a m 1 _73W1T 1' J Q AE S PE NINS55LAM AM PLANE TRIGONOMETRY PART I. WITH THE USE OF LOGARITHMS THE RIGHT REV. COENSO,.D. THE RIGHT REV. J. W. COLENSO, 1).D. BISHOP OF NATAL NEW EDITION LONDON LONGMANS, GREEN, AND CO. LONDON: PRINTED BY SPOTTISWOODE AND CO., NEW-STREET SQUARE AND PARLIAMENT STREET FACTS AND FORE3/ULIE To be committed to memory. r =.141 59 = 3 or 31, nearly circumference of circle = rr, area =- is. 180 w = -- = 57.29577: &~ (of arc) = radius; c~ (of angle) = unit of cire. meascure. The unit of circular measure is the angle which, in any circle, subtends at the centre an arc equal in length to the radius. sin & cosec cos & sec are reciprocals, and have the same sign. tan & cot J tan si- sin2 + cos2 =1 sin 30~= 1 cos cot - cot2-cosec2 = 1 sin 45~a = 1 /2 ~~sin ~~ s 45 C = 2 The sequence of signs in the four quadrants is for sine and cosecant ( + + - -), for cosine and secant ( + -- - ) for tangent and cotangent (+ - + -). sinA-= +sin (1800-A)=-sin (1800+ A)= -sin (-A) cosec A=.... =...... cos A -cos(180 -A)= -cos(180 + A)= + cos (-A) sec A=.......... tanA= -tan (1800~-A)= + tan (180 + A) = -tan(-A) cot.... =..... iv FACTS AND FORMULa. sin (AB)=sin Acos BcosA sinB tan (A +B) tan A ~ tan B cos (A + B) = cosAcos B sinA sinB 1 tan A tan B sin 2A = 2 sin A cos A 1- tan 2A cos 2A - cos2 A- sin2A cos 1 + tan 2A =2 cos2 A-1 2 tan A = 1 -2 sin2 4 tan 2 1 tan 2A 2 sin A = (2 - 2 cos 2A) 2 cos A4= /(2+ 2 cos 2A) sin A sin B sin C b2+C2-a2 — = =- cos A= a b 2bec s(s -a) (s-b) (s-c) cos A = sV F-c, sin 2A= be,wheres=+J(a+b+): area S = be sin A= /{s(s-a) (s-b) (s-c)}.,/2=1.41421..., 3=1.73205..., /5=2.23607... PLANE TRIGONOMETRY. CHAPTER I. ON THE MEASUREMENT OF LINES AND ANGLES. 1. TRIGONOMETRY, from TplJvovr, a triangle, and ftSTpSOo, I measure, means properly the science which treats of the measurement of triangles, that is, of their sides, angles, areas, &c.; being called either plane or spherical trigonometry, according as the triangles are drawn upon a plane or on the surface of a sphere, in which latter case the sides will be circular arcs and the angles curvilineal. But the word is now used in a much more extended sense, so as to include all algebraical reasoning about lines and angles (whether parts of a triangle or not), when carried on by means of certain quantities, that are called the trigonometrical Ratios or Functions of an angle. Of'these Ratios we shall speak presently, so far as the subject of plane trigonometry is concerned. But we must first make some remarks upon the mode in which lines and angles are represented for the purposes of algebraical reasoning. 2. A line is represented algebraically by some letter, as a, which denotes the number of times it contains a certain line, taken as the unit of measurement. Thus if the unit be afoot or an inch, then the line a would be one containing a feet or a inches. (T.) 'B 2 PLANE TRIGONOMETRY. 3. Here, however, we have involved reference to the idea of a line having a positive or a negative value. For a, we know, means -+ a; and, if in our reasonings we should happen to arrive at some negative result as the value of a line, (if, for instance, it came out, at the end of a problem, that the value of a certain line was-b), What would this negative sign mean? We shall show that it indicates contrariety of direction to some line which we have before represented by means of a positive algebraical symbol. Let AB be a line, measured from left to right, and C' A c0 n containing a units, which we express algebraically by a or + a. Then, if from the end B we measure off in the direction of BA a line B C containing b units, the length of the line A C will be a-b units, and, therefore, its algebraical expression will be a-b. Now, if b be less than a, this expression for A C will be positive, and C will also lie to the right of A, that is, A C will be measured in the same direction as AB. But, if b be greater than a, then C will fall to the left of A, as at C(, and A C' will be measured in the contrary direction to AB, its value in units being b- a; while, at the same time, the expression a -b will become negative, and may be written -(b- a). WTe infer, then, that, when the algebraical expression for a line is negative, the negative sign indicates contrariety of direction to a line which has been already assumed to be positive; and, conversely, that if a line drawn in any one direction from a given point, when represented algebraically, be assumed to be positive, then a line drawn from the same point in the opposite direction must be considered negative. PLANE TRIGONOMETRY. 3 4. By definitions of Euclid, an angle is the inclination of two straight lines to one another, and a right angle is one of two equal adjacent angles which one straight line makes with another. Hence an angle in Geometry must be always of necessity less than two right angles. And the same will be true in Trigonometry, properly so called, where we shall have only to deal with angles that are actually angles of triangles. But, in the wider sense of the word, an angle may be imagined to be of any magnitude whatever, being conceived to be described by the revolution of a straight line about a given point from one position to another. Thus the angle BAC may be conceived to have been described c by the revolution of AC about A from the poD sition AB to its present position: but this it ' - may have done by going one or more times ' completely round, describing in each complete revolution four right angles about A. 5. Here also we may show, as in (3), that if an angle, formed upon one side of a line, by a line revolving in one direction, be assumed to be positive, when expressed algebraically, then an angle, formed upon the other side of it, by a line revolving in the opposite direction, must be considered to be negative. For let BA C be an angle, formed upon one side of AB, which we denote algebraically by A or + A; and from AC take off in the opposite direction an angle CAD = B. Then the algebraical expression for the angle BAD will be A-B, and this will be positive or negative, according as A is greater or less than B, that is, according as BAD is formed on the same side of AB with the positive angle BA C, or on the other side of AB, as BAD'; in which latter case, its algebraical expression may be written- (B -A), where B-A represents the B2 4 PLANE TRIGONOMETRY. actual magnitude of the angle, and the negative sign indicates its position with regard to the positive angle BA C. 6. It will be seen more fully, as we proceed, that the above principle, by which we are led to express by algebraical symbols the direction as well as the magnitude of lines and angles in our figures, is one of the utmost importance. Its usefulness will appear by what follows. In solving a problem by common Geometry, it often happens that a difference in the figure may lead to a difference in the result. If, for instance, we have drawn a certain angle acute, when we might have drawn it obtuse; or upon one side of a line, when we might have taken it upon the other: we cannot always be sure that the result we have obtained with the one angle would also hold good with the other. Hence we are often obliged in pure Geometry to examine different cases of the same problem, and sometimes to state different modifications of the same result. An instance of this occurs in Euc. IT. 12, 13, where the square of the side subtending an angle of a triangle is greater or less than the sum of the squares of the sides containing it, according as the angle is obtuse or acute; and two different propositions are required to prove this, the latter also being divided into three cases. Now in Algebraic Geometry it is found that any figure that we draw, consistent with the conditions of the problem, will lead us to a result, that shall be true for all possible cases comprehended in it, provided we attend to the above Rules with respect to the signs of lines and angles, of the same kind but drawn in contrary directions, and interpret our results also in accordance PLANE TRIGONOMETRY. 5 with our assumptions. We are led to these Rules by such reasoning as that in (3) and (5): and their truth is confirmed by the perfect agreement of innumerable results, of all possible kinds and degrees of complexity, obtained by them, with results obtained by the more laborious processes of common Geometry. 7. In the annexed figure, BB', DD', are drawn, D c intersecting at right angles at A; cu s ~ and AC, AC, AC, AC4, are supposed to be different positions of the line A C, revolving about B[. s /a^^ A A from a position coincident with oc /a AB, in the direction from B to D, which it is usual to consider D' the positive direction. It is plain that the extremities of these lines will all lie in the circumference of a circle, whose centre is A. By the lines BB', DD', the whole angular space about A is divided into four equal parts, BAD, DAB', B'AD', D'AB, which are called respectively the first, second, third, and fourth quadrants. When the line begins to revolve from AB, it may be supposed to have made with AB an angle zero; when it reaches the position AC,, it will have described an angle BAC1 less than a right angle, which is said to be an angle in the first quadrant; when it comes to coincide with AD, it will have described the right angle BAD; when it reaches the position A C2, it will have described the angle BA C2, greater than one right angle and less than two, which is said to be an angle in the second quadrant; and, when it comes to coincide with AB', it will have described the trigonometrical angle BAB', or two right angles. PLANE TRIGONOMETRY. In like manner, the angle BA C3, viz. that subtended by the arc BD C3, greater than two and less than three right angles, is an angle in the third quadrant; the angle BAD' is an angle of three right angles; the angle BA C4, greater than three and less than four right angles, is an angle in the fourth quadrant; and, when the line becomes again coincident with AB, it will have described a trigonometrical angle of four right angles. And it may still be supposed to go on revolving, and so come again to the position AC, in the first quadrant, having described an angle greater than four and less than five right angles: and so on. So also, the above being all positive angles, the line may be supposed to revolve in the opposite direction, and form successively the negative angles BA C4, BA C3, &c. It will be found, however, that all the trigonometrical properties of the angle BA C4 will be the same, whether we suppose it to be the positive angle, greater than three right angles, or the negative angle, less than one, except such properties as may depend upon the actual magnitude of the angle, and not merely upon the position of the lines containing it. 8. Although it is usual, in treatises on trigonometry, to take AB as the starting, or initial, line, when, as in this case, we represent in one figure angles of different magnitudes, yet we might reckon our angles from any other line, or some angles from one line, and others from another. In all such cases, each angle would be in the first, second, &c. quadrant, reckoning the quadrants from its own initial line, and would be positive or negative, according to the direction which we might choose to regard as the positive direction for angles PLANE TRIGONOMETRY. 7 measured from that same initial line. Thus we shall sometimes find it convenient to measure angles from AD, and call them positive in the direction DB; then DA C, will be a positive angle in the first quadrant, and DA C2 will be a positive angle in the fourth quadrant, or a negative angle in the first. 9. In Euc. I. Def. 10. we are told that right angles are formed, ' whenever one straight line, falling upon another, makes the adjacent angles equal.' This defines the magnitude of a right angle (4), which, being thus known, may now be taken as a measure of other angles. 10. The right angle is divided into 90 equal parts, called degrees, each degree into 60 minutes, each minute into 60 seconds, angles less than a second being usually expressed as decimals of a second. The marks ~ ' " are used to express degrees, minutes, and seconds, respectively. Thus 59~ 14' 57".85 denotes an angle containing 59 degrees, 14 minutes, and 57.85 seconds. Hence it follows that in one right angle there are 90~, in two right angles 180~, in three right angles 270~, and in four right angles 360~, &c. So also ~ of a right angle is 30~, 2 of a right angle is 45~, ~ of a right angle is 60~, &c. 11. The complement of an angle is its defect from a right angle, or comp(A) = 90~ - A. The supplement of an angle is its defect from two right angles, or supp(A)= 180~-A. Thus comp(23~ 23' 27//)=660 36' 33", comp(123~ 13' 27") =-33~ 13' 27", supp(45C)=1350, supp(-45~)=225~ PLANE TRIGONOMETRY. Hence in the annexed figure, if BA C = 75~, I, c then its complement = 90~-75~ C, =: 150 = DAC1; and if BAC2 -\ 130~, then its complement = 90'- 130~= — 40= DA C2, the B /'^^ 'J angle being here formed upon 4 the negative side of AD, since, 1cN p / in taking DAC1 as a positive D' angle, we have made the direction from D towards B, in which DAC, increases, the positive direction for all complements, measured from AD. So likewise the supplement of BAC2 is B'AC2, and that of BA C3 is R'AC3, which latter is measured on that side of AB' which, though positive for the angle BA C3 itself, is negative for its supplement. To avoid the awkwardness of having the same direction positive for one set of angles and negative for others, it will be found hereafter generally more convenient to measure the complement and supplement, as well as the angle itself, from AB, all in the same direction. 12. Since the three angles of any triangle are together equal to two right angles, it follows that (1) Each acute angle of a right-angled triangle is the complement of the other; (2) Each angle of any triangle is the supplement of the sum of the other two angles. We see also that if A, B, C, represent the three angles of any triangle ABC, then A+ B+ C= 180~; and besides this, we only require two other independent equations to be given between the three angles, in order to determine them completely. Ex. 1. The angles of a triangle are in A. P., and the difference between the greatest and least is 20~: find them. PLANE TRIGONOMETRY. Let A the least =- 10 deg., C the greatest =x + 10 deg.,.'.B=x deg. A + B + -- 3x= 180;.'.x = 60~; and the angles are 50~, 60~, 700. Ans. Ex. 2. Suppose, in the triangle ABC, that A is 4 of B, and B is 4 of C; find the degrees in the exterior angle adjacent to C. Here, if the magnitude of C be represented by 7, that of B will be 4 of 7=5, and that of A 4 of 5=4. Then, since 7+5+4=16, it appears that C is -.7- of the sum of the three angles, that is, '7 of 180~;.'. its supplement will be 6 of 180~= 1010 15'. Ans. Ex. 1. 1. Write the complements of 15~ 37' 2", 142~ 15' 29", 1~1' 3".2. 2. Write down the supplements of 145~ 15' 20' and-27~ 17' 7". 3. If one acute angle of a right-angled triangle is 9~ 9' 9/, what will the other be? Express them both in degrees only. 4. Given the difference of the two acute angles of a right-angled triangle to be 24~ 24', find the two angles. 5. The base angle of an isosceles triangle is 59~ 59' 59": find the vertical angle. 6. The vertical angle of an isosceles triangle is a third as large again as either of the base angles: find the three angles. 7. The semi-sum of two angles of a triangle is 60~, and their semi-difference is 100: find the three angles. 8. The exterior angle of a triangle is half as large again as one of the two interior and opposite angles, and its supplement is half as large again as the other: find the angles of the triangle. 9. The angles of a right-angled triangle are in A. P.: find them. 10. The supplement of one angle of a triangle is double of the complement of another, and triple that of the third: find all three. 13. By Laplace, however, and some other French writers, the right angle was divided into 100 equal parts called grades, each grade into 100 minutes, each minute into 100 seconds, which were denoted by the marks ": thus 599 14' 577"85. It will be understood that the Foreign minute and second, though called by the same name, are not of the same magnitude as the English. 10 PLANE TRIGONOMETRY. The advantage of this decimal method of dividing the right angle, (which was introduced in France, with other decimal divisions for money, weight, length, &c. in the time of the French Revolution,) was that any angle, given in grades,minutes, and seconds, could immediately be written down as a decimal of a grade. Thus the above angle 59' 14' 57\.85 may be written down at once as 59.145785: for l'= —g, and, therefore, 14'=-o,1g=.14g, and l"-T-= —g, and, therefore, 57\.85= ---~5 g —5O-g-=.005785g. And so, conversely, an angle, given in grades and decimals of a grade, could be written down at once in grades, minutes, and seconds. Thus 43f.012345=43g 1' 23".45. The use of the grade, however, is now abandoned even in France, as it involved the sacrifice of such voluminous tables and records, already adapted to the sexagesimal division, or else of such a vast amount of time and labour, in reducing them to the decimal system. It may be observed that the sexagesimal method allows of the whole angular space about a point being divided into a greater number of aliquot parts than the decimal, the divisors of 360 being 24 in number, and of 400 only 15. Thus we can express exactly in degrees 1, 1,, 1-, &c. of a right angle, but not in grades. 14. To convert the English measure of an angle into the Foreign, and vice versa. Questions of this kind are now merely matters of curiosity; but they may afford the Student some useful practice. Since 100 grades=90 degrees, therefore 1 grade---- of 90 degrees=.9 of a degree. Ience, to reduce grades to degrees, multiply by.9; to reduce degrees to grades, divide by.9. PLANE TRIGONOMETRY. 11 Ex. 1. Convert 59' 14' 57".85 into English measure. Here 59r 14' 57".85=59.145785 grades.9 53.2312065 degrees 60 13.8723900 60 53~ 13' 52".34340 or, 53~ 13' 52". Ans. Ex. 2. Convert 59~ 14' 57".85 into Foreign measure. Here 59~ 14' 57".85=59.2494027 degrees, as appears from below;.9) 59.2494027 degrees 65.8326696 grades=65g 83' 26".69 60) 57.85000 =65g 83' 27". Ans. 60) 14.964166.2494027 Ex. 3. In the triangle ABC, the sum of the angles A and B contains 7.7 more grades than degrees. Find the angle C in English measure. Let A+ B = 9x degrees = 10x grades;.' x = 7.7; and 9x a 69~ 18'; hence the angle C= 180 - 69~ 18 = 110~ 42'. Ans. Ex. 4. One of two angles contains twice as many French minutes as the other contains English minutes; and the difference in magnitude of the two angles is one degree. Find each angle. Let the angles measure 9x and 9x -1 degrees. 9x deg. = 10xx 100 Fr. min.; 9x-1 deg. = 60 (9x —1) Eg. min. 10 x 50 = 60(9x —1); whence 9x = 13I, 9x -1 = 120~. Ans. 15. It appears, from Euc. I. 32, Cor. 1, that ' all the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.' Hence if n be the number of sides of any rectilineal figure, we have the sum of its n angles + 90~ x 4 =90~x 2n, or the sum of its n angles=(2n-4) 90~ =(n-2) 180. 12 PLANE TRIGONOMETRY. If the figure be also a regular figure, that is, equilateral and equiangular, then its n angles are all equal, and, consequently, each of them = n-180~. n Thus, if n=3, each angle of an equilateral triangle is 1 of 180~ =60~; if n=4, each angle of a square is 2 of 180~=90~; if n=8, each angle of a regular octagon is 8 of 180~=135~; &c. 16. Since (Euc. I. 15, Cor. 2) the sum of all the angles which the n sides of a regular polygon subtend at its centre, or at the centre of its circumscribing circle, is four right angles, or 360~, the angle subtended by each side is 3- at the centre, and (Euc. III. 20) half of n 1800 that, or -, at any point in the circumference. Thus the angle between the two lines which join the extremities of a side of a regular octagon with any one of its angular points is 180 — 8 = 22o0. Ex. 4. Find the number of sides of a regular polygon, each of whose interior angles contains 18 more grades than degrees. Each angle contains 18 x 10=180 grades; so that n-2 n2 200g=1808; whence n=20 sides. n Ex. 5. One polygon has 7 sides more than another, and the sum of the interior angles of the latter is 3 of the sum of those of the former. How many sides has each? Let the polygons have x and x-7 sides, respectively; then the respective sums of the interior angles are (x-2) 180~ and (x- 7 -2) 180~; and we have x-9=- (x-2), or 10x-90=3x-6; whence x=12; and the polygons have 12 and 5 sides, respectively. Ex. 6. The angles of a pentagon are in A.P. and the smallest is 60~. Find the largest. Here (n-2) 180~ amounts to 540~, the sum of the series. Let x~ be put for the greatest term;.'. a (x+60) is the mean term, and J r'-60) X 5=540, or x+4-60=216; whence x=156~. PLANE TRIGONOMETRY. 18 Ex. 2. 1. Write down the complements of —15, 279.25, 32' 1". 2. Write the supplements of g 2' 3", 899.0004, 225g.1001. 3. Convert 37g 15' 7" into English measure, find the complement of the result, and reconvert it to grades. 4. Express 1~ and 1" in Foreign measure, and 1I and 1' in English measure. 5. Find the interior angles of a regular pentagon and hexagon. 6. Shew that the interior angles of a regular octagon and dodecagon are as 9: 10. 7. If an isosceles triangle be inscribed in a circle on the side of a regular inscribed heptagon, find the ratio between its vertical and base angles. 8. The angles of a pentagon are as the numbers 2, 3, 4, 5, 6: find them. 9. In the figure of Euc. iv. 10, shew that AC is the side of a regular decagon inscribed in the larger circle, and of a regular pentagon inscribed in the smaller. 10. There are two polygons, such that the difference of the sums of their interior angles is four right angles, and the ratio of the same as 5: 3; find the number of sides in each. 11. One regular figure has twice as many sides as another, and an angle of the first is a third as large again as an angle of the second: find the interior angle in each. 12. The interior angles of a rectilineal figure are in A. P., the least is 120~, and the common difference 5~: find the number of sides. 17. In equal circles (Euc. vI. 33) 'the angles at the centre are proportional to the arcs which subtend them.' Hence in the figure of (7) the arcs BC1, BD, &c. of the same circle are proportional to the angles BzC1, BAD, &c.: and, if the whole circumference be divided into 360 equal parts, then BD will contain 90 of these, BDB' 180, &c.; and, generally, any arc will contain 14 PLANE TRIGONOMETRY. just as many of these parts, as the corresponding angle contains degrees. These parts of the circumference are also called degrees, and subdivided, as the angular degrees, into minutes and seconds, the same marks ~ / being used, as before, to denote them. And, since the number of degrees, minutes, and seconds in any arc will be the same as in the angle it subtends at the centre, (whatever be the radius, and, consequently, whatever may be the actual length of an arcual degree,) on this account the arc is often used as the representative, or, as it is called, the measure of the angle. So also BC4, &c., measured in the opposite direction to B C, will be negative arcs; and D C will be the complement, and B'C, the supplement, of BC1. 18. The circumferences of different circles are proportional to their radii. For let C, c, be the circumferences of two circles, whose radii are R, r; and suppose {idO - AB, ab, to be sides of two regular A\ polygons, each having n sides, in2xIfl Adz J^ scribed in the two circles. Then, A' - B' (16), AB, ab, subtend equal angles at the centres 0, o, and the triangles OAB, oab, being also isosceles, are, consequently, equiangular and similar; hence (Euc. VI. 4) AB: ab:: AO: ao, and, if P, p, be the perimeters of the two polygons, we have also P:p::n.AB: n.ab:: AO: ao:: R: r. Now let the number of sides be increased indefinitely, and their magnitudes be diminished; then the areas of PLANE TRIGONOMETRY. 165 the polygons tend more and more to equality with the two circles, and their perimeters with the two circumferences. Let P= C —X,p=c-x, where the differences X, x may be made as small as we please by increasing the number of sides; then C-X: c-x::R: r, or r C-rX= Rc- Rx, and, therefore, r C- Rc= rX- Rx, where each of rX, Rx, and, therefore, their difference rX-Rx, may be made as small as we please, and less than any conceivable magnitude, by increasing the number of sides. Now, if rC-Rc=any finite value a, then rX-Rx could not be made less than a, which is not the case: therefore, r C- Rc =0, or r C= Re, and C: c:: R: r, that is, the circumference of any circle ca its radius. 