadrant, i he student is referred to Art. 112. .1.1 . 262 PLANE TRIGONOMETKY. Examples. U Prove that 1 1 I log,a=7i{(l-a«)-fi(l-a«)2+i(l-a")3 + ....} 2. Prove that ^^"^ iif ■^1.21 M J ■^i.2;3l':atf~j ■*■•••• irhere N is any number and M the modulus. 3. J£ a, h, c be three consecutive numbers, prove that log 6=i(log a+log cHm{^^H^^^ + ....} and log c=2 log 6-log a-23f {-^^y-j+J^^^^-^ + ....|. ,11 1 1 4. Prove that — =— —4- ,^„^ -f , ,, „,^>. + . • .. e 1.3 1.2.3.5 1.2 3.4 5.7 , „ ,^ e 1 1+2 1+2+3 1+2+3+4 6 Provethat-=— + ^^^ + -^^^+--^- + .... 6. Provethat2e=l+| + f^ + ^ + --A-- + .... 7. Find the modulus of the system whose base is -- . Ans. -.91024. 8. Prove that log«(v/3i/^~'=l-i + i-l+A-A+.... 9. Prove that loge 101 - loge 99=— very nearly. 50 1 10. To what base is - 5 the log of 32768 ? Ans. — . o 11. The log of a number to one base is the same as that of its reciprocal to another base; find the relation of the bases to each other. DE moivre's theorem, 263 CHAPTER XV. DB moivre's theorem — EXPANSIONS OF CERTAIW TRIGONO- METRICAL FUNCTIONS. De Moivre's Theorem. 198. ' m he awy ratumal quantity j either integral or fractionclf positive or negative, then (cos B±\/ -1 sin 0)"*=cos md±\/ — Isin. md, (1) Let the index be a positive whole number. (cob d±\/-l sin 6/)2=(co8 e±V - 1 sin d) (cos B±V^-lBm 6) =co82 e - sin2 6±V^ 2 sin 6 cos 6 =cos26±^-l sin2«, (cos 0±n/ - 1 sin 0)='=(cos 2d±V - 1 sin 20) (cos e±\/-l rniO) =co8 20 cos 6 - sin 2# sin 6 ± y - 1 (sin 2ff 008 + cos 2fl sin 6) s=co8 S^iv/- 1 sin 3^, and so on. Suppose this law to hold for m factors, so that 3,. (cos e±V^ sin 6)"'=coa md±>/'- 1 sin w^, - then (cos ^±v/-l sin ^)*"+^=(coa m8±\/ -1 sin wifl) (cos 6±\/~^lBm 6) =:co8 m^ cos 6 - sin m^ sin ti i v/ - 1 (sin m^ cos #+cos m^ sin ^) . , . srscos (mH- 1)W±n/-1 sin (m+1) 0. If, then, the law holds for m factors, it also holds for m+1 factors ; but we have just shewn that it holds when m=:=3, therefore 264 PLANE TRIGONOMLTRY. it holds when m=4, and by sucoeasive inductions w© conclude that the formula is true for any positive integer. Therefore we have (cos d±\^-l sin (?)"'=cos md±\/ -iBinmB, (273) when m is any positive integer. (2) Let the index be a negative whole number. (cos 6/ ±>/ - 1 sin 0)-'»= -- (co8 0±v'-i sin ey* (cos« g + B in«6>) "* _ (cos ±v/^nr sill ^)"» =(cos eT v/TT sin &)•, by actual division s=cos mdT\/ -1 sin m^, by (273) =co8 ( - m#) ± v/ - 1 sin ( - m6>), which proves the theorem when m is a negative integer. m (3) Let the index be a fraction — , either positive or negative. n (co8 d±>/-l sin 0)"*=co8 imd± v/-l sin md '-e)±N/^sinrv(^ =008 » (— ©j ± V - 1 sm rv ( — 61 =(cos - 0± v^-l Bin - e)» , by (273) therefor© .wi ,. I / — T . w (cos e±y/ -\ sin ^)'^ =corf- (^±s/ - 1 sin - 6, (274) which proves the theorem for fractional indices. 199- A^9 long as m is an integer, both members of (273) can have only one value ; but in the case of (274), in which the index is a fraction, the first member has n difierent values in consequence of the vH'^ root, while the second member has only one of these. Th6 second member, however, may be transformed so as to exhibit the same number of values as the first. DE moivre's theorem. 265 Thus, since cos 6=cob (2rn+ff) and sin (^=sin (2r7r+^), we have (cos e±>/-Tsin ey*=:{coa (2nr + d)±^/^Bm {2rn-\-e)\» =cos - (2r7r+0)±vAri sin — (2r7r+&), (275) n n in which the second member has n different values corresponding to the values 0, 1, 2, — n-1, of r, according to Art. 171. There- fore the theorem is entirely general under the form (275). 200. To express an imaginary quantity of the form (a-^hj^if by means of Trigonometrical Functions. Assume h cos d=a and k sin d=6, whence tan 0=— and h^s/a'^ + h', a Then, by De Moivre's Theorem, {a+h\^~^y=1^ (cos 0+\/~l sin e)"» =(a2 +6^)2 (cos w^+v^^ sin mSy. In a similar manner we find (a + 6v/^)-=(a^+6=^)2-(co8 ^J^^W^ sin Hl!^^) • J^sc.— Find the three values of (y5" + 2v/"^)i. Hero, tan^=-y5 or •0=41" 48' 37* 5 and A;=3 ; then we have, by giving r the valur' 0, 1, 2 in succession, riK3(co8| + v/Tr8in |), ^3 (cos -j-+y^sin -J-), .^ / 47r + .—- . 47r + ^. f 3 (cos — — — + s/ - 1 sin — ^— ) 266 PLANE TRiaONOMETRY. 201. To express the sine and cosine of an Angle and of its multiple as Algebraic Binomials. Since (cos 0+\/-l ain 0) (cosy-v/-l sin y)=coa» O+sin* 6>=1, if we assume cos fl+v/ - 1 sin B=Xf then COB 6 - V -1 sin 6= — , » and by addition and subtraction we obtain 1 , I 2co8fl=x+— , and 2v- lain ^=05 ; • '• X X ' also, cos m^+ \/ - 1 sin wi0=(co8 6+\/ -1 sin 0)"'=x**, and cos md-V -1 sin m0=(cos 0- \/ - 1 sin 6)"*a= -- , »"* by De Moivre's Theorem, whence as above 2 cos m^=x"*+ — , and 2\/'^sinm0=w'" . (27C) ajm' y.m If the index be fractional, we have by Art. 199, ml ml 2 cos —(2r7r + is ( - 1)^ sin"* B. 268 PLANE TRIGONOMETRY. 203. To ejfpress the tangent of the multiple of an Angle in terms of the tangent of the Simple Angle. The quotient of (278) by (277) is m(m - 1) (m - 2) m cos"*-i sin - ^ ^^ cos"^ d sin" 6+..,, , _ 1. J.o tan md= ^ ^ m(m - 1) „ . cos™ d cos'"-^ d 8in2 e+,,,, m(m-l)(m-2) m tang- tans 0+ . . . . = r^T ■ m m(m-l) . 1 tan3(9+.... 1.2 by dividing numeratuv and denominator by cos"* 6. 204. To express the sine and cosine of an Angle in terms of its Circular Measure. a In (278) and (277) assume m6=^ or m=— , thea we have , sing a(a-e)(a-2d) ^^/sing^^ sin a=a 008*^1 g ^ ^ ^ cos»'^g ( — ) +.... 6 1.2.3 ^ d -6) „ / sin 1.2 '"'"-^"(t a(a-6) „ /Singv' cos 0=008"* - V - cos"»-2 g ( )+.... sm JNow, when 4Ji=« , 6=0, cos g and all its powers = 1 and 6 and all its powers = 1. (Art. 74.) Hence we have ultimately (280) .;,.r-. (281) a3 1 1.2.3 ' 1.2.3.4.5 a* COS a=l - h a< 1.2 1,2.3.4 By means of the last two series we may compute the sine and cosine of any angle. Thus, suppose we require the sine of 12". TRIGONOMETRICAL EXPANSIONS. 269 The circular measure of 12°=;^^=. 20943951, " 15 . .«o «««.o«.. (.20943951)' (.20943951)* then sxn 12-20943951 - '-^^ ^^^^^^ - =.20943951 - .00153117 +.00000335 =.20791169.... , .. : . agreeing with the tables which give .2079117. ' ' 205. To express the positive integral powers of the cosine of an Angle in terms of the cosines of its multiples. Assume 2 cos ^=xH — , X 1 .=v"... then 2 cos nd=x^ +— by (276). i 1 " 2''cos"^=(cc+— ) - . . .. /v, ,-• ;■, r. , n(n-l) _ n(w-l) 1 1 1 1.2 1.2 X»^ «n-2 x" ,7 / 1 \ / o 1 \ w(w - 1) / . 1 \ by placing together the first term and the last, the second and the last but one, and so on ; 11 ' but x" +— =2 cos n0, x»*-2+ — -=2 cos (n - 2)6, &c. , ; .> therefore • 2" cos" 0=2 cos n0+2n cos (n - 2)6+2. ^ cos (n - 4)6+ 1 ** In the expansion of (x + — ) by the Binomial Theorem, there are n+1 terms, therefore when n is even there will be a middle term, / n \th — 1 viz., the ( - + 1 j , which does not involve x, since x^ • =x*'=l. This term is ;.■'• ' '-^^ n(n-l)(n-2). ...(hn+l) 1.2.3....^ 270 PLANE TRIGONOMETRY. When n is odd there will be two middle terms of the expanded binomial, viz., the ^{n+lf^ and the ^(n+S)*^, whose coefficients, however, will be equal, involving x and — respectively ; their sum is X n(n-l){n-2) ^(n + 3) (-^) 1.2.3. ...^(n-1) ^ X Therefore we have generally n(n — V) 2^^ cos* 6— COB vJd+n cos {n - 2)g+ cos (w - 4)5+ . . . . , 1. a the last term being i(n-2) (^ + 1) \ — — — :; ; n even. (282) ^1. (^-l)(^- 2).. .-(i^ + l) . n(n-l)(n-2 )....i (n + 3) 1.2.3 ^{n-1) 2o6. To express the positive integral powers of the sine of an Angle in terms of the sines or cosines of its multiples. Assume 2>/ - 1 sin 5 ==x , X then 2>/ - 1 sin iud=x^ x» » 1 •• and 2^(-l)2 sin" <^=(x ) - n(n-l) =X»- «M5'^2+ - - x"-* - . . . . 1 . ^ ^n(7i-l) 1^1 1 1.2 x""^ x«-2^x'* the upper or lower sign being taken as n is even or odd., Arranging the terms as in the last Article we have 1, If n be even, it is of the form 4^ or 4w+2, since every even TRIGONOMETRICAL EXPANSIONS. 