A TREATISE ON ELEMENTARY TRIGONOMETEY BY THE REV. J. B. LOCK, M.A. FELLOW OF OONVILLX AND OAIUS COLLEGE, OAUBBIDOK ; FOBMEULY »IASTKB AT ETON. STE/iEOTYFED EDITION lonbon : MACMILLAN AND CO. AND NEW YORK. 1892 [The Eight of Trarulation it r^*rrivd.] Printed 1882. New edition 1884, new edition 1885, new edition 188 new edition 1887, new edition i888, reprinted 1889, i^ 1891, March and September 1892. FROM THE PREFACE TO THE FIRST EDITION. The present Work on Elementary Trigonometry contains that part of the subject which can con- veniently be explained without the use of infinite series. It is intended either for class-teaching or for private study. Accordingly the Examples are numerous and for the. most part easy. Those which are not original have been selected from the Cam- bridge and Array Examination Papers of the last few years. The Miscellaneous Examples are somewhat more difficult, and should in most cases be postponed until the student reads the subject for the second time. The order in which the chapters are read may be varied at the discretion of the Teacher. ^ . _^ J. B. L. Eton, March, 1882. In the Second Edition a short course has been indicated for the use of Students who wish to read the Solution of Triangles as early as possible. Such vi PREFACE. Students are advised to omit every article that is marked with an asterisk. A double asterisk has been placed before those articles which should be omitted by all Students until they are reading the subject for the second time. An Appendix containing a description of the Vernier, the Theodolite, the Sextant etc. has been added. A Key by Mr Carr is now ready. CONTENTS CUAP. I. On Measubement II. Om Incommensubablb Qdaniitiks . III. On the Relation between the Cibcumfebence of A ClBCLB AND ITS DiAMETEB IV. On THE Measdbemknt or Angles . V. The Tbiqonometbical Ratios VI. On the Ratios of Certain Angles Vn. On the Trigonometbical Ratios of the same Angle Vni. On the Use of the Signs + and - IX. On the Use of + and - in Tbigonomexuy X. On Angles unlimited in Magnitude. I. XI. On the Ratios of Two Angles XII. On the Ratios of Multiple Angles XIII. On Angles unlimited in Maonitudb. II. XIV. On Logarithms XV. On the Use of Mathematical Tables . XVL On the Relations between the Sides and Angles of a Tbla.nole XVn. On the Solution of Triangles XVni. On the Measurement of Heights and Distances XIX. On Triangles and Circles .... XX. On the Area of the Cibcle, the Construction OF Trigonometbical Tables, etc. . Appendix Examples fob Exercise Examination Papers Answers to Examples CHAPTER I. On Measurement. 1. It is usual to say that we have measured any con- crete quantity, when we have found out how many times it contains some familiar quantity of the same kind. We say for example, that we have measured a line, when we have found out how many feet it contains. We say that we have measured a field, when we have found out how many acres or how many square yards it contains. 2. To know the measurement of any quantity then, we must have two things. First, we must have a unit^ or standard of i*eference, of the scume kind aa the thing measured. Secondly, we must have the measure^ or the number of times the thing measured contains the unit, or standard quantity. 3. Hence, the measure of a quantity is the number, and the unit is the Concrete quantity, by means of which it is measured. Example 1. A line contains 261 feet Here the measure or number is 261 and the unit a foot Example 2. What is the measure of 2^ miles when a yard is the unit ? 2^ miles=f x 1760 yards, =4400 yards =4400 x 1 yard, therefore the measure is 4400 when a yard is the unit. L. E. T. 1 'I TRIGONOMETRY. Exwm/ple 3. What is the unit when the measure of a field of 10 acres is 242 ? 10 acres 10 acres = 242 X 242 ' , , ., . 10 acres -. . , .-, the unit IS — =200 square yards. Example 4. If the unit be a yards, what is the measure of h miles ? 6 miles =6 x 1760 yards, , ., 6x1760 , .*. miles = X a yards, . , . 6x1760 .-. the measure required is . EXAMPLES. I. (1) What is the measure of 1 mile when a chain of 66 feet is the unit ? (2) What is the measure of an acre when a square whose side is 22 yards is the unit ? (3) What is the measure of a ton when a weight of 10 stone is the unit ? (4) The length of an Atlantic cable is 2300 miles and the length of the cable from England to France is 21 miles. Express the length of the first in terms of the second as unit. (5) The measure of a certain field is 22 and the unit 1100 square yards : express the area of the field in acres. (6) Find the measure of a miles when h yards is the unit. (7) The measure of a certain distance is a when the unit is c feet. Express the distance in yards. (8) A certain sum of money has for its measures 24, 240, 960 when three different coins are units respectively. If the first coin is half a sovereign, what are the others ? ON MEASUREMENT. 6 '4. iiie measure of a quantity is the iiuuiber u£ times which that quantity contains the unit. We may express the same thing in different ways, (i) The measK/re of a quantity is the ratio of that quantity to the unit. (ii) The measure of a quantity is the fraction that the quantity is of the unit. * 6. This last statement is in the language of Arith- metic, and the word * fraction ' is to include whole numbers, and improper fractions. * 6. The following are therefore different forms of the same question : (i) What is the measure of 4 miles when 66 feet is the unit? (ii) How many times does 4 miles contain 66 feet? (iii) What is the ratio of 4 miles to 66 feet? (iv) What fraction of 66 feet is 4 miles? Example (i) What is the measure of a yards when b feet is the unit ? a yards = 3a feet=-T- x b feet, ,, . 3a .*. the measure is -i- . (ii) How many times does a yards contain b feet ? As in (i), a yards=-v- x b feet. Answer. -,- times. (iii) What is the ratio of a yards to b feet ? Afi in (i), flkyards=^x b feet, a yards _ 3a •'• Tf^t ~T' .•. the required ratio is -r- .^""'^ 1—2 4 TRIGONOMETRY. (iv) What fraction is a yards of h feet ? . . ,.... a yards 3a .*. the required fraction ia -r- . * EXAMPLES. 11. (1) The ratio of the area of one field to that of another is 20 : 1, and the area of the first is half a square mile. Find the number of square yards in the segond. (2) The ratio of the heights of two persons is 9 : 8, and the height of the second is 5 ft. 4 in. What is the height of the first? (3) The measure of a field with 3 acres for unit is 65. Find the ratio of the field to an acre. (4) - One field contains a second of '2,\ acres, 6| times. What is the measure of the first field in terms of the second ? What is the ratio of the first field to the second ? Express the first as the fraction of the second. (5) A certain weight is 3'125 of a ton. What is its measure in terms of 4 cwt. ? How many times does it contain 4 cwt. ? What is its ratio to 4 cwt. ? What fraction is it of 4 cwt. ? (6) The ratio of a certain sum of money to 3 guineas is ^". Find its measure in terms of one pound. Find the unit when its measure is 22. (7) What is the measure of a miles when h chains is the unit? How many times do a miles contain c chains ? What is the ratio of a miles to d chains ? What fraction is a miles- of h chains ? EXAMPLES. II. ^(8) What is the unit when the measure of £20 is ^ ? £25 contains a certrun sum ^^ times. What is that sum ? The ratio of £30 to a certain sum is 4^. What is that sum ? The fraction which £10 is of a certain sum is ^. What is that sum? 7. It is explained in Arithmetic, in the application of square measure, that the measure of the area of a rectangle is found in terms of a square unit, by multiplying together the measures of the sides in terms of the corresponding linear unit. Example. Find in square feet the measure of a square surface whase side is 12 feet. The area is 12 x 12 square feet = 144 x 1 square foot, .'. the measure required is 144. 8. We shall apply this result to Euclid I. 47. Example 1. The sides containing the right angle of a right- angled triangle are 3 ft. and 4 ft. respectively ; find the length of the hypotenuse. Let X be the number of feet in the hyx)otenuse. Then by Euclid I. 47, the square described on the side of X feet = the sum of the squares described on the sides of 3 feet and 4 feet respectively, b TRIGONOMETRY. .'. x^ square feet =9 square feet + 16 square feet = 25 square feet, .-. :p2=25, .*. x=h. Therefore the length of the hypotenuse is 5 feet. Example 2. Find the length of the perpendicular drawn from the vertex to the base of an isosceles triangle whose equal sides are 10 feet each, and whose base is 12 feet. Let ABG be the isosceles triangle such that AB is 10 feet, AG is 10 feet and BG is 12 feet. B ««• D Draw AD perpendicular to BC. Then because the triangle ABG is isosceles AD will bisect the base BG\tiD\ therefore BD is 6 feet. Let AD contain x feet.. Then by Euclid L 47, the square on ^5= the sum of the squares on BD and AD. .'. 102 sq. ft. = 62 sq. ft. +072 sq. ft. .-. 102 = 62 + ^, .♦. ;p2=ioo-36, /. ^=64, .-. a?=8. Therefore the required length of ^Z) is 8 feet. EXAMPLES. III. 7 ipLe 3. Find the length of the (li.imeter of the square me of whose sides contains a feet. ^•/ 7 \-f // ^y V e ABCD be the square, so that AB is a feet, and AD \s a Let the diameter ED be x feet. Then the square on 2)5= the sum of the squares on DA and AB. .'. x^ sq. ft.=a2 sq. ft. +0* sq. ft. .'. x^=2a% .'. x=,^2. a. Therefore the required length of the diameter is V2 • « feet. EXAMPLES. III. (1) Find the length of the hypotenuse of a right-angled triangle whose sides are 6 feet and 8 feet respectively. (2) The hypotenuse of a right-angled triangle is 100 yards and one side is 60 yards : find the length of the other side. (3) One end of a rope 52 feet long is tied to the top of a pole 48 feet high and the other end is fastened to a peg in the ground. If the pole be vertical and the rope tight, find how far the peg is from the foot of the pole. (4) The houses in a certain street are 40 feet high and the street 30 feet wide : find the length of the ladder which will reach from the top of one of the houses to the opposite side of the street. 8 TRIGONOMETRY. III. (5) A wall 72 feet high is built at one edge of a moat 54 feet wide ; how long must scaling ladders be to reach from the other edge of the moat to the top of the wall ? (6) A field is a quarter of a mile long and three-sixteenths of a mile wide : how many cubic yards of gravel would be required to make a path 2 feet wide to join two opposite comers, the depth of the gravel being 2 inches ? (7) The sides of a rectangular field are 4a feet and Za feet respectively. ' Find the length of its diameter. (8) If the sides of an isosceles triangle be each 13a yards and the base 10a yards, what is the length of the perpendicular drawn from the vertex to the base 1 ""* (9) Show that the perpendicular drawn from the right angle to the hypotenuse in an isosceles right-angled triangle, each of whose equal sides contains a feet, is ^ . a ft. (10) If the hypotenuse of a right-angled isosceles triangle be a yards, what is the length of each side 1 (11) Show that the perpendicular drawn from an angular point to the opposite side of an equilateral triangle, each of whose sides contains a feet, is ^ . a ft. (12) If in an equilateral triangle the length of the perpen- dicular drawn from an angular point to the opposite side be a feet, what is the length of the side of the triangle ? (13) Find the ratio of the side of a square inscribed in a circle to the diameter of the circle. (14) Find the distance from the centre of a circle of radius 10 feet, of a chord whose length is 8 feet. (15) Find the length of a chord of a circle of radius a yards, which is distant h feet from the centre. (16) The three sides of a right-angled triangle, whose hypo- tenuse contains 5 a feet, are in arithmetical progression : prove that the other two sides contain 4a feet and 3a feet respectively. * CHAPTER II. On Incommensurable Quantities. 9. Two numbers are said to be commensurable when their ratio can be expressed as an arithmetical fraction : that is, as a fraction whose numerator and denominator are both whole numbers. Example. 4'93 and 81| are two commensurable numbers. 444x4 Their ratio is 4f^-r81J= 90 X 327 10. Two numbers are said to be incommensurable when their ratio ca/nnot be expressed as an arithmetical fraction. Example. ^2 and T are two incommensurable numbers. For — cannot be expressed exactly as an arithmetical quantity. So V3 and »J2 are two incommensurable numbers. 11. One number alone is said to be a/n incommensv/r- ahle number when it is incommensurable with unity. So that an incommensurable number cannot be expressed as an arithmetical fraction. Example. ^2 and ^^3, and all surd nimibers, are incom- mensurable numbers. 1 TRIG ONOMETR Y. 12. Two quantities are said to be commensurable when their measures referred to a common unit are com- mensurable. Example. A mile and a thousand yards are two commen- surable quantities. Their measures with a yard for unit are 1760 and 1000 ; and these are commensurable numbers. 13. Two quantities are said to be incommensurable when their measures referred to a common unit are incom- mensurable. Example. The side of a square and its diameter are two incommensurable quantities. For if the side of a square contain a feet, the diameter (see Example 3, p. 7) contains v'2 . a feet, and therefore the ratio of their measures is 1 : ^1% So that their measures are incommensurable. 14. There is no praxiticcil difficulty in dealing with incommensurable quantities. We can always find for their measures arithmetical expressions sufficiently accurate for all practical purposes. 15. A little consideration will convince the student that no measurement can in practice be made with absolute accuracy. For example : A skilful mechanic is probably satisfied if in measuring some material two or three feet in length the error in his measurement is less than the thirty-second part of an inch. (The thirty -second part of an inch is less than half the height of the smallest letter on this page.) That is to say, he is satisfied if he make no greater error than about a thousandth pa/rt of the whole l&ngth to be measured. He would record such a measurement thus, 2 ft. 3fJ inches, which is 891 thirty-second parts of an inch. ON INCOMMENSURABLE QUANTITIES. 11 1 6. We will suppose that the length thus measured is the side of a square, and that the workman wishes to know to the same degree of accuracy as his measurement, what is the length of the diameter of the square. He can find it thus. The diameter of a squai'e = J2 x its side, .*. the diameter of this square = J2 x 891 thirty-second pai*ts of an inch. Also ^2 = 1*414 nearly, ^ .-. 72x891 = 1-414x891 = 1259-8. .-. the required diameter =1260 thirty-second parts of an inch, nearly. The error being less than one thirty-second part of an inch. 17. The student will be able to see from the above example, that if the value of an incommensurable number is found to 4 figures, a very considerable degree, of accuracy is attained. Also that a much greater degree of accuracy is attained for every additional figure. 18. It is no advantage in calculations such as the above to have the value of such quantities as ^2 calculated to any greater degree of accuracy than the observed measurement. For instance, our calculation being correct as far as the whole numbers are concerned we should gain nothing by using 1-4142 instead of 1-414 ior J2. This would give us the answer 1260*0 instead of 1259*8; the difference being a fifth of a thirty-second part of an inch, a quantity by hypothesis too small to be of any importance. 12 TRIGONOMETRY. Eoiample 1. The side of an equilateral triangle contains 2 feet ; find, correct to the ten-thousandth part of a foot, the length of the perpendicular drawn from an angular point to the opposite side. Here (as in Example 2, p. 6), let the perpendicular contain X feet, then ^2sq. ft. = 22sq.ft.-l2sq. ft., /. ^2=4-1 = 3, = 1-7320 «Scc.; .-. the length of the perpendicular =1 "7320 feet. Example 2. Find the length, correct to the ten-thousandth part of a foot, of the side of the square described upon a dia- meter whose length is a feet. Let X be the number of feet in each of the sides of the square. EXAMPLES. IV. ar'nii. ft. + ^sq.ft.= =a-8q. = o, .i'L .*. jr= =72 = aV2 2 r EXAMPLES. IV- 13 =|x(l-4142) = ax'7071..., /. the length of the perpendicular -ax '7071... feet. * EXAMPLES. IV. (1) Find, correct to the thousandth part of a foot, the length of the diameter of a square whose side is 7 feet. (2) Find, correct to a yard, the length of the diameter of a square whose side is one mile. (3) Find, correct to the hundredth part of an inch, the hypotetiuse of a right-angled triangle whose sides are 3 ft. 6^ in. and 3 ft. 4 in. respectively. (4) Find, to the nearest inch, the side of a square whose area is 1000 square yards. (6) Find, correct to the tenth part of a foot, the diameter of a square field whose area is ten acres. (6) Find, to the nearest inch, the side of a square field whose area is one acre. (7) Find the height of an equilateral triangle whose side is 10 feet. (8) Find the height of an equilateral triangle whose side is 5-32 feet. (9) The top of a table measures 24*6 inches and 41-3 inches along two adjacent sides. What should the diameter measure if the table is rectangular ? (10) Find, to the nearest inch, the diameter of a lawn-tennis coiuii whose length is 78 feet, and breadth 36 feet, supposing tliat it is properly marked out. ( 14 CHAPTER III. On the Relation between the Circumference of a Circle and its Diameter. 19. The circumference of a circle is a line, and therefore it has length. We might imagine the circumference of a circle to consist of a flexible wire ; if the circular wire were cut at one point and straightened, we should have a straight line of the same length as the circumference of the circle. 20. A polygon is a figure enclosed by any number of straight lines. 21. A regular polygon has all its sides equal and all its angles equal. 22. The perimeter of a polygon is the sum of its sides. 23. If we have two circles in which the diameter of the first is greater than the diameter of the second, it is evident that the circumference of the first will be greater than the circumference of the second. 24. It seems, therefore, not unlikely that, if the dia- meter of the first circle be twice that of the second, the "jj circumference of the first will be twice that of the second. THE CIRGUMPERENCE OP A CIRCLE. 15 25. And also not unlikely that whatever be the ratio of the circumference of the first circle to its diameter, the same will be the ratio of the circumference of the second '"ircle to its diameter. This suggests that it is not unlikely that the circum- ference of a circle = k times its diameter, where k is some nuTnher which is the same for all circles. We shall presently prove that this is the case. 26. But although we can prove that the circumference of a circle r. ^ . , ; TT = a nxed numerical quantity, its diameter ^ •' the method of calculating the value of this number is beyond the limits of an elementary treatise. 27. We shall therefore simply state here, what is proved in the Higher Trigonometry, (i) that this numerical value is incommensurable, (ii) that it is approximately 3*14159265 ^ 39 -0741 6666 etc. degrees •43415740740 etc. right angles Answer. '43415746 of a right angle. ON THE MEASUREMENT OP ANGLES. 29 52. An angle given as the decimal of a right angle may be expressed in degrees, minutes, and seconds by the con- verse of the above. Example. Express '43415746 of a right angle in degrees, minutes, and seconds. •43415740740 etc. right angles 90^ 39-07416666600 degrees The last two figures would be 66 if we were to write down the recurring part to more figures. This gives 39-07416666666 etc. degrees 60 4-44999999G0 minutes that is 4-449 minutes or 4-45 miiuites 60 27-00 seconds. The result is 390 4' 27". *63. We have seen that an angle expressed as the decimal of a right angle can be at once expressed in grades, minutes, and seconds. Hence an angle expressed in degrees, minutes, and seconds, can be expressed in grades, etc. by first reducing the angle to the decimal of a right angle. Example. Express 39® 4' 27" in grades, minutes, and seconds. This angle is '43415746 of a right angle, by Art. 51 and this = 43« 4 r 5 7 -407^' . * 54. An angle given in grades, minutes, and seconds can be expressed in degrees, minutes, and seconds by first ex- pressing the angle as the decimal of a right angle. 30 TRIGONOMETRY. Example. Express •43« 4r 57 ^Ot^' in degrees, minutes, and seconds. This angle is -43415746 of a right angle, which is 39° 4' 27" from Art. 52. * EXAMPLES. VIII. Express each of the following angles (i) as the decimal of a right angle, (ii) in grades, minutes, and seconds ; (1) 80 15' 27". (4) 160 14' 19". (2) 6M'30^ (5) 1320 6^ (3) 970 5' 15'^ (6) 490. Express in degrees, minutes and seconds, (7) 18 ZT 5a\ (10) 24« 0^ 25". (8) 88 75\ (11) 188 r 15". (9) 1708 45^35'\ (12) 358. 55. The decimal or centesimal system of subdividing a right angle was proposed by the French at the commence- ment of the present century; but, although it possesses many advantages over the established method, no one has been found willing to undertake the great expense that would have to be incurred in rearranging all tables and all books of reference, and all the records of observations, which would have to be transferred from the old system to the new, before the advantages of the decimal system could be felt. Thus the decimal system of angular measurement has never been used even in France, and in all probability never will be used in practical work. ON CIRCULAR MEASURE. 31 On Circular Measure. 06. By the following construction we get an angle of gi-eat importance in Trigonometry. On the circumference of a circle whose centre is let an mc RS be measured so that its leijgth is equal to the radius of the circle, and let R and aS^ be joined to the centre. 57. We are about to prove (Art. GO) that this angle ROS'iB B. fixed fraction of a right angle, so that all such angles are equal to one another. 58. We may state the same thing thus — We are about to prove that if we take any number of difierent circles, and measure on the circumference of each an arc equal in length to its radius, then the angles at the centres of these circles which stand on these arcs respectively, will be all of the same size. 59. Definition. The angle which at the centre of a circle stands on an arc equal in length to the radius of the circle is called a Radian. 60. To prove that all Radians are equal to one anotJier. Since the Radian at the centre of a circle stands on an arc equal in length to the radius, and an angle of two right angles nttbn gnntrjT^i circle stands on half the circuraferenoe, jr\ ^ ^ '^ ^^7^V ' £ OF THH X f XJNIVERSITY I 32 TRIGONOMETRY. and since angles at the centre of a circle are to one another as the arcs on which they stand (Euc. VI. 33), therefore the radian is to an angle of two right angles as the radius is to half the circumference; that is, as the diameter is to the whole circumference; that is, in the constant ratio 1 : tt. Therefore the radian = '' — — . TT That is, the radian is a fixed fraction of a right angle. But all right angles are equal to one another. Therefore all radians are equal to one another, q. e. d. 61. Thus the radian possesses the qualification most essential in a unit, viz. it is always the same. The student will find, in the theoretical part of Trigo- nometry, that many expressions can be written more shortly when a radian is used for the unit of angle, than when any other unit is used. 62. Thus the reasons why a radian is used as a unit are : (i) All radians are equal to one another. (ii) Its use simplifies many formulae in Theoretical Trigonometry. 63. The system of angular measurement in which a radian is the unit is called Circular Measure. Therefore the circular measure of an angle is the num- ber ofradiam^ which the angle contains. 64. A radian = — x 2 right angles, TT = ii| of 180° nearly, ^57-2957 .... degrees. ON THE MEASUREMENT OP ANGLES. 33 The student should notice that a radian is a little less than an angle of an equilatei'al triangle. 65. Circular measure is, as we have said, used in theo- retical investigations, in which the angle under consideration is almost always expressed by a letter. This is usually one of the Greek letters a, )3, y..., ^, ^, »/r.... Strictly speaking these letters represent numbers, i.e. measures; so that some unit of angle must be undei'stood in suc^h an expression as *the angle ^.' (Art. 2.) v,^ For this reason, when an angle is denoted by a Greek letter such as a, ft y, etc., 0, ,... a, (3, .. stand for mere numbers. Thus as we have said above (29) tt stands for a numb&r and a number only^ viz. 3*14159 , but in the expression *the angle ir' that is 'the angle 3-14159 ' there must be some unit understood. The unit understood here is a radian, and therefore 'the angle tt' stands for 3*14159 ', that is two right angles. Hence, whe7i an angle is referred to, iris a very convenient abbreviation for two right angles. 69. To express in degrees or grades an angle given in radians, we first express the angle in right angles, remem- bering that 2 right angles =77 radians. Example. How many degrees are there in the angle whose circular measure is 2 % ^, . , ^ 1. c^ 2 right angles 4 . , , - This angle =2 radians =2 x — ^ ^ — = - right angles, TT TT • _ 4x900 ^ 3600 TT TT .*. the angle contains — degrees. TT ■^70. If Z>, G and a be the number of degrees, grades and radians respectively in any angle, then 180 "200 "tt* For each fraction is tlie ratio of the angle to two right angles. EXAMPLES. IX. 35 Example. Find the number of degrees in two radians. Let B be the number, then 180 TT* ... dJ±^. EXAMPLES. IX 1. Express the following angles in rectangular measm'e. 4 (1) n. (2) ^. (3) 1' (4) 3=. (5) 3-141 59265«' etc. (6) -. (7) B. (8) •00314159'= etc. (9) IOtt. 2. Express the following angles in circular measure. (1) 1800. (2) 3600. (3) 600. (4) 22^0. (5) P. (6) 57-2950 etc. QOO (7) nO. (8) — . (9) A. IT * 3. Express the following angles in circular measure. (1) 33« 33^ 33-3". (2) 50«. (3) 16-6«. (4) 1«. (5) r. (6) 10". 9008 (7) n'. (8) ^^. (9) 1000«. tr * 4. Find the ratio of (1) 450 to 1^. (2) 600to60«. (3) 25«to220 30'. (4) 24' to 2*. 1000 (5) 1-75'to -^. (G) lotol'. TT 3—2 36 TRIGONOMETRY. *71. 7^0 prove that the measure of an angle at tJie centre of a circle in radians (i.e. in Circular Measure) is the ratio of the arc on which it stands to the radius of the circle. LOR Since angles at the centre of a circle are to one another as the arcs on which they stand (Euc. YI. 33), Therefore any angle ROP at the centre of a circle is to the radian as its arc RP is to the arc of the radian. But the arc of the radian is equal to the radius, „, . any angle ROP . , its arc RJ^ iherefore — —, — ^-^p is equal to .-=— — ..— . the radian the radius And therefore any angle ROP is equal to —j: — x (a radian). That is, the measure a of an angle in radians is the ratio arc arc or, a radius ' ' "" radius * Hence (cf. Art. 70) ^ = |g = ^ = ^^^^ x 1 . Example. Find the number of grades in the angle subtended by an arc 46 ft. 9 in. long, at the centre of a circle whose radius is 25 feet. The angle stands on an arc of 46| ft. and the radian, at the centre of the same circle, stands on an arc of 25 feet. ., , 46| ,. ,g„ 2 right angles .'. the angle =^ radians, =fg^ x — i 2008 = 1^ X ^ = 11 9« nearly. EXAMPLES. X. 37 ♦EXAMPLES. X. (In the answers ^ is used for n.) (1) Find the number of radians in an angle at the centre of a circle of radius 25 feet, which stands on an arc of 37^ feet, X^) Find the number of degrees in an angle at the centre of a circle of radius 10 feet, which stands on an arc of hir feet. (3) Find the number of right angles in the angle at the cen- tre of a circle of radius 3|\ inches, which stands on an arc of 2 feet. (4) Find the number of French minutes in the angle at the centre of a circle of radius 8 ft. 4 inches, which stands on an arc of 1 inch, f (b) Find the length of the arc subtending an angle of 4i ratlians at the centre of a circle whose radius is 25 feet. (6) Find the length of an arc of eighty degrees on a circle of 4 feet radius. (7) Find the length of an arc of sixty grades on a circle of ten feet radius. (8) The angle subtended by the diameter of the Sun at the eye of an observer is 32' ; find approximately the diameter of the Sun if its distance from the observer be 90,000,000 miles. (9) A railway train is travelling on a curve of half a mile radius at the rate of 20 miles an hour ; through what angle has it turned in 10 seconds ? (10) A railway train is travelling on a curve of two-thirds of a mile radius, at the rate of 60 miles an hour ; through what angle has it turned in a quarter of a minute? 38 TRIGONOMETRY. X. (11) Find approximately the number of English seconds con- tained in the angle which subtends an arc one mile in length at the centre of a circle whose radius is 4000 miles. (12) If the radius of a circle be 4000 miles, find the length of an arc which subtends an angle of 1" at the centre of the circle. (13) If in a circle whose radius is 12 ft. 6 in. an arc whose length is '6545 of a foot subtends an angle of 3 degrees, what is the ratio of the diameter of a circle to its circumference ? (14) If an arc 1*309 feet long subtend an angle of 7 J degrees at the centre of a circle whose radius is 10 feet, find the ratio of the circumference of a circle to its diameter. (15) On a circle 80 feet in radius it was found that an angle of 22^ 30' at the centre was subtended by an arc 31 ft. 5 in. in length ; hence calculate to four decimal places the numerical value of the ratio of the circumference of a circle to its diameter. (16) If the diameter of the moon subtend an angle of 30', at the eye of an observer, and the diameter of the sun an angle of 32', and if the distance of the sun be 375 times the distance of the moon, find the ratio of the diameter of the sun to that of the moon. (17) Find the number of radians in (i.e. the circular measure of) 10" correct to 3 significant figures. (Use ff^ for tt.) (18) Find the radius of a globe such that the distance measured upon its surface between two places in the same meri- dian, whose latitudes differ by 1^ 10', may be one inch. (19) Two circles touch the base of an isosceles triangle at its middle point, one having its centre at, and the other passing through the vertex. If the arc of the greater circle included within the triangle be equal to the arc of the lesser circle without the triangle, find the vertical angle of the triangle. (20) By the construction in Euc. I. 1, prove that the unit of circular measure is less than 60^^ EXAMPLES. X. ^^ (21) On the 31st December' the Sun subtends an angle of 32' 36", and on 1st July an angle of 31' 32" ; find the ratio of the distances of the Sun from the observer on those two days. (22) Show that the measure of the angle at the centre of a k .a , J circle of radius r, which stands on an arc a, is , where k r depends solely on the unit of angle employed. Find k when the unit is (i) a radian, (ii) a degree. * 72. Questions conceming angles expressed in different systems of measurement are easily solved by exjyressing each angle in right angles. Example 1. The sum of the measure of an angle in degrees und twice its measure in radians is 23i^, find its measure in degrees (7r=^). Let the angle contain x right angles. Then the measure of the angle in degrees =90x, radianB=|^. .-. 90x + 2.'^x=2Zf, .-. 90X + ^X=^^y .-. 652a: =163, • • '*' — 4- The angle is j of a right angle, that is 22^*', nearly. 40 TRIGONOMETRY. Example 2. The three angles of a triangle are in arithmetical progression, and the measure of the least in grades is to that of the greatest in circular measure as 120 : tt. Express each angle in degrees. Let the angles contain x-y^x^x-\-y right angles respectively; they are then in a. p. Their mm is Zx right angles ; but since they are the angles of a triangle, their 8um is 2 right angles; ,-. 3^7=2, .-. x = l. Again, the least angle contains {x-y)x 100 grades, and the greatest angle contains {x-^y) - radians, .:lQO{x-y) : |(:r+y) = 120 : tt. .♦. 100(^-y) = 60(:r + y), or, 40;r=160y, or, x=4y. because .a7=§, or,y=h Thus the angles contain I, §, and f right angles respec- tively ; therefore the angles are 45^, 60°, 16^. *^ EXAMPLES. XL (In the following examples the answers will be given in terms ofTT.) (1) The sum of the degrees and of the grades in a certain angle is 38 ; find its circular measure. (2) The difference of two angles is 20^, and their sum is 48<' ; find them. EXAMPLES. XI. ^^ (3) Oue angle is double of a secoud, and the sum of their measures in degrees and in grades respectively is 140 ; express the angles in degrees. (4) Two angles are in the ratio of 4 : 5, and the difference of their measures in grades and in degrees respectively is 2^ ; find the angles in degrees. (5) The difference between two angles is - , and their sum y is 56 degrees ; find the angles. (6) If the three angles of a triangle are in arithmetical pro- gression, show that the mean angle is eo**. (7) The three angles of a triangle are in arithmetical pro- gression, and the number of grades in the least is to the number of degrees in the mean as 5 : 6. Find the angles in degrees. (8) The three angles of a triangle are in arithmetical pro- gression, and the number of grades in the greatest is to the number of degrees in the sum of the other two as 10 : 11. Find the angles in degrees. (9) The three angles of a triangle are in arithmetical pro- gression, and the number of grades in the least is to the number of radians in the greatest as 200 : 3n. Express the angles in gi-ades. (10) If D be the number of degrees and G the number of grades in any angle, prove that 0-D = \D. (11) If i/" be the number of English minutes and m the number of French minutes in any angle, prove that (12) If Oy D and C be the number of grades, degrees and 20(7 radians in any angle, prove that G -D = — . (13) If an angle be expressed in French minutes, show that it will be transferred to English minutes by multiplying by '54. 42 TRIGONOMETRY. XI. (14) Divide 33^ 6' into two parts so that the number of Eng- lish seconds in one part may be equal to the number of French seconds in the other part. (15) Find the ratio of 9^ 27' to 122 50^ . (16) Find the number of radians in an angle of n English minutes. (17) Express in each of the three systems of angular mea- surement the angles (i) of a regular hexagon, (ii) of a regular octagon, (iii) of a regular quindecagon. (18) Show that the number of degrees in an angle of a regular decagon is to the number of grades in an aagle of a regu- lar pentagon in the ratio of 6 : 5. (19) Show that the number of grades in an angle of a regular pentagon is equal to the number of degrees in an angle of a regular hexagon. (20) Find in English minutes the difference between the angle of a regular polygon of 48 sides and two right angles. (21) If we take for unit the angle between a side of a regular quindecagon and the next side produced, find the measures (i) of a right angle, (ii) of a radian. (22) Find the unit when the sum of the measures of a degree and of a grade is 1. (23) What is the unit when the sum of the measures of 9® and of 58 is ^ ? (24) If the measure of h grades is a, find the measure of c degrees. (25) What is the unit when the sum of the measures of a grades and of h degrees is c ? EXA MPLEH. XI. 43 (26) The number of grades in a certain angle exceeds the number of degrees in it by ^ of the number of degrees in a radian. If this angle be taken as unit, what is the measure of a right angle ? (27) The three numbers which express the three angles of a triangle are all equal, and the units of angle in each are respec- tively a degree, a grade and the sum of a degree and a grade ; express each of the angles in circular measure. (28) The three angles of a triangle have the same measure when expressed in degrees, grades and radians respectively ; find this measure. (29) The measures of the angles of a triangle in degrees, grades and radians respectively are in the ratio of 1 : 10 : 100 ; find the number of radians in the smallest angle. (30) The interior angles of an irregular polygon are in a. p. ; the least angle is 120*^; and the common difierence 5^ : find the number of sides. ( 44 ) CHAPTER V. The Trigonometrical Ratios. 73. Let ROE be any angle (see the figure in Art. 83). In one of the lines containing the angle take any point P, and from P draw PM perpendicular to the other line OR. Then, in the right-angled triangle 0PM, formed from the angle ROE, (i) the side MP, which is opposite the angle under considera- tion, is called the perpendicular ; (ii) the side OP, which is opposite the right angle, is called the hjrpotenuse ; (iii) the third side OM, which is adjacent to the right angle and to the angle under consideration, is called the base. From these three, — perpendicular, hypotenuse, base, — we can form three diflferent sets containing two each. The ratios or fractions formed from these sets, viz. ... perpendicular .... base ..... perpendicular (i) ■, . , (n) .r 7 , (ill) , , hypotenuse ' ^ ' hypotenuse ' ^ ' base * and the ratios formed by inverting each of them, viz. . . hypotenuse hypotenuse . .. base perpendicular' ^ ' base ' ^ ^ perpendicular ' will be found to be of gi-eat importance in treating of any angle ROE. Accordingly to each of these six ratios has been given a separate name (Art. 75). THE TRiaONOMETRICAL RATIOS. 74. The student should observe carefully 45 (i) that each ratio, such as ^ ^ . , is a mere nuniber; ^ ' hypotenuse (ii) that, as we shall prove in Art. 83, these ratios remain unchanged as long as the angle remains unchanged ; (iii) that if the angle be altered ever so slightly, there is a consequent alteration in the value of these ratios. [For, let ROE^ ROE' be two angles which are nearly equal; rF Let OP=OF\ then OM is not = OM'y and therefore the ratios OM OM* ^ and jr-pi are not equal : also MP is not^M'F and therefore ,, ,. MP - M'F , , T the ratios jrp and -^p; are not equaLj (iv) that by giving names to these ratios we are enabled to apply the methods of Algebra to the Geometry of Euclid VI., just as in Chapter I. we applied the methods of Algebra to Euc. 1. 47. The student is recommended to pay careful attention to the following definitions. He should be able to write them out in the exact words in which they are printed. AG TRIGONOMETRY. 75. Definition. To define the three principal Trigono- metrical Ratios of an angle. M \M Let ROE be an angle. (^ ' In OE one of the lines containing the angle take any point P, and from P draw PM perpendicular to the other line OR, or, if necessary, to RO produced. Then, in the right-angled triangle 0PM, the side MP, which is opposite the angle under consideration, is called the perpendicular. The side OP, which is opposite the right angle, is called the hypotenuse. The third side OM (which is adjacent to the right angle and to the angle under consideration) is called the base. Then the ratio (i) (ii) (iii) MP perpendicular . hypotenuse OP OM _ OP hypotenuse MP perpendicular OM ~ base is called the Sine of the angle ROE. cosine tangent Note. The order of the letters, in MP, OM and OP, indicates direction and decides their algebraical signs. [Art. 132.] THE TRIGONOMETRICAL RATION. 17 ^6. If A stand for the angle ROE, these lutios are called sine A, cosine A and tangent A, and are usually abbreviated thus : sin J, cos -4, tan^. 77. There are three other Trigonometrical Ratios, formed by inverting the sine, cosine and tangent respectively, which are called the cosecant, recant, and cotangent respec- tively. 78. To define the three other Trigonometrical Ratios of cmy angle. The same construction and figure as in Art. 75 being made, then the mtio ,. . OP hypotenuse ^'""^ MP ^ pe rpendicular '"^ ^^"^^ ^^^^ COSecant c,f tlio angle ROK. , , OP hypotenuse <''> 0^" baae " ^^"^^ • ^ . OM base "^'^ F/>°. perpendicalar " Cotangent,, 79. Thus if A stand as before for the angle ROE, these ratios are called coS^cant A, secant A, and cotangent A. Tliey ai-e abbreviated thus, cosec -4, sec i4, cot^. 80. From the definition it is clear that 1 / cosec _ . , sin A . 1 secil = coayi cot A = tan A 48 TRIGONOMETRY. 81. The above definitions apply to an angle of any magnitude. (We shall return to this subject in Chapter X.) For the present the student may confine his attention to angles which are each less than a right angle. 82. The powers of the Trigonometrical Ratios are expressed as follows : (sin Af, i. e. ( , ^ ) , is written sin' A^ ^ ' \ hypotenuse / (base \ ' , ; ) , is written cos'^, hypotenuse/ and so on. The student must notice that *sin ^ ' is a single symbol. It is the name of a number, or fraction, belonging to the angle A ; and if it be at any time convenient, we may denote sin A by a single letter, such as s or x. Also sin^ A is an abbreviation for (sin A)\ that is for (sin A) x (sin A). Such abbreviations are used because they are convenient. 