19. In different circles, the arcs subtending equal angles are proportional to the radii. For, in different circles, the length of one degree, (being the.1th part of the circumference,) will vary as the circumference, and, therefore, (18) will vary as the radius; hence the arcs, which in different circles subtend at their centres the same angle A, (and which contain, consequently, each the same number of degrees,) will vary as the length of one degree, and, therefore, will vary as the radius. So that, for the same angle, the subtending arc s radius of the circle. 20. Since the circumference of a circle a radius, and, therefore, o diameter, it follows (Alg. 167) that, for * O i,i,. circumference all circles, the ratio circumference has a certain fixed diameter numerical value. This value will be shewn (in Part II) to be (as far as five places of decimals) 3.14159=37 nearly, or, still more nearly, 3-5-; so that the circum-r~ Io thttectun t6 PLANE TRIGONOMETRY. ference of any circle is rather more than three times its diameter. This number, 3.14159, is, consequently, a very important one to be remembered, as well as the ratios 3-, 3-, which approximate to its value. (The latter of these, it may be observed, is formed of the figures 113355 in order, beginning with the denominator.) The number itself, 3.14159 &c., it is usual to denote by r; but, for most common purposes, we may take for xr the approximate value 3-= 22. 21. Hence, if C represent the circumference of any circle and r its radius, we shall have C = ~r, and, therefore, C= 27rr = the length of 360~; hence the length of 180~, or the semicircumference, is Ir, that of the quadrantal arc, 90~, is 2r, that of the octant, 45~, is 2r, &c. 2 4 The radius, however, is often referred to as the unit of measurement, and the arc in such cases is said to be expressed, or measured, to radius unity. The expressions for the lengths of 360~, 180~, 90~, &c., are then written 27r, 7r, ~Tr, &c.; in which cases it must be remembered that 27 really means 27r times the radius, whatever that may be, and so of the rest. 22. Generally, if the length of an arc of A~ be Or, (as A that of 180~ is 7r,) then Or: 7rr:: A: 180, or -- 7r 180 wher e, (ill be the measure of the ar the contains unity. radius,) will be the measure of the are to radius unity. Hence, if A be given, then we know the measure of the A arc to radius unity, for 0 = 1-; and, conversely, if a 180 be given, then we know the number of degrees in the PLANE TRIGONOMETRY. 17 0 arc, for A= - 180 = Ow, if we denote by c the number 180 =57.29577 &c. 7r COR. I. If 0 be taken = 1, or the arc (Or) be taken equal in length to the radius, then, by the formula just 180 found, we have A = -- = w; so that, in any circle, 7r the length of an arc of c (=570.29577) is equal to the length of the radius. COR. II. If the length (a) of an arc be given (in feet. yards, &c.), then 0 may be found by expressing the arc as a fraction of the radius; for a=Or, and, therefore, 0 a arc r rad Ex. 1. Express to radius 1 an arc of 27~. Ans. 0 = -127o = 9- T r, which answer, in terms of the known number r, will be generally sufficient in such a case, without putting for 7r its numerical value. Ex. 2. Find the measure of an arc of 10 feet to a radius of 4 feet. Ans. 0 = -~-= 21, by which is meant that the length of the arc is (not 2~feet, but) 2~ times the radius. Ex. 3. Find the length of an arc of 10~ to a radius of 10 feet. Ans. a = Or -= ~rr=r -= Trft =- X ' 2ft =1.74ft. Ex. 4. Find the radius, when an arc of 1~ measures an inch. a 180 Here a=0r;.. r -=1 (T7r)= —=5-7.29577in. = 4.77ft. Ex. 3. 1. Find the length of a degree of the meridian upon a globe of 18 inches diameter. 2. Obtain the number of degrees in a circular arc, 30 feet long, whose radius is 25 feet; and measure it to radius unity. 3. Find the measures of two arcs of 40 feet and 40 degrees, in terms of a radius of 2J yards. (x.) c 18 PLANE TRIGONOMETRY. 4 Find the thickness of a tree, which a man can exactly clasp round with three grasps of his arms, supposing him to stretch with his arms 6 feet each time. 5. Taking the Earth's circumference at 25,000 miles, find its diameter. 6. A mill-sail Is 7 yards in length, and goes round uniformly ten times in a minute. At what rate per hour does its extremity move?,7. The hour-hand of a watch is 5 inch long, the minute-hand i inch, and the second-hand -3 inch; compare the rates at which their extremities move. 8. The Earth is distant from the Sun 95 millions of miles, and describes her orbit round him (nearly) in 365 days; at what rate per minute does she travel?.9. The Earth being supposed a sphere, with a radius of 4000 miles, find the length of an arc of 1~, and the distance round the world on the Equator. 10. At what rate per minute does a point on the Equator move by reason of the Earth's rotation on her axis? 23. It will be noticed that in (22 Cor. in), a represents the length of the arc in question, or the number of feet, &c. in it, while A represents the number of degrees in it, and 0 the number of radii in it, or the number of times it contains the arc of w~, whose length is equal to the iadius. Hence also A will be the number of degrees in the angle which the arc subtends, and co will be the numbel of degrees in the angle which, in any circle, is subtended at the centre by an arc equal to the radius, while 0 will express the number of times the angle A4 contains the angle w~. So then, for the angle as well as for the arc, we have the equations of (23), 180 arc a 4= w, c = —, where 0 -a =- or a=r. 7r rad r PLANE TRIGONOMETRY. 24. It may be as well, however, to derive the above results directly for the angle, without referring to them as previously proved for the arc. Let there be any angle of A~, subtended by an arc a, at the centre of a circle whose radius is r: then, since the semicircumference rr is the arc which subtends the angle 180~, and (Euc. vI. 33) 'angles at the centre of a circle are proportional to the arcs which subtend them,' we have A a a 180 A: 180:: a: rr, or -- =- whenceA-=-x —l8 180 6 rr r 7r if (as before) we put 0 for the fraction a or and r rad 180 for the number 0 57.29577. 25. Theanglew (=570.29577=57017'45" or206265"), which is thus easily defined as the angle, which, in any circle, is subtended by an arc equal to the radius, may now, like the right angle, be used as a unit to measure other angles, and will be found to be an angle of great importance, more especially in the higher parts of the subject. It is commonly called the unit of circular measurement, or the circular unit; and the fraction arc rad (or 0) is called the circular measure of the angle. 26. We may represent, therefore, either the are or the angle by the circular measure of the angle, if we understand the unit ow, which, in the former case, is an arc of the circle of the same length as the radius, and, in the'latter case, is the angle subtended by that arc. We shall find it convenient to call the arc or angle 180~ 1 (= 57~ 29577), simply, a measure; and it is usual to 2 20 PLANE TRIGONOMETRY. denote the number of measures in an arc or angle, (or the circular measure of the angle,) by some Greek letter, as a, p, 7, &c., 0, P, 4, &c. Thus, if, for any angle, the length of the subtending arc in any circle be three times that of the radius, then the angle (or arc) may be expressed by 0, where the value of 0 is 3; meaning that the angle (or arc) contains 3 measures=3 X57~.29577=1720, nearly. Hence, since 180~ of arc=7r=7rcv~, we have 180~ of angle = rw~; and, consequently, the circular measure of 180~ is x, and those of 360~, 90~, 60~, 45~, 30~, &c. are 2r, 17, 1-r, tr, 6X, &c., the same, of course, as the measures (22) of the corresponding arcs in terms of the radius. 27. As the number of degrees (A) in an arc, or!80 angle 0, is given by A-= x cow= x - = 0 x 57.29577, 7r so the value of 0 for any angle of A~ is given by 80 18 A 0= A w =A_ -- 180=- n, just as in (22). 7S 0 180 Since 200 grades = 180 degrees, the number of grades 200 in a measure, will be 200; and, if A be the number of 7r grades in an arc or angle 0, then A=0 200 A = -- A T. tD 7r 200 Ex. 1. Find the circular measure of 35~. Ans. 0= —7r=n ^r. Ex. 2. Express the angle To-Uu in degrees, &c. Ans. A =20~ 2 5/' — 206".265 = 3' 26".3. Ex. s. 1. Obtain the circular measures of 72~, its complenment, and its supplement. 2. Express 6~ 3' 36" in circular measure. 3. The angles of a triangle are as 3, 4, 5; express them each in measures. 4. Express, both in measures and degrees, the angle subtended by an arc of 7 feet to a radius of v, yards. PLANE TRIGONOMETRY. 21 5. Find the number of degrees, minutes, and seconds in the angle.1. 6. Express 2' 18' 75" in circular measure. 7. Find the angle whose circular measure is a. 8. Find the complement of r'0, and the circular measure of the supplement of 27r~. 9. Express 1g 3' 75", in measures and degrees. 10. Find in degrees the angle whose circular measure is.7854. 28. In different circles, angles are directly proportional to the arcs which subtend them and inversely to the radii. This follows at once as a Corollary from (23); since, if', 0', be the circular measures of two angles A, A', subtended by arcs a, a', to radii r, r', then 0-, -a a, a r rrr an,4 ',4'" ' ~"a a' aar and A: Al:: 0:: a: a, or angle ocr. r r' rad But it may be well to give the following direct Proof, Let B C, B' C', be arcs subtending angles BA C, B'A C' to radii AB, AB', respectively, the circles Ho being supposed concentric without affecting \Zc.' the demonstration; and let the larger radius, A C' cut the inner circumference at c; then (Euc. vi. 33) BAC: L BAc (or / B'AC'):: are, BC: ar Be..arc B C arc ee..arB C. arc B' C' radAB radAB radAB rad AB' 29. It is proved, in Euc. xII. 2, that 'the areas of circles are as the squares of their diameters' or radii; and the same may be thus demonstrated, as in (18). For, referring to the figure in (18), let AB, ab, be sides of regular polygons, each having n sides, inscribed in two circles, whose centres are 0, o, and radii R, r; 22 PLANE TRIGONOMETRY. then the triangles OAB, oab, being similar, we have (Euc. vI. 19) area OAB: area oab:: A 02: ao2; and if A, a, be the areas of the polygons, we have also A: a:: n. OAB: n.oab:: AO2: ao2:: R2 r2. But, when the number of sides is increased and their magnitude diminished indefinitely, the areas of these polygons approximate more and more to those of the circles; and hence, reasoning exactly as in (18), area of circle ABC: area cf circle abc:: R: r2. 30. It will now further appear that the area of a circle, whose radius is R, is XrR2. For let A'B', in the same figure, be the side of a regular polygon, circumscribing the circle whose radius is R, and subtending the same angle at the centre as AB. Draw the radius OD to the point where A'B' touches the circle, and, therefore, perpendicular to A'B'. Then, since the area of any triangle (Euc. I. 41)= 2 area of rectangle of the same base and height= =base x height, we have area ofA'OB'= A'B' x OD= A'B' x R and, if P be the perimeter of the polygon, we have also area of polygon= - (A'B'+ &c.) R= P.R. But, when the number of sides is increased indefinitely, the polygon becomes a circle, and P= 2R; and, therefore, area of circle= I (27rR) R = 7 R12. Ex. The area of a circle, whose radius is 1 foot, will be rr, that is, 7r square feet, or 3+ sq.ft. The circumference will be 27r, that is, 2rr linear feet, or 6 ft. 31. Since (Euc. VI. 33) 'sectors of circles are proportional to the arcs on which they stand,' or to the angles which subtend them, it follows that the area of a sector whose are, or angle, is 30~, 45~, 60~, &c., will be XT1r2, 1 r2, 7r2 r2 &c. respectively, (since the whole be T'V P 6 PLANE TRIGONOMETRY 23 circumference contains 360~); and, generally, the area of A a sector, whose angle is A~, will be r- r2, or, if the 360 circular measure of A be 0, it will be r?2 =r Or2. Ex. 5. 1. Compare the magnitudes of two angles, when the arcs, which subtend them, are inversely as the radii and directly as the numbers i and i. 2. The measure of an arc is 33 to radius 2-; what would be the measure of an arc of the same length to radius 21? Express also in each case the angle at the centre in degrees. 3. What length of cord will it take to tether an ass, so that he may graze over an acre of ground? 4. The inscribed square is cut out of a circle whose diameter is a yard: find the area of the remainder. 5. Compare the areas of two circles, when an arc of 60~ in the one is equal in length to an arc of 45~ in the other. 6. Find the area of a section of the tree in Ex. 3. 4; and the quantity of wood in a length of 30 feet, supposing it of uniform thickness. 7. If a foot-square of card-board be divided into four equal squares, by lines bisecting the sides at right angles, how much of it will be left, when the four circles, that can be inscribed in these squares, are cut out? 8. The length of an arc of 45~ in one circle being equal to that of 60~ in another, find the circular measure of the angle, that would be subtended at the centre of the first by an arc equal to the radius of the second. 9. If a be the arc which measures the complement of an angle to radius r, find the arc which measures the supplement of the same angle to radius r'. 10. The diameter of one circle is equal in length to the semicircumference of another; how many degrees of the first are equal in length to the diameter of the second? ( 24 ) CHAPTER II. ON THE TRIGONOMETRICAL RATIOS OF AN ANGLE. 32. THE following are the quantities, which have been already spoken of under the name of the Trigonometrical Ratios. c Let BAC be any angle (A) in the first quadrant; from any point Pin AC, one of the lines containing the angle, draw PN perpendicular upon the other AB. Then (using hyp. for the A N B word hypotenuse), PN Eperp.N (1)PN, orP, is the sine of LA, or sinA-=; /A1P hyp. AP (AN or bae is the cosine of LA, or cos A=. AN (2)AP hyp.' A (3)N or rp,is the tangent of L A, or tanA= AN AN base AN (4 ", or ase, is the cotangent of L A, or cotA = pN; 4PN perp. PN AP hy AP (6) AP or hyp, is the secant of L A, or sec A= A v AN base AN ()AP ho yp. AP (6) or yp, is the cosecant of L A, orcosecA AP. perp. /2V Let the student notice distinctly that the values of these quantities are mere numbers, being the ratios of certain lines to one another. PLANE TRIGONOMETRY. 25 Ex. If the length of PAT be 3, (that is, of course, 3 units of length, whatever the unit may be,) and that of AN be 4, then AP = V(16 +9) = 5; and, therefore, sin A, cos A=o tan A =. 33. As PN, the side opposite, is the perpendicular, and AN, the side adjacent, is the base, in forming the Ratios for the angle BA C or PAN, so AN is the perpendicular and PN the base, in forming the Ratios for the angle APN, which (11) is the complement of P'AN or A. Hence we have AN sin APN= p = os PAN, or sin (90 -A)= cos A PN cosAPN= - = sinPAN, or cos (90~-A)= sinA. And so, generally, the sine, tangent, and secant of the complement are the cosine, cotangent, and cosecant of the angle itself; and, conversely, the cosine, cotangent, and cosecant of the complement are the sine, tangent, and secant of the angle. We may express this by saying that any Ratio or Function of an angle is the corresponding Co-ratio or Co-function of its complement. 34. It is plain that(1) and (6),(2) and (5), (3)and(4), are the inverse, or reciprocals, of each other; so that we shall have at once 1 1 1 sin A -- cos A tan A = cosec A secA cot A 1 I I cosecA = secA= --, cotA=-: sin A' cos A tanA and, consequently, whenever we have found the sine, cosine, and tangent of an angle, we may consider that we have found all its Ratios. 35. It will be easily seen that the numerical values 26 PLANE TRIGONOMETRY. of the different Ratios depend only upon the magnitude of A, and not at all upon the position of the assumed point P'. For if in A C, as above, we take, besides P, some other C point P', and draw P'N' perpendicular to AB, then P/ \ the triangles APN, A PN', /\ \ being similar, will (Euc. vi.4) 'have the sides about A N N', pi/\ B their equal angles propor~A. ~~~ N N/ /PN P'N' tional.' Hence the ratios N will be eql, or A-' AP' will be equal, or sinA will have the same value, according to our definition, wherever we take the point P in A C, and similarly for the cosine, tangent, &c. So, likewise, if we take a point P" in AB, (instead of A C,) and draw P"N" perpendicular on A C, the triangles APN, AP"N", will also be similar, and, as before, the PN P "N" two expressions for sinA, namely AP' AP" ' will still be equal: and so of the other Ratios. It appears then that for any one angle there is but one set of trigonometrical Ratios; so that, if the angle -be given, the numerical values of the corresponding Ratios are determznate, and, as will be shewn hereafter, can in all cases be found. Such values have been calculated for all angles from 0~ to 45~, (by which, as will appear, those of all other angles can be immediately obtained), and are registered in Tables for use. 36. It was formerly customary to consider the trigonometrical functions as belonging to the subtending arc rather than to the angle, and to represent each of them by a line rather than by the ratio between two lines. PLANE TRIGONOMETRY. 27 Let PAB be an angle at the centre of a circle, and subtended by the arc PB: the line PN, drawn from one extremity of the arc, perpendicular to the line AB, passing through the D other extremity, was called the sine Mof the opposite angle PAB, or of the arc PB; and, similarly, the line PM A -"B being regarded as the sine of the angle DAP, the complement of PAB, and being equal to AN, therefore AN was called the cosine of PAB. Thus when the hypotenuse was made radius, the perpendicular was the sine, and the base the cosine, of the angle at the centre. NB, = radius minus cosine of PAB, was called the versed sine of PAB, and MD, = radius minus versed sine of DAP, was called the coversed sine of PAB. Again, let PAN be an angle at the centre of the circle FcN, and subtended by the arc Nc, and let NP be a perpendicular to AN, drawn from one extremity of the arc to meet the f line AP passing through the other extremity: PIN, according to this construction, was called the tangent of A the angle PAN to which it is opposite, and AP the secant of that angle. Thus when the base was made radius, and A taken as centre, the perpendicular was the tangent, and the hypotenuse the secant, of the angle at the centre. Similarly, as the angle APN is the complement of PAN, then, P being taken as centre, and PN as radius, AN would be the tangent of APN, and was therefore called the cotangent of PAN, while 28 PLANE TRIGONOMETRY. AP would be the secant of A PN, or cosecant of PAN. Thus when the perpendicular was made radius, and P taken as centre, the base was the cotangent of the adjacent acute angle, and AP its cosecant. 37. Hence may be seen the origin of the names given to the trigonometrical functions. The name sine, from sinus, bosom, is the mere Latin translation of an Arabic word, PN, in the first of the above figures, being half the string, as it were, of the arcus, or bow, PBQ, of which PB is half, and being the part brought up to the breast of the archer in discharging it. The names tangent (touching line) and secant (cutting line) explain themselves from the second figure. The cosine, cotangent, &c., as already explained, were so called as being the sine, tangent, &c. of the complement. The name versedsine is not so easy to explain: but BN was formerly called, from its position, the sagitta, or arrow, of the bow, of which PQ is the chord, or string. We may here add, that the versed sine of the supplement of an arc was called its suversed-sine (suvers.). 38. It will now be seen how the old definitions of the sine, tangent, &c. as lines, may be converted into the modern definitions of them as ratios, by making the radius always unity, or the unit of measurement. Thus, AP being taken as radius, with A as centre, the line PN was the sine, and AN the cosine of the angle PAN; but to eApress the modern sine and cosine, which are not lines but ratios, we divide AP by AP, to make radius unity, then PN becomes the sine, a AN A the sine, and A- the cosine, to radius unity. And so with other AP functions. 39. Hence, if, in any case, the radius be not taken as the unit of measurement, let r be its numerical value, expressed in terms of some other unit, as feet, yards, &c.; then if we use Sin A (with a capital letter) to denote the line PN, and sin A to denote PN, the ratio -f, we shall have PLANE TRIGONOMETRY. 29 sin A=Sin A, or Sin A = r sin A; r and so with the other functions. 40. According to the line-definitions, then, it is plain that, unless the radius be taken for unity, the trigonometrical functions of any given angle are not fixed in value, as they are with the ratiodefinitions, but are always, like the arcs themselves, proportional to the radius employed. It becomes necessary, therefore, to mention the radius in any such case, or, at least, to refer to it as understood. It is this inconvenience which has led to the general disuse of these lines. 41. Let a line of unlimited length revolving about B, from an initial position coincident with the touching line BT, always pass through the extremity of the arc BP: then the T portion of this line intercepted between the two extremities of the arc, that is, the straight line D BRP, is called the Chord of the arc BP. This is A/ B \ a strictly correct definition of the Chord; but for all ordinary purposes, it will be sufficient to say, / that, If BP be joined, then BP is the Chord of the arc BP, It is easily shewn that chord A 2 Sin I A. For let BP be any arc A, and draw the radius ADC bisecting the chord BP (at right angles) at D, and its arc at C: then we have BP=2PD=2 Sin PC, or Chord A=2 Sin I A. We may define the chord as a ratio, as follows: Take equal lengths AB, AP, in the lines containing the angle PAB, or intercepting A; then the ratio A-, or BP, is called the chord of A: and, as before, chord A = BP-2Pp =2 sin ~ A. AP AP This function, however, is not very frequently used. 42. The Student should now accustom himself to express with ease any side of a right-angled triangle in terms: of any other side, by means of a trigonometrical Ratio; and it will be found that each side may be ex 30 PLANE TRIGONOMETRY. pressed, in terms of the other sides and angles, in four different ways. Thus if ABC be a triangle, right-angled at C, we have BC B A =sinA or cosB, and BC==AB sinA or AB cosB; - ACj =tanA or cotB, and BC=AC tanA or AC cotB. In like- manner,,C=AB cosA=AB sinB =BC cotA =BC tanB, AB=AC secA AC cosec B=BC cosecA-BC secB 43. By reference to the definitions in (32), the following results may be easily obtained: / A A PN PN AN sinA P ta-n A AN=- AP AP cosA' A AN _AN PN cosA cot A N — p AP AP =sinA A N B in2A + o A =PN2 AN 2 PN2 + AN2 sin2A cos2A= (-) + ( - AP2 \AP} \AP} AP2 whence sin2A = 1- cos2A, cos2A = 1 - sin2A, or sinA=.V((1-cos2A), cosA = V(1 -sin2A): secA _ =( 2_ _- AN+ = 1 + tanYA; ACN AN'2 A(-N)' A whence secA= V(1 + tan2A), and tanA = /(sec2A — 1): cosec +A= (AP)4 PN';+AN2 (AN' co N} E2 — pN2=1 + 1 + cot2A;N whence cosecA= A(l +cot2A), andcotA= /(cosec2 —1). 44. We will here collect the preceding results, all of which must be carefully remembered, so as to be forthcoming, when required, without a moment's hesitation. PLANE TRIGONOMETRY. 31 For shortness' sake we give them in the following form, without mentioning the angle to which the Ratios belong, since A in the proof stands for any angle whatever: 1 1 1 1 (i) sin= -, cos=-, tan=-, cot= — cosec sec cot tan 1 1 sin cos sec=-, cosec=. -. (ii) tan =, cot-=-. cos sin cos sin (iii) sin2 + cos 2=1, sin= V(1 - cos2), cos= V(1 -sin2). (iv) sec2= 1 + tan2, tan2 = sec2- 1. (v) cosec2 -1 + cot2, cot2= cosec2- 1. And to these, of course, must be added sinA = cos (90~- A), cos = sin (90~-A), &c. 45. By means of the above formulae we may express any Ratio of an angle in terms of any other Ratio. Ex. Express all the trigonometrical Ratios in terms of the sine sin sin Ans. cos= ( - sin2), tan - cos - V(1 - sin2) 1 V/(1-sin2) 1 1 1 cot== sec- -- cosec -. tan sin ' cos /(1-sin2)' sill 46. The following will also be found a very useful result to be remembered. If tan=, then sin= a bs V(a + b2) (a2 + b2) a2 a2+b2 1 b Forsec2-1+tan2=1+ —b _;.'