271 number is either exactly divisible by 4, or divisible by 4 with a remainder 2. If 7i=4m, (V^l)<»* = 1=(-1)2'« =(-1)2. n If w=4m + 2, (^^)4«+2= - 1=( - iym+i—(^ _ 1 jT — + 1) of the preceding one, does not involve x, and is n{n-l) (n-2) {^n+ 1) 1.2.3...T^ii ' n according as — is even or odd; it has therefore the samo sign as n ( - 1)2. Hence we have, n being even, 2n-i( _ i)Tsin» 0=co8 w9 - n cos (n - 2)0 + '^^^^^1 cos (rv - 4)5 - . . . . 1. a ,.(_1)t K^-1)(;-2).--.(^+1) . (283, 1 . J . o . . .. . ^ (2) If n be odd, it is of the form 4m+l or 4m 4- 3, since every odd number is divisible by 4 with a remainder y^hich is either 1 or 3. n—X If»i=4m+1, (v/-l)4'»+i=(-l)a >/-!. n— 1 Ifw=4m + 3, (\/^)*'"+3=( - 1)2 v/-l. The sum of the two middle terms of the first of the above series is n{n-l){n-2) ^(n4-3) . 1. " 1.2.3. ..4(n-l) ^*""x^ according as \{n-\-l) is odd or even; it has therefore the same sign^ n— I as (-1)2. Hence we have, n being odd, , • ., n— 1 (-1)2 2'»v/^sin''0=2v/^sinn(J-»i2\/"^sin(n-2)O r.f.. n{n-l) ^ , :. ^: >. ' + -j-^2v/-l sin(n-4)0.... n'. n— 1 +' ^' i.2....ior-ir,.^^,""'*' 272 PLANE TRIGONOMETRY. or dividing by 2\/ - 1, ( - iJ2-2«-i sin" e=sin nd - n sin {n - 2)0+ -.^?^^ sin (n - 4)5+ , . . . n— 1 + ( - 1) 2 'o ,, %\ sin 5. (284) 1.2.t}. . . .^(?i-l) 207. To express cos nB in a series of descend- ing powers of cos 6, n being a positive integer. Assume 2co8 0=a + — , then 2 cos n0=a** H — , a a" Now, (l-a»)(l- — )=l-(a + — )a;+xa =l-!B(6-a;), \ih=a-\ — =2 cos 5. a Then log (l-ax) + log (l--)=log {l-a;(6-«)|. Expanding both members of this equation by (260) we have ax I • • • 2 n I jp, a 2a2 na" x**~2 x"^^ x* Equating the coeflQicients of «" in this identity, we find on th» 112 left-hand side the coefficient of x" to be — (a*» +— ) or — cos nd n^ a*» ' n ' while the coefficient of x** on the right-hand side is found by taking x" out the coefficient of x" from the expansion of — (6 - x)" and of all n the terms that precede it. Thus, the coefficient of x*» TRIGONOMETRICAL EXPANSIONS. 273 in — (6 - ae)" is -I- — , n n a;**~^ 1 in — :(6-iK)"-i is -.(n-l)&«-', w-1 n-1 w-2 n-2 1.2 * n-r^ ' 1.2.3.... r Therefore we have 2 6" 71-3 — cosn0= 6"-^+ -— r 6«^-.... r,,- r- T n «i 1.2 (-ir(n~r- l )....(n-2r+l) _,, . ^ 1.2.3....r ^ ' or writing 2 cos B for 6 and multiplying by w, , 2 cos »i^=(2 cos e)** - w(2 cos <9)*^H--7^ (2 cos e)"-* - . . . , 1.2 •,•'', (-l)'-n(n-r-l)....(w-2r+l) , ■" , _ ' + 1.2.3.. ..r ^-(2cose)-'^ (285) ' r_.. • , •-• . - .,",... Here we must observe that since none but positive integral powers of (6 - x) appear, the index of b in the general term must also be a positive integer, that is, r must not be greater than — ^ 2o8. In (285) write — - ^ for 6, then we have when n is even, ( - 1)=^ 2 cos n^=(2 sin 0)« -n{2 sin 0)n-2+ -(*^^) (2 sin )^ - ... . . -» nin-r-V) (n-2r4-l) +(-!)--. -^ rixT^T =^^2sin^)-2r. (286) and when n is odd, ( - 1)2 2 sin n^=(2 sin 0)«- n(2 sin 0)«-2+!!^-ri (2 sin ^)'*-* - . . . . : : .+(.1)^ . «('>-'- l)--(>»-«^+l) (2 ,i„ ,)^. (287) 274 PLANE TRIGONOMETRY. 209. To express cos nO in a series of ascend- ing powers of cos d, n being a positive integer. (1) Let n be even In (285) r is limited to values not greater than — , therefore writing for r in the general terra the values — , — -1, — -2, .... 2 2 2 n ..,,3, 2, 1, 0, in succession, the number of terms will be — +1, and as r is diminished successively by 1, the terms are alternately n positive and negative, the first term having the same sign as ( - 1)^. Therefore we have 2 cos n&=(- 1)-* { -^ '-^ —^ (2 cos ey K 1.2.3 \n n.^(^-l)....(n-2(^-l)+l) , 1.2.3....(i«-l) (2cos«)« n(in+l) ^{\n - 1) in -2{\)i- 2)+l) + 1.2.8....(^-2) (2 cos »)• «(l«+2)(Jn+l)....(ri-2(i«-3)+l) \ .1.2.3....(^-3) ^(2«o.»)' + &<^| ^" ^ \ 1.2.3....^ «.in(|n-l)... .4.3 ,^ .. . ,, ?_A2 !. (2cos0V ' V 1.2.3.... (^-1) ^ ^ ■ " n(in+l)in(^-l)....6.5 ^ , . : 1.2.3.... (iii- 2) "^ ' ■• > f'l .=(-l)T|2.!^.(2oos.)'+!iS^|M(2.o..). _ nan+2)(i«+l)X^-l)(i,.-2) (g ^, ^y, ^ 4^ K 1.2.3.4.5.6 ^ ) WA TRiaONOMETRICAL EXPANSIONS. 275 dividing by 2, and reducing coefficients we have cos rK? =(-1)2 {1- -— cos«g+ /^ „ / cos^g ^ ' I 1.2 1.2.3.4 n» (n''-2') ( 7i'-4«) ) '• (2) Let n be oc2dL Since r cannot be greater than — , it may be i(n - 1), the integer next less than — ; the terms are alternately positive and negative, the first term having the same sign as ( - 1) *(**-^>, and the number of terms will be^(M-l)+l or \{n+l). Writing forr, J(rt-1), i(n-3), \{n - 5) ... .3, 2, 1, successively in the general term of (285), we have as before, by reducing the coefficients and dividing by 2, ^, cos n5=: - 1)2 {nco&e- \ ^ ^ cos^ g+ \ ^ »\ ^ cos* ' ( 1.2.3 1.2.3.4.6 ,, 1:2.3.4.6.6.7 ^^^^'Y (289) 2Z0. To expand sin oo and cos x in series con- taining the ascending powers of Xy independently of De Moivre's Theorem. ^m! The series for sin x must vanish when a5=0, therefore it can contain no term independent of as, nor can the even powers of x enter into the series ; for suppose sin a=^x+5a;2+(7x3+Da;*+&o.; substitute — a for sc and this becomes sin (-«)=- -4a+J9x2- 0*3+ J[)x< - &c. ; but ;/ . sin (- 05)= - sin a, (Art. 41) 'f::" "^'' ,^^ =-Ax-Bqc^'Cx^-Dx*-&o., therefore '-, ^ B=-B, D=-D. &c.. which is absurd unlesg^uv iJ=0, D=0, &c. , "■•'ih 276 PLANE TRIGONOMETRY. therefore and but if iJierefore Bin x=Ax+Cx^+Ex'^+&o., sin » . _ _ ■=A+Cx*+Ex*+&o.i X y.=o, -^^^=1, (Art. 74) X A=l, and we have ein oc=x+Cx^+Ex'^+&c. (a) Again, the series for cos x must =1 when x=0, therefore its first term is 1, and it can contain no odd powers of x ; for suppose " ■ ' ' cob«=1+^x+Bx»+Cx'+jDxH&o., ; ' ■ -• then rj ooa(~x)=l~Ax+Bx^-Cx^+Dx*-&o., : a • * but (i cob(-»)=cosx (Art. 41) ' r' ..^ ; "'• f 'v =:l+Ax+Bx^+Gx^+Dx*+&o.f therefore ' "' ' A=-A, C=-G, &o., which is absurd unless -4=0, C=0, &o., therefore cos x—l+Bx^+Dx*+&o. Adding and subtracting (a) and (h) we get cos X + sin x=l + x+ J5x« + Cx'^+Dx* + ^x*+&a cos X - sin x=:l - x+jBx' - Cx^+Dx* - Ex^+&o. In (c) write x+h for x, then it becomes ^' "^ - "" -' * • ' " cos (x+/i)+sin {x+/i)=l+(x+^)+5(x+^)«+C(x+^)'+ .\ ' . (&) (c) (d) ' f '.'■ but cos (x+A.)+sin (x+?i)==co8 x cos fc - sin x sin ^+sin x cos h+cosxamh =«os h (cos x+sin x)+sin h (cos x - sin x) =(1+B;i2+DA<+ . . . . ) (l+x+BxH0x3+ . . . .) +(fe+C7i3+jE7iH. . . .) (1-x+j5x2-Ox3+. . . .). Equating the second members of the last two equations and expanding we have 1+X+ Bx^+ 0x9 + Dx* + h'{-2Bhx+ZChx^+U)}ix^ + Ch? +4Z)/i3x + Dh* + otf = -- / l+x+JBx'+Ox" +Dx< + h-hx +Bhx^ -Chx^ + Bh^+Bh^x +B*h^x'>+ Ch? 'Ch'x + SINES, ETC., OF SMALL ANGLES. 277 Cancelling the terms common to both members of this equation and equating coefficients of hx^ hx^f &c. , we have 8(7 = 5 -B = - 1.2 C = - 1 1.2.3 D = H 1.2.3.4 B = 1 1.2.3.4.5' &0, Hence we have by substituting in (a) and (h) ■ .1 : , ' ■ ■ ■ ■'.-.'.'•;. 1. J) a* X* — ■ + ■ ■ ~ • ••• 1.2 1.2.3.4 ax I. Sines and tangents of small Angles. The last two formulae furnish us with the means of finding the sine and tangent of a small angle, and conversely, of finding a small angle from its sine or tangent J , , .... Thus, when x is small -' . ging;=x~ , very nearly, . ■> ,>' 1.2.3 ' Let a be an angle containing n*, i. f . ui;«i ; ui: ; . «(;) t\ then n,.,.^,. '*="7rT; or «=nsinl', ,; ,, ^: m-I i^i.i jd sin X since sin l"=3circular measure of 1" very nearfy; therefore sin x=m> sin 1 cos* x or Log sin n"=log n+Log sin l^+^Log cos «- 10) '•^'^.*' -^ =log n+Log sin 1"- 1(10 -Log cos «) =logw+4,6855749-i(Logsec»-10). (290) 278 PLANE TBIGONOMETRY. And also log n=Log sin n"-Log sin l"4-J(Log sec x- 10) =Log sin n"+Log cosec l"+i(Log sec x- 10) =.Log sin n" +6^ 3144251+i(Log sec x - XO). (291) That is, to find the sine of a small angle we have the following rule: " To the logarithm of the angle reduced to seconds add 4. 