83. The Trigonometrical Eatios are always the same for the same angle. THE TRIGONOMETRICAL RATIOS. 49 Take any an«5le EOE ; let F be any point in OE one of tlie lines containing the angle, and let P', /'" be any two points in OB the other line containing the angle. Draw PM perpendicular to 07?, and F'M'y F'M" perj^endiculars to OE. Then the three triangles OMP, OM'F, OM"F' each" ' oatain a right angle, and they have the angle at com- mon ; therefore their third angles must be equal. Thus the three triangles are equiangular. MP M'P' M"F' Therefore the ratios .yp , -rpp, , TfpTT are all equal. (Eu. VI. 4.) But each of these ratios is -~ with ref&rence hypotenuse to the cmgle at ; that is, they ai-e each sin ROE, Thus, sin POE is the same whatever be the position of the [)oint P on either of the lines containing the angle ROE. Therefore sin ROE is always the same. 84. A similar proof holds good for each of the other ratios. 85. Also if two angles are equal, it is clear that the numerical values of their Trigonometrical Ratios will be the sama We have already shown (Art. 74), that the values of these ratios are different for different angles. Hence for each particular value of A, sin A, cos Aj tan A, etc. have dejinite numerical values. Example. We shall prove (Art. 92) that L. E. T. sm300=i = '5, co8 30«='^ = -8660..., tan 30«=-^=-577... 50 TRIGONOMETRY. 86. In the following examples the student should notice (i) the angle referred to, (ii) that there is a right angle in the same triangle as the angle referred to, (iii) the perpendicular, which is opposite the angle referred to, and is perpendicular to one of the lines contain- ing the angle, (iv) the hypotenuse, which is opposite the right angle, (v) the base, the third side of the triangle. Example. In the second figure on the next page, in which BDA is a right angle, find sin DBA and cos DBA. In this case (i) DBA is the angle. (ii) BDA is a right angle in the same triangle as the angle DBA. (iii) DA is the perpendicular, for it is opposite DBA and is perpendicular to BD. (iv) BA is the hypotentise. (v) BD is the ha^e. Therefore sin DBA, which is ^J^ , , = v,-t , ' hypotenuse BA COS DBA, which is r — — i , =17-; • ' hypotenuse BA EXAMPLES. XII. (1) Let ABC be any triangle and let AD be drawn perpen- dicular to BC. Write down the perpendicular, and the base when the following angles are referred to : (i) the angle ABD, (ii) the angle BAD, (iii) the angle ACD, (iv) the angle DAC. FXAMPLES. XII, .4 ;^\ Write down the following ratios in the aijuve tigure ; (i) Bin BAD, (ii) coa ACD, (iii) tan I) AC, (iv) ain ABB, (v) i&nBADj (vi) sin D AC, (vii) cos BCA, (viii) tan />C^, (ix) cos ABB, (i) sin^Ci). (3) Let ACB be any angle and let ABC and i5Z)C be right angles ; (see next figure). Write down tux) values for each of the following ratios ; (i) sin ACB, (ii) cos ACB, (iii) tan ACB, (iv) sin BAC, (y) cos B AC, (vi) tan BAC (4) In the accompanying figure BBC, CBA and' E AC are right angles. Write down (i) sin DBA, (ii) sin B£JA, (iii) sin CBD, (iv) cosBAE, (v) cos BAB, (vi) coa CBD, (vii) t&u BCD, (viii) tan DBA, (ix) tan BEA, (x) tan Ci?2>, (xi) sin DAB, (xii) sin BAE. 4—2 52 TRIG ONOMETR Y. XII. (5) Let ABC be a right-angled triangle such that ^2?-- 5 ft. BG=Z ft., then AC will be 4 ft. B Find the sine, cosine and tangent of the angles at A and B respectively. In the above triangle if A stand for the angle at A and B for the angle at B, show that sin^^ + cos^ ^ = 1, and that sin2 5 + cos2^=l. (6) If ABC be any right-angled triangle with a right angle at C, and let A, B, and C stand for the angles at A, B and C respectively, and let a, b and c be the measures of the sides oppo- site the angles A, B and G respectively. a Show that sin ^ = ■ cos^==-, tanJ.=^ c b Show also that sin^ A + cos^ A = l. Show also that (i) a=c . sin A, (ii) b=c.ainB, (iii) a—c. cos B, (iv) 6 = c . cos ^, (v) sin A = cos B^ (vi) cos A = sin B^ (vii) tan A =C0t5. (7) The sides of a right-angled triangle are in the ratio 5 : 12 : 13. Find the sine, cosine and tangent of each acute angle of the triangle. (8) The sides of a right-angled triangle are in the ratio 1:2: >/3. Find the sine, cosine and tangent of each acute angle of the triangle. (9) Prove that if A be either of the angles of the above two triangles sin^ A + cos^ A = l. ( ^3 ) CHAPTER VI. On the Trigonometrical Ratios of Certain Angles. 87. The Trigonometrical Ratios of an angle are numeri- cal qitayiiities simply^ as their name ratio implies. They are in nearly all cases incommensurable numbers. Their practical value has been found for all angles between and 90°, which differ by V ; and a list of these values will be found in any volume of Mathematical Tables. The student is recommended to get a copy of Chambers' Mathematical Tables for instruction and reference, 88. Tlie finding the values of these Ratios has involved a large amount of labour; but, as the results have been published in Tables, the finding the Trigonometrical Ratios does not form any part of a student's work, except to ex- emplify the method employed. 89. The general method of finding Trigonometrical Ratios belongs to a more advanced part of the subject than the present, but there are certain angles whose Ratios can be found in a simple manner. 90. To find the aine^ cosine and tangent of an anrjle of 45". If one angle of a right-angled triangle be 45", that is, the half of a right angle, the third angle must also be 45*! Hence 45" is one angle of an isosceles right-angled triangle. 54 TRIQONOMETRY. F^ /i^ \ "i J Let ~M Let POM be an isoscel^. triangle, such thatPi^O is a right angle, and OM = MP, '^Then POM= 0PM =^ 45". Let the measures of OM and of MP eacl^ be i^. the measure of OP be x. Then x^=^m'-¥m' = 1m'; .'. x= J 2 . m.: MP _ m _ JL_ OP "727^" 72' OM _ m _ 1 OP ~ 727^ " 72 ' Hence, sin 45" = sin POM Gos i5' = cos POM: ^ tan45" = tanPOJf=.^=^=i L 91. To find the sine, cosine and tangent of 60'^. In an equilateral triangle, each of the equal angles is 60", because they are each one third of 180". And if we draw a jDerpendicular from one of the angular points of the triangle to the opposite side, we get a right-angled triangle in which one angle is 60". Let OPQ be an equilateral triangle. Draw PM per- pendicular to OQ. Then OQ is bisected in M. Let the measure of OM be m ; then that of OQ is 2m and therefore that of OP is 2m. TRIGONOMETRICAL RATIOS OF CERTAIN ANGLES. 55 P Hence, sin 60" = sin POM ■■ cos60°-cosPO.lf: M Q Let the measure of MF be x. Then x» = (2m)' - m» = 4m' - m' = 3»i^ .-. x= JTTm. . MP ^ JT7^ _ JZ OP 2m 2 0J/_ m_ _ 1 OP " 2^ ~ 2' /3" ^"^^'=^-^^^=S=^!r- 1 -7 A 92. To find the sine, cosine and tangent of 30°. With the same figure and construction as above, we have the angle OPJf =30°, since it is a half of OPQ, Le. of 60°. Hence, 8in30° = sin (9Pif=^= ^ =\, PO 2m 2 tan 30° = tan OPif- MO PM JZ.m J3 OG TRIGONOMETRY, **93. To find the sine of 18". In the 10th Prop, of Euclid IV. a triangle is described such that each of the angles at the base is double of the third angle. [See also Example (6) p. 140.] Let POQ be such a triangle, and let the vertical angle PO^ contain n degrees ; then 7i+ 2^ + 271=180, ri=36. Draw OM perpendicular to PQ, bisecting the angle POQ. Then since QOP = 36", MOP = 18". Also PM= MQ. Let the measure of MP be m, and the measure of OP be X. From OP cut off OR = PQ. Then by Euclid lY. 10 PO . PE^PQ". .-. x{x-2m) = {2my, .'. x^ — 2mx + 7)i' = im^ + m" = 5m", . *. x — m= J 5 . m, .'. x= Jb . m+ m, .*. sin 18" = sm MOP = pm = —r^ = tk^ — r = ^i — • OP ^5.m-\-7n v^5 + 1 4 OMETRICAL RATIOS OP CERTAIN ANGLES. 57 ^ I 94. To find the sine, cosine and tangent o/O''. By this is meant, — 2'o find t/ie valines, if any^ to which Trigonometrical Ratios of a very small angle approach, icM the cmgle is contintially diminished. Let BOP be a small angle. Draw FM perpendicular to OB, and let OF be always of the same length, so that F lies on a circle whose centre is 0. Then if the angle BOF be diminished, we can see that MF IMF is diminished also, and that consequently j-^ , which ■ sin BOF, is diminished- And, by diminishing the angle mOF sufficiently, we can irmke MF as small as we please, ■id therefore we can muke sin ROF smaller tJian any assign- mle number however STnall that number may be. I Thus we see that the value to which sin/?OP ap- Iroaches as the angle is diminished, is 0. I This is expressed by saying, sin 0° = i. i Again, as the angle BOF diminishes, OM approaches ' . . . OM OF in length; and cos BOF, which is ^r-p , approaches in value to jyp^ ^'^' ^ • This is expressed by saying, cos 0" = 1 ii. MF Also, tan BOF is y^^ ; and we have seen that MF ap- proaches 0, while OM does not ; .*. tan J?0/* approaches 0. This is expressed by saying, tan 0" ^ ... iil 58 TRIGONOMETRY. 95. To find the sine, cosine and tangent of 90". By this is meant, — To find the values, if any, to which tlie Trigonometrical Ratios of an angle approach, as the angle approaches a right angle. Let liOUhe a right angle = 90". M Draw EOF nearly a right angle ; draw PM perpen- dicular to OE, and let OP be always of the same length, so that P Kes on a circle whose centre is 0. Then, as the angle EOP approaches to EOU, we can see that MP approaches OP, while OM continually diminishes. MP Hence when EOP approaches 90°, sin EOP, which is -^-p , OP . 1 . approaches in value to ^^n > ^^^* is to y , i.e. to 1. Hence we say, that sin 90° = 1 i. Again, when EOP approaches 90°, cos EOP, which is , approaches in value to -rr-p , t Hence we say, that cos 90° = ^Yp , approaches in value to -rr-p , that is to 0. )Nu.\JLiJi/OAL RAT/OS OF CERTAIN ANGLES. 59 MP Again, when ROB approaches 90*, tan ROP which is OP OM )proaches in value to : ... , rv • a quantity which approaches But in any fraction whose numerator does not diminish, smaller the denominator the greater is the value of that jtion ; and if the denominator continually diminishes the tlue of the fraction continually increases. Hence, torn. ROP can be made larger t/utn any assigruihle imber by making the angle ROP approach 90" near enough. This is what we mean when we say, that tan 90" is infinity, or, tan 90° = oo iii. 96. The followiug table exhibits the results of this Chapter. angle 0« 180 30« 450 COO 90« 1 sine 4 1 2 1 ^2 V3. 2 1 cosine 1 V3 2 1 V2 2 1 j tangent 1 1 7^ ' N^3 The student may notice mat the sine increases with the angle, while the cosine diminishes as the angle increjisea Also that the squares of the sines of 0*, SO^, 45", 60« and 90« are re8t)ectively 0, ^, f, J and J, and that the squares of the cosiues of the same angles are }, ^, f, \, and 0. GO TRIQONOMETR Y. EXAMPLES. XIII. If ^1 = 900, ^ = 600, (7=300, 2):^450^ ^=18°, prove the following : (I) 2.sini).cGsZ)=sm^. (2) 2. sin C. cos (7= sin J5. (3) cos2 B - sin2 ,8=1-2 sin^ B. (4) sin B . cos C+ sin C . cos 5=sin A. (5) cos2i)-sin2Z)=cos ^. (6) 4.sin2^+2. sini^^l. (7) sin2 5 + cos2 5=l. (8) cos2 (7+sin2{7=l. (9) cos2i) + sin2i)=l. (10) sin 5. cos C- sin (7. cos i?= sin C. (II) 2 (cos B . cos Z) + sin ^ . sin Df = 1 + cos C. (12) 2 (sin D . cos C- sin G. cos Df=\ - cos C. (13) sin 300 = -5. (14) sin 450= -7071.... (15) sin 600 = -8660.... (le) tan 600=1-7320508.... (17) tau 300 = -5773.... (ig) siu 180 = -3090... 97. The actual measurement of the line joining two points which are any considerable distance apart, is a very tedious and difficult operation, especially when great accu- racy is required; while the accurate measurement of an aTigle can, with proper instruments, be made with compara- tive ease and quickness. 98. A Sextant is an instrument for measuring the angle between the two lines drawn from the observer's eye to each of two distant objects respectively. A Theodolite is an instrument for measuring angles in a horizontal plane ; also for measuring * angles of elevation ' and ' angles of depression.' 99. The angle made with the horizontal plane, by the line joining the observer's eye with a distant object, is called ^VMhTlUCAL RATIOS OF CERTAIN AiSyLE.s. 61 (i) its angle of elevation, when the object is cihoce the observer ; (ii) its angle of depression, when the object is belcrw the observer, t 100. Trigonometry enables us by measuring certain ]le8, to deduce, from one known distance, the lengtJis of distances : or, by the measurement of a convenient ie, to deduce by the measurement of angles the lengths of ines whose actual measurement is difficult or impossible. [In the Trigonometrical Survey of England, made by the Ordnance Department, the only distance actually meaaured was one of about seven miles on Salisbury Plain.] 101. For this purpose we require the numerical values of the Trigonometrical Ratios of the angles observed. Ac- cordingly mathematical tables have been compiled, contain- ing lists of the values of these Ratios. These Tables constitute a kind of numerical Dictionary, in which we can find the numerical value of the Trigonometrical Ratios of any required angle. Example 1. At a point 100 feet from tlie foot of a tower the ingle of elevation of the top of the tower is observed to be 60**. Find the height above the point of observation of the top of the tower. P t In measuring the angle of depression the telescope is tamed from a horizontal position downwards. See Ex. (3) p. 63. 62 TRIGONOMETRY. Let be the point of observation ; let P be the top of the tower ; let a horizontal line through meet the foot of the tower at the point i/. Then OM =100 feet, and the angle MOP^m"^. Let MP contain x feet. Then MP OM' :tan JfOP=tan 600=^/3 100 =V3. .-. A'=].00. x/3 = 100 X 1-7320 etc. = 173-2. Therefore the required height is 173-2 feet. Example 2. A flagstaff', 25 feet high, stands on the top of a cliff*, and from a point on the seashore the angles of elevation of the highest and lowest points of the flagstaff* are observed to be 470 12' and 45° 13' respectively. Find the height of the cliff*. Let be the point of observation, PQ the flagstaff*. Let a horizontal line through meet the vertical line PQ produced in M. Then : how far is the top of the house from the tower? ^'4 TRIGONOMETRY. XIV. (5) A man 6 ft. high stands at a distance of 4 ft. 9 in. from a lamp-post, and it is observed that his shadow is 19 ft. long. Find the height of the lamp. (6) The shadow of a tower iji the sunhght is observed to be 100 ft, long, and at the same time the shadow of a lamp-post 9 ft. high is observed to be 3^/3 ft. long. Find the angle of elevation of the sun, and the height of the tower. ^ (7) From a point P on the bank of a river, just opposite a post Q on the other bank, a man walks at right angles to PQ to a point R so that PR is 100 yards ; he then observes the angle PRQ to be 320 17' . ^^ the breadth of the river, tan 320 17' = -6317667. - (8) A flagstaff 25 feet high stands on the top of a house ; from a point on the plain on which the house stands the angles of elevation of the top and bottom of the flagstaff are observed to be 60^ and 45" respectively : find the height of the house above the point of observation. (9) From the top of a cliff 100 feet high, the angles of depres- sion of two ships at sea are observed to be 45*^ and 30^ respectively ; if the line joining the ships points directly to the foot of the chff, find the distance between the ships. (10) A tower 100 feet high stands on the top of a cliff; from a point on the sand at the foot of the cliff the angles of elevation of the top and bottom of the tower are observed to be 75^ and 60<) respectively ; find the height of the cUff. (Tan 75^ = 2 + a/3). (11) A man walking along a straight road observes at one milestone a house in a direction making an angle 30^ with the road, and that at the next milestone the angle is 60° : how far is the house from the road 1 (12) A man stands at a point A on the bank AB of a straight river and observes that the line joining J. to a post C on the opposite bank makes with ^^ an angle of SO*', He then goes 400 yards along the bank to B and finds that BG makes with BA an angle of 60°; find the breadth of tha river. EXAMPLES, XIV. C5 '*(13) A building on a square base A BCD has two of its sides, B and CD, parallel to the bank of a river. An observer, stand- Iat B on the other side of the river so that DAE is a straight >, finds that AB subtends at his eye an angle of 45*^. Having ked a yards parallel to the bank, he finds that DE subtends angle whose tangent is v2. Show that DB = a yards. (14) From the top of a hill the angles of depression of the and bottom of a flagstafi" 25 feet high at the foot of the hill observed to be 45^ 13' and 47^ 12' respectively ; find the height of the hill. \J^m tan 450 13' = 1-0075918, l^r tan 470 12' =1-079901 8. (15) From each of two stations. East and West of each other, the altitude of a balloon is observed to be 45**, and its bearings to be respectively N.W. and N.E. : if the stations be 1 mile apart, determine the height of the balloon. (16) The angle of elevation of a balloon from a station due south of it is 60^; and from another station duo west of the former and distant a mile from it it is 45<>. Find the height of the balloon. {11) An isosceles triangle of wood is placed on the ground in a vertical position facing the sun. If 2a be the base of the triangle, 6 its height, and 30^ the altitude of the sun, find the tiingent of half the angle at the apex of the shadow. (18) The length of the shadow of a vertical stick is to the length of the stick as '^3 : 1. If the stick be turned about its lower extremity in a vertical plane, so that the shadow is always in the same direction, find what will be the angle of its inclina- tion to the horizon when the length of the shadow is the same as before. (19) \{]ia.i distance in space is travelled in an hour in conao- quence of the earth's rotiition, by a person sitaated in latitude 00»? (Earth's radius = 4(;O0 miles.) L. K. T. C ( 66 ) CHAPTER VTI. On the Relations between the Thigonometrical Ratios of the Same Angle. 102. The following relations are evident from the definitions : 1 - 1 - 1 cosec 6 -^ — - , sec - sin 6 ' cos sm cot tan^ 103. To prove tan ^ - "^^^ perpendicular Wc have «i^^ = 1^otenuse base and cos 6 = hypotenuse sin ^ _ perpendicular _ ^^^ ^^ • ' cos " base cos B 104. We may prove similarly cot 6/ = -r-^ . 1 cos 6 Orthus,cote = ^^^=-^ m ^RIGOSOMETRICAL RATIOS OP TIIE SAME AyOLE. G7 tl05. Euclid I. 47 gives us that in any right-angled Qgle the square on the hypotenuse = the sum of the ires on the perpendicular and on the base, or, (hypotenuse)^ - (perpendicular)' + (baae)^ (i) Divide each side of this identity by (hypotenuse)®, and we get ( hypotenuse Y /perpendicularx^ / base \* hyj>ot«nuse/ \ hypotenuse j \hypotenuse/ ' it is, 1 = sin' d + cos' 0. (ii) Divide each side of the same identity Ijy (base)', and we get (hypotenuseV _ /perpendicularX' /base\' base / ~ \ base / Vbase/ ' (lii) Divide each side of the same identity by (perpendicular)', and we get / hypotenuse \' _ /perpendicularX' / base \' Vperpendicular/ ~ \perpendicular/ Vperpendicular/ ' it is, cosec'^ = cot^^ + 1. 106. Thus the three results (i) cos' e + sin' $ = 1 (ii) l+tan«^=8ec'd (iii) 1 + cot? 6 = cosec' each a statement in Trigonometrical language of Euc. 1. 47 5—2 G8 TRIGONOMETRY. 107. We give the above proof in a different form. To prove that cos' + sin' 6 = 1. Let EOU be any angle 6. \ In OE take any point P, and draw PM perpendicular to OR. Tben with respect to 0, MP is the perpendicular, OP is the hypotenuse, and OM is the base ; .-. sin-'6>=^^, cos OM- 'OP' We have to prove that sin" 6 + cos" ^ - 1. Now co&'O + bin = ^ ijg- + ~q'jp^ MP' + OM' _qp^^i - OP' ~ OP' ' since by Euclid I. 47, MP'+ 0M'= 0P% Therefore cos^'^-f sin*^^ = 1. Similarly we may prove that l+tan^'^^sec^^, and that 1 + cot' = cosec"-' 0. , WONOMETRICAL RATIOS OP THE SAME ANGLE, GO 108. The following is a List of FoRMULiE with which student must make himself familiar : /) ^ • cosec o sin. 6' sec^ = -L. ^^ coaO cot^--.J--, 3 ^ tan^ co&O coj^ sin 6 cos COB 6 sin' ^ + cos" ^=1, tan*^+l=sec'^, "^ cot' ^ + 1 = cosec' 0. O 109. In proving Trigonometrical identities it is often ^nvenient to express the other Trigonometrical Ratios in 'ms of the sine and cosine. Example. Prove that tan A + cot A=secA. cosec A. Since . sin A , , cos A um A = , , cot A = -; — . , cos A ' sm il ' sec A = 7 and cosec A = -; — , . con A sm J have to prove that sinil that sin A cos J ' sin ^ sin* A + cos* A 1 cos A . sin A cos A . sin .1 ' lid this is true, because sin* A + cos* .4 = 1. 70 TRIGONOMETRY. 110. Sometimes it is more convenient to express all the other Trigonometrical Ratios in terms of the sine only, or in terms of the cosine only. Example (i). Prove that ,^ sin* ^ + 2 sin2 ^ cos2 (9 = 1 - cos* ^. Here the right-hand expression does not contain sin 6 ; hence for sin2^ in the left-hand expression we substitute (1 — cos^^)- thus •sin* e + 2 sin2 e cos^ 6 = (sin2 0)^^ 2 (sin2 6) cos- = (1 - C0S2^)2 + 2 (1 - C0S2 ^) C0S2 ^ = 1-2 cos2^-l-cos*<9-f2 cos2^-2 co.s'^ ^l-COS*^. Q.E.D. Example (ii). Prove that sinc (9 + cos« ^ = 1 - 3 cos2 6 + 2 cos* 0. Here sin*^ e + cos" 6 = (sin2 $ + cos^ 6) (sin* - sin2 cos^ 6 + cos • ^) = 1 X (sin* 6 - sin2 ^ cos- 6 + cos* ^) == (sin2 (9)2 _ (sin2 ^) cos^ <9 + cos*^ = (1 - cos2 ^)2 - (1 - cos'- 6) cos''^ <9 -f cos* e = 1-2 cos2^ + cos*^ - cos2 e + cos*d + cos* (9 = 1-3 cos2 (9 + 3 cos* ^. Q. E. D. l^OTE. (1 - cos 0) is called the versed sine of 6; it is abbreviated thus versin 6. /■y^^rprES. XV. cot J^ . tail .1 = 1. Bee ^ . cot ^ =ico.sec A . ^ EXAMPLES. XV. e the following statements. cosJ .tan^=sm J, (2) (3) cos J=sini4.cot^. (4) (')) cosec A . tan A = sec ,1 . (C) (tan ^ + cot J) sin -ii .cos ^1 = 1. (7) (tan ^ - cot ^) sin .4 . cos ^ =sm- A — cos' A. .(8) cos2 A - sin- .4 = 2 cos'-^ . 1 - 1 = 1 - 2 sii i'-^ .4 . (9) (sin A + cos A )- = 1 + 2 sin ^ . cos .4. (10) (sin J -cos ^)2=1 -2 sin J . cos J. (11) cos*5-sin^J[?=2cos2/J-l. (12) (sin2i; + cos'-i?)- = l. (13) (sin2 i; - cos2 Z?)2 = 1 - 4 cos2 B vA cos» B. (14) l-tan*i5=2sec-i5-sec''i?. (15) (secJ5-timZ;)(socZ? + tan5) = l. ( 1 6) (cosec 6 - cot 6) (coseu ^ + cot ^) = 1 . ( 1 7) sin^ e + cos3 e = (sin ^ + cos ^) (1 - sin 6 coa 6). (18) cos3^-8in3^=(cos^-bin6^) (l + sin^cos6>). (19) 8in« e + cos« ^ = 1 - 3 sin' 6 . cos-' 6. i^IQ) (sino 6 - cos" 6) = (2 sin' ^ - 1) (1 - sin- 6 + sin* 6). . . tan^ + tani^ (21) (22) (23) (24) (25) (2fn / cot ^ + cot 5 cota+tan/3 = tan^. tani?. tana + cotj3 1 - sin yl =cot a. tau/y. = (sfcc.4-tan^l)'. = (cosec A + cot A )-. 1+siu J 1-f cos^ 1 -cos J 2 versin $ - versin' ^=sin' 6 versin ^ (1 + cos ff) = sin' 6. 72 TRIGONOMETRY. 111. All the Trigonometrical Ratios of an angle can be expressed in terms of any one of them. Example 1. To express all the trigonometrical ratios of an angle in terms of the sin^. Let ROE be any angle A. We can take P anywhere in the line OE; so that we can make one of the lines, OP, OM, or MP any length we please. Let us take OP so that its measure is 1, and let s be the MP measure of MP ; so that sin A, which is -^73 , Let X be the measure of OM. Then since (91/2 _ Qp2 _ j/p2^ .-. x'^=l-s\ = sin^. x=s'T^ Hence , OM co,A=^^. vr tan^: MP OM' = \h -siii^A, sin^ x/f v/r sin^ A and so on. Note. The solution of the equation 'x'^=l- s^, gives and therefore the ambiguity ( ± ) must stand before each of the root symbols in the above. This ambiguity, as will be explained later on, is of great use when the magnitude of the angle is not limited. When we limit A to be less than a right angle we have no use for the nec^ative sign. ""RIGONOM ETHICAL RATIOS OF THE SAME ANOLt. /.j Example' 2. To express all the other trigonometrical ratios ' an angle in terms of the tangent. this case tau^= MP OM Take P so that tlie measui'e of OM is 1, and let t be the jasure of MP ; so that tan ^, which is ^ > ~ 7 5 *^^» ^ = tan 6. Then we can show that the measure of OP is v 1 + fi. „ . ^ MP t tan^ Hence, Bin ^ = -^ = -^^ = - =- . . OM 1 1 OP Vn-<2 Vr+tan2^ Vuid SO on. 112. The same results may be obtained by the use of the formulae on patre 69. Example cos^ ^ + si n- ^ = 1 , .-. cos2^=l-sin2^, .-. cos^=\^l -sin'-*^. sin 6 sin 6 Again and so on tan ^ = cos ^ Vl - sin2 d ' 74 TRIGONOMETRY, ' EXAMPLES. XVI. (1) Express all the other Trigonometrical Ratios of A in terms of cos A . (2) Express all the other Trigonometrical Ratios of A in terms of cot A . (3) Express all the other Trigonometrical Ratios of A in terms of sec A. y (4) Express all the other Trigonometrical Ratios of A in terms of cosec A. (5) Use the formulae of page 69 to express all the other Trigonometrical Ratios of A in terms of sin A. (6) Use the formulae of page 69 to express all the other Trigonometrical Ratios of A in terms of the tan A . 113. Given one of the Trigonometrical Ratios of an angle less than a right angle, we can find all tlie others. Since all the Trigonometrical Ratios of an angle can be expressed in terras of any one of them, it is clear that if the numerical value of any one of them be given, the numerical value of all the rest can be found. Example. Given sin^=f, find the other Trigonometrical Ratios of 6. Let ROE be the angle 6. Take P on OE so that the measure cf OP is 4. Draw PM perpendicular to OR. w y? M TRIGONOMETRICAL RATIOS OF THE SAME ANGLE, i (MP \ so that j-p = ^ ) > and since the measui lof OP is 4, therefore the measure of MP must be 3. Let X be the measure of OM; then OM'=OF^-MP^, .: a,-2=4-- 3-= 10-9 = 7. I Therefore the measure of OM is >/"• Hence, ^ OM v/7 tan^ = MP^3 OM ^/■; 3v/7 cot ^ = ^7- ^ ^rrilXAMPLES. XVII. (1) Given that sin A = 'j, fintl tair Jnand cosec A.j^ (2) Given that cos B=^, find sin B and cot /^, (3) Given that tan A = ^, find sin A and sec il (4) Given that sec ^=4, find cot 3 and sin 0. (.")) Given that tan $=J3j find sin B and cos 6. (6) Giv^that cot ^= -"^ , find sin ^ and sec 6. ■^^- (7) (8) (9) (10) .(11) h Ml X', Given that sin ^=-, find tuu^. Given that tan Q- 6' find sin Q and cos 6. Given that cofc=- , find sin 6 and cot Q. If sin b^^cL, and km ^=6, prove that (1 - a*) (1 + 6'-^) = 1. If co8d=/i, and tan B—h^ find the equation connecting 76 TRIGONOMETRY. 114. The following propositions are important. PROP. I. To trace the changes in the magnitude of sin A as A increases from 0<* to 90^. Then sin J. Take a line 0-R, of any length; and describe the quadrant RPU of the circle whose centre is and radius OR, Draw the right angle ROU, cutting the circle in U. Let OP make any angle ROP {=A) with OR ; draw PM per- pendicular to OR. MP OP' When the angle A is 0^, MP is zero, and when A is 90", MP is equal to OP ; and as A continuously increases from 0" to 90^, MP increases continuously from zero to OP ; also OP is always equal to OR. Therefore, when A = O^', the fraction -y^ is equal to -y^ , that MP OP is 0; when ^ = 90" the fraction -y^j is equal to -r^, that is 1; and as A continuously increases from 0*^ to 90", the mmierator of MP the fraction jyp continuously iticreases from zero to OP, while MP the denominator is unchanged, and therefore the fraction jr.^ , which is sin J, increases continuously from to 1. t as t RIGONOMETRICAL RATIOS OF THE SAME ANGLE. 77 PROP. II. To trace the changes in the magnitude of tan A as A increases from QP to 9(/^. With the same construction and figure as in the last article, have tan^ = ^. When the angle A is 0", 3fP is zero ; when A is 90*', MP is equal to OP ; and as the angle continuously increases from 0** to '0*', MP increases continuously from zero to OP. I When the angle A is O^', 03f is equal to OP ; when A is 90^ M is zero; and as A continuously increases from 0" to 90", '■"■"■""""■ Hence, when A is 0^, the fraction ^yry is equal to jjTf , that is 0; when A is 90'', the fraction j-^^ is equal to -rr-, that is 'in- finity' (see art. 95) ; and as A continuou.sly increases from 0** to 90'', the numerator continuously increases from zero to OP, while the denominator continuously diminishes from OP to zero ; >o that the fraction -rrrr^ which is tan^, continuously increases from until it is greater than any assignable numerical quantity. EXAMPLES. XVIII. (1) Show that as A continuously increases from 0° to 90", cos^ continuously diminishes from 1 to 0. (2) Trace the changes in the magnitude of sec ^ as ^ increases from to ^ . (.3) Trace the changes in the magnitude of sin Ab& A dimin- ishes from 90<' to 0". (4) Trace the changes in the magnitude of cot 6a&6 increases from to ^ . 78 TRIGONOMETRY. 115. Since the hypotenuse is the greatest side in a right- angled triangle, it is clear ..V ^, ^ . . , . , . j)erpendicular (i) that sin Ay which is ~-^ , is never hypotenuse greater than unity, base (ii) that cos A, which is hypotenuse greater than unity, /•••\ ^^ L. i 1 • T- • hypotenuse (ill) that cosec A. which is — ^— — - — ^ — , is never ^ perpendicular numerically less than unity (we shall see later on that it may be negative), /• X .^ , A II- hypotenuse . (iv) that sec A, which is — ^^-V j is never nume- base rically less than unity (it may be negative). Hence such a question as ' Find an angle whose sine is ^' has no solution, since there is no angle whose sine is greater than unity. And such a question as * Find an angle whose secant is I ' has no solution, because there is no angle whose secant is numerically less than unity. 116. If ^ be small, the perpendicular is smaller than the base: and tan A. which is - — ^ , can be made as base small as we please (see figure in Art. 114). Also if the angle A be nearly a right angle, the perpendicular is greater than the base : and tan A, which is -— ^-^ , can be made base ' as large as we please. 'JOONOMETRICAL RATIOS OF THE SAME ANGLE. 79 So that an angle whose tangent is a can always be tound be positive ; that is, there is always an angle A between „nd 90", such that tan ^ = a, if a be a positive number. *[The student will find as he proceeds that the equation x = a can always be solved if a be any arithmetical intity]. Example 1. To draw an angle A whose sine is J. Draw any line OR^ and take R such that the measure of OR is unity ; describe the quadrant RPU of the circle whose centre is and radius OR. In OU which is perpendicular to OR^ take O^V so that the measure of ON is f . Draw NP parallel to OR cutting the quadrant in P. (The student should observe that if the measure of ON were greater than unity y N would be outside the circle altogether, and the line parallel to OR would not cut the circle at all.) Join OP, and draw PM perpendicular to OR. Then ROP is the angle required. Therefore an angle POR has been drawn whose sine is J. 80 TRIGONOMETRY. Example 2. To draw an angle whose cosine is b ; where 6 is a proper fraction and positive. M R Draw any line OR, and take OR so that its measure is unity. Describe the quadrant RPU of the circle whose centre is and radius OR. Take M in OR so that the measure of OM is b. OM will be less than OR because & is a proper fraction. Draw MP perpendicular to OR cutting the quadrant in P. (The student will observe that if b were an improper fraction, M would be outside the circle altogether.) Then the angle ROP is the angle required. For cos ROP ■ OM b , 0P = 1 = ^- Therefore an angle ROP htis been described such that the co^ ROP =b. Example 3. To draw an angle A such that tan A = a, where a is any positive numerical quantity. P TRIGONOMETRICAL RATIOS OF THE SAME ANGLE, 81 Draw any lino OR, and take M in it so that the measure of !/ is unity. Draw MP perpendicular to DM, and take P so that the measure of MP is a. Join OP. Then the angle ROP is the angle required. FortanjR(?P=^ = ^=«. Thus an angle ROP has been drawn so that tan ROP=cu 117. Definition. One angle is called the complement of another, when the two angles added together make up a '-^'jht angle^ Example 1. The complement of A is (90*^ - A). Example 2. The complement of lOO^ is (90^ - 190°) = - lOO^'. For 1 900 + (900 _ igQO) = 900. Example 3. The complement of-^^^lo'Tj"" T^ 118. To prove tJuit the sine of an angle A. is equal to the cosine of its complement (90" — A). [This is true whatever be the magnitude of A, as will be proved later on.] If ^ be less than 90', let ROP be A (see last figure). Draw PM perpendicular to OR. Then since PMO = 90°, therefore POM+OPAf= 90" and therefore 0PM =(90"- A). Now, sin 4 = ^p = cos 0PM = cos (90° - A), q.e. d^ L. E. T. S2 TRIGONOMETRY. -. EXAMPLES. XIX. \ (1) Draw an angle whose sine is \. (2) Draw an angle whose cosecant is 2. (3) Draw an angle whose tangent is 2. (4) Can an angle be drawn whose tangent is 431 ? (5) Can an angle be drawn whose cosine is f ? (6) Can an angle be drawn whose secant is 5 ? (7) Find the complements of (i) 300. (ii) 1900. (iii) 900. (iy) 3500. (v) -250. (vi) -3200. (vii)^. (viii) -|. (8) Prove that sin 70° = cos 20<^. (9) Prove that cos 47<^ 1 6' = sin 420 44'. (10) Prove that tan 79<> = cot ll^. (11) Prove that sec 360=cosec 540. (12) If ^ be less than 90^, prove ^ (i) cos^ = sin (90<>-^). (ii) tan^ = cot (90*>-ui). (iii) sec ^ = cosec {^0^ — A). (iv) cot ^= tan (900-^). On the Solution op Trigonometrical Equations. 119. A Trigonometrical equation is an equation in which there is a letter, such as ^, which stands for an angle of unknown magnitude. The solution of the equation is the process of finding an angle which, if it be substituted for 6, satisfies the equation. SOLUTION OP TRIGONOMETRICAL EQUATIONS. 83 Example 1. Solve cos 6=^. This is a Trigonometrical equation. To solve it we miust find some angle such that its cosine is ^. We know that cos600=^. Therefore if 60'' be put for $ the equation is satisfied .*. ^=60*^ is a solution of the equation. Example 2. Solve sin B - cosec 6 + ^=0. The usual method of solution is to express all the Trigono- metrical Ratios in terms of one of them. Thus we put -; — ;, for cosec 6. and we get This is an equation in which ^, and therefore sin 6 is unknown. It will be convenient if we put x for sin 6, and then solve the equation for a; as an ordinary algebraical equation. Thus we get x-- + 3=0, X ' Whence x= - 2, or j;=^. But X stands for sin $. Thus we get sin 6= - 2, or sin 6=^. The value - 2 is inadmissible for sin dy for there is no angle whose sine is numerically greater than 1 (Art. 115). .*. sin^=^. But sin 300=^. .-. sin^=sin30*'. T]}erefore one angle which satisfies this equation for 6 is 30". 6—2 84 TRIGONOMETRY. Example 3. To solve the equation cosec 6 - cot^ ^ + 1=0. We have 1 + cot'^ 6 = cosec2 6. [Art. 1 08] .-. cot2^=cosec2^-].. and we get cosec 6 - cosec^ 6 + 2=0. Let X stand for cosec 6. Then x-x^ + '2.=0. .: x^-x=+% or .-. x'^-x + l= +1 + 2 = 1. .'. x-l=^^. Whence a^=2, or ;p= - 1, .•. cosec ^ = 2, but cosec 300 = 2, .'. cosec ^ = cosec 30". therefore 30^ is one angle which satisfies the equation for 6. ' EXAMPLES. XX. Find one angle which satisfies each of the following equations. ■A 1 (I) sm6=-j-o' (2) 4 sin ^= cosec (9. (3) 2cos^=sec^. (4) 4sin^-3 cosec^=-0. (5) 4 cos ^-3 sec ^=0. (6) 3tan^=cot^. (7) 3sin^-2cos2(9=0. (8) V'2sin^=tan <9. (9) 2cos^=x/3cot^. (10) tan ^=3 cot (9. (II) tan^ + 3cot^=4. (12) tan (9 + cot ^=2. (13) 2sin2<9 + V2cos^=2. (14) 4sin2^ + 2sin(9=l. (15) 3tan2<9-4sin2^=l. (16) 2sin2^ + ^2 sin^=2. (17) cos2 ^ - V3 cos ^ + 1=0. (18) cos2 e + 2 sin2 ^ - f sin ^ = 0. MISCELLANEOUS EXAMPLES. ^^ ♦* MISCELLANEOUS EXAMPLES. XXL (1 ) Prove that 3 sin GO^ - 4 sin^ eO® = 4 cos^ 30° - 3 coa 30». (2) Prove that tan 300 ( 1 + cos SO^ + cos eo^) = sin 3(y» + sin QCfi. (3) If 2 cos* ^ - 7 cos ^ + 3 = 0, show there is only one value of cos 6. (4) Find cos 6 from the equation 8 cos' ^-8 cos ^ + 1=0. (5) Find sin 6 from the equation 8 sin^ ^ - 10 sin ^ + 3 = 0, and prove that one value of ^ is - . (6) Find tan 6 fi-om the equation 12 tan* ^-13 tan 5 + 3=0. (7) If 3 cos* 5 + 2 . V3 . cos 5 = 5^, show that there is only one value of cos 5, and that one value of 6 is ^. (8) Prove that the value of sin* 6 + cos* 5 + 2 . sin* 6 . cos^ S is always the same. V (9) Simplify cos* A + 2. sin* A . cos* A. (10) Express sin® A + cos® A in terms of sin* A and powers of *iin*i4. v' (1 1) Express 1 +tan* in terms of cos and its powers. /,nN r> i.1 . COS ^+co«i? sin^+sin^ (12) Prove that - — -. ; — „+ -: o=^' ^ ' sm^-sm^ cos J- cos ^ ( 1 3) Express (sec A - tan A )* in terms of sin A. (14) Trace the changes in the magnitude of cosec 5 as 5 iu- lases from to - . (15) Trace the changes iri the magnitude of cot 5 as 5 de^ creases from „ ^ 0- (16) Solve sin(5+<^)=y,co6(5-0) = — . ( 86 ) CHAPTER YIIL On the Use of the Signs 4- and — . 120. The student is probably aware that, in the appli- cation of Algebra to Problems concerning distance, we some- times find that the solution of an equation gives the measure of a distance with the sign — before it. Example. Let M, iV, be places in a straight line ; let the distance from M to iV be 3 miles, and the distance from N to 0, 3 miles. /W n P ^ I ^ 1 1^ One man A starting from M, rides towards at the rate of 10 miles an hour, while another man B starting simultaneously from JV, walks towards at the rate of 4 miles an hour ; If Q be the point at which they meet, ho2V far is Q beyond ? Let P be any point beyond 0, and let x be the number of miles in OP. We wish to find x, i. e. the measiu-e of OP, so that P may coincide with Q, the point at which A overtakes B. When A arrives at P, he has ridden 6 + a: miles. The time occupied at the rate of 10 miles an hour is ^j-^ hours. When B arrives at P, he has walked 3 + x miles. The time occupied at the rate of 4 miles an hour is -j- hours. gt' USE OP THE SIGNS + AND - 87 When P is the point at which they meet, these times are [oal, so that 6 + ^ 34-a: , -— — = —J— ; whence x= - 1. Thus the required number of miles has the sign - before it ; id we have failed to find a point beyond at which A over- :es B. 121. Such a result can generally be interpreted by tering the statement of the problem, thus : The line whose measure appears with the sign — before , must be considered as drawn in the direction opposite to t in which it was drawn in the first statement of the problem. ^ M y p 9 Example. Taking the former example, let us alter the ques- on as follows : If $ be the point at which A overtakes B, how far is Q to the left of 01 Let P be any point to the left of 0, and let x be the number of miles in OP. We wish to find x (i.e. the measiu-e of OP), so that P may coincide with Q, the point at which A overtakes B. When A arrives at P, he has ridden 6 - x miles. When B arrives at P, he has walked 3 -x miles. Proceeding as before, we get %-x Z-x , [q"^~a'' whence .r= + l. Therefore if P is to coincide with Q (the point at which ii overtakes Z?), OP must be one mile to the left of 0. a 8 TRIGONOMETRY. 122. The consideration of such examples as the above has suggested, that tlie sign - may be made use of, in the application of Algebra to Geometry, to represent a direction exactly opposite to that represented by the sign +. Accordingly the following Rule, or Convention, has been made. IIULE. Any straight line AB being given, then lines drawn parallel to ^^ in one direction shall be posi- tive; that is, shall be represented algebraically by their mea- sures with the sign + before them : lines drawn parallel to BA in the opposite direction shall be negative; that is, shall be represented algebraically by their measures with the sign - before them. 123. We may choose for the positive direction in each case that direction which is most convenient. Example. Let LR be a straight line parallel to the printed lines in the page. and let lines drawn in the direction from Z to i2 in the figure that is, from the left-hand towards the right, be considered positive. Then by the above rule, lines drawn in the direction from Rio L, that is, from right to left, must be negative. 124. In naming a line by the letters at its extremities, we can indicate by the order of the letters, the direction in which the line is supposed to be drawn. Example. Let and P be two points in the line LR as in the figure, and let the measure of the distance between them be a. Then OP, i.e. the line drawn from to P, which is in the positive direction, is represented algebraically by + a. While PO, i.e. the line drawn from P to 0, which is in the negative direction, is represented algebraically by - a. USE OF THE SIGNS + AND - 89 125. Hence in using the two letters at its extremities represent a line, the student will find it advantageous rays to pay careful attention to the order of the letters. Example. Let X/2 be a straight line parallel to the printed los in the page. Let A, B, C, By E be points in Z/2, such that the measures of 15, BC, GDy DEy are 1, 2, 3, 4 respectively. Find the algebraical representation of (i) AC+CB (ii) AD+DC-BC. A B C '-' D - £ C-" ' ' ' "—R (i) The algebraical representation of AC ia +3, the algebraical representation of CB is - 2. Hence that of AC+ CB in + 3 - 2 ; that is, + 1 * (ii) The algebraical representation of AD is + 6, that of DC - 3, and that of BC is + 2. Therefore that of AD -[-DC- BC= 6-3-2=+ 1. This is equivalent to that of AB. EXAMPLES. XXII. In the above figure, find the algebraical representation of (1) AB + BC+CD. (2) AB + BC+CA. (3) BC + CD + DE+EC (4) AD -CD. (5) AD + DB + BE. (6) BC-AC+AD-BD. (7) CD + DB + BE. (8) CD-BD + BA+AC+CE. * By AC+CB (attention being paid to direction), we mean 'Go from A to C and from C to B.' The result is equivalent to starting lirom A and stopping at B, i.e. equivalent to AB. ( 00 ) CHAPTER IX. On the Use op the Signs + and - in Trigonometry. 