. cos= = b2 b2 see (a+ b2) and sin2= =l cos2= 1 — =;. sina= 2 b 2ab 72 a2+) Ex. 1. Given tan = 2, find the sine and secant. 2 2 2 1 1ec Here tan —;.', sin= /(441) --- cc —, se -- - cos 32 PLANE TRIGONOMETRY. tan tan 1 Ex. 2. Since tan- -,.'. sin=,(l cos+tn) = C (S+t-an2)' 47. To find the sine, cosine, Arc. of 45~. Let ABC be a triangle, right-angled at C, and having equal sides, CA, CB, and, therefore, also equal angles, CAB, CBA. Hence, since these two angles make up together 90~, each -A c of them must be an angle of 45~. Now AB2=AC2+BC2=2AC' or 2BC2, that is, AB= V2.AC or = V2.BC: BC 1 BC.'. sin 450= B C=- = cos450, and tan 45~ = C 1; AB V2 A C and.. also cot45~= 1, sec45~= /2 = cosec45~. If we express the angle 45' in circular measure, these results may be written sin-= — =cos tan-1 = cot-, sec -= 2=cosec4 V2 4' 4 4 4 N.B. Observe that, if A be greater than 45~, and, therefore, greater than B, B C> A C, and sinA > cosA; whereas, if A be less than 450, B C< A C, and cosA > sinA. 48. To find the sine, cosine, ec. of 30~ and 60~. Let ABC be an equilateral triangle, and, therefore, c also equiangular: then, since its three angles make up together 180~, each of them must be an /angle of 60~. Draw CD perpendicular to AB, and, therefore, (Euc. I. 10) biA D B secting both the side AB and the angle ACB, so that L ACD= 30~. i;,, PLANE TRIGONOMETRY. 33 ThenA C= AB = 2AD, and CD2 = A C2 - AD2 =3AD:. sin 30~ D =1 cos 60~, AC 2 COS30 - CD 3.AD /3=sin 60o, 30AC 2A — 2 AD 1 V/3 tan 30~ =CD_ -3 cot60~;.-.cot30~= V3=tan 60~,sec 30~=- 2 2 V3= cosec 60~, /3 3 cosec 30~= 2 =sec 60~. It will be found sufficient to mark well the values of the sine and cosine of 30~, from which all the other Ratios, both of 30~ and 60~, may be at once obtained: sin 300 1 1 thus tan 30=sin 30- = - 1 3, &c. cos 30~ V3 3 If we use the circular measure, the above results become. T 1 r X7T /3. 7 1 l- = =cos-, cos= -- =sin, tan 3= - =cot-, & 6 2 3 6 2 3 6 V3 3 Ex. 6. 1. Express the sine in terms of the secant and cotangent. 2. Express the cosine in terms of the cosecant and cotangent. 3. Express the tangent in terms of the sine and cosecant. 4. Express the cotangent in terms of the cosine and secant. 5. Express the secant in terms of the cotangent, and the cosecant in terms of the secant. 6. Express the sine in terms of the versed sine. 7. Given tan=i, find all the other Ratios. 8. Given sin=-, find the cotangent and secant. 9. Given tan=-, find the cosecant and cosine. 10. Given sec=4, find the sine and versed sine. 11. Shew that (sin -7-+cos 1 r) (sin — 7-cos -7r)= sin Jr. 12. Shew that 1-tan2 30~ sin 45~- sin 30~ 1+tan2 30-= cos 600, and sin 45~ + sin 30- (sec 45~-tan 45~)2. 49. We have hitherto been considering the Ratios of angles in the first quadrant only, or less than CO0: (T.) l) 34 PLANE TRIGONOMETRY. but the definitions of (32) hold equally for all angles whatsoever, as seen in the figure annexed. Here BB', DD', are drawn, as usual, intersecting at ~D ~ right angles at A; the K 1 3?~position of the revolving line is taken in the first, B' N } N | second, &c. quadrant reB' — - -- Bspectively, and PN is drawn perpendicular on BB'. Now in each case /P D we have, as before, Pn AN PN sinA = cos A, tanA = &c. AP A-PI AN' Here, however, in accordance with (6), we shall consider PNto be positive in the first and second quadrants, where it is drawn upwards from BB', and negative in the third and fourth quadrants, where it is drawn downwards; and so, likewise, we shall take AN to be positive in the first and fourth quadrants, where it is drawn to the right of A, and negative in the second and third quadrants, where it is drawn to the left. AP is in all cases to be positive, because it is always drawn from A in the direction of the revolving line, being measured off upon it. N.B. In other cases of algebraic geometry, cases occur where AP is negative, being measured, not on that part of the revolving line which is actually describing the angle, but upon the line produced backwards. The Student, however, will perceive that the word "direction" in (6) is not to be understood of afixed direction only, as in the case of AN, or even of a direction parallel to a fixed line, as in that of PN, but of any determinate direction whatever, defined by the conditions under which the figure has to be drawn, as, in the case now PLANE TRIGONOMETRY. 35 before us, along a revolving line. And this instance will further illustrate the meaning of the words in (6), "lines of the same kind, &c.": for AP, we see, is positive, though it may lie in the direction AB' or AD', in which directions AN or PN respectively will be negative - the reason of course being that the Law, according to which AP is drawn, is not contradicted by its being drawn in the direction AB' or AD'. 50. Now, strictly, it would be most correct to denote the value of any of these lines by some algebraical letter, supposed to express the number of units of length it contains. Thus, for instance, we might use p, b, r, for the perpendicular PN, the base AN, and the revolving line, or radius, AP, respectively; and then we should say that, in the first quadrant, PN= +p, AN= + b; in the second, PN= - p, AN= -b; in the third, PN= -p, AN= -b; in the fourth, PN= -p, AN= + b; while, in each quadrant, AP= + r. But it is more usual, (as it is often more convenient,) to use the geometrical symbols PN, &c. themselves in an algebraical way, namely, so as to express by them, not the lines themselves as they actually stand in the figure, but only their positive values; so that PN, AN, express merely the lengths of the corresponding lines, without regard to their position, and thus, if the lines are drawn positively, their values will be expressed by + PN, + AN, or, if negatively, by -PN, -AN. PN. Hence we shall no longer have sinA= in every \PN- P NV PtPN PN quadrant, but +PNor - N, that is + N or - EN AP AE A I AP' according to the position of PN: and so with the other 1tatios. 36 PLANE TRIGONOMETRY. 51. To trace the changes in sign of the different Ratios in the four quadrants. (1) sin A = ARin the first and second quadrants, AP — PN. = AR in the third and fourth; AP hence the sine is positive in the first and second quadrants, but negative in the third and fourth. Similarly for the cosecant. (2) cosJ2.. A- A:i the first and fourth quadrants, AP -AN in the second and third; AP hence the cosine is positive in the first andfourth quadrants, but negative in the second and third. Similarly for the secant. (3) tanAs + N in the first quadrant, = + in + AN — AN the second, — A in the third,=- in the fourth; - AN + AN hence the tangent is positive in the first and third quadrants, in which PN2and AN have both the same sign, but negative in the second and fourth. Similarly for the cotangent. These results will be best remembered by observing the sequences of signs in the four quadrants, namely, for the sine and cosecant, - + - —, for the cosine and secant, + - - +, for the tangent and cotanqent, + - + -. It must be observed, however, that all the results obtained in (43), with the angle in the first quadrant, where the Ratios are all positive, will, by (6), hold PLANE TRIGONOMETRY. 37 equally good for all angles whatever, and might be separately proved, though with more complexity, for each particular case. So, likewise, in all future reasonings, we shall be justified in drawing the simplest general figure we can, that is, with the lines and angles concerned all positive, if possible, and assuming the result thus obtained to be universally true. 52. To trace the changes in magnitude of the different Ratios through the four quadrants. Whenever the revolving line AP coincides with BB', (as it does at first, and again when it has described 180~,) then the perpendicular PN vanishes entirely, and the base AN becomes equal in magnitude to the hypotenuse AP: and so, likewise, whenever AP coincides with DD', then AN vanishes, and PN becomes equal in magnitude to AP. It is plain also that we need only follow the changes of magnitude of the different Ratios through the first quadrant, since they go through exactly the same changes in the other three quadrants, alternately increasing and decreasing. Observe that we are not now considering the signs of the Ratios, which will have to be separately determined for each quadrant as in (51): in the first quadrant, however, we know they are all positive. Now, in the first quadrant, or as A changes from 0~ to 90~, sinA (= -P) changes from - to r, that is, from 0 to 1; AP r r AN rt 0 cosA (=-p)........... o...... 1 to 0; AP r r PlN 0 tr tan (= )................. 0 to-....... OtoCO; ANr )'............... PLANE TRIGONOMETRY. t4 N r 0, cotA (= W-N) changes from - to, that is, from c to 0 AP r r cosecA (= — )................ to.. to.cose (PN *.. *.a * t. * e to --.to 1. 53. Hence for all the four quadrants we may form the following scheme, to express the changes in sign and magnitude of the trigonometrical functions of an angle A, as the angle changes from 0~ to 360~. A sinA cosA tanA cotA secA cosecA 0~ to 90~ Oto to 0 to 1 1o to 0 O to oo to 1 ________ (+) (+) (+) (+) 90~to 180 1 toO O to o~ to 0 O to oo oo to I I to o (+) (-) (-) (-) (-) (+) 180~ to 2700 tol toO to o to O O to oo o too _______ (-) (-) (+) (+) ( (-) (-) 270~ to 3600 1 to O Oto 1 oo O O Oto oo oo to to oo (-) (+) (-) ( -) (+) (The above Table must be well studied, and may be best remembered, by remarking carefully the results of the first line, on which all the rest depend. 54. It will be observed, that the values of the sine and cosine range between 0 and + 1, so that these always lie between + 1 and — 1; and those of the secant and cosecant range between + 1 and + mo, and between - 1 and- oo, so that these never lie between + 1 and - 1; while those of the tangent and cotangent range between 0 and t+ oo, and, therefore, may be of any magnitude whatever, positive or negative. PLANE TRIGONOMETRY. 39 Notice also that each of the ratios changes its sign, whenever its value passes through zero or infinity. In fact, the same Ratio may be considered as having at the same moment the value +0 and -0: thus cos 90~= +0 or -0, according as we consider D to be the end of the first quadrant, in which the cosine is positive, or the beginning of the second, in which it is negative; and the same may be said of the values + oo and - oo; so that we have tan 90~= + oo or -. The real meaning, however, of such statements, is this: at the point D, where the angle becomes 90~, there is actually no such thing as a cosine or tangent at all, for no triangle APN can be formed, and the definitions, therefore, cannot be applied. But, on either side of D, however close to it, there are values of the cosine and tangent, those of the cosine exceedingly small, and those of the tangent exceedingly great; and these, however small in the one case, or however great in the other, will be positive or negative on different sides of D. 55. Hence also the versed-sine (=l —cos) changes, in the four quadrants, from 1 —1 to 1 —0, from 1 —0 to 1- (-1), from 1-(-1) to 1-0, and from 1 — to 1-1, or from 0 to 1, from 1 to 2, from 2 to 1, and from 1 to 0. The versed-sine, therefore, is always positive, and its values increase gradually from 0 to 2 in the first two quadrants, and then diminish from 2 to 0. 56. The following is a Miscellaneous Exercise upon all the foregoing portions of the Book. Ex. 7. 1. Find the number of grades in the angle whose circular measure is -. 2. Shew that the areas of two circles, in which the circumference of the one is equal to an arc of 60~ in the other, are as 1:36 PLANE TRIGONOMETRY. 3. The distances of Venus and the Earth from the Sun are as the numbers 7 and 10, and they revolve in 224 and 365 days respectively: compare their rates of motion. 4. Compare the angles subtended by an arc of 3~ feet to a radius of 2~ feet, and an arc of 23 feet to a radius of 31 feet; and express the first in degrees and the second in grades. 5. Shew that sin'2r s in2- 7, s sin,2r, sin27r, are as 1, 2, 3, 4. 6. Find tanA from the equation sin2A+ 5 cos2A=3. 2 sinA-cosA 2 7. Given2 sinA+3 cosA-3 find the sine, cosine, and tangent of A. 8. Determine sinO and tan0 from the equation tan0+3 cot0=4. cosA cosB 9. If cosx —s= and cos (90~-x)= csB shew that sinC"'C ( — ) sinC sin2A sin2B+ sin2 C=2. 10. If tan2A+4 sin2A=6, find the value of versA. 11. If tan=ab, find the value of a sin0+b cos9 sin29 cos20 and 1 -a sin-'-b cosO 1l-cot l+-tan0' 12. If sinA=/V2 sinB, and tanA=/'3 tan B, determine A and B. 57. To express the TrigonometricalRatiosof 180- A, 180~+A, and -A, in terms of A. Draw BB', DD', cutting at right angles at A: make L BAP, = B'AP2 = L BAP3 2M \ /I -= /- BAP4= A: then we shall have ^ -— Bum --— >/ --- Bi3 the positive angles ^. BAP1 = A, BAP2 = 180~-A, BAP3 3P ] ~is =180~+ A, BAP4 = 360~- A, and the negative angles BAP4= -A, BAP3= — (180~-A), &c PLANE.lAIGONOMETRY. Take AP, = AP=A P3= AP4; and draw P P4, P2P,, cutting BB' at right angles in N, N'. Then the triangles AP<N, AP2N7, AP3N', AP4N, are equal in all respects, so that P1N=P2N',=P3N'=P4N, and AN= AN'. Hence it is manifest that any function of either of the above angles is equal in magnitude to the same function of A, whether differing, or not, in sign: thus, sin (1800- A)-+PN' + PA t1N + sinA, AP2 A=P, - AN' AN cos (180~ + A) -- - - A = -- cos A AP, APr tan (-A) = - P t —P --- tan A, &c. - +AN AN The only difficulty then will be to know when the sign, which connects the two equal Ratios in any case, is to be (+) or (-). And for this we observe that, since all the ratios of A are positive, the sign to be used in any case must be that which belongs to the equal Ratio of 180~-A, 180~+A, or-A. Hence, attending to the signs of the different Ratios in the four quadrants, as laid down in (51), we have sin (180~-A)= +sinA, sin (180 +.A)=-sinA, sin (-A)= -sin A; cos(180~-A)= -cos, cos (180~ +A)= -cos A, cos (-A)= + cos A; tan (180~-A)=-tanA, tan (180~ +A)= +tan A, tan (-A)= -tan A: and the cotangent, secant, and cosecant, follow the law of the tangent, cosine, and sine, respectively. Hence also, as in (51, suversA = vers(1800-A = 1 -cos(180 —A)= 1 + cosA. 42 PLANE TRIGONOMETRY. 58. Of course, it follows that the converse of these equations are true; thus sin A= — sin (-A), cos A=+cos (- A). It is plain also, from the figure, as observed in (7), that the Ratios of the angle BAP4 will be the same, whether we regard it as the positive angle (360~- A), or, the negative angle (- A); and, generally, that the Ratios of any angle will not be at all affected, by supposing the angle increased or diminished by any multiple of 360~ (or 27r), since this only brings the revolving line to the same position exactly as before. 59. It is useful to notice the pairs of quadrants in which the different Ratios are the same, in sign as well as in magnitude, with the corresponding Ratios of A, namely, the sine and cosecant in the first and second, the cosine and secant in the first and fourth, the tangent and cotangent in the first and third,-in fact, the same pairs of quadrants as those in which (51) the respective Ratios are positive. 60. Of course we may write the above results, if we please, in circular measure: thus sin(7r-0)=+sin0, sin(7r+0)=-sin0, sin(-0)= -sinO, &c. 61. In like manner, if we now make Z DAP1 = Z DAP2= A D'AP3= D'AP4= L BAP=A, then we shall have the posiP2 P/iP tive angles BAP, = 90~- A, \ B-AP2 = 900+ A, BAP3 = s_, \ -" --- B 2700~-A, BAP4 = 270 -A, A s &c. and the negative angles BAP = - (90~ -A), BAP3 /,:>/' a = - (90~ + A), &c. PLANE TRIGONOMETRY. 43 Take, as before, AP, = AP=AP3= AP4=- AP, and draw PN, PN'P4, PN'P3, perpendiculars on BB'. Then the triangles APPN, &c. are, each of them, equal in all respects to APN, but so that PN' = PN" = &c. = A N, and AN' = AN" = PN. Hence any function of either of the angles BAP1 &c. will be equal in magnitude to the corresponding co. finction of A, whether differing, or not, in sign: thus in (90~ —A) = + P N' AN c AP, + AP= - AN" PN cos(90~+A)= A - - - = - sinA, &c. So that, determining the signs as before, we have sin (90~-A)= + cosA, cos (90~-A) = + sinA, tan (90~-A)= +cotA, as in (35); sin (90~+ A)= + cosA, cos (90 + A)= -sinA, tan (90~ + A) -cotA; &c. 62. We may now shew, as stated in (35), that any function of any angle whatever may be expressed as a function of an angle less than 45~. First leave out of the angle any multiple it may contain of 360~ (or 2r); for by (58) the given function of the remaining angle will be the same as that of the original angle. Then put this remainder in the form n.90~+A, which may always be done so as to have A less than 45~, by taking that multiple of 90~ which is nearest to the angle in question, whether greater or less than it. And now, according as n is even or odd, the given function of n.90~+_A may be written down at once as the corresponding function or cofunction of A. 44 PLANE TRIGONOMETRY. with that sign which belongs to the given function of n.90~+ A. N.B. It will be observed that these propositions, Arts. (57-62), only refer to the six Trigonometrical Ratios, defined in (32), not to the chord, versed-sine, &c. Ex. 1. sin 150~ = sin (90~ + 60~) = + cos 60~ = -: here, however, we should most naturally have used the formula sinA = sin (180~ -A), or sin 150~ = sin 30~= -, as before. Ex. 2. cot 330~ = cot (3.90~ + 60) = - tan 60~ = - V3; or =cot(4.900-300)= —cot300=-,-/3 where 300~45~. Ex. 3. sin 369~= sin 9~, by merely leaving out 3600; tan 500~= tan 1400=tan (180~-40~)= -tan 40~; cot 660~= cot 300~=cot (3.90~ 430~) =-tan 30~; sec 820~= sec 100~=sec (900 +10)= —cosec 10~ cosec 900~= cosec 180~= o. Ex. 8. 1. Write down the values of the cosine and cosecant of 150~, the sine and tangent of 135~, the tangent and secant of 120~. 2. Write down the values of the sine, cosine, and tangent of 240~. 3. Find the chord and versed sine of 120~ and 300~. 4. Find the coversed sine of 330~, and the suversed sine of 315~. 5. Express sin 100~, cos 200~, tan 300~, cot 400~, sec 500~, cosec 600~ as functions of angles less than 45~. Ex. 9. Simplify the following expressions: 1. sin(7r + 0),cos(27r - 0),cos(2 + ),sin(- + 6),tan(7r + 0),tan(3 - 0) 2. sin ( + 0), cos (-), cos ( 6), sin (0 — ), tan (0 + ). 3. tan (A -180~) + cot (90~ + A) + tan (180~ - A) + cot (A - 270~) + tan (180~ + A) + cot (270~-A). 4. Express sin {(4n + 1) 2 + 0} and tan {(4n + 3) + 0 in terms of 0. 5. Shew that vers(90~ + A) vers (A - 90~) + covers (270~- A) covers (A - 270~) 1. PLANE TRIG{ONOMETRY. 45 6. Given sec(7r- 6) cos( + a) cos (r-a) = tan (7 -a) sin (r + 0), determine sec 0. 63. To obtain general expressions for all angles which have the same given function as A. (i) sinA= +in(180~-A),andcosecA= +cosec(1800-A): hence by (58) all angles will have the same sine and cosecant as A, which are included in either of the expressions n.360~+ A or n.360~+(180 —A), where n.360~ represents a multiple of 360~, n being any integer positive or negative, including zero, and A being either positive or negative. Now these expressions may be written 2n.180~+A and (2n+ 1) 180~-A, where, we observe, the sign is + or -, according as the multiple of 180~ is even or odd: hence they may be both included in the single formula, n. 180~ ( —1)" A, since (- 1)n = + 1 or — 1, according as n is even or odd: hence sinA= sin {n. 180~ + (- 1) A}, cosecA =cosec {n. 180~ + ( - 1) A}. (ii) cos4= + cos ( —A), and secA= + sec ( —A): hence all angles will have the same cosine and secant as A, which are included in either of the expressions n.360 + A or n.360~-A, or, in one formula, 2n.180~ A, the multiple of 180~ being here always even: hence cosA = cos(n.360~+A), secA =sec(n.360~+ A). (iii) tanA= +tan( 80~+A) andcotA=+cot(180~+A): hence all angles will have the same tangent and cotangent as A, which are included in either of the expressions n.360~ + or n.360~0+(1800+A), which PLANE TRIGONOMETRY. may be written 2n.180~+A, or (2n+1) 180~+A, or, in one formula, n. 80~+A, where n may be even or odd, the sign being always +: hence tanA=tan(n.180~+ A), cotA=cot(n,180~ + A). 64. For the circular measure, the above results become sin 0= sin{n+ ( —1)n 0}, cosec =cosec {n (- 1)" 0}, cos 0 = cos(2nr + 0), sec 0 =sec(2n77r+ ), tan = tan(nr + 0), cot 0 =cot (n-r + 0). Ex. Since sin 30~ = sin Xr = 1, therefore the general value of all angles, whose sine is,, will be n7r+(-l1)' 17-={6n+ ( —)"}i7r; from which expression, by giving n the values 0, 1, 2, &c. successively, we get the series of angles tr, A-, 1w 7r, &c. 65. It should be observed that sin0, when determined from cosO, has algebraically two values, equal in magnitude, but of opposite signs, namely, sin0= -+ A(1-cos2): and this agrees with the principles of Trigonometry. For, when cosO is given, 0 itself is not given, but may be any of the angles expressed by the formula 2nr+ cc, usinog a for the least of those angles which has the same given cosine. Now sin (2nz7r +)= + since, which agrees with the algebraical formula, when sinO is expressed in terms of cosO. And so in other similar cases. Ex. To shew that the sine, when determined trigonometrically from the tangent, will have two values, in accordance with the algebraical equation, sin9= -+ /tani2 Here tan9 is given us to find sin9; and tanO=tan (n7r+a); our result, therefore, will give us the value of sin (n7r-+a),and this will be + sina, according as n is even or odd. Thus, if tanA=-, then (32 Ex.) sinA=-3, cosA=4; but these should properly be -, -,+4 since the denr. is v/(16+93)=-+5 For, although tanA=i, we have nothing to tell us whether A is PLANE TRIGONOMETRY. 47 to be an angle in the first or third quadrant, in either of which the tangent is positive, the sine and cosine being both positive in the one, and both negative in the other. So, if tanA = —, we shall have sinA= —i, cosA= 45. 66. The preceding results may now be practically employed, in certain simple cases, to find the heights, &c. of objects, as in the following Example. Ex. A person, standing at a distance of 80 feet from the foot of a tower BC, observes the angle BAC, which a line from his eye to the summit of the tower makes with the horizontal line AC, to be 60~. Find the height of the tower. Here, BC=AC tan A=80 tan 60~=80A/3=80x 1.7320 &c= 138.56=139 feet nearly, to which adding the height of the observer's eye at A, say 6 feet, we shall have the height of the tower 145 feet. The angle BA C in the above is called the angle of elevation, or, simply, the elevation of the tower. And, in like manner, if an object at A were viewed from a higher position at B, the angle which the line BA makes with the horizontal line through the eye at B, or its equal, the angle BAC, is called the angle of depression of the object at A. Such angles would be commonly observed with an instrument called a Sextant, held in the observer's hand. Ex. so. 1. Given tanA=l, write down the general value of A. 2. Find the general value of A, when tan 2 A=+1. 3. Find the general value of A, when secA= -2. 4. Given cos 2A=-, find the general value of A. 5. Given tan 5A= /3, find the general value of A. 6. A ladder, 9O feet long, inclined at an angle of 60~, rests against a wall; find the height of its top from the ground, and the distance of its foot from the base of the wall. 7. From the top of a ship's mast, 90 feet above the surface of the water, the angle of depression of the hull of another ship was found to be 30~: find the distance between the ships. 8. At a certain distance from the foot of a tower, its elevation was observed to be 60~, and, 80 yards further off, it was found 48 PLANE TRIGONOMETRY. to be 30~: find its height, allowing 6 feet for the height of the observer. 9. CD is the perpendicular from the right angle C of a triangle upon the hypothenuse AB, AC is 108 yards, BC is 144 yds.: find CD, BD, and DA. 10. A May-pole being broken off by the wind, its top struck the ground at an angle of 45~, and at a distance of 21 feet from the foot of the pole: what was its whole height? 11. The shadow of a church-tower extends 56 yards from its base; find its height, it being observed that a two-foot rule, held vertically, casts a shadow of 4 ft 3 in. 12. A ladder, 45 feet long, being placed in a street, will exactly reach to a window 27 feet from the ground on one side; and upon being turned over without moving the foot, so as to lie at right angles to its former position, it will just reach a Aindow on the other side of the street: determine the height of this latter window, and the breadth of the street. ( 49 ) CHAPTER III. ON THE TRIGONOMETRICAL RATIOS OF THE SUM AND DIFFERENCE, MULTIPLES AND SUBMULTIPLES, OF ANGLES. 