6855749, and from the sum subtract ^ of its logarithmic secant, the character- istic of the latter logarithm being previously diminished by 10 ; the remainder is the logarithmic sine."— (Chambers's Logarithmic Tables, Art. 27.) To find a small angle from Log sin, we have this rule : '* To the given Log sin add 5.3144251 and ^ of the corresponding Log sec, the characteristic of the latter logarithm being previously diminished by 10, and the sum will be the logarithm of the number of seconds in the angle." In like manner a formula may be established for finding the tangent of a small angle, and conversely. nn. , . , A sin X X COS» X X Thus tanx= = = , '' - ...,..0 ; ... ...,.:. COSX COSX ^^^^ ^ ■ , . and if the angle x contain n% we have as before x=n sin 1", and Log tan n"=log n+Log sin 1" - f (Log cos x - 10), =:log «+Log sin 1" + f (Log sec x - 10). (292) and log n=Log tan n" - Log sin 1" - f (Log sec x - 10), =:Log tan n" + Log cosec 1" - |(Log sec x - 10). (293) This method of finding the sine and tangent of a small angle is generally known as Maskelyne's method, and was first given by him in his Introduction to Taylor's Logarithms. It is not as convenient •as that of Art. 112, which is known as Delambre'a method. « Examples. _ , L Prove that sin 5x=5 cos* x sin x - 10 cos» x sin' x+sin* x. !«; . cos 5x=coB *x - 10 cos' x sin« »;+5 cos x sin* x. EXAMPLES. 279 2. Prove that 2* sin* a=10 sin x - 5 sin 3ac+sin 5ac 2* cos^ ac=cos 5x + 6 cos 3ic+10 cos x. 3, If tan 6=— , shew that a m (a+bv/ - 1)" +(a - 6v/ - 1 ) » = 2(aH6f " cos - 6. 4. Prove that cos 4x=l - 8 COS* x+8 cos< as. sin 5x=lC sin/ x - 20 sin^ x+5 sin x. 5. By means of (280) prove the following rule for finding the length of a small arc: *' From eight times the chord of half the arc subtract the chord of the whole arc : one-third of the remainder ia equal to the arc very nearly." 6. Prove that when 6 is small 2 sin 6/+ tan 6=Sdf nearly. 7. Shew that 3 S^-l 3 3*-l S'"-! y "^'^^=4 ..i:^'' 172.3:475^ ■^••••±1.2.... (2n+l)^ ^^r 8. Find the three values of (-1) by De Moivre's Theorem. 280 PLANE TBIGONOMETRY. .',!■• ' .' :■■'. I ' CHAPTER XVI. EXPONENTIAL FORMULiE — COMPUTATION OP THE NUMERICAL VALUE OF TT — TRIGONOMETRICAL SERIES. Exponential values of the sine, cosine and tangent. aia. From (258) we have .. t 1.2 1.2.3 1.2.3.4 ' in which let ds/ - 1 and - 6v--l be successively substituted for x, then we find the sum and difference of which are =^2cos^, by (281). (294) ^ 1.2.3 1.2.3.4.6 ^ =2\/^8in0, by (280). (295) The quotient of (295) by (294) is >/~ltan6=-—zz — =- —, (296) e +e e +1 by multiplying numerator and denominator by e ~ , ft;.' ■ ; . EXPONENTIAL FORMUL-^. ^81 These formulae, which are due to Euler, axe reckoned amongst the moat useful in Modern Analysis. By the addition and subtraction of (294) and (296) we get ,.,., ,: /^^=cos9+v/^sin0 X -^^ (297) ,^d :e-*^=:cos0-N/^sin0; •-: (298) or introducing the notation of Art. 201, we have * J x= /^-^=cos0+v/^8ind """^^ (299) ^ l=e-'^-'=ooBd-^^Bme. "^ (300) Hence (294) and (295) may be written 1 \ Zcoa6=x + — X 2v/^sin0=x- — X ' (301) and if we .ubatitute «rf for fl in (299) »nd (300) we find ty addition and subtraction , ,; 1 . 2 cos m0=x"*+-- -i^ . '.:> 'J v a"* I (302) 2N/^sinme=»'"- — 213. To express the Circular Measure of an Angle m terms of its tangent. .^, - ^A'.y(259). 282 PLANE TEIGONOMETRY. Equating the real and imaginary parts of this equation we have \^ ■ ^ tan2(9 tan*^ tan' ^ . ' " 0=Joga cos 6+—- -— + -— &a, (303) ■ ^ tan8 tan* 5 tan' tf . A. .} 03= tane--^+-— ^+&c., (304) the last of which is known as Gregory's series, and is convergent for all angles whose tangent is not greater than 1, . . If tan d=x, so that ^=stan~^ x, the series may be written x^ x^ x"" x^ taii-ix=w--+---+- -(&o. ; (305) 214. To find the numerical value of ir, TT TT In Gregory's series let 6=--, then since tan — =1 we have 4 4 TT ^ 1 1 1 1 ■' '=^(n"+^+"^+*^^-)- (306) This series converges too slowly to be of much use. To obtain a rapidly converging series, let TT — =tan~^ m+tan~* n ■ > 4 Of tan~U=tan-iJ^^, by (253) ^, , ^ m-\-n , 1 — n thereforb 1=, and m=:- 1 1-mn li-n If we make n=:^ , we find wi=^ t : \. , TT 1 1 • therefore --=tan-i— +tan-^ — - ^ 4^0 ^ ^-' ;;.'<>: . ;■ > i I ■J 1 l/lx» l/lx" I/I7 ^ 2 - 3(2) -^5 (2) -7(2)+*^- 1 1/1\' 1/lv" 1/lJ ^ (307). SERIES FOR COMPUTING tt. 283 which are Euler's series for the computation of tt. They converge much more rapidly than (306), , ^ , t ;,. 215. Machin's Series for computing w. 1 ./. .": '. 1 5 2 tan-* 1 - tan-^ — =taii-i ^ =tan-i — 6 ^1 o . I i ' ,■ • J / u V 2 J^ 3 ~ 5 ^ - ««x g -v«- 2 17 2 1 3 5 7 tftn-i — -tan-* — =tan-* — — -=tan-i — ■^•"3:6 7 1 tap-* - - tan-* -=tan ^=:tan-* - 17.5 9 1 46~ 5 . , /-I %J J'i;i n^jt . n >.,.'.,U7 X -^'vv^ =-t«n-* — . ■' •'■l^^.i.a . ^ 239 _ Adding these equations and cancelling the terms common to both sides, we have ■ '"f;\ , ,5..:v -■•■-. ^-v tan- * 1 - 4 tan-* ^= - tan-* 1 . ' ", t : v^ ^■- ) -i -i '1--' 239 1.^-11=4 tan-* --tan-*—, 'V:\ «,i^ ih«.for. -= ^ ^ , ^ h (308) 284» PLANE TRIGONOMETBY. These serien converge quite rapidly, especially the latter. If we take eigh! terms of the first and three of the second we find 7r=3. 141592653589793 .... For other series for computing tt, see the examples at the end of this chapter. Expansion of certain Trigonometrical Equations into Series. 216. Cfiven sin p'=sin P sm (z+p), it is reqmred to express p in a series of multiples of z. [See (a), Art. 151.] Multiplying both members of the given equation by 2v^^, we have 2v/ - 1 sin p=sin P . 2v/^ sin (z+p) or e^^'^-e-^^'^sinPCe^'+^'^-e-"-^^'^), by (295) whence 6^^=^"""^^"^^', l-8inPe'^-*' and taking the Napierian logarithms of both members we have 2pv/rr=log (1 - Bin Pe-^"^) - log (1 - sin Pe' ^) - sin Pe-*^ - ^ 8in« Pe'^^^-i 8in3 Pe" > + ainPe'^-'H-i8m«Pe''^ +i8in3Pe'' by (260) =8in P(e' ^-e-*^^)+ J sin« P(e^^-e-^^) ^ +*8in3P(e'*^^-e-^^=^)4-.... =8inP.2v/^sin a+^sin* P.2\/^sin 2« therefore / ^ . +J ain3P.2v/Tr8in3.+.... p=Rin P sin z+^ sin« P sin 2»+i sins P sin 3«+. . . . (309) Here, p is expressed in circular measure ; to find p in seconds we must divide both members by sin 1", according to (128), and since e - .... I TRIGONOMETRICAL SERIES, 285 2 Bin l''=8in 2", 3 sin l"=Bin 3", &c., approximately, the last equation may be written thus, „ sin P sin z sin^ P pin 2z , sin^ P sin 32 , .„- ..v '^--^v- + -Iter- + -STs" +•••• ^^^'^ ^ac. —Given P=58' 10" and a=40% to find p. (Same as Ex. 64, Chapter X.) Bin P=8. 228380 sin' P=6. 4567 sin^ P=4. 685 sin z=9. 808067 sin 2z=9. 9933 sin 32=9. 937 cosec 1"=5. 314425 cosec 2"=5.0134 co3ec3"=4837 2243". 23=3. 350872 29". 07=1. 4634 0". 29=^. 460 |)=2243".23+29".07+0".29=37' 52 .6. * Here we omit the symbol Log for the sake of brevity. '• "'■ '■' 217. CUven tan x=n tan y, to express xma series of multiples of y. Multiplying both members ol the given equation by v/-l, wo have by (296) . • • e -1 e -1 • ... ;.. • , e — — : =n -pz — » *,V^ (l + ^)e''^ ^-^4-(l-n) . ..V':. A.-: (l-n)e +(! + ») =- ' li.' r'lt 3y^. 1-n ^■•'-- ^•.•'- " e +-;; 1-W 2yV-l , t ..... \ 1 4-1::^ e'^^-' ■:\\-<-rVyK>'- 1- \ 14-— 6 1+W ^r. ,..„...-..;,,•;.• .-,{^- '■ ' ayV=i/l +me' 2i/>/^ / --^ ssse"' ' 'I — ^ 1 1 I V S^r - ■ >: \v Jiiio ,^...;; ,.•?;, %!; « • • u+me ' . =7:^ it;- ■;; >a V 286 PLANE TRIGONOMETRY. 1-n by putting m=- •^ *^ ® 1+n or e ,i^,2,^^lj-me___ ^ ^^ .j^ logarithms 1+me (a; - y) 2v/ - 1 =log (l+me"^"^) - log (l+we^^^) -I me -^m'e +-J m^e -.... » =s-m2\/-l BinJ2t/+im^2s/-l sin 4y - J m« 2\/"^ sin 6y+.... therefore «=i/ - m sin 2y +^ m« sin 4i/ - J m^ sin 6j/ + . . . . (311) ^*_,/' wsin2y m«Bin4y m» sin 6y sin 1 sin 2s sin o o ^» ** "1^ y ^ ... 