126. In Trigonometry in order conveniently to treat of angles of any magnitude, we proceed as follows. We take a fixed point 0, called the origin ; and a fixed straight line OR, called the initial line. The angle of Avhich we wish to treat is described by a line OP, called the revolving line. This line OP starts from the initial line OR, and turns about through an angle ROP of any proposed magnitude into the position OP. Initial Line 127. We have already said in Art. 41 (i) that, when an angle ROP is described by OP turning about in the direction contrary to that of the hands of a watch, the angle ROP is said to be positive ; that is, is re- presented algebraically by its m.easure with the sign + before it. (ii) that, when an angle ROP is described by OP turning . about in the same direction as the hands of a watch, the angle is said to be negative; that is, is represented alge- braically by its measure with the sign - before it. ^^^^i E OP THE SIGNS + AND - IN TRIGONOMETRY. 91 Example. (180**-^) indicates (i) the angle described by OP turning about from the position OR in the positive direction until it has described au Igle of (180 - A) degrees. Or, (ii) the angle described by OP turning about 0, from I he position OR^ in the positive direction until it has described u angle of 180" (when it has turned into the position OL), and pen turning back from OL in the negative direction through he angle - A into the position OP. Or, (iii) the angle described by OP turning about from the position OR, in the negative direction through the angle - A, Pd then turning back in the positive direction through the gle 180^, into the position OP. The student should observe that in each of these three ways regarding the angle (180*^-^), the resulting angle ROP is the me. EXAMPLES. XXin. Draw a figure giving the position of the revolving line after it has turned through each of the following angles. I (1) 270«. (2) 3700. (3) 4250. (4) 5900, (5) -30«. (6) -3300. (7) -4800. (8) -7500. (9) 'i-. (10) 27i,r + ^. (11) (2n + l),r + |. (12) (27i+l)Tr -^.(13) 2n.-^. (14) (2n+l)7r-|. Note, ntr always stands for a wholo number of two right angles. 92 TRIGONOMETRY. 128. It is often convenient to keep the revolving line of the same length. In this case the point P lies always on the circumference of a circle whose centre is 0. Let this circle cut the lines LOR^ UOD in the points X, i?, U^ D respectively. The circle EULD is thus divided at the points R, U, L, D into four Quadrants, of which RUh called the first Quadrant. UL is called the second Quadrant. LD is called the third Quadrant. DR is called the fourth Quadrant. Hence, in the figure, ROPi is an angle of the first Quadrant. ROP^ „ „ second Quadrant. ROP, „ „ third Quadrant. POP. fourth Quadrant. I USE OF THE SIGNS + AND - IN TRIGONOMETRY. 93 129. When we are given the initial and the final line an angle, we can at once decide the Quadrant of which is. We cannot however decide the magnitude of the Igonoinetrical angle. For we do not know how many complete revolutions the revolving line may have made. ^^ In other words, when the geometrical representation of I^B angle (consisting of the positions of the initial line and ^^W the final line) is given, we are oiot (without further information) given the Trigonometrical angle. The Geometrical angle is here taken to be the amount of turning in the positive direction, which will transfer the revolving line from the initial to the final Une. 130. Let il" be the magnitude of a given geometrical angle JiOP ; then, every trigonometrical angle having OR for its initial line and OP for its final line, is included in the expression I^m nx360° + A°; or, 2nxlB0^ + A\ ^^^ere n is some integer positive or negative. For n X 360° indicates a number of complete revolutions of tho revolving line. When we know the magnitude of the trigonometrical angle, then we know what integer n is. Example. In the figure on p. 92 the geometrical angle ROP^ con- ios 135°. This angle is the geometrical representation of the angles 495°, 855°, -225°, etc., for 495° = 3G0° + 135° ; 855° = 720° + 135° ; - 225° = - 360° + 135°. N.B. Lot 6 be tho circular measure of a geometrical angle BOP ; then, every trigonometrical angle represented by POP, is included in the expression n <2n+ d; or, 2n7r + 0* 94 TRIGONOMETRY. EXAMPLES. XXIV. State in which Quadrant the revolving line will be after describing the following angles : (1) 120°. (2) 340°. (3) 490°. (4) -100°. (5) -380°. (6) -1000°. (7) y . (8) IOttH--. (9) ^ir-^. (10) 2n7r-'^. (11) (2n + l)7r + ^. (12) 7i7r + ^. Shew that the following angles are represented by the same diagram geometrically : (13) 365°, 725°, -355°, 1085°, -715°. (14) -65°, 295°, 655°, -425°. (15) 199°, 559°, 3799°, -521°, 7001°. (16) 6, 47r + ^, -(27r-^), -^n + S. (17) TT-a, Stt-q, -TT-a, (2w + l)7r-c:. (18) TT + a, 37r + a, -7r + a, {2n + l)7r + a. 131. The principal directions of lines with which we are concerned in Trigonometry are as follows ; I. that parallel to ORy the initial line. OR is usually drawn from towards the right hand, parallel to the printed hnes of the page. II. The direction at right angles to OB. III. The direction parallel to the revolving line OP. U ID I USE OF THE SIGNS + AND - IN TRIGONOMETRY. 95 Accordingly we make the following rules ; I. For lines parallel to the initial line 07?, the posi- tive direction is from to R; consequently the negative l^lurection is from R to 0. I^B II. 06^ is the position of the revolving line I^Kter it has described from OR a right angle ■^R)out in the positive direction of revolution; jnd for lines perpendicular to ORy the positive direction ^Hfrom toU 'j consequently for lines perpendicular to OR ^Hb negative direction is from to U to 0. ^B III. Any line drawn parallel to the revolving line in ^the direction from to P is to be positive, and consequently I any line drawn in the direction from P to is to be gative. Note. The student must notice that the revolving line OP carries its positive direction round with it, so that the line *0P' is always positive. The revolving line, as it turns about 0, nowhere undergoes •ny sudden change of direction such as would be indicated by a ange of sign. 132. We said, in Art. 81, that the definitions of the Trigonometrical Ratios (on pp. 46, 47), apply to angles of any magnitude. We have only to remark that it is gene- A\y convenient to take F on the revolving line; that is dmwn j^erpeudicular to the other line produced if saury; and that the order of the letters in i/P, OP, 0\[ is an essential part of the definition. ^mce, MP The order of the letters P, J/, in the expressions ^j, , ., b therefore of great importance. 96 TRIGONOMETRY. 133. We proceed to show that the Trigonometrical Ratios of an angle vary in Sig^ according to the Quadrant in which the revolving line of the angle happens to be. From the definition we have, with the usual lettei-s, mn BOP = ^^~, cos i?OP = ^, tan i?OP= ^. fig.r. Fig.lh I. When OF is in the first Quadrant (Fig. I.). J/P is positive because from M to F is upwards (PvuleiL p. 95.) OM is positive because from to Mis towards the rigid (Rule I. ) OF is positive (Rule iii.) OF THE SIGNS + AND - IN TRIGONOMETRY. 97 Hence, if A be any angle of the^rs^ Quadrant, . ^ .. .. MP . sm A, which is yrp , is positive ; . .... OM . cos J, which IS Y)pi ^^ positive ; MP tan J, which is ^^, is positive. II. When OP is in the second Quadrant, (Fig. ii.). positive^ because from Jf to P is wpwards^ OM is negative, because from to if is towa/rds the left. OP is positive. L^ OP IS j ^^H Hence, ^P ni. When OP is in tlie third Quadrant (Fig. in.) ^^TP is negative, OM is negative, OP is positive. So that, if -4 be any angle of the third Quadrant, sin A is negative, cos .4 is negative^ tan .4 is positive. Hence, if .4 be any angle of the second Quadrant, . , .. ,. MP . sin A, which is ^rp, \s positive) COS A, which IS jyp j is negative; , , . , . MP . tan A, which is ^yrv, is negative. IV. When OP is in the fourth Quadrant (Fig. iv.) ^fP is negative^ OM is positive, OP is positive. So that, if i4 be any angle of iYiQ fourth Quadrant, sin A is negatirr. cos i4 is positive, tan ^ is negative. L. E. T. 7 98 TRIGONOMETRY. 134 The table given below exhibits the results of the last article. Quadrant ... I. II. III. IV. Sine + + - - Cosine + - - + Tangent . . . + - + - The student should notice that in any particular Quadrant the three signs of sine, cosine, and tangent are unlike their signs in any other Quadrant. EXAMPLES. XXV. State the sign of the sine, cosine, and tangent of each of the following angles : (1) 600. (2) 1350. (3) 2650. (4) 2750. (5) -100. (6) -910. 0) - 1930. (8) - 3500. (9) - lOOQO. 10) 2nn + l. (11) 2n..'l. (12) 271 TT- 135. The NUMERICAL VALUES through which the Trigo- nometrical Ratios of the angle BOP pass, as OF moves through the first Quadrant, are repeated as OF moves through each of the other Quadrants. Thus as OP moves through the second Quadrant from U to L, Fig. II, {OP being always of the same length) MP and OM pass through the same succession of numerical values through THE ^JUNH + AND - IN TUinuNoM Lllii . 09 ich they pass, as P moves through the first Quadrant in the ite direction from U to R. Example 1. Find the sine^ cosine and tangent of 120". 12(y is an angle of the second Quadrant Let the angle ROP be 120^' (Fig. ii. p. 96.) Then the angle P0L = l8Qf^ - 120° = 60*>. MP Hence, sin I2(fi=y.-^ =8in 60" numerically^ and in the second Irant the sine is positive. Therefore sinl20o = '^| (i). in, cos 120"=-yy, = cos 60", numerically ^ and in the second it the cosine is negative. Therefore cos 120"= -\ (ii). 2 ^^ Similarly, tan 120"= -Vs (iii). ^^V Example 2. Find the sine, cosine and tangent of 225", ^H 225" is an angle in the third Quadrant ^P Let the angle ROP he 225" (Fig. iii.) Here the angle P{?Z= 225^ - 180" =45". Therefore the Trigonometrical Ratios of 225"= those of 46" numerically ; and in the third Quadrant the sine and cosine are each negative and the tangent is positive. Hence, sin 225"= - -^ ; cos 225" -- - j^; tan 225"= 1. 136, The cosecant, secant and cotangent of an angle A have the same sign as the sine, cosine, and tangent of A respectively. 7—2 100 TRIGONOMETRY. EXAMPLES. XXVI. Find the algebraical value of the sine, cosine and tangent of the following angles : (1) 1500. (2) 135». (3) -2400. (4) 330°. (5) - 450. (6) -3000. (7) 2250. (8) -135'^. (9) 3900. (10) 7500. (H) -8400.- (12) 10200. (13)2n7r + j. (14) (2;i+l)7r-^. (15) (27i-l)7r + J. ■^ Example. To trace the changes in the magnitude and sign of sin A, as A increases from QO to 3600. Take the figure and construction of page 96. As A increases from 0^ to 90'', MP increases from zero to OP, and is positive. Therefore sin A increases from to 1 and is positive. As A increases from 90° to 180", MP decreases from OP to zero, and is positive. Therefore sin A decreases from 1 to and is positive. As A increases from 180'' to 270", MP increases from zero to OP^ and is negative. Therefore sin A increases numerically from to 1 and is negative. As A increases from 270" to 3600, MP decreases from OP to zero, and is negative. Therefore sin A decreases numerically from 1 to and is negative. ^EXAMPLES. XXVII. Trace the changes in sign and magnitude as A increases from QO to 360" of (1) cos A. (2) tan A. (3) cot A. (4) sec A, <5) cosecA (6) 1-sin^. (7) sinM. (8) sin JL . cos A (9) sin A 4-cos A. (10) tan ^ + cot A. (11) sin A - cos A. CHAPTER X** On Angles Unumited in Magnitude. 1^^ 137. Just as the definitions of the Trigonometrical ^Ktios apply to angles of any magnitude whatever, so every ^■leral Formula involving these Ratios is true for angles PBany magnitude whatever. It is most important that the student should examine for himself into the truth of this statement. 138. TheformulflB cosec A = -. — -. , sin^ . 1 secil = , co&A' cot A =- . , tan J' are really definitions ; and since the definitions apply, there- fore 1 of J. fore these formulae are true, whatever be the magnitude The formulae tan^ = . , cot^ = - — , , cos-d sin^ are deduced immediately from the definitions, and therefore they are true whatever be the magnitude of A. 139. The formulae sin* -4 -f cos* .4 = 1, 1 +tanM =sec'^, 1 +cot'^4 =cosec'^, are each a trigonometrical statement of Euc. i. 47, and depend only on the fact that MP, OM^ and OP are the sides of a right-angled triangle. That this is the case, whatever be the magnitude of the angle Ay is evident from the figures on page 96. • * To be omitted on a first reading,', except pp. 104, 105. 102 TRIGONOMETRY. 14:0. In Art. 118 we proved that the sine of an angle is equal to the cosine of its complement, provided the angle lies between 0" and 90". We now give some examples of a method of proving the truth of this and other like formulae, whatever be the magnitude of the angle concerned. Kx'M/ijjle 1. To prove that the sine of an angle = the cosine of its complement. That is, to prove sin A = cos (QO*^ - A) and cos J. = sin (90*^- J). We take two revolving lines OP and OF. OP starting from OR is to describe the angle A ; OP' starting from OR is to describe (90^-^). As usual, PM^ P'M' are perpendiculars on OR and P' N' is a peipendicular on OTJ. In describing (90^ — ^) we shall consider that OP' starting from OR^ turns first through 90*^ into the position OU, and then turns back from OU through the angle UOP' = {-A). p u u L I P' 1 y / JJ_ ~~\ '^ A/' \ k J \ P 7= D p'L^ U N' \ M \ \m' D ■> So that ROP^ the angle which OP describes from OR, is always equal to UOP', the angle which 0/" describes from OU m the opposite direction. r l^" Hon ON ANGLES UNLIMITED IN MAQNITUDE. 103 Hence, N'P'y that is 0M\ is alioays equal to MP m magni- tude. Also it will be seen that when P is above LOR, F is to the of UOD\ when P is hdow LOR, P* is to the left of ^/O/). Ift: Hence, OM' and J/P have always the «awie ai^^n. Therefore ^-^ = ^^ always, 1^" Again, 0^', that is M'F, is always equal to Oi/ in magnitude. And /*' is above or below LOR according as P is to the right or to the left of C/'OZ>. ^^^KSo that J/'P and OM have always the saTTia sign. I Therefore -y5 «= ^fpT always, cos il =8in (90® - ^) for all values of A. EXAMPLES. XXVm. Prove, drawing a separate figure for each example, that (1) sin300=cos6(y>. (2) sin650=cos250. (3) sin 1950=co8(- 1050). (4) cos 275=8in(- 1850). (5) cos (-27«)= sin 1170. (6) cos 3000= sin (-2100). ' If A, B, Che the angles of a triangle, so that A + B + C= IBO", i)ve _ A . B + C ,„, B . A + C (7) cos-^=8in — ^.^ (8) cos 2= sin 2 . ,^. . C A+B „^, . A B + C (0) 8in-=co8— ~ . (10) sin-=cos— ^. 104 TRIGONOMETRY. 141. Def. Two angles are said to be the Supplements the one of the o€her when their sum is two right angles. Thus (1800 - A) is the supplement of A If A, B, C be the angles of a triangle, {A + B + C) = 180^, so that {B + C)ia the supplement of A. ExaTnple 2. To prove that the sine of an angle=the sine of its supplement; and that the cosine of an angle = -{the cosine of its supplement). That is, to prove sin J. = sin (ISO^^ - ^) and cos ^ = - cos (180^ - A). We take two revolving lines OF and OF. OP starting from OR describes the angle A ; OP' starting from OB describes the angle (1800 -J.). In describing (180^ - ^) we consider that OP' starting from OR turns first through 180^ into the position OL^ and then back from OL through the angle LOP' = {-A). So that ROP^ the angle which OP describes from ORy is always equal to LOP', the angle which OP' describes from OL in the opposite direction. ON ANGLES UNLIMITED IN MAGNITUDE. 105 Hence, MP and M'P* are always equal in magnitude. Also, P and P' are always both above, or both below LOR. So that J//* and M'P' are always of the «ame «^n. Therefore ^ = -^ always, sin ^ =8in (18(y> - ^), for all values of A . Again, OM and OM' are always equal in magnitude. Also it will be seen that when P is on the ?n^A< of UOI), F is on the left of i70i>; when P is on the left of C/'OZ), 7^ is on right of iT^OZ). So that OM and Oi/"' are always of opposite sign. Therefore OP^ ~ ~0F ^^^^^' cos ^ = - cos (180*^ - A\ for all values of A. EXAMPLES. XXIX. Prove, drawing a separate figure in each case, that (1) sin 60<'=sin 120«. (2) 8in3400=8m(- 160«). (3) sin »,- 400)= sin 2200. (4) co8 3200= -cos(- 1400) (5) cos ( - 3800) ^ _ COS 5600. (6) cos 1950 = - cos ( - 150). If A^ By Che the angles of a triangle, prove (1) 8inil=Bin(5+C;). (2) sinC=8in (^ +^). (3) cosi?=-cos(il + C7) (4) co8 4= -cos((7+5). 100 TRIQONOMETR Y. Example 3. To prove sin xi = - sin (180*' + ^;, and COS ^ = - cos (180*'+^). As before, we take two revolving lines OP and OF. OF starting from OR describes the angle A ; OP' starting from OR describes the angle (180*' + ^). In describing (180^ + ^) we consider that OP' starting from OR tm-ns first through 180° into the position OL, and then on from OL through the angle A. So that ROP, the angle which OP describes from OR^ is always equal to LOP*^ the angle which OP' describes from OL in the same direction. Hence, M'F aliuays—MP in magnitude. Also it will be seen that when P is above LOR, F is below LOR ; and vice versd*. So that MP and M'P' are always of opposite s-lgn „ MP M'F , Hence, OP " ~ 'OF ^^^^^^ or, sin ^ = - sin (180<' + ^), for all values of A. Similarly, OM always =0M' in magnitude. And F is to the left or to the right of UOD according as P is to the right or to the left of UOD, * This will be more clear if the student observes that POP' is always a straight line. ON ANGLES UNLIMITED IN MAGNITUDE. 107 OM OM' , lence, OP ^ ~ 'OF ^^*^''' cos A= - cos (180® + -4), for all values of A. Example 4. To prove aiii ^ = - cos (90*' + .1), co8^=8in(900 + ^). As before, we take two revolving lines OP and OF. OP ting from OR describes the angle A ; OF starting from OR bribes the angle 90^ + A. In describing (90<>+^) we consider that OF stari;ing from OR Tims first through 90** into the position OU^ and then on from U through the angle A. U So that ROP, the angle which OP describes from OR, is always equal to UOF, the angle which OF describes from OU in the same direction. Hence, N'F, that is 0M\ always = MP in magnitude. Also, P' is to the left or to the right of UOD according as P s above or below LOR. So that MP and OM' are always of opposite sign. MP OM' , Hence, OP ^ ~ ^F ^^^^^^ or, sin i4 = - cos (90*^ + ^), for all values of A . 108 TRIOONOMETR 7. Similarly, OM always=M'P' in magnitude. And P is above or below LOR according as P is to the right or to the left of UOD. So that OM and M'F are always of the saTnie sign. OM M'F - Plence, ^ = -^ always, or, cos ^ =sin (90° + ^), for all values of A. MISCELLANEOUS EXAMPLES. XXX. Prove, drawing a separate figure in each case, that (I) sin 600= _ sin 240°. (2) sin 170°= - sin 3500. (3) sin ( - 200)= - sin leO^. (4) cos 3800= - cos 5600. (5) cos ( - 225) = - cos ( - 450). (6) cos 10050= - cos 11850 (7) sin 600= _ cos 1500. (8) cos 600=sin 1500 (9) sin225= -cos3150. (10) cos ( - 600) = sin 300. If ^ + 5 + C be the angles of a triangle, prove that (II) sin ^=- sin (2^ +5 + (7). /ION • A ZA + B^G (12) sm A= - cos ^ (13) cos5=sin ^r (14) cos (7= -cos (^ + 5 + 20- ._, B-G . ^ + 25 ( 15) cos -g— = sm - 2 — ,_, . G-A B + W (16) sm -g— = - cos — g- I I A^ul.n^ viSLlMlTED IN MAGNITUDE. 109 e the following statements for all values of A and of a. (17) 8m^=-8in(-^). (18) cos^=cos(-^). (19) 8mi4=cos(^-90<»). (20) cos /i = - sin (^ - 900). (21) 8ina = cos f-^ + a J . (22) cos a = - sin ( — + a j (23) sin a = - cos i-^ " " ) • (24) C08a= - sin( -^ - a j . (25) sin (I - « ) "=sin (l "'" ") * (26) COs(7r + a) = cos(7r-a). (27) tan (900 -^) = cot ^. (28) tan^= - tan ( - ^). (29) tan(900 + ^)=-cot(^). (30) tana= -tan(7r-a). (31) tan ^ = tan (1800 + ^). (32) cot (^| - a^=tan a. 142. We have seen (Art. 135) that there are many angles of different magnitude which have the same sine. If two such angles are in the same Qudd/rard they are represented geometrically by the same position of OP, so that they differ by some multiple of four right angles. 143. If we are given the value of the sine of an angle it is important to be able to find, Geometrically and Alge- braically, all angles which have that value for their sine. 110 TRIGONOMETRY. 144. To find tlie complete Geometrical Solution of the equation sin 6 = a. With the usual construction, let the radius of the circle RTJLD be the unit of length ; then the measure of OF is 1. From draw on Of/ a line ON so that its measure is a. \0N will be drawn upwards (Fig. i.) or downwards (Fig. II,) from according as a is positive or negative.] Through N draw P^P^ parallel to LOR to cut the circle in Pi and P/. Join OP,, OF^, Draw P^Jf,, FJd^ per- pendicular to LOR, Then M^F^ = 0N= 2^^ for OP ^^^' "" ^"^" and hyp. OF, 1 I a 1 M,F,, 0F„ ^ for OP,, hyp. Hence every angle whose initial line is OP, and whose final line is either OPj or OF^, is an angle whose sine is a. And no other angle has its sine equal to a, for there is uo other possible position for N. 145. To find an expression to include all angles^ having the given value Chfor sine. With the usual construction, let the measure of the radius of the circle RTJLD be 1 ; onOU take N so that the measure of ON is a ; through N draw F^NF^ parallel to * If a were greater than unity, ON would be greater than OU, and the construction would fail. Uiy yli>L ri.r.o 1> J. iiuAsr.u IN MAGNITUDE. Ill IL ; join OP^^ OP^. Then every angle whose initial line )Ii and final line, either OP,, or OP,, and no othevy is an i^le whose sine is o. Let KOP^ contain a radians ; then ROP^ -IT -a. ^1\ a Flg.o.y"" TT-a R angle whose initial line is OR and final line OP, 8 of those included in the expression 2m7r + a, ^where m is some integer, positive or negative. [Art. 129.] ^^^P Every angle whose initial line is OR and final line OP^ ' ^ one of those included in the expression ^^ IVTT + TT - a, or (2r + 1) TT - a, ^^pere r is some integer, positive or negative*. [Art. 129.] ^^ Both of these expressions are included in n7r + (-l)"a, where n is some integer, positive or negative*. This is therefore the required expression. Example. Find six angles between —4 right angles and + 8 right angles which satisfy the equation dnA^=8in 18®. We have r^=7i7r + ( - 1)* t^, or] A^ = n x 180« + ( - 1)" W. Put for n the values - 2, - 1 , 0, 1, 2, 3, 4 8ucce.ssively and we get the six angles -3600 + 18«, -1800-180, 180, 18OO-I8O, 3600-f 18®, 6400-180, i.e. -3420, _ 1930^ 150^ 1620, 3780, 5220. • For if n be even, let it be 2in, when (- 1)^= +1; if n be odd^ let it be 2r+l, when ( - l)^^^ \. 112 TRIGONOMETRY. The student is recommended to draw a figure in the above example. Also to draw a figure in each example of this kind which he works for exercise. EXAMPLES. XXXI. (1) Find the four smallest angles which satisfy the equations 1 /3 (i) sin ^ = ^. (ii) sin J. = -^ . (iii) sin J. = ^ . (iv) sin ^ = - ^. (2) Find four angles between zero and +8 right angles which satisfy the equations 1 (i) sin ^ = sin 20^ (ii) sin^= - V2- (iii) sin^= -sin (3) Find the complete algebraical solution of (i) sin ^ = - ^. (ii) 2 sin2 ^ + 3 sin ^ = 2. (iii) sin^ e = cos^ 6, (4) Prove that 30^, 150^, -3300, 390°, -210^ have all the same sine. 146. To find the complete Geometrical Solution of the equation COS 6 = d. With the usual construction, let the radius of the circle RTJLB be the unit of length, so that the measure of OP is 1. From draw on OR a line OM^ such that its measure is a. OM^ will be drawn towards the right or towards the left according as a is positive or negative. iroi ?he] IS a. Let BOl Every angle whosi. one of those included i Every angle whose initial lin >ne of those included in the exp. 2 'TT-a. Both of these expressicns are includ* Thus tlie sohition of Uia equation cos d 2n7r ± a. OM^, M^ a line J/,P^ /gPj whose measure = v=«, a I whose initial line is OR^ and final g, is an angle whose tangent is a. .ngle has its tangent = a. id the complete Algebraical Solution of the J = a. i, be the initial line; from draw on OR two .1, OM^ wliose measures are + 1 and — 1 respec- ^ from M^ , J/g draw perpendicular to LOR lines . M^^ whose measures are a and - a respectively; Join OPg. Then every angle whose initial line is OR and J line either OP^, or OP,, and no other ^ is an lingle nose tangent is a. r ON ANGLES UNLIMITED IN MAGNITUDE. 1 1 Let ROP, = a ; then, EOP^ = tt + a. Every angle whose initial line is OP and final line OP^ me of those included in the expression 2m7r + a. [Art. 129.] Every angle whose initial line is OP and final line OP^^ ... one of those included in the expression ^fc 2r7r + (77 + a); or, (2r + 1) tt + a. [Art. 129.] ■ Both of these expressions are included in 7177 + a*. ^^xnm Thus the solutiou of the equation tan = tan a is 0=7111 + a. EXAMPLES. XXXII. (1) Write down the complete Algebraical Solution of each of the following equations : (i) cos^=i. (ii) tan^=l. (iii) tan^=-l. (iv) tan^= - v/3. (v) cos^=cos— . (vi) tan^=tan — . 4 (2) Show that each of the following angles has the same cosine : -120'', 2400, 4800, _4800. (3) The angles 60^ and - 120^ have one of the Trigono- metrical Ratios the same for both ; which of the ratios is it ? (4) Can the following angles have any one of their Trigono- metrical Ratios the same for all ? - 230, - 1570 and 1570. (5) Find four angles which satisfy each of the equations in (1). * For if n be even, this is the first formula ; if n be odd it is the seoond. 8—2 1 1 6 TRIQONOMETR Y. 150. We can now point out the use of the ambiguous sign ± in the formula cos ^ = ± ^1 - sin^ 6. If we know the numerical value of the sine of an angle 6, without knowing the magnitude of the angle, we cannot from the identity, cos^ 0-=l - sin^ 6, completely de- termine cos 9y for we get cos^ = ±,^1 - sin^^. This is a general formula, and we shall find that it represents an important Geometrical truth. 151. Given sin^ = a, we can say that 6 is one of the angles represented by one or the other of the positions OP,, OP^ of the revolving line in Fig. I. on page 110. If we attempt to find the cosine of these angles we get two different values for the cosine ; for ^pp^ ^^^ TJiT although equal in magnitude, are opposite in sign. Hence, if a be the least angle whose sine is equal to a, we have cos^ = ± cos a = ^Jl- sin^ a. 152. The same result may be obtained from the formula = mr + ( — l)"a. For cos {mr + ( - l)"a} is of different sign according as n is even or odd. EXAMPLES. XXXIII. (1) If ^ be found from the equation cos^::^a, show geo- metrically that there are two values of sin 6 and of tan 6. (2) If 6 be found from the equation tan^=a, show geo- metrically that there are two values of sin 6 and of cos 6. (3) If A be the least angle without regard to sign such that sin J.=a, show that cos J.= 4-Vl -sin^iL. (4) If A be the least positive angle such that cos A = a, prove that sin ^ = -1- */l - cos^ A. ( 117 ) CHAPTER XI. THE Trigonometkical Ratios of Two Angles. 153. foi-mulM I We proceed to establish the following fundamental sin (A-\- B) = sin A . cos B + cos -4 . sin ^ COS (A +B) = cos A . cos B — sin A . sinB sin (il - ^) = sin ^ . cos ^ - cos ^ . sin ^ cos (A-B) = cos il . cos -ff + sin il . sin J5 ^ Here, A and B are angles; so that (A + B) and {A -B) are also angles. Hence, sin (^ + B) is the sine of an angle, and must not l)e confounded with sin X. + sin B. Sin (A ■\-B) \& 2t. single fraction. Sin A + sin B is the sum of two fractions. 154. The proofs given in the next two pages are per- fectly general, as will be explained below (cf. Art. 169) ; but the figures are drawn for the simplest case in each. The student should notice that the words of the two proofs are very nearly the same. 118 TRIGONOMETRY. To prove that sin {A + B) = smA. cos B + co^A. sin B, aitd that cos {A + B) = co& A .co&B-BmA . sin B. Let ROE be the angle A^ and EOF the angle 5. Then in the figure, ROFis the angle (A+B). In Oi^, ni^ ' OM OK+KM B)=cosROF==^=—^j^ = ^r.+ OP OP iTjy.NP ON. OP^ NP. OP ~ ON' OP ^ NP'7>P = cos ROE. cos PO^+sin HPN. sin i^O^" = co8 .4 . cos J9+sin ^ . sin i?. • Or thus. A circle goes round OMPN, because the angles OMP and ONP are right angles ; therefore MPN and MON together make up two right angles ; so that HPN = MON = A. 120 TRIGONOMETRY. Example. Find the value of sin 75°. sin750=sin(450 + 300) =sin 450. cos 30° + cos 450. sin 30® _ V3 + l ^ N^2(v/3 + l ) 2^2 4 EXAMPLES. XXXIV. (1) Show that cos 750= 2^2 (2) Show that sin 150= —^^ • \/3- l 2^2 (3) Show that cos 150= -^-^ . (4) Show that tan 750 = 2 + ^3. (5) If sin ^ =1 and sin B=% find a value for sin {A + B) and for cos {A - B). (6) If sin J. = -6 and sin.B=j^^, find a value for sin (^-5) and for cos {A + B). --' (7) If sin^=-T- and sin5=-yY^, show that one value of {A-^B)mAb\ (8) Prove that sin 750= -9659... - (9) Prove that sin 150= -2588... (10) Prove that tan 150=-2679... 155. It is important that the student should become thoroughly familiar with the formulae proved on the last two pages, and that he should be able to work examples involving their use. EXAMPLES. 121 * EXAMPLES. XXXV. Prove the following statements. (1) 8m(^ + 5) + sm(^-5) = 2sm^.cosZ;. (2) 8in(^ + 5)-cam(^-i?)=2cos^.sm^. 1(3) C08(^ + 5)+C0S(^-5) = 2C08^.C08^. f(4) cos(^-jS)-cos(^ + 5)=^2sin^.sin^. L. sin(^ + i?) + siD(^ -^) ^ *^ ^ cos(^ + ^)+cos(^-J)~^'*- [(6) tana+tan/3=i^^L(£m. ! cos a * cos /3 i(7) tana-tan|3.= °'°<°-^t cos a . COS /3 (8) cota + tan^=^^2l(^) sin a . cosp (9) cota-taa0=-?i(i±« . Sin a, cos /3 (10) tana + cot^=^^"^°-^) cos a . sm ^ . tan ^ + tan _ sin {6 + (^) ^ ' tan ^-tan^ ~ Bin"(^ - 0) ' (12) tan 6 . tan + 1 _ cos (^ - 0) 1 - tan ^ . tan ~ cos (^ + 0) * , . tand + cot<^ ,. ,. ,a , ^\ \J (14) (15) cot dj+cot 0_ _ sin {6 + <^) cot (J-coT^" sin (d- 0) * tan ^ . cot + 1 _ sin (^ + 0) tan ^ . cot - 1 ~ sin {6 - 0) , 1+coty.tand . , .. (16) r — -T — r- =tan (y + 3), ^ ' cot y- tan d ^' ' . . 1-coty.tand , .. (17) — .— -i— * =tau(y-d). coty + tan^ ^' ' 122 TRIGONOMETRY. <-_. tany.cotd-l ^ , '''' Jy + cot8 =tan(y-8). /._. tany.cotS + l . , ^, ('') coU-tan, -tai.(y + a). (20) cot«-coty ^ ' COty.COtS + 1 ^'^ ' (21) ten'a-taa^^="'°(°+f-^'V°-^^ COS^a.COS^^ (24) sin (a + ^) . sin (a - ^) = sin^ a - sin^ /3 = cos^ /3 - cos^ a. (25) cos (a + /3) . cos (a - jS) = cos^ a - sin^ ^ = cos^/S - sin^ a. /o£?\ • /^ ^cn\ sin J. -cos J -^ (26) sin(^-45*')= j- . /- ^ (27) V 2 . sin {A + 45°) = sin J^ + cos A . (28) cos^-sin^=V2.cos(^ + 450). (29) cos {A + 450) + gin (^ _ 450) = 0. (30) cos {A - 450) = sin {A + 450). (31) sin {d + ). cos + cos {d-S' - sin T= 2 sin — ;r- . cos Q.E.D. 2 • "■' 2 In the above figure the angles VKQ, ROK are each the com- S+T 2 ' meutofZ^O .-. VKQ = ROK=- Hence sin S - sin T= -^rn - 7^ = MP NQ MP-NQ 2VK OQ ' OP 0(^ OQ ' (LK+HP)-{LK- WH) OQ ?^|^ = 2 cos VKQ . sin ^OA', - . S-T S+T ■ 2 sin — ;r— . cos — X— . (4) To prove 2 Q.E.D. *S+7' S-T cos 7"- COSTS' =2 sin —jr— . sin — . Z 2 ; With the same figure and construction, we have ^ ^ ON OM ON-OM cosr-cos^=-^--^=— ^^-, (OZ + XiVQ - {PL - ML) 1LN OQ ~ OQ ' - . 5+7^ . S-T -28m -^ .am-g-. Q.E.D. 1 30 TRIQONOMETR Y. EXAMPLES. XXX Vn. Prove the following statements : (1) sin 600 + sin 300= 2 sin 45f» . cos I50. (2) sill b'QO + sin 20° = 2 sin 40<> . cos 20^ (3) sin 400 - sin 100=2 cos 250. sin I50 (4) cos- + cos- = 2cos — .cos — • (5) coSg-cos - = 2sin— .sin — . (6) sin 3^ + sin 5^ = 2 sin 4^ . cos A . (7) sin 7 A - sin 5^ = 2 cos 6 A . sin ^. (8) cos 5 A + cos 9 A = 2 cos V^ . cos 2 A. 9A A (9) cos 5 A - cos 4A= -'2 sin -^ . sin ^. (10) cos J. - cos 2^ = 2 sin '-^ . sin — . . sin2^+sin^_ 3^ ^^^^ cos^ + cos2'^~^^''2"- „„, sin 2^- sin ^ ^3^ 5^1. cos^-cos2r ^"^2- sin 3^ + sin 2^ _ 6 ^^'^^ cos 2^ -cos 3'^-'^''^ 2- (14) sin^+sin0 cos ^ + cos cos ^- cos sin<^-sin^" (1 5) cos (600 + ^) + cos (600 _ ^) = cos .1. (16) cos (450 + A) + cos (450 -A)=s/^.cobA. (17) sin (450 + A)- sin (450 - ^) =^/2 . sin J. (18) cos(300-^)-cos(300 + ^)=sin^. ,,-. sin ^- sine/) ,0 + 4> (19) ^ r = cot — TT^ . cos^-cos^ 2 TRIGONOMETRICAL RATIOS OP TWO ANGLES. 131 161 . It is important that the student should be thoroughly liliar with the second set of formulae on p. 126. Written as follows, they may be regarded as the invei-se the 'S, T' formulae. 2sm A .cosB = mi{A + B) + sin (^4 - B)y ' 2coaA .sin B = sin (A ■{■ B) - sin (A - Ji), 2 cos -4 . cos i5 = cos (^ + -6) + cos (A - B), 2 sin ^ . sin B = cos {A -B) - cos (A + B).. IV. EXAMPLES. XXXVm. Express as the simi or as the difference of two trigonometiical ios the ten following expressions : (1) 2 sin ^. cos <^. (3) 2 sin 2a . cos 3fi. (5) 2 sin S3 . cos 56. (2) 2 cos a . cos /3. (4) 2cos(a + /3).cos(a-/3). (6) 2 cos ^ . cos . . (7) 8in4d.sin^. (8) cos^.sin^. (9) 2co8l00.9in5()". (10) co8 450.sin IC^. (1 1) Simplify 2 cos 2^ . cos ^ - 2 sin \0. sin 6. (12) Simplify siu -^ . cos ^ - sm — . cos -^ 3^ (13) Simplify sin 3^ + sin 2^ + 2 sin — . cos 2 3^ (14) Prove that sin . sin ^ + h'^' ^^" - =Hin 2^. sin 6. 9—2 1 32 TRIG ONOMETR Y. * ^ MISCELLANEOUS EXAMPLES. XXXIX. / (1) If tai} o=^ and tan ^ = ^, prove that tan (a + /3) = 1. (2) If tana=f and tan/3=^, prove that one of the values of a+/3 is TT (3) If tana=— — and tan^ = _ — -— , shew that one value of(a + ^)is|, -v.^ OLz-W^f (4) Simplify ^^-^^?^J^^ ^- ^ ^ ^ -^ sina + sin5a /r\ ci- vi. sin 5^ -sin 3a; (5) Simplify — . ^ ''' cos5a7 + cos3.r ,^, T-, , , , cos A + cos 3 J. cos 2.4 (6) Prove that — --.— — . = — r . ^ ' cos3J. + cos5J. cos 4^ /«N o- i-i? sin 3^- sin .r sin3^+sina; (7) Simplify — ^ — ; H ^ . ^ ' -^ cos 3.r+cos 0? cos Zx - cos a? „ . , . „ (sin 4.4 — sin 2A) (cos J. - cos 3^) (8) feimpiity ^^^ ^^ _^ ^^ 2^y(sin A + sm'sll) * (9) Prove that 2 sin 2a . cos a + 2 cos 4a . sin a=sin 5a + sin a. (10) Prove that ^ cos 2a . COS a - sin 4a . sin a = cos 3a . cos 2a. {t\) tan 2 J. . tan ZA . tan bA = tan 5 J. - tan 3^1 - tan 2 J . (12) Solve 4 sin (^ + 0). cos (^ -<#)) = 3] 4 cos (^ + 0) . sin (^ - 0) = Ij • (13) Prove that sin A . sin 2^ + sin 2 A . sin 5^ + sin ZA . sin 10 J. cos J. . sin 2^ + sin 2^ . cos bA - cos 3 J. . sin 10^ ,, ,, ^ ^ + ^ , ^-i^ 2sinJ5 (14) tan — ^ tan — ^_— = j— rj . ^ ' 2 2 cos J + cos i? = - tan lA. CHAPTER XII. On the Trigonometrical Ratios of Multiple Angles. l62. To express the Tiigononietrical Katies of tlie 2 A in terms of those of the angle A. lince sin (A + B) = sin A . cos B + cos A . sin i5 ; .*. sin {A + A) = sin A .COB A + cos A . sin A ; .'. sin 2^1 = 2 sin ^ . cos ^ (1). since cos {A + ^) = cos A . cos B - sin A . sin ^ ; .*. cos {A + A) = cos A . cos A — sin A . sin ^ ; . •. cos 2A = cos' A - sin- A (2). But 1, = cos' A + sin' A ; • .-. 1 + COS 2i4 = 2 cos"^, and 1 - cos 2^1 = 2 sin" A. ^^ The last two results are usually writton cos 2vl = 2 cos" /I - 1 (3), ■ ami cos 2i4 = 1-2 sin' yl (4), Again, tan (^^1 + Z>) - .-. tan(yl + ^) = .-. tan 2/! = tan A + tan B 1 - tan yl . tan ^ ' tan A + tan A 1 - tan A . tan A ' 2tany l 1 - tanM (5). 134 TRIGONOMETRY. 163.* ■'^ 7^0 pi'ove the '2A' formulae geometrically. Let ROP be the angle 2^. With centre and any radius describe the semicircle RVL. Draw PM perpendicular to OR. Join RP . PL. Then the angle RPL in a semicircle is a right angle. The angle R0P = 0LP-\-0PL = 2 0LP [since OL = OP]. .: OLP= a half of ROP=A. Also i/P^ and OZP are each the comple- ment of if PZ. .'. MPR = OLP = A, Hence . „, MP 2MP ^MP 2MP.FB, (1) ^^^^^=op^2or-^;r-tw7lr =2 cos J/Pi2 . sin PZi2 = 2 sin A . cos A. „, OJ/ LM-LO 2LM LO ^(3) cos2^ = ^ = -^^^-= ^op-OP ^TT^R ■ ^ = ^ cosi/ZP . cos PLR - 1 = 2cos2^-l. (2) hei DM' = DM. Then WM=M'M=LM-LM' = LM-MR. Hence, _2qM_LM-^R_LM MR cos2A-^^^- ^^ -LR~LR' Zif.LP JfP.PR _ . ,, = cos'^J. -sm^>4. (5) tan2il = LP. Zi2 PR.Zi^ 2JfP 2JfP 20J/ LM-MR 2MP LM 2tan^ 2tan^ LM MR ^ M)R.M'P ~ 1 - tan2 J LM LM MP.LM fOONOMETRICAL RATIOS OF MULTIPLE ANGLES. 135 164. These five formulae are very inmortant. sin 24 = 2 sin -4 . cos .i . . cos 2 A = cos' A - sin^ xi cos 2A = 2 cos" A - 1 cos2^ = l- 2 sin" J 2tan^ tan 2^ l-tan«^ (2), (3), (5), 65. The following result is important, sin2il ^^inii.cos.4 1 +008 2^1 % cos" A ~ tani4. 166. The student must notice that A is any angle, and jfore these formulae will be true whatever we put for A . Example. Write — instead of A^ and we get sm 4 = 2 sm — . cos ^ . cos A =cos2 -^ - sin* — \\ so on. (1), (2), ^ EXAMPLES. XL. Prove the following statements (1) 2co8ec2il=seCi4.cosec J. coaec'^ (2) (3) =8ec2^4. - = C08 2J. C08ec*-4-2 2-860*^ ~s©?T (4) co9M(l-tan«J)=cos2J cotM-1 (5) cot 2.4 2coti4 136 TRIGONOMETRY, (7) tan ^ + cot J5 = 2 cosec ^B. (9) cot ^ - tan ^= 2 cot 2^5. ,,^. cot2^ + l (^ /9\2 sin- + cos2J =l+sin(9. (12) fsin^-cos-j =l-.sin^. (13) cos2|(^l + tan|J=l + sinA (14) sm2|(^cot|-lJ=l-sm(y, 1 + sin ^ 1 - sin ^ . . sin^ /3 -^^'^^ iT^<^^=*^^-2- (17) -^^=eotf. ' l-cos/S 2 (20> cosec i3 - cot /3 = tan ^ . 2 (21) ^Q^2 a? ^ 1 - tan ;g l + sin2^ l + tano?' '^ 1 + tan zi (22) ^^'^ 2 l-sin4? ^ , X 1 - tan - EXAMPLES. XL, X cot-^-l 137 1 + am 4? . -P . 1 cot - + 1 cot^ + 1 H(25) (26) (27) 1 - sin 0? ,£ . cotg-l l+sinx + cosJT .X , =cot 5 . I + sm a: - COS J? 2 cos^g + sin^g _ 2 -si n 2g coeg + sina ~ 2 cos' a - sin^ a 2 + sin 2u COB a - sin a (28) cos* g - sin* a = cos 2g. 1(31) (32) (33) -(34) (29) oos«g + sin«g = 1+3 COB* 2g . , (3 + coa2 2o)co8 2g (30) cos" g - sin« a = ^ j-^ sin 33 cos 3/3 sin 3 cos /3 COB 3/3 sin 3/3 ^ ^ „^ ^^ + r = 2 cot 2/3. 8in/3 cos/3 sin 43 = 2 cos 2/3. I (35) sin 2/3 sin 5/3 cos 6/3 sin/3 cos/3 ' Stt 5rr 4 cos 2/3. 2^3. ""^12 ^12 (36) tan (450 + ^) - tan (45® - .1) = 2 tan 2il. (37) tan (450 - ^ j + cot (450 - 4) - 2 sec 2^4. 138 TRIGONOMETRY. ^^^^ tan2(450 + ^)--"'^^^- = tan Ub^ + ^) . cot (^450 - ^ (39) (40) sec A + tan ^ sec A - tan ^ cos (A + 45*^) o ^ X o .1 )- — ;7^; =sec 2 J. - tan 2 A. cos (^ - 45*^) ,A^^ J. D sin5 + sin25 (41) tan^=, 5- prp. ^ ' l + cos^ + cos25 /.^N ^ 71 sin 25- sin J5 (42) tau5=- ^ ^Td- ^ ' 1 - cos 5 + cos 25 167. The following two formulae should be remem- bered : sin ZA = 3 sin yl - 4 sin' Ay ,. cos 3J =4 cosM -3 cos^ ) ^ '' Note. The similarity of these two results is apt to cause confusion. This may be avoided by observing that the second formula must be true when ^^ = 0*^; and then cos 3^ = cos 0^ = 1. In which case the formula gives cos 0^ = 4 cos 0^-3 cos 0®, or 1=4-3, which is true. The first formula may be proved thus : sin 3^ =sin {^A + ^)t^sin 2^ . cos ^ + cos '2 A . sin A = (2 sin JL . cos A) cos J. + (1 - 2 sin^ A) sin A = 2 sin A . cos^ ^ + sin ^ - 2 sin^ A = 2 sin ^ (1 - sin^ jl) + sin ^ - 2 sin^ A = 2 sin J. - 2 sin^ ^ +sin ^ - 2 sin^ A = 3sin^-4sin' J. The second formula may be proved in a similar manner. w OONOMETRICAL RATIOS OF MULTIPLE ANGLES. 139 3tan^-tanM Example. Prove that tan 3^4 = tan3il = tan(2^4-^): l-3taii«^ taD 22t-t-taD^ l-t&n2Ai8LnA 2tan^ l-tanM + tan^ 1 - , — . ---/ X tan A 1-tan'^ 2tan^4-tan^(l-taD^ii) l-tan2^-2tanM 3tanJ_-tanM ° l-3taD2^ • EXAMPLES. XLI. I Prove the following statements (1) -^ — ;. =2 cos 2^ + 1. i co83^^2 cos 2^1-1. COSil 3 sin ^ - s in 3ii 3 cos3^ + 3"w^~^" '^- ,_. sin 3 J - sin ^ ^ ^^) cos3^-Kco8.r ^^- , . sm 3^ - cos 3 .1 „ . o^ 1 W „• J , r- = 2 sm 2i4 - 1. sin^+cos^ .». 8in3^+coa3^ « • « / (0 ^^^^3S^-=2sm2^ + l. ^®^ tan 3^ - tan J + cot ^- cot 3.4 °^^^^ (9) / 3 sin A - sin 3^1 y ^ /sec 2^ - IV- \3co8i4+co8 3^/ \sec2^ + iy * (10) ^r^^_(,+2co8^)« 1 - eo8 .1 ^ ' 140 TJRIGONOMETRY. ^* MISCELLANEOUS EXAMPLES. XLIL Prove the following statements : v,v sin J. + cos J. , ^, (1) , — ^ ~. =tan2^+sec2^. oosA- sm A tan — + 1 (2) . =tan^ +sec J. 1 - tan — (3) sin (n + l)a.cos(n- l)a-sin2a = sin(n- l)o.cos(w+ l)a. ,^, sina + sin/3 , a + i3 (4) ^ = tan — ^ . ^ ' cos a + cos/3 2 . . cos 2a + cos 12a cos 7a - COS 3a o ^^^^ ^a _ ^ ' COS 6a + COS 8a cos a - COS 3a sin 2a ~ (6) If J. = 18^, prove that sin 2^4 = cos ZA ; hence prove that V, . sin a + sin j3 + sin (a + )3) _ . a ■ i3 ^^^ sina + sin3-sin(a + ^)~*'''^2-''''^2- (8) sin 2^ . sin 25 = sin^ (^ + 5) - sin2 (^ - B). (9) cos 4^ = 8 cos* ^ - 8 cos^ ^ + 1. (10) tan 500 + cot 50^ = 2 sec lO^. (11) sin 3^ = 4 sin ^. sin (600 + ^) sin (60*^-^). (12) ^cot^-tan^") (cot^ -2cot2yl) = 4cot ^. .,„, cos3a-sin/3.sin5a-cos7a . . , j x p o (13) -z—z, . 'X — ~ —^ IS independent of /3. ,^ '^ sin3a + sini3.cos5a-sin7a (1 4) (cos X + cos yf + (sin :r + sin yf — 4 cos^ — -^ . (1 5) 2 cos2 ^ . cos2 5 + 2 sin2 ^ . sin2 5= 1 + cos 2vl . cos 25. (16) cot I -tan ^ = 2. O O 4 tan ^ (1 - tan^ Q) l-etan^^ + tan^^* ^17) tan 4(9: (22) (23) MISCELLANEOUS EXAMPLES. XLII. 2coSg=V2 + V2. 2 cos IP 15'=V2 + V2 4-V2. sin J . sin 2 J + sin A . sin 4^ + sin lA . sin lA sin ^ . cos 2.4 + sin 2vl . cos 5^4 + sin u4 . cos 8^ 8in^ + 8in(-^ + 0) + sin(^ + 20) _ cos ^ + cos (^ + <^) + cos (^ + 2<^) ~ '^'^ "^ ^^* 2 cos8 i4 - 2 8in8 J = cos 2 J (1 + cos^ 2 J). (3 sin il - 4 sin^ ^)2 + (4 cos' ^ - 3 cos ^)2 = 1. sin 2a . cos a 141 'tan 5^. (1 + cos 2a) (1 + cos a) tan 2* _ cot(n-2)a.cotna + l 2 — :^ ^- : =C0ta tana. cot ^^71 - 2) a - cot na If tana = | and tan/3=i2^, prove tan(2a + /3) = J. A A Prove that tan -^ and cot - are the roots of the equation [28) If tan -ff = - , prove that 2x. cosec^ + 1=0. a + b U-h _ 2 cos 5 Vcos 2/? 