67. To find the sine and cosine of the sum and difference of two angles in terms of the sines and cosines of the angles themselves. Let BA C and CAD be two given angles, and let BA C be called A, and CAD be called B: then, according as we measure CAD forwards or bachwards from A C, the angle BAD will be expressed by (A + B) or by (A- B). (i) To find the sine and cosine of (A + B). Take any point P in AD, and draw PM, PQ, perpendiculars on AB, A C,:m[I QN, QR, perpendiculars on AB, PM: A —Yr then we have Z QPR=90~- LPQR = BR QA = / QAN=A. Hence sin(A + B)= sinBAP= PM QN+ PR QN+ PR AM AP -APq AP _-QN. AQ + PRQ PQ= sinA. cosB + cosA. sinB; AQ-AP PQAP AM AN-QR_ AN QR cos(A + B) - cosBAP= AMAN-QR AN_ QR AP AP AP AP. A A Q-Q PQ cosA. cosB-sinA. sinB. A.) AP Q AP (I.) E 60 PLANE TRIGONOMETRY. (ii) To find the sine and cosine of (A - B). Drawing the figure as before, except that QR will in this case fall on PALfproduced backwards, we have LQr s QPR=900 — Z PQR= LRQC= BAC=A. PM QN —PR Hence sin(A - B)= sin BAP =- p - AP QN AQ PR PQ QNA AP PQ ' PA =- sinA. cosB-cosA. sinB; AM AN+ QR and cos(A —B) =cosBAP=A- AP AN AQ QR PQ =- Q' AQ - Q A- A * cosA. sB+sinA. sinB. 68. We have given in (67) the complete geometrical proofof each of the cases: but the results of the firstinclude those of the second by merely writing - B for + B; thus, sin(A-B)=sin {A+ (-B)} =sinA.cos(-B)+cosA.sin(-B) =sinA.cosB - cosA.sinB, cos(A-B)=ccos tA+(-B)) =cosA.cos(-B)-sinA.sin(-B) =cosA.cosB + sinA.sinB. And these results we might have anticipated from the figure itself, since in the latter case the angle CAD and the lines PR, QR, are all negative, being drawn in opposite directions, with respect to the line AC and the points P, Q, to those which in (i) we assumed to be positive. Hence, looking at the results for (i), we should expect to get for (ii), (as we actually do,) QN-PR AN+ QR sin(A-B)= Ap =&c, cos(A-B)= &c. In short, we have drawn the first figure in (67) for the most simple case; but, by the principle laid down in (6), the results are applicable to all cases whatsoever. PLANE TRIGONOMETRY, 51 69. We shall here, however, give the strict geometrical proof in one other case, in order to shew how it may be conducted in similar instances. Suppose then that A lies between 180~ -,9\ r and 225~, and A-B between 45~ and 90~: to find the sine and cosine of A-B. 3B3 —g' / J B The figure will be as annexed, where / | BAC=A, CAD=B, and PQ falls on AC produced backwards: hence we have sin (A-B)=sin BAD PM QN+PR QN A Q PR PQ -AP= AP -AQ.'AP - PQ' APj cos (A-B)=cos BAD AM AN-Q R AN AQ QR PQ ~~AP~~ AP A2Q AP PQ'AP' QN But A=sin BAC'=sin (A-180~))= -sin (180~-A) ---sin A, AN AQ cosBAC'=cos(A-180o)=+cos (180 —A)=-cosA; AQ PQ A-=cosDAC'=cos (180~-B)= —cosB, A- =+sinB; PR~R QR Q-=sinPQR=cosR QA=cos BA C'=- -cosA,-p Q -sinA:: hence sin (A-B-)=(-sin A) (-cos B) +(-cos A) (+sin B) = sin A cos B-cos A sin B; cos (A-B) = (-cosA) (-cosB) - (-sinA) (+sin B) =cosA cosB+sinA sinB. 70. So too we may derive cos(A + B) from sin(A + B); forcos(A+B)=sin {90 ~-(A+B)- =sin {(90~-A)+(-B)} = sin(900 — ) cos (-B) + cos(90- A) sin (-B) = cos. cosB - sinA. sinB: and, generally, in the same way, from any one of the four formulae the others may all be deduceLc E2 52 PL XNE TRIGONOMETRY. 71. To express tan (A + B) in terms of A and B. sin(A + B) sinA cosB + cosA sinB tan A ~ Bs) = os (A + B) cosA cosB T sinA sinB tanA + tanB - 1 tn tan tanB ' by merely dividing each term by cosA. cosB. Ex. 1. sin 750=sin (45~+300) =sin 45~ cos 30~-+cos45~ sin 80~ - I_ +3 1 /3 1< (6+ V2)cos 15/; A/2' 2-+ /-2'2 ' 2V2 so also, sin 15~-sin(450-300) = /3 (V6- -2)=cos750; 2 a/ —2 whence tan75~=2+ v/3=cot15~, and tall 15=2-V/3=cot75~. tan 45~+ tanA 1+tanA Ex. 2. tan (45~+~A)-= 1 + tan 45~ tan A-1 f- tanA' since tan 45~= 1. Ex. IM. Prove the following results: cos (60~-A) 2(cosA + 3 sinA); sin (45~+A)= — (sinA+cosA) 2. vers (A+30~) -vers (A-300) = sin A. sin (45~+A)-cos (45~+A) sin (45~ A) +cos (45~ + A) tanA tan (45~ +A)-tan (45~-A) 4. - =2 sin.4 cos A. tan (45~0-A) + tan (450-A)2 sin cos 5. Shew that sin(n+1) A+sin(n-1)A=2 sinnA cosA, and cos(n+l) A+cos (n7-1) A-2 cosnA cosA; and hence express sin 2A, cos2A, in terms of sin A, cos A. 6. Express cot (A+-B) in terms of cotA and cotB 7. Express sec (A~B) in terms of secA and secB. 8. Express cosec(A~B) in terms of cosecA and cosec B. 9. Given the expression for sin (A-B), deduce that for cos (A-B). 10 Given the expression for cos (A +B), deduce thatfor sin (A+ B). PLANE TRIGONOMETRY. 53 72. To express the ratios of 2A in terms of A. If in the formula for sin (A + B) and cos (A - B) we write A for B, we get immediately sin 2A = sinA cosA + cosA sinA =2 sinA cosA, cos 2A-=cosA cosA- sinA sinA = cos2A-sin2A. For the latter formula, however, it is often convenient to use one of its equivalents; viz. cos2A = cos2A-(1 -cos2A)= 2 cos2A -- 1, or =(1 - sin2A)-sin2A - 1-2 sin2A. So also, writing A for B in the expression for tan (A + B), we get tanA + tanA 2 tanA tan 2A - tanA tanA tan 2A 73. We may derive other results from the above, by giving different values to A. Thus, writing } A for A, we get sin A=2 sin A cos 4A, cosA=cos2'A-sin2'A=2 cos 2A-1=1 —2 sin 2A; &c. Hence 1 +sinA (cos 2A +sin 2-A)- 2 sin -A cos -A H-lnc~ -=-r-===- --. A * - cosA c- os-A-sin A (cos A + sin A)2 cos- A+sin A 1+tan!A cos 'A-sin21A -cos ^A+sAsiA1 -tan At (45 -2) N.B. Observe well the step in the above, by which 1+sinA is changed to (cos IA+ sin- A)2, or (which is the same thing) to (sin -A+ cos -A)". So, likewise, 1 +sin 2A=(sin2A+cos'A)+2 sinA cosA=(sinA+cosA)2. 74. Notice also the following expressions for tanlA. sinA 2 sinA cosA sin2A tn cosA 2 cos2A = +cos2A' 2 sin2A 1-cos 2A or = 2 sinA cosA sin 2A sinA 1-cosA and so tan IA=,' or n --- 2 1+cos A sIn AL PLANE TRIGONOMETRY. 75. Hence also we may get sin 3A, cos 3A, tan 3A, in terms of sin A, cos A, tanA, respectively. (i) sin 3A = sin (2 A + A)= sin 2A cosA + cos 2A sin A =(2 sinA cosA) cosA + ( -2 sin2A) sinA = 2sinA(1 - sin2A) + (1-2 sin2A)sinA = 3sinA-4sin3A. (ii) cos 3A = cos (2A + A)= cos 2A cosA - sin 2A sinA (2cos2A-1 )cosA-2(1- cos2A)cosA = 4cos3A-3cosA. (iii) tan 3A= tan (2A + A) 2 tan A tan 2A + tanA 1 -tan t A 3 tan - tan3A 1 -tan 2A tanA 2 tan A ta 1 -3 tan2A 1- - tanlA 1- tan2A A Hence also, as before, by writing -A for A, we get sinA=3 sin LA-4 sin 3A, cos A=4 cos 3A-3 cos-1A, &c. 76. To express sin A and cos A in terms of cos2A. By (72) 2 sin2A= 1-cos2A, 2cos2A= +cos2A; -.sinA= V {(1- cos 2A)}, or2 sinA= /(2 - 2cos2A), cosA = V {( (+ cos 2A)), or 2 cosA = /(2 + 2 cos2A). Hence sin2A 1 - cos 2A 1 -tan2A tan2A cos= 1 cos2A whence cos 2A= tan2A Cos2A - I + cos 2A1 tan 2A which last result, however, may be obtained more directly as follows: 2 2A c=n os2A - sin2A 1 - tan2A cos 2A = os2A-cos2A + sin2A= 1 + tan2A' if we divide every term by cos2A. Hence also, as before, 2 sin2LA=1 —cos A tan 1 sA A 1-tan'A 2 I COSA- - 1zy COS A A_1 —eosA 1 — tan2}A 2 cos2'A=1 +-ossA '2 1 - + cosA 1 +tan2.A' 77. To express sin A and cosA in terms of sin 2A. In order to this, we might square the equation sin2A = 2 sinA cos A = 2 sinA /(1 - sin2A), and then we should get a biquadratic for determining PLANE TRIGONOMETRY. 55 sin A in terms of sin2A: but the following is a more simple method of arriving at the same result. Since (sinA + cosA)2 =sin2A - cos2A + 2 sin A cos A = 1 + sin2A, '.sinA +cosA= + V (l + sin 2A), sinA - cosA =+ / (1 - sin 2A): whence, adding and subtracting these equations, we get 2 sinA = + /(1 + sin 2A)~ /(1 —sin 2A), 2 cosA = ~+ V(1 + sin 2A) (1 - sin 2A). 78. In the above results it is seen that the signs of the square roots are not determined: nor can they be, until we know the value of A, or of 2A, in any case. We now proceed to fix them for different values of A. Draw, as usual, BB', DD', intersecting at right angles in A; and draw also PP', A\ i ~s Q Q', so as to bisect these angles, and, ' I \^ therefore, making angles of 45~ with BB' or DD'; these we shall call the octant-lines. Measuring, as before, all angles from AB, it is plain that for any angle, whose bounding line lies any where in either of the spaces, PA Q', QAP', that is, between BB' and either of the octant-lines, AN will be greater than PN, or the cosine will as to magnitude (without regard to sign) be greater than the sine; whereas for any angle, whose bounding line lies between DD' and either of the octant-lines, PN will be greater than AN, or the sine will be greater than the cosine. Now sin A + cos A is positive, (1) when sin is(+) and cos (+), that is, in the whole of the first quadrant; or (2) when sin is ( + ) and cos (-), if sin > cos, that is, in the former half of the second quadrant; or (3) when sin is (-) and cos ( + ), if cos > sin, that is, in the latter half of the fourth quadrant: 56 PLANE TRIGONOMETRY, collecting which results, we may say that sinA + cosA = + V/(1 + sin 2 A), when A is an angle in the first quadrant, or in either of the two adjacent half-quadrants. Similarly, sinA + cosA = - (1 + sin 2A), when A is an angle in the third quadrant, or in either of the two adjacent half-quadrants. Again, sinA-cosA is positive, (1) when sin is (+) and cos(-), that is, in the whole of the second quadrant; or (2) when sin is ( +) and cos ( + ), if sin > cos, that is, in the latter half of the first quadrant; or (3) when sin is (-) and cos (-), if cos > sin, that is, in the former half of the third quadrant: collecting which results, we may say that sinA- cosA = + V(1 -sin 2A), when A is an angle in the second quadrant, or in either of the two adjacent half-quadrants. Similarly, sinA-cosA - /(1 -sin 2A), when A is an angle of the fourth quadrant, or in f;iher of the two adjacent half-quadrants. It will be seen that each of these formuai holdks good for 180~ together, between alternate octant-lines, and that any pair of them, (namely, one for sin A + cos A, and one for sinA —cosA,) holds good for SO') together, between successive lines of octants: that is, for instance, all angles, whose bounding lines lie between AP and A Q will have the same pair of formulk, namely, sinA cosA= + (1 +sin 2A),sinA - cos i = + ( 1 -,in2A, 'PLANE TRIGONOMETRY. 57 When, therefore, any value of A is given, and we are required to find for it the corresponding pair of formulae, we must consider in what quadrant the given angle is, and select the proper signs accordingly. Ex. If A=30~, then sinA+cosA=+/- /(1 +sin 2A) sinA-cosA= —/(1- sin 2A); sinA= { /(1 +sin 2A)-/(1- sin 2A)}, cosA== { V/(1 + sin2A) + V(1-sin 2A)}. We may prove the truth of this result by trial: for sin 2A= -V/3; 1 VA3 V3 1 4+223 4 2 /3 and.'.sinA= { /1 )-V/(1-2)l= — 4 1 V33+1 VJ3-1 1 2 2 2 2 I V3+-1 V23 and cosA ={- 2 -+ } =2 - as it should be. 79. The reason why in (77) the expressions for 2 sinA and 2 cos A appear with so many undetermined signs is, that, as in (65), we cannot help including in the steps by which we arrive at the result, a great number of angles which agree in certain of their trigonometrical properties, those, namely, expressed in the formulae with which we start, but not in all, nor, in fact, in those for which we are seeking expressions. Thus there is a multitude of angles, for which the equation (sinA + cosA)2= 1 + sin 2A is true; but these angles may differ in their expressions for sinA +cosA, which in some of them may = + V(1 + sin 2A), and in others may = - V(1 + sin 2A). Ex. 12. Prove the following formula: 1. cosA=cos44A-sin4fA. 2. cotA-tanA=2 cot2A tanA 1 + cot2A 3. tan 'A-= tanA 4. cosec2A=- + cotA 2. +secA= 2 cotA 5. secA=l - tanJ tan{A. 6. cosecA= (tangA -- cot-A), PLANE TRIGONOMETRY. 2 sinA + sin 2A 8 sn2A cosA 2sinmA-sin2A cOtA. l+cos2A l+csA- 1 sinA cosec2A 1 +tan2A 91 +cosA=(ltnA )2 101 +cosec 2A (1 +tanA)2' 1. cotA —cot 2A= cosec2A. 12. tan-A+ secl-A=tan(45~+ A) 80. The formule of (67) may be combined in various ways, so as to produce other formulae, which are of great use in trigonometrical operations. Thus we have given sin (A - B)= sinA cosB + cosA sinB, sin (A - B) = sinA cosB —cosA sinB, cos (A - B) = cosA cosB - sinA sinB, cos (A- B) = cosA cosB + sinA sinB. Hence sin (A + B) + sin (A - B)= 2 sinA cosB... (1), sin (A + B)- sin (A -B)=2 cosA sinB... (2), cos (A + B) + cos (A-B)= 2 cosA cosB... (3), cos (A-B) —cos (A + B)=2 sinA sinB... (4). Again, sin (A + B) x sin (A -B) = sin2A cos2B -cosA sin2B = sin2A(1 -sin2B) -(I -sin2A)sin2B = sin2A -sin2B; or = (1 -cosA)cos2sB cos2A( 1 -cos2B)=cos"B-_cos2A: cos (A + B) x cos (A - B) = cos2A cos2B - sin2A sin2B = cos2A(1 - sin2B) - (1 - cos2A)sin2B=co2A-sin2B; or = (1 - sin2A) cos2 -sin2A( 1 - cos2B)= cos2B - sin2A. The preceding results are very useful, and may be easily remembered by observing that each product is expressed by the difference of the squares of two functions of A and B, those functions being taken, the first out of the first term of the expressions for the factors multiplied, and the second out of the second term. Thus sin(A+B) sin(A —B)=sin2A-sin2B, where PLANE TRIGONOMETRY. 59 sin A is taken out of sin A cos B, the first term of the expressions for sin (A+ B), and sin B out of cos A sin B, the second term: but this product also = cos2B —cos2A, where cos B is taken out of the first term, and cos A out of the second. Of course A and B in the above may stand for any angles whatever, as in the following Examples. Ex. 1. sin (2A+3B))+ sin (2A-3B) = 2 sin 2A cos 3B. Ex. 2. 2 sin (A+B) cos (A-B) =sin {(A+B) + (A-B)} + sin {(A+B)-(A-B)} = sin 2A+ sin 2B.,x.3. cos(A-B)cos(B —C)= {cos(A-B+B-C)+cos(A-B-B + C) =t {cos(A-C)+cos(A-2B+ C)}. Ex.4. cos(A-+30~)cos(30~-A), = cos (A+ 30~) cos (A —300)=cos2A — sin300 =cos2A —=- (1+cos2A) —= (1+2 cos2A). Ex. 13. Prove the following formula: 1. sin (30~+A)+ sin (30~-A)= cosA. -2. cos(30~ —A)-cos (30~+ 'A) = sin'A. 3. sin (45~+A) sin (45~-A)= cos 2A. 4. cos(60~+A) cos (60~-A)= ~ (2 cos 2A-l). 5 1+cos2 (A+B) cos2 (A-B) =cos22A+cos22B. 6. 2 sin I (A +B) sin (A-B)=cos (A-2B)-cos (2A-B). 7. 4 sinA sin (60~+A) sin (60~-A)= sin 3A. 8. sin (A+B) sin (A —B)+sin (B+ C) sin (B-C) +sin (C+A) sin (C-A)=0. 9. cos (A+2B) cos (A-2 B)+sin (2B+ C) sin (2B-C) = cos (A+C) cos (A-C). 10. sin (A- B) sin C+sin (B- C) sinA+sin (C-A) sin B = 0. 11. sinA cos (A+ B) —cosA sin (A-B) = cos 2A sin B. 12. vers (A+B) vers (A-B)=(cosA-cosB)2. 81. All the preceding formulae, in which functions of A+ B and A —B are expressed in terms of A and B, may now be used to lead to others, in which functions of A and B are expressed in terms of -(A + B) and (A - B), that is, of the semi-sum and semi-difference of A and B. 60 PLANE TRIGONOMETRY. For, in (80), write (A + B) for A, and (A —.B) forB: then, since - (A + B) + -(A- B) = A, and -(A + ) -;(A-B)=B, 'we get sinA+sinB=2 sin -(A+ B) cos (A-B)...(1), sinA- sinB = 2 cos (A + B) sin (A —B)...(2), cosA + cosB =2 cos (A + B) cos (A - B)...(3), cosB-cosA= 2 sin (A + B) sin -(A- B)...(4). So also sinA sinB =sin2A(A + B) - sin2(A- B), or =cos21(A-B)-cos2o(A+ B); cosA cosB = cOS (A + ) - sin 2(A- B), or = cos2(A - B) — sin2-(A + B). Hence also we get sinA + sinB _ 2 sin (A + B) cos-(A -B)_ tan (A + B) sinA -sinB 2 cos (A + B) sin-(A B) tan(A-B) with other similar results. 82. From the above it will be evident that any formula whatever, obtained for functions of A+B and A-B in terms of A and B, may be converted at once into a corresponding formula for functions of A and B in terms of (A + B) and (A -B), by merely writing in it A for +- B, and B for A- B, '(A+ B) for A, and ~(A-B) for B. Thus sin (A + B) = sinA cosB + cosA sinB;.*.sinA- sin'(A + B)cos (A-B) + cos-(A + B)sin(A-B). 83. Since sinA sinB= {cos (A-B)- cos (A + B)}, or = sin2I (A + B) - sin2l (A - B), and cosA cosB= {cos (A + B) + cos (A-B)}, or = cos2 (A +B)-sin 2 (A-B), we see that we may always resolve a product of two sines or two cosines, in two different ways, namely, PLANE TRIGONOMETRY. 61 either by means of a sum or difference, or by means of the difference of two squares; and the same is also true of the product of a sine and cosine, since sinA cosB= {sin (A + B) + sin ( A- B)}, and also -sinAsin(90-B) =sin2(A + 90-B)-sin 2 (A 900 +B) =sin2 {45~ - (A-B)) -sin2- {45~-0 1(A +B)} =sin2 450 -- (A-B)} -si n2 45{~-0-(A +B)}. Ex. IL. Express (i) by means of a sum or difference, (ii) by means cf the difference of two squares, the following products: 1. sin 5A sin A; sin 3A sin2A; sinA sin A. 2. cos3A cos A; cos 2A c os A; cos A co; A cos IA. 3. sin 2A cos 3A; sin IA cos 2A; cos 2A sin A. 4. sin(A-IB) sinC; cos (A-B) cosC; sin (A+B) cos (A-B) Prove the following formula: 5. cosA+cos (A +2B) = 2 cosB cos(A+B). 6. cos2A (1+sec-A+-tanLA) (1-sec-A+tan,A)=sinA. cosA —cos 2A 7. tan 2A tan A=cosA-cos 2A cosA4-cos 2A 8. tan22A-tan2A- sin 3A sinA cos 22A costA 9. tan 2A-tanA —2 sinA cosA — cos 3A 10sin A+sin HA =taA 1 lsinA+sin A an 2 10. = tanA. 2 A. cos-j-A, - cos 4A cosA+ cos 3A 1 sinA+sin 3A+sin 5A tan 12. tan 3A. cosA+cos 3A+ cos5A 84. Conversely, the sum or difference of any two sines or cosines, or of a sine and cosine, or the difierence of their squares, may be transformed at once into a product. PLANE TRIGONOMETRY. Ex. 1. sin(A +B)+sin(B+ ( C)=2 sin-(A-2B+ C)cos (A-C) Ex. 2. cosA-cos (A+2B) = 2 sin I (2A- 2B) sin I (2B) = 2 sin (A+ B) sinB. Ex. 3. sin (A-B) + cos (A - B) = sin(A —B)+ sin (90~-A —B) =2 sin-(90~-2B)cos (90~- 2A)=2-sin(45 —B)cos(450-A). Ex. 4. sin2(A-B)-cos2B- - {cos2B-sin2(A-iB)} =-cos(B+-A-B) cos(B-A+B)=-cosA cos(2B-A). These last transformations are in fact more practically useful than the former, as they enable us to adapt the given quantity in each instance to logarithmic computation, for which, as will be seen in the chapter on logarithms, it is necessary to express it in factors. On this account, and because it is very desirable that the Student should be able readily to perform these operations, which are required frequently in Problems, we shall make them the subject of another Exercise. Ex. 1s. Express in factors the following quantities: 1. sin 5A- sin 3A; cosA-cos i-A. 2. sinA+ sin(A-2B) cosA- sin(A -B). 3. sin(A+B) -sinB; 2 sin i-A+2 sin A. 4. vers 2A-versA; cos(A-B) -cos(B-C). 5. sin2A-sin2(A-2B); cos'A-cos'2 A. 6. cos(2A +-B)-cos(A-2B); sinA+ sin 2A. 7. vers(A+- C)-vers(B+ C); cos2B-cos2(A-2B). 8. 4 sin2 A —4 sin2 I (A- B); sinA+cosA. 9. sin2(A+B)-cos2(A+B); cos(A —B)-sin2(A+.B). 10. sin (A+.B)- cos(A —B); sin~A-cos~ A. 85. We may now go on to expand sin (A+B+ C) and cos (A + B + C), by making use of the formula for sin (A + B) and cos (A + B). Thussin(.A + B C)==sinAcos B+ C) + cosAsin(B + C) =sinA (cosB cos C- sin B sin C) + cosA (sin B cos C cosB sin C\ PLANE TRIIGONOMETRY. 63 Hence, if A, B, C, are t th three angles of a triangle, then sin(A + B + C) = sin 180~=0;.sinA cosB cos C+ sinB cosA cos C+ sin C cosA cosB = sinA sinB sin C; or, dividing every term by cosA cosB cos C, tanA + tanB - tan C= tanA tanB tan C. Like results may be found by expanding cos (A + B + and observing that cos(A + B + C) = - 1, when A,B, C, are the angles of a triangle. So too, if A + B + C= 90~, or A, B, C, be the semi-angles of a triangle, other results may be obtained from the above expansions by observing that sin (A+B + C)= 1, cos (A+B + C) = 0. In like manner, if A, B, C, be the angles of a triangle, so that A + B + C= 180~, or -A + -B + C=90~, then we have cosA +- cosB + cos C =2) cos ( ) cos -(A- B) +cos C =2 sin -C cos (A- B)+(1- 2sin2C) =1 +- 2sin C {eos - (A-B)-sin 1 C} =1 -- 2 sin C {osI(A-B) —os}(A+B)} = 1 + 2 sin ~ C {2 sinIA siniB} = 1 + 4 sin -A sin-B sin- C. Again, tan (A + B + C) tnA tanB + tan C tanA + tan (B + C) 1 - tanB tan (C l-tanA tan(B4+ C) _tanA(tanB +tan C 1 -tanB tan C tanA + tanB + tan C- tanA tanB tan C 1- tanA tanB - tanA tan C- tanB tan C' Hence, if A-+ B + C= 180~, or tan (A +B+ ) = 0. we have the numerator= 0, or (as before) tan4 4- tanB + tan C= tanA tanu tan C: 64 PLANE TRIGONOMETRY. and if A+B+ C= 90~, or tan (A B+ C)= oo, we have the denominator= 0, or tanA tanB + tanA tan C+ tanB tan C= 1. Ex.. 1. Prove the following formulae, where A+ B+ C= 1800: 1. sin(A+B) sin (B+ C) = sinA sinC 2. cotA cotB+cotA cotC-cotB cotC= 1. 3. sin(A+B- C)+sin(A+B-C9+sin(A+ C-B)+sin(B+ C-A) — 4 sinA sinB sinC. 4. sinA+sinB+sinC= 4 cos A cos -B cos C. 5. cos2A +cos2B+cos2C+2 cosA cosB cosC= 1. 6. sin2 -A+sin2 B+ sin2 C+2 sin-A sin}B sin C= 1. Prove the following formule, where A+B + C= 90~: 7. cotA+cotB+cotC= cotA cotB cotC. 8. tanA +tanB +tanC= tanA tanB tanC+secA secB secC. 9. sin 2A +sin 2B+sin 2C= 4 cosA cosB cosC. 10. cos(A+ 2B) +cos(B+2 C)+cos(C- 2A) = 4 sin (A —B) sin I (B - C) sin (C-A). For all values of A, B, C, shew that 11. cos(A+B+ C)-cos(B+C-A)+cos(A+ C-B)+cos(A+B-C) = 4 cosA cosB cosC. 12. sin(A+ B) sin (B+C) == sinA sinC+sinB sin(A+B+ C). 86. To shew that tan-'m + tan-'n=tan-' rn + n 1 - fmn By the expression tan-'m is meant the angle whose tangent is m; thus if tanA=-m, then A =tan-'m. (The origin of this notation is easily seen; for, if we could for a moment conceive the symbol tan separated from the angle A, we might write the above equation A= m or by the Theory of Indices A =tan-.m.) tan Let tanA = m, tanB =n; then A = tan-1?, B =-tan- n, but tan (A B tanA + tanB _ m + n 1 + tanA tanB 1 - mn;.t,'tam + taan'n = A B =-. tan-l1 - n 1 4- +nM PLANE TRIGONOMETRY. 65 2m Ex. 1. 2 tanlm tan-'; 3 tan- an- =2 tan-'m+tan -_ tan.1 2m +tanr'm -tana'l-m2 + 3m-r3 l-~m~ 2Nm2 — = tan' — 3m2 1 —m2 Ex. 2. tan'l + tan'l —tan-l, ~+ ---=tan' 6 - tan- 1: 1.2' hence, if tan a =, tan/3 =, we have a+/ =tan-1 1 = 7-r, or, in degrees, A+ B = 45. 87. We may here observe that the expression sin2A, which we have used all along for (sinA)2, should be more correctly employed to denote sin (sinA), that is, the sine of an angle whose value (in circular measure) is the number, sinA. Thus, if sinA -=- then sin2A=sin (sinA) =sin3, that is,= sin( -cw)= si 34~.4; and, if sin 34~.4 be found from the Tables, = lnearly, then sin3A= sin(sin2A) = sin = sin32~.7, &c. In like manner, sin-2 m is the angle whose sine is sin-' m: thusif m=4-, then sin-l m=Tr=.523; and sin-2 m would be the angle whose sine is.523. But instances of this kind so rarely occur, that no mistakes are likely to arise from using generally the more convenient notation sin2A, sin3A, &c. for(sinA)2, (sinA)3, &c. 88. By means of the formulae for sinA, cosA, and tanA, in terms of sin2A and cos2A, we may obtain the numerical values of the functions of those angles, which are the halves of 30~, 45~, 60~, &c. and so again, of the halves of these angles. Ex. 1. sinA=V-/(2-2 cos 2A) by (76). sin 22~ 30'= A(2 -2 cos 45~) = j/(2- /2). Ex.2. 2 cosA = /(1-+sin 2A)-V(1-sin 2A), if A <45; 2 cos 11l 15' = VJ(1 +sin 22~ 30') —/(1-sin 220 30') -=V{l+ i(2- %/2)}-/v1-/IV(2- 2)} (I.) F 66 PLANE TRIGONOMETRY.. -cos 2A 10 2-1 i Ex. I 7. Prove the following results: 1. 2 tan 1=tan-'1. 2. 2 tan —= cots 13. 3. tan-l++2tan4l =17r. 4. cot-i +cot"4 7Ad. 5. sing tan-l'. 6. sin-i+ tan1=_ tan2 32. 7. sin- 1 ot-13 = 4. 8. tan-+tan +tan- +tan-1 Obtain the values of the following trigonometrical functions 9. cos22 30n'.. 10. cos. 11. tan 37~ 30'. 12. sin 730 3,. vers 15. 14. cos 11~ 15'. 15. sin 56~ 15' 16. chdl185~ ( 67 ) CHAPTER IV. ON THE CALCULATION OF THE NUMERICAL VALUES OF THE TRIGONOMETRICAL RATIOS. WE shall now proceed to shew how the values of the Trigonometrical Ratios are found for all angles from 0~ to 45~, at intervals of 10": but, in order to this, we must first prove the following propositions. 89. The perimeter of any convex curvilinear figue, contained within a rectilineal figure, is less than that of the contaiinng fiure. Let AEB be a convex curvilinear figure; and let A CDB be any exterior rectilineal f' e~/~ ~ figure, formed by lines touching the A Z curve at A, E, B. At any intermediate point between A and E, let the tangent FHG be drawn. Then (Euc. I. 20) CF and CG are together greater than FG, and, therefore, the perimeter AFGDB will be less than the perimeter A CDB. And, by drawing other tangents at intermediate points, the perimeter of the figure so formed will be continually diminished, and will, therefore, be least when the number of tangents is increased and their length diminished without limit, in which case the circumscribing figure coincides ultimately with the curve. Hence the perimeter of the curve AEB is less than that of any circumscribing rectilineal figure. COB. In like manner it may be shewn that the perimeter of the curve AEB is greater than that of any inscribed rectilineal figure, and, therefore, a fortiori, greater than the line AB. 68 PLANE TRIGONOMETRY. 