21o. GwcH rm «;=; , fo emress x vn, a senea of nrnti- 1-ncosy pUs oj % * . -. . , The given equation may be easily reduced to sin aj=n sin (w+y), ,. ... , , . (a) which is of the same form as the equation of Art. 216. Hence writing x for p, n for sin P, and y f or « in (309), we have at once «,=n sin y+l ri« sin 2i/+J n' sin 3i/+.... (313) Formulee (309)-(313) are very useful in Spherical Astronomy. The left-hand members are limited to acute angles, but entire gener- ality may be conferred upon them by observing that the equation tan x=n tan y is true when we write n'n+x for x and wV+t/ for y, f^d therefore for x-y we may write ac-y-(m'-n.')7r or x-y-pn TKIGONOMETRICAL SERIES. 287 where p is, like nf and m', any integer or ze 'o. Therefore (311) may be written thus : x=pTr+y - m sin 2i/+^m« sin 4i/ - . . . . In the same manner the other series of Articles 216-218 may be generalized. „ , 219. In a triangle ABG, given two sides a cmd b and the included angle C, to express either of the other angles by a series of multiples ofC, 5 sinB sin (A+G) ..'-.1 Since — =-: — 7= : — z a sin A sin A , . •'. ... .'-..,,.'■ ..,., we have sin -4=— sin (^+0), which, compared with (a) of the last Article, gives by (313) t a . ^ a^ sin 2(7 , a^ sin 3(7 ,„ a sin . a' sin 2(7 a' sin 30 , -o,.. which will be convergent when — is a proper fraction. - ^ ' ■ r' ■'■•' ■ ' 220. In a triangle, given two sides a and b and the included a/ngle C, to express cby a series of multiples of C, ' From Art. 122 we have • r. •''■ ' •; c2=a2+62-2a6cos (7 .-V'tV- .=ss b ^ b ^ =a2(l-2-cos (7+-) ' =a2(l --(«+-)+-), by (301) -_^ ^^.^. ==aMl--aO(l--) ^ a ^ oo"/ ..r' h b 2 U>ge C=2 loge a+loge (1 - - a5)+loge (l - -) a tttC 288 PLANE TRIGONOMETRY. 2a'' 62 3a^ 63 ax ^a^x"^ 2a^x3 -...., by (260) =2iogea — («+-)-7r^pH-i)--rT(» +-i)-"" a ^ X' 2a^ ^ x^' Sa^ ^ x^' b 62 63 =2 logs a 2 cos G-—~ 2 cos 2C- —-- 2 cos 3(7- a 2a^ ^'»3 3a'' loge C=logg a - (— COS C 62 2a2 63 COS 20- 63 3a= 63 cos 3(7+....) or logc=loga-iltf(— cosO+— — cos20+— — cos3(7+....), (315) ^a 2a2 3a3 /' ^ where ilf is the modulus. , re88 r in a series of rmdtiples 1+e cos d of 6; e being Uss than 1. (The Polar £Jquation to the Ellipse.) 26 Assume e=- — — • , which is always possible, since 1+6^ > 26. then 1+6 I_e2^/Ll^f and 6= = h + b^f l + v/f- _ ^ 1 V Let 2co8^==»+— , then 1+e cos ^=1 + 1+6 -. (-i) therefore =1TP <'+^> (^4)' B=a +6= (1-62)2 1+63 + 6a _1 1 + (1 - 6x+62a;3 - b^x^+b*x* -...,)\ .^ b b^ b\b* x X* x^ as' TRIGONOMETRICAL SERIES. 289 ^(l::^l!|(i +&2+6*+?>«+b" +....) + ( • ) •••• J* 1 But (1 +62+6*+6« + ....)= JTi? b -(6 +63+b^+?>' + ..-.)=-][r6i - (63 H-b"^ +6^ +6" + . . . .)= - ^TftJ + ( )= •••• W:^^ ''" Therefore __« ^^^'^ / 1 - 26 cos 0+262 cos 2d - 26^ cos 36+....} =a v/i^'^ (1 - 26 cos 0+262 cos 26 - 26^ cos 30+ .... )• (316) 222. To shew that iV i''^ . .Vrr/^ cos - cos^ COS -....ad W:=-^^. ^v V- 8inx=2 sin -cos- .^_ ., =2« cos I sin I cos I , since sin |=2 sin | cos ^ , r 20 ■ ■ 290 * ■ PLANE TRIGONOMETRY. ^, X X , X X =2 cos — cos - sin —:: COS — =2'» COS I cos ^ COS ^ . . . .COS ~ sin ^. _, . XXX X mnx Therefore cos — cos — cos r^ , . . .cos — = 2- sin -- . a; Bin — X 2** but 2" sin — =x —x, when n=oc , by Art. 74. TT « 05 jc , . , sin aj ^ ^ Hence cos "^ cos — cos — ad inf.= . (OJ 7 4 2^ 2" X Whence we also get • • ' XXX aj=sin (B sec — sec — sec — ... .ad inf. (318) 2 2^ 2" Multiplying (317) by cos x and expressing in logarithms we haV j 1 .1 ^1 * -• • .. 1 /sin 2«x log cos a+log cos — +log cos — + ad mf.=log (-— — ) . 2 , . }L ^ 2x ' Summation of Trigonometrical Series. 223. To find the mim of the sines .v-^„ ^ , >. - (1) sin x+sin (a;+i/)+sin (a!+2|/)+&c., to n terms. By (54) we have cos (a5--jr) -cos («+— ) =2 sin — sm x, / 1/ \ 3 y cos (a5+-T) - cos («+-^ l/)=2 sin ^^ sin (x+y), 3 5 « '''"* -~*^^?^ cos (»+— y) - COS (x+— y)=2 sin — sin (x+2y) , .. ►• ^ 4 Ji TRIGONOMETRICAL SERIES. * 291 cos (x+?^y)-c5«(x+?^iy)=2 sin fain (x+(n-l)i/), Lot -S denote the sum of the series, then we have by addition cos (» - ^) - cos (x+ -y- y) s=2 sin " (sin x+sin {x+y)+am (x+2y)+ . . . .) s=2Bin|--5 cos («5 - "I) - COB (36+-^- y; , . ,. whence S= 2'^ a (319) sin(x+— ^l/)8ln- (2) cos X+C08 (x+y) +cos (x+2y)+&c. , to n terms. By (52) we have .^ rin(.+|) -.in(«-p =2.in|co,«. ^^^.^^^^ •fa (a+|i/)-Bin (x+l) =2 sin | cos (x+y), *--^ Bln(x+|i/)-flin(x+2^)=2 8in|co8(x+2t/), \ "■*»» am (x+?^ y) - sin (x+?^ y)=2 sin ^ cos (x+(u - l)y). Let 8 denote the sum of the proposed series, then we have by addition Bm(x+?^-v)'-«in (x-|)=2Bm | (co.»+cos(x+y)+&o....) 292 PLANE TRiaONOMETRY. =2 sin I ' 8' 2 sin (x+— ^ y) - sin [% - -) whence 5— 2si„| W — 1 WW COB («+— ^ 1/) sin — -J (320) . V sin — 2 224. Tf t/=x, 2a5, &c. , in succession, we find from (319) and (320) Sin — - — X sin — • Bina+8in2ac+8in3a8+&c...+sin»Mi; — (321) sin — ('!;*.) ,— '■•'•- -^ ■ 2 . >~ -V sin'** ■na y«^«v sin x+sm 3a54 am 5a;+&c. . . +sin (2»i- l)a;= -; . (322) sm a &c., &a« &o. . , cos — - — X sm -- ' ' * 2 2 oos«+co8 2a5+co8 3a+&c...+costM5 — - (323) . « r . . ,; e'.n — . . ^ ■ .w I',,;- :;:'■■ ( •.. t":, - •■ 2 ■ ,- . /« ^v sin 2n» ^„^^^ cos jc+cos 3«+cos oas+ftc... +co8 (2n-l)a;=— — : . (324) 2 sin « &C., &c., &c. 225. The formuliB of the last two Articles enable us to find the sum of the squares of the sines or cosines of a series of angles in arithmetical progression ; thus, let sin^ ac+sin^ (x+i/)+8in2 (a;+2i/)+&c., to n terms be the proposed series. ' ; ^ TRiaoNOMETlUCAL SERIES. 293 Since 2 sin^ x=l - cos 2x, 2 sin« ix+y)=l - cos (2x+2i/) Ac. , we have by substitution 2S=l-co8 2x+l-co8(2x+2i/)+l-co8(2x+4i/)+l-cos(2a;+6i/)+&o., to n terms =» - (cos 2X+C08 (2x+2i/)+cos (2x4-4i/)+&c. , to n terms) cos(2x+(n-l)y)8inny ^^ ^^^Q) ""*''" siny * and n cos (2x-\-{n-l)y) sin ny *^~2 2 amy In a similar manner we may find the sum of the series cos" oj+cos^ {x+y)+coa^ {x+2y)+&G., to n terms by using 2 cos* x=l+coa 2x, &o. • - ^ ' 226. To find the sum of n term of a aeries of the form i y ^ sin X cos y+svn2x cos Zy+dkc .... +sin nx cos (2n - l)l/. .By (61) we have : , ^ ^ r ^ ^ , . j^ -. ^ ^.^ ' Bin (x+y) +sin (x-y)=2sinx cosy • ' ' sin (2a+32/)+sin (2x - 3y)=2 sin 2x cos 3y, &c., &c- 1^^ . Thus by addition the proposed series is resolved into two others which can be summed by the method of Art. 223. 227. TofindthemmofnUrmofth^seriAS ,, ,,^ ^ tan x+2 torn 2x+4 tan 4x+tti'i.> t»V» (a a* a' % — I 1 I-&C., to n terms.) X x« x3 ' ax (a*» x" - 1) ax-* (a" x-** - 1) - ax-1 ax-*-l »(Colenflo'sAlg.,Art.l57) an+2 (x» - X-") - a»+* (x"+Hx-^*) -t- a (x - x-*) "" a8-a(x+x-i)+l * therefore a/;:.! o-* <^?K: ■■'. a«+» sin nfl - a"+i sin (n+l)0+a sm * 2s/^S=cos e (X-C^V'-^' (x^ -x-)+'-^' (x»-x-)-&0, ^o^ x^_co^^^^'^^ log (1+x cose) 2 3 x-^cos0+^^— ^ 3 + &c. = -log (1+xr-i cos d) 1+XCO8 ,, . 2SV:=I _1+X£OS0 -'^^ 1+^^ cose * ''''''''''' /=U^^e* .>..o and by composition and division, .^Uiu: • ■ •- e'^^^-l _co3_g_(g-ar-^) _ 2 cos sin v^^ - ■ " 2SV-1 ^""S+coseCx + x-O 2+2cos2 * 2 cos e sin vA-1 --- ' thert-ore by (296) v^-ltanfe=- a(i+cos''e) " and .- - ^-tan-^ (^(I^). ' ^ 231. The investigation of Trigonometrical Series cannot be fuUy carried on without the aid of the Differential Calculus. The • student must therefore consult the treatises on that branch of mathe- matics for further information on this subject. (See Todhunter's Diff. Cal., chaps, vi., viL ; Clark's CaL, chaps, v., vi; Williamson's Diff: Cal., chap, iii.) i v' • ■ ^^' "''''' Examples. 1. Prove by the aid of (294), (296) and (296) the foUowing identities : .:. .. /^ V X ^ 1-CO80 \ , .\ - ^ /2\ cos 2&=-u-^+J^(|;._|_(^;,^.,^^. 6. Prove that versin «= - Ke'* - « "a" "^"^^a, 6. Prove that (l+cos e+v^TIsin 0)" +(l+cos 5- v/TTsin e)»=2«+icos»- cos- 2.2* 7. If aH6«=2, and a6=v/rrtan 20, shew that a=cos 0^2 sec 2^. a If 2 cos a„ =a„ +- , and ai+aa+a,+ . . . . +a, =2»r, \SC?S- ^n then will sCjaCaXa Xn =1. 9. Prove that 2 sin (aj±y v/^)=(ev +erv) sin x±(ev. e-*) >/^cos aj. 10. If 5i=tan ^+tan J5+tan 0+&c.