168. The following examples are symmetrical, and each )lve more than two angles : Example 1. Prove that I (a + /3 + -/) = sin a . cos /:i . cos y +sin /3 . cos y . cos a 1+ sin y . cos a . cos /3 - sin a . sin Q . sin y. a + jy + y) = 8in (a + i3) . cos y + COS (a + ^) sin y =8in a . cos ^. COS y + COS a.sin ^ . cosy + COS a . COS j3 . sin y - sin a . sin /3 . sin y =sin a . cos ^ . cos y + sin ^ . cos y . cos a + sin y . cos a . COS /3 - sin a . sin ^ . .sin y. Q.E.D. 142 TRIGONOMETRY. Example 2. Prove that sill a + sin /3 + sin y - sin (a + /3 + y) = 4 . sm^^ . sin -^ . sin -g- . Now sina-sm (a +3+y)= - 2cos. 1- — ^ . sin — ^ . And sin /3 + sin y = 2 sin ~-^ . cos —^ , [Art. 1 58] .-. sin a f sin /3 + sin y - sin (a + /S +y) „ . /3 + y /ii-y „ 2a + ^ + y . /3 + y = 2 sm ^^^ . cos'^^ - 2 cos ^^ — ' . sm "^y^ „ . iS + yf ^-y 2a+^ + y\ = 2 sin '^-^jcos ^ - cos — g— | =. 2 sin ^^ . 2 sin ^-p . sin " J^ [Art. 158] , . ^ + y . y + a . a + B = 4 Sin ^— ^ . sm '-^— . sm - . q.e.d. A ^ £t * EXAMPLES. XLin. Prove the following statements : (1) cos(a+^+y) = cosa . cos^. cosy-cosa. sin/3, sin y - cos /3 . sin y . sin a - cos y . sin a . sin ^. (2) sin(a + ^-y) = sina.cosi3.cosy + sin/3. cosy.cosa -siny ,cosa. cosjS + sinasin^siny. (3) cos (a - /3 + y) = cos a . cos /3 . cos y + cos a . sin jS . sin y - cos/3 . sin a . sin y+cos y . sin ^ . sin a. (4) sina + sin/3-siny-sin(a + /3-y) , . a-y . /3-y . a + /3 = 4sin — ^ . sin ^-2"- . sm -Y" • (5) sin (a - ^ - y) - sin a + sin i3 + sin y . . a-/3 . a-y . ^ + y = 4 sm — ^. Sin — ^ . sm -— ^ . (6) sin 2a + sin 2/3 + sin 2y - sin 2 (a + ^3 + y ) =4 sin (/3 + y) . sin (y + a). sin (a + ^). EX A MPLE6. XLIII. "03 - y) + sin (y - a) + sin (a - /9) ^ ^ ^ 11:3 ^H (8) 8in09 + y-a)+8in(7 + a-/3) + sin(a+/3-y) ^^P -sin (a + /3 + y) = 4sina.8in/3.siny. (9) sin (n + ^ + y)+8in(/3 + y-a)+8in(y + a-/3) - sin (a + /3 - y) =4 cos a . cos /3 . sin y. (10) cosa: + cosy + cos« + cos(^+y + 2) = 4COS'^-^r- .cos— 5- .cos— ^- . 2i Z £, (11) cos2a: + cos2y + cos22 + cos2(^+y + «) =4cos(y + «).co8(«+a:). cas(j;+y). (12) cos(y + «-x) + cos(z + j:-y) + co8(a:+y-2) + cos (a: + y + 2) = 4 cos X . cos y . cos 2. (13) cos* X + cos^y + cos* z + cos^ (or + y + 2) = 2 {1 + cos (y + 2) . cos (2 + ar) . cos {x + ^)}. (14) sin* X + sin* y + sin* z + sin* (x + y + 2) = 2 {1 - cos (y + 2) . cos (2 + a:) . cos (^ + y)}. (15) co8*4?+cos'y + co8*2 + cos*(;r+y-2) =2 {1 + cos {x-z). cos (y - 2) . cos (j? +y)}. (16) cos a . sill ()3 - y) + cos /3 . sin (y - a) + cos y . sin (a - /3) = 0. (17) sin a . sin (/3 - y) + sin j3 . sin (y - a) + sin y . sin (a - /3) = 0. (18) co8(a + ^).cos(a-/3) + sin03 + y)sin03-y) - cos (a + y) . cos (a - y) = 0. (19) cos (d - (i) . sin (/3 - y) + cos (3 - /3) . sin (y - a) - cos (^ - y) . sin (/3 - a) = 0. (20) 8 cos J — ^.cos^- — J — .cos^ — - — ^.cos — — * 2 ^ ^ 2 «: cos 2d + cos 2 /> + cos 2x + 4 cos ^ . cos <^ . cos X + 1 . ( 144 ) CHAPTER XIIL** On Angles Unlimited in Magnitude. II. 169. The words of the proofs (on pages 118, 119) of the '-4, -6' formulae apply to angles of any magnituda The figures will be different for angles of different mag- nitude. 170. The figure for the 'A-B' formulae on page 119 is the same for all cases in which A and B are each less than 90°. The figure given below is for the proof of the ^ A + B^ formulae, when, A and B being each less than 90°, their sum is greater than 90". The words of the proof are precisely those of page 118. We may notice however that cos (A + B) = OM OP -MO' OP MK+OK OK MK OP ~ OP OP and the rest follows as on page 118. I ON Ai\uLh.^ i\\ LIMITED IN MAGNITUDE. 1 i5 171. Thus we have proved that the *J, B' formulfie i e true provided A and B each lie between 0" and 90°. The student can prove them for any other values by drawing the proper figure. The 'Ay B' formulae are therefore true for any values whatever of the angles A and B. 172. By the aid of the 'A,B* formulae we can prove the formulae of Art. 140. Example. Prove thai sin (90'' + A) = cosA. Bin (90® + J ) = sin 900 cos il 4- cos 90® sin i4 , = 1 xcosil+Oxsin/i, =cos A. Q. E. D. EXAMPLES. XLIV. <'" Draw the figures for the tii-ht four of the following examples. (1) For the {A + B) formuloe, when A is greater than 90^ and'(^ + 5) less than 180o, (2) For the {A - li) formula?, when A and B eiU'li lie between 900 and ISO®. (3) For the {A-k-B) formulae, when A lies between 90® and ' -Qo, and (.4 + B) lies between 180« and 270®. (4) For the {A - B) formulae, when A lies between 180^ and 270», and {A - B) lies between ISO® and A. Deduce the six following fonnulfe from the M, B' foimulre. (5) cos(90°+v4)= -siuA (6) sin (900- J)=co8 J. (7) cos(90«-^l)=sinyl. (8) sin(1800- J)=sin J. (9) cos(180«-yl)= -cosJ. (10) din(180« + J)= -sin J. (11) Assuming that the foraiula sin (.1 + 5) = sin J . cos 7^ + co3u4 . sin D is true for all values of A and B, deduce the rest of thn < 4 JV ff.nnnlrn }.v the aid of the results on p. 107. 10 146 TRIGONOMETRY. 173. We may also conversely prove that the 'A,B' formul89 are true for angles of any magnitude by the aid of the result of page 107. (This is a very convenient method of proving the 'A, B' formulae to be true for all values of the angles.) For, assuming that the ^A,B' formulae are true for certain values of the angles A and B, we can show that they are true if either of the angles -4 or ^ be increased by 90". Example. sin(900+^+^)=cos(^+5) [p. 107.] =cos A . cos ^ - sin ^ . sin .S^ =sin (900+^) . cos jB - { - cos (900 + ^)} sin B, = sin (90*' + ^) . cos ^ -r cos (90<^ + A) . sin B. Or, writing A' for 90*^ + ^, we have sin {A' + ^) = sin ^' . cos B + cos A' . sin B. We have proved (Art. 170) that the '^, jB' formulae are true for all values of A and B between 0" and 90°. And therefore, by what we have said above, they are true for all values of A or B between 0" and 180°. And so on. Therefore the ^A, B^ formulae are true for any values whatever of the angles A and B. 174. It follows that all formulae deduced from the ^ A, B' formulae are true for angles of any magnitude what- ever. Thus the ' S, T' formulae (page 126) are true for angles of any magnitude. Also the formulae of the last Chapter for multiple angles, and all general formulae in the Ex- amples, are true for angles oi any magnitude. 1(1) Deduce the values of sin 18(y and cos 180** from those of sine and cosine of QO'. EXAMPLES. 147 EXAMPLES. XLV. (2) The angle A is greater than 180" and less than 270" and tan^ = ^ : find sin 2^ and sin 3^4. (3) The angle 6 lies in the fourth quadrant and cos^=J, find sin 2^ and sin 3^. Find also cos 3^, and hence determine in ch quadrant 26 lies. (4) Prove that the different values of 6 which satisfy the equation cosjo^ + cos^^-O, form two series in A. P. with com- i^^kou differences — , and respectively W On Submultiplk Angles. 175. We have now proved all the formulae of the last two Chap tore to be universally true. We may expect therefore that any result, which can be ijtained from these formulae by algebraical transformation, vill have a complete geometrical interpretation. [See p. 116.] [Art. 166.] 176. Since, ^1 cos ii = 1 - 2 sin' — , aid, cos ii - 2 cos* -n - 1 ; we have, "" 2 - 2 • and, ^A 1 + cos ^ 10—2 148 TRIOONOMETRY. Or, sm cos^ ~-^J 1 - COS A 1 V yl + cos ^ I J Thus, given the vahie of cos A (nothing else being known about the angle A), we get two values for sin — , one posi- A tive and one negative, and two like values for cos ^ . 177. To prove geometrically that, given the value of cos A (nothing else being known about the angle A), there A A are two values each of sin ^ and of cos jr . 2i Z u yPl ,.,-'-6 D N. -^. ^ Let a be the least positive angle which has the given cosine, aud let ROP^ and PfiR in the figure each = a. Then A is one of the angles described by the revolving line OP when, starting from OR, OP stops either in the position OP or in the position OP^; i.e. any one of the angles 2ii7r=ta. [Art 147 ) m ON SUBMULTIPLE ANGLES. U9 e angle described is (some multiple of four right gle8"fca); a half of this is (some multiple of two right s^les »*• i a) ; so that a half of the angle whose cosine is equal that of a, may be any one of the angles indicated by Ofi the four dotted lines OP, and 0P„ or OF^ and OP, in the figure. And it will be seen that sin EOP^ = sin POP^ = - sin HOP^ = - sin ROP^ ■ Also, cos BOP^ = - cos BOP, = - cos BOP, =- cos BOP,. From these it is clear that there are two values of sin -^ , equal in magnitude and opposite in sign. Also, that there are two like values for cos — . EXAMPLES. XLVI. en A lies between - 180" and 180", prove that A . /T+i ■cos A 2 (2) When A lies between 180" and 540*>, prove that A /l+cos^i (3) Find sin — iu teruib of cos Af when A lies between 180" (4) Prove that when A lies between (4h+ 1) «• and (4?H-3)7r, ... . ^ A /l+co»A ►emg a positive integer, <^os g "^ ~ \/ o — • (5) Find sin ^ in terms of cos A^ when A lies between inn and (4n + 2) tt, where n is a positive integer. 150 TRIGONOMETRY. A A 178, Since, 2sin -^ . cos - =sin yl, [Art. 166.] and, mn^ -^ + cos^ "o ^ -^ ' we obtain by addition and subtraction A A A A sin* -^ + cos' - + 2 sin — . cos — = 1 + sin A, 2i It Z it • 9A „ A ^ . A A ^ . sin -jr- + cos — - 2 sm -^ . cos -- = 1 — sin A. 2i A Z 2i Therefore ( sin - + cos — j = 1 + sin A, and ( sin - - cos ^ j = 1 - sin yl. Whence, sin -^ + cos -^ = ± vl + sin ^ (i), ^ " \ sin-jr-cos-^ =± Vl^-sin^ (ii)- Z Z J Adding we get, 2 sin = ± Vl+sin^i J\ -sin.4 (iii), Subtracting we get, 2 cos ^ = ± n/1 + sin ^ t V 1 - sin ^ (^^0- Z Thus if we are given the value of sin ^, (nothing else being known about the angle A\ we have four values for A A sin — and four values for cos ^ . 179. To prove Geometrically that, given tlie value of sin A (nothing else being known about the angle A), there A A are four values for sin -^ , and four values for cos j . Let a be the least positive angle which has the given sine, and let IWF^, 1\0L in the figure each = a. ON SUBMULTIPLE ANGLES. 151 Then A is one of the angles described by the revolving; line OP, when, staiting from OR, OP stops either in the posi- tion OP, or in the position 0/*,. The angle described is either (an even multiple of two right angles -f- a) or (an odd multiple of two right angles — a); a half of this is either (an even multiple of one right angle + J a) or (an odd multiple of one right angle - J a); so that a half of the angle whose sine is equal to that of a may be any one of the angles indicated by OR and the dotted lines OP^ and OP,, or OP^ and OP, in the figure. In the figure, ROP^ = PfiU= LOP, = P,OD. And sin— may have any one of the values sin POP,, sin POP^, sinPOi^, sin POP,. These values are all dif- ferent, and are those given by the solution (iii). Hence, there are four values for sin — , of the nature 2 indicated by the solution (iii); also foxir values for cos ^ (iv). 152 TRIGONOMETRY. 180. If we know the magnitude of A, we can decide which sign to take in the formulae lin — + cos TT = ± vl + siu A. iS). . A A r. ^-. sm ^ - cos — = ± V 1 - sin J (u). and Example. When -^ lies between - 45*^ and + 45*^, A . .u • -1 1 • cos — IS greater than sm — and is positive. So that (sin — + cos ^ j is positive, and = + v^l + sin A^ / A A\ — ____ (sin - - cos n ) is negative, and = - Vl - sin ^. When - lies between +45^' and + ISS^*, sin - is greater than cos ^ , and is positive. So that f sin - + cos - j and ( sin - - cos ^ J are both positive^ And so on. The following diagram completes the above results. (i)Js \ (iijis (i^)is4l 'iiJisM\ do yO \ A (i)h (iiJis :] \ ("Jis -l -) ON SUBMULTIPLE ANGLES. 2 urn 2 181. Since tan A = — - 153 1 2tAn tan' 4 lan' 2 UnA 1=0, lence A -1 *>/l +tan*i4 tan -jr = 7 -. 2 tan^ Thus, given tan A we find two unequal values for tan 2 one positive and one negative. ^H 182. The student will be able by the aid of the fol- ^^wing figure to verify this result geometrically. 183. We may remark that in this figure P^OP^ and PftP,, are sti-aight line» at right angles to each other. So that tan PfiR = - cot PfiR ; or, tan PfiR . tan PfiR = - 1. |: Hence, one value of tan ^ is the reciprocal of the other, II d of opposite sign. So that there is always one positive ue of tan-^, and one negative; one numerically greater unity and the other numerically less than unity. 154 TRIGONOMETRY. Example. If A = \ 90^, prove that A - 1 - v/l+tan^^ ^"^2= ti^^ • Now tan J =tan 190*^, which is positive ; tan -=tan 95°, which X- Ai ^ -l±Vl+tanM .- X. 1 i. IS negative. Also tan — = r -^ J the negative value of , . , . -l-\/l+tan2^ which is -, ; . tan^ EXAMPLES. XLVn. (1) State the signs of f sin — + cos — j and (sin — - cos ^ when — has the following values : (i) 220, (ii) 1910^ (iii) 2900, (iv) 3450, (v) -220, (yi) _2750, (vii) -4700, (viii) lOOQO. (2) Prove that the formulae which give the values of sin — and of cos — in terms of sin A, are unaltered when A has the values (i) 920, 2680, 9000, 4n7r + f7r, or (4M + 2)7r-f7r. (ii) 880, _880, 770", - 77O0, or 47i7r±^. (3) Find the values of (i) sinQO, (ii) cos 90, (iii) sin SP, (iv) cos 1890, (v) tan 202^0^ (vi) tan 97^o. (4) If ^ = 2000, prove that (i) 2 sin- = +\^l+sin^ + Vl-sin^. .... . A -(1 + Vl + tanM) (11) tan-=-i-^^^ . (5) If A lie between 270" and 3600, prove that (i) 2 sin — =Vl -sin^-Vl+sin J.. (ii) tan -x= - cot A + cosec A. (6) \i A lie between 450*^ and 6300 prove that 2 sin ^ = - VT+smX- v^l - sin J. ■ (8) ^H (9) If A lie between n x 360o - 90® and w x 360® + 9(y> wliere n IS a positive integer, prove that ^ A -l+Vr+tan2j ^2= -te^2 ' "and that when A lies between 71x3600+900 and nx 360®+ 270", A - 1 - \^1 + tan2 A ON SUBMULTIPLE ANGLES. 155 (7) Find the limits between which -^ must lie when 2 2 sin ■^=Vl + sinyl - Vl -sin^^, (8) Given that A lies between AbQP and 630<>, prove that 2 cos -= -Vl+sinA + Vl -sin^. then tan 2 tanil (10) Prove geometrically that if v/e are given the value of sin A^ there are three different values for sin - , and six different o values for cos _ . (11) Prove geometrically that, if we are given the vahie of cos Af there are three different values for cos - , and six different u values for sin - . 3 (12) Prove that if we are given the value of tan A there arc three different values for tan - . (13) Given the value of tan A^ prove that there are four A A values each for sin _ and cos - . /> Ji (14) Given the value of sin J, prove that there are two value-s for tan -- 2 1 5 6 TRIQONOMETR Y. 184. It is important if possible, in solving Trigono- metrical equations, to avoid squaring both sides of the equation. Example. Solve cos 6=k sin 6. If we square both sides of the equation we get cos2 6=B sin2 ^=F (i _ cos2 $). B k .'. cos2^=- — .- , or cos^=db . 1+P' v/l+p k Now if a be the least angle such that cos 3 = , , then the above gives us O — mr^a (i). But the equation may be written cot 6 = k, whence 6=11/17 + a (ii). (ii) is the complete solution of the proposed equation, while (i) is in fact the solution of both cos ^=^ sin ^ and also of co^6= -k sin 6. So that by squaring both sides of the equation we obtain solutions which do not belong to the given equation. 185. We can often avoid squaring by the use of a Subsidiary Angle. Example. Solve a cos 6-{-b sin ^ = 1. That is a ( cos ^ + - sin ^ j = 1. Find in the tables the angle whose tangent is - ; let it be a. a Then -=tan a; and the equation becomes a ^ a (cos ^+ tan a. sin^) = l, /cos ^ . cos g + sin ^ . sin a\ ^ ^ \ cos a / , cos a cos id -a) = . ^ ' a ON SUBSIDIARY ANGLES. 157 Fiud from the Tables the value of cos a. Next find from Tables the magnitude of the angle /3 whose cosine = — " , a we get cos (d - a) = cos /3 ; 186. Def. When an angle a is introduced to facilitate calculation it is called a Subsidiary Angle. =-EXAMPLES. XLVm. Solve the following equationn. — (1) 2sintf + 2co8^=\'2. - (2) sin ^ + v/3 . cos^ = l. — (3) v^2sintf + V2cos^=V3. —(4) sin ^- cos ^ = 1. (6) sintf+cos^ = l. (6) V3sin^- cos ^ V2 = 0. (7) 2 8ina: + 5co8:r=2. [2-5 = tan680 12'] (8) 3 cos a: - 8 sin X- 3. [2-6 = tan 69® 26' 30"] (9) 4 sin a: - 1 5 cos j: - 4. [375 = tan 750 4'] ^ (10) cos(a + x) = sin(a+a:) + v^cos^. ♦On the Inverse Notation. 187. The equation sin 6 = a means that 6 is an angle whose sine is a. d = 8in~'a is a convenient way of writing the same equation. Thus sin"' a (is an angle, and) is an abbreviation for an angle whose sine is a. Example 1. Show that 30" t» one valtie of sin~* ^. We know that 8in30*'=^. Therefore 30^ is an angle whose siino i« I or J^fV^ -^ Qin ~' 1 158 TRIQONOMETR Y. Example 2, Prove that 45^* is one value o/tan-^ ^4-tan-i ^ tan"^ ^ is one of the angles whose tangent is ^. Let a = tan-i ^, so that, tan a = \. Let ^ = tan-i .^, so that, tan /^ = 3j . We have to prove that 45*^ is one value of a + ^. *^^ (°+^) = r- tana, tan /S-f^T^-^- Buttan45« = l. Therefore 450 is one value of a+^, i.e. of tan-' ^+tan~i^. ^EXAMPLES. XLIX. .J.- Prove that the following statements are true when we take for sin~i a, etc. their least positive value. (1 ) sin-i % = cos-^ I = tan-i |. Vs /- (2) sin-i \ = cos-i -zr- = cot-i v 3. , a (3) sin-i a = cos-^ v i - a^ = tan-^ / _ ^ * (4) If a=sin-i | and j3=cos-if , then a+^=|. (5) If ^ = sin-i a and jB = cos-i a, then A+B= 90^. (6) tan-if + tan-i^=|, (7) tan-i ^ + 2 tan-^ ^ = tan-i ^. (8) tan-i Wj + tan-i Wg = tan^i -^^ . (9) sin (2 sin-i a) = 2 a Vl - aK (10) 2cos-ia=cos-i(2a2-l). (11) cos-iH2sin-ii=1200. (12)2sin-ii -sin-i|^ = 2cos-if*. ,, « V X ,/cot2a-tan2a\ (13) 2tan-i(cos2a)=tan-M ^ — 1. (14) tan-i ;r + tan"! y + tan-^ ^^^^^..^^ = 4 • (15) 4 tan-i } - tan'^ ^-ig =^^ . ^^^^^^ ON THE INVERSE NOTATION. 159 ^^^^H^wn-i t + sin-i ^ + 8in-i H=2' ^^^ (17) tau-iV5(2-V3)-cot-W6(2 + V3)=cot-'vA'-i. (18) If sm~*m + 8iii~^n=-, prove that m Vl - n2 + n\^l -m2=l. 188. The student must notice carefully that such a itement as »tn~' ^ = co«~' -^ is not an identity. For sin"' J is one of the values of wtt 4- (- 1)" 30", /3 and cos"' -^ is one of the values of 2mr «*= 30". Thus 150" = sin"* J, but 150° is not = cos"' ^ . ♦* MISCELLANEOUS EXAMPLES. L. 1 12 (1) Prove that tan~' , f-tan~*- htan~* -»=n7r. ^ 1 + a \-a a* (2) Prove thui uui ^ +tau"' „ - =n7r+ . . ^ ' a 2a - 1 4 (3) Prove that sin"^ X - sin-* y = cos~* {xy ^>I\-x^-y^+ xh/\ (4) Iftan-i^l,+tan-i^-=^,showthatx»=* ^ ' ar+2 x-2 4 (5) Prove that tan"! a + cot" » a = (2n + 1 ) | . (6) If tan-»a4-tan-i/9 + tan-»'y=7r, prove that a + ^ + y = a^. (7) Solve the equation tan"* — - -tan~* "TT'^' fE* I GO TRIGONOMETRY. L. (8) Solve the equation tan-i {x+\)- tan-i {x - l) = cot-i {x"- - 1). (9) Solve the equation sin"^ ^ 2 +tan~i - — -^ = - , k '\' X 1 — X^ ii (10) Prove that .a->Jo^'A . , 1 ^ , a+\/a2-4 . tt (11) If a be positive and less than unity, and if a be the least value of sin-^ a, then 8in-ia + cos~ia = 7i7r + (-l)"a±f|-aj . (12) Prove that tan A and sin '2, A have always the same sign- Solve the six following equations. (13) cos ^ + cos 3^ + cos 5^=0. (14) sin5^ + sin3^ + sin^ = 3-4sin2^. (15) 2sin2 3^ + sin2 6^ = 2. (16) a(cos2:r-l) + 26(cos^ + l)=0. (17) sin(m + 7i)^ + sin2wi^4-sin(m-w) ^=0. (18) siii{'»'^('^+y)Hsin{7ry(a;+y)}=0, 1 ^ ' sin7ra;^ + sin7ry2=0. J (19) Trace the changes in the sign and magnitude of the following expressions, as 6 changes from to tt. (i) 2 sin ^ . cos 6. (ii) cos^ 6 - sin^ B. (iii) sin 3^. (iv) cot 2^. (v) sin(^ + a). (vi) cos (2^ -a). (20) Explain why the equations ^=W7r + (-l)"a and |- ^ = 2?^7r± (|-a) have exactly the same series of solutions. (21) Explain why exactly the same series of angles are given by the two equations ^ + ^ =^(117 + ( - 1)" - , and ^ - t = ^titt ± .-7 . ( ici ) CHAPTER XIV. On Logarithms. In Algebra it is explained (i) that the multiplication of different powers of the same quantity is effected by adding the indices of those powers; (ii) that division is effected by subtracting the indices ; (iii) that involution and evolution are respectively effected by the multiplication and division of the indices. Example 1. If m=a'', n = a*, then my.n = a'^ y a* •(ii), ;m»=(a*)»=a'*, (iii). Example 2. If 347 - 10«"«328o» ^nd 461 = I02«ww prove that 347x461 = 10»-»»03M. We have 347 x 461 = 102 M03a>6 j, loauesrow _ 10264032»6+2«J3700J> „106-JM09(H. Q,E.D. ♦ The number 347 lies between 100 and 1000, i. e. between 10» and 10*. Hence, if there is a power of 10 which is eijual to 347, its inio3oo^ 3 = 10*^i»« and 7 = 10-8«<»8o, find the ^^■dices of the powers of 10 equivalent to the quantities in the ^^pet 6 examples. (I) 2*, 32, 23, 2x3, 2^ 7f (2) 14, 16, 18, 24, 27, 42. (3) 10, 5, 15, 25, 30, 35. (4) 36, 40, 48, 50, 200, 1000. (5) 3»« X 71° -r 2«>, 2^2 X 320 -1. 7". (6) v^ X 4^, y-'49T4» X -^S^TM (7) Find approximately the numerical value of ^42 having given that \0-^«^»=l'45Z2 nearly. (8) Find approximately the numerical value of 4^(42)* x sf (42)' liaving given that 103 38177=24086. (9) Find the value (i) of 4^6 x ^^7 x ^"9, (ii) of '^2 x 3"* x 7ti having given that 10««i*<»7 = 4.5368 and 10" <««5094 = . 93646^ (10) Find the value of (67-21)1 x (49-62)y x (3'971)-ff having given that 67-21 = 10i-8274339^ 49-62 = 101'^*^^, 3971 = 10 "^s^^' and 10 6»:i3io = 3-9549. (II) Find the area of a square field whose side is 640-12 feet ; having given that 640-12 = 102-8o«'^«" and that 10S-8i262-^ =40975-3. (12) Find the edge of a solid cube which contains 42601 cubic inches; having given 42601 = 10* «294i98 and 10^ "31399=, 34.925. (13) Find the edge of a solid cube which contains 34*701 cubic inches; having given that 34-701 = 101*^3*20^ and 10^»3**78=3.2ui7. (14) Find the volume of the cube the length of one of whose edges is 47-931 yards; having given that 47*931 = 10' ««»>"^ and Uuit 10*'W>»**-«=110115. 11—2 164 TRIGONOMETRY. 191. The powers of any other number than 10 might be used in the manner explained above, but 10 is the most convenient number, as will presently appear. 192. This method, in which the indices of the powers of a certain fixed number (such as 10) are made use of, is called the Method of Logarithms. Indices thus used are called logarithms. The Jioced nv/mber whose powers are used is called the base. Hence we have the following definition : DEF. The logarithm of a number to a given base, is the index of that power of the base, which is equal to the given number. Thus, if I be the logarithm of the number n to the base a, then a^=n. 193. The notation used is log.w = ^. Here, log„ n is an abbreviation for the words ' the logar rithm of the number n to the base a.' And this means, as we have explained above, 'the index of that power of a which is equal to the number n.' Example 1. What is the logarithm of a^ to the base a ? That is, what is the index of the power of a which is a^ ? The index is f ; therefore f is the required logarithm, or log„a^=f. Example 2. What is the logarithm of 32 to the base 2 ? That is, what is the index of the power of 2 which is equal to 32 ? Now 32 =2^ .'. the required index is 5 ; or log2 32 = 5. ON LOGARITHMS. 165 e""n8e of Logarithms is based upon the following propositions : — I^H I, The logarithm of the product of two numbers is IM^al to the logarithm of one of the numbers + the logarithm of the other. I^KFor, let \og^m=x and log, n=y, I^Hd, log, {mxn) = log. (a* x a*) = log, (a* "^ ") = ;r 4-y I^B = log. m + log, n. ^^H H, The logarithm of the quotient of two numbers is \^^^ logarithm of the dividend - the logarithm of the divisor. ^^HFor, log. (^j = log, (^ = loga {a*-9) = :c - y [as above] ^^B = loga m- log. n, ^^K m. The logarithm of a number raised to a power k is ^Kraes the logarithm of the number. For, log, (m*) = log„ {(a*)*} = log, (a**) = Jb: =/fc times logaWi. Examples. Given logio 2 = -3010300, logio 3 = -4771213, logjo 7 = -8450980 j(/id the valites of the following : (i) logio 6 = logio (2x3) = logio 2 + log^o 3 [by I.] = •3010300 + -4771213 -•7781513. (ii) Iogio5 = logio7-logio3 = -8450980- 4771213 [by II.] = •3679767. (iii) logio 3^=5 times logio3 = 5x^l77UM:i [by ill. J = 2-3856065. (iv) log,o^^^=logio (^)* = J of log,o ^^ [by III.] = J of (log 3 + log 4 - log 7) [by I. and II.] = \ of {•4771213 + twice •30103- '8450980) = i of '2340833 = 0780278. (V) logio 5 = logio ^ = logio 10 - log,o 2=1- -3010300 = -6989700. 166 TBIGONOMETUY. ^ EXAMPLES. LIII. - (1) Find the logarithms to the base a of a^, a-^, ^a, sla\ -^ - (2) Find the logarithms to the base 2 of 8, 64, ^, '125, •015625, 4^64. ,;w (3) Find the logarithms to the base 3 of 9, 81, ^, -^, '1, ^. (4) Find the logarithms to base 4 of 8, \^16, ^-5, \/'-015625. (5) Find the value of loggS, log2-5, log3 243, logs (-04), logiolOOO, logio'OOl. (6) Find the value of log^at, logft^P, logs 2, loggia, logioolO. Given that logio 2 = '3010300, logio 3= -4771213 and log^o 7 = '8450980, find the values of (7) logio6, logio42, logioie. (8) logio49, logio36, logio63. (9) Iogio200,logio600, logio 70. (10) logjoS, logio3-3, logio50. (11) logio35, logiolSO, logio -2. (12) logio3-5, logio7-29, logio'OSi. (13) Given logio 2, logio 3, logio7, find the value (i) of ^6x4/7x4^9, (ii) of jy2x3-^x7T^ [•6615067 =logio 4-5868 ; - '0285094 = logio "93646]. (14) Prove that (i) log {^2 x v^7 ^4^9} = i log2 + ilog7 - f log 3, (ii) log f^2 X 3-T X 7 A} =T^ log 2 - f log 3 + 3^ log 7. (15) If logio a =2-6560982 ^^^ l^^io &=2'8129134, show that (i) logioa&=6'4690116, (ii) logio a* =10-6243928, (iii) logio a2?,3 = 13-7509366, (iv) logio ^a= '8853661. (v) logio (a36)i =1-7968680, (^) logio ^^ ^^ = 8*9699598. (16) Show (i) that logjo 4/21 x 4/18 = '7545579, (ii) that logio ^("49 X 4'^) X 4/(3* x 2^^) = 2-989843. ON LOGARITHMS, 167 Common Logarithms. 194. That System of Logarithms whose base is 10, is tiled the Common System of Logarithms. Li speaking of logarithms hereafter, cormnon logarithms are referred to unless the contrary is expressly stated. 195. We shall assume that a power of ten can be found ^which is practically equivalent to any number. ^H 196. The indices of these powers of 10, i.e. the Com- ^^■>n Logarithms, are in general incommensurable numbers. Their value for every whole number, from 1 to 100000, has been calculated to 7 significant figures. Thus any cal- culation made with the aid of logarithms is as exact as the most carefully observed measurement (cf. Arts. 17, 216). 197. Now, the greater the index of any power of 10, the greater will be the numerical value of that power; and the less the index, the less will be the numerical value of the power. Hence, if one number be less than another, the loga- rithm of the first will be less than the logarithm of the second. But the student should notice that logarithms (or indices) are not proportional to the corresponding numbers. Example. 1000 is less than 10000 ; and the logarithm to base 1 of the first is 3 and of the second is 4 But 1000, 10000, 3, 4 are not in propt^rtion. • 168 TRIGONOMETRW 198. We know from Algebra that 1 -= 10", 10-10^ and that 'l = A =10"' 100 = 10^ -01= ^ =10-^ 1000 = 10^ OOl = Tir?rT7 = 10-' 10000 = 10^ OOOl - ^J^ = 10-^ and so on. Hence, the logarithm of 1 is 0. The (common) logarithm of any number greater than 1 is positive. The logarithm of any positive number less than 1 is negative. 199. We observe also that the logarithm of any number between 1 and 10 is a positive decimal fraction ; that the logarithm of any number between 10 and 100, i.e. between 10' and 10*, is of the form 1 + a decimal fraction ; that the logarithm of any number between 1000 and 10000, i.e. between 10' and 10*, is of the form 3 4- a decimal fraction ; and so on. 200. We observe also that the logarithm of any number between 1 and -1, i.e. between 10° and 10"^ can be written in the form - 1 + a decimal fraction ; that the logarithm of any number between -1 and -01, i.e. between 10"' and lO"', can be written in the form - 2 + a decimal fraction; and so on. ON LOGARITHMS. 169 1 . How many digits are there in the integral part le number whose logarithm is 3-^7192? We know that 3= log 1000 and 4= log 10000 ; .-. 3-67192 = log of some number between 1000 and 10000; kt is, the integral part of the number has 4 digits. Example 2. Given that log 3= '4771213, find the number of the digits in the integral part of 3^. Here, log (3^) = 20 times log 3 = 9-542426 ; hence, log {3^) hes between 9 and 10 ; therefore, as in Example 1, the number Hes between 10** and 10^® j that is, its integral part has 10 digits. Example 3. Supposing that the decimal part of the logarithm is to be kept positive, find the integral part of the logarithm of •0001234. This number is greater than -0001 i.e. than 10"* and less than •001, t.e. than 10"'. Therefore its logarithm lies between - 3 and - 4, and there- fore it is - 4 + a fraction ; the integral part is therefore - 4. EXAMPLES. UV. Note. The decimal part of a logarithm is to be kept positive. (1) Write down the integral part of the common logarithms of 17601, 361-1, 401, 723000, 29. (2) Write down the integral part of the common logarithms of -04, •0000612, -7963, -001201. (See Note above.) (3) Write down the integral part of the common logarithms of 7963, -1, 2-61, 79-6341, 1-0006, -00000079. (4) How many digits are there in the integral part of the numbers whose common logarithms are respectively 3-461, ^020300, 5-4712301, 2'67101(X)? 170 TRIGONOMETRY. LIV, (5) Give the position of the first signijlcant ligui'e in tlie numbers whose logarithms are -2 + -4612310, -1 + -2793400, -6 + '1763241. (6) Give the position of the first significant figure in the numbers whose common logarithms are 4-2990713, -3040595, 2-5860244, - 3 + -1760913, - 1 + -3180633, -4980347. (7) Given log^o 2 = -30103, find the number of digits in the integral part of Q^\ 2^2, 1620, 2ioo. (8) Given that log 7 = -8450980, find the number of digits in the integral part of 7^^ 49«, 343^, (V)20, (4-9)i2, (3-43)io. (9) Find the position of the first significant figure in ^2, {\r, (¥)^ (•02)S (-49)6. (10) Find the position of the first significant figure in the numerical value of 207, (-02)7, (-007)2, (3-43)T^, (-0343)8, (-0343)1^. 201. Prop. To prove that if two numbers expressed in the decimal notation Jiave the same digits {so that they differ only in the position of the decimal point), their loga- rithms to the base 10 loill differ only by an integer. The decimal point in a number is moved by multiplying or dividing the number by some integral power of 10. Let the numbers be m and n; then m = n x 10*, where /fc is a whole number positive or negative; and log m = log {n X 10*) = log n + log 10* = log n-}-k. Example 1. Let the numbers be 1-2345 and 1234-5. Then, 1234-5 = 1-2345 x 10^, therefore log (1234-5)=log (1-2345) -v 3. ON LOGARITHMS. I 7 I invvAi log 1*7692 = -24776, find (i) log 17692, (ii) log -0017692, (iii) 176*92. [Here, log 17692 = log {(1-7692) x 10*} =log (l-7692) + 4 = •24776 + 4 = 4-24776. log -001 7692 =log {(1-7692) X 10-3} = -3+ -24776. logl76-92 = log {(1-7692) X 102} =2-24776. 202. It is convenient to keep the decimal paH of com- mon logarithms always positive, because then the dednial part of the logarithms of any numbers expressed by the same digits will be always the same. 203. The decimal part of a logarithm is called the mantiBsa. 204. The integral part is called the characteristic. 205. The characteristic of a logarithm can be always obtained by the following rule, which is evident from page 167. HXTLE. The characteristic of the logarithm of a number greater than unity is one less than the number of integral figures in that number. The characteristic of a number less than unity is nega- tive, and (when the number is expressed as a decimal,) is one more than the number of cyphers between the decimal point and the first significant figure to the right of the deci- mal point. 206. When the characteristic is negative, as for ex- ample in the logarithm - 3-1- '1760913, the logarithm is abbreviated thus, 3'1760913. Example 1. The characteristics of 36741, 36-741, 0036741, 36741 and '36741 are respectively 4, 1, -3, 0, and 1. 1 7 2 TRIG ONOME TR Y. Example 2. Given that the mantissa of the logarithm of 36741 is 5651510, we can at once write down the logarithm of any number whose digits are 36741. Thus log 3674100 =6-5651510, log 36741 =4-5651510, log 367-41 =2-5651510, log -36741 =1-5651510, log-00036741 =4-5651510, and so on. 207. In any set of tables of common logarithms the student will find the mantissa only corresponding to any set of digits. It would obviously be superfluous to give the clia/rao teristic. 208. It is most important to remember to keep the mantissa always positive. Example. Find the fifth root of -00065061. Here logio -00065061 = 4-8133207, .-. logio (•00065061)^=i(4-8133207)=K-4 + -8133207) =i(-5 + l-8133207)=-l + -3626641 = 1-3626641, and 1-3626641= log -23050, .-. the fifth root of -00065061 = -23050 nearly. EXAMPLES. LV. (1) Write down the logarithms of 776-43, 7-7643, -00077643 and 776430. (The table gives opposite the numbers 77643, the figures 8901023.) (2) Given that log^o 59082 = 4-7714552, write down the logarithms of 5908200, 5-9082, -00059082, 590-82 and 5908-2. ON LOGARITHMS. LV. 173 (3) Find the fourth root of '0059082, having given that log 5 -9082 = -77 1 4552 ; 4-4428638 = logio 27724. (4) Find the pixxluct of -00059082 and -027724, having given tt -21431 =log 16380 (of. Question 3). (5) Find the 10th root of -077643 (cf. Question 1), having fen that -8890102 =log 7-7448. (6) Find the product of ('27724)2 and '077643. (See Ques- tions 1 and 3; 7758288 = log 59680.) * * 209. To transform a system of logcvrithm^ having a given base, to another system vnth a different base. If we are given a list of logarithms calculated to a given base, we can deduce fi-om it a list of logarithms calculated to any other base. Let a be "the given base ; let 6 be any other base. Let m be any number. Then the logarithm of m to the base a is in the given list. Let this logarithm be I. Then m = a\ We wish to find the logarithm of m to the base 6. Let it be X. Then m = b'. Butm = a'; .-. a' = 6*; or, 6 = a*; or, - is the logarithm of 6 to the base a. Now the logarithm of 6 to the base a is given. For it is in the list of logarithms to the base a. Thus, ^ = log,6; or,a: = j-^^; K)\\ xooKm = . — J- . 174 TRIGONOMETRY. Hence, to calculate the logarithms of a series of numbers to a new base b, we have only to divide each of the logarithms of the numbers to any given base a, by a certain divisor, viz. log^ b. If then a list of logarithms to some base e can be made, we can deduce from it a list of common logarithms, by multiplying each logarithm in the given list by r^ . Example. Show how to transform logarithms, having 5 for base to logarithms having 125 for base. Suppose m = 5 *, so that I = log., m. Now 125 = 53, so that 3 =log5 125, and m = 5^^ = 125 s , so that - = logigg wi. Thus the logarithm of any number to base 5, divided by 3 (i.e. by logg 125), is the logarithm of the same number to the base 125. *'^210. The student will find that the logarithms of numbers cannot be calculated to the base 10 directly. They are first calculated to the base 2-7182818, etc., which is the sum of the series ^ +1 + o + m + i-o:4+"*''- '^ '''■^- This number is called e. And the constant divisor in this case is log^ 10, When this constant divisor is transformed into a multiplier, this constant multiplier is called a modulus. Example. Given that logio 12 = 1-0791812, shew how to transform common logarithms to logarithms having 12 for base. Here 10^-^^^^^^^= 12; 1 . •. 10=1 2i^iffl2 = 12 ■S'"'^fi«2. etc. EXAMPLES. " ' EXAMPLES. LVI. [Show how to transform Logaritlima with base 2 to logarithms with base 8. Logarithms with base 9 to logarithms with base 3. Common logarithms to logarithms with base 2. Logarithms with base 3 to common logs. Common logs to logarithms with base 3. Given log^o 2 = 3010300, find loga 10. Given logio7 = '8450980, find logy 10. Given log^o 2 = -3010300, find logglO and log^ 10. ** MISCELLANEOUS EXAMPLES. LVH. (1) Find log2 8, log, 1, logs 2, log; 1, log.^ 128 (2) Show that the logarithms of all except eight of the ibers from 1 to 30 inclusive, can be calculated in terms of 2, log 3 and log 7. (3) Show that the logarithms of the numbers 1 to 10 inclu- ive may be found in terms of the logarithms of 8, 14, 21. (4) The mantissa of the log of 85762 is 9332949. Find the ■rr of iy -0085762. Find how many figured there are m (85762)^^, when it is multiplied out. (5) Find the product of 47*609, 476-09, -47609, -000047609, having given that log 4-7609 = -6776891 and •7107564=log 5-1375. (6) What are the characteristics of the logarithm of 3742 to the bases 3, 6, 10 and 12 respectively. (7) Having given that log 2 = -3010300, log 3 = "477 1213 and log 7 = -8450980, solve the following equations : (i) 2'x3*' = 72, (ii) 3**= 128 x 7*"*, (iii) 12*=49, (iv) 2*' = 21««. (8) Given logio7, find logy 490. ' (9) Given log^o 3, find logj 270. (W) Given log,. 2, find log. 10. 176 TRIGONOMETRY. LVII. (11) Given log89 = a, \og^b=h^ log57 = c; find the logs to base 10 of numbers 1 to 7 inclusive. (12) How many positive integers are there whose logarithms to base 2 have 5 for a characteristic ? (13) If a be an integer, how many positive integers are there whose logs to base a have 10 for their characteristic ? (14) Given log 2 and log 7, find the eleventh root of (39*2)2. log 1-9485 = -289688. (15) Prove that 7 log i| + 6 log f + 5 log f + log ff =log 3. (16) Prove that 2loga + 2loga2 + 21oga3 ._ +2loga"=n (w+ l)loga. (17) Prove that log„6 . log^a = 1 ; and that \ogJb . logjC . logea= 1. (18) Prove that log^r =loga6 . logjC . log^o?. . .log/. (19) Given that the integral part of {Z-^b^y^^^^ contains 53856 digits, find log 345*6 correct to five places of decimals. (20) Given that the integral part of (3*981)ioo<^ contains sixty thousand digits, find log 39810 correct to five places of decimals. (21) If the number of births in a year be ^ of the pppu- lation at the beginning of the year, and the number of deaths j^, find in what time the population wiU be doubled. Given log 2, log 3, and that log 241 = 2-3820170. (22) Prove that log s + log (s - a) - log h - log c=2 log /v/*"^^ • (23) Prove that log (a2 + ^) + log {a + x) + log {a-x) = log (a* - x/^). (24) Prove that log sin AA = log 4 + log sin A + log cos A + log cos 2il. CHAPTER XV. On THE Use of Matuematical Tablks. 213. The Logarithms referred to in this chapter, and m re throughout the book, are Common Logarithms. 214. Books of Mathematical Tables usually give an ex])lanation of their own contents, but there are some points mmon to all such Tables which we proceed to explain. Bo: 215. The student will be supposed to have access to a k containing the following : (i) A list of the logarithms of all whole numbers from 1 to 99999, calculated to seven significant figures; (ii) A list of the numerical values, calculated to seven lificant figures, of the Trigonometrical Ratios of all jle.s, between 0" and 90°, which difier by 1' ; (iii) A list of the logarithms of these Ratios calculated seven significant figures. These will be found in Chambers' MatJiematiccd Tables. At the end of this Chapter we give the logarithms to figures of all numbers from 100 to 999. This Table 11 be found useful in questions involving certain kinds of lerical calculations. L. t:. t. 12 178 ' TRIGONOMETRY. 216. We liave said that logarithms are in general incommensurable numbers. Their values can therefore only be given approximately. If the value of any number is given to seven significant figures, then the error (i. e. the difference between the given value and the exact value of the number) is less than a millionth part of the number. Example. 3'141592 is the value of ir correct to seven signifi- cant figures. The error is less than '000001 ; for tt is less than 3-141593, and greater than 3-141592. The ratio of -000001 to 3-141592 is equal to 1 : 3141592. The ratio of -000001 to tt is less than this ; i. e. much less than the ratio of one to one million. 217. An actual measurement of any kind must be made with the greatest care, with the most accurate instru- ments, by the most skilful observers, if it is to attain to anything like the accuracy represented by ' seven significant figures.' Therefore the value of any quantity given correct to * seven significant figures ' is exact for all practical purposes. 218. We are given in the Tables the logarithms of all numbers from 1 to 99999 ; that is, of any number having Jive significant figures. A Table consisting of the logarithms of all numbers from 1 to 9999999 (i.e. of any number having seven significant figures) would be a hundred times as large. 219. There is however a rule by which, if we are given a complete list of the logarithms of numbers having ^ve sig- nificant figures, we can find the logaiithms of numbers having six or seven significant figures. ^^ Exa ON THE i^h Ul' MATHEMATICAL TABLES. 179 Example. Suppose we require the logarithm of 4'804213. From the Tables we find log 4-8042= -6816211, i.e. 4 8042= 10-«8»«2"-, log 4-8043 = -6816301, 4-8043= 10<»i630i..., The number 4-804213 lies between the two numbers 4-8042, 43 whose logarithms are found in the Tables, so that the Hired logarithm must lie between the two given logarithms. Therefore we suppose that log 4-804213=-6816211+rf, i.e. 4804213= 10-«8i«2ii..+ corresponding differences in the •0002491) angles are 60" and 25"; .-. r/=^^ of -0002491 -=•0001038... Hence, from (iii) sin 3106'25" = •5166333 + -0001038 = •5166371. 