90. The circular measure of any angle between 0~ and 90~ is greater than its sine and less than its tangent. Let BAP, BAP', be equal angles, whose circular measure is 0: with any A <-> radius AB describe the circular are \ / PBP'; draw PNP perpendicular to AB, and PT, P T, perpendicular to p/ AP, AP'. Then arc PBP' > line PNP', or PB > PN, PB PBN and, therefore, Ap > P-, that is, 0 > sin; also (89) arc PBP' < line PTP', or PB < PT, PB PT and, therefore, A-p < AP, that is, 0 < tan0. sinG tanG 91. As 0 is diminished, each of the ratios --, 0 tends more and more to unity, and has unity for its limiting value. For 0 lies between sin and tan0, and, therefore; sin0 0 1 sin,, cos- ' or (dividing each by sin0,) 1, 'in0 ' cos, ' are placed in order of magnitude, that is, 0Q 1 n lies between 1 and -o 0 But, as 0 is diminished, the value of cos0 tends continually to 1, and ultimately, when 0=0, becomes 1; hence ~~0 ~sin 0 hences-in (and therefore also -a) ultimately becomes 1; tan 0 sin 0 1 and hence also -- (= -6 x o- ) ultimately, when 0 =0, becomes 1. The Student will understand, from the remarks in (48), that the real meaning of such expressions as these is that, as the angle is diminished, its circular measure becomes more and more nearly equal to its sine or its tangent, and, by diminishing the PLANE TRIGONOMETRY. 69 angle, the difference between 0 and sin0, or tan 0, may be made as small as we please, without its ever becoming actually zero, because it is impossible that the angle should actually vanish, and its cosine become 1. CoR. Hence also cos-=(1 —sin20) =1 -- sin20-&c.=l — 602, nearly, when 0 is very small. It should be carefully remembered that this equation holds only when circular measure is supposed. If 9 denoted the degrees in an arc, instead of the circular measure of the arc, the limiting value of sin- would not be unity. 0 92. If 0 be the circular measure of an angle between 0~ and 90~, then sin > 0 —103. sinl-0 0 For tan -O > -O, or s 2 > or 2sin - > 0cos 0;. 2sin'0 cos10 > cos2- O, that is, sin0 > cos 2 ->O(1 —sin2~0), and, a fortiori by (91), >0(1 -42); that is, sinO > 0- 403. N.B. It will be seen (in Part II) that, when 0 is small, sin O = 0 -— 03, nearly. 93. The properties just proved may be at once applied to obtain some interesting results, as in the following Examples. Ex. 1. A church-tower, seen along a horizontal plain, at a distance of 2 miles, subtends an angle of 1~ 5' 6": to find approximately its height. In the figure of (90), let A be the observer's place, PN the tower, L PAN=0; then PN=AN tanO: but Z PAN=1~ 5' 6"= 1~.085; 10.085 1085 22 341 -- X Xr 180~ 1800000 7 18000 hence PN (=AN tan ) = ANx 0 nearly = 2 m. X T3 4 =200 ft. We shall presently see (Ex. 4, Cor. 2) that the true height is about. 2 ft. 8 in. more. Ex. 2. The Earth's radius (4000 miles) is found, by Astronomical calculations, to subtend at the centre of the Sun an angle of 8".57116: determine the Sun's distance from the Earth. In the same figure, let A represent the centre of the Sun, and PV the Earth's radius: then, as before, PN=ANX 0 nearly, PLANE TRIGONOMETRY. 8"//.57116 206265 or AA=PN — =40 - =4000- =40X by (26) 570.29577 8.57116 =4000 x 24065.004=96,260,016 miles. In fact the Earth's radius may be regarded as a small circular arc PN, subtending at the Sun the angle PAN, to which arc we may apply the formula, a=rO. Hence, if the Sun's distance from the Earth, or the radius of the Earth's orbit about the Sun, be taken to be 96 millions of miles, it will be 24,000 times the Earth's radius. Ex. 3. Given the Sun's apparent diameter to be 31-': determine his actual diameter in miles, and compare his bulk with that of the Earth. By the Sun's apparent diameter is meant the angle, or arc (to radius 1), which his actual diameter subtends to an observer upon the Earth, or, rather, to one supposed to be placed at the Earth's centre. Hence if, in the figure of (90), A be taken to represent the Earth's centre, and PNP' the Sun's diameter, then Z PAP' is his apparent diameter; and PNP'-arc PBP' nearly =APxOe=APx "^:xWr=APx 31 x =96,,ooox 22 11 x -96,000,000X 12 1800 X 60 7 1200 =880,000 miles, or the Sun's diamr=ll0 times the Earth's diam'. Hence, since the volumes of similar solids are as the cubes of homologous lines, we have volume of Sun: volume of Earth::(diamr)3 of Sun: (diamr)3 of Earth:: (110)3: 1:: 1,331,000: 1; so that the Sun is more than a million times as large as the Earth. Ex. 4. The top of a mountain can be just seen at sea at a distance of d miles: determine its height (h) above the level of the sea. In the figure of (90), let PBP' represent a portion of the Earth's surface, BT the mountain, whose, top T cal be just seen at the point P, where the line TP touches the circle: then BT=h, PB=d=rO, if r be the Earth's radius AP, and 0 the circular measure of Z PAB: hence 0= d, and (91 Cor.) P r I (1+ AT rl-h l-icos(l1) 1 — '-1 — &c; r._ h nearly=, or d-2 1 H d2=2r, a h 2r Cou. 1. Hence d2=2rh, and hoc d2, or the height varies as the square of the distance. PLANE TRIGONOMETRY. 71 COR 2. If d=1, and r=4000 miles, we have h-=8-v mile= 8 inches nearly, so that, over the surface of still water, an object, 8 inches high, would be hid, by the curvature of the earth, from an eye on the level of the water at the distance of a mile. This is expressed by saying that the dip or depression of the horizon is 8 inches for one mile of distance; hence, since the dip increases as the square of the distance, therefore h': h:: d2: d2, so that a hill, 600 feet or 7200 inches high, would become hid at the distance d'=d/hi= /7200 =30 miles. hT 8 Hence also, since at the distance of 2 miles, the dip would be 4 X 8 in.=332 inches, the height of the tower as found in Ex. 1 above should be increased by 2 feet 8 inches, to get its true height. Conversely, from the formula d2=2rh, may be found the distance when the height of the object is given; or the radius of the Earth may be found, if we know by other means the values of d and h in any case. Ex. 18. 1. Two towers, seen along a horizontal plane, at distances ot 21 and 38 miles respectively, subtend angles of 1~ 26' 33" and 1~ 36' 10,; compare their heights, neglecting the dip of the horizon, 2. The Earth's radius (4000 miles) subtends at the Moon an angle 57' 1".8: shew that the Moon's distance from the Earth is about 60 times the Earth's radius, or 240,000 miles. 3. The Moon's apparent diameter being 31' 15", determine its actual diameter, assuming the result of the last question: and shew that the Earth appears to the Moon about 13 times as large as the Moon appears to us. 4. The diameter of the Earth's orbit subtends at the nearest fixed star an angle of about 1": shew that the star's distance from the Earth must be nearly 20 billions of miles. 5. Shew that the dip of the horizon for three miles of survey is nearly a fathom. 6. At what distance may the Peak of Teneriffe, which is 21^ miles above the level of the sea, be just seen from the deck of a ship? 7. The summit of Dhawalagiri, one of the highest points of the Himalayas, is 28,000 feet above the level of the sea: at what dis. tance would it first appear in the horizon? 72 PLANE TRIGON OIMETRY. 8. If a mountain, 6600 feet high, can be seen at the distance of 100 miles, what must be the Earth's radius? 9. Two points, each 10 feet above the surface of still water, cease to be visible from each other at a distance of 8 miles: find the Earth's diameter. 10. The hull of a ship is 40 feet above the water, and the masts reach 80 feet above this: at what distance will the hull be hidden from the sight of a person standing on the sea-shore? and at what distance will the whole vessel be lost to view? 94. To find the numerical value of sin 10". Let the circular measure of 10" be ac: then, since the circular measure of 180~ is 7r, the value of which number is, more accurately, 3.141592653590, we have a 10 1 7r-180 x 60x 60 64800' 7r 3.141592653590 648or a 00 — 64800 ---.00004,84813,68. Now sin < a > -c3, a fortiori, > c — (.0005 3 > a —.00000,00000,00032, where the quantity to be subtracted from a does not at all affect the first twelve places of decimals. Hence to twelve places the value of sin a (or sin 10") coincides with that of a, so that sin 10"=-circ. meas. of 10 = 64800 =.00004,84813,68. CoR. I. Hence, a fortiori, sin 1"=circ. meas. of 1 = 648000- =.00000,48481,36. CoR. II. Having determined sin 10", we obtain cos 10' by the formulacos0" V(1 - sin210")=. 99999,99988,24. 95. If 0 be the circular measure of n", we have 0 -=n x circular measure of 1" = n sin 1", and n= s PLANE TRIGONOMETRY. 37 This agrees with the Rule in (24); for, if A=0w, where A is the number of degrees in the angle, whose circular measure is 0, then 60x60x 180 n, the number of seconds in it, =60 X 60 X 0w=0 X 648000' 0 O —X - -sm 1 7 as above. 96. Knowing now the sine and cosine of 10", the sines and cosines of all angles between 0~ and 90~, at intervals of 10", may be computed as follows. In the formula, sin(A + B) = 2sinA cosB -ain(A- B), put A = n. 10", B=10"; then sin(n + 1) 10"=2 sinn. 10" cos 10-sin(n- 1) 10" = (2 -) sin n.1 0"-sin(n —1) 10", if for 2 cos 10" we write 2- h, where K is a very small quantity,.00000,00023,52. Hence {sin (n q- 1) 10" -sinn. 10"} = {sinn.10"-sin (n — 1) 10"} - sin n. 10" where by putting, successively, n= 1, 2, 3, &c, we get the values of sin 20", sin 30", &c: thus (sin 20"-sin 10")= (sin 10"-sin 0')-h sin 10", (sin 30"-sin 2 3)= (sin 20"- sin 10")-k sin 20", &c.=&c. The formula is put in the above form, because it is seen that the quantity in brackets on the second side is merely a repetition of that which stands upon the first side in the preceding line; so that the only additional labour in each line will be to multiply by k the value of sinn.10" found from the preceding line. Having thus computed the sines up to 60~, we may proceed to find the rest more simply by mere addition of sines already found, as follows: sin (60~ + A) - sin (60~ -A)= 2 cos 60~ snA =sinA;. in(6 1- A) = sin (60- - A) + sinA 74 PLANE TRIGONOMETRY. thus sin 60~ 10"/sin59 59' 50" +sin 10", sin 60~ 20" =sin 59 59' 40" + sin 20", &c. And, lastly, having found the sines from 0~ to 90~, these give us immediately the cosines from 90~ to 0~, since cosA = sin (90~- A). 97. The tangents and secants from 0~ to 90~ may sin 1 now be found by the formula tan = -, sec= -; Cos Cos and these will give the cotangents and cosecants from 90~ to 0~. But when the tangents have been found up to 450, the rest may be found more simply by the formula tan (45~ + A)-tan (45~- A) = 2 tan 2A: thus tan 45~ 10'"= tan 440 59' 50" + 2 tan 20", &c. Also, since 1 sin2A - cos2 I A. cosec A - sinA - 2 sin -A cos-A -- tan A + cot iA), we have cosec20" = I (tan 10" + cot 10"), cosec 40" = - (tan 20" + cot 20"), &c. so that the cosecants, and therefore also the secants, may be found for all even multiples of 10", by mere addition from the Tables of tangents and cotangents. 98. As, by the preceding methods, an error, intro duced at any point, would be carried on throughout the series of operations, and probably before long begin seriously to affect our results, it is desirable to calculate directly some of the Ratios for particular angles, by comparing which with the values for the same Ratios obtained as above, we may check our work at different points, and be certain that we have attained a sufficient degree of accuracy. The Ratios employed for this purpose are the sines and cosines of angles at intervals of 9~ from 0~ to 90~. PLANE TRIGONOMETRY. 75 99. To find the value of sin 18~. Let A = 18~; then sin360= cos 540, or sin 2A = cos 3A;. 2 sin A cos A = 4 cos3A-3 cos A, or 2 sinA=44 cos2A-3 =1-4 sin2A;.4sin2A + 2 sinA- 1 = 0, whence sinA = -(/5 - 1), where we take the positive sign of the root, because we know that sin 18~ is positive. 100. We have now sinl 8~= (V5 - 1) = cos72~, whence cos18~= (1 —sin2180)= V {1-1 (6-2 /5)} = I/(10+ 2 /5) sin 72~; cos360~- -2sin218= 1 — (3 -/5)=( (/5+1)= sin54~; sin36~= V(1 -cos2360)= / {1 — 1(6 + 2 /5)} =.V/(10-2 V/5)=cos54~. Also putting 9~ and 27~ for A in the formulam sinA = { V/(1 + sin2A)- V/( 1-sin2A)}, cosA -= { /(1 + sin2A) + V( 1-sin2A)}, we get sin9~= { v(3 + V5)- / (5- V5)} =cos81~, cos9~= { (3+ V/5)+ /(5- /5)} = sin8 1~, sin27~0= {V/(5 + 15) — /(3- V5)} =cos63~, cos27~=- { /(5 + V/5) + / (3- v/5)} = sin63~. And thus, if we include sin45~= - /2= cos45~, we have the values of the sine and cosine for all angles at intervals of 9~ from 0~ to 90~, which, by extracting the roots indicated, may be expressed as simple decimals. 101. The following are called Formule of Veriication. sin(360+ A) -sin(360-A)=2 cos360sinA=(,/5 + 1)sinA, sin(720 + A)-sin(72~-A) =2 cos72~ sinA= 2( v/5 - l)sinA;.'.sin(36+ A)-sin(36-A)-sin (72 + A)+sin(720-A)= sinl, a relation between the sines of five angles, which, if found to be satisfied by our tabulated results, will prove them to be sufficiently accurate. 76 PLANE TRIGONOMETRY. In like manner cos(36+A)+ cos (360-A)-cos (720+A)-cos (72~-A)-=cosA. COR. Since sin(360+ A) =cos(90~-360-A)=cos (540-A), &c, these two formulae may also be written sinA = cos (540-A)-cos (540+A)-cos (18~-A)+cos (18~ +A) cosA=sin (54~-A)+ sin (54+ A) - sin (18~-A)-sin (18~+A). 102. The Tables do not go beyond 45~: the values of the sines, tangents, and secants for angles above 45~ being the same as those of the cosines, cotangents, and cosecants of their complements, which are less than 45~. Thus the page which contains the values of the Ratios from 19~ to 20~, is marked at the top with 19~, and on the left-hand is a descending column of minutes; while at the bottom it is marked with 70~, and on the right is an ascending column of minutes; and the columns, which are marked sin, cos, tan, &c. at the top, are marked cos, sin, cot, &c. at the bottom: so that (for instance) the same numerical value would be read from the top as that of sin 19~ 54', and from the bottom as that of cos 70~ 6'. Since the values of the sine and cosine, or of the tangents of angles less than 45~, are always less than 1, and therefore will be decimal fractions, sometimes with two or three cyphers after the decimal point, they are printed in some Tables with the decimal point moved four places to the right, (as if they were each multiplied by 10,000,) while in others the decimal point is omitted altogether. No mistake, however, can arise in taking out the true value, if the Student is only made aware of this fact. All the above values are called the Natural sines, PLANE TRIGONOMETRY. 77 cosines, &c., to distinguish them from what are commonly called the Logarithmic sines, (or shortly, logsines,) &c., which, however, would be more properly called the Logarithms of the (Natural) sines, &c. These latter are by far the most frequently required in prag tice: more will be said about them in the chapter on Logarithms. 103. All Tables, however, are not constructed for so small intervals as 10"; but, in such cases, the values of the sine, &c. for any angle not found in the Tables, may be obtained with sufficient accuracy for all ordinary purposes by means of a simple Proportion: and, conversely, the value of any angle, whose sine, &c. cannot be found exactly in the Tables, may be obtained, in like manner, by means of the same proportion. Thus, let the interval at which the angles are given in any set of Tables be a", and let D be the difference between any two like functions of two successive angles, A and A+a", D being obtained by subtracting the function of A from that of A + a"; so let d be the difference between the same functions of A and A + n": then, assuming that D: d:: a: n, which (as will be presently shewn) is in most cases very nearly true, we n d get d=xD, n=- xa. If a=60, as in Hutton's Tables, (which we shall use,) where the angles are given at intervals of 1', we get n d d= —ox D, n=. x 60. Ex. 1. To find the value of sin 37~ 23' 47". Here sin 37~24'=.6073758 a sin 37 23'=.6071447 as taken out from the Tables; Diff.forl'or 60"= +.0002311:.. diff. for42 -= 4 of.0002311 -.0001 618,. sin 37023' 42 =.6071447 +.0001618 =.6073065. 78 PLANE TRIGONOMETRY. We may omit the cyphers in the values of D and d, as below Ex. 2. To find cot 55~ 27' 48". Here cot 55~ 28' =.6881379 cot 55~ 27'=-.6885666 Diff. for 60"= -4287;.. diff.for48'"/= of-4287 —3430;. cot 55~ 27' 48" =.6885666 -.0003430 =.6882236. N.B. D and d will always be negative for the cosine, cotangent, and cosecant, which decrease as the angle increases from 0~ to 90~. Ex. 3. Given secA'= 9.7654237: to find A' or sec-' 9.7654237. Here, by the Tables, it would be found that secA' lies between gec 84~ 7 = 9.7557944 and sec 84~ 8' = 9.7834124, so that A' = 84~ 7' n" = A + n", suppose. Now sec (A+ 1') = 9.7834124 and sec(A +n") = 9.7654237 secA = 9.7557944 secA = 9.7557944..Diff.forltor60= +276180; diff. for n" — +96293: hence n = i9-,,,O of 60 - 20.9 = 21 nearly, and A' =840 7 21" Ex. 4. Given cosA'=.8241657: to find A' or cos-1.8241657. Here, by the Tables, cosA' lies between cos 34~ 29' =.8242909 and cos 340 30'=.8241262, so that A'34~ 29' n/=A -n/, suppose. Now cos(A+1) =.8241262 and cos (A+n") =.8241657 cosA =.8242909 cosA =.8242909. Diff. for 1' or 60"= -1647 diff. for n" -1252: hence n = -5of60= 45.6 =46 nearly, and A'= 34~ 29' 46". Ex. 39. 1. Given sin 24~ 37' =.4165453, sin 24~ 38'= 4168097; find sin 240 37 15", and sin-1.4166287. 2. Given cos 68~ 12'=.3713678, cos 68~ 13'=.3710977 find cos 68~ 12' 24", and cos-1.3711999. 3. Given tan 45~ 1'=1.0005819; find tan 45~ 0' 35", and tan-' 1.0002345. 4. Given cot 30~ 1'= 1.7308878; find cot 30~ 0' 17", and cot-11.7312123. 5. Given sec 59~ 59'=1.9989929; find sec 59~ 59' 25", and sec11.9990678. 6. Given cosec 60 1'= 1.1545067; find cosec 60~ 0' 37", and cosec-11.1546025. PLANE TRIGONOMETRY. 79 104. We shall here prove the truth of the principle assumed in (103), namely, that, for a very small increment of the angle, the increment of its sine, cosine, &c. is, except in particular cases, proportional to the increment of the angle. Let 0 be any angle, and let a (expressed in circular measure) be any very small increment of 0. Then we have sin (G + )-sin 0=sin 0 cos S-cos 0 sin — sin0 =cos 0 sin — sin e (1-cos a) ( 1 —cos ) =cos 0 sin - 1-tan sin - =cos 9 sin (1 -tan 0 tan If) by (74): (i) tos (0 + )-cos 6= cos cos S-sin 0 sin - cos 0 =-sin0 sin a-cos 0 (1-cos ). = - sin sin S (1 +cot 0 tan ): (ii) sin (6+) sin 0 sin (-+-) cos 9 —cos (0+5) sin0 a(0+)-tan -cos (0+3) cos 0 cos 0 cos (0+6) sin{t(-6)- }_ sin a cos J cos (0 )-cos (cos0 cos -sin sin ) sin c sec2 9 tan 6 - cos2 0cosa (1-tan0 tan) 1 —tan 0 tan (iii Now, if 6 be very small, sin 6 and tan 6 are by (91) very nearly qual to 6, and are both, therefore, very small; hence, unless tan 0 e large in (i) and (iii), or cot 0 in (ii), we may write the above sin (-0+ J)-sin ' = 6 cos e, cos ( 0+- )- cos 0 — - sin0, tan (0+ 6)-tan C= 6 sec2 0; Whence we see that, for any given value of 0, a small increment (6) f the angle will produce a small increment of the sine or tangent, roportional to that of the angle, provided that tan 0 be not very reat, or 0 near 90~; and similarly for the cosine, provided that AtO be not very great, or 0 near 0' or 180~. Similarly, we have t (0 +O )-cot 0 cos(0+6) cosO sin0 cos (O-6-cos sin (03a) sin (.+6) sin - sinO sin (0+-) sin 10-(C+6J)} -tan 6 cosec 0 'sinm cos6 (1-+cot tan) — 1+cotO tan 6 8'0 PLANE TRIGONOMETRY. sec (0+8)-see 0 1 1 cos 0-cos(9-a) cose(l —cosS)q-sin0 sil cos (0+3) cosd - os0 cos(0 + a) - cos0 cos(0+f) — cos a sin3 cosO ( sin- +tanG) tan sec (tan + a tan0) (v) cos 0 cos (1-tan tan ) 1 -tan tan cosec (9+ ) —cosec 0 1 1 sin —sin (+3) sin 0(1 -cos ) -cosO sin sin (0 + ) sin sin sin (0- ) - sin sin ('+ ) 1-cos J sin sin 0 ( siCn — cot 0) tan a cosec 0 (tan — cot (vi) - sinm cos (1-cot tan d) 1-cot' tan So that here also, if 8 be very small, we may write cot (0+e)-cot0= — cosec23, sec (0+-) —sec 0-= sec 9 tan 0, cosec (O + 3)- cosec 0= - cosec 0 cot 0, unless 0 be near 0~ or 180~ for the cotangent and cosecant, or near 90~, for the secant. It follows then that, for all the trigonometrical functions, (except in the particular cases above noticed,) a small increment (8) in the angle will produce a small proportional increment in th( function. Hence, if D, d, be the increments of any function corresponding to small increments of a" and n", respectively, for an) angle A, and if a and v be the circular measures of a" and n", ther (by the above) we have D: d:: a: v::a: n, the same result as we assumed in (103). Col. Of course, we are not at liberty to employ the method ii (103), in order to determine with greater accuracy the value of ai angle (or its function), when not exactly given in the Tables, if th, case be one of those above excepted. Hence, if the angle be nea 90~, we should not be able to find it with very great accuracy fron its sine, tangent, or secant; nor, if near 0' or 180~, from its cosine cotangent or cosecant. ( 81 ) CHAPTER V. ON THE TRIGONOMETRICAL PROPERTIES OF TRIANGLES, QUADRILATERALS, AND POLYGONS. 105. To shew that in any triangle the sides arc proportional to the sines of the opposite angles. In future we shall use the letters a, b, c, to denote the sides BC, AC, AB, opposite to the angles A, B C(, respectively, of any triangle ABC. C C2? 3 A 9 S -D - A Let ABC be any triangle, and from C draw CD perpendicular on AB, or on AB produced: then sinA BC a CD =A C sinA = BC sinB, or s in —4C _ b sinB A C b Similarly, by drawing a perpendicular from B on sinA BC a AC, we may shew that s ---- = a sinC AB c Hence we have a: b: c:: sinA: sinB: sinC; sinA sinB sin C or, as it may be also written, = - - a b c 106. To express the cosine of an angle of a triangle in terms of its sides. Let ABC be any triangle, and from C, as before, draw CD perpendicular on AB, or on AB produced. Then in figs. 1 and 2, BC2= A C2 + AB2 - 2AB.AD, (Euc. II 13) in fig. 3, 8 C2 = A C2 + AB2 + 2AB.AD: (Euc. II 12) (i.) Q ~8~2 PLANE TRIGONOMETRY. but AD = A C cos CAD= A C cosA in figs. 1 and 2, =-A C cosA in fig. 3;. in each case, BC2= A C2+ AB2 - 2 AB.A C cosA, b' + ct - a2 or a2 = 2 + c2 - 2bc cosA, whence cosA = 2 + - 2 2bc Of course, it was not strictly necessary to have proved (as we have done) the truth of the above in all cases; since the result with the first figure would have served for all the rest by the principle of (6). In like manner we should get b2 =a2 +c2 -2ac cosB, 2 = a2 + b2 - 2ab cosC, a2+ c- 2 a2 + b2 c2 and cosB = a2Oc CosC= 2ac 2ab 107. To shew that, if s= (a b- + c), then cosA= / s(s ), si A= /( s - b) (s - c). b " be and thence to deduce expressions for tan'A and sinA. We have b2 + c2-a2 (b + c)2 —a2 2 cos A = 1 + cosA = 1 + -a = ( )2 — 2bc 2bc _(b + c a)(b+c-a)_ 2s(2s-2a). oA s(s-a). 2bc 2b c be 2 sin2 ' A = 1 - cos= -- I 1-m~ - - ' ~ 2be 2be 2 sin 1= (s-b) (s - c) The positive roots are taken, because A, being an angle of a triangle, must be < 180~, and IA < 90~. PLANE TRIGONOMETRY. 83 In like manner cos-B= s(s - b) cos C= v/s(s - C) ac ab and sin B= V(s-a)(s-c) sin C=,(s-a)(s-b) ac ab Hence tan A = =sinA /(s-b)(s-c). cos-A s (s-a) and sinA= 2 sin -A cos-A = V {s(s-a )(s-b)(s-c)}. In like manner the expressions for tan IB and sinB, tan7 C and sinC, may be written down at once, by Symmetry, from those for tan'A and sinA. COR. From the last result it appears, as in (105), that sinA 2 {s(s-a)(s-b)(sc)} = sinB sinC a abe b e 108. The formulse, which have been proved above for all triangles, when applied to a right-angled triangle, will be found to agree with our former expressions. Thus, if C= 90, then sin C= 1, and s sin a c b or a=c sinA, b=c sinB: a2 +b 2 - C2 -s80ocoC=O= a2+ c;.a2 -t b2 = c2, (as in Euc. I 47); b^2 -c22 b2 2 and.*. cA - -, or b=c cosA, 2bc 2bc c and, similarly, a-=c cosB. 109. To express the Area of a triangle in terms of the sides and angles. We have the area of every triangle = rectangle of same base and height - base x height: G- 84 PLANE TRIGONOMETRY. hence, referring to the figures in (105), we have Area of triangle AB C= AB. CD~AB.A CsinA = Ibc sinA = 2ac sinB = -ab sin C. Hence also, Area= Ibc sinA =by (107) he 2 bex 2 /{s(s —a)(s-)(s-c)}=V /{s(s —a)(s-b)(s-c) aosinB asinC si 1 sinBsinC or=bc 2 sA= in A sinA sin(B+ C)' since the sine of A = the sine of its supplement ( B + C). CoR. The expression for the area /V{s(s-a) (s-b) (s-c)} = v'{(a+b+c) }(b+ec-a) (a+c —b) -(a+b-c)} = } V(2a2b2 - 2a2c2 + 2b2c2 — a4-b4 —c4). 