=sum of the tangents, ^2=sum of their products taken 2 and 2, Sg=avim of their products taken 3 and 3, &c., ^ &o., shewthat >. '.w-'.') <:,■-•:-; :;i-- iji'. tan(J[+^+C+&c.)=^^=#^~. ■ " ■' 11. In any triangle shew that log —==.¥{ (cos 2Jl - cos 2B)+^(cos 4A - cos 4B) - +i(cos 6ui - cos 65)+&c. } iV •UT EXAMPLES. ; 297 12. Prove that loge sec 6=^ tan* d-^ tan* d+i tan' - i tan^ 0+&c . . . . 13. Prove by Gregory's series that 3eo-i-(-+-)=-(--l)(- + l)-— -(--l)(-+l)+&c 14. Prove that log tan* ( j+ d) =tan e+\ tana e+i tan^ 0+&c.... 15. Shewth»t|=i+-L+^^+.... 16 Prove that — =4 tan-^ — - tan""^ — + tan"^ — : •" ' ' 4 5 70 99 and — =4 tan~^ — - 2 tan-* — — -+tan-i 4 5 408 1393' 17. K tan 2e=8in 2^, shew that • ; • ■' .. ^ ■ - ^} .^- 6=cos 2^ tan ^ - i cos 60 tan^ 0+i cos 10^ tan* ^ -> • • • • Sum to n terms the following four series : -. ';' t '- n Q a a a ..>..'. ^ 18. tan - sec e+tan — sec — +tan — sec — +. ... » • 2 4 a o 4 B ' Ans. tan d- tan — . 2* 1 1 '• 1 19. — sec 0+— ace d sec 20+ rr. sec 6 sec 20 sec 22 0+ 2 2* 2'' ... : ' .4m. cos - sin cot 2" Q. 20. sin («-y)+sin (2a;-3i/)+sin (3ae-5y)+.... ' - sin {\{n + l)x-ny\ 8in(|fix-ni/) Am. —— • sm i(x - 2i/) 21. — log tan 2x+— log tan 22x+— log tan 2»x+ .... 1 « • o log 2 sin 2«+ia5 ilna. log 2 sin 2x — • Sum to infinity the following twelve series: ^^ cos 6 cos^ A cos' .. 1 / . ^ .\ 22. — ~- + — r-^ H — r— + . . . . Aim. log (cot — coseo fj, \ i O A 298 PLANE TRIGONOMETRY. cos 20 cos Ad cos 6d 23. 1 + 1.2 ^1.2.3.4' 1.2.3.4.5.6 Ans. ^ cos (sin 6) (c*-"'* Ve"""*). 24. sin a sin a+^ »in^ a sin 3a+i sin^ a sin 5a+ .... i •-" Ans. ^ tan-^ (2 tan* a). or or 25. X cos (a+i3)+-— r cos (a+2/3) +r-r-^ cos (a+3i3) + ...., .4rw. e* "^^ ^ cos (a+oc sin /?) - cos a. - - Tj'ii-. a 26. sin a-^ sin 2a+^ sin 3a- J sin 4a+.... .4m. - - . 27. sin ac sin ac-^ sin 2x sin*«+| sin 3x sin' «— .. .. . , ,, Ans. cot~^ (1+cot cc+cot* aj). 28. cos 2^+^ COS* 20+i cos^ 26+ .... -. .. Ana. log cot d. 1 + 23 1 + 2' 29. (1+2) log 2+-p^ (log 2)3 + j-^ (log 2)» + . . . . ^m. 4. 30. 2 {ain^ ac - ^ sin* 2»+J sin* 3a; - i sin* 4«;+ }. Ana. log sec x. 31. K cos x+i cos* x+J cos* x+ . . . .=/S>, and cos x+i cos 2«+^ cos 3x+ . . . .=», u ji-. u.v, shew that "" " 4 ^' . cos 9 , ,^ „. An>s. - log (1 - cos 6). 34. Prove by Art 222 that - - .*iu, (l - tan" — ) (l - tan* -^ (l - tan* rj) . . . .ad inf.=x cot x. 35. Shew that - n n(?i-l) „ „ n(n-l)(n-2) , 1+— a cos wxH — r-,r- ine By completing the square and extracting the square root we liave a:» =co8 0± \/ - 1 sin 0=co8 (2r7r+(^) ± v' - 1 sin (2r7r+0) whence aj=|cos (2r7r+^)±v -1 sm (2/'ir+0)| » sxcos ^±v/-lsm ^, „ *'(330) by De Moivre's Theorem. 1.* Aa.*.\ I'. '• J . ■•..xs\.t\*it% i..*\\ Therefore the 2n values of x will be found by assigning to r the values 0, 1, 2, .... n - 1, in succession ; thus, using first the upper sign, we have :■ ; ,'■. 'v^nn ,i.i^- ^^ '^'i- '^^ imrA ^^n tk'iu^ (1) if r=:0, 8C=co8 — + v/'^8in— , .i?b'x.K ■^% n n "it + d> / — — - . 27r + (2) rx=l, a;=cos |-v-lsm , (3) rs=2, a5=cos hv-lsm , * 2(n-l)7rH-^ . /— -- . 2(n-l)^+0 (n) r=n - 1, «=co8 h v - 1 sm — ; ^ ' n n 300 PLANE TRIGONOMETRY, and using the lower sign. (1) if r=0, «=cos -^ - \/rr sin ^ , n n (2) r=l, ai!=cos - n/ - 1 sin ^ , n n (8) r=2, x=cos ^ - V" 1 sm , n n (n) r==n- 1, a;=cos -^ — ~ - v -1 sin -^ L_Lr ^ n n Now, from the first of each of these groups we have the first two ■imple lectors {as-cos s/^BUi — ) and f jc-cos — + \/^sin — ) , the product of which is «2-2a; cos — +1, the first quadratic factor; from the second of each we have the second two simple factors (a-cos ^->/-lsm ^) and (a-cos ^+\/-lsin — ~^\, the product of which is a;8-2« cos — — ^+1, the second quadratic factor ; and so on, there being 2w simple factors in all. Therefore we have for the w quadratic factors a6*»-2»"cos^+l= (a;2_2xcos-^+l) w • ' • "J , 27r + <6 x(a52-2a;cos -^-1) n ' r-n X (X^ - 2x COS -^^+1) • ': n / , « 2(n-l)7r+(4 ;•,' X (a;2 - 2x COS ^-^+1 . (331) RESOLUTION OF EQUATIONS. 301 Since the simple factors of a;2'»-2oc'* cos ^4-1 are obtained from (330) by making r=0, 1, 2, m- 1 in succession, it is evident that they are all included in the form / 2/'7r + , — — . 2r7r + 0v .„„^- % - (cos ~~- ± \/ - 1 sin -\ (332) Ex. Besolvc the equation x*-x^+l=0 into its quadratic and simple factors, ^ere n—2, 2 cos ^=1, therefore <}>=—, and by (331) o If ^TT fl6*-a2+l= (x2-2acos — +1) (»2-2xcos — +1) 6 o = (x - cos — - v/ - 1 sin — -) X -COS — -H-v -1 sm — I ■■.'.., .... •; :■ ..n X (x +COS -+ v' - 1 sin — ) .. ^ ^ p o ' ' - •■;■'•» :/ '■'"■'■'■''■' '' ' ■' - > ,, . X (x +C08 — -\/^sin — ), by (332) X (x +iv/3"+K^) X (X +^v/"3-^\/"^. -w.^ .,•..■ , ,,'7' ,;;,fc. Hence the four roots of x<-x2+l=0, are " • r\-^^^^*Pi: , Kv/3"±v/^ and ^(-v/3"±v/^. . m! 233. In (331) let x=l, then we shall have 2 - 2 cos <6=(2 - 2 cos — ) (2 - 2 cos -^^—^) (2-2 cos -^^) , &o., Of . , . to n factors, 23 8in2 ^=2«» sin' /- sin^ V^ si^^ — ^ , &c., 2 2ii 2n 2n ' ' 302 PLANE TRIGONOMETRY. md writing 2nB for (p, dividing by 2" and oxtrafl>ting the square ro(>( , we have Q O sin tie=2'»-i sin d sin (#+— ) sin (e+—) sin (d+—) , &c. , (333) , • - > , ; to n factors. Let n^=— or fl=— ; — , then we have 1=2'*"^ sm — — sin — — sin — — , &c., to n factors. (334) 2n 2n 2n ' * ^ Again, in (331) let jc=— 1, then we shall have .: (n even) 2-2 cos =tan(9tan (0+— ) tan (0+ ), &c., 3; C) (rj. odd)itan w0 j ^ n^ ^ n ^ to 11 factors. If In (336) let n0=— , then we have, whether n be even or odd, l=*an — tan — tan — , &c. , to n factors. (337 ) 4n- 4>i- 4ri, , , 234. To resolve the Equation as" -1=0 into ity Quadratic Factors, n being odd. >, .t m In (331) let ^=0 and it becomes (»»- 1)2=(X - 1)2 X («2 _ 2X COS - +1) n ■ ■ V , ,*''••• . , X (i«2 _ 2x cos — +1) . ,,' .., "-. }i{9a9-2xco^^ -^+1), •.;. (338) RESOLUTION OF EQUATIONS. SOIj Now, as n is odd, and as there are in all n factors, the number of (juadratic factors in (338), exclusive of (x-iy, is even; and since 2{n-l)ir , 2k. 277 COS =C08 I2vr )=cos — ; n ^ n' n 2(?i-2)ir / 47rv cos =C08 n ^ n' n I 47rv 4n- {2n ) =cos — , and so on : the first and the last of these factors are equal ; the second and the last but one are equal, and so on ; hence uniting these equal factors and extracting the square root, we have, when n is odd, 27r a^ - l=(x - 1) X (x2 - 2x cos - +1) 91- , 47r X (x* - 20* cos - +1) n '■•■'■ ""-" ■■•''• ^' (n-l)fl- ,, •-■ X (x« - 2k cos +1). . } (339) n 235. To resolve the Equation aj»-l=0 into its Quadratic Factors, n being even. • I ■ .1. . ^1 1 (•••«• When n is eoen the number of quadratic factors in (338), exclusive of (x- 1)2, is odd, and there is a middle factor, the — , which will not combine with any other. This factor ia ^ n 2 «• - 2x cos -«»— +I=x2+2x+l=fa;+l)«. n ' '/ '^ Hence uniting the other factors and extracting the square root, we have, when n is even, 27r x" - l=(x - 1) (x+1) X (x2 - 2x cos — +1) n L.1 1 s) :> '] t '; - J •' 'tr X (x3 - 2x cos - +1) ' - '^^ -^~' '^' 47r n - • - .,/,'f^-'£i '^ x(x2_2xcos^^ ^+1), (340) 304 PLANE TRIGONOMETRY. 236. To resolve the Equation 05" + 1=0 into its Quadratic Factors, n being odd. In (331) let (j)=Tr and it becomes («" +1)»= (x2 - 2x cos -+1) x(x* -2» cos — 1-1) n x(x2-2a;coB — +1) 71 ^(^..2xcoB-^?^?^^^+l). (341) Since there are n quadratic factors and n is odd, there is a middle factor which will not combine with any other. This factor is evidently x« - 2x cos ^-fl=(x+l)«, f... , n ■. r- , ,-::' and it ia easily shewn, as in Art. 234, that the factors equally dis- tant from the first and last, are equal; hence uniting the equal factors and extracting the square root, we have, when n is odd, flc" + l=(x + 1) X (x* - 2x cos —- f-1) n ■ I X (x' - 2x cos — hi) n x(x'-2x cos — hi) n . . {n-2W X (x» - 2x cos ^ ^ 4-1). (342) n 237. To resolve the Equation a;"+l*0 into its Quadratic Factors, n being even. When n is even the number of quadratic factors in (341) is even ) therefore we have ^^ RESOLUTION OP EQUATIONS. 