184 TRIGONOMETRY. Exam/pie 2. Find the angle whose logarithmic cosine is 9-7858083. The Table gives 9-7857611 =X cos 520 22' (j^^ 9-7859249 =Z cos 520 21' ^^^^ The cosine diminishes as the angle increases. Hence corre- sponding to an increase in the angle there is a diminution of the cosine. Hence, let 9-7858083 = L cos (520 22' - i)) (iii). Subtracting the first tabular logarithm from the second the difference is -0001638. Subtracting the first tabular logarithm from the third, the difference is -0000472. Subtracting the first angle from the second, the difference is - 60". Subtracting the first angle from the third, the difference is -D. By the Rule these four differences are in ijroportion. Therefore 7> = iV^j^^ of 60" = 17-3". Hence 9-7858083=Z cos (52»22' - 17") =Z cos 620 21' 43" EXAMPLES. LIX. (1) Find sin 420 21' 30" having given that sin 420 21' = '6736577 sin 42" 22' =-673^727 (2) Find cos 470 38' 30" having given that cos 470 38' = -6738727 cos 470 39'= -6736577. (3) Find cos 210 27' 45" having given that cos 2l0 27'= -9307370 cos 210 28'= -9306306. EXAMPLES. LIX. 1^^ (4) Find the aagle whose sine ia '6666666 having given that 6665325 = sin 410 48' 1-6667493 =8in 41« 49'. (5) Find the angle whose cosine is *3333333 ving given that '3332584 = cos IQf^ 32' •3335326= cos 700 31'. (6) Find the angle whose cosine is '25 having given that -2498167 = cos 75<> 32' -2500984=cos750 31'. §(7) FindZsin450l6'30" ing given that Zsin 450 I6'=9-8514969 Z sin 450 17' =9-85 16220. (8) Find L tan 27° 13' 45" ^^ving given that L tan 27" 13' = 971 12148 ^f L tan 27" 14'=9-7115254. (9) Find X cot 36® 18' 26" having given that L cot 360 18' = 101339650 Zcot36<' 19' = 10-1337003. (10) Find the angle whose Logarithmic tangent is 9*8464028, 1 laving given that 9-8463018 =X tan 35® 4' 9-8465705 =Z tan 35» 5'. (11) Find the angle whose Logarithmic cosine is 9-9448230, having given that 99447862 = Z cos 28" 1 7' 9-9448541 = Z cos 28« 16'. ( 12) Find the angle whose Logarithmic cosecant is 10-4274623, having given that 10-4273638 =Z cosec 2P 57' 10-4276774 = Zcosec2P 56'. 186 TRIGONOMETRY. 227. Problems in which each of the lines involved contains an exact number of feet, and each angle an exact number of degrees, do not occur in practical work. As from time to time the skill of observers and of in- strument-makers has increased^ so also has the number of significant figures by which observations have been recorded. Thus the want was felt of some method by which the labour involved in the multiplication and division of long numerical quantities could be avoided. At the end of the Seventeenth Century a celebrated Scotch mathematician, John Napier, Baron of Merchiston, proposed his method of * Logarithms ' ; i. e. the method of representing numbers by indices; 'which, by reducing to a few days the labour of ' many months, doubles, as it were, the life of an Astronomer, * besides freeing him from the errors and disgust inseparable ' from long calculations.' Laplace. 228. We shall now give a few examples of the practical use of logarithms. Example 1. The sides containing the right angle C in a right angled triangle ABC contain 3456-4 ft. and 4543-5 feet respec- tively; find the angles of the triangle, and the length of the hypotenuse. b Let a, 6, c be the lengths of the sides of the triangle opposite the angles Ay B, C respectively. ON THE USE OF MATHEMATICAL TABLES. 187 Then, a = 3456 4 feet, 6=45435 feet, taii^ = r . lerefore logtan J = log r=loga-log 6 = log 3456-4 -log 4543-5. In the Tables we find log 3456-4 = 3-5386240 log 4543-5 = 3-6573905. ,•. log tan ^ = 3-5386240 - 3-6573905 = 1-8812335. .•. Z tan ^ = 9-8812335. the Table we find 9-8813144=Ztan370l6' 9-8810522 = Xtan370 15'. lence by the rule of Proportional Difierences 9-8812335 = Z tan 370 15' 42". Therefore A = 37® 15' 42") and 5=900-.4 = 52«44' 18"/ .(I). To find c, we may take the square root of a^-\-b\ or we may ;.ud it by the aid of logarithms thus : - = cosec A = cosec 37<> 15' 42", .*. log c = log a + log cosec 37*' 1 5' 42" =loga + X cosec 370 15' 42" -10 = 3-5386240 + 10-2179174 -10 = 3-7565414 = log 5708-8, .'. the hypotenuse contains 5708 8 feet Thus we have found the angles and the third side of the triaiii/le. 188 TRIGONOMETRY. 229. There are some formulae which are seldom used in practical work, because they are not adapted to logarithmic calculation. They are those in which powers of quantities are connected by the signs -i- or -. Example. In the above example we might have found the length of the hypotenuse by means of the formula c2 = a2 + 62. But we should have had to go through the process of calcu- lating by multiplication the values of a^ and 6^. For this reason, a formula which consists entirely of factors is always preferred to one which consists of terms, when any of those terms contain any power of the quantities involved. If in the above example, the lengths of the hypotenuse c and of one side a were given, then the formula 62 = c2-a2=(c_a)(c+o) will give the length of b. For log62=log{(c-a) (c + a)}, or, 2log6=log(c-a) + log(c + a). And the values of {c-\-a) and {c-a) are easily written down from the given values of c and a. V EXAMPLES. LX. In the following questions A^ B, C are the angles of a right angled triangle of which (7 is a right angle, and a, 6, c are the lengths of the sides opposite those angles respectively. (1) Given that a = 1046-7 yards, c= 1856-2 yards, (7=90°, find A. log 1046-7 = 3-0198222, log 1856-2 = 3-2686248, Z sin 34^19' = 9-7510991, Z: sin 340 20' = 9-75 12842. EXAMPLES. LX. 189 (2) Giveu that a - 843-2 feet, C= 900, and ^ = 34« 15' ; liud c. log 843-2 = 2-9259306, L cosec 34^ 15' = 102496421, log 1-4982 = -17557. (3) Given that a = 4845 yards, 6 = 4742 yards, and C''^90, d^l. log 4845 = 3-6852938, log 4742 = 3-6759615, L tan 450 36' = 100090965, L tan 450 37' = 100093492. (4) Given that c=8762 feet, C=90, and ^=37M0', find a and b. ^_ log 8762 = 3-9426032, Z sin 37M0' = 9*781 1344, ^■^ Z cos 37M0' = 9-9013938, log 5-2934 = -72373, ^^^^ log 6 9823 = -843997. ^^^TO Given that 6 = 1694-2 chains, (7=90", and J = 18M7', ..ad a. log 1694-2 = 3-2289647, L cot 18<» 47' = 10-4683893, log 5-7620 = -76057. (6) Given that a = 1072 chains, c=4849 chains, and (7=90^ find 6. log 5921 = 3-7723951, log 3777 = 3-577 1470, log 4-729 = -67477. (7) Given that 6 = 841 feet, c=3762 feet, and (7=90®, find a. log 4603 = 3-66304 10, log 2921 =34655316, log 3-6668 = -56428. (8) Given that a = 7694-5 chains, 6 = 8471 chains, (7=90®, find A and c. log 7694-5 = 3-8861804, log 8471 =39279347, L tan 420 15' = 995824, L cosec 42° 15' = 101723937, log 1-1 444 = -05857. 230. In the following examples the student must find the necessary logarithms from the Tables. lyO TRIGONOMETRY. * MISCELLANEOUS EXAMPLES. LXI. (1) . A balloon is at a height of 2500 feet above a plain and its angle of elevation at a point in the plain is 40° 35'. How far is the balloon from the point of observation ? (2) A tower standing on a horizontal plain subtends an angle of 37*^ 19' 30" at a point in the plain distant 369*5 feet from the foot of the tower. Find the height of the tower. (3) The shadow of a tower on a horizontal plain in the sim- light is observed to be 176"23 feet and the elevation of the sun at that moment is 33° 12'. Find the height of the tower. (4) From the top of a tower 163 '5 feet high by the side of a river the angle of depression of a post on the opposite bank of the river is 29° 47' 18". Find the distance of the post from the foot of the tower. (5) Given a=673-12, 6=415-89 chains, C=90°, find A and B. (6) Given a=576-12, c=873-14 chains, (7=90°, find h and A. (7) From the top of a light-house 11 2 '5 feet high, the angles of depression of two ships, when the line joining the ships points to the foot of the light-house, are 27° 18' and 20° 36' respectively. Find the distance between the ships. (8) From the top of a clifi" the angles of depression of the top and bottom of a light-house 97-25 feet high are observed to be 23° 17' and 24° 19' respectively. How much higher is the cliflf than the light-house ? (9) Find the distance in space travelled in an hour, in con- sequence of the earth's rotation, by St Paul's cathedral. (Lati- tude of London =51° 25', earth's diameter =7914 miles.) (10) The angle of elevation of a balloon from a station due south of it is 47° 18' 30", and from another station due west of the former and distant 671-38 feet from it the elevation is 41° 14'. Find the height of the balloon. DIGRESSION ON LOGARITHMS. 190 (i) ^^^ On the four following pages are given a Table of the I^Bntissse of the logarithms of all numbers from 100 to 1000 ©orrect to 5 significant figures. By the aid of the principle of proportional parts this ible can be used to work correctly to 4 significant figures. It is only in observations or measurements, made with the most elaborate care and under special circumstances, that a greater degree of accuracy is attained than that indicated by 4 significant figures [see Lock's Arithmetic, chapter viii.] ; hence the Table here given will be found sufficient for most numerical calculations in which Arithmetic only is solved. Example i. Find to three significant figures the length of the diagoiial of a cube whose side contains 147 inches. Let X be the number of inches in the diagonal, then a;2 = 3x (14-7)2, .-. x=^/3xl4•7; /. log :c=-^ log 3 + log 14-7 = ^ (-47712) + 1-16732 [from the Table] = -23856 + 1 -16732 = 1 -40588 =log 25-45 nearly [from the Table]. Hence the diagonal is 25*45... inches. Example ii. Find the value of (43-72)^. ^tt^t!?!^! difference for K)l=00099, log 4-38 = -64147/ .-. difference for -002 = -00020; .-. log 43-72=1-64068; /. log(43-72)* = f xr64068 = -9115... = log 8-156... [from the Table]. Therefore (43-72)* = 8-156... 190 (ii) TABLE OF THE LOGAiilTHMS OF NUMBERS FROM 100 TO 1000 ALL No. Log. No. Log. No. Log No. Log. No. Log. lOO 00000 143 15534 186 26951 229 35983 272 43457 lOI 00432 144 15836 187 27184 230 36172 273 43616 I02 00860 145 16137 188 27416 231 36361 274 43775 103 01284 146 16435 189 27646 232 36549 275 43933 104 01703 147 16732 190 27875 233 36736 276 44091 105 02119 148 17026 191 28103 234 36922 277 44248 106 02531 149 17319 192 28330 235 37^07 278 44404 107 02938 150 17609 193 28556 236 37291 279 44560 loS 03342 151 17898 194 28780 237 37475 280 44716 109 03743 152 18184 195 29003 238 37658 281 44870 110 04139 153 18469 196 29226 239 37840 282 45025 III 04532 154 18752 197 29447 240 38021 283 45179 1 12 04922 155 19033 198 29667 241 38202 284 45332 1^3 05308 156 19312 199 29885 242 38382 285 45484 114 05690 157 19590 200 30103 243 38561 286 45637 "5 06070 158 19866 201 30320 244 38739 287 45788 ii6 06446 ^59 20140 202 30535 245 38917 288 45939 117 06819 160 20412 203 30750 246 39094 289 46090 118 07188 161 20683 204 30963 247 39270 290 46240 119 07555 162 20951 205 31175 248 59445 291 46389 120 07918 163 21219 206 31387 249 39620 292 46538 121 08279 164 21484 207 31597 250 39795 293 46687 122 08636 165 21748 208 31806 25^ 39967 294 46835 123 08991 166 22011 209 32015 252 40140 295 46982 124 09342 167 22272 210 32222 253 40312 296 47129 125 09691 168 22531 21 1 32428 254 40483 297 47276 126 10037 169 22789 212 32634 255 40654 298 47422 127 10380 170 23045 213 32838 256 40824 299 47567 128 10721 171 23300 214 33041 257 40993 300 47712 129 11059 172 23553 215 33244 258 41162 301 47857 130 ii394 173 23805 216 33446 259 41330 302 48001 J31 11727 174 24055 217 33646 260 41497 303 48144 132 12057 175 24304 218 33846 261 41664 304 48287 133 12385 176 24551 219 34044 262 41830 305 48430 134 12710 177 24797 220 34242 263 41996 306 48572 '35 13033 .78 25042 221 34439 264 42160 307 48714 136 13354 179 25285 222 34635 265 42325 308 48855 137 13672 180 25527 223 34830 266 42488 309 48996 138 13988 181 25768 224 35025 267 42651 310 49136 ^39 14301 182 26007 225 35218 268 42813 3" 49276 140 14613 183 26245 226 35411 269 42975 312 49415 141 14921 184 264S2 227 35602 270 43136 313 49554 142 15229 185 26717 228 35793 271 43297 314 49693 LOGARITHMS. VJO (iii) ^.. Log. No. Log. No. Loij. No. Log. No. Log. 315 49831 361 55751 407 60959 453 65610 499 69810 316 49969 362 55871 408 •61066 454 65705 500 69897 317 50106 363 55991 409 61172 455 6^801 501 69984 3»8 50243 364 561 10 410 61278 456 65896 502 70070 3»9 50379 365 56229 411 61384 457 65992 503 70157 320 50515 366 56348 412 61470 458 660S7 504 70243 3" 50651 367 56467 4'3 61595 459 66181 505 70329 3" 507«6 368 56585 414 61700 460 66276 506 70415 323 50920 369 56703 415 61S05 461 66370 507 70501 3^4 51055 370 56820 416 61909 462 66464 508 70586 325 51188 371 56937 417 62014 463 66558 509 70672 326 51322 372 57054 4.8 62118 464 66652 510 70757 327 51455 373 57171 419 62221 465 66745 511 70842 3^8 51587 374 57287 420 62325 466 66839 512 70927 329 51720 375 57403 421 62428 467 66932 513 71011 330 51851 376 57519 422 62531 468 67025 5'4 71096 331 51983 377 57634 423 62634 469 67117 515 71181 332 52114 378 57749 424 62737 470 67210 516 71265 333 52244 379 57864 425 62839 471 67302 '"'l 71549 334 52375 380 57978 426 62941 472 67394 518 71433 335 52504 381 58093 427 63043 473 67486 519 71517 336 52634 382 58206 428 63144 474 67578 520 71600 337 52763 383 58320 429 63246 475 67669 521 71684 338 52892 384 58433 430 63347 476 67761 522 71767 339 53020 385 58546 431 63448 477 67852 523 71850 340 53148 386 58659 432 63548 478 67943 524 71933 341 53275 387 58771 433 63649 479 68034 525 72016 34' 53403 388 58883 434 63749 480 68124 526 72099 11343 53529 389 58995 435 63849 481 68215 527 72181 I344 53656 390 59106 436 63949 482 68305 528 72263 1345 53782 391 59218 437 64048 483 68395 529 72346 I346 53908 392 59329 438 64147 484 68485 530 72428 I347 54033. 393 59439 439 64246 485 68574 531 72509 348 541$ 394 595. "^o 440 64345 486 68664 532 72591 349 54283 395 59660 441 64444 487 68753 533 72673 350 54407 396 59770 442 64542 488 6884a 534 7^754 351 54531 397 59879 443 64640 489 68931 535 72835 35^ 54654 398 59988 444 64738 490 69020 536 72916 353 54777 399 60097 445 64836 491 69108 537 72997 354 54900 400 60206 446 64933 492 69197 538 73078 355 55023 401 60314 447 65031 493 69285 539 73159 356 55145 402 60423 448 65128 494 69373 540 73239 357 55267 403 60530 449 65225 495 6946 1 541 73320 358 55388 404 60638 450 65321 496 69548 542 73400 359 55509 405 60746 451 65418 497 69636 543 73480 360 55630 406 60853 452 65514 498 69723 544 1 73560 j L. E. T. 1:; 190 (iv) LOGARITHMS. No. Log. No. Log. No. Log. No. Log. No. Log. 545 73640 591 77159 637 80414 683 83442 729 86273 546 73719 592 77232 638 80482 684 83506 730 86332 547 73799 593 77305 639 80550 685 83569 731 86392 548 73878 594 77379 640 80618 686 83632 732 86451 549 73957 595 77452 641 80686 687 83696 733 86510 550 74036 596 77525 642 80754 688 83759 734 86570 551 74115 597 77597 643 80821 689 83822 735 86629 552 74194 598 77670 644 80889 690 83885 736 86688 553 74273 599 77743 645 80956 691 83948 737 86747 554 74351 600 77815 646 81023 692 8401 1 738 86806 555 74429 601 77887 647 81090 693 84073 739 86864 556 74507 602 77960 648 81158 84136 740 86923 557 74586 603 78032 649 81224 695 84I98 741 86982 558 74663 604 78104 650 81291 696 84261 742 87040 559 74741 605 78176 651 81358 697 84323 743 87099 560 74819 606 78247 652 81425 698 84385 744 87157 561 74896 607 78319 653 81491 699 84448 745 87216 562 74974 608 78390 654 81558 700 84510 746 87274 563 75051 609 78462 655 81624 701 84572 747 87332 564 75128 610 78533 656 81690 702 84634 748 87390 565 75205 611 78604 657 81757 703 84696 749 87448 566 75281 612 78675 658 81823 704 84757 750 87506 567 75358 613 78746 659 81889 705 84819 751 87564 568 75435 614 78817 660 81954 706 84880 752 87622 569 75511 615 78888 661 82020 707 84942 753 87680 570 75587 6r6 78958 662 82086 708 85003 754 87737 571 75664 617 79029 663 82151 709 85065 755 87795 572 75740 618 79099 664 82217 710 85126 756 87852 573 75815 619 79169 665 82282 711 85187 757 87910 574 75891 620 79239 666 82347 712 85248 758 87967 575 75967 621 79309 667 82413 713 85309 759 88024 576 76042 622 79379 668 82478 714 85370 760 8808 r 577 76118 623 79449 669 82543 715 85431 761 88138 578 76192 624 79518 670 82607 716 85491 7|2 763 88196 579 76268 625 79588 671 82672 717 85552 88252 580 76343 626 79657 672 82737 718 85612 764 88309 581 76418 627 79727 673 82802 719 85673 765 88366 582 76492 628 79796 674 82866 720 85733 766 88423 583 76567 629 79865 675 82930 721 85794 767 - 88480 584 76641 630 79934 676 82995 722 85854 768 88536 585 76716 631 80003 677 83059 723 85914 769 88593 586 76790 632 80072 678 83123 724 85974 770 88649 587 76864 633 80140 679 83189 725 86034 771 88705 588 76938 634 80209 680 83251 726 86094 772 88762 589 77012 635 80277 681 83315 727 86153 773 88818 590 77085 636 80346 682 83378 728 862.3 774 88874 LOGARITHMS. 190 (v) No. Log. No. Log. No. Log. No. Log. No. Lor. 775 88930 820 91381 865 93702 910 95904 955 98000 776 88086 821 9H34 866 93752 911 95952 956 98046 777 MD4a 822 91487 867 93802 912 95999 957 98091 778 8^098 823 91540 868 93^52 913 96047 95S 98137 779 89154 824 91593 869 93902 914 96095 959 98182 780 89209 -"^25 91645 870 93952 915 96142 960 98227 781 89266 826 91698 871 94002 916 96190 961 98272 78a 893^1 827 91751 872 94052 917 96237 962 98318 783 89376 828 91803 873 94101 918 96284 963 98363 784 89432 829 91855 874 94151 919 96332 964 98408 785 89487 830 91908 S75 94201 920 96379 965 98453 786 8954a 831 91960 876 94250 921 96426 966 98498 787 89597 832 92012 877 94300 922 96473 967 98543 788 89653 833 92065 878 94349 923 96520 968 98588 789 89708 834 92117 879 94399 924 96567 969 98632 790 89763 835 92169 880 94448 925 96614 970 98677 791 89818 836 92221 881 94498 926 96661 971 98722 79a 89873 837 92273 882 94547 927 96708 972 98767 793 89927 838 92324 883 94596 928 96754 973 98811 794 89982 839 92376 884 94645 929 96801 974 98856 795 90037 840 92428 885 94694 930 96848 975 98900 796 90091 841 92480 886 94743 931 96895 976 98945 797 90146 842 92531 887 94792 932 96941 977 98990 798 90200 843 92583 888 94841 933 96988 978 99034 799 90255 844 92634 889 94890 934 97035 979 99078 Hoo 90309 845 92686 890 94949 935 97081 980 99123 801 90363 846 92737 891 9498S 936 97128 981 99167 80a 90417 847 92788 892 95036 937 97174 982 9921 1 803 90473 848 92840 893 95085 938 97220 983 99255 804 90526 849 92891 894 95134 939 97267 984 99300 805 905 So 850 92942 895 95182 940 97313 985 99344 806 90634 851 92993 S96 95231 941 97359 986 99388 807 90687 852 93044 897 95279 942 97405 987 99432 808 90741 853 93095 898 95.^2 8 943 97451 988 99476 800 90795 854 93146 899 95376 944 97497 989 99520 810 90849 855 93197 900 95424 945 97543 990 99564 811 90902 856 93247 901 95472 946 97589 991 99607 8n 90956 857 93298 902 95521 947 97635 992 99651 l'^ 91009 858 93349 903 95569 948 97681 993 99695 814 91062 c^9 93399 904 95617 949 97727 994 99739 I'i 91116 860 93450 905 95665 950 97772 995 99782 816 91 169 861 93500 906 95713 951 97818 996 99826 817 91222 862 93551 907 95761 952 97864 997 99870 8 18 91275 863 93601 908 95809 953 97909 998 99913 819 91328 864 93651 909 95856 954 97955 999 99957 13—2 190 (vi) TRIGONOMETRY. EXAMPLES. LXI.a. Prove that the following statements are correct to four signi- ficant figures : (2) ^802 = 3-809. (4) (451)^ X (231)^ = 55460. (6) 2iZ?)!=.04023. (41-25)^ ^^) :^>< (89130)^ = 1165. ^-) ;/{3^}=-- S13 X 455 \i 27x39 ) (12) (?^iii5!^* = 823-6. (1) 4/451 = 7-669. (3) (273)^ X (234)^ = 47-32. (5) (1^7 = 1.03. (7) (^^8281. (•0045)^ 4^3' Solve the equations correct to 4 figures : (13) 10«=421 [^=2-624]. (14) (fi)^=3 [.r=22-61]. (15) (|^§)2»==2 [^=23-28]. (16) (ff)-=3 [.2;=28-01]. (17) log37»'+3 = 3-4i2 [^= - -8243]. (18) a?=10^(31-2)[:i7=31-48]. Example i. i^iTic? ^Ae amount at Compound Interest on £,\ for 8 years at 5 per cent. To find the amount for 1 year we multiply by ^, i.e. The amoimt for 2 years will be ^|^ x f^ and the amount for 8 years = (|^)8. Let X be the required amount in pounds, then .'. log J?; = 8 (log 21 - log 20) = 8 (1-32222 - 1-30103)= 8(-02119) = -16852 =log 1-474... Hence, to find the amount at Compound Interest for 8 years at 5 per cent, we multiply the Principal expressed in pounds by 1-474+.. . ^^^^sample DIGRESSION ON LOGARITHMS. 190 (vii) mple ii. In how many years will the Principal he doubled at 5 per cent. Compound Interest ? Let X be the number of years, then »(f^)* is the amount at the end of x years, ce (5J)'=2, ,(.og.X-Iog.O)=log.,.-. .= 1^ = 1^=14.. Example iii. Find the present value of an annuity of £100 a. year payable yearly for 10 years, the first paymemX to be at the end of a year. Compound Interest being reckoned at 4 per cent, per annum. The principal which will amount to £1 at the end of n years at 4 per cent. Comp. Int. is [See Lock's Arithmetic^ p. 193, Ex. iv.] Hence the required present value is II £100{r+r2+r3 + ...+rio}, where r=^8J oc £100 i-^' - £100=^100 {1 - (i^)"} X 20 - £100. Tog (^^)" = 11 Gog 100 - log 104) = (2 - 201703) x 11 -=(1-98297) X 11 = 1-81267 = log -6496... .-. the required present value =£100 x -3503... x 26 - £100, that is jCSIO about. EXAMPLES. LXLb. (1) Find the Compoimd Interest on £100 for 10 years at 4 per cent [Result, £48.] (2) Find the Compoimd Interest on £1 for 8 years at 5 per cent [Result, 9«. 6^.] (3) In how many years will a sum of money be doubled at 3 per cent Compound Interest? [Result, 234.] 190 (viii) TRIGONOMETRY. (4) In how many years will a sum of money be doubled at 4 per cent. Compound Interest? [EesiUt, 17"7.] (5) Find the present value of £100 to be paid 8 years hence reckoning Compound Interest at 4 per cent. [Resvlt, £73*07.] (6) If the number of births in a town are 25 per 1000 and the deaths 20 per 1000 annually, in how many years will the population be doubled ? [After 140 years.l (7) On the birth of an infant £1000 is invested at Com- poimd Interest in the Funds (3 per cent, payable half-yearly) ; calculate what it will be worth when the child is 21 years old. [Result, £1869.] (8) In what time will a sum of money treble itself at 3 per cent. Compound Interest payable half-yearly ? [Result J 36-9 ^/edrs.] (9) A sum of 1 shilling lent on condition of 1 penny interest being paid monthly, accumulates at Compound Interest at the same rate for 12 years ; what will be then the amount? [Result, about £5066.] (10) A man puts by 2d. at the end of the second week of the year, 4c?. at the end of the fourth week, 8c?. at the end of the sixth week ; what sirna would be put by for the last fortnight in the year ? [About 67,100,000 pevice.] (11) A train starting from rest has at the end of 1 second velocity -001 ft. per sec. and at the end of each second its velocity is greater by one-third than at the end of the preceding second; find the velocity in miles per hour at the end of 25 seconds. [*679 miles per hour.] (12) The volume of a sphere is ^x(cube of the radius); find the diameter of the sphere which contains a cubic yard. [Result, 1'24 i/ards.] (13) Find the present value at 4 per cent, per annum Com- pound Interest of a Fellowship of £250 a year for six years, payable half-yearly, the first payment being due in six months. [Remit, £1322.] CHAPTER XVI. THE Relations between the Sides and Angles OF A Triangle. 231. The three sides and the three angles of any triangle, are called its six paxts. By the letters -4,-6, C we shall indicate jeometrically, the three angular points of the triangle ABC ; jd>raically, the three angles at those angular points re- jctively. By the letters a, b, c we shall indicate the measures of the sides BC, CA, AB opposite the angles A, B, C respec- tively. 232. I. We know tiiat, A + B + C= 180". [Euc. i. 32] 233. Also if -4 be an angle of a triangle, then A may have any value between 0" and 180°. Hence, (i) sin A must be positive (and less than 1), (ii) cos A may be positive or negative (but must be numerically less than 1), (iii) tan A may have any value whatever, positive or negative. 192 TRIGONOMETRY. 234. Also, if we are given the value of (i) sin A, there are two angles, each less than 180", which have the given positive value for their sine. (ii) cos A, or (iii) tan A, then there is only one value of Ay which A^alue can be found from the Tables. 235. :f + f + ^ = 90^ Therefore ^ is less than 90", A and its Trigonometrical Ratios are all positive. Also, -^ is known, when the value of any one of its Ratios is given. Similar remarks of course apply to the angles B and G. Example 1 . To prove sin (^ + -S) = sin G. A + B + G=im .-. A + B=im-C, and .'. sin(^ + J5) = sin(1800-(7)=sina [p. 104.] Example 2. To prove sin — ^ — = cos - , ^+|±^=90«. .-.^^ = 900-1 Now ^^^ =900. ... _ZL =9oo_^, and .•. sin^^ = sin ^QO^ - |V cos|. [Art. 118.] Example 3. To prove ■ . ' jy - ^ . A B C smJ[ + smi5+sm(;=4cos^ . cos- . cos-. Now sin^+sin^=2 sin — — .cos ^^a~^l^]/l'';1W-^- ^^"^-^ f = 2 cos I . cos ^ig-^. ^* '■ /JArt. 118.] O O and sin (7=2 sin - . cos - . ;/'ii[Art. 166.] Z A ;l ^^ = 2 cos ^^ . cos ? /■[Art. 118.] A A '1' I SIDES Avn ANGLES OP A TRIANGLE. 193 A + ain5-j-sm(7 = 2co8 - . cos - +2 cos ^. cob , C{ A-B ^ A+B\ = 2 003-2 (cos— 2 - +COS— y-j. = 2 cos - (2 cos — . cos - j . [Art. 157.] ABC = 4 cos — . cos -r^ . cos ^ . Q.E.D. Z 2s i2 EXAMPLES. LXn. Find A from each of the six following equations, A being an ;le of a triangle. (1) cos.l=i. (2) co8^=-i. (3) ain^=^. (4) tan^=-l. (5) sin^ = 4= (6) tan^=-V3. Prove the following statements, A^ B, C being the angles of a triangle. (7) mn(A + B+C) = 0. (8) coa(A + B + a)=^ -I. (9) »in^-±?=,. (10) cos4±|±^=o. (1*1) tan(^ + 5)=-tana (12) cot -■^-?= tan 4- ^ 2 (13) co6(^+^)=-co8a (14) 008(^ + ^-0= -cos 2a (15) tan^-cot5=cosC sec^ . cosec^. ,,_. sin^-sin5 . C . A-B „,, BID 3j5 - sin 3(? ^ 3A ABC (18) aiii^+8Uii^-i4mC*=4sii:.- . am- . CO8-. 194 TJtIQONOMETRY. LXII. ABC * (19) sin J. - sin ^ + sin (7= 4 sin — . cos— . sin- . A A 2i /'^n\ ' ^ A . B B . G G (20) sin - . cos 2 +sin - . cos ^ +sin - . cos - o A B G = 2 cos — . cos -^ . cos — . Z Z 2 A B G (21) cos ^ + cos5 + cos(7-l=4sin— . sin-^ . sin^; AAA (22) cos2-+cos2- -cos2- = 2cos- .COS -. sin-^. (23) sin2:^ -sin2|+sin2 ?=1 -2 cos^ . sin| . cos? f^A\ • B+G- A . . G+A-B^ . A+B-G , (24) sin + sm +giii — I-^ 1 ^ . A . B . G = 4sin- .sm-.sm-. (25) sin 2^ + sin 25+ sin 2C= 4 sin ^ . sin5 . sin (7. (26) sin J. . cos^ -sin5 . cos5+sin(7. cos(7 v =2 cos J^ . siniS . cosC. (27) sin(J5 + C-^)-sin(C+^-5) + sin(^ + 5-(7) = 4 cos J. . sin5 . cos (7. (28) cos 2A + cos 25 + cos 2(7= - 1 - 4 cos 4 . cos B . cos G. (29) sin^ A - sin^ B + sin^ C= 2 sin ^ . cos J5 . sin G. (30) cos(i? + {7-^) + cos(a+^-5)-cos(^ + 5-C0 + l = 4sinJ[ . sin5 . cosC. ,^,. .A B G . B G A (31) sin - . cos -3- . cos -^ +sin - . cos - . cos - . G A B . A . B . G , + sin - . cos 2" . cos — = sm 2 . sin - . sin -+ 1. (32) tan A + ta.nB + tan C= tan ^ . tan 5 . tan G. (33) tan— . tan - + tan - . tan - +tan — . tan -^ = 1. SIDES AND ANGLES OP A TRIANGLE. 195 16. UABC be a right-angled triaugle having C = 90°, (A + B) = 90°. A Hence, sin il = sin (90° -B) = cobB. Also, sin A= — = cos Bj c and so on. (See page 52.) b EXAMPLES. LXin. In a right-angled triangle ABCy in which C is a right angle, prove the following statements. (1) tan^=cotA (3) 8in2il=sin25. (6) sin 2^=^. (2) tan 5 = cot ^-h cos C. (4) cos2^+cos25:=0. a ^ (6) co8ec25=5T + (7) co8 2il (9) 8in« 6«-a« B _c-a 2" 2c" (8) cos 25=- 26 • 2a sin* A - sin* B sin*il + sin*5 (10) coe*^=-^, (^2) ^6=*^-2-' ,„. / J^ . Ay a+c (11) (^cos-^sin^j =--. (13) 8in(.4-5) + co82^ = 0. (14) 8in(^ -5)-|-sin(2^ + C)=a (15) (sin 4 - sin 5)2 + (cos J + cos 5)2 =2. ;sini4 ^cos25' (16) A±ft + ^/^^4Si V a -6 V a-l-6 Vcos 196 TRIGONOMETRY. 237. II. To prove a=bcoaC + gcos B. From Af any one of the angular points, draw AD per- pendicular to BC, or to BC produced if necessary. There will be three cases. Fig. i. when both B and C are acute angles : Fig. ii. when one of them (B) is obtuse ; Fig. iii. when one of them (B) is a right angle. Then, Fig. i. -^— = cos A CD ; or, CD = h cos C; CA DTi and -j-rr = cos ABD ; or, DB - c cos B, AB . '. a = CD + DB = bcosC + c cos B. Fig. ii. ^ - = cos AGD ; or, GD = h cos G, (^ A. ^ = cos ABD • or, BD = c cos (180» - 5), .'. a = GD-BD = bcosG-ccoB(180'-B) = 6 cos G + c cos J5. Fig. iii. a=GB = b cos C = 6 cos C + c cos B. [For, cos ^ = cos 90" = 0. ] Similarly it may be proved that, b = c cos A + a cos G ; c = a cos ^ + 6 cos ^. ■■238 thai - SIDES AND ANGLES OP A TRIANGLE. rj7 238. III. To prove that in any triangle, Uie sides a/re tional to tite sines of the opposite angles ; or, To prove a b c sin A sin B sin 6* * From A, any one of the angutar points, draw AD per- idicular to BC, or to BC produced if necessary. TJien, Fig. L AD AD = b8mC; for, -^ = sin C [Def] AC/ AD also AD = c8mB; for, -j-^ = sin B. . b sin C=csinB ; b c ' sinjB sin 6^* AD^bainC, aud ^i> = c sin ABD = c sin (180° - B) .'. AD = csin B ; \ 6 sin C = c sin B ; b c sin B sin C ' Fig. iii. AB = AC . sin C ; or, c = 6 sin C; -r^ = -Ay, . [For sin 5 = sin 90" = 1.1 sin 6 sin i? ■- ■• ^U sin ^ ~ sin ^ ' Q.E.D. 108 TRIGONOMETRY. 239. IV. To prove tlmt d? = h^^c^- ^hc cos A. Take one of the angles A. Then of the other two, one must be acute. Let B be an acute angle. From G draw CF perpendicular to BA^ or to BA produced if necessary. There will be three figures according as A is less, greater than, or equal to a right angle. Tlien, B e A B e A F B c A Fig. i. BC = OA' + AB' ~\.BA.FA', [Euc. ii. 1 3] or, a'^Jf + c'-^c.FA = b' + G'- 2cb cos A. [For FA = h. cos A.] Fig. ii. BC ==CA' + AB' + 2.BA.AF; [Euc. ii. 1 2] or, a' = b' + c' + 2cbcosFAC ^b' + c'-2bc cos A, [For F AC =^ ISO' - A.] Fig. ui. BC'= CA' + AB'; . [Euc. i.. 47] \ or, a' = b' + c'- 2bc cos A. [For cos A = cos 90" = 0.] Similarly it may be proved that b' = c' ■{■ a^ — 2ca cos J5, and that c^ = a^ + b^ - 2ab cos C. 240. V. Hence, cos A TT^ , cos n = h' + c'-a' ^ c' + a'-b' ^ a' + b' 26c 2ca cos C = 2ab SIDKS AM) A.\LiL£S oF A TlilANGLE. 199 Let 8 stand for Imlf the sum of ct, 'v '• ^'> '^^ '< {a + b+c) = 28. JTlieii, (b+c-a) = (b+ c-\-a-2a) = (28-'2a) = 2 (« - a), (c + a - 6) = (c + a + 6 - 26) = (2« - 26) = 2 (« - 6), (rt + 6 -c) = (« + 6 + c - 2c) = (2« - 2c) -2 (« -c). ,., ,. , . .1 /(8^b)(8-c) 24l'. \ 1. /- prove that sin - = ^ ^ '^^ \ A /s(8 — a) tjud cos2 = w'-^-^ 8 stands for lialf the sum of tlie sides a, b, c. 1 = 1- [Arts. 240, 166] Now, since cos A = • ^, , and cos -4 = 1 - 2 snr - ; loc 2 • 9 oi«2 i'^ />- f c- — a- 26^ = 26c-(6' + c»-a») _ aa-(6^-26c4-c^) 26c ~ 26c ^ ^ (6 -c)2 ^ {a-(6-c)}{a + (6-c )} 26c " 26c .'. Sin yl ^ / (2«-26)(2g-2c) ^ /(8~b)(8-c) 2 V 46c ~V" "6c" Again, since." i o J 6c [Art. IGGJ 2 cos -^r = 1 + cos ^ = 1 4- -^ 2 26c ,,^ _(6 + c)°-a°_ (6+c + a)(6 + c-a) ■•"^^2- 46c 46c /2 g . (2g - 2 o) _ Is . (s -a) \ 4Sr~ " y ~ be 200 TRIGONOMETRY. Similarly it may be proved that sm and that . C sm C_ {s-a){s-b) C /s{s-c) 2 - V ^ ' ""'^ - V ^6^ • NOTE. In taking the square root the positive sign only is admissible befon the V, because the half of an angle of a triangle is less than 90°. 243. VII. Since sin ^ = ./5E45E£), A /sis — a) and co«2=/^^^, COS- V^(^-^«) 244. VIII. Again, A A since, sin ^ = 2 sin — . cos ^ ; [Art. 166] .in._-2y(ir^)./j^ 2 , The letter *S' usually stands for Js {s — a){s — b) (s - c) ; so that the above may be written — — = ^ . a abc „. .- , sin B 2S sin C similarly, — ; — = —j- = . •^ b abc c It should be noticed that this result gives an independent proof of III. ; for 2>S' _ sin ^ _ sin ■B _ sin C abc a b c ' SIDES AND ANGLES OF A TRIAM.LE. L'Ol 245. IX. To prove that z . cot — = tan — -zr- ■ 6+c 2 2 Since , let each of these fractions = d. sin B sin C Then 6 = ci sin B, and c = c? sin C. b- c dsinB- d sin C sin B — sin C b + c d^nB + damG~ smB + BinC 2 sin 2 sin tan B-C 2 BT^ 2 5+a tan cos B-C tan 2 cot^ [Since tan —^ - tan ('9O" - -^Y tan B-C cot cot2- = tan-^ Q.E.D. 2 Similarly, c-a B ^ C-A . cot ^= tan — ^- c + a 2 2 a-6 C ^ A-B —- . cot -^ = tan — ^ — a + b 2 2 246. Note. In the next chapter we shall prove, that given three of the parts of a triangle (one of which must be a side), we can find the remaining three parts by means of the formulfiB of this chapter. Hence, these formula} cannot be equivalent to more than three independent equations. It is an instructive, but somewhat difl&cult problem to take three of these formuljc {e.g. .4 + B + C = 180*, b = c cos A-\-a cos C, e = h cos A-it-a cos B) and deduce all the others from them. L. E. T. 14 202 TRIGONOMETRY. 247. The following examples are important. Example 1, Suppose we are given that a = b cos C+c cos 5, j b =c cos A + a cos (7, I c =a cos^+6cosJ[. I Then, taking a times the first + b times the second - c times the third, we get a'^ + b^-c^= (ab cos C-h ac cos B) + {be cos A +ba con C) - {ca cos B + cb cos >4), = 2ab cos C. Example 2. Suppose we are given that A + B-\- (7= 180^, and that -^ — -. — . — =5 = .— 7^ f = c?l , 52^^ _ ^2 ^ c^2sin^.g+c^^sin^O-c^2sinM 26c ~ 2(^2sin^sin(7 _sin2^ + sin2(7- sin^ J. _ 2 cos^ . sin5 . sin (7 "~ 2sin^sin(7 " 2sini?.sinC^ [Ex. 29, p. 1 94, since {A + B + a) = ISO".] = cos^. 248. The formula a b = d sin A sin B sin is very frequently of use in solving examples. Example. Prove that acosJ^ + 6cos5+ccos(7=2asin^ sinC Since a = c?sin^, b = dainB, c — dainCy the above may be written o?sin^. cosJ. + c?sin5.cosjB + c?sin(7.cos(7 = 2c? sin A . sin 5 . sin C, or sin 2 A + sin 2B + sin 2C= 4 sin ^ . sin B . sin C, which is Example (25) on page 194. SIDES AND ANGLES OP A TRIANGLE. 203 249. The student is advised to make himself thoroughly liliar with the following formulse: a = 5 cos C + c cos J? (ii), ^ "-r rq^^f'^ sin A sin B sin C b' + c'-a' COS A sin 26c .(V). A_ / {s-b){s- c) 2 " V be A _ I B (« - a) 2"V 'W' (vii), I 2^ sin il = ^- Vs (s - a) (s - 6) (« - c) = T- (viii)} B-C b-c A (ix). P - EXAMPLES. LXIV. In any triangle ABC prove the following statements sin ^ + 2 sin 5 _ sin C a + 26 c~ * sin'* A-m. sin^ B _ sin^ C a^-m.b^ ^ ' (3) a cos yl + 6 cos B-c cos C= 2c cos A . cos 5. (2) (4) (a + 6) sin 5" = c ( A^B 2 (5) (6 - c) cos — =a sm —5 — . (6) .,Tl^+ ./.^^:|-T + cosC sin i^ . sin C sin C. sin A sin ii . sin i? = 2. 14—2 204 TRIGONOMETRY. LXIV. (7V asin (i? - (7) + 6 sin {G- ^)+csm (yl - i5)=0. I a—b cos B -cos A (9) c 1 + cos (7 6+c cos^+cosC a 1 - cos 2 62 sin (7+ c2 sin 5 (10) s/bc sinB . sin (7= t • ' ^ b + c (11) a + 6 + c=(6 + c)cos^4-(c+a)cos5 + (a + 6) cos (7. (12) b + c-a=(b+c) cosA-{c-a)co3B+{a-b)co3C. /-.ox X ^ a sin (7 (13) tan ^ = 7 ?y. ^ ' b-a cos (7 (14) a(62 + c2)cos^ + 6(c2 + a2)cos5 + c(a2 + 62)cos(7=3a6c. (15) acos(^ + 5 + (7)-6cos(^+4)-ccos(^ + CO = 0. ,_-. cos^ cos^ cos (7 a2^52 + c2 ,,^, tan^ a2 4.52_^2 ,,0x1. 2^. 2^ (19) taii^.tan^^j:^^:^^. J. -5 (20) tan-(6 + c-a)=tan-(c+a-6). (21) c2=(a + 6)2sin2^+(a-6)2cos2^. ** MISCELLANEOUS EXAMPLES. LXV. ^ "^ (1) Up is tlie length of the perpendicular from A on Z*(7, sin ^ = r^ . be (2) If 2 cos ^ . sin C= sin A, then i?= a 36^ a (3) If A = 3i?, then sin ^ = | ^ — ^ I EXAMPLES. LXV. 205 (4) If she sin j8 . Bin C = jH > t^en B=C. B G A , C A B (5) o cos ^ . cos ^ . cosec ^ =o cos ^ . cos ^ . cosec ^ ABC =ccos g .cos g. cosec ^=8. (6) Given cos ^ =f and cos 5= ^, prove that cos C= - \^. B(7) If sin«5 + 8in«(7=sinM, then ^ = 90°. (8) If sin 2^ + sin 2C= sin 2^, then either 5 = 90° or C= 9(V>. (9) If ^ : 5 : C=l : 2 : 5, then 1 +4cos ^ . cos5. cosC=0, and a^, 6^, c^ are in a. p. .,^, . ^ . B-C , . B . C-A ■(10) asing.sin-^ + ftsing.sm-g— I . C . A-B ^ + C8in -. sin - _ -«=0. ^^^ (11) If Z) is the middle point of BC^ prove that H 4.4i)»=262+2c2-a2. (12) Given that a = 26, and that A = 35, prove that C7= 60". (13) a6c (a cos ^ + 6cos5 + c cos (7) =8*S^. (14) If h cos^ ^ + cco8* - = — , then 6, a, c are in a.p. (15) If D, E, F are the middle points of the sides BC, CA, . itfl, then IB 4(JZ)« + J5^«+C/^2) = 3(a2 + J2 + c2). (16) If D is the middle point of BC, ooi ADB=^^~S • 4o (17) If rf, c, / are the perpendiculars from A^ B, C on the oppoaite sides of the triangle, then osin^+isin^ + csin (7= 2 (rf cos ^ 4- e cos 5 +/cos C). [Some of the Examples in the Appendiz might be worked by the student at this stage.] LL ( 20G ) CHAPTER XYII. On the Solution of Triangles. 250. The problem known as the Solution of Triangles may be stated thus : When a sufficient number of the parts of a triangle are gi'ven, to find the magnitude of each of the other parts. 251. When three parts of a Triangle (one of which must be a side) are given, the other parts can in general be determined. There are four cases. I. Given three sides. [Compare Euc. I. 8.] II. Given one side and two angles. [Euc. I. 26.] III. Given two sides and the angle between them. [Euc. I. 4.] IV. Given two sides and the angle opposite one of them. [Compare Euc. YI. 7.] Case I. 252. Given three sides, a, b, c. [Euc. I. 8; YI. 5. J We find two of the angles from the formulae ^ V s{s-a) tan|=y(^-^)(^-^) s{s-b) The third angle C =1%0-A-B. ON THE SOLUTION OP TRIANGLES. 20? 253. In practical work we proceed as follows : Iogtan^=log /(£Z.*HiZf); ^ V 8 [8 -a) L tan ^ - 10 = ^ { log (« - 6) + log (s - c) - log s - log (s - a)]. Similarly, L tan TT - 10 = J {log (s - c) + log (s - a) - log « - log (s - 6)} . i 254. Either of the formul® sin :^ = / (^ - ^) (^ - gj _ / s{8 — a) j^ ^ig^ 1^ ^gg^ ^g above. 2 V 6c "^ . -4 A The sin -jr and the cos -^ formulae are either of them as j^ venient as the tan -^ formula, when one of the angles only < to be found. If all the angles are to be found the tangent formula is convenient, because we can find the L tangents of two half angles from the 8a.me four logs, viz. log s, log (s - a), log (» — 6), log («- c). To find the L sines of two half angles we require the six logarithms, viz. log (s - a), log (s - 6), log (8 - c), log a, log 6, log c. Example. Given a=275-35, 6 = 189-28, c = 301-47 chains, find A and B. Here, a =383 -05, 5- a=107-70, «- 6 = 193-77, «-c=81-58. Then /.tan ^ = 10 + ^ {log 193-77+ log 81-58 - log 383*05 -log 107-70} = 10 + i {2-2872865 + 1 91 15837 - 25832555 - 2-0322157} =9-7916995 [from the tables], whence ^ = 3l0 45'28-5"; .-. J = 63«30' 57". 208 TRIGONOMETRY, Also Ztan| = 10 + Hlog81-58 + log 107-70 -log 383-05 -log 193-77} = 9-5366287 = Ztan 180 59' 9-8"; .-. 5=370 58' 20"; (7=1800- ^-J5 = 78^30' 43". 255. This Case may also be solved by the formula h' + c'-a' But this formula is not adapted for logarithmic calculation, and therefore is seldom used in practice. It may sometimes be used with advantage, when the given lengths of a, b, c each contain less than three digits. Example. Find the greatest angle of the triangle whose sides are 13, 14, 15. Let a = 15, 6 = 14, c = 13. Then the greatest angle is A. ,^ , 142+132-152 140 , __.^-_ Now, COS^=^r — T7i- = ^ — Tl — rr = A = -384615. ' 2x14x13 2x14x13 ^^ = cos 670 23', nearly. [By the Table of natural cosines.] .-. the greatest angle = 67^ 23'. EXAMPLES. LXVL (1) If a = 352-25, 6 = 513-27, c = 482-68 yards, find the angle A, having given log 674-10 = 2-8287243, log 321*85 = 2-5076535, log 160-83 = 2-2063401, log 191-42 = 2-2819873, L tan 200 33' = 9-5758104, L tan 200 39' = 9-5761934. ON THE SOLUTION OF TRIANGLES. 209 (2) Find the two largest angles of the triangle whose aides are 484, 376, 522 chains, having given that ^R log 6-91 = -8394780, log 3-15 = -4983106, ^H log 2-07 = -3159703, log 1 -69 = -2278867, Hp L tan 360 46' 6" = 98734581, L tan 3P 23* 9" = 97853745. (3) If a =6238, 6 = 5662, c=9384 yards, find the angles A fjB, having given log 1-0142 = -0061236, log 4-904 = '6905505, . log 4-48 =6512780, log 7*58 = '8796692, Ztan 140 38' = 9-4168099, Z tan 15^57' =9-4560641, L tan 140 39' = 9 '4 173265, L tan 150 58' = 9-4565420. (4) If a = 4090, h = 3850, c = 381 1 yards, find ^, having given log 5-8755 = -7690448, log 3*85 = '5854607, log 1 -7855 = -2517599, log 3'81 1 = '5810389, Zco8 320l5'=9'9272306, Zcos320 16' = 9'92715O0. (5) Find the greatest angle in a triangle whose sides are 7 feet, 8 feet, and 9 feet, having given log 3 = -4771213, Zcos 360 42' = 9-9040529, log 1 '4 = -146128, difi". for 60" = -0000942. (6) Find the smallest angle of the triangle whose sides are 8 feet, 10 feet, and 12 feet, having given that log 2 = -30103, Z sin 200 42' = 9-5483585, difil for60"= -0003342. I I (7) If a : 6 : c=4 : 5 : 6, find C, having given log 2 = '3010300, log3 = -4771213, Zcx)S 4 10 26' = 9-8750142, diff. for60"= -0001115. (8) The sides of a triangle are 2, »/6, and 1 + ^3, find the angles. (9) The sides of a triangle are 2, ^2 and V3- 1, find the igles. 210 TRIGONOMETRY. Case II. 256. Given one side and two angles, as a, B^ G. [Euc. I. 26; VI. 4.] First, A = 180° -B-C; which determines A. _ ^ , h a , a. sin B Next, - — ^ = - — -. , or, 6 = —. — j- ; sm B sin A sm A J c a a. sin and, -; — ^ = -^ — . , or, c = — ; — .— . sm 6 sm A smA These determine b and c. 257. In practical work we proceed as follows : a.sin^ Since b = . '. log b = log sin^ ' a. sin^ sin A . •. log 6 = log a + log (sin B) + 10- {10 + log sin A). or, log 6 = log a + L sin B - L sin A. Similarly, log c = log a + L sin G — L sin A. Example. Given that c = 1764-3 feet, C= 180 27', and ^5 = 660 39', find b. From the Tables we find log 1764-3 = 3-2465724. Z;sin 180 27' = 9-5003421, Z sin 660 39' = 9-9^28904 ; .'. log 6 = 3-2l65724 + 9-9628904 - 9-5003421 = 3-7091207 = log 5118-2; .-. 6 = 5118-2 feet. ON THE SOLUTION OF TRIANGLES. 