110. Since sinA, sinB, sinC, are respectively proportional to a, b, c, we may substitute the former quantities for the latter (or vice versa) in cases where, being involved homogeneously, they occur either on opposite sides of an equation, or in the numr and den' of a fraction. This step will often be found of great use in the solution of Problems, and will be best illustrated by the following Examples. (See Alg., Part I 85-88, and Part ii 26-28.) Ex. 1. Since sinC=sin(A+B)-sinA cosB+sinBcosA, w( should have, by the above substitution, c=a cosB+b cosA: an( this would be true; for sinC sinB c b sinA =cosB + s-A cosA, or -=cos B+-cosA; sinA smnA a a.'. c=a cosB+h cosA. a" a sinA a2 (b+c) sin2A Ex2 b+c sinB+sinC' or (b+c) (b+c)2(sinB+sinC)2. We may best prove the truth of such transformations by puttin a b c;si-A=x-=;si —sinC or a=x sinA, b=x sinB, c=x sinC, a2 ax sinA a sinA * * -snBx sinBC+x sinC B+ siinC' Ex. 3. Area=bec sinA be sinA sinA sinB sinC =i(,+CZ) +-f c =~(b++ c ) sin2B+sin2C PLANE TRIGONOMETRY. 85 Ex.. 20 In any triangle, right-angled at C, prove the following formula. 2ab bM —ao 1. cos(A-B)=- 2. cos2A= b=+a 2ab a-b 3. tan2B= ---a-_ 4. tan(A-B) — b a c-b 5. cos(2A-B)= (3c2 —4a2). 6. tan'A=,/ c-b c c+b' In a right-angled triangle, obtain these expressions for the area: 7. (a-+b+c) (a+4b-c). 8. ~c2sin2A. In any triangle prove the truth of the following formulae: a2 —b2 versA a(a+c-b). 9. sin(A-B)= 2 sin(A+B). 10. versB-b(b+c-a) tanB a2+b —c0 a-b 1,. t-an — a-+-c2-b2. 12. sin- (A-B) — cosiC. a+b-c tanAA+tan-B c 13. tanA tanB=+b 14. tan-A-tanBa — 2 12 a-tb+c- tan1A-tan1 a-_ by In any triangle, obtain these expressions for the area: sinA sinB 2abe 15. }(a -2 b) snA ) 16. b cos-A cosLB cos 111. There are six parts in every triangle, three sides and three angles; and there are three independent relations necessarily existing between these parts, or they could not belong to a triangle at all. These are expressed algebraically by the following equations, A+ + = 1 80 s and sinA sinB sin C A+ B + C= 180~, and- = --- =-. a b c All other results which we have obtained are in fact only modifications of these. Thus the expression for cosA might have been derived as follows: In (110 Ex. 1), we found, merely using the above formulae, that c=a cosB+o cosA: 86 PLANE TRIGONOMETRY. hence e-b cosA= a cosB, or c2-2be cosA+b2 cos2A=a2 cos2B, whence c2-2bc cosA+b2-a'==b2 sin2A-a2 sin2B=0 by (105); b2 —c2 —a2 and thus we get cosA= --, as before. 112. Generally then it will be sufficient, for the determination of all the parts of a triangle, if three of the six parts, or three independent equations between them. be given: since then (with the above three equations) we shall have six equations for determining the six parts. But there are two exceptions to this statement. (1) If the three angles alone are given, we have in reality onlyfive equations given us; for when two angles are given, the third is known from A + B + C=180~, and is therefore useless as a third datum. In this case the ratios, but not the magnitudes, of the sides, will be a sinA a sinA determinate from the equations -_sin, sin= b sinB' c sin C Hence the triangles, corresponding to our data, will be all similar, but indeterminate in magnitude. Consequently, one side, at least, must be given, in order to determine a triangle. (2) If a, b, A, are given, that is, two sides and the angle opposite to one of them, then, if a < b, or the side opposite to the given angle be less than the other given side, we shall now shew that there are two triangles, which have the same given data, and the problem, therefore, is ambiguous, or admits of a double solution. This, however, can only happen when A< 90~; for, since b > a, therefore also B > A, and, consequently, 4 must be < 90~, or else we should have two angles of a triangle together greater than two right angles. PLANE TRIGONOMETRY. 87 Take A C b, and / CAB ( < 90~)= A; with centre C and radius = a, describe a circular arc /4\ then, if a<b, it is plain that the circle / \ will cut AB in two points B, B', on the a J g- g g same side of AC as that on which is the angle A, and that each of the triangles CAB, CAB', will have in common the three data, a, b, A. If, however, b be such that the points B, B', coincide, then CB will be perpendicular to AB, and there will be but one triangle in which CB=AC sinA, or a=b sinA. Hence, if we see that the given data satisfy this equation, we may infer at once that the case is not ambiguous, although a be less than b, since the triangle is a right-angled triangle. COR. Observe that L CBA= 180 - L CBB'= 180~- L CB'B; that is, the angle B in one triangle will be the supplement of the angle B' in the other. 113. If in (112) the radius (a) be less than the perpendicular (b sinA) from C upon AB, the circle will not cut AB, and no triangle can be formed. In this case the solution is impossible. So also, if two angles arc given, whose sum is greater than two right angles, or three sides, any two of which are not together greater than the third, the solution is impossible. Ex. 21. 1. One angle of a triangle is 30~, and the sides including it arc 2j yards and 3t yards respectively: determine its area. 2. Find the area of an equilateral triangle, one of whose sides is 60 yards. 3- Find the area of an isosceles triangle, whose hass is 50 feet, and each of its equal sides 30 feet. 88 PLANE TRIGONOMETRY. 4. How many trees can be planted in a triangular piece of ground, the lengths of whose sides are 130, 120, and 50 yards, if two yards in length and breadth be allowed for each tree? 5. Find the area of a triangular field, whose sides are 216 yards, 270 yards, and 162 yards. 6. What length (AE) must be taken along the diagonal of a square field (ABCD), whose side is 100 yards, so that the triangle AEB may be exactly a fifth part of the whole square? 7. Find the area of a parallelogram, whose adjacent sides are 28 and 30 feet, and the included angle 75~. 8. The sides of a triangular field measure 189, 169, and 42 yds: find its value at ~150 an acre. 9. The hypothenuse of a right-angled triangle is 5.25 feet, and one of its angles is 30~: find the remaining parts, and the area of the triangle. 10. The angles of a triangle are in A.P., the least being 30~, and the opposite side is 100 yards: find the area. 11. Given a=10, c=20, A=30~; solve the triangle. 12. The sides of a triangle are as 1, 1*, lf; find the greatest angle, and the sines of the other two. 114. To find the radius of the circumscribed circle of a triangle, in terms of its sides and angles. Let A BC be any triangle, OA = OB= 0C=R, the I radius of its circumscribed circle, whose // A centre O is found (Euc. IV 5) by bisecting ( I,) any two sides of the triangle AB, A C, in I' )m E, F, and drawing the perpendiculars EO, FO; in which case the perpendicular OD, from O on BC, will bisect the third side BC in D. Then 1BD = BO sinBOD, or a = -R sinA, since L BOD = BOC = LBAC=A; hencewe haveR= -a - or(107, Cor.), 2sinA -2sinB 2sinC' abc abc in sides only = -- 4 / Vse(s-a s-b (s-c )} 4' if S represent the Surface, or area, of the triangle ABC(" PLANE TRIGONOMETRY. 89 115. Tofind the radius of the inscribed and escribed circles of a triangle, in terms of its sides and angles. Let ABC be any triangle, and let OD= OE= OF= r be the radius of its inscribed circle, whose centre 0 is found (Euc. Iv 4) by / -s bisecting any two angles of the triangle wJ^_ii' ^ by the lines BO, CO, meeting in 0; in which case the line A 0 will bisect /7 - the third angle A. Then we shall have area ABC= area OB C+ area OA C+area OAB, or -bc sinA = -ar + Ubr + Icr; be S.be sinA = (au b + c)r = 2sr, and r=2s sinA=Similarly, if the exterior angles at B and C be bisected by B O', CO', then A O' will bisect the angle A, and 0' will be the centre of a circle, which will touch B C and the other two sides AB, AC, produced, and which is called an escribed circle of the triangle. Now if the radius of this circle O'D'= O'E'= O'F'=r, we shall have area AB C= area O'A C+ area O'AB - area O'BC, or -bc sinA = br, - + cr - ar,; hence b sinA (b + c-a) r=,2 (s —a) r, be S orrl=2(s-a) ln s-a In like manner, if r2, r3, be the radii of the escribed circles, touching the sides b and c respectively, then we should obtain S S i=a-dz r3=0.e 90 PLANE TRIGONOBMETRY, N.B. Notice, for the sake of problems, that, since AE = AF, BD= BF, CD= CE,. AE+BF+ CD = I (perimeter) = s, and AE=s-(BF+CD)=s-(BD+CD)=s - it: hence AE or AF=s-a, BD or BF=s-b, CD or CE s-c. 116. To find the area of a quadrilateral, whose opposite angles are supplementary, or (Euc. III 22) of a quadrilateral which may be inscribed in a circle..Dz? Let ABCD be a quadrilateral, such / that the angles A and C, and, therefore, also B and D, are supplementary: then sin C= sin(180~- A)=sinA,cos C= - cosA. Let AB=a, BC=b, CD=c, DA=d, and join BD: then area A B CD = area ABD + area CBD = ad sinA + 2 be sin C= 1 (ad + bc) sinA. Now we have by (106) 2ad cosA = a2 +d2 - BD2 and 2bc cos C or -2be cosA = bc2 + - BD2; hence, subtracting, 2 (ad + be) cosA = a2 + d2 — b c2, a2 + d2-b2-c2 or cosA = 2(ad+bc) a2+ d2 - b2 _ c2 (b + )2 _ ( _ -d)2 ' cosA=- 1 -2 (ad + bc) 2 (ad + bc) a2+-d2-b2 - C2 (a +d)2 - (b- C)2 1 - cosA = 1 + 2(ad+b) 2 (ad+ be) 2 (ad - be) and sin2A= 2A (b+C)2 (tI-d)2 (a +d)a-(b-c) aldsnA=-csA- 2(ad+be) x 2(ad+bc) (b+ c +a-.d) (b-c+d-a) (a+d-+b-c) (a+d+c —b 4 (ad + bc)2 (2s-2d)(2s-2a) (2s- 2c)(2s- 2b) =(if 2s= a+b+c+d,) 4( -b) 4 (ad bC)2 2 hence sinA =- a V^/ {(s- a) (s-b) (s-c) (s-d)}, and area AB CD= (s -- V ias -^ s-~)b. PLANE TRIGONOMETRY. 91 117. To find the radii of the circumscribed and inscribed circles of any regular polygon. Let AB (a) be the side of a regular polygon of n sides, 0 the common centre of its inscribed Q ((e7\ and circumscribed circles. Draw OC perpendicular on AB: then, since the sum of all the angles, subtended at O by the sides of the n polygon, is 360~ or 27r, we have Z. A OB= -. Hence, R= radius of circle circumscribed = OA = A C cosec A 0 C= a cosec r. n Again, r= radius of circle inscribed = OC=AC cot A O C = a cot-. COR. Conversely, if R or r be given, we may find a. For, if R be given, then a=2R.-+-cosec I=2R sin; n R and, if r be given, then a=2r - cot 2r tan n r 118. Hence we obtain the following results: (i) Area of any regular polygon, in terms of a, =n x area OAB=n x 'AB. OC= na2cot, n (ii) Area of inscribed polygon, in terms of R, =n x area OAB=nx A O. OB sinAOB= nR2sin 2-; n (iii) Area of circumscribed polygon, in terms of r, =n x area OAB=n x A C. OC=nr tan. n If the circle be the same in (ii) and (iii), then the sides of its inscribed and circumscribed polygons will be r and their areas nasin 2rsn 2r tan -, and their areas am -rtn, 9' ~ *- % n 92 PLANE TRIGONOMETRY. which may be written nr2sin - x cos -inr sin- -cos - n n n n 119. Hence also we may deduce the area of a circle in terms of its radius. For, as above, the area of a regular circumscribed polygon of n sides=nr2tan = (multiplying and dividing by - ) rr2 x tan -_-: and since, by increasing n n 2z the number of sides, the polygon tends continually more and more to equality with the circle, and (91) the ratio tan --- tends more and more to 1, therefore, as n is n n increased indefinitely, the Limit of the geometrical figure is the circle, and the Limit of the corresponding algebraical expression for its area is Xrr2; hence, we have the area of the circle, = 7rr2, as in (30). Ex. 22. 1. Shew what the expressions for R and r in terms of the sides become, when the triangle is equilateral, or a = b = c. 2. Obtain the same when the triangle is isosceles, or a = b. 3. In the figure of (115), shew that the products of the alternate segments of the sides are each equal to I (a+b+c) r2. 4. Shew that the areas of all triangles, described about the same circle, are proportional to their perimeters. 5. If a, f, y, denote the distances from the angular points of a triangle to the points of contact of the inscribed circle, express the radius in terms of a, 3, y. 6. Compare the areas of two regular hexagons, described in and about the same circle. 7. Express the radius of the inscribed circle of a triangle in terms of the radii of its three escribed circles. 8. One side of a right-angled triangle being 18 in., and the radius of the circumscribing circle 15 in., determine the radii of the inscribed and escribed circles. 9. Shew that R= 1 V abc and2Rr abc 2 sinA sinB sinC a+b+c 10. If the triangle be right-angled at C, then R + r = (a+b). t 93 ) CHAPTER VI. ON THE SOLUTION OF TRIANGLES. WE now proceed to the general solution of triangles by the aid of Trigonometry. But the Student should have read Arts. (131-144, 155) in the Chapter on Logarithms, before entering on this Chapter. 120. If the triangle be right-angled, we have here one datum, namely, that one of the angles, C suppose, = 90~, and we require only two other parts to be given, in order to solve the triangle completely. One of these, however, (112) must be a side. Hencefour cases may occur. (1) Given a, A, or one side and one angle. Here B = 90 - A, b = a cotA or = a c^^ a c = a cosecA or= sin; A b, e1 so that, in logarithms, logb=loga+log cotA =oga+L cotA - 10, or =loga-log tanA =loga-L tanA + 10, log c = log a + log cosecA =loga + L cosec A - 10, or =loga-log sinA =loga-L sinA + 10. This case includes, of course, that in which B, b are given, or in which A, b, or B, a, are given; since, when one of the angles A or B is given, the other is given also: and similar remarks apply to the other cases which follow. (2) Given c, A, or the hypothenuse and an angle. Here B =90~ - A, a = c sinA, b = c cosA;4 so that, in logarithms, loga = loge + L sinA - 10, logb = logc + L cosA- 10 PLANE TRIGONOMETRY. (3). Given a, b, or two sides. Here tan A=, or L tanA =10+loga-'logb; then, having found A, we have B=90~- A, and c=a cosecA or = b secA. We might have found c from its value /(a2 + b2); but, unless a and b should be members of one or two figures only, this would be more laborious, as the above expression is not adapted to logarithmic computation. (4) Given a, c, or one side and the hypothenuse. Here sinA = -, or L sinA= 10 +loga-logc; c and now, as before, B=90'~-A, b=c cosA or = a cotA. Here also b v/(c2 - a2) / {(c +a) (c-a)}.. logb =lo ((c+ a) + log(c -a). Ex. 23. Verify each of the above methods upon the following triangle:a=123.4567, b=234.5678, c=265.0721, A=27~ 45' 31"; obtaining the required logarithms from the Tables on p. 141. 121. If the triangle be obliqwue, there will also be four cases, one side at least being given in each. (1) Given a, A, B, or one side and two angles. sin B B Here C= 180~-(A + B), = a sin sin C / \|c = a.; so that, in logarithms, fA c sinA logb = loga + L sinB -L sinA, logc= loga + L sin C- L sinA. This, of course, includes the case when a, A, C, are given, and also the case when a, B, C, are given, since, if B, C, are given, A is also eiven. PLANE TRIGONOMETRY. 95 (2) Given a, b, A, or two sides, and an angle opposite to one of them. Iere sinB=b sinA, or L sinB = L sinA + log b- log a and now C= 180~- (A4 + B), =a sin Here, however, if a < b, we light upon the ambiguous case (112); and this ambiguity appears also from the above algebraical expression for sinB. For, if a < b, then A< B; and, therefore, since A must he < 90~, (otherwise, we should have A + < 180~,) B may be either greater or less than 900, that is, B may be either b of the two angles whose sine = - sinA, one of thein a being supplementary to the other. If, however, a>b, then A >>B so that B must be < 90~; and we have here no uncertainty, but must take b for B the less of the two angles whose sine = - sin4o Ex. za. Verify each of the above methods upon the following triangle: A=44~, B=66~, C=70~, a=765.4321, b=1006.62, c= 1035.43; and for the ambiguous case, 1=220,. c'=412.7725. 122. (3) Given a, b, C, or two sides, and the incnldcd angle. Let a be the greater of the two given sides. a sinA Then b-sin B' whence (Alg. Part I 85) a-b sinA-sinB tan (A-B) Ga + b' inA + sinB tan (A B) by (81) PLANE TRIGONOMETRY. a-b a-b hence tan ~(A-B)= b tan (A +B) = b cot C, (since (A+B)=(180~-)= 900;) o. L tanj(A —B)=L cot C+ log a-b)-log(a + b), = 20-Ltan- C+log (a-b)-log(a+ b); and ~(A —B) is, therefore, known: whence A=-(A + B) + (A-B), and B= (A + B)- (A - B), sin C sin C and then c= a sinA or = b sin (i) N.B. If b should be the greater of the two given sides, instead of the formula above, we should use b-a tan (B- )= b cot_ C. 123. We may, however, obtain a different expression for c as follows: c sin C sin(A + B) 2sin-(A + B)cosf(A + B) a+b = sinA + sinB sinA + sinB 2sin (A + B)cos (A- B); co(A + B) sin! C c. c=(a + b)cos -( B= (a + ) cos(A —B) (ii) Or otherwise: we have c2=a2+ b2-2ab cos C - (a2 + b2) (cos2I C+sin2 C)-2ab (cos2- C- sin2} C) = (a + b)2 sin2- C (a- b)2 cos21 C a-b =(a+b)2 sin2 C { +(a cot C)2} Now, since the tangent of an angle may be ol any sign or magnitude, a-b put tan; = b cotI C; then c2=(a + b) sin21 C(1 +tan2) = (a + b)2 sin2 Csec2p, sin c C or c=(a + b) sin C sec=( + b)sin Co (iii) Q~~~~~~~~~~~~~~~~~~O co PLANE TRIGONOMETtRY 97 which result, however, coincides with (ii), since a-b tan p=a b cota C= tan (A — B), as appears from(122);. = = (A - B), and the two expressions coincide. 124. (4) Given a, b, c, or the three sides. Here we must employ the formulxa in (107), for sin-A, cos}A, tan-A, each of which requires only four logarithms, whereas that for sinA requires seven, and is, therefore, less convenient for use. If two angles A, B, are to be found, it will be best to use the formulat fortan tan - iB, as the same logarithms are required for each of them, which would not be the case if we used the other formula. In other cases any one of the three formula may be used; except that, when 'A nearly equals 90~, it is better not to determine it by the sine or tangent, nor when near zero by the cosine; inasmuch as (104) we should not be able, in such cases, to determine 2A with sufficient accuracy. Ex. as. Verify the methods (3) and (4) on the triangle in Ex. 2Zi. i25. The angle p, introduced in (123), is called a subsidiary angle. Such an angle may often be employed, as it is there, in order to change an expression of more than one term into factors, and so render it suitable for logarithmic computation. We shall here give a few additional instances of' the use of such subsidiary angles. Ex. 1. To adapt V(a+b) to logarithmic computation. Here V(a+b)=-/a^/(l+). (I.) 3H PLANE TRIGONOMETRY. For the upper sign, put tan20=-, and, for the lower, put sin2- -; Ut (1 then /(a+b) = /a. sec and /(a-b)= V/a. cos 0: but, in the latter case, b must be less than a, or we could not make the assumption, nor would the given expression V (a-b) be a possible quantity. Ex. 2. To adapt a+b-+c to logarithmic computation. Here, putting tan20=b, we get a+b = a (1 +tan2%) = a sec23; a and a-+b =c- asec2+ -c, whence, putting tan2 = c -- CosOB, a sec a a + b + c =a sec23 (1 + tans) = a sec"^ sec2>. Ex. 3. To adapt a cos a+-b sin a to logarithmic computation. Here, if k sin 0=a, K cos '=b,. *a cos a +b sin o-k sil (0~-t); where k2 (sin2'-+cos2 ) or k2=a2+b2, that is h= /(a'+b2), and sin or tan = a whence k and 0 are determined. k cos d b The expression for h, however, may be adapted to logarithmic computation, for k= A/ (a2 + b) - ca^/V( + cot2 )=a cosec (', a result which we may obtain immediately from k sin 60=a. These examples are sufficient to shew the method of proceeding in most ordinary cases. Thus, if we had to adapt to logarithmic computation such an expression as a+-b cos 0+-c cos (0-a) +d sin (+/3), this may be written a+(b+c cos a+d sin f) cos 0+(c sin a+d cos f) sin 0; which, by Ex. 2, may be converted to a+b' cos0+c' sin0, (where b', c', will have been found in forms adapted to log. computation;) this again, by Ex. 3, may be expressed as a-+ sin (0+y), which last may be converted by Ex. 1 to a sec2o, if tan29 = sin (04-). ( 99 ) CHAPTER VIL ON THE MEASUREMENT OF HEIGHTS AND DISTANCES. 126. We have already seen some instances of the application of Trigonometry to the actual measurement of heights and distances. We shall devote this chapter to illustrate the practical application of the Science to such cases as may commonly occur of a similar kind - that is, where the distance required cannot be measured, either on account of its great magnitude or of some intervening obstacle, and cannot also be more simply obtained by the ordinary rules of common Geometry. For, in actual surveys of land or countries, there are many results which may be obtained very easily without the help of Trigonometry; and of these we shall first give some instances, 127. Besides the property proved in Eue. I. 47, namely, that the square upon the hypothenuse in a rightangled triangle is equal to the sum of the squares upon the two sides, and the proposition deduced in (30) that the area of any triangle — = base x height, it is very useful to notice the following geometrical properties, which are much employed in common field-surveying. (1) The area of any quadrilateral AB CD D = area AB C - area AD C A 2 ACx BE+ 1A Cx DF =IA C(BE+ -d) c if BE, DFE be the perpendicilars f ron t~ angles B, D, on the diagonal A U, H2, 100 PLANE TRIGONOMETRY. (2) The area of a quadrilateral AB CD, two of whose A p D sides AD, BC, are parallel, Z/X -- I area AB C area AD C E, C 1 -BCxAE+ ADx CF= 1(BC+AD)AE =2(sum of parallel sides) x (perp. distance between them). -r C, (3) The area of any quadrilateralAB CD = area ABE+ area BEFC+ area CFD, A -E -/ F \ (if BE, CF, be perpendiculars on AD,) -AE x BE+!(BE+ CF) EF by (2) + CFx FD - (AE+ EF)BE+ (EF+FD)CF=- AFxBE+ ~ DEx CF. N.B. The same result will be found to hold good, if the perpendiculars, either or both of them, should fall without the side AD: thus, if CF fall beyond D from A, as at D', we shall have trea ABCD'= area ABE+ area BEFC —area CFD' -= AEx BE+} (BE+ CF) EF- CF x FD' -- (AE+EF) BE+ (EF-FD') CF AF x BE+ 1 D'E x CF, as before. 128. By means of the above properties, fields, &c are usually measured, being broken up into triangles or quadrilaterals, for which the necessary lines are measured, and the areas found as above. The following Example will give an idea of the mode of proceeding in such a case, and the way in which the operation is conducted. Let ABCDE be a five-sided field, AC its longest diagonal: 5C the surveyor measures along from A till he comes Bto a, where he sees the angle E in a direction at / t /D right angles to AC: he measures aE, and then n easures on ab in the same line as before, to the A E point b, rwhere B appears in a direction at right angles to A C; in like manner he measures bB, be, cD, and finally cC. The line AC, joining the two extreme stations, is called the chab?-line, and tie perpendiculars aE, bB, cD, are called offsets. 'The lengths of these lines are measured by Gunter's Chai/, wl'icl, is 22 yards or 4 poles long, and divided into 1CO links. rLAN1E TRIGONOMETRY. 101 Hence 10 chains make a furlong, and an acre, which contains 4840 square yards, will contain 10 square chains or 100,000 square links: and, therefore, square links are converted into acres by merely cutting off five figures to the right. The perpendicularity of the lines aE, &c. may be ascertained by the use of the cross-staff, which is a circular piece of wood, about six inches across, on the face of which two slits are made. along diameters at right angles to one another. It is supported upon a vertical staff, which is thrust into the ground when the surveyor reaches a point, as at a, where the angle E appears to lie at right angles to the line AC. Then, the circle being so turned that one slit is in the direction AC, an eye will look along the other in the perpendicular direction, and the staff must be moved, if necessary, along AC, until the angle E can thus be seen. Suppose now Az=480, aE=480, Ab=660, bB=200, Ac=860, cD=460, and AC=-1340 links. Then these results would be thus Left Chain Right entered in the field-book, beginning.........- - from the bottom, the measures on the 1340 to C Chain Line being set down in the 860 460 middle, and the offsets to the right or 200 660 480200 480 left on each side. from A Hence the area of the field = areaAB C + areaAEDC = A C xbB + LAc x aE + I Ca x cl) [127 (3)] 2= {1340 X 200+860 X 480 +-860 X 460}sq. links=538200 sq. links 5.382 acres =5A 1R 21p nearly. 