805 IT as* + 1= (x^ - 2x cos -+1) n x(«'-2x cos — +1) 11 x(x'-2»cos — +1) n x(«»-2xco8^^^^^7r+l). (343) n 238. The simple factors in each case may be found by resolving each of the quadratic factors ; or, they may be found from (332) by putting 0=0 for the form x" - 1=0, and ,f . <.». By (343) we have aj*+l=(xa _ 2x cos 45''+l) (x^ - 2x cos 135'+1) ' '■" . =(x2-\/2'x+l)(x»+v^x+l). .--r, Bence x== -= — and -= . v/2 v/2 "if' 239. In (339), divide both sides by x - 1, thus 27r x'^i+af-»+ x+l= (x^- 2x008 — +1) X '.':'' , x(xa-2xcos-^^^^^7r+l) n . n « W-1 ^. X (x^ - 2x cos o"-!-!), n-1 ''' uiiere being — r — quadratic factors. 21 ,::. - ,• ' ,;; 'Vila *••^"l^- T<> m'H ■■■.. 't .r T -i) 306 PLANE TRiaONOMETRY. Put x=l, then we have n— 1 n=2 » (l-ooB -)(l -COB -)...(! -cos iili7r)(l- cog -!LLl;r). o«_i • « 27r 47r n - 3 vt - 1 3*2**"^* flin* -- sin* — ... sm* — r — tt sin' ir, 2n 2n 2n 2*1 ' or T iTT" • 2ff . 47r . n - 3 . n - 1 ,, , , n^=s2 ^ Bin — sin — .... sin — - — ir sin ir. (344, 2n 2n 2n 2n ^ 240. In (340), divide by ao-l and put x=l', then we have, w — 2 there being —r — quadratic factors, "~ 27r\ /- 47r\ ,, n-4 v,^ M-2 , n=s2.2 2 (X-cos —)(l-cos —)...(! -cos — ^^7r)(l-cos rt -^;.- 47r .«»i-4 ..n-2 ^«-«,«27r.,47r .„»i-4 ., =2.2*^' sin' — sin' -- .... sm' — r — tt sin' ■IT, 2w 2n 2n 2»i or 1 fl-i 2^ . 47r . n-4 . m-2 n'=2 ' sin — sin — .... sin — - — «• sin —- — tt. ' (346) 2n 2w 2n ^r ^ 241. In (342), divide by x+1 and put x=l; then we have, there being quadratic factors, nr-\ 1=2 ' (l-cos— ) (l- COB — )...(! -cos — ^^7r)(l-C0S — - — tt), ,.„'r.o3^ .-n-4 .-n-2 =2^1 Sin* — sin* —....sm* — — tt sin* -- — tt, 2n 2n 2n 2» , .1 .p • or 1=2 * Bin — sin — .... sin -— — re sin —- — tt, (346) 2n 2n ^ 2n 242* In (343), put x=l and we have 2=2^ (l-COS— ) (l-COB — )...(1-C0S — - — 7r)(l-C08-- tt) \ n^ ^ ' n' n n or - r-j— . ir . Sir . n — 3 . n — 1 r . . /njt-,\ 1=2 ' sm — Sin —....sin — - — ir sin — - — ir. ; ^ (347) 2n 2n 2n 2n c >'^ RESOLUTION OF EQUATIONS. 307 343. To resolve sin x and cos x into factors. The series for ain x, Art. 210, may be written thus .i„,=,(l._|_^_|_.4e.), (348) from which we see that x is one factor, and the factors of the series within the parenthesis must evidently be of the form x« a where a is constant but has a different value in each factor. For suppose the factors to b«) 1- — , 1-—, 1- — , 40., ^he product of these will give a series of the same form as that vithin the parenthesis. Now, the required factors must reduce /he second member to zero when the first member is zero, that is, nrhen x=0 qx^=^±w^\ therefore the general form of the factor is 1 ; a and since this must be equal to zero when x=±n7r, we have a , whence a=n'7r«, ;« ^ , - o' x« therefore the general factor is 1 — --; ^ . ' =• n«7ra * -■ ■■' - '1 :,<.; then making n=l, 2, 3, &c., in succession, (348) becomes «nx=x(l-— )(1-— )(1-— ).... "(349) Again, the factors of the series for cos x, Art. 210, must also be of the form -. ■' 1--; :, - , ' • • a IT but the first member is zero for «=±(2n+l) — , where n is any iteger including zero ; therefore we have ' 308 PLANE TRIGONOMETRY. l_<>:±}ll!=^ 22a whence (2n+l)^7r« 2^ 22x2 therefore the general factor is 1 - 72~TTTa^ ' and making n=0, 1, 2, &c. , in succession, we have 22x2 2^x2, 22x2, COS x=(l- — ) (1- — ) (l-gi;;:,) .... ^50) Logarithmic sines and cosines. 244. The last two series furnish us with the means of calcu- lating the logarithmic sines and cosines without first computing the natural sines and cosines. . -. . j. . . In (349) and (350) put x=m—, then m'x •V C08m-=(l--)(l-^)(l-g2).... mK and expressing these by logarithms we have log sm m^=log ^+log m+log (l- ^) +log (l- ^; +*c-i logco»m-=log (1-^J +log (l-^)+log (l-g^)+*c- Developing the second members of these equations by the loga- rithmic series, and arranging according to the powers of m, we obtain WITT «■ , ^ / 1,1 1 t \ ^ logsin— «=log-+logm-m2. J-(^ + ^ + g-2 + *c.) , M,l 1 1 ^ V 2 ^2* 4* 6* ' • Ac. '»r:j; Fl ii <:/■■ ."..■^■i '^ \' a (851) ■i * -" Examples. 309 log cos -^= Mfl 1 1 ^ V 'S 1 ua 32 S^* ' M,l 1 1 ^ V -m*' — I— H — + — +&C. I ^/l 1.1 ^ \ &o. (352) By summing the constant numerical series, substituting the value of the modulus M and giving m different values, the loga- rithmic sine and cosine of any angle may be easily computed. Of course 10 must be added to these expressions to give the tabular logarithmic sine, &c. 1. Prove that Examples. . fe-*=2(l + -—){l +-—).... and e* -e-«=2«(l +— ) (1 +;;;:— ) 2. Shew that sin 20" sin 40° sin 60° sin 80°=- . lo 3. If 0=cos 0, shew that 0=42° 20' very nearly, 4. Eliminate d between the equations '^^"^ l-a2 a2 cos* 6= — - — , tan a=tan3 — . 3 . 2 jf.'.. -'r '. i< ■: • . u 'fir. Ans. sin' a+cos* o=(2a)S. 'H>i\-i) II 310 PLANE TRIGONOMETRY. s^> MISCELLANEOUS EXAMPLES. 1. Shew that cos^ (45° - e)—\ (1 + cosec 2B) sin 2d, 2. If m tan {a~e) sec^ e=n tan H sec^ (a - «), find ft ^n». e=i(a-tan-^ — — tan a), 3. In any triangle prove that coa--+co8--+cos — =4cos (45°-—) cos (45°-—) cos (45°- -). a a u ' 4 * 4 * 4 4. From the figure of Art. 121 prove formulce (186) and (187) geometrically. . , , 5. In any triangle prove that tan — +tan - =cos — sec — sec — . u a M A A 6. Solve the equation sin 5x=sin x+cos 3x. TT , . Am. «=(2n+l) - . 7. If a, 6, c, d be the sides of a quadrilateral circumscribed about a circle, shew that the area of the quadrilateral is v'ahcd sin 6, where 26 is the sum of two opposite angles. ,. ,.. ,,,..: A B C 8. In any triangle shew that sin — +sin 77 > cos — . 2 J 2 9. In the ambiguous case, o, 6 and A being given, if Cj , c^ are the third sides of the two triangles, shew that the distance between the centres of the circumscribing circles is ^(c^ - c,) cosec A. 10. Prove that (1+tan — +sec — ) (1+tan — -sec ^)=2 tan — . * 2 2* A '=180°, prove that « '-' 3 y , a+3 p+y a+y cos — +C0S --+COS — =4 cos — — cos — .— CCS — — , 2 2 2 4 4 4 20. Prove that sin 5° sin 15° sin 26°. . . .sin 85°=2 K 812 PLANE TRIGONOMETRl. 21. In any triangle prove that J2 _ g2 g2 _ ^2 ^2 _ 52 sin 2A+——- sin 2B+ sin 2(7=0. a2 6' c2 22. If ^1 , rj , rg denote the radii of the circles inscribed between the inscribed circle and the sides containing the angles A, B, C respectively, prove that v/ r-ir, + v/ri'i-j + v/r^,=:r. 23. If sin-i e+sin-i -=^ , then ». MISCELLANEOUS EXAMPLES. 313 32. Through the centre of the circumscribing circle of » tri- angle ABCi AOD is drawn to meet BG in D, prove that OD'.BD: CD=coa A : sin 20 : sin 2B. 33. In any triangle prove that cot A= ; • abc 2ww w' — ti' ff 34. Prove that sin-^ + sin-^ =— • m^+M* m^ + n^ 2 35. If p, a, r be the perpendiculars from the angles of a triangle ABC, upon the opposite sides a, h, c respectively, shew that a sin A+h sin B+c sin 0=2(p cos A+q cos jB+r cob 0). 36. Shew that cos (9 - i cos 20+^ cos Sd - &c. , ad inf. =loge {2 cos -- j • 37. If cos" j3 tan (a+d)=6m^ (3 coi (a - 6), find 0. J.?i«. tan &=\/tan (a+/3) tan (a-fi), 38. If logo 6=m and logj, a=n, shew that ^" = I— • logft w \n 39. Prove that cos {^+x>/^)= — ;=-{(e* +e-*)-v/^(e« -e"*)} • V4 ^ 2^2^ :<1 and the angle ^ of a triangle, find a. a* - A cot — Aim, q — I *-■ 48. Given the area= A > the angle C and a+h=my find the sides of the triangle. Ans. a=|(m-f K m'"^— 8A cosec (7). 49. Given JB, r and p the perpendicular from G on c, to solve the triangle. . , . ^v »' sec fl , ^ . ^v (i> - >*) sec ^«.. cos 4(^+B)=;7= . 00. i(^-5)=^JL^^. where sin 0= -Jf- 60. Given the perimeter=25, C and jp the perpendicular from G ou c, to solve the triangle. V G Ans. c=s cos^ 6, where tan* 0=_- cot — . 51. If a )3, y be the distances between the centres of the escribed MISCELLANEOUS EXAMPLES. 