211 EXAMPLES. LXVn. (1) If ^ = 53<» 24', ^ = 660 27', 0=338 65 yards, find C and a, having given that Zsin 530 24' = 9-9046168, log 33865 = -5297511, Z8in60« 9' = 9-9381851, log 3-1346= '4961821, log 3-1347 = -4961960. (2) If vl = 48°, 5 = 540, and c = 38 inches, find a and b, iiug given that log 38 = 1-5797836, log 2-88704= -4604527, log 3-14295= -4973368, X sin 54^ = 9-9079576, L sin 78»= 9-9904044, L sin 480 = 9-8710735. (3) Find c, having given that a = 1000 yards, A = 50®, C= 66^, and that L sin 50« = 9-8842540, L sin 66^=9 9607302, log 1-19255 = -0764762. (4) Find 6, having given that 5 = 32n5', C= 210 47' 20'. a = 34 feet. log 3-4= -531479, Zsin 320 15'=9-727228, log 2-241 = -350442, Z8in540 2' = 9-908141, log 2-242 = -350636, X8in540 3' = 9 908233. (5) Find a, 6, (7, having given ^ = 720 4', J5 = 410 56' 18", = 24 feet log 2-4 = log 1-755 =^ log 1-756 = log 2-4995 = 3802112, Zsin 720 4' =99783702, 2442771, Zsin 410 56' 10" = 9-8249725, 2445245, Z sin 410 56' 20" = 98249959, 3978531, Zsin 650 59' =9-9606739, log 2-4996 = -3978705, Zsin' = 9-9607302. 212 TRIGONOMETR Y, Case III. 268. Given two sides and the included angle, as h, c,A. [Euc. I. 4; VI. 6.] First, B + C=^ 180" - A. Thus {B + G) is determined. JN ext, tan — r: — = , cot -^ . ' 2 6 + c 2 Thus (jB - G) is determined. And B and (7 can be found when the values of [B + G) and (5 - G) are known. -. ., a h 6. sin -4 Lastly, -T — 7 = - — jc , or a = — . ~ -„ . *" sm ^ sm jB ' Bin B Whence a is determined. 259. In practical work we proceed as follows ; B-G h-c ^A Smcetan— 2-=5^^««t2, (ij / '\ tan — 9- )+ 10 = log {h-c)- log (6 + c) + log ^cot -^ j + 10, B — C A or, L tan —r — = log {h-c)- log (6 + c) + Z cot — . Z> . sin ^4 Also, since a - — -. — ^ , ' sin^ .*. log a = log 6 + Z sin ^ - Z sin By as in Case II . ON THE SOLUTION Of TRIANGLES. 213 Example. Given 6 = 45612 chains, c=29&'8^ chains, and A = 74" 20', find the other angles. Here, 6- c= 159-26, 6 + c = 752-98. From the Table we find, ^ . . log 159-26 = 2-2021067, and log 752- 98 = 2-8767884, L cot 37»'l%' = 10-1202593 ; L tan ^^^=2-2021067 - 2-8767834+ 101202593 = 9-4455826 = X tan 15" 35' 18". .-. B-C=Zin(yz&\ and 5 + (7=1800-74° 20'. Thus 5+ (7=1050 40'; .-. 2^= 1360 50' 36" ; 2(7= 74° 29' 24", i? =680 25' 18" ; or, (7= 370 14' 42". 260. Tlie formula a*=b' + c'- 2bc cos A may be used in simple cases; or it may be adapted to logarithmic calculation by the use of a subsidiary angle. [CI Example LXXVI. (20), p. 256.] Example. If 6 = 35 feet, c=21 feet, and ^ = 500, find a, given that cos 60« = -643. Here 02 = 352 + 21-^-2x35x21x008500; .-. ^^=52 + 32 - 2 X 5 X 3 X cos 500, =25 + 9 - 30 X -643, =1471. .-. 1=3-82 nearly ; or, a = 26-74=about 26J feet. EXAMPLES. LXVm. (1) Find B and C, having given that ^ = 40^ 6 = 131, c = 72. log 5-9 =-7708520, Z cot 20" =10-4389341, log 2-03= -3074960, Z tan 380 36'= 99021604, Z tan 380 37'= 9-9024105. 214 TRIGONOMETRY. LXVIII. (2) Find ^ and B, having given that a = 35 feet, 6 = 21 feet, C=500. log 2= -301030, Z tan 28° 11' = 9729020, Z tan 650= 10-331327, Z tan 280 12'== 9-729323. (3) If 6 = 19 chains, c=20 chains, ^ = 60^, find B and (7, having given that log 3'9= '591065, Z tan 2^ 32' = 8-645853, Z cot 300=10-238561, Ztan 20 33' =8 -648704. (4) Given that a = 376*375 chains, 6 = 251-765 chains, and C= 780 26', find A and B. L cot 39^ 13' = 10-0882755, log 1-2461 = -0955529, Ztan 13^39'= 9-3853370, log 6-2814 = -7980565, Ztanl30 40'= 9-3858876. (5) If a= 135, 6 = 105, (7=600, find A, having given that log 2 = -3010300, Ztan 12^ 12' = 9-3348711, log 3 = -4771213, Ztan 12^ 13' = 9-3354823. (6) If a = 2 1 chains, 6 = 20 chains, C= 60^, find c. (7) Find c in the triangle of example (5). (8) In a triangle the ratio of two sides is 5 : 3 and the in- cluded angle is 70° 30'. Find the other angles. log 2 = -3010300, Z cot 350 15' =10-1507464, Z tan 190 28' 50" = 9-5486864. Case IV. 261. Given two sides and the angle opposite one of them, as 6, c, B. [Omitted in Euc. I; Euc. YI. 7.] T^. ^ . c 6 . ^ csin^ Jb irst, since - — 7; = -. — = : . *. sm C - — 5 — . sm (7 sin -g G must be found from this equation. ON THE SOLUTION OP TRIANGLES. 215 When C is known, il = 180" - J5 - (7, 6sin^ sin^ Wliich solves the triangle. 262. We remark however that the angle C, found from le trigonometrical equation sin C = a given qucmtity^ rhere C is an angle of a triangle, has two values, one less 90", and one greater than 90". [Art. 234.] The question arises. Are both these values admissible? This may be decided as follows : If ^ is not less than 90°, C must be less than 90* ; and le smaller value for C only is admissible. If ^ is less than 90° we proceed thus. V * J) 1 . If 6 is less than c sin J?, then sin C, which = — - — , — b greater than 1. This is impossible. Therefore if h is less than c sin .6, there is no solution whatever. 2. If 6 is equal to c sin B, then sin C = 1, and therefore C = 90°; iiud there is only one value of C, viz. 90°. 3. If h is greater than c sin Bj and less than c, then B is less than C, and C may be obtuse or acute. In this C may have either of the values found from the equa- on sin C = — , — . Hence there are two solutions, and the 6 ■iangle is said to be ambig^OUS. I^^io ^^^ 4. If 6 is equal to or greater than c, then B is equal to ^^ftr greater than C, so that C must be an acute angle ; and the smaller value for C only is admissible. 216 TRIGONOMETRY. 263. The same results may be obtained geometri- cally. ConstructioiL Draw AB = c', make the angle ABD = the given angle B; with centre A and radius = 6 describe a circle; draw AD perpendicular to BD. Then AD = c sin B. 1. If 6 is less than csin J?, i.e. less than AD, the circle will not cut BD at all, and the construction fails. (Fig. i.) 2. If h is equal to AD^ the circle will touch the line BD in the point Z), and the required triangle is the right-angled triangle ABD. (Fig. ii.) 3. If h is greater than AD and less than AB, i.e. than c, the circle will cut the line BD in two points C„ C^ each on the same side of B. And we get two triangles ABC^, ABG^ each satisfying the given condition. (Fig. iii.) 4. If b is equal to c, the circle cuts BD in B and in one other point C; if b is greater than c the circle cuts BD in two points, but on opposite sides of B. In either case there is only one triangle satisfying the given condition. (Fig. iv.) ON THE SOLUTION OP TRIANGLES. 217 4. We may also obtain the same results algebrai- y, from the formula 6* = c* + a* — 2c a cos B. In this 6, c, B are given, a is unknown. Write x for a d we get the quadratic equation X* -2cco^B .x = h^ - c'. Whence, x* -Icco^B .x + c' cos' B = b'-c' + c^ cos' B = b'-c'sm*B; .: a; = c cos ^ ± ^6' - c* sin* j5. Let a, . a, be the two values of x thus obtained, then a, = c cos -5 + y/b* — c' sin* B\ a, = c cos j5 - ^6' - c' sin* B) Which of these two solutions is admissible, may be jided as follows ; 1. If 6 is less than csin B, then (6* - c* sin* -B) is nega- ive, so that a„ a, are impossible quantities. IM| 2. K 6 is equal to c sin B, then (b' - c' sin* B) = 0, and ^ ^^i = o,; thus the two solutions become one. 3. If 6 is greater than c sin B, then the two values ai, Oi are different and positive unless V6* - c* 8in*i? is > c cos jB, Xe. unless 6' - c* sin* ^ >c* cos' B, I* a unless 6' >c'. 4. If 6 is equal to c, then aj = 0; if 6 is greater than c, len a, is negative and is therefore inadmissible. In either of these cases a, is the only available solution. L. E. T. 15 218 TRIGONOMETRY. 265. We give two examples. In the first there are two solutions, in the second there is only one. J Example 1. Find A and C, having given that 6 = 379-41 chains, c= 483-74 chains, and B=M^ 11'. L sin C=log c + Z sin ^ — log h = 2-6846120 + 9-7496148 - 2-5791088 = 9-8551180 = Zsin45M5'; .-. (7=45045', or, 1800 -450 45' = 1340 15'. Since h is less than c, each of these values is admissible. When (7= 450 45', then .4 = 1000 4'. When(7=134«15', then J = ll0 34'. Example 2. Find A and (7, when 6=483-74 chains, c = 379*41 chains, and J5 = 340ll'. L sin C= log c + Z sin 5 - log 6 = 2-5791088 + 9-7496148 - 2-6846120 = 9-6441116 = Z sin 260 9'; .-. 6^=260 9', or, I8O0 -26*'9' = 1530 51'. Since h is greater than c, C must be less than 900, and the larger value for C is inadmissible. [It is also clear that (1530 51' + 34^ 11') is > I8OO]. .-..(7=260 9', ^1 = 119040'. EXAMPLES. LXIX. ^^^ ( 1 ) If i? = 400, 6 = 140-5 feet, a = 1 70-6 feet, find A and C. log 1-405 = -1476763, Zsin400 =9-8080675, log 1 -706 = -231 9790, Z sin 510 18' = 98923342, Zsin 510 19' = 9-8924354. EXAMPLES. LXIX. 19 a-97 (2) Find D and C, having given that A =50°, 6 = 119 chains, 97 chains, and that log 1 -19 = -075547, L sin 50^ = 9-884254, log 9-7 =-986772, Xsin700 =9-972986, X sin 700 1' = 9-973032. (3) Find B, C, and c, ha\ang given that .^=50^, 6 = 97, 119 (see example (2)). log 1*553 = -191 169, Z sin 380 38' 24" = 9-795479, L sin 880 38^24" = 9-999876. (4) it Find A J having given that a = 24, c=25, C=650 59', and log 2-5 = -3979400, L sin 650 59' =99606739, log 2-4 = -3802112, Lain 61* 16' = 9-9429335, Z sin 610 17' = 9-9430028. (5) If o = 25, c = 24, and C= 650 59', find A, B and the iter value of b. log 1-755 = -2442771, Z sin 72^4' = 99783702, log 1-756 = -2445245, Z sin 720 5' = 9-9784111, Z sin 410 56' iq" = 9-8249725, Z sin 410 56' 26" = 9-8249959 ; other logs see example (4). (6) Supposing the data foi the solution of a triangle to be ^ in the three following cases (a), O), (y), point out whether ihe solution will be ambiguous or not, and find the third side in the obtuse-angled triangle in the ambiguous case : (a) ^ = 300, a = 125 feet, c = 250 feet, 03) ^ = 300, a = 200 feet, c = 250 feet, (y) J = 300, a = 200 feet, c = 1 25 feet. log 2 = -3010300, Z sin 380 41' = 9*7958800, log 6-0389 = -7809578, Z8in80 41' =9-1789001, log 6-0390 = -7809650. 15—2 220 TRIGONOMETRY. 266. It saves a little trouble in practice when using the formula a = — ; — ^ , to write it thus ; , c=J\d, a = 2, find h. (5) Given ^4 = 600, 6 = 4^/7, c= 6 VT, find a. (6) Given A = 45®, ^ = 6(y and a = 2, find c. — (7) The sides of a triangle are as 7 : 8 : 13, find the greatest iigle. _ (8) The sides of a triangle are 1, 2, ^7, find the greatest igle. (9) The sides of a triangle are as a :b : »J{a^ + a5 + 6'), id the greatest angle. (10) When a : 6 : c as 3 : 4 : 5, find the greatest and least gles ; given cos 36<> 52' = -8. J (11) If a=5 miles, 6=6 miles, c= 10 miles, find the greatest gle. [cos 490 33' =-65.] 220 (ii) TRIGONOMETRY. :=r(12) If a = 4, b = 5, c = 8, find C; given that cos 54° 54'= -576. rs (13) The two sides a and 6 are ^3+1 and 2 respectively; C= 30* ; solve the triangle. (14) Given C=W, a = V5 + l, c-^5 - 1, find the other a.ngles. (15) If 6 = 3, (7=1200, c=Vl3, find a and the sines of the other angles. (16) Given A = 105^, B=45\ c= ^2, solve the triangle. - (17) Given B = 75^, 0= 30^, c = ^/8, solve the triangle. (18) Given ^ = 45^, c=y/1b, b = s^50, solve the triangle. (19) Given ^ = 30^ c=150, 6=50^3, show that of the two triangles which satisfy the data one will be isosceles and the other right-angled. Find the third side in the greatest of these triangles. (20) Is the solution ambiguous when J5 = 30**, c= 150, 6 = 75 ? (21) If the angles adjacent to the base of a triangle are 22^<^ and 112^^, show that the perpendicular altitude will be half the base. (22) If a = 2, 6 = 4 - 2V3, c=V6 (v/3 - 1), solve the triangle. (23) If ^ = 90, i? = 45", 6 = V6, find c. (24) Given B = Ib^, 6 = ^/S - 1, c = ^3 + 1, solve the triangle. (25) Given sin 5 = '25, a = 5, 6 = 2-5, find A. Draw a figure to explain the result. (26) Given C=15", c = 4, a =4 + ^48, solve the triangle. (27) Two sides of a triangle are 3^6 yards and 3(^3 + 1) yards, and the included angle 45^, solve the triangle. (28) If C= 30^, 6 = 100, c = 45, is the triangle ambiguous ? (29) Prove that if ^ = 45^ and ^=60*> then 26'=a(l + ^3). (30) The cosines of two of the angles of a triangle are j and |, find the ratio of the sides. ( 221 ) CHAPTER XVIII. [E Measurement of Heights and Distances. 268. We have said (Art. 97) that the measuremeiit, rith scientific accuracy, of a line of any considerable length ivolves a long and difficult process. On tlie other hand, sometimes it is required to find the irection of a line that it may point to an object which is )t visible from the point from which the line is drawn. for example when a tunnel has to be constructed. By the aid of the Solution of Triangles re can find the length of the distance between points which inaccessibl*' ; we can calculate the magnitude of angles which cannot be practically observed ; we can find the relative heights of distant and inaccessible points. The method on which the Trigonometrical Survey of a country is conducted affords the following illustration. 222 ON THE SOLUTION OF TRIANGLES. 269. To find the distance between two distant objects. .^\ \ '5 Two convenient positions A and B^ on a level plain as far apart as possible, having been selected, the distance between A and B is measured with the greatest possible care. This line AB is called the base line. (In the survey of England, the base line is on Salisbury plain, and is about 36,578 feet long). Next, the two distant objects, P and Q (church spires, for instance) visible from A and B, are chosen. The angles PAB, PBA are observed. Then by Case II. Chapter xvii, the lengths of the lines PA, PB are cal- culated. Again, the angles QAB, QBA are observed ; and by Case II. the lengths of QA and QB are calculated. Thus the lengths of PA and QA are found. The angle PAQ is observed ; and then by Case III. the length of PQ is calculated. 270. Thus the distance between two points P and Q has been found. The points P and Q are not necessarily accessible ; the only condition being that P and Q must be visible from both A and B. MEASUBBMSNT OP HE 10 NTS AND DISTANCES. 223 1271. In practice, the points F and Q will generally be cessible, and then the line FQy whose length has been cal- culated, may be used as a new base to find other distances. 272. To find the lieight of a distant object above the point of observation. P M m Let B be the point of obsen'ation ; P the distant object. From B measure a base line BA of any convenient length, in any convenient direction; observe the angles PAB^ PBA^ d by Case II. calculate the length of BP. Next observe at the * angle of elevation ' of P ; that is, the angle which the line BP makes with the horizontal line BM, i/ being the int in which the vertical line through P cuts the horizontal lane through B. Then PM^ which is the vertical height of P above B can be calculated, for PM= BP . sin MBP. Example 1. The distance between a church spire A and a milestone B is knoxcn to be 1764'3 feet ; C is a distant spire. The Kvngle CAB is 94° 54'. and the angle CBA is m^ 39'. Find the listance of C from A. ABC is a triangle and we know one side c and two angles A and B), and therefore it can be solved by Case II. The angle ACB= ISQO - 94» 54' - 660 39' = 180 27'. Therefore the triangle is the same as that solved on page 210. erefore ^(7=5118'2 feet. jKThe 224 TRIGONOMETRY. Example 2. If the spire C in tlie last Example stands on a hill J and the angle of elevation of its higJiest point is observed at A to he 4° 19'; /ind how much higher C is than A. The required height x=AC.&m 4*^ 19' [Art. 268], and AC in 5118-2 feet, .-. log j7=log {AC. sin 40 19') = log 51 18-2 + X sin 40 19' -10 = 3-7091173 + 8'8766150- 10 = 2-5857323 = log 385-24. Therefore x = 385 ft. 3 in. nearly. - EXAMPLES. LXXI. (Exercises xiv. and lxi. consist of easy examples on this subject). (1) Two str&ight roads inclined to one another at an angle of 60^, lead from a town A to two villages B and C ; ^ on one road distant 30 miles from A, and C on the other road distant 15 miles from A. Find the distance from B to C. Ans. 25-98 m. (2) Two ships leave harbour together, one sailing N.E. at the rate of 7^ miles an hour and the other sailing North at the rate of 10 miles an hour. Prove that the distance between the ships after an hour and a half is 10-6 miles. (3) A and B are two consecutive milestones on a straight road and is a distant spire. The angles ^jB(7 and BAG are observed to be 120'^ and 45° respectively. Show that the dis- tance of the spire from A is 3-346 miles. (4) If the spire C in the last question stands on a hill, and its angle of elevation at A is 15^, show that it is -866 of a mile higher than A. (5) If in Question (3) there is another spire D such that the angles DBA and DAB are 45® and 90® respectively and the angle DAC is 45^ ; prove that the distance from C to i> is 2f miles very nearly. ana MEASUREMENT OP HEIGHTS AND DISTANCES. 225 6) A and B are two consecutive milestones on a straight and C is the chimney of a house visible from both A and B. CAB and CBA ai-e observed to be 36" 18' and 120" 27' »ly. Show that C is 2639-5 yards from By log 1760 =3-2455127 Z sin 36« 18'= 97723314 log 2630-5 =3-42152 L cosec 23« 15' = 10-4036846. (7) A and B are two points ou opposite sides of a mountain, and C is a place visible fiom both A and B. It is ascertained t C is distant 1794 feet and 3140 feet from A and B respec- vely and the angle ACB is 58** 17'. Show that the angle which the line pointing from A io B makes with AC is 86** 55' 49", J log 1346 = 3-1290451 X cot 29« 8' 30" = 102537194 I log 4934 = 3-6931991 /> tan 26° 4' 19"= 96895654 (8) A and B ai-e two hill-tops 34920 feet apart, and G is the I top of a distant hill. The angles CA B and CBA are observed to be 61® 53' and 76** 49' respectively. Prove that the distance from k to C is 51515 feet, I log 34920=4-5430742 Z8in76M9'= 9-9884008 I log 51515 = 4-71193 Z cosec 41M8' = 10-1804552. .,^(9) From two stations A and B on shore, 3742 yards apart, a ship C is observed at sea. The angles BAC^ ABC are simul- taneously observed to be 72'^ 34' and 81** 41' respectively. Prove that the distance from A to the ship is 8522-7 yards, 1^^ log 3742 = 3-5731038 Zsin8P41'= 99954087 ^^B log 8522-7 = 3-9305774 Z cosec 25*> 45' = 103620649. ^^^K^ (10) The distance between two mountain peaks is known to ^^^H)e 4970 yards, and the angle of elevation of one of them when ^^^Been from the other is 9** 14'. How much higher is the first than ^Bhe second ? Sin 9® 14' = -1604555. An*. 797 5 yards. ,.^ (11) Two straight railways intersect at an angle of 60**. From their point of intersection two trains start, one on each line, one at the rate of 40 miles an hour. Find the rate of the second train that at the end of an hour they may be 35 miles apart. Aiu. Either 25 or 15 miles an hour. (Art. 264.) 226 TRIGONOMETRY. LXXI. (12) A and B are two positions on opposite sides of a mountain ; C is a point visible from A and B \ AG and BG are 10 miles and 8 miles respectively, and the angle BGA is 60*^. Prove that the distance between A and ^ is 9*165 miles. (13) In the last question, if the angle of elevation of C at ^ IS 80, and at B is 2^ 48' 24" : show that the- height of B above A is one mile very nearly. sin 80= -1391731 sin 2^ 48' 24" = -0489664. (14) Show that the angles which a tunnel going through the mountain from A to B^ in Questions (12) and (13), would make (i) with the horizon, (ii) with the line joining A and (7, are respectively 6^ 16' and 49^ 6' 24". sin 60 16' = -1091 ; tan lOO 53' 36"= '192450. ■**^15) A and B are consecutive milestones on a straight road; G is the top of a distant mountain. At A the angle GAB is observed to be 380 19' ; at B the angle CBA is observed to be 1320 42', and the angle of elevation of (7 at ^ is IQO 15'. Show that the top of the mountain is 1243*5 yards higher than B. Z sin 380 19'= 9.7923968 log 1760 = 3*2455127 L cosec 80 59' = 10*8064659 log 1243*5 = 3*09465 X sin 100 15' = 9*2502822. (16) A base line AB, 1000 feet long is measured along the straight bank of a river ; (7 is an object on the opposite bank ; the angles BAG, and CBA are observed to be 650 37' and 530 4' respectively. Prove that the perpendicular breadth of the river at G is 829-87 feet ; having given X sin 650 37'= 9.9594248, Z sin 530 4' = 9*9027289 Z:-cosec 610 19' = 10*0568589, log 8*2987 = '91901. ^MEASUREMENT UF HEIGHTS AND UJSTA/yvi. * MISCELLANEOUS EXAMPLES. LXXn. r ^^B (1) A man walkmg along a straight road at the rate of three ^^iles an hour sees, in front of him at an elevation of 60", a balloon which is travelling horizontally in the same direction at the rate B8i:!C miles an hour; ten minutes after he observes that the vation is 30*^. Prove that the height of the balloon above the id is 440^3 yards. (2) A person standing at a point A, due south of a tower built on a horizontal plain, observes the altitude of the tower to be 60*'. He then walks to a point B due west from A and observes the ^^^titude to be 45*^, and then at the point C in AB produced he ^^p^rves the altitude to be 30<^. Prove that AB^^BC. V) The angle of elevation of a balloon, which is ascending uniformly and vertically, when it is one mile high is observed to be 35<> 20' ; 20 minutes later the elevation is observed to be 55'' 40'. How fast is the balloon moving ? Ans. 3 (sin 20<> 20') (sec 550 40') (cosec 35° 20') miles per hour. (4) The angular elevation of a tower at a place A due south of it is 30<*; and at a place B due west of A^ and at a distance a from it, the elevation is 18*>; show that the height of the tower is jy\ V2(l+\/5) (5) The angular elevation of the top of a steeple at a place due south of it is 45^, and at another place due west of the former station and distant a feet from it the elevation is 15*'; show that the height of the steeple is - (3* - 3 ~ *) feet. (6) A tower stands at the foot of an inclined plane whose in- lination to the horizon is 9^; a line is measured up the incline rom the foot of the tower of 100 feet in length. At the upper xtremity of this line the tower subtends an angle of 54®. Find the height of the tower. Ans. 114'4 ft. 228 TRIGONOMETRY. LXXII. (7) The altitude of a certain rock is observed to be 47^, and after walking 1000 feet towards the rock, up a slope Inclined at an angle of 32^ to the horizon, the observer finds that the altitude is 77^. Prove that the vertical height of the rock above the first point of observation is 1034 ft. Sin 47<> = -73135. (8) At the top of a chimney 150 feet high standing at one corner of a triangular yard, the angle subtended by the adjacent sides of the yard are 30*^ and 45^^ respectively ; while that sub- tended by the opposite side is SO*'. Show that the lengths of the sides are 150 ft. 86*6 ft. and 106 ft. respectively. (9) A flag staff h feet high stands on the top of a tower. From a point in the plain on which the tower stands the angles of elevation of the top and bottom of the flagstaff are observed to be a and ^ respectively. Prove that the height of the tower Atan/3 „ , . Asin^.cosa „ iSr 7 pi leet, I.e. — : — -. r-r- leet. tan a - tan )S sm (a - ^) V (1Q/ From the top of a cliff A feet high the angles of de- pre^on of two ships at sea in a line with the foot of the cliff are a and j3 respectively. Show that the distance between the ships is h (cot /3 - cot a) feet. (11) The angular elevation of a tower at a place due south of it is a, and at another place due west of the first and distant d from it, the elevation is ^. Prove that the height of the tower is d . d sin a • sin B I.e. Vcot2 ^ - cot2 a ' " ' N/sin(a-/3).sin(a+/3)" -W (12) A man stands on the top of the wall of height A, and observes the angular elevation (a) of the top of a telegraph post; he then descends from the wall, and finds that the angular elevation is now ^ ; prove that the height of the post exceeds the height of ,, , , sin/?, cos a the man by h -r~^ — . . •^ sm (y8 - a) I MEASUREMENT OP HEIGHTS AND BISTANCEH. 229 (13) Find the height of a cloud by observing (i) its elevation and (ii) the depression {fi) of its reflexion in a lake A feet ow the point of observation. Ans. Height above the lake n (a + /3) cosec O - a) feot. ^(14) If the angular elevation of the summits of two spires rhich appear in a straight line) is a, and the angular depressions their reflexions in a lake, h feet below the point of observation, are /3 and y, then the horizontal distance between the spires is 2A cos* asin (/3 - y) . cosec (/3 — a) . cosec {y-a) feet. (15) A statue AB on the top of a pillar BC situated ou level imd is found to subtend the greatest angle (a) at the eye of an rv^er E when he is distant c feet from the pillar. Prove t the height of the statue is (2 c tan a) feet ; and, if h be the ight of the observer's eye from the ground find the height of pillar. (See Note in the Answers.) (16) A man walking along a straight road observes that the greatest angle which two objects subtend at his eye is a. From the place where this happens he walks a yards ; the objects then appear in a straight line making a right angle with the road. Prove that the distance between the objects is (2 a tan a) yards. ^«ia (17) A man standing ou a horizontal plain at a distance h feet from a tower, observes that a flagstaff on the tower sub- ds an angle a at his eye, and that on walking 2 k feet towards e tower the flagstaff subtends again the same angle. Prove at the height of the flagstafl' is 2 (A - k) tan a feet (18) A man walking along a straight road observes that the greatest angle which two objects subtend at his eye is a ; from the place where this happens he walks a yards, and the objects then appear in a straight line making an angle /3 with the road. Prove that the distance between the objects is — ^ yds. '' cos a + cos /3 *' ( 230 ) CHAPTER XIX. On Triangles and Circles. 273. To find the Area of a Triangle. The area of the triangle ABC is denoted by A. f/, ,4 ^ //.„__ K Through A draw HK parallel to BG^ and through ABC draw lines AD^ BK^ CH perpendicular to BC. The area of the triangle ABC is half that of the rectan- gular parallelogram BCHK [Euc. i. 41]. BG.GH BC.DA Therefore A = 2 a . h sin G (i). But sin (7 = -7 . \/s (s - a) (s — b)(s — c); .'. A = Js(s-a)(s-b)(s'-c) = S (ii;. ON TRIANGLES AND CIRCLES. 231 4. ToJl7id the Radius of tJie Circumscribing Circle. Fig. I. Fig. II. Let a circle AA'CB be described about the triangle ABC. Let R stand for its radius. Let be its centre. Join BO, and produce it to cut the circumference in A'. Join A'G. Then, Fig. i. the angles BAC^ BA'G in the same segnienl re equal; Fig. ii. the angles BAC^ BA'C are supplementary; also the angle BCA' in a semicircle is a right angle. CB Therefore -77-7, = sin CA'B = sin CAB = siu yl , A u •2R = sin ^ ; 2i2 = a sin A ' 275. Similarly, it may be proved that 2R = sin/^ and that 2R = .-— . sin C Hence, sin ^ ~ sin ^ ~ sin C 2R. IS df the value of each of these fractions, is the leterof tlio circumscribing circle. L. E. T. iQ 232 TRIGONOMETRY. 276. To find the radius of the Inscribed Circle. Let D, E, F be the points in which the circle inscribed in the triangle ABC touches the sides. Let / be the centre of the circle ; let r be its radius. Then ID = IB = IF= r. The area of the triangle ABC = area of IBG + area of ICA + area of lAB. And the area of the triangle IBC = ^ID , BC = ^r . a, .-. area oiABC = \ID . BG + \IE . GA + \IF . AB ^^ra+ ^ rb + ^rc; or, A = Jr (a + 6 + c) = Jr . 2s = rs. A S 277. A circle which touches one of the sides of a triangle and the other two sides produced is called an Escribed Circle of the triangle. ON TRIASOLES AND CIRCLES. To find the radius of an Escribed Circle. 233 Let an escribed circle touch the side BC and the sides \ AB produced in the points />„ E^, F^ respectively, jt /, be its centre, r, its nidius. Then 'A-IA = I.F,-r, The area of the triangle ABC = area of ABIfi - area of I^BC, of IfiA + area of T,AB - area of I,BC, A = J/,^, . CA + ^I/^ .AB- \I^D^ . BC = ir^(b + c-a) = ^r, (28 - 2a) = r^{8- a). A S ^rr = 8 — a 8— a 279. Similarly if r, and r, be the radii of the other two escribed circles of the triangle ABO, then ^ 8-b' ^» 8-C ■ * 280. The following results are often useful With the construction of the last two articles ; the lines /.I, IBf IC bisect the angles A, B, C respectively j the lines I^A, I,B, Ifi bisect the angles BAC, CBF^, BCE^ respect- ly; so that AII^ is a straight line. IG— 2 234 TRIGONOMETRY. (i) Two tangents drawn from an external point to touch a circle are equal ; therefore AE = AF', BD = BF', CE^CD; AE, = AF, ; BD^ = BF^ ; GE^ = CB^. (ii) 2AE^ = AE^ + AF, = (AC + CB,) + (AB + BB,) = AC+CB + BA = a + b + G=^2s. (iii) 2AE ^AE + AF={AC-CB) + (AB - BB) = AC + AB-BC = b + c-a = 2{s-a). Similarly, BF = BB =(s-b); CB = GE={s- c). (iv) BB^ = BF^ = AF, -AB = {s- c). Similarly, CE^ = CB^ = {s- b). (v) Hence, BB^ = CB, and CB^ = BB. (vi) FF^=AF^-AF=s-(s-a) = a. Similarly, EE^ = a. ON TRT ANGLES AND CIRCLES. 235 (vii) //, = AI^-AI = AF^ sec ^ - AFsec ^ = ^^x sec ^ z z z A ^ cos-. (viii) //'=^i?^tan^; or, r = («-a)tan|, = //^ cot ^ , and i?/'= /i?' cot :? ; .-. c = uii^ + ^i^=r(cot| + cotf^; ^ . B .'. r = . A . B csin— .sin-^ csin ^.siu — A B . A+B C cot ^+ cot 2 81^ -y- cos- (x) F^BC =IS0'-B, . ♦. I^BC = 90° - :? . Hence, z ./eot(90°-D^cot(90»-f)} = a; B C a cos — . cos -^ '•' = A • cos 2 i) r, = AF^ tan ^ , and r ^AFixa - ; therefore rtr, . -A. A r, - r = i^i^, tan - - a tan - . 23G TRIGONOMETRY. ■^281. 2^0 find an expression for the area of a quadri- lateral inscribed in a circle, in terms of the sides. Let DEFG be the quadrilateral ; and let d, e,f g repre- sent the sides. The angle F= 180" - D. [Euc. iii. 22.] The area of GDE= \dg sin D. [Art. 273.] The area of EFG = |e/sin F= Je/'sin (1 80" - D) = Je/sin D. Thus the area of the quadrilateral is I (dg + ef) sin D. We must find an expression for sin D in terms of the sides. From the triangle EGD we have EG'' = d'^g'- 'idg cos D. [Art. 239.] From the triangle EGF we have EG^ = e''+f-2efco^F = e'+f' + 2efcosD; [For i^= 180° -i>] .-. d' + g'-2dgcosD=e'+f' + 2efcosD; ON TRIANGLES AND CIRCLES. 237 •. 2{dg + e/)co8D = d* + g*^^-/'; 4 (dg + e/Y \ Bin' D 4(dg^efy-(d^^g'-e'-/r I i{dg + e/y The numerator of this expression is equal to + 26/) + ((/•+/ - e'-D) {(2dg + 2e/)-{d'+g'-e^ -/')} {(d^gy-(e-/y].{{e^fy-(d-gy] {{d^g) + (e-/)}{{d-^g)-(e-f)] ^{{e+/)-^{d-g)}{(e-^/)-(d-g)} ^« (d + g + e-/)(d + g-e+f) {e +/+d~g) (e +f+g - d). ^H Let 28 stand for (d + e+/+ g). ^^B Then the above expression may be written ^B' (28 - 2/) (2s - 2e) (2s - 2g) (2s - 2d). ' Therefore sin'i> = ^s - ^) (s -eKs -/) (s - ,) {dg + e/y ^p_ 2>/(s-cf)(8-6)(8-/)(8-y) dg-¥e/ sin Hence the area of the quadrilateral DEFG which is \ (d^ + ef) sin D = V(8-0?)(s-«)(8-/)(8-^). 238 TRIGONOMETR Y . ■^ 282. To prove tliat if I and O (we the centres of tlie inscribed and circumscribed circles of a triangle^ then Or = K-1Rr. Let 01 produced cut tlie circumscribing circle in //and K. Join lA^ IG ', produce I A to cut the circle ABG again in F. Draw the diameter VOW and join CW. Then the angle VIC = I AG -^IGA^IA ^\G = \{A + G). The angle IGV^IGB + BGV=IGB + BAV= ^G + iA = ^{A + G). Therefore the angle VGI= VZG, and .-. VI ^ VG. NoAv OK' - OP = III. IK [Euc. II. 5] =:VI.IA= VG.IA. [Euc. III. 35.] But VG= Fr.sin(7TrF=2/i'.sinC^r=2A\sin^, and lA [See Figure on page 234.] sm Hence or OK' 'OP 2R sin — X , ^ .A R'-^Or^Uir. ON TRIANGLES AND CIRCLES. 239 EXAMPLES. LXXm. (1) Find the area of the triangle ABC when (i) a =4, 6 = 10 feet, C=300. (ii) 6=5, c=20 inches, ^ = 60". (iii) c=66|, a= 15 yards, B=17'^ 14' [sin IT^ 14' = -29()2G]. (iv) a=13, 6=14, c= 15 chains, (v) a = 10, the perpendicular from A on jBC=20feet. (vi) a = 625, 6 = 505, c= 904 yards. (2) Find the Radii of the Inscribed and each of the Escribed farcies of the triangle ABC when a = 13, 6 = 14, c=15 feet. (3) Show that the triangles in which (i) a = 2, y!=60"; (ii) 6=§. "/s, i?=30° can be inscribed in the same circle. . (4) Prove that R=-^; find /J in the triangle of (2). i(5) Prove that if a series of triangles of equal perimeter arc cribed about the same circle, they are equal in area. : (6) If J = 60^, a = V3, 6 = v/2, prove that the area = j(3 + V3). (7) Prove that each (jf the following expressions represents the area of the triangle ABC: I' (iii) r». (iv) Rr (sin ^ + sin 2?+ sin C). (v) ^a* sin Z? . sin C . cosec A. (vi) ra cosec — cos ^ cos . :-- — (i) j^. (ii) 2/2*8in J.sinjS.sinC 240 TRIGONOMETRY. LXXIII. Prove the following statements : (8) If a, 6, c are in A. P., then ac=6ri2. (9) The area of the greatest triangle, two of whose sides are 50 and 60 feet, is 1500 sq. feet. (10) If the altitude of an isosceles triangle is equal to the base, jK is five-eighths of the base. (11) jB(sin^ + sin^-|-sin(7) = s. (12) 6c = 4i22(cos^-l-cos^.cosC). (13) If (7 is a right angle, 2r + 2i2 = a + 6. (14) r^r^^r^r^^-r^r. ^1 i 1 1 _ 1 ^ ' he ca ah 'irR' (16) r^cot - = r2Cot -=r3Cot -=r cot -- . cot - . cot-. Q (17) ri + r2 = ccot-. (18) \.\^IA- (19) !i_r + !J__!: = £-. (20) ri + r2 + r3-r = 4i2. (21) a.6.c.r = 4/?(s-a)(s-6)(s-c). (22) The distances of the centres of the escribed cil-cles of the triangle ABC from that of the inscribed circle are 4jR sin - , 4i2 sin - , 4i2 sin - . (23) If ^ is a right angle, rg + ^3= a. (24) In an equilateral triangle 3i2 = 6r = 2r, . r25^ ^i + !2 . !>_ i__L ^^ he ca^ ah r 2/2* ♦ CHAPTER XX. On the Area op the Circle, the Construction op Trigonometrical Tables, (fea 283. Let r be the radius of the circle described about a regular polygon of n sides. Let be the centre ; HK one of the sidea From dmw OD perpendicular to HK^ bisecting both UK and the angle HOK. Then, since the polygon has n equal sides, n times the angle 110 K make up four right angles .'. 110K=~, and . •. DOK = - . Hence, n n the perimeter of the polygon, which is n times HK^ = 2n.Dn=2n. OR sin DOE = 2nr sin the area of the polygon, which is n times the area of HOK^ nOD . UD = nr cos - . r sin - nr sin - . cos - . n n 242 TRIGONOMETRY. 284 Let r be the radius of the circle inscribed in a regular polygon of n sides. Let be the centre, HK one of the sides. From draw OM perpendicular to HK, bisecting both HK and the angle HOK. Then since the polygon has n equal sides, n times the angle HOK make up four right angles, Hence '. HOK= — and .-. MOH ^"^ . n n the perimeter of the polygon, which is n times HK, = 2n . MH = 2n . OJf tan MOH — 2nr tan - . n The area of the polygon, which is n times the area of HOK, =^n.OM. MH= n.OM. OMtam MOH - nr'tan- 71 = the radius x half the perimeter. 285. Let a cirfcle oi radius r have a regular polygon of n sides circumscribed about it, and also a regular polygon of the same number (n) of sides inscribed in it. ON THE AREA OP THE CIRCLE. 243 The perimeter of the circumscribing polygon is 2wr tan The perimeter of the inscribed polygon is 2nr sin TT 71* IT n ' The ratio of these perimeters is 1 : cos - (1). The area of the circumscribed polygon is nr* tan - . The area of the inscribed polygon is nr^ sin - cos - . n n The ratio of these areas is 1 : cos^ - (ii). n ^ ' 286. We must assume the two following axioms : (i) The circumference of the circle lies in magnitude between the perimeter of a circumscribed and that of an (scribed polygon, (ii) The area of the circle lies in magnitude between e area of a circumscribed and that of an inscribed polygon. Now, when n is increased, - is .liminished, and therefore )y Art. 94) cos - approaches 1. m- Hence, as the number of the sides of the two polygons in t. 285 is increased, their perimeters approach to equality. And since the circumference of the circle always lies in magnitude between them, each of these perimeters must approach the circumference of the circle. Therefore, the circumference of a circle is that, to which ' he perimeter of a regular inscribed (or circumscribed) poly- ^^^on approaches as the number of its sides is increased, ^Hhd from which it can be made to differ by less than any 244 TRIGONOMETRY. 287. In like maimer it follows, from (ii) Art. 285, that the area of the circle is the ' limit ' of the area of a regular inscribed (or circumscribed) polygon when the number of the sides is indefinitely increased. Now, twice the area of a polygon circumscribed about a circle is equal to (the radius x the perimeter). [Art. 284] This is true however great be the number of the sides. It is therefore true when the number of the sides is in- definitely increased. Therefore it is true for the circle itself Hence, twice the area of a circle = the radius x the circumference. Or, the area of a cii-cle = \r {2irr) [Art. 34] EXAMPLES. LXXIV. (1) Prove the surd expressions of Art. 37 for the ratios of the perimeters of certain regular polygons to the diameters of their circumscribing circles. (2) Prove that the area of a regular polygon of twelve sides described about a circle whose radius is 1 foot is 3*215 sq. ft. (3) Prove that the area of a square described about a circle is ^ of the dodecagon inscribed in the same circle. (4) Find the perimeter of a regular polygon of 100 sides (i) when described about a circle of 1 foot diameter, (ii) when inscribed in the same circle. Ans. (i) 3'14263, (ii) 3'14108 ft. (5) An equilateral triangle and a regular hexagon have the same perimeter : show that the areas of their inscribed circles are as 4 : 9. OA TilL AULA OP THE CIRCLE. 245 Prove that the area of a regular polygon of n sides, each I., . . na^ .\m M whose sides is a, is -r- . cot . ^™^ (7) If the areas of a regular peutagon and decagon are equal, the ratio of their sides is \/20 : 1 . (8) If a be a side of a regular polygon and R and r the radii of the inscribed and described circles respectively, then m = a cosec and 2r = a cot - . n 7i (9) U R and r be the radii of the inscribed and circuiJ>- ribed circles of a regular polygon of n sides, each = a, (10) The triangle formed from one side each of a regular l)entagon, hexagon and decagon inscribed in the same circle, is right-angled. (11) Prove that the area of an irregular polygon described about a circle is equal to the product of the radius and half the perimeter of the polygon. (12) The area of an irregular polygon of an even number of sides circumscribed about a circle is equal to the radius x the sum of every alternate side. (13) The diameter of the dome of St Paul's is 108 feet ; prove that it covers an area of 1018 sq. yds. (14) The radius of the circle whose area is one acre is 39} yds. (15) A length of 300 yards of paper, the thickness of which IS the hundred and fiftieth part of an inch, is rolled up into a solid cylinder. Find approximately the diameter of the cylinder. Ans. 9'575 in. (16) The diameter of a roll of carpet is 2 feet and the thick- ness of the carpet is the eighth of an inch. What is the length of the carpet 1 A ns. 301 6 ft. 246 TRIGONOMETRY. On the Construction of Trigonometrical Tables. 288. To prove that, if 6 he the circular measure of an angle less than a right angle, sin 6, 6^ tan 6 are in ascending order of magnitude. IP Let EOF be an angle (6) less than a right angle. Make the angle HOF' on the other side of OR equal to BOP. With centre and any radius OB describe the arc P'BP. Draw PT, FT to touch the circle at P and P'. PT and FT will meet on OB. The line PF is bisected by OR at right angles at M. Then, since the circumference of a circle lies betwe&n the perimeter of the inscribed and circumscribed polygons, it follows that the arc P'BP lies in magnitude between FMP and FT + TP. In other words, FMP, the arc P'BP, and FT+ TP are in ascending order of magnitude. Therefore also their halves MP, BP, TP are in ascending order of magnitude. M^ BP TP OP' OP' OP' And so also are That is, sin 0, 6, tan 6 are in ascending order of magni- tude, where angle refen'ed to, radius is the circular measure of the CONSTRUCTION OP TRIGONOMETRICAL TABLES. 217 6 1 289. Hence, 1, - — -^ , ^ are in ascending order of sin ^ cos ^ ° magnitude. 290. To prove that the ^limii* of - — ^ , when ^ = 0, w 1 : '^ *^ sin ^ 6 being the circular measure of tJte angle referred to. 6 1 [The value of -r— ^ lies between 1 and 7,. As 6 sin d cos 6 HThe ^Hiinished, cos 6 approaches 1 ; and the smaller 6 becomes, the more nearly does . approach 1. [Art. 94] Therefore, by diminishing 6 sufficiently, we can make Q -r—r. differ from 1 by less than any assignable quantity however small. This is what is meant when it is said that * the limit of Q - — -r. when 6 = 0, 18 1.' 8in d ' 291. The student must notice carefully that 6 here is the number of radians in the angle refeired to.* I Example. Prove that the limit of —. — 5 , when n=0, w - . Let 6 radians =n®, then TT ~ 180* * * ~ TT ' " sin /i** tr * sin ^* When n is diminished, $ is diminished also, and the limit of -; — -r. when ^=0, is 1. Therefore the limit of -: — ^ (which Bin 6^ ' smw'' \ 180 ^ \ , ^ . 180 = — . -: ;: I , WhCU 71 = 0, 18 . TT Sin ^y ' * JT * It is on this acoonnt that a radian is nsed as a unit of angle ). L. E. T. 17 I 248 TRIGONOMETRY. 292. To prove that if 6 he the circular measure of a positive angle less than a right a/ngUy sin B lies between 6 and It has been proved that sin is less than (i). And that 6 is less than tan 6. Therefore ^ is less than tan « ; or, ^ cos^. is less than sin ^. B B Now, sin ^ = 2 sin - . cos ^ ; [Art. 166] /B B\ a .'