129. Without entering further into the details of Land-surveying, which may be found in books specially devoted to the subject, we shall now give a few instances to illustrate the applications of Trigonometry. It must be taken for granted that it is always possible, by means of proper instruments, to measure the angle subtended at the eye between two visible points, (that is, the angle between the lines joining the eye with the two points,) and also the angle at which an object appears elevated above, or depressed below, the horizontal plane through the eye of the observer. And it will be remembered that, in order to determine any 102 PLANE TRIGONOMETRY. triangle, wt must have three of its parts given, one of them being a side. As a general rule then, in order to determine any particular side or angle which may be required in a Problem, we must first consider, if it be not a part of some triangle, for which we have already the three necessary data; and, if not, we must seek to determine, by means of the given data, three such parts for some triangle, to which the line or angle in question belongs. Ex. 1. To find the height of a visible object, whose foot is accessible. This problem we have solved in some simple cases in (66): B more generally, having measured a horizontal base, AC, and observed the angle of elevation BAC, then, in the right-angled triangle BAC, we have the two A c parts AC and A, and the height BC=- AC tanA. Ex. 2. To find the height of a visible object, whose foot is inaccessible, and its distance from the place of observation. From A, the place of observation, measure a base AD in any convenient direction; and at each end of the base observe the angles which the base makes with a line drawn to the object, that is, observe the A -- -C angles BAD, BDA. Then, in the triangle BAD, we have the three ID necessary data, from which the distance AB may be determined; and, by observing also the angle of elevation, BAC, we have the height BC= AB sinA. Ex. 3. To find the distance between any two visible, but inaccessible, objects, and the distance of either from the place of observation. Let B, C, be the two objects, A the place of observation; B measure a base AD in any direction, but so that the two points, B, C, may both be visible from D; and observe the angles BAC, BAD, CAD, BDA, CDA. A (c Then, in the triangle ABD, we have given AD, and angles BAD, BDA; hence we may determine D the side AB: in the triangle ACD, we have given AD, and angles CAD, CDA; hence we may determine the side PLANE TRIGONOMETRY. 103 AC G and now, in the triangle BAC, we have given AB, AC, and angle BAC; hence we may determine the side BC. If BC be supposed to be vertical, we may in this way determine the height BC, when the line AC is not horizontal. 130. As the bearings by the Mariner's Compass are Q a ^ + often referred to in Trigonometri-,4~g^ ^'. cal Problems, it is well that the. /\\\\'7///>x,;i Student should make himself acw. n V//~vzs quainted with the lines of the twvrsy -- F / s Compass, and the names given to -!,^///1/\~? the different Points. It will be ~s e s*, seen that there are in all 32 Points, ~ " ~ X the angle between two adjacent points being 36o = 110. Sometimes, however, the bearings of objects are referred to in Problems without the use of the Points, as so many degrees from the North (or South) towards the East (or West). Thus, if to a spectator P one object A appears to the NE and another B to the SSW, the angle between the lines drawn from his position to the two objects, that is, the angle APB would be 10 points = 1122~; and A may be said to have a bearing N 45~ E, B a bearing N 157~~ E, or, better, S 22~~ W, reckoning from the nearest of the two points, N and S. N.B. The logarithms required in the following Problems, (besides those which may be given in any of them,) will be found in Art. 137, or may be obtained from those there printed. The following square roots are also given for reference: /2 = 1.4142..., 13 = 1.7320..., V5 = 2.2360..., /6 = 2.4494.. Ex. 26. 750 to C 1. Calculate the;reaa of a field from the 420 540 125 280 annexed extract from a field-book, from A 2. A river AC, whose breadth is 200 feet, runs at the foot of a tower BC, which subtends an angle BAC of 25~ 10' at the edge of the bank. Required the height of the tower, having given L tan 25~ 10' = 9.6719628, log 9.397 =.9729928, 104 PLANE TRIGONOMETRY. 3. Two observers, 1000 yards apart, in the same vertical plane with a balloon, but on opposite sides of it, take its angles of elevation at the same moment, 36~ and 54~: determine its height. 4. A person observes the elevation of a tower to be 60~, and, on retiring from it 100 yards further, he finds the elevation to be 30~ determine the height of the tower. 1045 to B 5. Find the area of a four-sided field from 435 710 the annexed extract. 147 225 from A 6. Two men are surveying, and, when each is at a distance of 200 yards from the flag-staff, one of them finds the angle between it and the other's position to be 36~. How far are they apart? 7. From the top of a rock, 500 feet above the level of the sea, the angle of depression of a ship's bottom is observed to be 18~: find the ship's distance from the foot of the rock, having given L tan 18~ =9.5117760, log 1.5388 =.1871940. (See Art. 155.) 8. At the top of a tower, 108 feet high, the angles of depression of the top and bottom of an upright column, standing on the same horizontal plane with the tower, were found to be 30~ and 60~: find its height. 9. In a right-angled triangular field, given one side to be 75 yards, and the perpendicular from the right angle upon the hypothenuse to be 60 yards, determine the other sides of the triangle and the area of the field. 10. To determine the distance of a ship at anchor at C, 1 measured a straight li:le AB of 1000 yards along the shore, and observed the angles CAB =32~ 10', CBA=83~ 10'. Find the ship's distance from A, having given L cos 6~ 50' = 9.9969040, L cos 25~ 20'= 9.9560886, log 1.0985 =.0408154. 11. Determine the sides and area of a right-angled field, when one angle is 60~, and the line joining the right angle with the middle point of the opposite side is 100 yards. 12. At the top of a pillar, 100 feet high, the elevation of the summit of a tower is observed to be 30~, and at the foot of the pillar it is 60~. Find the height of the tower. 13. A is a point inaccessible to an observer at B: shew that he may find the distance AB by measuring a base BC (300 yards), and observing the angles BCA (30~) and BAC (54~). 14. The elevation of al object, standing upon a horizontal PLANE TRIGONOMETRY. 105 plane, is observed, and, 80 feet nearer, the elevation is found to be the complement of the former, and now, on retiring 30 feet, it is double of its original value. Find the height of the object. 15. A light-house was observed from a ship to bear N 45~ E, and, after sailing due South for 6 miles, it bore N 30~ E. Find its distance from the ship at each observation. 16. Find the area of a four-sided field, two of whose sides are parallel, their lengths 230 and 480 yards, and their distance 200 yards. 17. The length of a road, in which the ascent is 1 foot in 5, is 1 mile, from the foot of the hill to the top. What will be the length of a road up the same hill, in which the ascent shall be 1 foot in 12? 18. Find the area of a quadrilateral, the diagonal being 108k feet, and the perpendiculars on it 564 feet and 603 feet, respectively. 19. A person, standing on a river's bank, observes the angle of elevation (a) of a tower on the opposite bank: going backwards a feet, he then finds the elevation to be la. Determine the height of the tower, and the breadth of the stream~ 20. An object B is invisible from A: but a base AC (80 feet) is measured, and the angle ACB (144~) observed; and again, at D in the same line, 40 feet further on, the angle ADB is found to be the supplement of ACB. Find the distance AB. 21. From the top of a hill I observe two objects, in a straight horizontal line, directly before me, and find their angles of depression to be 45~ and 30~ respectively. I know that the objects are T of a mile apart. What shall I find to be the height of the hill? 22. Two objects, P and Q, were observed from a ship to be at the same instant in a line bearing N 15~ E. After sailing NW for 5 miles, P bore due E, and Q bore NE. Find the distance between P and Q. 23. B is a point which cannot be seen from A: but a surveyor measures a line AC (300 yards) and, going on still in the same direction, he measures CD (100 yards), and observes the angles ACB (54~) and ADB (36~). Determine AB. 24. The elevation of a tower, 100 feet high, when due N of an observer, was 60~: what will it be after he has walked due E 400 feet, given nat. tan 13~ 53'=. 2471663, nat. tan 13~ 54'=.2474750? ( 106 ) CHAPTER VIII. ON LOGARITHMS, AND THE EXPONENTIAL THEOREM. 131. DEF. The logarithm of a number to a given base is the index of that power to which the base must be raised, in order to become equal to the number: so that, if a= any number N, then x is called the logarithm of N to the base a, which is denoted thus, x=log^N, or (if there be no occasion to mention the base) x = log V. Thus, since (Alg. Part I. 106) 10~=1, 101=10, 102= 1O, &c.,..logo 1 = 0, log,0 10 =1, log,0100= 2, &c.; and so likewise, since a~=l, a1-=a, we have log, 1 = 0, loga == 1, whatever a may be. CoR. Hence, if ax=n, then x=logn, and a'~g, =n. 132. By taking any positive number (except unity) for base, we may express any positive number as some power of it. Thus, take a = 10, as above, and let V= 2; then, since 10~= 1 and 101 = 10, there is some value of x, if we could find it, between 0 and 1, such that 10'= 2: and, in point of fact, it may be shewn that this value is (to five places of decimals).30103, so that log,02 =.30103. It would of course be possible, though tedious, to verify 30103 this statement by expanding 10'30103 or (1 + 9) Yxovo to a sufficient number of terms by the Binomial Theorem: but we shall see below that such would certainly be the case. 133. If we give to N the successive values 1, 2, 3, &c., and register the corresponding values of x, (which may be found by methods to be given hereafter,) the table thus formed is called a 'Table of Logarithms to the base 10'. The integral part of any logarithm is called the characteristic, the decimal part is called the mantissa, or handful, as it were, thrown in over and above the characteristic. Of course when there is no integral part, the characteristic is 0. PLANE TRIGONOMETRY. 107 134. If the base a be>l, then since a- =0, a~=l, F - oo, we haveao log0= - o, log =, loo = oo; and thus the log. of any number between 0 and 1 will lie between — oo and 0, and the log. of any number between 1 and oo will lie between 0 and + oo; that is, the log. of a number will be positive or negative, according as the number itself is > or < 1. Of course, the contrary will be the case if we suppose a to be < 1. 135. Although, in reasoning generally about Logarithms, we may consider the base to be any positive number other than unity, yet in actual practice we shall have only to deal with (i) Logarithms to the base 10, which are called Common Logarithms, and (ii) Logarithms to the base e, where e denotes a certain number 2.7182818, (of which more hereafter,) which are called Napierian Logarithms, from the name of Lord Napier of Merchiston in Scotland, who first invented Logarithms. 136. It will be seen below that series may be found, by means of which logarithms to the base e may be readily calculated. Among these will be logel0= 2.3025850928; and we shall now shew that, if the logarithm of any number to base e be multiplied by the factor 1 - logl = 1 -- 2.302 &c. =.4342944, it will become the corresponding logarithm to the base 10. For let x, y, be the logarithms of any number Nto any two bases a and b, that is, let x = logaN, y = logbN, and.,.a'"=N=bY: then, since (131 Cor.) b=alo~', we have a = (alogab)y = alog b*, and.. x = logzb.y, or logaN= logb.logjN 108 PLANE TRIGONOMETRY. Hence, of course, it follows that logN= log lO.loglN, or logN 10 xlogN. log, 10 This constant multiplier, by which the two systems of logarithms are connected, is called the Modulus of the system to base 10, and will be denoted in future by M. COR. LogaN= logab.logbN= logb.logc.log~CN=- &c., and so on, through any number of bases. 137. We will suppose then that, by means of the Napierian logarithms, we have formed a Table of Common logarithms, and from these we will extract for our present use the following. Logarithms, to base 10, of all Prime Numbersfrom 1 to 100. No. Logs. No. Logs. No. Logs. No. Logs. 2 0.3010300 19 1.2787536 43 1.6334685 71 1.8512583 3 0.4771213 23 1.3617278 47 1.6720979 73 1.8633229 7 0.8450980 29 1.4623980 53 1.7242759 79 1.8976271 11 1.0413927 31 1.4913617 59 1.7708520 83 1.9190781 13 1.1139434 37 1.5682017 61 1.7853298 89 1.9493900 17 1.2304489 41 1.6127839 67 1.8260748 97 1.9867717 Also log 5=.6989700: see Art. 138. Ex.. 138. Now the following are the properties of logarithms, which make them of singular value in diminishing the labour of Arithmetical calculations. (i) Log (mn) = log m +log n, or the logarithm of any product= the sum of the logarithms of the factors. For, if m = a', and n = av, then mn = a+Y, and o n)=. y=log()=x+y= loglogn. Hence also log(mnp)= log(mn) + logp = logm + logn + logp, &c. (ii) Log = logm —logn, or the logarithm of any PLANE TRIGONOMETRY. 109 quotient = the remainder obtained by subtracting the logarithm of the divisor from that of the dividend. m For m For -=- = ax-S and.. log( )=x- y logm-log n. Hence log ~ = log 1 - logm = - log m, since log 1 = 0. (iii) Logm" -n logm, or the logarithm of any power of a number is obtained by multiplying the logarithm of the number by the index of the power. For mn"=(aq)n==anx, and.. log (m") = nx= n logtrc. As the index in (iii) may be either integral or fractional, we see that, by means of the above results, all Arithmetical operations of Multiplication, Division, Involution, and Evolution, may be converted into Addition and Subtraction of Logarithms. Ex. 1. log6=log2+log3=.7781513; - log 5=log 10-log2=1-log2=.6989700. Ex. 2. log 00-loglO 102 log 10=2, log 1000=3, &c. Ex. 3. log 4=-og 22=2lo g2-.6020600; log 18=log 2 +log 9=log2+2 log 3=1.2552726. Ex. 4. log.07=log —1T=log7-log100=.8450980-2, which is written thus, 2.8450980, it being understood that, in this position of the negative sign, it belongs only to the characteristic 2, and not to the mantissa, which is still positive. Ex. 5. log 2.4=log,=-log 3 +log8-1l=.4771213 -.9030900 —1 =.3802113; log.0023=-log- rL = 1.3617278 —4=.3617278-3=3.3617278. Ex. 6. log -=.4771213 —1.9867717 —1.5096504, which may be written thus, -2 + (1-.5096504)=2.4903496. 139. It will be seen from (138 Ex. 6) how we may readily convert a logarithm which is wholly negative into one that shall be negative only in its characteristic, viz. by increasing the given negative characteristic by unity, and subtracti7 gfrom unity the given mantissa. PLANE TRIGONOMETRY. Thus, generally, if -( C+ m) represent a logarithm in which the characteristic (C) and mantissa (m) are both negative, then -(C+ m)= - C-m= -(1+ C) (1- ), where the mantissa is now positive. The decimal thus obtained, by subtracting another from unity, is called its Arithmetical Complement, and is most readily written down in practice by subtracting its last figure from 10 and the others from 9. A logarithm thus modified we may call a Complementary Logarithm, and denote by colog. By the use of Complementary Logarithms the Subtraction of a logarithm may be turned into Addition. Thus log =-log 3+colog97=.4771213 +2.0132283=2.4903496, as before. But it is necessary to notice some peculiarities which occur in the use of such logarithms, whose characteristics only are negative, and of which instances are given in the following examples. Ex. 1. log:-+log -=log2+colog 17+log 3+colog19 =0.3010300 +2.7695511 Here there was an integer -1 2, arising from the +0.4771213 sum of the positive mantisse, which, combined with +2.7212464 the sum of the characteristics-4, leaves-2 or 2. 2.2689488 Ex. 2. log (T)3=3 log 1=3 colog 13=2.8860566 3 4.6581698 Here there was an integer + 2, arising from the product of the positive mantissa by 3, which, combined with 3X2=-6, leaves-4, or 4. Ex. 3. log V/-r=-+ colog 71=- (2.1487417)=-.7355345. Here it was necessary to imagine the logarithm thrown into the equivalent form + (7+5.1487417), so that the negative characteristic may become a multiple of the divisor 7. PLANE TRIGONOMETRY. 111 Ex. 27. Obtain the logarithms of 1. 8,9,1.2. 2. 20,25,60. 3.,, - 4..03, 1,.0033. 5. 1.8, 140, 1.44. 6..0625, -1- 1.05 7. 10.6, 45, 42. 8..0111, V/1, l 4Y. 9. V.1, 4.02, (1.2)4 -10. (3)A, i/1.1, (.069). 1. X3/ / X^,i'. 12. /{/(J 4/_a(V)}. 140. There are, however, two observations to be made, which greatly facilitate the finding of Common logarithms. (i) In the Common system having given log N, we can find immediately log (Nx 10") or log(Nl- 10") = log (Nx 10-"), n being an integer, that is, we can find the logarithm of any number, which differs from N only in the position of the decimal point. For log (Nx 10-)= logN + n log 10 = log N~ n, and consequently, (since n is an integer,) will differ from logN only in the characteristic, which will be increased or diminished by n, while both logarithms will have the same mantissa. Thus, suppose that we have given log 1362=3.1341771: then log 136200=log (1362 X 10)=log 1362+2=5.1341771, log 1.362=1og (1362 1]03) =log1362-3=.1341771, log.001362=log (1362- 106)=log 1362 -6=3.1341771. (ii) In the Common system, the characteristic of the logarithm of any number may be written down at once by inspection. For, if the number lie between 1 and 10, that is, between 10~ and 101, its logarithm must lie between 0 and 1, and, therefore, its characteristic will be 0; so, if it lie between 10 and 100, that is, between 101 and 10", its characteristic will be 1; if it lie between 100 and 112 PLANE TRIGONOMETRY 1000, that is, between 102 and 103, its chalact ristic will be 2; and generally, if it have n digits, that is, if it lie between 10"' and 10, its logarithm will lie between n- 1 and n, and its characteristic will be n-1. Again, if the number lie between 1 and.1, that is, between 1 and -l = 10-, its logarithm must lie between 0 and -1, and, therefore, (with positive mantissa) its characteristic will be 1; so, if it lie between.1 and.01, that is, between 10-' and 10-', the characteristic will be 2; if it lie between.01 and.001, that is, between 10-2 and 10'3, the characteristic will be 3; and generally, if it have n - 1 cyphers after the point, the characteristic will be n. It will be seen that both cases are comprised in the following Rule: Reckon the distance of the first digit of the number from the units-place: if it be n places to the left, the characteristic will be + n; if n places to the right, n. 141. Hence it appears that, in the Tables of Common Logarithms, it is only necessary to register the mantissue, corresponding to certain sequences of figures, which look like numbers, but are not really so, because the place of the decimal point is not fixed in them. Thus, in the Table given below, we have, opposite to the value 3570 for V, the mantissa 5526682, where 3570 is no number, but merely a sequence of figures, and the Table shews that for all such sequences, wherever we insert in them the decimal point, the mantissa is still 5526682: thus, determining the characteristic in each case by (140 (ii)), we have log3.570 or log 3.57=.5526682, log 35700=4.5526682, log.00357=3.5526682. 142. The mantissa of logarithms are registered in the best Tables to 7 decimal places, corresponding to PLANE TRIGONOMETRY. 113 sequences of 5 figures, as in the lines below, extracted from Hutton's Tables. N. 0 1 2 3 4 5 6 7 8 9 D. Pro. 570 5526682 6804 69257047 7169 7290 7412 75347655 7777 1 s 2 71 7899 8020 8142 82638385 850718628 8750 8871 8993 122 3 7 72 9115 9236 9358 9479 9601 9722 9844 9965 0087 0209 67 731 5530330 045210573 0695 0816 0938 105911181 1302 1424 8 8 Thus, to find the mantissa for the sequence N=35725, we look horizontally along the third column for 3572, and vertically down under the figure 5, and thus find it to be 5529722, including the figures 552, which, being once printed, (as in the first line) at the beginning of the line in which they first occur, are understood to be repeated before each mantissa, until (as in the fourth line) they are replaced by 553. This change, however, actually begins with the last two mantissa of the third line, where the dotted figures are introduced to mark this: thus, the mantissa for 35728 is 5530087. In some books, instead of a dotted figure, a figure with a line above it, as 0, or a smaller figure, is used to mark the point where this change begins, if it happen not to be at the beginning of a line. 143. Although the mantissa are only given in the Tables for sequences of five figures, yet they may be readily found for sequences of six or seven figures by the following considerations. It will be shewn hereafter, that, when the difference of two numbers is small compared with either of them, the difference of their logarithms is very nearly proportional to the difference of the numbers. Now let m, m + D, be the mantissa for two consecutive numbers, Nand 1N+ 1, each of five figures, m+d the mantissa for a number N+ n, lying between them, that is, having the same integral part as N, but one or more figures after the decimal point. Then, since the three numbers have all the same number of integral (I.) I !14 PLANE TRIGONOMETRY. digits, they will have all the same characteristic, and so the difference of their mantissae will be the same as the difference of their logarithms: hence, by the above statement, D: d::1: n, or d =nD, by means of which result we may find d when n is given, or, conversely, n when d is given. Ex. 1. To find 1og35.7235 and log.003572357. Here N=35723, N+1=35724, N+n-35723.5, and.'.n=-; and D=the Difference of the mantissae of 35723 and 35724=-122 hence for 35723.5, d=-5 of 122=5 x12.2=61; and thus, the whole mantissa for 35723.5 being 5529479+61=5529540, we have log 35.7235=1.5529540. So for 35723.57, d= —il of 122=57 X 1.22=69.54=70 nearly; and thus the whole mantissa for 35723.57 being 5529479+70 =5529549, we have log.003572357=3.5529549. Ex. 2. To find the number corresponding to the log2.5528797. Here, referring to the Table on page 119, the next lower mantissa is that for 35717, viz. 5528750, and.'.d=47: hence, since the difference (D) between the mantissae for 35717 and 35718 is 121, we have n=d.-D=-47=.38, (it being useless to go beyond two decimal places, for a reason that will appear hereafter;) and so if the number of five figures, referred to as N in the proof, be 35717, the number N+n will be 35717.38; from which we see that the given mantissa corresponds (141) to the sequence 3571738, and therefore the given logarithm (the characteristic being 2) corresponds to the number.03571738. 144. But the columns headed D. and Pro. (Proportional Parts) are intended to facilitate the calculation of such logarithms, and the converse operation of finding the corresponding number from the given logarithm. The column D. shews that the prevailing difference between two consecutive mantissae in this neighbourhood is 122, as will be seen at once to be the case by looking PLANE TRIGONOMETRY. 115 at them, the difference being sometimes 121, but generally 122: further back in the Tables, it would have been 123, 124, &c., and further on, 121, 120, &c. Now, for each value of D, there is formed what is called a Table of Proportional Parts, by multiplying DO that is, in this case, 12.2, by 1, 2, 3, &c., which products give 12.2 = 12 (nearly), 24.4 = 24,36.6 = 37,&c., (as in the extract on page 119), the integer only being retained in each product, but (14 Ex. 2) increased by unity when the rejected decimal part equals or exceeds or.5. The use of this will now be apparent: for since d = nD, let a, 3, &c. be the figures in order of the decimal n; then (ro P D D d= (-+ + &c.)D = a - + D- +&c. 10 100 10 100 Now, in the Table of Pro. Parts, the first column contains the successive values 1, 2, 3, &c. which a or P might have, and the second contains the corresponding D DD values of a -. In order, therefore, to find a -,0 we 10 have only to glance at the Table, and take down the number opposite to the value of a; and, in order to find P -, we have only to take down Ioth of the number, opposite to the value of,3 Ex. 1. To find log 357.2357. Here we have mantissa for 35723 = 5529479 diff. for 5 a= 61 diff. for ' =8.5 = 9 5529549 and, therefore, the logarithm required is 2.5529549. Ex. 2. To find the number corresponding to the log. I.5529/89. Here the mantissa next below the given one being that for 35725, viz. 5529722, we have d= 67, from which taking 61, which 116 rLANE TRIGONOMETRY. we see, by the Table of Proportional Parts, gives the sixth figure 5, we have still remaining 6, which, being? of 60, shews that the seventh figure is 4 (nearly 5); hence the sequence required is 3572554, and the number (the characteristic being 1) is.3572554. Ex. 28. In these Examples the Tables on pp. 114, 119, are to be consulted. I. Find the logs of 35.70925,.3572739,.003571246, 3.572804. 