315 circles, and a', (i', y' the distances between the centre of the inscribed circle and the centres of the escribed circles, prove that • =— , and that a^Y s , , , ri-r A r^-r B r.-r C a : /T : y : : cos — : — ; — cos — : oos — . ' ' a 2 6 2 c 2 52. If z and »' be the meridional zenith distances of the moon or a planet as Been from two observatories on the same meridian, and whose latitudes are and cp' respectively, one being N and the other S, r the earth's radius, D the distance of the moon or planet and P the horizontal parallax, shew that sin «+ sin «* («+«') -(0 + 0')' (2 + 2') -(0 + 0') sin «+sin «/ 63. 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. 54. Shew that, if Q and G be the centre of the circumscribed circle and the intersection of the perpendiculars of a triangle, QG'^=B^ {1-8 cos A ooa B ooa G). ^r 55. If a, Py y, d be the angles of a quadrilateral, prove that cos o+cos ;9+cos y+cos rf=4 cos i(a+/3) cos K/^+y) cos ^(a+y). 56. Shew that (1 +Bec 2x) (1 +sec 4x) (1 +sec Sas) .... (1+sec 2*» a5)=tan 2**« cot x, 57. If tan-i (»+ 1) v^ - tan-i -^^=cot-i 4 v^^ find x, V 2 . . Ana. 6 or - 2. 58. In any triangle if a ^, 6 2, c^ be in arithmetical progression, shew that Bin3.B / aa-c' \a aia B V 2oc / 59. Prove that 2» cos 6 cos 26 cos 2« cos 2" =cos + cos 3d + cos 50 + + cos (2'»+» - 1)^. 316 PLANE TRIGONOMETRY. / 60. If tan A sec -4 f tan B sec 2?-ftan C sec (?-|-2 tan A tan B tan (^=0, prove that sec^ A-\-9fyfi^ B + sec* 0=1 ±2 sec A sec B sec 0. 61. Straight lines -40, £0, CO are drawn bisecting the sides BGy AGf AB of a triangle in the points D, E, F respectively ; if »*i> *'a> ^8 *^6 the radii of the circles circumscribed about the tri- angles EOF, FDD, DOE, shew that OA^.r]^OB^.j>^_OC^^Tl^l a2+5«+c2 o2 6^ c2 ' 3 ■ a3 6'-' c2 ' a ^ a "^ a y ^ y 1 a 3 a 62. If cos 6= ■■■ ^ r-=i , shew that v/a2 + 6a f(a + 6v/-l)+ f (a-6v/ri)=2«/(a3+&a)oos|-. A« « .-/... sin ^ sin 20 sin 3(9 63. Sum to infinity — - — 23 2* Ans. 6 d ' 9 tan — hcot — 2 2 64. If -4, 5, 0, D are the angles of a quadrilateral, shew that tan -4-ftan B+tan O+tan D cot ^-l-cot jB + cot + cot D :tan A tan B tan G tan D. 65. If a and /? be the roots of the equation jb*-|>x+9=0, prove that tan-^ a+tan-^ )3=tan-^ - — . 1-q 66. Find 6 from the equation "■ 8in(30+a)+cos (30-o)=cos(45°+0) ■ * ' Ans. 0=45° and sin 20=j(^ cosec (45'*-f-a) - 1) • TT 67. If i>=rA } shew that . COB fH-COS 3^i-COS 5^+ +COS 11^=— . 'f ' I'r MISCELLANEOUS EXAMPLES. 317 68. Prove the formulfe, ^ , /TT , 0v . cos 3/9 8in*30 sin 4/? l±8m 0=2 8m2 (— ± —) , sin2 6 +coa' —-—=—- — . 4 ^ o «> 4 69. If sin (x+y\/ ~ l)=/3(co8 a+sin a\/- 1), shew that tana= oot X and B^=k(e''y+e-^ -2 cog 2x). 70. The area of a regular polygon circumscribed about a circle is a harmonic mean between the area of an inscribed regular polygon of the same number of sides and a circumscribed polygon of half the number of sides. 71. If 3 sin 0=sin (a—B), shew that •0=-- sin a - - • — sin 2a+- • -- sin 3a -&o. 72. Shew that e-* cos - i e-^ cos 30+^ e"^ cos 5(9 - ad inf. =h tan"^ , e* — e-* 73. Prove that loga m=loga b . logs c . logo d logj m. 74. Shew that / e 27r + Att + Bs/ e 2n + B 4:Tr+B\ ' (tan — +tan — - — +tan — — W cot — +cot — - — +cot — — — j=9. 75. A common tangent is drawn to two circles which touch each other and whose radii are r and 3r ; shew that the area of the curvilinear triangle bounded by the common tangent and the two circles is n 6 (4v/3--7r)r^. :, s, m: ^,. 76. Shew that sin* {x+y)+cos^( «- i/)=l+sin 2x sin 2y. 77. Given tan (tt cot 6)=cot (tt tan 6), find B. Ans. tan 0=^(2n+l± v/4n(n+ 1) - 15). 78. Shew tlMit when n is an odd integer, coan (—-«?)=( -1) " sinwc. r;> e/r . 318 PLANE TRiaONOMETRY. 79. Shew that (X * ^ / * * \* sin - +008 — ) = (sin --- - coi - ) . 80. In a right-angled triangle, given a+b- c=m and the angle A^ shew that ^~~7^ ^^^ 2 ^^^^^ \^° ""2 /* 81. K straight lines be drawn from any point in the circum- ference of a circle to the angular points of an inscribed regular polygon of n sides, shew that the sum of the squares of these lines =2nx (radius)*. 3 y/T + l 82. Solve the equation x^ — -7iX= — 7:^=—» 2 4^2 Ans. >/¥ cos 12% - y/2 cos 48% - v^ cos 72*. 83. A circle is inscribed in an equilateral triangle ; an equilateral triangle in the circle ; a circle again in the latter triangle, and so on : if r, fi, rj, r,, &c., be the radii of the circles, prove that r=ri+r,+rg+&c., ad inf. 84. The radii of two circles which intersect are r, r^ , and a the distance between their centres ; prove that the common chord =— I (r+ri+a) (r+Vi-a) (r-ri-\-a) (-r+r.+a) } . a ^ ' ' A C 4- & A 85. Prove that in any triangle tan (B+—) = — - tan — . ^ 2' c-o 2 86. Find the value of cos-^ J( - 1)*", m being any integer, and express the different values by one formula. Ans. {6n±i(3-(-l)-)}|. 87. If p be the perpendicular from the angle A oi & triangle upon the opposite side, shew that he a sin J. 4-6 sin ^+c sin (7 «= — . ^ 2 6c cos A+ac cos B+ah cos C 88. The straight lines which bisect the angles A and ^ of a tri- MISCELLANEOUS EXAMPLES. 319 angle ABC, meet the opposite aides in D and E respectively ; shew that the area of the triangle CED is A B A sin — sin -- sec \{C-A) sec |(C-5), where A is the area of the triangle ABC. 89. In any triangle prove that g' cos JjjB-C) b^ cos i^{C-A) c" cos iU-B) _ COB ^{B+C) COS i{C+A) COB ^{A+B) ^ o+o -t-oc;. 90. If the straight lines which bisect the angles A, B, C of a, triangle ABC meet the opposite sides in D,E, F respectively, shew that the area of the triangle DEF is „, . A , B . C B-C C-A A-B 2 A sm — - sm — sm — sec — - — sec — - — sec — - — , « ^ 2 '' 2 Jt 2 where A is the area of the triangle ABC. 91. If Pii Pii Pi denote the perpendiculars drawn from the centre of the circumscribed circle of a triangle to the sides a, 6, c respec- tively, shew that . /a 6 c \ ahe 4(-+-+-)= , ^Pl Pi Pt' PiPiPz and A:(p^ ^-p^ +p\)=a^ cot* A+h"^ cot* B+)' MISCELLANEOUS EXAMPLES. 321 1111 7r» 104. Shew that p+^+^+^a+^c., ad inf.=-. 105. In any triangle prove that A =sin — sm — sm — ( -;; — - 4- -; — r- + -; — t: I * 2 2 2Vsin^ sin 2i sin / 106. If a and /? be two different values of B which satisfy the cos ^ sin ^ 1 , , equation 1 ; — = — , shew that a c a cos ^ (a+(3)=h sin ^ (a+/9)=:c cos ^ (a - /3). 107. The angles .4, 5, (7 of a triangle are in descending order of magnitude; if another triangle be constructed having two of its angles (A - B), {B - C) and sides m, w, p, then will an-\-cm=bp. 108. In any right-angled triangle, G being the right angle, prove that a' cos A-\-h^ cos B=abc. 109. Three circles are so inscribed in a triangle that each touches the other two and two sides of the triangle ; prove that the radius of that lide^ which touches the sides ^£, J. (7, is 3 5 115. In any triangle prove thalf 1 2 «i 116. In any triangle, the square of the distance between the centre of the inscribed circle and the intersection of the perpen- dicularsis 4Ba-2Ur . a+b+e 117. If h and h be the diagonals of a quadrilateral and ^ their angle of intersection, shew that the area is ^hk sin /2 4 119. Eliminate 6 from the equations < . , ,. r 3a cos + a cos 35=4m 3a sin ^ - a sin 30=4*i ^'^ • i Result, a^ - m^ - n'^=3a^ m^ n* 120. Solve the equation tan-^ ( — -) - tan-^ ( — r)=-:r . ' " • V3-I' 121. Prove that e+tan-» (cot 2d)=tm-^ (cot ff). ' MISCELLANEOUS EXAMPLES. 323 122. In any triangle prove that A B C ^ r C082 — +C0S8 — +cos« - -1 *+-r-» 2 2 2 OB 123. Sum to w terms t»n-i «H-tan-i , , ^ +tan-^ - — T-r-+&c. l+1.2aja 1+2.3x2 Ans. tan~* rue. 124. Shew that sin^ 6 cos^ 0=- sin 0- - sin 50+- sin 3d. 8 16 16 126. Shew that sin 9°=| {Vz + y/b -Vb^ y/~d] - . 126. Given sin 2(a+0)+sin 2a=2 sin 26, find d. Ans. 0=tan~^ (3 tan a) - a. 127. In any triangle prove that A ^ B 0,11. acoss — +6 cos* — +c cos' — =A It:"' — I* 2 2 2 ^i( r' 128. In the figure of jLrt. 152, jgijove that - V OA.OB.OC^.