. sin is greater than 2 f- cos - j cos - , B B i. e. greater than cos^ ^ , i. e. greater than B-B sin' ^ . But ( - I is greater than sin* ^ . Still greater therefore is sin B than B-\^ (ii). Hence, sin B is less than ^, and greater than B- -r. 293. To find sin 10''. In the above, B is the circular measure of the angle. The circular measure of 10", correct to three significant figures, is -0000484.... [Examples X. (17).] Let ^=-0000484.... Then, ^_j(98=^_i (-0000484. ...)'= ^--000000000000028.... Hence, B and {B — ^B^) are decimal fractions which agree in their first twelve figures at least. I CONSTRUCTION OP TRIGONOMETRICAL TABLES. 249 And since sin 10" lies between these fractions, therefore the first twelve decimal places of sin 10" are the same as those of the circular measure of 10". Hence, if the value of tt be given to a sufficient number of decimal places, we can calculate the circular measure 110", and therefore also sin 10", to 12 decimal places. 294. To show how to construct a Table giving the Trigo- metrical Ratios of angles which form an arithmetical pro- gression having \0" for tJie cortimxyn diffefrence. I In the identity sin (n + 1) a + sin (n - 1) a = 2 sin na . cos a, a= 10", and suppose a table of the sines of all angles at itervals of 10" to have been calculated up to w. 10". ^^H Then, sin (n - 1) a, sin no. and cos a [ = Vl — sin" 10"] are ^^Tnown. Therefore by the above formula sin (w + 1) a can be found. ^^B Hence, since we know sin 10", the sine of 20" can be ♦ound ; and then the value of sin 30"; and so on. 295. When the sines of angles up to 45" have been tculated, the rest may be found from the formula, sin (45* + id) - sin (45'» -A) = ^2. sin A. Also, when the sines of angles up to 60" have been cal- Ated, the remainder up to 90" can be found still more lily from the formula, sin (60" + vl) - sin (60" - ^1) = sin i4. The other ratios may be found from the following : cos ^= sin (90" -.4), tan^=^^, cot .1 = tan (90"-^), oosec/l , , secil =cosec(90''-yl). 250 TRIGONOMETRY. 296. Since 3° = 18" - 15", the sines of angles differing by 3", or by any multiple of 3", can be found independently. (See p. 120.) These values may be used to test the accuracy of the Tables calculated as above. 297. The following formulae may be also used to test the accuracy of the Tables, sin(36°+^)-sin{36"-J^)-sin(72''+ii) + sin(72''-^)=sin^, cos(36V^)+cos(36"-^)-cos(72''+^)-cos(72°-^)=cosJ. They are called formulse of verification. On the Limit of the Visible Horizon. 298. The surface of the sea is very nearly that of a sphere whose radius is 3957 miles. The height of the highest mountains on the globe is less than 6 miles. Thus a point must be considered to be at a very considerable height above the surface of the sea if its height is a thousandth part of the earth's radius. 299. In the figure, let be the centre of the earth, PRP' part of the surface of the sea, T a point of observation, TR its vertical height. Draw TP, TF tangents to the earth's surface. Then PMF is parallel to the 'horizontal plane' at R. The angle TPM is called the dip of the horizon at T, It is the angle of depression of the most distant point on the horizon seen from T. It must obviously be a very small angle, since TR is so small compared with RO. }N THE LIMIT OF TUB VISIBLE UOlilZON. 251 If RO be produced to cut tlie circle again in L, TP'^TR. TL [Euc. iii. 36] =^TR{RL+TR) = 2TR,R0 + TR\ But TR will in general be much less than a thousandth part of ROj and therefore TR!^ will be much less than a thousandth part of 2TR. RO. Hence, the formula TP* = 2TR,R0, i.e. rP«= twice the earth's radius x vertical height, will give the value of TP correct to at least three significant figures. Example. Three times the Jieiyht in feet of the place of obser- vation above the sea is eqtial to twice the square of the distance of the horizon in miles. I Here, TP^^RLx 7914 miles. Let / be the number of feet in JRX, then the number of miles in RL is ^i^ ; let x be the number of miles in TP^ then Xx7914=3/ 6280 X 7914= -^ nearly. Q.B.D. 252 TRIGONOMETRY. EXAMPLES. LXXV. (1) Show that the limit of - iZ^sin — (i.e. the area of a poly- gon of n sides inscribed in a circle of radius B\ when n = oo is (2) Prove that the limit of nr^ tan - , when n= oo , is nr^. (3) Given that 7r = 3*141592653589793... prove that the cir- cular measure of 10" is -00004848136811... (4) Prove that 2 sin (720 + ^ ) - 2 sin (72° - J ) = (^5 - T ) sin A, and that 2 sin {Z& + ^) - 2 sin (360 - ^) = (^5 + 1) sin A. (5) If a mast of a ship be 150 feet high, show that the greatest distance seen from its top is 15 miles nearly. (6) Prove that if the dip of the horizon at the top of a mountain is 1^ 26' [ = tan~i -025], the mountain is about 6530 feet high. NOTE. The definitions given in Arts. 75, 78 of the Trigonometrical Eatios are now used exclusively. The NAMES tangent, secant, sin^, were given originally to quantities defined as follows. Let ROP be any angle. With centre and any radius describe the arc RP. Draw PM perpendicular to OR and PT perpendicular to OP. (See Figure pn previous page.) Then PR is called an arc, PT is the tangent of the arc PR, OT is the secant of the arc PR, MP is the sine of the arc PR. The name sine is derived from the word sinus. For, in the figure, PMP' is the string of the "bow" {arcus), and the string of a bow when in use is pulled to the archer's breast. The co-tangent, co-secant and co-sine are respectively the tangent, secant and sine of the complement of the arc or of the angle. The sine, tangent, etc. of the angle are the same as the measures of the sine, tangent, etc. of the arc, when the radius of the circle is the unit of length. t, THE Level, the Theodolite, the Sextant, THE Mariner's Compass. 301. The practical Surveyort has to measure distances and angles, and has also to make plans or pictures, re- 1>rding the result of his measurements. For the measurement of distances the surveyor uses ther rods, or chains, or tapes. Rods used in measurement are made of woody or of metal or sometimes (when extreme accuracy is required, as in the case of the measurement of the base line of the t I nance survey on Salisbury Plain) of glass. All these instruments, when exposed to changes of temperature, liable to change of length ; hence for great accuracy, a surveyor must know the exact length of his measuring rod at all ordinary temperatures; and when making a measurement, must note the temperature of his rod at the instant of observation. The change of length caused by change of temperature is greater in a rod of metal than in a rod of wood. Hence wood is a very suitable material for measuring rods under ordinary circumstances. A tape made of cotton or hemp if used for measurement must be carefully protected ^^djpom moisture by varnish or otherwise ; as such tapes sensibly shrink ^^Bien allowed to become damp ; also, if of any considerable length, they ^^atretch sensibly under tension. A tape of 66 feet can be easily stretched an inch or so. t It must not be supposed that any verbal or pictorial description, such as the following, can iji any wav take tlie place of a practical explanation of the instruments themselves. A study of these figures may perhaps tell the student what to look for when he actually has the instrument in his luuids. TBIGONOMETBY. The Vernier. 302. A vernier is a simple instrument for increasing the accuracy of the measurement of a small distance by one significant figure. 303. Description of a Vernier. Suppose we have a rule (i.e. a measuring rod) of brass graduated! to tenths of an inch. The vernier is a little slip of brass which slides along the rule. This slip of brass is a little more than 1^^ inches long, and a portion of its length ly^^ inches in length is divided into ten equal parts, by fine scratches on the surface of the metal. Thus the distance between each scratch and the next is yY^ of an inch or ( j-^ + t^) oi an inch ; i. e. this distance exceeds the distance between two scratches on the rule, by an hundredth part of an inch. 304. To read the Vernier. This will be best explained by an example. Suppose the length to be measured is ascertained to be 3 ft. 11-5 inches and a little over. This can be ascertained by the use of the rule (or measuring rod). Now let the rule be so placed that one end exactly coincides with one extremity of the length to be measured ; then the other extremity K of the length to be measured will be between the scratches on the rule indicating 3 ft. 11*5 in. and 3 ft. 11-6 in. Now slide the CD vernier on the rule till its extremity D coincides with the extremity K of the length to be measured. t i.e. having fine scratches upon it, each the tenth part of an inch from the next on& TTTh: vf<:nNn':n. Rule. It will be observed that one of the scratches on the vernier coincides with one of the scratches on the rule more nearly than any other. Suppose this to be the scratch marked 6 on the vernier. Then the length to be measured is 3 ft. 11-56 inches nearly. For the length exceeds 3 ft. 11-5 inches by just as much as 6 spaces on the vernier exceed 6 tenths of an inch, that is by 6 hundredths of an inch. 305. A vernier may be used to read the graduations of a circular arc ; in which case it is made curved so as to follow the line of the arc. 306. The student should notice that the advantage gained by the use of the vernier depends on the fact that the eye is able to judge with considerable accuracy when two scratches are or are not, coincident. Note. French instrument makers make their verniers (^ - xi^) of an inch and divide it into ten equal parts. We leave it as an exercise to the student to discover how such a vernier would be used. IV TRIGONOMETRY. The Level. 307. A level is an instrument used for ascertaining whether a given straight line is or is not horizontal. F The essential part of it consists of a glass tube ABC, in the form of the arc of a circle, which is closed at both ends and is nearly full of water. The tube, being not quite full of water, will have a bubble D in it. This bubble, when the tube is at rest, will only rest at the highest point of the tube. The tube is fixed in a case of wood or metal which is so made that when the base of it FG is horizontal the highest point K of the tube is visible, and the bubble can be seen at rest in it at the highest point of the arc of the tube. This highest point is carefully marked on the tube. To ascertain whether any given line is horizontal it is only necessary to put the base FG of the case of the * level' in the position of that line and watch the position which the bubble takes up when the tube is kept at rest in that position. If the bubble rests at the position marked on the tube the line is horizontal and not otherwise. 308. The student can easily purchase for himself an ordinary carpenter's level and can make experiments with it. TUE THEODOLITE. V The Theodolite. . )9. A Theodolite is an instrument for measuring the izontal angle subtended at the position of observation two distant visible objects. P, Q be two visible distant objects seen from a place of obser- kion 0. Let PM, QN perpendiculars be let fall from P and Q respectively the horizontal plane passing through O. Then the horizontal angle subtended at by P and Q is the angle ION [See Examples lxxvi. (16)]. [The angle subtended at by P and Q is the angle POQ.] §310. The essential part of a Theodolite may be de- ribed as follows. Suppose two circular brass plates to be laid one on the her so that they are concentric and both are free to turn about an axis througli their centre. ILet the rim of the lower plate be graduated. That is on the rim will be marked 3G0 Unes at equal distances dicating degrees subtended at the centre. Each degree will be bdivided into minutes etc. according to the size of the circle and the degree of accoraoy to which the instrument is to be read. Let the rim of the upper plate have inscribed on it a rernier suitable for reading the graduati ons of the other rim. ^""^ B RA^'pSv _ f or THU ^ I { UNIVERSITY I VI TRIGONOMETRY. Now suppose the lower plate can be fixed in a hori- zontal position and that a telescope or other means of pointing at a distant object is mounted centrally on the upper plate. Then by first turning the upper plate till the telescope points to one distant object and reading the vernier; and again by turning the upper plate till the telescope points to another distant object and again reading the vernier, we shall obtain from the difference of these readings, the Jwrizontal angle between the two distant objects. 311. The following figure is a picture of a part of a theodolite, shewing the arrangements usually made for fixing the lower plate in a horizontal position. li THE THEODOLITE, vii EFD is the upper plate on which is engraved a vernier E, AB is a portion of the rim of the lower plate. IHK are the upper parts of a tripod stand whose feet placed on the ground. LM are two of the three screws which connect the d of the tripod stand with the axis round which the lates can turn. These screws are fastened to the tripod head by a clip which is indicated near M. When the tripod stand has been firmly placed on the ground so that the tripod head is fairly horizontal the plates can be made accurately horizontal by judicious turning of the three screws X, if, N. I The screw at C allows the lower plate to be sliglitly ned with reference to the axis and tripod head. The arrangement FGII is a 'clamp' and a 'tangent w.' When the screw head F is loosened, the upper plate can 1)0 turned quite freely round its centre; so that the telescope which is mounted on it, can be turned freely through any angle. V When the screw F is tightened, the upper plate can only be turned with reference to the lower plate very slightly, by turning the screw liead at B which is a tangent screw. viii TRIGONOMETRY. 312. On the following page is a picture of a complete * Transit ' Theodolite. The upper plate is here shewn carrying a telescope TF which is fixed to an axle of which one end H is seen. This axle is kept fixed parallel to the plate by the two pairs of upright legs which are firmly screwed to the upper plate. The graduated circle XYZ is fixed to the axis of the telescope and turns with it. The verniers b and c are fixed to the upright legs. a is a small magnifying glass for reading the verniers, c? is a level fixed to the telescope and parallel to its axis. Atyj e are the clamp and tangent screw which respect- ively fixes the graduated circle XYZ and slowly turns it. ^ is the end of a level fixed parallel to the plane of the upper plate. On the top of the upper plate itself is seen a mariner's compass. 313. Suppose now that we are about to use the theodolite. We arrange the tripod; we loosen the clamp F ; we turn the screws X, M, N so that the plate ABC is horizontal; testing this by turning the upper plate into various positions and observing the level g in each position. The instrument is now ready to make an observation. 314. To make an observation. Suppose in the tig. on page v the observer is at 0. THE THEODOLITE. IX X TRIGONOMETRY. He turns the upper plate of the theodolite and moves the telescope until he can see F through it. He then clamps both circles UFD and XYZ. And by means of the tangent screws brings the point P exactly into the centre of the field of view. He then reads both verniers. He then unclamps both circles. And repeats the operation for the point Q. The difference between the reading of the horizontal vernier gives the horizontal angle MON. By unclamping the vertical circle and reading the vernier when the level d shews that the telescope is hori- zontal he can obtain each of the angles of elevation POM, QON (from his reading of the vernier of the vertical circle when it pointed first to P and then to Q.) The Sextant. 315. There are three experimental facts connected with Optics which the student must understand who wishes to understand the principle of the sextant. I. When a telescope is pointed at a plane mirror (or looking-glass) the eye sees exactly what it would if the telescope were placed on the other side of the mirror as in the following diagram. A ^0' THE 6LXTAM\ plane of the mirror be perpendicular to that of paper. Let ADB be the line in which the mirror is )po8ed to cut the plane of the paper. Let EF be a section of the telescope and CD the lino \ Bight. Make the angle C'BB equal to CDB. And draw U'F' a dotted section of a telescope round BO that C'D = CD. Then an eye placed at C and looking directly at the me mirror in the direction CO' would see in the direction DO, exactly what an eye placed at C, the eyepiece of a Iescope whose line of sight is CD, would see if the rror were removed. i It follows that the atigle BDC between the mirror the axis of the telescope is half ODO', the angle ;ween the direction of the axis of the telescope and the jction from D of the object 0. II. When the /utlf of the object glass (or large glass) a telescope is covered over, then an eye looking through escope will see exactly what it saw before except t the image will be half as bright. The effect ia very similar to the effect of looking at a picture in the first case by the light of two candles and in the second by the light of only otie caudle. 18 xii TRIGONOMETRY. III. When looking through the eyepiece of a telescope the eye looks at an image (which is a small picture formed by the rays of light coming from the object looked at, inside the tube of the telescope) of the distant object at which the telescope is pointed. The eyepiece or small glass of a telescope forms in fact a microscope with which the eye sees this image magnified. ^ Thus in the figure, is the object-glass, U is the eyepiece or microscope ; / is the position of the image. / is called the focus of the object-glass. 316. To describe the arrangement of a Hadley's Sextant. Let the axis AB of a telescope be directed to the edge G of a plane mirror perpendicular to the plane of the paper whose section is CI>. And let FGF be the section of another plane mirror arranged so that light coming from a distant object Q THE SEXTANT. xiii on EF and on being retiected by tlie mirror falls on and on being again reflected is in the direction of the of the telescope AB. Then an eye looking through the telescope would see :,he images of two distant objects P and Q. These two images would be superposed. the 11 I ^' ^^H One image that of a distant object F in the direction ^^K^/> the light from which is direct. H '^^ ^nrect The other image is that of a distant object Q in the ;tion GQ the light from which is reflected. The mirror EF is arranged so that it can turn about G keeping always perpendicular to the plane of the paper and l^^e angle through which it turns can be observed. When the mirror FF is turned so as to be parallel to CD the reflected image and the direct image would be that of the same very distant object. The angle through which the mirror EF is turned, from the above position, until the image of Q is visible in the telescope is half the angle between CF and CQ ; that is, half the angle subtended by P and Q at the observer's eye. [P and Q are always very distant objects such as the horizon at sea or the sun or a star.] For, produce QC to A'B* ; then the angle PCB' is fixed. Draw GK perpendicular to EF ; the angle turned through by EF in any moment, is equal to the angle turned through by EK ; and the angle QEC is always double of KED' ; therefore the amount of turning of EQ relative to EC (which is fixed) is double that of EK relative to EB' ; that is is double of the amount of turning of EF. 18—2 XIV TRIGONOMETRY. 317. The following is a picture of a Hadley's Sextant. AJB is the telescope; (7 is the fixed mirror; i^is the mirror which can be turned about an axis perpendicular to the plane of AJ3CF. FN is an arm by which the mirror F is turned. LL' is a graduated arc and iV is a vernier for reading its graduations with the aid of the microscope M; K and T are the usual clamp and tangent screw ; II is a handle for holding the instrument. ^ THEsEXtaNt. XV ^^^^^tre three blackened glasses to interpose between He light and the telescope when looking at a bright object Hch as the sun ; each glass is on a hinge so that it can bo Hought into the line of sight or turned back at pleasure. H At X are blackened glasses which can be placed in the ^bc of the reflected light or turned out of it at pleasure. H 318. Thus, the sextant is an instrument for observing He angle subtended at the observer's eye by two distant Hjects. Its peculiar advantage is that it does not require ^meady platform as the Theodolite does. The observer holds B in his hands, and the observation consists in his noting when he h;is so moved the mirror [EF'\ that the images of le two distant objects, whose angle he wishes to take, are irposed in the field of the telescope. This he can do even if he cannot get the images to rest any length of time in the field of view. Accordingly the angle subtended by the edge of the sun the horizon can with a sextant be observed with con- lerable accuracy by an observer standing on the deck of ft ship in motion. Also, at night, the anajular distance be- the moon and a stai- can bo observed uiidci- liko sumstauces. xvi TRIGONOMETRY. 319. Below we give a figure of a Mariner's Compass. It consists of an ordinary magnetic compass with a card attached to the needle, this card is so arranged that when the needle is pointing along the magnetic meridian the pointer on the card is pointing due North. The Points of the Compass are figured on the card. They are N. = North. N. by E. = North by East. N.N.E. = North North East. N.E. by N. = North East by North. N.E. = North East. N.E. by E. = North East by East. E. by N. = East by North. E. = East. And so on. The angle subtended at the centre of the card by two consecutive points is = ^ of 90° = 12J°. EXAMPLES FOR EXERCISE. LXXVIa. Define the terms sine, cotangent; and prove that if -4 be any angle, ain^^i + cob^A = 1. I^H If tani4=|, find einA and cos^. I^H 2. Trace the changes in the sign and magnitude of cos d-aecd I^H>r values of d between and x. 3. Prove geometrically that cos (18(y Find A if 2 sin ^= tan il. A) = -co3A. Prove (1) (2) sin {A + B). sin {A-B) = sin^A - sin^ J] ; sin ^ + sin £ tan J (-dt +£) ^i&ni{A-B)' BinB I^~' sin A 6. Prove that cosM - COS A cos (600 + A) + sin^ (30" - .4) = f . 6. Find the greatest side of the triangle of which one side is 2183 feet and the adjacent angles are 78" 14' and 71" 24'. 1^ log 2183 = 3-3390537, log 42274 = 4-6260733, ^K L sin 78° 14' = 9-9907766, log 42275 = 4-6260836. ^^1 L sin 30" 22' = 9-7037486. 7. Express the other trigonometrical ratios in terms of the cosine. »8. Prove sin (180 + ^4) = - sin ii ; tan(90 + J)=-coti!l. Write down the sines of all the angles which are multiples of 30" and less than 360". 10. Prove t&D?A = 1 - cos 2 A 11. If tauii + seOil 90». 1 + cos 2^1 2, prove that sinil I, when A is less than K If sin A = ^, prove that tan -4 + seCil = 8, when A is less than 90". 254 TRIGONOMETRY. 12. The length of the greatest side of a triangle is 1035-43 feet, and the three angles are 44'*, BG**, and 70°. Solve the triangle, having given i sin 440 =9 -841771 3, I, sin 660=9-9607302, I, sin 700 = 9-9729858, log 1035 •43 = 3-0151212, log 765432 = 5-8839067, log 10066 = 4-0028656. 13. Express the other trigonometrical ratios in terms of the cotangent. 14. Prove that cos (1800-^) = -cos^; cosec (1800 + A)=.- cosec A, 15. Write down the tangents of all the angles which are multi- ples of 300 and less than 3600. 16. If tan ^ + sec ^ = 3, prove that sin ^ =4 when A is less than 900. If sin 4=1, prove that tan A + sec A =2, when A is less than 900. 17. Find the sines of the three angles of the triangle whose sides are 193, 194, and 195 feet. 18. Investigate the following formulae : SA (1) cos — = {2 COS A- 1) cos ^A ; (2) cos 6 - cos (^ + 5) = sin 6 sin 5 (1 + cot 9 tan 1 5). 19. Define the secant of an angle. Prove the formula — 5-7 + -„— 7 = 1. sec^^ cosec^^ If sin 4 = J, find sea A. 20. Trace the changes in sign and magnitude of sin ^ + cos sin - cos 6 as 6 changes from t to 2ir. 21. Find a formula to include all angles that have the same cotangent as the angle a. Solve the equation tan ^=scot 6. 22. Prove the formula to express the cosine of the sum of two angles in terms of the sines and cosines of those angles. Express cos 5a in terms of cos a. EXAMPLES FOR EXERCISE. LXXVI a. 255 Prove the formala t2mii^A=±^{l + BinA)=i=J{l-B'mA). Acoonnt for the double signs in this formula, and examine which signs must be taken if wi be an angle between 540*' and 630°. 24. A ring 10 inches in diameter is suspended from a point 1 foot Iove its centre by six equal strings attached to its circumference at ual intervals ; find the cosine of the angle between two consecutive rings. 25. Define 1*. Assuming that y is the circular measure of two ;ht angles, express the angle A^ in circular measure. Find the number of degrees in the angle whose circular measure •1- 26. Find the trigonometrical ratios of the angle whose cosine iBf. 27. Prove that t(l) COS (1800 + ^) = cos (180»-^); (2j tan (90" + J ) = cot (180" -A). 28. Prove sin x (2 cos x - 1) = 2 sin = cos — , a a 129. Trace the changes in sign and magnitude of 2 sin d - sin 2d 2 8ind + 8in2. 53. Two sides of a triangle are 10 feet and 15 feet in length, and Iie angle between them is 30°. What is its area? 54. Given that Bin 40° 2^=0 -6492268, sin 40° 30' = 0-6494480, Dd sin-i (0-6493000). : igul ami I 65. Express in circular measure (1) 10', (2) J of a right angle. If the angle subtended at the centre of a circle by the side of a lar pentagon be the unit of angular measurement, by what umber is a right angle represented ? 66. If sec a = 7, find tana and coseo a. 67. Prove the formulsB : (1) cos2 (a - /3) - sin' (a + /3) = cos 2a cos 2/3 ; (2) l + tanatania = seca. 68. Solve the equations: (1) 5tan3x + sec3x=7; (2) cos 5^ + cos 35 =^2. cos 4^. 59. The lengths of the sides of a triangle are 3 feet, 5 feet, and ifeet. What is its area? Qiven that sin 38° 25' = 0-6213757, sin 38«26' = 0-6216036, sin-i (0-6215000). 258 TRIGONOMETRY. 61. Which is greater, 7Ck or the angle whose circular measure is 1-2? 62. Determine geometrically cos 30** and cos 45^ If sin^ be the arithmetic mean between sinB and cos I?, then cos2^=cos2{5 + 450). 63. Establish the following relations : (1) tan"^^ - sin^^ = tan2^ sin"-*^ ; (2) cot A - cot 2A = cosec 2A ; ._, sin(a: + 3i/) + sin(3a; + v) „ , 64. Show that for certain values of the angles 2 sin ^A = sjl + sin ^ - Jl - sin A. Is the formula true when ^=240*'? If not, how must it be modified ? 65. Prove that sin {A+B)= sin A cos B + cosA sin B, and deduce the expression for cos (A+B). Show that sin A cos (B + G)- sin B cos {A + G) = sin {A - B) cos C. 66. One side of a triangular lawn is 102 feet long, its inclinations to the other sides being 70° 30', 78° 10' respectively. Determine the other sides and the area. L sin 70° 30' = 9-974, log 102 = 2 -009, L sin 78° 10' = 9 -990, log 185 = 2-267, L sin 3P 20' = 9' 716, log 192 = 2 -283, log 2= -301, log 9234 = 3 -965. 67. Which is greater, 126° or the angle whose circular measure is 2-3? 68. Establish the following relations: (1) cotM -cos2^=cotM cosM; (2) tan ^ + cot 2^ = cosec 2^; cos(a.--3y)-cos(3:.-y)^ .^ ^ ' sin 2a; + sm 2?/ ^ ^' 69. Show that for certain values of the angles 2(iOB^A = sJl + QmA-\-Jl-B\nA. Is the formula true when ^=300°? If not, how must it be modified? 70. Prove that sin 30° + sin 120° = ^2 cos IS®. 71. Establish the identities : (1) l + coSi4 f-sin^ = ^2(l + cos^)(l + sin^); /rtv « i cosec^^ (2) cosec 2 i: 2 ^cosec'^-4 - 1 EXAMPLES FOR EXERCISE. LXXVI a, 259 (8) Bin ^ + smy-8my=4Bm^8myBmy . 72. The sides of a triangular lawn are 102, 185, and 192 feet in igth, the smallest angle being approximately 3P20'. Find its ler angles and its area. log 102 = 2 009, Lain 31» 20' = 9-7 16, log 185 = 2-267, L sin 700 30' = 9-974, log 192 = 2-283, L sin 78^0' = 9-990, log 2 = -301, log 9234 = 3965. If the circumference of a circle be divided into five parts in rithmetical progression, the greatest part being six times the least, IfixpresB in radians the angle each subtends at the centre. 74. Define the sine of an angle, wording your definition so as to include angles of any magnitude. Prove that sin (90" + A) = cos A, and cos(90<> + -4) = -sinJ, and by means of these deduce the formula sin (180° + ^) = - Bin ^, cob (180" + A) = - cos A, 75. Prove the formulae : (1) cot:^A=cosec^A-l; (2) cot* ^ + cot^A = cosec*il - cosec'il. Verify (2) when .i= 30". 76. Show that all angles satisfying the equation cos d= cos a re included in the formula 6 = 2nir ± a. Solve completely the equation 2 cos'^ + sin-0 - 1=0. 77. If sin B be the geometric mean between sin^i and cob A, then co8 2iJ = 2co8»(i4 + 45«). 78. The lengths of two of the sides of a triangle are 1 foot and L^^^2 feet respectively, the angle opposite the shorter side is 30''. Prove I^^Ehat there are two triangles which satisfy these conditions; find their I^Hunglcs, and show that their areas are in the ratio ^3 + 1 : ^3 - 1. I^H 79. If the circumference of a circle be divided into six parts in I^^Lrithmetical progression, the greatest being six times the least, '^"express in radians the angle each subtends at the centre. 80. Define the tangent of an angle, wording your definition so as to include angles of any magnitude. Prove that tan (90° + ^) = - cot J, and by means of this formula deduce the formula tan (180° + ^4) = tan id. !■" 260 TRIGONOMETRY. 81. Show that all angles satisfying the equation tan 0=tana are included in the formula e=nir + a. Solve completely the equation sec^^ - 2 tan^^^ 2. 82. Prove that cos {A + B) = cos AcosB- sin A sin B, and deduce the expression for sin (A+B). Show that cos A cos {B + C)- cos B cos {A + C) = sin {A - B) sin G. 83. Establish the identities : (1) l + cos4-sin^ = ,y2(l + cos^){l-sin^); sec''* A (2) sec 2 A = 2 - sec2^ .„v 27r 47r Gtt . v Stt Stt ^ . (3) COS — + cos — + COS ^ + 4 cos- cos --cos — + 1 = 0. 84. Two adjacent sides of a parallelogram 5 in. and 3 in. long respectively, include an angle of 60*^. Find the lengths of the two diagonals and the area of the figure. 85. Investigate the following formulae : ^A (1) sin -=.(1 + 2 cos ^) sin ^4; (2) sin [6 ^■ 5) - sin ^ = cos ^ sin 5 (1 - tan d tan 1 5). 86. Prove that (1) sin 10» + sin 50^ = sin TO^ ; (2) .^3 + tan 40o + tan SQO = ^3 tan 40" tan 80« ; (3) il A+B + G=1%0\ sin ^ - sin 5 cos C _ sin J5 - sin A cos G cosB ~ coaA 87. Find the value of sin 18®, and deduce that 4 sin 180 cos 360 = 1. 88. The length of one side of a triangle is 1006*62 feet and the adjacent angles are 44" and 70". Solve the triangle, having given L sin 440 = 9-8417713, i sin 700 = 9-9729858, L sin 660 ^ 9-9607302, log 1006-62 = 3-0028656, log 7654321 = 6 -8839067, log 103543 = 5-0151212. EXAMl'LKS FOR EXElWlUh. LWVIa. 2G1 Find the length of the arc of a circle whose radius is 8 feet rhich subtends at the centre an angle of 50<>, having given x=31416. 90. Prove that sin i4 = - sin (^4 - I8OO) . Find the sines of 30® and 2010<*. 91. Investigate an expression for all angles which have a given cosecant. Write down the general value of cosec"! ( - ,^2). (92. Prove that (1) co82ii + co82I?-2co8-4co8Bcos(it + B) = sin« (A-^B)\ (2) cos'^ + sinM cos 2B = cos^B + siu'B cos 2A. Prove that in any triangle a" cos 2B + Ifi cos 24 = a» + i' - \ah sin A sin B. If a = 123, B = 29n7', C=135o, find c, having given log 123 =2-0899051, log 2 =-3010300, log 3211 = 3*50GG403, diff. for 1 = 1302. L sin 150 43' =9-4327777. 95. Define the unit of circular measure, and prove that it is an ivpriable an^le. If an arc of 12 feet subtend at the centre of a circle an angle of 50®, rhat is the radius of the circle, tt being equal to 3-1416 ? 96. Express the cosine and cotangent in terms of the cosecant. If cotii + coseCii = 5, find cos4. 97. Investigate an expression for all angles which have a given int. Write down the general value of sec~^ ( - 2). Prove that (1) sin' A + sin'B + 2 sm i4 sin 5 cos (A+B) = sin' [A-\-B)', (2) Bin»4 - cos'il cos 2jB = 8in2B -cos^^cos 2 A. )9. Prove that 2co8i-4 = :izJl-^%\nA^Jl -sin 4, and deter- le the proper signs of the roots when A = 2 100®. 100. Prove that in any triangle cog 24 cos 2B _ 1 1 __ _^^ _ _____ . Prove (i) 2 cot 24 = cot A -tan A, (ii) sin-i I - sin-^ ^ = sin-» H. 4 cos 2 A (iii) cot (A + lo«) - tan {A - 15")=-^^ ^^-^ 262 TRIGONOMETRY. 102. Solve the equations cos(2a; + 3?/) = i, cos (3a; + 2i/) = J^3. 103. Trace the changes in the sign and magnitude of sin lir sin x) as X increases from to 2ir. 104. Solve the equation tan-i X + tan-i (1 - a;) = 2 tan-^ ikjx - x'K 105. A ship sailing due north observes two lighthouses bearing respectively N.E. and N.N.E.; after sailing 20 miles the lighthouses are seen to be in a line due east ; find the distance between the light- houses in miles accurately to four places of decimals ; having given, log 2 = -3010300, L tan 22030'= 9-6172243, log 11-715 = 1-0687423, and log 11-716 = 1-0687794. 106. Prove (i) tan|= ^ /(l^l^V .... tan 5^ + tan 3^ , (") r Ka 4. — ^ = 4 cos 2d cos ie, ^ tan 6^ -tan 3^ * (iii) sin-i | + sin-i j^= gin-i 7|., 107. Solve the equation 2sin2 ^- (1 + ^3) sin 2^ + 2^3 cos2^=0. 108. Show that in any triangle (i) a sin {B - C)+hsin{C- A) + c sin {A-B) = 0, (ii) c(cos.4+cosJ5)=:2(a + &)sin2iC. 109. Prove that if D, E, F are the feet of the perpendiculars from A, B, C upon the opposite sides of the triangle ABC the diameters of the circumscribing circles of the triangles AEF, BDFj CDE are a cot A, b cotB, c cot C respectively. 110. A man who is walking on a horizontal plane towards a tower observes that at a certain point the elevation of the top of the tower is 10*^ and after going 50 yds. nearer to the tower the elevation is 15" ; find the height of the tower in yards to 4 places of decimals, having given L sin 150=9-4129962, i cos 50 = 9-9983442, log 25-783 = 1-4113334, log 25-784=1-4113503. 111. Prove that when sin 4 is a geometric mean between sin B and cos B, then cos 2^ = 2 sin (45" - B) cos (45" + B), 112. Prove (i) sin A (cos 2 A + cos 4^ + cos QA) = sin 3.4 cos 4^4, (ii ) 2 sin-i ^ = cos-^ J. 113. In any quadrilateral ABCD prove that AB cos A-BC cos {A+B) + CD cos D= AD, r EXAMPLES FOR EXERCISE. LXXVI a. 263 114. Prove that in any triangle tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C, hence show that if xyz are numbers such that x + y + z = xyz, then x(l-y^){l-t^}-¥y{l-z'){l-x^)+t{l-x^{l-y*) = ixyz, 116. From each of two ships a mile apart the angle is observed which is subtended by the other ship and a beacon on shore, these angles are 52" '25' 15" and 75° 9' 30"; find the distances of the beacon from each of the ships, having given L sin 75° 9' 30" = 9-9852635, L sin 52« 25' 15" = 9-8990055, log 1 -2197 = 0862530, log 1-2198 = -0862886. l^el^ The cosines of two angles of a triangle are f and f| respec- ely ; find all the trigonometrical ratios of the third angle. ,,„ _, ,., sin 4 + 2 sin 3i4 + sin 5 J 4 sin ^ - 3 cosec ^ 117. Prove (i) — g-j— =-t = — ; -. — ^ 7- , ^ ' cos A -2 cos dA + cos oA 4 cos A-dBecA (ii) cot-4=cot-i3 + cot-if. 118. The lengths of the side of a triangle are 242, 1212 and 1460 show that the area is 6 acres. 119. Prove that adi'' + bd^^ + cd^^ = abc, where d^, d^ d^ are the distances of the centre of the mscribed circle from A, B, C. , 120. A man observes that when he has walked c feet up an inclined plane, the angular depression of an object in the horizontal plane through the foot of the slope is a ; and that when he has walked a farther distance c feet the angular depression of the same object is /3; show that the inclination of the slope to the horizon ia cot-' (2 cot )3 - cot o). I: l^r 122 121. Prove tan-i (i tan 2vi) + tan-' (cot i4) + tan-'(cot» J) = 0. 122. The general solution of tan (45- ix)+cot (450- .U)=4 is (6n±l) 60°, n being any integer. 123. Prove that ^(1 + sin 4 ) = 1 + 2 sin i J ^^(1 - sin i J), when A lies between - 90° and 180°. 124. If asinil+6sinB+C8inC7=0, and aco8il+6co8 5 + ccosC = 0, then a:h: <' = 8in(/?-C) : %\n{C-A) : sin {A - B). L. E. T. 19 264 TRIGONOMETRY. 125. A man ascends a mountain by a direct course, the inclina- tion of his path to the horizon being at first a and afterwards chang- ing suddenly to /S, which continues to the summit; the mountain being a feet high and the angle of elevation of the summit from the starting point being y, show that the length of the ascent is ^ cos{^(a + /3)-7} ^ sin 7 cos i (j3 - a) * 126. cosec ^ cosec 2^ + cosec 2 A cosec SA = cosec A (cot ^4 - cot 3^ ) . 127. Solve the equation cos 3^ + cos 50 + ^2 (cos 6 + sin 6) cos ^ = 0. 128. In any triangle 4 sin ^ sin B sin^ ^C = (sin £ + sin C - sin ^) (sin C + sin ^ - sin B). 129. Find (without tables) to 3 places of decimals the numbers of which 1-6, '3 and 1-3 are the common logarithms. 130. Two chords diverging from the same point on the circum- ference of a circle are to each other as the sines of the angles they respectively make with the tangent at that point. 131. Prove that there are eleven, and only eleven, pairs of regular polygons which are such that the number of degrees in an angle of one of them = the number of grades in an angle of the other ; and that there are only four pairs when the number is an integer. 132. cos 114 -h 3 cos 9^-1-3 cos 74 + cos 5^ = 16 cos^ A cos (44 + iw) cos (44 - iir). 133. If am-^x + sm-'^^x = ]^7r,thenx = i^{^\{5-2,J2}}.. 134. Transform ^/(2 sec 4 ) = ^ ^/(cos^ B cosec G) into an equation between tabular logarithms. 135. The distances of 4 one of the angular points of a regular octagon ABCDEFGH from the sides BC, CD, BE respectively are as 1:1 + ^2:2 + ^2. 136. sin (2a + /3 + 7) sin (/3- 7) + sin (a + 2^ + 7) sin (7-0) + sin (a + 18 + 27) sin (a - /S) : 137. If log f8^i=a and log 2=/S, then log 4100 =a+ 12^. 138. sin 810 _ gi^ 390 _ gjn 210 + sin 99*= sin 90*. 139. Solve sin {n + 1) ^ + sin (w - 1) 5 = sin 2d. |Bho. tne aiu EXAMPLES FOR EXERCISE. LXXVI a. 2G5 140. ABCD is a quadrilateral and AB^ DC produced meet in E; e angle DAG = DBC=a; the angle CAB=fi; the angle CBE=y', )ve that BE = AB sin /3 sin (a + jS) 8in(7-/9)8in(a + /3 + 7)' If ■H 141. cos 2il + sin 2B = 2 sin {^ir -(A- B)) cos {^tt - (^ + B)) IH cos 2^ - sin 2B = 2 sin {^Tr -{A+ B)\ cos {4ir -{A- B)\. 142. (i) cos 55« + cos 650 + cos 175" =0, (u) 8ina240-Bin»Go = i(V5-l). 1143. If A +B + C7+D = 180», then Hb 2/1 - cos 25 + cos 2(7 - COB 2D = 4 sin (B + C7) cos (C + A) sin (J + B). ^m 144. Given log 35 = a, log 325 = 6, log 245 = c, find log 5, log 7 and log 13. 145. A train is going due East at the rate of 24 miles an hour; when will it be 18 miles distant from a town which is ou the N.E. of e train at a distance of 24 miles ? If 146. then 147. then 148. then 149. 150. If If 4 sin il tan (i< - B) + 3 sec ^4 =4 sin' A sec A, tan ^ tan B = 3. If tan (t cot 6) = cot (t tan d), 4tan5 = 2n+l±V{4n2 + 4n-15). « = cos 2a + cos a, y = sin 2a + sin a, 2x=(x« + yV-3(x2 + y2). In any triangle 2i2 8in C = 6co8i4 + ^(a'- 6^ sin-/l). In a triangle right-angled at C tan- r^ — +tan-i ; 6 + c a + c 4* 151. One angle of a quadrilateral is 60*, another 60«, another f^r ; express all four angles in degrees. 162. 153. If sin-^m + Bin-*n=i7r, prove that 8in-*m=co3-^ n. If cos I cos /3 - « then tan^a: V(^:) tan i/S. 1 - « cos /3 , (008 d + sin e) (cos 26 + sin 20) = cos ^ + cos (3^ - Jt). . In any triangle a^b^c^ - Ab-S"^; a^-c^ coa A - ibcS- and Ac^S^ are in o. p. lii-2 266 TRIGONOMETRY, 156. Simplify coss {A+B) + cos^ {A - B) - cos 2A cos 2B. 157. With two units differing by 10'' the measure of an angle is as 3 is to 4 ; find the units. 158. cos 70 30' = I { - 1 + V2 + x/3) ^(2 + ^/2). 159. In any triangle sin ^A + cos ^B - sin ^G _ 1 + tan |B sin ^A + cos ^G - sin ^B ~ 1 + tan ^G ' 160. In any triangle a'^, b^-c^,b^+c^- 26c cos (B-C), are in g. p. 161. Solve cos3^ + 8in3^=cos£> + sin^. 162. If a; = 3 cos + cos 30, y = 3 sin - sin 30, then 163. cos2 18« sin2 36o - cos 36^ sin IS" = ^. 164. Given that 003^=-^ is one solution of the equation cos0+cos3^=i, find the others. 165. If the diagonal BD of a quadrilateral inscribed in a circle passes through the centre its area = {s-a){s-d) = {s- b) (s-c). 166. n tan(4+B)=3tan^, then sin(24 + 2B) + sin24 = 2sin2B. 167. sin 180 + cos 180=^2 cos 270. 168. Solve tan 8^ + tan 2^-!- tan = 0. 169. U A+B + G= 90°, then cos^ 2 A + cos^ 2B + cos^ 2(7 = 3 sin ^ sin 5 sin C - sin 3A sin 3B sin 3(7 + 1. 170. In any triangle be cos^ ^4 + ca cos^ ^B + ab cos^ ^G = (s)^ 171. Prove that 7d=ir satisfies the equation 8 cos (9 cos 20 cos 30 = 1. 172. If ^+J5 + C=90^ then cos ^ cos J5 cos G cannot be greater than I ^B. 173. sin-i^v'5 + cot-i3 = i7r. 174. In any triangle sin 3^ + sin BB +sin 3(7 + 4 cos ^A cos |J5 cos f C=0. 176. If the circle circumscribing an isosceles triangle is equal to the escribed circle touching one of the equal sides the triangle is right angled. EXAMPLES FOR EXERCISE. LXXVIa. 267 176. If 8m4; + 8in'x = l, then co82x + co8*a;=l. 177. sinlO'J8in50«8in70o=i. 178. Sunplify tan-» ( z ^—, ) - cot"! ( v—z ] , ^ "^ \l-XBmdJ \« - sin 6 J 179. A, Bf C, D are the angles of a quadrilateral ; prove that the sum of the products of the sines of J. 4, ^B, ^(7, i^D taken two together is equal to the sum of the products of their cosines taken two together. 180. At each end of a horizontal line of length 2a the altitude of a certain tower is a and at the middle point of the line it is /3 ; show that the vertical height of the tower above the horizontal plane is a sin a sin /3 ^{cosec (P + a) cosec (/9 - a)}. 181. Prove that sin (o - /3) cos 2/3 + cos (a - /9) sin 2/3= sin 03 - a) cos 2a + cos (j3 - a) sin 2a. 182. Find the values of sin 26 cos ^9 and tan d when ^ = sin-i ^. 183. Prove geometrically that sin 26 and 2 sin 5 cos 6 are always of the same sign. 184. If D is the middle point of the side BC, then 4JD»=&«+c« + 26ccosJ. 185. If I, m, n are the distances of the centre of the inscribed circle from ABC, then Imn __ 2ahc r ~a + b + c' 186. If a sin 0=6 sin 0, c = a cos ^ + 6 cos 0, then tan {6 + c „ 2 2 2 "■ (6 + c)(c + a)(a + 6)~ B^ TTTa TTB' ^ '^ ' ^ ' cos — jr . cos — - — . COS — z z ^ (32) If a, /3, y, d are the angles of a quadrilateral inscribed in a circle, then 008 (a + ^) . cos (/3 + y) . cos (y + d) . cos (d + a) = ( 1 - cos* a - cos* /3)'. (33) If $1 and dj be two values of found from the equation 008^+6 sin d=c, then - . cos --x— ^ = I . sin -*-7i— = - . cos * * . a 2 b 2 c 2 + CCOS0) + 6cos^) (34) If a= 6 cos x + c cos 1 6=ccos^ c=ocos0 prove that cos2^ + co8 20 + cas2x + 4coe^. co8 . cos^ + l^O, hence [see Ex. (20), p. 143], prove that (^db0*;j)=:(2n+l)7r. 274 TRIGONOMETRY. (35) A circle is drawn touching the inscribed circle and the sides ABy AC (not produced) of the triangle ABC. Another circle is drawn touching the circle and the same side. Prove that the radius of the n* circle thus drawn is r I I aod \l + sin-/ that the sum of the radii of all the circles, when n = oo , is ^cosec^-l). (36) If AO, BO J CO pass through the centre of the circum- scribing circle and meet the opposite sides on DEF^ then 1 1 1 _ 2 ad"" BE'^ CF~ R' (37) If h and k be the lengths of the diagonals of a quadri- lateral and 6 the angle between them, prove that the area of the quadrilateral is \ hh sin 6. (38) If -T^ = • ^ = -^ and X + Y+ Z^ ISQO, prove that sm ^ sm 1 sm Zi X =y cos Z+ z cos Y. (39) Two regular polygons of n and 2n sides are de- scribed such that the circle inscribed in the first circum- scribes the second; also the radius of the circle inscribed in the second is to that erf the circle circumscribing the first as 3 4- Vs is to 4 \/2. Prove that n = 6. (40) Prove by the aid of a figure like that of Art. 282 that I^0^==R^ + 2Rri. (41 ) U2/z + zx+a?7/ = lf prove by Trigonometry that ^^ , y , ^ 4^2; + , o "I l-«;2 l_y2 l_;g2 (l_^)(l_y2)(l_22)- [See Ex. (33), (32), p. 194.] EXAMINATION PAPERS. tl. PREVIOUS EXAMINATION. December, 1886. PAPER I. Define the secant and cotangent of an angle, and prove that (i) sec dcosecd=tand + cot&, /••\ xo ., sec ^ (u) cot^ d 1 --Sin 5-1 - 1 + Bin 5 1 + aeo 5 PAPER U. 1. From the definitions of the Trigonometrical Functions, prove that sin'^: I Prove that ;l-co8'ii and sin^i tan J=Beo^-co8^. i (co8«^ + sinSJ) - i (coB^^ - sin'-' A)^= ^. 2. Investigate the values of tan 45° and sin &0^. Two adjacent sides of a parallelogram are of lengths 15 and 24, and the angle between them is 60°; find the lengths of both diagonals. 1^^ 1 rjofir PREVIOUS EXAMINATION, CAMBRIDGE. June, 1887. PAPER I. 1. Define the sine and tangent of an angle, and shew how to find the sine and tangent whose cosine (m) is given. If sin A = tan J5, prove I cos'^co8'£=s(co8£ + 8inJ5) (cosJ3-Bin£). t The Additional Subjects in the Cambridge Previous Examination, which are required of Candidates for any Tripos, are now (1887) either Mathematics, or French, or German. The Mathematical subjects are (i) The Trigonometry of one Angle, (ii) Elementary Dynamics [see Lock's Dynamics for Beginners], (iii) Elementary Statics. The Trigonometry is set in the first one or two questions in each of the two papers. The questions quoted above were set in December, 1886, and in June, 1887. 276 EXAMINATION PAPERS. 2. Trace the changes in the tangent of an angle as the angle changes from 180° to 2700. If sin = - f , find tan 6 ; and explain by means of a figure the reason why there are two answers to the question. PAPER n. 1. Explain the mode of measuring angles in degrees, minutes and seconds. Find the number of seconds of angle through which the earth revolves about its axis in a second of time. 2. Find the value of sec 600, and of sec 45*'. Solve the equation 2 cosec A = 2smA + cot A. III. CAMBEIDGE LOCAL EXAMINATIONS. Dec. 1886. Junior Candidates. 1. Prove that the angle subtended at the centre of a circle by an arc equal to the radius is the same for all circles. Express the angle as a fraction of a right angle. 2. Define the sine, cosine and tangent of an angle. Prove that these trigonometrical ratios are always the same for the same angle. Find these ratios for an angle of 45*'. , / 8. Prove that ^ co8{A+B)=co8AQOsB-ainABinB. Prove that the sum of the cosines of two angles is equal to twice the cosine of half their sum multiplied by the cosine of 'half their difference. 4. Prove the following relations : /. T^v tan^ + tanB (ii) (l + sin4+cos^)2=2(l + sin^)(l + cos J). ,.... sin3ii + sin5^ ^ , (m) ^-. --=cot^. ^ ' cos SA - cos bA 5. Prove that the logarithm of the product of two numbers is equal to the sum of the logarithms of the numbers. Having given log 2 = -3010300, log 7 = '8450980, find the logarithms of (1-75)^, (24 5)"3. I EXAMINATION PAPERS, 277 6. Prove that in any triangle ABC (i) a=6co8(74-c cosB. (ii) 2bc COB A = b^ + c^-a'^. 7. Shew how to solve a triangle when one side and two angles are known. Find the side 6 in the triangle ABC from the following data : a=156-22, B=67»25', C=63042', log 15-622 = 1-1937366, L sin 57»25' = 9-9256261, log 16-37652 = 1-1808297, L sin 58053' = 9-9325330. 8. The angles of elevation of the top of a tower on a horizontal plane observed at two points distant a feet and b feet respectively from the base and in the same straight line with it are found to be complementary. Shew that the height of the tower is Jab feet. If e be the angle subtended at the top of the tower by the line joining the two points, prove that . ^ a~b a +b IV. OXFOBD LOCAL EXAMINATIONS. July, 1887. Junior Candidates. 1. What is T? ' What is 'the angle whose circular measure is t*? In a triangle ABC the angle ^ is x degrees, the angle B x grades, and the circular measure of C is — ; find the number of degrees in y each of the angles. 2. Prove that 2 sin ^ = ± Jl + sin A ± Jl - sin A^ and determine which are the correct signs when 2700>.4>180«. 3. Obtain the following formulae : (1) cos {A-\-B)=(mbA cos B - sin ^ sin B ; (2) tan A + tan B = sin (J. + JB) sec ^4 sec U ; /3) Bip( P-Hg)-2 BiPI> + Bu^ ( J>19) = tan n • ^' cos(2? + 5)-2cosp + coR(p-g) (4) 8in20 8in20 = sin2(^ + ^)-sm»(tf-0). 278 EXAMINATION PAPERS, 4. Given log 2 = -3010300, log 3 = -4771213, log 24668 = 4-3921339, log 11 = 1 -0413927, log 24669 = 4-3921515 : find the logarithm of 30-25 and calculate the value of {165 X (30)9 x\/24}-- (121)7. 5. In a triangle ABC, a, b, c are the sides, s is the semi-perimeter, A the area, R, r the radii of the circumscribed and inscribed circles : prove that ^,, , B-G b-c A (1) tan^- = ^^cot-; (3) 2i? = acosec^ = 6 cosec J5=ccosec C; (4) 2A = casing. 6. In a plane triangle the sides a, b and the angle A are known ; shew that in general two values of c can be found, and that the differ- ence of these values is 2 Ja^-b^sin^A. 7. A ladder placed at an angle of 750 just reaches the sill of a window 27 feet above the ground on one side of a street. On turning the ladder over without moving its foot, it is found that, when it rests against a wall on the other side of the street, it is at an angle of 15*'. Find the breadth of the street. 8. If 5 = 360 46', 6 = 311-8785, c = 521 -05, find C. [L sin 5 = 9-7771060, log 31187=4-4939736, log 521-05 = 2-7168794, log 31188=4-4939875.] V. CAMBEIDGE LOCAL EXAMINATIONS. December, 1886. Seniob Students. 1. Explain the method of measuring angles in circular measure, and find the circular measure of the angle of a regular pentagon. Prove that tan'^ cot^ A _ 1 - 2 sin^ A cos^ A 1 + tan'^ 1+ cotM sin A cos A 2. Establish the identities : \ ^ ^ 3 - 4 cos 2il + cos 4J. (1) tan ^-3.,.4cog 2^+008 44 * (2) sin4-sin5 = 2 sin — ^ — cos . I EXAMINATION PAPERS. 279 U A+B+ C' = 180^ shew that A B C (sin^-sinC) cot- + (8in (7-Bmil)cot jr +(8mi4 -sinB) cot- = 0. Z Z 2 3. Find an expression for all the angles which have a given sine. Solve the equations : (1) cos^ + tan^ = sec0. (2) 8ine-2 8in2^co8e + cos35=co8 2^. 4. Prove that in any triangle if a, b, c be the sides opposite to the angles A, B,C -. sin .i _ sin B _BinG ^^ '~a~~ b ~~c~ ' (2) cos^+cosi? + cosC=l+ jz, where r and R are the radii of the inscribed and circumscribed circles respectively. If the line joining A to the centre of the inscribed circle meets the opposite side in D, prove that tan^Z)B = ^^an|. h-c 2 *5. If d is the circular measure of a positive angle less than a right angle, shew that sin^ lies between 6 and ^- -s- . D Find the limiting value of f- sin- | when n is indefinitely in- creased. *6. Prove that, if 6 lie between - and - - , 4 4 d=iQ.nd-\i2inH-\-\t&\i^0- Shew that log (a + 6 V^) = i log (a' + i=) + \/^ tan-i - . * See Higher Trigonometry. VI. OXFORD LOCAL EXAMINATIONS. July, 1887. Sbniob Candidates. 1. Prove geometrically that the sine of 90^ + A is equal to the cosine of A for all values o(A. Find all the values of B, less than 180<>, for which sin 5B = i»J'2. L. E. T. 20 280 EXAMINATION PAPERS. 2. (1) Prove ttiat 2 8in^= ± ^^l + 8in 2A±s/l- sin 2A ; and determine the signs* of the roots when 2 A is greater than ISO*' and less than 2700. (2) Verify the identities : (a) sin 4 A = 4: cos A (sin A -2 sin^ A) ; (^) (tan iA + tan 2A) (1 - tan^ 3 A tanM) = 2 tan 3 J sea^A. 3. Prove the following properties of a triangle ABC, of which the sides are a, b, c, and the semi-perimeter is s : (1) &rea.=ijs{s-a){s-b){8-c); bccoa A + ca cos B + ab COB C ^ ' bc + ca + ab a sin ^ + & sin P + c sin G {b + c) sm A + {c + a)sin.B + {a + b) sm G 4. ABG is an isosceles triangle, having the vertical angle A = 5Q^ 30' ; on AB as base an isosceles triangle DAB is described, having its vertical angle D = 38" 40': the perpendicular from ^ upon J5C = 22*75 inches: find the length of the perpendicular from D upon AB to four places of decimals. L cot 19° 20' = 10-4548807 ; L cos 28» 15' = 9*9449220 ; log 11375 = 4-0553514; log 86754 =4-5653046, diE. for 1 = 119. 5. (1) If in a triangle the radius of the circumscribed circle is double of the radius of the inscribed circle, the triangle is equi- lateral. (2) Compare the areas of two equilateral and equiangular polygons, each of n sides, one inscribed in and the other circum- scribed about the same circle. VII. WOOLWICH. December, 1886. [N.B. — Great importance will be attached to accuracy.] 1. Explain what is meant by the circular measure of an angle, and shew that the circular measure of an acute angle is intermediate in value between the sine and the tangent of the angle. The perimeter of a certain sector of a circle is equal to the length of the arc of a semicircle having the same radius. Express the angle of the sector in degrees, minutes, and seconds. 2. Prove that the secant of any angle will be either greater than -i-1 or less than - 1. Shew that sec4= sfc^yi + tan^^, and explain the appearance of the double sign. 3. Express sin [A - B) and cos (^ - JB) in terms of the sines and cosines of A and B. Find the values of sin 15", sin 105", cos 165", and tan 195". B" EXAMINATION PAPERS. 281 4. Obtain a formula including all angles which have a given gent. If tan A + tan 2A = tan SA^ prove that A must be a multiple of 60* ' or 900. 6. A ring, 10 inches in diameter, is suspended from a point 1 foot above its centre by six equal strings attached to its circumference at equal intervals. Find the cosine of the angle between two consecutive strings. 6. Prove that ii\ ' n A 2 tan A 11) em 2^ = - — - ,, , . ^ ' l + tan-*^ h(2) cos* 4 - cos ^ cos (60« + A) + sin^ (30 -A) = l. 7. State and prove the rule for finding, by inspection, the racteristic of the logarithm of a fraction. Find from the tables suppUed an approximate value of the seventh root of -000026751. 8. Express the cosine of any angle of a triangle in terms of the three sides. Prove that Bm{B-C) : Bin{B + C) :: b^-c^ : a^. 9. The sides of a triangle are 17, 20, and 27. Find from the tables supplied all the angles. 10. A man travelling due west along a straight road observes that when he is due south of a certain windmill the straight line drawn to a distant church makes an angle of 30" with the direction of the road. A mile further on the bearings of the windmill and tower are N.E. and N.W. respectively. Find the distances of the tower from the windmill, and from the nearest point of the road. 11. Find the area of a regular polygon of n sides inscribed in a circle of given radius. If a regular pentagon and a regular decagon have the same perimeter, prove that their areas are as 2 : ,J5» VIII. SANDHURST FURTHEB. December, 1885. 1. Define the two common units of angular measure. Find the oiroular measure of 42<', and find the angle whose circular ire is |. Assuming that sin {A+B) = BmA cos B + cos AainB} coB{A+B) = coBAcoBB-BinAH'uiB^ ' find in terms of the ratios of A, the values of 8in2^, co32.-(, t&n2A, sin J^, cos^A &nd t&n^A. 20—2 282 EXAMINATION PAPERS. 3. Prove that and from the equation 2tan2^ = sec-^ find a general expression for 6. 4. Prove that in any triangle ^ 2(a + 6) . „ cos A + cos B = — ^^ 8in2 40. A lighthouse appears to a man in a boat 300 yards from its foot, to subtend an angle of 6° 20' 27-7"; find in feet the height of the lighthouse, having given i tan 60 20' = 9-0452836, difference for 1 = -0011507; log 3 = -4771213. 6. Prove that in any triangle tan 2^ + tan 2B + tan 2 C = tan 2^ tan 2 J5 tan 2 (7. Hence shew that ii x, y, z are three numbers such that x + y+z=xyz, then x{l- ?/2) (1 -z^)+y{l- z^) (1 - a;2) + ^ (i _ ^2) (1 _ y) = Axijz. IX. SANDHUEST FUBTHEE. June, 1886. 1. Express the value of the tangent, secant, and cosecant in terms of the sine of the angle, and also in terms of the cosine of twice the angle. Find the values of the tangent, secant, and cosecant of 22^30'. 2. Prove the following: (sin 2^)2 = 2 cok2 ^ (1 - cos 2^), „ , cos 5A - cos 7 A tan QA = -^——- ; — -~ ; sin 7 A - sin 5 J. J .- a cos A and if - = , 6 cosB prove that a tan A + b tan B = {a + b) tan ^{A+B). 3. Shew that in any triangle (6 + c - a) tan ^^ = (c + a - b) tan JjB = (a + 6 - c) tan | G (b + c-a) {c + a-b) {a + b-c) )h^ a+b+c ) If jR, Ta, Ti, re are the radii respectively of the circumscribed and three escribed circles of a triangle, shew that JlVa {s-a)= Rr^ {s-b)= Rr^ {s-c)= ^abc, where S is the semi-perimeter of the triangle. m EXAMINATION PAPERS. 283 4. Find the area of a triangle in terms of (i) two sides and the igle between them, (ii) two angles and the side between them. 6. Define the characteristic and the mantissa of a logarithm, d the logarithm of 5 when the base is 3, and the logarithm of ^ when the base is 5 ; logjo 3 = -4771213, logj^ 2 = -3010300. 6. Two sides of a triangle are 9 and 7, and the included angle is SS** 56' 32*8" ; find the base and remaining sides L tan 19<'29'=9-5487471, L tan 19" 28'= 9-5483452. X. COLLEGE OF PKECEPTORS, PUPILS' EXAMINATION. I July, 1886. Pi. In a quadrilateral ABCD, the angle ^ = 30<', B-QO grades, f^=z^ir; find tlie number of degrees in D. Find what number of -legrees must be taken from the angle D and added to A so that the ure may be inscribed in a circle. 2. If sec^=7n, what will cot^ be? Shew from your result it when ^ =0 or 180** the cotangent will be infinite in both cases. i I^H 8. Prove sin (A+B) = sin A cos jB + cos ^ sin B. W^m Shew that if A+B + G=180^, I^^V sin iii + sin £ + sin C = 4 cos i^A coa i^B coa J^C ; l^^pd that this is true numerically if A^ B, C the angle of a triangle be ■^^0, 60, 90 degrees respectively. 4. Solve cot^^ + tan' A=^\ giving the general value of A. 16. If log. 243 = 5, find a. Given logjo 30= 1-4771213, log^o 20 = 13010300, find logio72 and ; 6. In any triangle, prove the formula * ^ ^ /i(«-6)(«-c) id shew from it that if a =5, 6 = 3, c=4, then A will be a right Can you give any explanation of the double sign ? 7. Find the area of a regular polygon of n sides, each side being a. 8. Two angles of a triangle A and J5, and their included side C being given, find an expression for the area; shew from the result that if A and B each remain constant while the side C varies, Euclid's statement is true ; ' that the areas of similar rectilineal figures vary in the duplicate ratio of their homologous sides.' 284 EXAMINATION PAPERS. 9. Standing at a certain point, I observe the elevation of a house to be 15<* 15', and that the sill of one of its windows, known to be 20 feet above the ground, subtends an angle of 20^^ at the same point ; shew that the height of the house is nearly 22 yards ; logio 200 = 2-3010300, Ltan 500 15' = 10 -0800379, X tan 200=9-5610659, XI. OXFORD AND CAMBRIDGE SCHOOL EXAMINATIONS, 1884. 1. Define a degree, and the unit of circular measure ; and find the ratio of the first to the second. In a triangle ABC, the circular measure of one angle (A) is a, and the number of degrees in another is B. If the ratio of the number of degrees in the third angle (C) to the circular measure of B he B : a, prove that A, B, G are in continued pro- portion. 2. Define the secant of an angle, and trace its changes in sign and magnitude as the angle increases from 90" to 270^^ SimpUfy ( 1 1 ) sec ^ cosec ^ — 1 ) [cos 6 + tan^ d sin d cos 6 cot^ + sin ^ ) cosec ^ - sec ^ ) * 3. Prove that if two angles have the same tangent, their difference is a multiple of two right angles. Solve the equation sec e (sec2 6+2) (cosec d - sin 6) = 4. 4. Prove geometrically that when .4 +J5 is less than 90", cos {A + B)~ cos ^ cos B - sin A sin B. Prove that 5. A hill is inclined at a angle 36^ to the horizon. An observer walks 100 yards away from the foot of the hill, and then finds that the elevation of a point halfway up the hill is 18". Find the height of the hill ; and find the ratio of an error in measuring the distance walked to the consequent error in the height of the hill. 6. In any triangle prove that sin A _ sin B _ sin G 1 a ~ b ~ c ~2R* where R is the radius of the circle described round ABG. Find b, c, R in the triangle for which i5 = 45", = 60", a = ^2. EXAMINATION PAPERS, 285 7. Find the area of a regular polygon of n sides ; and deduce the area of a circle. ^^^^ 8. Find the cosines of the half angles of a triangle in terms ^^H[ the sides. ^^^■^ If the bisectors of the angles A, 7>. C of n triando meet the ^^^Bes in D, iU, F respectively, prove tha* I, Questions 9, 10 in Higher Trigonometry. [I. OXFORD AND CAMBRIDGE SCHOOL EXAMINATIONS, 1885. 1. What is the unit of circular measure? Wliy is this unit used? The angles of a triangle are in A. P. ; prove that the mean angle is 60". If the number of right angles in the greatest is equal to the circular measure of theileast, express the angles in degrees ('=f) 2. Define the tangent of an angle less than 90", and extend tlie definition to angles unlimited in magnitude. Find tan 30", tan 120", tan 240o. 3. Find the general solution of the equation sin 0-- sin a. Solve the equations : (i) 2 cos2 6 + yJ2 sin 6 = 2. (ii) (1 + tan ^) (1 - sin 2^) = 1 - tan 6. 4. Prove geometricaUy : (i) sin (.4 - B) = sin A cos B - cos A sin B. cos (j4 + B) _ 1 - tan ^ tan /? ^"^ e^n{A~B) ~ tan .4- tan iF" Find the relation between a, /8 and 7, in order that cot a cot/3 cot 7 - cot a - cot /3 - cot 7 should vanish. 5. Prove the following statements : 1 + sin 2x + cos 2x ^' i + sin 2x - cos 2x = cot .r. (ii) 3 co8-a = 4 cos« - + 4 sin" n ~ !• (iii) cos 3o cos 2a + sin 4a sin o = cos a cos 2tt. 28G EXAMINATION PAPERS. 6. Prove that in any triangle -^iih ^/(:-^)■ Having given that a =12270 ft., & = 11550 ft., c=: 11433 ft., log 1-76265 = -2461661, log 5-3565^-7288811, log 1-155= -0625820, log 1-1433 = -0581602, L cos 320 15'= 9-9272306 diff. for 1'= -0000797 ; find the angle A. 7. Find an expression for the radius of one of the described circles of a triangle in terms of the sid€s. Prove that be ca ah „„ ^& c c a a b A - + — + — =2R - + - + - + -+ -. + --31. f'l ^2 r^ {a a b b c c J 8. Prove that if 6 be the circular measure of an angle, the limiting value of — -r— as d is diminished ii unity. 6 What is the limiting value of when n is diminished? Questions 9, 10, 11 in Higher Trigonometry. XIII. TEINITY COLLEGE. June 6, 1885. 9—12. 1. Define the sine and tangent of an angle. The base BG of a triangle ABC is trisected in Q and B. Prove that sin BAR sin CAQ = 4: sin BA Q sin GAR. 2. Give a geometrical proof that (i) cos {d + (p) = cos 6 cos - sin ^ sin (p, (ii) cos - cos ^ = 2 sin ^ (^ + (p) sin ^{d- 0), where -^ and ^ are each less than a right angle. Show that if sin2 d cos - cos^ ^ sin ^ _ sin^

l ft. (3) 150? in.. (4) 3^ ft. (5) 7fTft. (6) 560. (7) 15i nearly. (8) 33600. (9) 32f. (10) 7 ft. (11) 553i,13-8in. (12) 339? ft. (13) 443 in. (14) 235 in. (15) 203 in. (16) 274 in. (17) 1886 in. Vn. Page 28. (1) -632118 of a right angle. (3) -021827 of a right angle. (2) 1-0426991 „ (4) -03294894 292 TRIGONOMETRY. (5) -006241 of a right angle. (6) 10-000812 (7) -3204052 (8) -0102034 (9) -6900071 of a right angle. (10) 1-19030045 „ (11) 10-061801 „ (12) -0226048 (13) 36s 78' or. (14) 1048 30^21". (15) Is 20' 3". (16) ia20". (17) 68 25\ (18) 3028 12' 50\ (19) 100« 10\ (20) 18 r -r. (2I) 6456 10'. (22) 28 30'. (23) riO". (24) 10". VIII. Page 30. (1) (2) (3) (4) (5) (6) (7) •09175 of a right angle =98 17' 50". •0675 1-07875 -180429012345679 1-467 •54 10 14' 15". (8) (10) 210 36' 8-1". = 68 75'. = 1078 87' 50". = 188 4' 29", etc. = 1468 IT lTr\ = 548 44' 44-4". 70 52' 30". (9) (11) 160 12' 37-26''. 1530 24' 29-34". (12) 310 30'. IX. Page 35. 1. (1) 2 right angles or I8O0. (2.) § of a right angle. (3) - right angles. IT (5) 2 right angles. 2d (7) — right angles. If (9) 20 right angles. 2. (1) TT. (2) 27r. (4) - right angles. IT 4 (6) right angles. (8) -002 of a right angle. (^) I (4) (5) 3. (1) ^^^ 20000 4. (1) i- IT 180* TT 6 IT (6) 1<= ^7) rs-o- (8) r (9) 180 (2) 7 (3) ^ 12 * (4) 200' (6) 20-(W- ^^) 2-00- (^) ^'' (^^ '-• (2) V- (3) 1. (4) Stt 50 (5) W (6) ,33 I I A.\>^i\t.ii.> ju THE EXAMPLES. 293 X. Pages 37—39. (1) \. (2) 90. (3) 4|. (4) W. (6) 112ift. (6) 6Hft. (7) 9fft. (8) 838000 miles. (9) J radian = 6^V degrees. (10) 21H degrees. (11) 5\^\". (12) about 34 yds. (13) 1 : 3-1416. (14) 3-1416. (15) 3-1416. (16) 400:1. (17) -0000484.... (18) 49^ in. (19) ^i.e. a right angle. (21) 473:489. 180 (32) (i) i = l, (ii) k = ^^. XI. Pages 40—43. (1) ^ . (2) 330, IS**. (3) 90», 450. (4) 18«, 22i». (5) 38", 18". (7) 45", 60", 75". (8) .39", 60", 81". (9) 33-3' <^">b2- (^^iD- , ., i)C , ... CD , .... DA ,. , ^D , , DA- <^)jc- <^^>c3- <^>CD- ^^^^52- ^^>cj- ,,, ,., DB Bil ,.., CD OB ..... DB BA ^^^ ^'^CB' CA' <">CB' CA' ^"^^CD» CB' . ■ DB BC , . AD AB , .. DB BG ^''^ AB^ AC ^""^AB^ AC' ^^^AD* AB' 294 trig0n0metr7 ..^ ... da .... ba ac ,.... dc ,. , ab .ad ab , ., bd , ... db ba ae ^'^ab'^'Tc ^''^bo' ^'''^cd''''cb^'''ca' , .... da .. , ba ag , .dc , ., db bc (vui) ^. (IX) ^^ or _ . (X) ^. (XI) — or _ . <^) IE' (5) sm^ = |, cos4=|, tan^ = |; sinB = |, cosJ3 = f, tanjB = ^. (7) Of the smaller angle, the sme = T^5-, cosine = ^§, tangent =t^V. Of the larger angle, the Bine=^|, cosine =y\, tangent = V- (8) Of the smaller angle, the sine=i, cosine = ^ , tangent = —^ . Of the larger angle, the 8ine=-^ , cosine = ^, tangent =\/3. XIV. Pages 63—65. (1) 179 ft. (2) 346 ft. (3) 86-6 ft. (4) 138-5 ft. (5) 7ift. (6) 600,173 ft. (7) 63-17 yds. (8) 34-15 ft. (9) 73-2 ft. (10) 86-6 ft. (11) -866 miles = 1524 yds. (12) 173-2 yds. (14) 373 ft. (15) 3733 ft. (16) ^ miles = 6465 ft. (17) ^|^-. (18) 300. (19) about 623^ miles. \ XVI. Page 74. „ V . . I^r- TTT X A vl ~ cos^ A ^ COS A (1) sm^ = Ayi-cosM, tan^ = ^^^ — , cot ^ = -y- -^^^ ^ cos^ 7r_cos*2' sec A = -, , cosec A = COS A' J1-cos^A' ,«x • . 1 . cot4 ^ ^ 1 (2) sm A = , — , cos A = —. — _ , tan A = — - — , ^yi + cotM ^1 + cotM GOtA seGA=— i— , — , cosec ^=v/l + cof^ ^. cot^ ^ /sec^ A — 1 1 (3) sin^=^^^^^ -. — , cos^ = -, tan^ = ^/sec2^-l, ^ ' bbgA sec .4 ^ 1 . sec^ cot.i^= , , cof^ec ^ VsecM-l' Jseo^A-l' I ANSWERS ^ TO THE EXAMPLES. ii) BinA = -^ , , co8^ = >/^«^'4-\ tan^: ^' ooseoA* coseoA 295 x/coseo' A -I cot A = v/cos^^TJ-T, BecA = cos A= Jl- sin' ^, tan i4 = sec^ coseCil AjooBed' A-l' Bin A . . J\ - sin' A - , COtA = : am A Vl-sinM —pr- — r — , cosec A=- — - . Jl-Bin^A 8U1 ^ .. - , cot^ = . ., Jl + t&n^A tan^ tan A (6) sin ^ = . - — , cos ^ = r_ M r . /, — I — i—r . x/l + tan'^ A = »Jl + t&n* A, cosec A=^ - » tan^l (1) f. \ (4) (7) (9) XVII. Page 75. 3 • 2^2* (2) (3) Vl6 (8) « ' ^^ (11) /i2(l + fc') = l. XVni. Page 77. (2) #ec ^ increases continuously from 1 to oo . (3) sin A diroinishes continuously from 1 to 0. (4) cot 6 diminishes continuously from oo to 0. XIX. Page 82. (4) Yes. (5) No. (6) Yes. (7) (i) 60». (ii) -1000. (ui) O". (iv) -260«. (v) 115° (vi) 410". (vii) -^. (viii) ^. XX. Page 84. (1) 450. (2) 30«. (3) 450. (4) GO*. (5) 300. (6) 30*. (7) 30". (8) 0», or 46". L. E. T. 2t 296 TRIGONOMETRY. (9) 90", or 600. ^iq) 60". (11) 45«. (12) 45». (13) 900, or 450. (14) I80. (15) 450. (16) 450. (17) 300. (18) 300. XXI. Page 85. 2 =fe /2 (3) the value 3 is inadmissible. (4) — -^ • (5) f, or^. (6) i,ori. 7 /3 (7) the value — ^ is inadmissible. (9) 1 - sin^ d. 1-2 cos^ 04-2 cos^ ^ (10) l-3rin'« + 3 6in^«. (11) ^5^i . (13) i^'-^4. ^ ' 1 + sm 4 (14) cosec 6 decreases continuously from 00 to 1. (15) cot 6 increases continuously from to 00 . (16) 6=1, = f2- XXII. Page 89. (1) +6. (2) 0. (3) +2. (4) +3. (5) +10. (6) 0. (7) +7. (8) +7. XXIV. Page 94. (1) The second. (2) The fourth. (3) The second. (4) The third. (5) The fourth. (6) The first. (7) The second. (8) The first. (9) The first. (10) The fourth. (11) ' The fourth. (12) The first, if n be even, the third, if n be odd. * XXV. Page 98. (1) +, +, +. (2) +, -, -. (3) -, -, +. (4) -, +, -. (5) -, +, -. (6) -, -, +. (7) +, -, -. (8) +, +, +. (9) +, +, +. (10) +, +, +. (11) +, -, -. (12) -, +, -. XXVI. Page 100. ANSWERS TO THE EXAMPLES. 297 I (5) (7) (9) (11) (13) (16) -i- -1 +1 V2' V2* + i. + V3 2 ' 2. 1^ 1 i. - V8 V3* . +1. 1 (6) (8) (10) (12) (14) \/3 1 1 "V2' + n/3. V2' + 1. + i V3 1 ^ 2 • ^V3 2 . +*' V3. 2 i -V3. ■ V3- XXVII. Page 100. Each of these expressions changes continuously as the angle A increases from 0° to 360°, and their values are repeated at each com. plete revolution of the revolving line. Their values are given below when they are zero and when they cease to increase and begin to decrease, and vice versa. The first table gives also the sign of each between the values. 0» 900 180« 270° (1) cos^ (2 tan A + 1 + _ -1 _ + + 00 — + 00 — (3) cot^ 00 + - 00 + - (4) sec .4 + 1 + 00 _ -1 — 80 + (5) cosec A 00 + +1 + 00 _ -1 — (6) \-BinA (7) Bux'A +1 + + +1 + 2 + + + 1 + + + 1 + 00 450 90* 135« 1800 2250 270«l +k -h + k + 1 + n/2 + 1 -1 -V2 -1 00 + 2 00 -2 00 + 2 00 -1 + 1 + ^/2 + 1 -1 (8) sin A . cos A (9) sin i + cos A (10) tan^ + ootii (11) sin ^ - cos A -h -2 -V2 The following figures exhibit the changes in the sign and magni- tude of sin $ (fig. i.), cos 6 (fig, ii.), and tan 6 (fig. iii.). The measure of the distance from along the line OX = the circular measure of the angle ; the vertical distance from OX measures the Ratio. 21- 2 298 TRIGONOMETRY. Fig. i. The Curve of the Sine. x' A 5 +1 ^V IT '? f^ "^^ ^f -1 ly ? V y 2"ir X Fig. ii. The Curve of the Cosine. x' /^- 5 N ? IT .- r- i-K \^ '¥ / J -IT 2 e v^ y V Fig. iii. The Curve of the Tangent. IT T +c / ? - ; '? >- 4031 M 9 l^ l^ /^ XXXI. Page 112. (1) (i) 300,1500, -2100, _33oo. (H) 450,1350, -225o, -315". (iii) 600, 1200, _ 2400, - 300o. (iv) - 300, - 150o, 2100, 33(y\ (2) (i) 200,1600,3800,5200. (ii) ^ , ^, 1|^ . l^L. ,.., 8jr 13t 227r llir ~1 ' 7 ' 7 * "t"' 7 ' (3) (i) ^ = n;r + (-l)«(-^y (ii) ^ = r77r4 ( - l)«^. (iii)?i7r±'!". I ANSWEJiS TO THE EXAMPLES. 299 XXXII. Page 115. (1) (i) ^ = 2uT=fcJir. (ii) ^=wir + jT. (iii) d = 'mr-^Tr. (iv) d = nx-^x. (v) 5 = 2nx±|T. (vi) e=7Mr + |ir. (3) The tangent. (4) No. (5) (i) 600, _ 600, 3000, _ 3000. (U) 450, 225«, 405», - 136». (iii) -450, -225». 135», 315". (iv) -OOO, -240", 120o, 300«. (v) 144'', - 1440, 216", - 2160. (vi) 135°, - 45®, - 2250, 315*> XXXIV. Page 120. (5) i,H. (6) U,U- XXXVIII. Page 131. (1) an{d + 4>) + Bin{d-(p). (2) co8(a-/S)fco8(a + /9). (3) sin (2a + 3/3) + sin (2a -3/3). (4) cos 2o + cos 2/3. (5) sin 8^- sin 2^. (6) cos ^ + cos 2^. (7) i (cos 3^ -cos 5^). (8) i (sin 45- sin d). (9) sin 60° + sin 400. (10) i (sin 60° - sin 300). (n) 2 cos 35 cos 25. (12) -cos 45 sin 25. (13) 4cos»|8in25. XXXIX. Page 132. (4) tan 2a. (5) tanx. (7) -2 cot 2a:. (8) tan^.!, (12) 8in25 = l, .•.5 = 45"; sin20 = i, .-.0 = 150. 3UiV. Page 147. (2) Bin2i4=$, 8in3^=-U \/5. (3) sin 25= -|V15. sin 35=^,^/15, cos35= - |J ; in the second. XLVI. Page 149. ._, .A /l-cos^ ,_, . A /l - cos A (3) 8in2 = + y 2 . (5) Bm^=+ ^ . XL VII. Page 154. (1) (i) +, -. (ii) -,+. (iii) -, -. (iv) +, -. (V) +, -. (vi) +, +. (vii) -, -. (viii) -, -. (3) 8in90=i{V(3 + V5)-V(6-V5)}, co8 90=i{V(3 + V5) + V(5-V5)}, sin81o = co890, cos 189"= -co8 90, tan 202^0 = ^2- 1, • tan 97^0 = - (n/3 + ^2) (V'2 + 1). (7) Between (2n x I8O0 - 450) and (2n x I8O0 + 450). 300 TJIIGONOMETRY. XL VIII. Page 157. (1) e=^-\ + nn + {-\rl. (2) ^=:-| + «. + (-l)H^. (3) ^=-| + 7i7r + (-l)«|. (4) e=^1 + nir+[~ir'^, (5) ^=:+| + 2n7r=fcj. (6) O^l + nir^'^. (7) x=- 680 12' + n X ISO" + ( - 1)" (21o 48'). (8) a; = - 690 26' 30" + 2n x ISO^ ± (690 26' 39"). (9) a; = 750 4' + wxl80o+(-l)»(U0 56'). (10) a;=- a- J + 23i7r±/3. L. Pages 159, 160. (7) i{V3 + l). (8) ±n/2. (9) il. (13) i (2ri + 1) TT ; or, ^ (3n ± 1) tt. (14) wtt ± ^ ; or J (2/2 + 1) ir. (15) i(2n + l)T; or-g- + (-l)»j2. (16) (2« + l).;0r,cos-ii^'. (17) ^ ; or '^ . .. (18) Both equations are satisfied if [x^-y^)=iSiU odd integer ; or, if (fl!+2/)2=2n, and [x-yY=^m - 2n, m and n being integers. (19) (i) 2 sin ^. cos ^ = sin 2^, and therefore it goes through the same changes as sin 6, while 6 changes from to 2ir. (ii) cos^^-sin^^ = cos2^. (iii) sin 3^ goes through the same changes as sin ^, while 6 goes from to Stt. (iv) cot 2A ; compare with cot 4. (v) sin {d + a) goes through the same changes while d goes from to tt, as sin 6 goes through while d goes from a to 7r + a. (vi) cos [26 - a) goes through the same changes while 6 goes from to tt, as cos 26 goes through as d changes from - - to tt - ^ • (See Ans. to XXVII.) (20) They are the solutions of the equations sin ^= sin a, and cos ( ^ - e\ =cos ( ^ - a j ; and we know that cos f ^ - ^ j =sin d. (21) They are the solutions of the equations sin f ^ + ^ J =sin ^, andcosT^-^Wcos^; also cos T^ -|j =sin ( ^ + |) . sin^=cos|. ANSWERS TO THE EXAMPLES. 301 LI. Page 162. (1) (i) a=*+»*. (ii) a**-", (iii) a» ». (iv) a«"*"«. (2) (i) 6'4690116. (ii) 10-6243928. (iii) 13;75093H6. ^^(iy) -8853661, (v) 1-7968680. (vi) 8-9699598. (vii) 2-7345058. (3) 2«, 26, 2-\ 2-\ 2-\ 27. (4) 3«, 3^, S-\ S-\ S-\ 3"*. LII. Page 163. (1) -60206, -9542426, -90309, -7781513, 1-20412, 1-690196. (2) 1-146128, 1-20412, 1-2552726, 1-3802113, 1 -4313639, 16232493. (3) 1, -69897, 1-1760913, 1-39794, 1-4771213, 1-5440680. (4) 1-6563026, 1 60206, 1-6812413, 1-69897, 2-30103, 3. (5) 7-201593,3-858708. (6) -7545579,2-989843. (7) 1-4^32. {8j 2408-6. (9) (i) 4-5868. (ii) -93646. (10) 3-9549. (11) 40975-3 sq. ft. (12) 34-925 in. (13) 8-2617 in. (14) 110115 cub. yds. LIII. Page 165. (I) 3, »/, i, I, -|. (2) 3, 6, -1,-3,-6, 2. (3) 2,4, -1, -3, -2, -4. (4) 4, |, -1, -1. (5) 3, ^1,5, -2,3, -3. (6) M, i, i, i- (7) -7781513, 1-6232493, 1-20412. (8) 1-6901960, 1-5563026, 1-7993406. (9) 2-30103, 2-7781513, 1-845098. (10) -69897, 5228787, 1-69897. (II) 1-544068, 2-1760913, - 1 + -30103. (12) -6440680, -8627278, - 2 + -9084852. LIV. Pages 168, 169. (1) 4,2,0,6,1. (2) -2, -5,-1,-3. (3) 3, -1,0,1,0, -7. (4) 4, 1,6,3. (5) the second decimal place, the first dec. pi., the sixth dec. pi. (6) ten thousands, units, hundreds, third dec. pi. , first dec. pi. , units. (7) 10, 4, 25, 31. (8) 9, 11, 85, 4, 9, 6. (9) units, fourth dec. pi, , thousands, seventh dec. pi., second dec. pi. (10) tenth integral pU twelfth dec. pi., fifth dec. pl.^ units, twelfth dec. pi., first dec. pi. 302 TRIGONOMETRY. LV. Pages 171, 172. (1) 2-8901023, -8901023, 4-8901023, 5-8901023. (2) 6-7714552, -7714552,47714552, 2-7714552, 3-7714552. (3) -27724... (4) -00001638... (5) -77448... (6) '005963. LVI. Page 174. (1) Divide each log by 3. (2) Multiply each log by 2. (3) Divide each log by -30103. (4) Multiply each log by -4771213. (6) Divide each log by -4771213. (6) 3-32190... (7) 1-183... (8) 1-10730... -66438... LVII. Pages 175, 176. (1) 3, 0,i, 0,|. (5) -51375. (7) (i) .. ''"'' (iii) X- (4) 1-8121177,55. (6) 7,4,3,3. 7 log 2 + 4 log 7 log 2 + 4 log 3 21og7 '21og2 + log3 (11) 0, (9) 3a (u) X (iv) x = 21og3+log7 * 4 (log 3 + log 7) 2 b 3a+2 81og2 + 3(log3 + log7) (10) ^ l-logio2 be b + 1* 26 + 2' 6 + 1' & + 1' 2& + 2' & + 1* (12)63-31 = 32. (13) (a"- flW) integers. (14) 1-9485 nearly (19) 2-53855. (20) 4-59999. (21) 167 years. LVIII. Pages 181, 182. (1) -8839066. (2) 2-7513738. (3) 4-9413333. (4) 6-8086920. (5) -5710750. (6) 3-70404. (7) 45740-26. (8) 2492837. (9) -000439658. (10) 5-689158. (1) -6737652. (4) 410 48' 37". (7) 9-8515594. (10) 350 4' 23". LIX. Pages 184, 185. (2) -6737652 (5) 700 31' 43-6". (8) 9-7114477. (11) 280 16' 27 -5". (3) -9306572. (6) 750 31' 21". (9) 10-1338768. (12) 210 56'41" ANSWERS TO THE EXAMPLES. 303 LX. Pages 188, 189. (1) 34«19'31-8". (2) 1498-2 ft. (3) 450 36'56''. (4) 5293-4 ft., 6982-3 ft. (5) 576-2chainB. (6) 4729 chains. (7) 3666*8 feet. (8) 42« 15', 11444 chains LXI. Page 190. (1) 3842-9 ft. (2) 281-74 ft. (3) 115-3 ft. (4) 285-6 ft. (5) 580 17' 24", 3P 42' 36". (6) 6561 chains, 410 17' 12". (7) 81ft. (8) 1942 ft. (9) 646-7 mUes. (10) 1000 ft. LXVI. Pages 208, 209. (1) 41« 16' 51-5". (2) 73» 32^ 12", 62P 46' 18". (3) 290 17' 16", BV 55' 31". (4) 64» 31' 58"., (5) 730, 23' 54-4". (6) 41»24'34-6". (7) 82»49'9". (8) 750, 600, 450. (9) 1350, 30«, 15«. LXVII. Page 211. (1) 313-46 yds. (2) 28-87 inches, 31-43 inches. (3) 1192-55 yds. (4) 22-415 ft. (5) 24-995=25 ft. nearly, 17 -559 ft., 650 59^" 42". LXVIII. Pages 213, 214. (1) 108» 36' 30", 31° 23' 30". (2) 93° 11' 49", 360 48' 11' . (3) 57" 27' 25 -4", 620 32' 34-6". (4) 640 26' 47", 37«, 7' 13''. (6) 72*12' 59". (6) 20-5 chains. (7) 122-7. (8) 74* 13' 50", 350 16' 10". LXIX. Pages 218, 219. (1) .i=5in8'21",C = 88041'39"; or4 = 1280 41'39",C = lP18'21". (2) JB = 700 0' 56", (7= 59" 59' 4"; or, B = 109° 59' 4", C= 20° 0' 56". (3) B = 38»38'24", (7 = 91<'21'36", c = 155-3. (4) 6in6'10". (5) .4 = 720 4' 48", B = 41" 56' 12" ; or,A = 107° 55' 12", B =6° 5' 48", b = 17-56. '6) /S is ambiguons ; 60-3893 ft 304 TRIQONOMETRY. LXX. Page 220. The angles are given correct to the nearest second. (1) 28" 35' 39". (2) 1040 44' 39". (3) 320 20' 48". (4) 430 40^. (5) 1280 23' 13". (6) 106531ft. (7) 3437-6 yds. (8) 1728-2 chains. (9) 25376 yards. (10) ^ = 660 27'48",J5 = 120 55'12". (11) ^ = 92012' 53", B = 350 37' 7". (12) jB=2901'40",C=740 55'50". (13)1?^70035'24"; or, 109024' 36". (14) J5 = 51056'17";or,1280 3'43". (15) 5 = 620 6'10";or,1170 53'50". (16) Very nearly 900. (17) 1319.6 yds. LXX b. P. 220 (i), (ii). (1) cos A=\, cos i^ = i V3. (2) 450, 600, 750. (3) 1350,300,150. (4) 3. (5) 14. (6) 1+V3. (7) 1200. (8)1200. (9)1200. (10)900,360 52'. (11)1300 27'. (12)1250 6'. (13) 1200. (14) .4 = 540 or 1260, 5 = 1080 or 360. (15) «=!, (16) C=300, a=V3 + l, &=2. (17) ^ = 76o, a=&=2^3 + l. (18) (7=600 or 1200. (19) 100^3. (20) no. (22) ^ = 1050,0=600,5 = 150. (23) iV3(N/5 + l). (24) A =900 or 6OO, O=750 or lOoO, a = 2yJ2 or ^6. (25) 300 or 1500. (26) A = 45^ or 1350, B = 300 or 120o, & = 2 ^/2 (1 + ^3) or2V6(l + >y3). (27) 60", 7oO, 6 yds. (28) It is impossible. (30) 15:8^3:4^3 + 9. LXXII. Page 229. (15) To find the point E in an unlimited straight line CE at which a finite straight line AB subtends the greatest angle, a circle must be described passing through A and B, and touching the line CE in the point E. In (15) the centre of the circle lies vertically above JB, and in the horizontal line through the middle point of AB. LXXIII. Page 239. (1) (i) 10 sq. ft. (ii) 43-3 sq. in. (iii) 148-13 sq. yds. (iv) 84 sq. chains = 8-4 acres. (v) 100 sq. ft. (vi) 151872 sq. yds. (2) 4, lOi, 12, 14 ft. (4) 8^ ft. ANSWERS TO THE EXAMPLES. 305 LXXVI a. Pages 253—268. (1) BinA = i,co3A = ^. (3) .4 = n x ISO" ; or, 360» =fc 60«. (6) 4227-47 feet. (9) 300, 60°, 90», 1200, etc. have for sine i, Vf. 1. s/h h, 0, -i, •\/f-l» -\/f. -i respectively. (12) The other sides are 765-4321 ft. ; 1006-6 ft. (15) 300, 600, 900^ etc. have for t^n ^ ^3, ^3, <» , - J^, - W^, 0, i >/3, 00 , - ,^3, - J V^ respectively. ,,_, 168 168 32592 ,,.. . . ,_ <1^ i93'l95'193x-195- <^^^ Bec^ = W2. (21) e=i{2n + l)ir. (22) cos6a = 16cos'a- 20co83a + 5coso. (23) Both signs negative. (24) fU (25) fl^^ A radians; 6-72956". (26) sine, ^ -, tan, f ; cot, | ; cosec, f ; sec, |. (31) W^deg.; 19-098540. (36) C =18\ a = c= j^^^^^-^-j^y (37) - 13200. (38) - f ^2. (39) (2n + 1) ir ; or, 27iir ± Jtt. (41) 1 foot, 1200, 300; or 2 feet, 60°, 900. (42) 1040 28' 39". (43) -6300. (44) -\J5. (46) njr; and 2n7r± jx. (48) i y6±V2} and 150, 1350; or, lOo", 45o. (49) 90; 2860. 28'. 41- 16"; |, (52) (i) nV ± ^TT. (ii) ^nw ± ^t, ornTr + ( - 1)" Jr. (53) 37i sq. ft. (54) 40° . 29' . 1985". (55) roW^; T^^; f- (56) tan a = 4^3, cosec a = ^^^^3. (58) (i) n7r±iT. (li) JnTri^T; or, 2mr± Jtt. (59) 2 ^14 sq. ft. (60) 38° . 25' . 32-725". (61) 1-2 radians = 76-39416«. (64) The proper formula is 4-^/(1 + sin ^) +^(1 - sin 4), (66) 192 ft., 185 ft. and 9234 sq. ft. (67) 2-3 radian8= 131-7799260. (69) The proper formula is - ^{1 + sin A) - ^{1 - sin A). (72) 780 10', 700 30', 9234 sq. ft. (73/ A^; A'r; Mtt; ^tt; Hr. (76) (2n±i)7r, or (2n±J)fl-. (78) 1350, I50; or 450, lOoO. (79) A'; tjVt; ^x"-; Att; Ht; Hir. (81) (2nii)T. (84) 7 ft. ; ^19 ft. ; V v/3 sq. ft. (88) 1035-43 ft.; 765-4321 ft.; 660. (89) 6-981 feet. (90) i;-i. (91) n7r + (-l)''i7r. (94) 321-0793. (95) 13-751 ft. (96) H. (97) 2nr^ir. 306 TRIGONOMETRY. (99) 2 cos iA=+ ^{1 + sin ^ ) - ^(1 - sin A), (102) a; = |r7r±^V'r±AT, 2/ = f rr ± ix ± ^V^. (104) x = ^. (105) 11-7157 miles. (107) nr + |7r ; or, wtt + ^tt. (110) 25-7834 yds. (115) 1-219714 miles and 1 mile. (116) sin = |^, 008= - ^f, tan^ - ff. (127) 4^±^ = 27i7r±f7r; or, 4(2« + l)7r. (134) 5 + ilog2 + |Lsec^ = l-log2 + |Lcos5 + ^icosec. (139) m7r; or, 7i^::=r7r+(-l)''(i7r-^). (144) log5 = 2a-c, log7 = c-a, logl3 = &-4a + 2c. (145) In i (2 ^2 il) hours. (151) 60«, 45o, 135o, 1200. (156) 1. (157) 300 and 400. (161) e = mrOT2d = mr + iT. (164) cos ^=: J (1±^5). (168) sin2^=0orcos2^=ior -i. (178) 6. <^®^) ll^6^°^'^- (^^^^ m«+i>«=n« + g». /ino\ A B + C-A G + A-B A+B-G A+B + G (192) - 4 cos cos jr cos ^ cos ^ . ^ z z z (196) i7r-e=n7r + (-l)«(j7r-a). ncio\ A 90^-B-G + A W-G-A + B (198) - 4 cos ^r cos jr 90^-A-B + G . ^+B + O-90» cos ^ sin . LXXVI b. Pages 269—275. (10) - {x + y + z) {y +z - x) {z + x -y) (x + y - z). (11) (i) ^{6n7r-27r + (-l)",r}. (ii) ^^^y (ill) :r = °^+cot-i|tan^.^??:^±^ ^ ' 2 I 2 nsina-msm/S (iv) ^^^ + ^ (V) sin 2^ (8 cos 2^-1) (2 cos 2^ + 1) = 0. ( vi) sin 4^ . cos -5- . sin g- = 0. (vii) sin 8^ . sin 4^ = 0. (21) 223-17. ^^^^'^C^BrT^ ^ OF THE XJNIVERSITY CAMBRIDGE : PRINTKD BT C. J. CLAT, M.A. AND SONS, AT THB UNIVERSITY FKESS. Messrs Macmillan & Co.'s Mathematical Books. WORKS BY THE SAME AUTHOR. Fourth Edition Revised. Globe 8vo. ARITHMETIC FOR SCHOOLS. Complete with An- swers, 45. 6d. Without Answers, 4J. 6ii. Part I., with Answers, IS. ; Part II., with Answers, y. Fourth Edition with a new chapter on Commercial Discount, KEY. Crown 8vo. loj. 6d. *^^* The complete book and both parts can also be obtained without answers at the same price, though in different binding. But the edition with answers will always be supplied unless the other is specially asked for. ATHEN^UM : — •• Mr Lock shews that he is not only a proficient in arithmetic, but a master of the art of teaching it, and his work may be confidently recommended to both teachers and scholars." CAMBRIDGE REVIEW :— *' The best arithmetic for its purpose which we have seen." 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The subject throughout is treated in the author's best style, and the book can be cordially recom- mended for the use of beginners.... The working of examples, as every teacher knows, is the only way to obtain a thorough knowledge of any subject which requires mathematical treatment, and Mr Lock has fully recognized the importance of this. Typical examples, excellently se- lected, are worked out at full length, and numerous others are given as exercises. There is also a selection of papers from some of the Oxford and Cambridge examinations. A new departure is the introduction of a short chapter on graphic statics, which we highly approve of. The teaching of this subject has made rapid strides during the last few years, and the methods are so simple, and applicable in cases which would involve laborious calculations, that the introduction of the subject into text-books is very desirable. 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