^F+BD4^)=^4.}^mT.BD.CE. 129. In the figure of Art. 153, prove that if JBj, Uj, iZ, denote the radii of the circles described about the triangles OiBC, O^ACy O^AB respectively, BiB^B^—2R^r, 130. A circle whose radius is r is inscribed in a quadrilateral ; if tift^fta, ft denote the tangents into which the sides of the quadri- lateral are divided at the points of contact, prove that . ya j—1 . ^—1 . J— 1 . ^—1 131. The line joining the tops of two towers of unequal height makes an angle a with the horizontal plane on which they stand, and the distance between the extremities of their shadows when the sun is in the same vertical plane as the towers is h ; if /3 be the sun's altitude, shew that the distance between them is h cos a sin fi coscc (a - /?). 324 PLANE TRIGONOMETRY. 132. If a, p, y denote the perpendiculars drawn from the centre of the circumscribing circle of a triangle to its sides, shew that the radius of this circle is the positive value of li in the equation 123 _ („2 4.^2 +y2) jB _ 20/37=0. Shew also that ^+f^+^=- , where a, 6, c are the sides of ao oc ac 4 the triangle. 133. In any triangle shew that • tan2 —+tan2 — +tan'» "^ > ^» (a+6) cos (7+(a+c) coa P+(&+c) cos A > 2xany side. 134. Find d from the equation 6 , e — = (cos - a AvLit. fan /45''4— W± „ V2 5cos2 ~ = (cos -- v/2 sin -) - (v/2-1), 2 ^ is '•> d 3 Ans. tan (45°+-) =± 135. In any triangle prove that & - o cos a* - 6^* JR= 2 cos A sin 2c sin {A-B) 136. If a, &, c be in arithmetical precession, the arithmetical mean between the logs of a and c is 137. I'ind a and /3 from the equations 2tani(a4-/8)=-J-t— ^, cos a+cos j8 l-v/2 » , 2tan*(a-i8)— -. "' ,_ -^V cos a+cos /3 .' ^ , . . Ans. 0=80", i8=:45°. 1 . , 1 T 138. Shew that 2 sin-* — =+ am"^ -7==,=T • V^IO v50 * MISCELLANEOUS EXAMPLES, 825 139. Sum the series 1 o 1 « 1 . , . . l+coe x+— cos 2x+ cos 3ae+— — -r— cos 4x+ . . . .ad inf. 2 1.2.3 1.2.3.4 Ans. e*^* cos (sin x). 140. If a, /?, y be the distances of the centre of the inscribed circle of a triangle from the angles, and if a~^+Y~^=2^^, then **!> *'a> ^1 3^® "^ arithmetical progression. TT 11111 141. Shew that -z^=l+-^ --=---=- + -77 + 7::- &o. 2\/2 3 5 7 9 11 i42. If the bisectors of the angles of a triangle ABC be produced to meet the circumference of the circumscribed circle in the points D, E, F, and if p, q, t denote the sides of the triangle DEFj prove par that R'^= — , where It is the radius of the circumscribed circle. a+h + c 143. li a, Pf-y be the three values of x (unequal) which satisfy the condition a b - I + _ +c=_o ;: ; > ■■ . vi cos X sm X ;, tana tan^(/3 + y) -"i "■■■ shew that = ;, — rr* ,, . ,. tan 7 tan^(a + jS) m' ' 144. If a circle be inscribed in a triangle and the points of con- tact A', B', G' joined, and if JSi, iJj, i?, be the radii of the circum- scribed circles of the triangles AB'G'y BA'G'f GA'B' respectively, prove that 4 B ■ jBj : JB, : jB,=cosec — : cosec — : cosec — . a ^ S 145. If ABG be a triangle, and the equation x^+'i/+z'^+2yz cos 2A+2xz cos 2B+2xy cos 20=0 be satisfied by real values of a, y, z, then V y z ' I ain 2A sin 2B sin 2(7 326 PLANE TRIGONOMETRY. 146. In any triangle prove that sin 2 A {h cob G-c cos By +anal. + - • . =2 cos A cos B cos G {a^ tan A +anal. + ). a 147. Adapt cot — =a(l+cos o+cos /?+cos y) to logarithmic com- putation by the use of an auxiliary angle. Ans. cot — =4a cos — cos — (a+<^) cos —{a - '==4i, find B. Ans. 8^. 152. In any triangle shew that , . ^ ab ' ^ he ' "" ac ' 2 153. Three circles whose radii are r^, r,, r, touch ©ne another, Oj, Oa, 0, being their centres and A the point of intersection of their common tangents at tlie points of contact ; if a^ , a,, a, denote MISCELLANEOUS EXAMPLES. 327 the distances AOi, AO^, AO^ respectively, and R the radius of the circle circumscribing the triangle O^O^O^, prove that 154. A circle is inscribed in a triangle, and any three triangles are cut off by tangents to the circle ; if r^ , rj , r^ be the radii of the circles inscribed in these three triangles, then the sum of the areas of the triangles is ■^ ( - ^i+'Ti+r^) +-^ (»*i - rj+rj +- {r^+r^ - r,), . : where a, b, c are the sides of the original triangle. 155. Sum the series ad infinitum • • cos d cos^ 6 cos' 6 cos 0+ — - — cos 20+— -— cos 80+— —r cos 40+.... Ans. e cos (0+| sin 2/9). 156. If a, /?, Y be the lengths of the lines which join the feet of th« perpendiculars from the angles of a triangle on the opposite sides, shew that *^: a P y _ a'^-]-h'^+c^ '' ^"'"i^'^c^" 2ahc * V? where a, h, c are the lengths of the sides opposite to the lines a, /3, y respectively. '''^ 167. In any triangle if a, h, c, be in arithmetical progression, shew that r^, r^, r^, the radii of the three escribed circles which touch the sides a, b, c respectively, are in harmpnical progression. f". ■\ 328 TLAJfE TRIGONOMETRY. APPENDIX. Geometrical Demonstrations of (45) and (46). I. We will here add a few geometrical demonstrations of the very important formulae of Arts. 45-5^. > I. — When (x+y) is an aiHgle in the second quadrant In Fig. 1 let the angle BAG=x and the angle CAD=y. Con- struct the Figure as in Art. 45. ; , sin {x+y)= PM .•>'k '■•;;,(' AP JQN PK ~AP'^AP ii'"' QN AQ PK PQ^ ^AQ AP PQ AP =sin X cos y+Gos x sin y. coa(x+i/)=— 2p AN KQ ■/■i 'j'l^'i , {I'/fi,.--,!- 'AP AP AN AQ KQ, PQ AQ AP PQ AP r=:cos X cos y - sin x sin y. APPENDIX. 329 II, — When (x+y)is an angle in the third quadraaU, In Fig. 2 let the angle BAC=x and the angle CAD=y.. Construct the Figure as in the preceding dia- gram, then we have sin (x+i/)=- PM AF QN PK ~" AP AP ~ AQ ' AP PQ ' AP' =: - sin QAN cos PAQ - cos QPK sin PAQ = - sin (180" - x) cos (180° -y)- cos (180" - x) sin (180° - y) = sin X cos y+coa x sin y. (Art. 39.) cos (ic+j/)=- AM AP AN KQ AP AP AN AQ^_^ PQ_ ' ' ^ !: ~AQ ' AP PQ' AP =cos QAN cos PAQ- sin QPK sin PAQ =co8 (180°- x) cos (180°- y) -sin (180°- x) sin (180°- y) =co8 X cos y - sin x sin y. (Art. 39. ) The student will observe that in the quadrilateral AMPQ, the opposite angles QAM, QPM are together equal to two right angles, and since the angle QAM=x, the angle QPK=1S0° - x. , f^i ■;■;■;■»' 330 PLANE TRIGONOMETRY. III. — When (x+y) is an angle in the fourth quadrant. In Fig. 3 let the angle BAC=*:x and the angle CAD=y. Construct the Figure as in the two preceding diagrams, then* we have sin (x+y)= - PM AP QN PK -n' • * COS {x+y)= AP AP QN AQ^ PK PQ^ AQ ' AP PQ' AP sin QAN cos PAQ +cos QPK sin PAQ sin (360° -x) cos y+cos (360° -x) sin y sin X cos y+coB x sin y. AM AP AN NM 'AP AP AN AQ KQ PQ AQ AP PQ AP =cos QAN cos PAQ+&m QPK sin PAQ =cos (360° - x) cos y+sin (360° - x) sin y =co8 X cos y - sin x sin y. Here the opposite angles QAM, QPM of the quadrilateral AMPQ, are together equal to two right angles ; therefore the exterior angle QPK is equal to the angle QAN, but the angle g^i^=360° - X, hence the angle QPE:=360° - x. APPENDIX. 331 2. The geometrical proof of sin {x - y) and cos (x-y) in each of the quadrants is similar to those of sin (x+y) and cos (x+y) which we have just given. We will, however, give the proof of one case, viz. , when x is an angle in the fourth quadrant and y an angle in the first quadrant, x-y being, in th Figure, an angle in the third quadrant. In Fig. 4 let the angle BAC=x and the angle DAC=y. Construct the Figure as in Art. 48, then wo have PM 6m(x-y)=- — NQ KQ AP NQ AP AQ_ AP KQ PQ AQ AP PQ AP = - sin QAN cos PAQ - cos QAN sin PAQ = - sin (360° - x) cos y - cos (360° - ac) sin y = sin a cos y - cos x sin y. coa{x-y)=~ AM AP , PK-AN ! ■ AP AN PK AP AP * AN AQ PK PQ AQ AP PQ AP = cos QAN cos PAQ - sin PQK sin PA Q = cos (360° -- x) cos y - sin (360° -x) siay = cos X cos y+sin x sin y, since the angle PQK is equal to the angle QAN. 332 PLANE TRIGONOME'J'RY. i