2. Find the numbers whose logarithms are 3.5528742, 1.5530895, 4.5527777,.5530199. 3. Given log 2000.1=3.3010517, construct a Table of Pro. Parts; and find log 20.00094, and the number whose log is 2.3010489. 4. Given log 31.001 = 1.4913757, find the logarithms of 3100023, 310004, and 31 4o~. 5. Find the value of V/21 x / 3~ x t/4 X / 5 given log 4.4985 =.6530677, log 4498.6 = 3.6530774. 6. Find the value of /357.1328 - 35712.75, given log 57.386=- 1.7588060, log.057387 = 2.7588135 145. We shall now explain the method of calculating logarithms to the base e. Exponential Theorem: To expand a' in a series of ascending powers of x. ax= fl+(a-1)}' =+x(ax-1)+ ( - ) (a- 1)2+ x(x -] )(x- 2) (a-1)3 - &a 1~ a-) 1.2 1.2.3 l+x{(a-1) + X — (a 1)2+ ( 1)(x-2)(a-1)3 + &c. 1.2 1.2.3 = 1 x {A + Bx + Cx2 + &c.} suppose, where A, (the sum of those terms within the bracket which are independent of x,) is easily found by putting r=0 within the bracket, when there remains (a-l) + -1 (a-l)2+ (1)(2) a-I)+&c., 1.2 1.2.3 PLANE TRIGONOMETRY. 117 or a-i-I(a- 1)2+ -(a )3- c which is therefore the value of A. We have then a- 1 + Ax + +B2 + Cx3 + &c., where A, B, C, &c. are functions of a, altogether independent of x, and will therefore remain the same for the same value of a, however we change that of x. Ience ax+a = 1 +A(x + ) + B(x + h)2 + C(x +h)3 + &c.. but ax = ax.a = { 1 - Ax - Bx- CX3 4- &c.} a; therefore, equating coefficients of x in these identical values of a"+h, we have + 2Bh+3Ch2 +&c.= ah=A {l +Ah + Bh2+ Ch3+ &c.) hence, equating coefficients of different powers of h in the above, we get A2 AB A3 2B= A2, 3C=AB,&c., orB= ' C= - 1-3' &c.; A2x2 A3'3 so that we find a= 1+ Ax +. + 1.-2.3 + &c., where A=(a-1) — (a-1)2+ (a-1)3 —&c. 146. Since the above result is true for all values of x, take x such that Ax= 1; then 1 1 1 1 x= and a= 1 +.+ &c.2.7182818 &c., 1 which number it is usual to denote by e; hence aA = e, or a e A, and therefore A = logea, and 2x2 X3 a = ] + (logea) x + (loga) )2 + (logea)3 1.2.3 - c.: whence also, writing e for a, we get, since(131)loge= 1, x2 x- ~ e=l + x+ + 1 2.3 +&c. 147. We have n _ (n —l) x2 n-__ x2 ( + x)-=l+nX + + &c.=+x —l + &C. n 1.2 n - n 1.2 118 PLANE TRIGONOMETRY. Now, as n increases, it is plain that I decreases, and tends to zero as its Limit: hence we see that, as n increases, the Limit of (1 -)is 1++x+ - -+&c. ore'. It 1.2 1.2.3 148. By (146) we have (e'-l)"=(x+-x2+&(e&.)"=-r+&c. (i): butwe have also (e'-l)"=e'-ne(-"l)'-n(n —1) e("-) —&c. (ii), in 1.2 which each of the quantities e, e'(n-', &c. may be expanded by (146), so that the whole coefficient of x' in (ii) will be n' n (n-l)'+n(n —1) (n-2)'&c.: 1.2...r 1.2...r 1.2 1.2...r but in (i) the coeff. of x' is zero, if r< n, and it is 1, ifr==n; hence, since the two expressions for (e'-1)" must be identical, n' —n(n —l)'+ n (n2) (n-2)'-&c.=O, if r<n, 1.2 and n -n(n-1"+ n(n —1) (n-2)"-&c.= 1.2.3...n. 1.2 149. In the value of log,a =A(a- 1)- (a- 1)2+ &c., write 1 + x for a, and.. x for a-1; then log.(l + x)=-x-2-X2 + x3 -1x4 + &c. Hence we might proceed to find the logarithms of numbers; thus I 1 1 1 1 1 log,2 = 1 - + — + c. =..2 A + 5+.6 + &c.: but the series thus obtained are not sufficiently convergent, as it would take very many terms to obtain the logarithm accurately to six or seven decimal places. 150. A more convergent series, however, may be obtained, as follows, from the above. Since log,(l + x) = x- x2 + ~x3- x4 + &c., ~* log,(1-X)=-x-~X, - - 3-4-&c., by writing -x for x: and hence 1 + x log,(l + x)-log,(1 - x) or log,Tr- = 2 {xr+ + Y+ &C.. "/ "~s\` "/"'"bC1 -r2: 3'~ k~) PLANE TRIGONOMETRY. 119 In this series write m — for x, and.. _ for - m+-n n 1-x lm (m-n 1 fm-n 3 1{m-n, )+c *\loge -=2<J —I -- +^ ( — + &c....()a.n m+n 3\mn/ 5-m+ nIf n=1, then logm=2 {m — +l (f-1)3+ f(m- 1)5+ &c.( 5e (m+1 3 (m + 5&m+ l. (1) a convergent series, from which the logarithms of low primes may be found. Thuslog,2=2 i+. 13+1 ~ 5 1+&c. =.6931471806, 3 3 33 5 35 5 as may be seen by taking nine terms of the series. 151. Again, if in (a) we put m=n+ 1, we get lon+1 1 1 1 log. =log(n+l) —2log=2 + 3(2 1 &c. ( n m2n+ 13(2n+ 1)3 a very convergent series, by means of which, having given the logarithm of one of two consecutive numbers, n and n+ 1, we may find that of the other. Thus log,9- log, 8 =2 log3 - 3 log,2 = 2 1 7+ - + &c.; and by means of four terms of this series, having given log, 2 above, we may find log,3= 1.0986122884. Again log, -log,4= log,5 -2 loge2-=2 19 +39 + &c., by taking five terms of which series we get log, 5 = 1.6094379122, and, adding log, 2 to this, we have log, 10 = 2.3025850928, and thence, by common division, the modulus M -- 1 —log,10-.4342944819. 152. Once more in (y) write xa for n+ 1, and, therefore, 2x- 1 for 2n + 1; then log - = 2 log,x - log.(x+ 1)- log( -1) =22 2 1 +3(22 1) &cil,...(a) 120 PLANE TRIGONOMETRY. a very rapidly converging series, by means of which, having given the logarithms of any two of three consecutive numbers, n + 1, n, n-l, we may find that of the other. Thus, taking the numbers 5, 6, 7, of which the logarithms of 5 and 6 (= 2 X 3) are now known, we have 1 1 2 log,6-log,7-log,5 = 2 { + (1+&c.}, 71 3 (71)' whence we may find loge7, &c. Or the same may be found by (y) as follows: log, 50-log, 49, or (log,2+ 2 log, 5)-2 log, 7=2{ 1+ &c.}. 153. By means of the above (or other similar formulae) a table of Napierian logarithms may be formed; and then these may be converted into logarithms to the base 10 or to any other base, by multiplying each by the proper modulus. Thus logo, 2=M log, 2=.4342944819 X.6931471806=.3010300; and so we might find log32 =(1 -loge3) xloge 2=loge2 — loge3,&c. Or, having once obtained logel0 and, by means of it, the modulus, we may write at once in (a) lom Mlog m -2 — - 3 log10 _ Mlog-=2Mm + (+ &c;} n n m+n m +na and so we get, corresponding to the formula (3),(7), (s), loglom=2M{+ + &c.}, log10(n +1 )-logon=2{2n - + &c.) and 2glogx-log,(x+ 1)-log1,(x-1)=2M {2 1+ &c.+: and thus we may calculate immediately the common logarithms, without finding the Napierian. 154. We are now able to prove the statement made in (143), viz. that, when the diff. of two numbers is small compared with either of them, the diff. of their logs is very nearly proportional to that of the numbers. PLANE TRIGONOMETRY. 12] 1 _ N+ nn)log(1N For d-lloglo(N+n)-logoN=lg Mlog (l + n 1 n2 M =by (149) M{ -- + &c. = n nearly when n is small compared with N: that is, d c n, or the increase of the logarithm is proportional to the increase of the number. Also D=logo,, = M log, (1 +~ )= nearly; now M=.43 &c. and is, therefore, <, while N, if a number of five places, is > 10000; hence we have D < o<.00005 in every case, and.D ---would at 0D~~,,~~,,~ 1000 the utmost only have its first significant figure in the D eighth place of decimals, and so the term 10 in (144) could only, if at all, affect the seventh or last figure of the mantissa, but would not generally affect it at all: thus to 7 places of decimals the logs of 35714.23 and 35714.232198 &c. would probably coincide, the difference appearing in the 8th and following places. Hence with Tables, which give mantissa only to 7 figures, we can only expect to find the numbers, which correspond to given logarithms, correct in their first seven figures. 155. As the sines and cosines of all angles, and the tangents of angles less than 45~, are all less than 1, their logarithms would be negative. To avoid the awkwardness of printing these, the logs of all the Ratios are increased by 10, and the true log of any Ratio is, therefore, to be found from the Tabular Log by diminishing it by 10: so that, using Log sinA, Log-cosA, &c., or L sinA, L cosA, &c., for the Tabular logarithms, we 122 PLANE TRIGONOMETRY. have log-sin A= L sinA -10, &c. Here likewise, as in (102), the same column, that serves, with a descending column of minutes, for L sinA, will serve, with an ascending column of minutes, for L cos(90~- A), &c.; so that the Tables of Log-functions need not be carried beyond 45~. Again, since sinA= -- 1 we have cosecA' log-sinA = log, - log-cosecA = -log-cosecA; whence (L sin A-10) = -(L cosecA -10), or L sinA = 20- L cosecA, L cosecA = 20- L sinA. And similarly for the other pairs of reciprocal functions. 156. We shall now shew that the principle of (104) applies also to the logarithmic, as well as to the natural functions, of angles, namely, that for a small increment of the angle the increment of the log-sine, &c. is proportional to that of the angle, with the same cases of exception as before. For let f denote the value of any of the natural functions of 0; then, by (104), f+ k will be the approximate value of the same function of 0+, where k is some constant quantity depending only upon 0, not upon, [thus for the sine, k is cosO, for the cosine, k is -sinO, &c., as appears in (104),] a being supposed very small, and 0 itself not in the excepted cases. Now log (f+ kh)-logf= log- f = log(l+ + ) f f k lk2 2 1 =M( ---- + &c.) = M8, nearly, where M= l10 f 2ft f log.10. So that, whatever be the function in question, we see that the increment of the log-function, and, therefore, also of the Log-function, of any angle is proportional to the small increment of the angle, provided that the PLANE TRIGONOMETRY. 123 angle itself be not in one of the excepted cases, namely, near 90~ for the sine, tangent, and secant, (that is, for those functions which increase from 0~ to 90~,) and near 0~ or 180~ for the cosine, cotangent, and cosecant, (that is, for those functions which decrease from 0~ to 90~). Hence we may apply precisely the same method as in (103) to find, approximately, the Log-function of any angle not exactly given in the Tables, or, conversely, to find the angle from its Log-function. 157. Moreover, since L cosec(A + n") - LcosecA = log-cosec(A + n") - log-cosecA = -log-sin(A + n") + log-sinA = - {Lsin(A + n") - LsinA', it appears that the differences for the Log-cosecants will be the same as for the Log-sines, being increments for the one and decrements for the other. Accordingly, in any set of Tables, the successive values of D for the sine and cosecant, for the intervals at which those Tables are constructed, will be found registered in the same vertical column. Similarly for the cosine and secant, tangent and cotangent. 158. If in (104 (i)), we put S for sine, and ~* for tan V;, we get sin (0+-)-sin =j cosO (1 — tan 8)=- cos 8o- 2 sin 8, sorin ( =1 + cot 0 —2), and Lsin (0+ ) -Lsin C=log-sin (0+a) —og-sinO sin (0+a).. =log sinL =log1 —+(cot -- 2)} = M1 (( cot e_ — a2) —(i cot 0-~62)2+&c.} by (149) =Mcot — iMS2(l +cot'2), if we include 6 and 62. Hence, though 6 may be small, yet if cot0 be very large, or 0 near 0~, we cannot omit the term involving 62, that is, we cannot assume the principle that the increment of the Log-sine is proportional to that of the angle: and the same may be shewn to 124 PLANE TRIGONOMETRY. be true for the tangent. N1ow it very frequently happens in practice that a very small angle has to be determined from its sine or tangent. Accordingly, in the best sets of Tables, a table is given of Log-sines and Log-tangents for every second in the first two degrees. 159. If we have to determine accurately an angle, which is given by some one of its functions in one of the excepted cases, we must replace the given equation by some other not liable to the same objection. Thus, if we have given sinA=a, where a=l nearly, or A is near 90~, we may write fdn(450~-A)= V/[ { 1-cos (90~-A) } ]= {(1-sinA) }= / (l-a) where the angle 45~ —A will be near 0~, and can be computed accurately from its sine. So if we have given cosA=a, where A=1 nearly, or A is small, we may write siniA=-/{i-(l-cosA)}-=V{l(-a)}, where 'A is near 0~: or, if tanA=-a, where a is very great, or A is near 90~, we may write tanA-1 a-1 tan(A-45~)= tanA+ 1 a+ 1' where A-45~ is near 45~. Ex. 29. 1. Given L sin 11 24'=9.2959129, and L sin 11~ 25'=9.2965390; find L sin 11~24' 25", and Log-sin-' 9.2959875. 2. Given L cos60~ 1'=9.6987511, find L cos60~ 0' 20" and L sin29~ 59' 15" 3. Given L tan 44~ 59'=9.9997473, find L cot44~ 59' 20" and L cot45~ 0' 35". 4. Given L cosec30~ 1'=10.3008113, findL sin 30~ 0' 45" and Log-sin-19.6990999. 5. Given L cos30~ 1'=9.9374577, find Lcos 30~ 0' 20", L sin 59~ 59' 25", and L sec 30~ 0' 25" 6. Given L cot44~ 59'= 10.0002527, find L tan 44~ 59' 20", L tan 45~ 0' 30", and Log-cot"'10.0001234. ~IIC~~L'CIIII V in ( 125 ) QUESTIONS WITH SOLUTIONS. *** These worked Examples are designed as a general introduction to the Miscellaneous Examples of Parts I. and II. Ex. 1. Eliminate 0 and 0 from the equations sinO+sinp=a, cos0+cosp=b, cos(0-p)=-c. 1st squared, sin20+8in2p -2 sinesin=a2, 2nd squared, cos2 + cos2 + 2 cos0 cosp= b2; By addition, 1 + 1 +2(sin9 sin0+-cos9 cosp)=a2+b2, or, 2+2 cos(0 —0)=a2+b2;..2+2c=a2+b2, or, a2+b2-2c=2. Ex. 2. The hypotenuse of a given right-angled triangle is horizontal, and the plane of the triangle is inclined at a given angle 3 to the horizon. Shew that the sines of the inclinations, to the horizon, of the sides CA, CB, which include the right angle, are sinCAB sin3, and sin CBA sin$, respectively. Ac> ~ Here, the angle CDE=3; CDsin3= CE; CAsinCAB=CD; '.. CA=CD +/ sinCAB; CAsin CAE = CE;.. sin CAE=CE - CA=CD sin/3 x si CAB CD =sinCAB sin3; similarly, sinCBE will be found=sinCBA sink3. A Ex. 3. Two rocks, A, B, at sea, are at a given distance, d, from each other, and from their summits, which are in a horizontal line, the dips of the horizon are observed to be a and 3, respectively. Shew that the Earth's radius is d cos-' (seca cos,3) n AF, BG, the perpendicular altitudes of r.......m the rocks; AB=d; EGm, DFn the angles of depression, a and /3 respectively;.'. also angle ECG=a, and angle DCF=3. CG — CE=seca; CD- CF=cos,3;. as CD= CE, we have CG -- CF=secacos3 C f =cosine of arc AB to radius unity;.'. arc AB to radius unity=cos-'(seca cos/); and the actual radius=d — cos-'(seca cos/3). 126 QUESTIONS WITI SOLUTIONS. Ex. a. On a horizontal plane I observed the angles of elevation, 190 20' and 21~, of two towers in front of me, the latter being partly hidden by the former. After walking 75 feet in a direct line towards them, the remoter tower became entirely hidden, and 100 feet farther on, the elevation of the lower tower was observed to be 50~. Shew that the heights of the towers are 87 and 121+ feet, nearly. G Here, DE, FG, the towers; Xi. angle GAF=21~, EAD =1220', ECD=50~, AB=75 feet, AC= 175 feet; AEC-ECD-EAD -30~ 40'. A B i sinAECA AC sinA EsinEC; and E CE siECD;.ED ACsinCAEsinECD; which is a form adapted to sinAEC logarithmic computation; but the present problem may be worked conveniently enough without logarithms, as follows: AD=EDcotEAD; CD=EDcotECD;.. AE-CD=175 =ED(cotl9~ 20'-cot500); hence ED= 175 — (2.8502-.8391)=87 feet, nearly. Again; AF= FG cotGAF= 75 + FG cotGB F=75 + FG X BD 75 BD BD BE. *%+ cotGAF; but BD=DEcotEAD-75;. DIE' FG DE DE _ _ _ _ '75 1 cotEAD — 75.5 cotEAD — =cotGAF; or F ~DE G DE FG 1 cot EAD-cotGAF DE 75 whence FG= 7 xD 75-DE(cotl90 20'-cot21~) =6526- {75 —87(2.8502-2.6051)} =121 ft, nearly. Ex. 5. A boy flying a kite at noon, when the wind was blowing a~ from the south, and the angular distance of the kite's shadow from the north was /3, the wind suddenly changed to al~ from the south, and the shadow to /31~ from the north, and the kite was raised as much above 45~ as it had before been below that elevation. Shew that, 6~ being the angular elevation of the sun, and 450- ~ that of the kite at first, QUESTIONS WITH SOLUTIONS. 127 ta2_0- sin3 sin,31 tan s~=. %Osin (_ sin(a-3) sin(a1-s)' tan2(45_0-0) — sin(a-/3) sin3l sin(cl —( 31) sin/ N Let B denote the position of the boy, K the kite, N the kite's shadow. BKO and NKO are A\ I\ triangles in vertical planes, the direction of the the latter being north and south, V, the shadow, being in the north at noon, while the direction of BKO is at an angle ABO, =a~, from the south. ABN=BNO= f~; NBO=a0-/; ONK=0, u the angle of the sun's elevation; OBK=45~ -0, the angle of the kite's elevation; KO, the perpendicular altitude of the kite, making KON and KOB right angles. BK is the r \ string. B OK=BK sin OBK BKsin (45 ~-0-), OB=BKcosOBK=BKcos(45~ —0); ON= OKtanNKO= O K cotb~= BK si (45-,) cot0~; also, ON= OB sin(( =)BKcos(45o-_-) sin(a-o0) sin30s. * sin(450-oO)cot0o=cos(45~ o) sin(t —'~) sin/3o sin(45~ —) 1 _sin(a~ —~) ocos(45~ —09) tan0 sinO or tan (450-~) =s i ) tano. (i) sinoa Similarly, for the second condition, tm (450 + ) =sin(1 -/)tan0o. (ii) But tan(450~+~)=cot(450~ —)=-tan(40 o) hence, multiplying (i) by (ii), 1sin (a —3) sin (a —1) tan2O, sinp sinp/ or tano20 n sini3 sin/3 sin(a ---) sin(a -/31) hence, also, dividing (i) by (ii), tan2 (45_ 0o) sin (a —3) sin31..-An (ax - -) sinS 128 QUESTIONS WITH SOLUTIONS. Ex. 6. Eliminate a from the equations x tan (a-/)=-y tan (a + ),. (i) (x + y) cos 2a + (x-y) cos 2 =z. (ii) From (i) -x tan ( + ) _ sin 2a + sin 2/3 y tan (a-/3) sin 2a-sin 2/3. x+y_sin 2a y _sin 2; or, sin 2a = sin 2/3; **x-y sm2si3 X-Y U —ysin 2c = 1- cos2 2. x — sin2 2,d;. cos 2a=- (zY) sin 2i -t_ 1/ ( + y)2 cos2 23-4xy; +Y,-{( + Y( y)22 c2 2 -4xy + (x-y) cos 2/ x-.y MISCELLANEOUS EXAMPLES: PART I. '* The following examples are of a most simple character, and have been constructed expressly to illustrate the text of Part I. At the end of Part II will be found a large collection of more general miscellaneous Examples, arranged progressively in order of difficulty. 1. If one angle of a triangle be three-fourths of another, and be half as large again as the third, determine the three angles. 2. One angle of a triangle is greater than another by 30~, and less than the third by 30q: express them all in degrees. 3. The base-angle of an isosceles triangle is three-fourths of the vertical: find the three angles. 4. ABCDE is a regular pentagon: join BD, CE, cutting in F; and shew that BAEF is a parallelogram, whose angles are in the ratio of 2: 3. 5. What will be the length of a railway-curve, which subtends an angle of 45~ to a radius of a mile? 6. One angle of a triangle is 45~, and another is 45g: find the third both in degrees and grades. 7. Find the measures of two arcs, of rrfeet and ir degrees, to a radius of 56 ft. (7r=-3 I). 8. Determine the angles of a triangle, when the exterior angle of one of them=two-thirds of that of the second=four-fifths of that of the third. 9. Express the complement of a measure in degrees, and the supplement in grades. 10. One angle of a triangle is a fourth as large again as the sum of the two others, and half as large again as their difference: find the largest angle in degrees. 11. The sum of the angles of one regular polygon: the sum of the angles of another:: 7: 8; but each angle of the first: each angle of the second:: 35: 36: find the number of sides in each polygon. 12. The vertical angle of an isosceles triangle is 7r~: find one of the base angles. (I.) E 130 MISCELLANEOUS EXAMPLES. 13. The angles of a quadrilateral are as, 1) 1, 2: express them all in measures. 14. Express in measures, degrees, and grades, the angle subtended by an arc of an inch to a radius of a foot. }5. The exterior angle of a triangle is double of the difference of the two interior and opposite angles, and is also half the sum of their complements. Determine the angles of the triangle. 16. Find the number of sides in a polygon when there are as many degrees in the sum of its angles as there are grades in the sum of the angles of a polygon, which has one side fewer than the former. 17. If the circumference of a circle were divided into 315 degrees (instead of 360), shew that an arc equal to the radius would contain 50~ nearly. (7r=3-) 18. One angle of an isosceles triangle is 2: determine its angles in degrees. 19. Two cylinders of equal bulk have their lengths as: 4; compare their radii. 20. If tanA=1-, find the value of sinA+cosA, and shew that versA=2 coversA. 21. A cocoon of silk was unwound by 378 turns of a reel of 10 inches diameter: what was its length? (r=3-) 22. One angle of a triangle is aT: what must the second be, that the third may be? Express it in measures and degrees. 23. How many cubic yards of earth will be taken out in digging a circular well 42ft deep, with a diameter of 6ft? 2ab 24. Given sine- b, find cos0 and tan9. a2+b2 25. If 1=sin2Acosec2B+cos2Acos2C, shew that sinC=tanAcotB. 26. Three angles of a quadrilateral are 7r~, 7rg, and 7r, respectively; express the fourth in degrees and grades. 27. Prove that the circular measure of 23~ is.401425... 28. If an arc of 2L, to a radius of 22, be formed into the circumference of a circle, what will be the radius of the circle? 29. Shew that (sec- r+tlan 1r) (cosec-7r+tan -r)=5. 30. Determine sinA from the equation 9 sin2A-4 tan2A=-. 31. If m versC —n vers.(7r-0) r+pvers(7r+- )=0, find cos0 MISCELLANEOUS EXAMPLES. 131 32. A is the centre of a circular plot of ground, whose radius AB=50 yards: with what radius AC must a circle be described, so as to be half the size of the given one? 33. How many degrees, &c. would the arc in Ex. 28 make of a circle whose radius is 3j? 34. Express sin 110~, cos 220~, tan 330, cot440~, sec550~, cosec660~, as functions of angles less than 45~. 35. Write the general value of cosec'2. 36. If m secO-tan0=l=n cosec0+cot0, eliminated 37. If tan0 = 1}, obtain the numerical value of (sin61+tan0+sec0) (cos + cot + cosec0). 38. Find the continued product of sin (A-90~) + cos (A-] 80~), cot (A-90~)-tan (A-180~), sec (A-90')- cosec (A-180~). 39. Write down the general value of 0, when (vers0)2=-. 40. If secA+cosecA=m and secA-cosecA=n, shew that tanA=+m; and eliminate A. m-n 41. Shew that vers(7r+ 0)=)covers(7r+6)=suvers(-7r+ 0. 42. Given sin(7r-0) cos( —37r)- sec(,r —O) cosec(p-27r), find sin 0. 43. Write down the general values of x in the equations (i) (3 tan x)2=1. (ii) cos x= —. 44. The angles of a triangle are as 1, 2, 3, and the greatest side exceeds the least by 100 yards: find the area. 45. If sin (A-C) cosB+sin (B-C) cosA=O, then tan A+tan B =2 tan C. 46. Prove that cos60~ l=-cosl't-os59~ 59'. Prove the truth of the following formulae: 47. suvers (60~ +A)-covers (30~+A)=cosA. cot -A sin + si 2A 48. cotA 1 +secA. 49. sin + si2A cotA osA+cosAc2A 50. vers (A+45~)-covers (B+45~)=2 sinI(A+B) sin {45~ + (A-B)} 51. (sin A sin B)2+ (cosA +cosB)2=4 cos2 (A ~ B). 52. (l-cot A)2=2 inA 53. 2 sin3A =cotA+cot2A. 1 l-cosA cosA-cos3A 54. cos(A+ B) cos (A-B)-cos (B+ C) cos(B-C) +cos (A+ C) cos (A- C)=cos A. K2 132 MISCELLANEOUS EXAMPLES. 65. tan(450+-4A)+cot(45~+-A)=2secA 56. tanA 1 tan2A. 1 +tanA 1-cotA 57. vers(A-B) suvers(A +B)=(sin A sinB)2. 58. 2sin(45~+A)sin(450+B)=cos(A-B) +sin(A+B). 59 sin22A-4 sin2A 4 60 cot(60 +A)_2-sec2A sin22A+-4sin2A-4 tan(60~-A) 2+ sec2A' 61. One side of a right-angled triangle is half the hypothenuse, and the perpendicular from the right angle on the hypothenuse is 60 feet: determine the perimeter. 62. Shew that cos47 —cos61~-cos ll+cos250=sin 7~ and sin47~+ sin 61~-sin 11~-sin 25~-=cos7~ 63. Shew that ta1tan'-ltan'-sTr. 64. The Sun's distance from the Earth being 24,000 times the Earth's radius, find (in seconds) the Earth's apparent diameter as seen from the Sun. 65. Assuming the height of the great Pyramid to be 486 feet, how far off may it be seen across the desert? 66. If A, B, C, are in A.P., shew that sin A-sinC=2 sin (A-B) cos B. 67. Shew that, to four places of decimals, tan22~ 30 =.4142. 68. Given sin 29 59'=.4997481, determine sin 29 59' 30", 'and sin-.4997777. 69. In any triangle, right-angled at C, shew that tanA+secA _ b c-b cotA+cosecA -a 'c-a 70. Taking the Sun's mean apparent diameter as 31-', and his distance from the Earth 96 millions of miles, shew that, if his centre were coincident with the Earth's, his body would extend in all directions (nearly) 200,000 miles beyond the Moon, whose distance from the Earth is 240,000 miles. 71. Shew that tan-' (2 + V3) —tan' (2- /3)=sec-' 2. 72. Chimborazo, among the Andes, can just be seen from the surface of the sea at a distance of 177 miles. Determine the height of the mountain. 73. In any triangle, right-angled at C, shew that versA: versB:: a(a+c): b2(b c). 74. Taking the Sun's diameter as 880,000 miles, and the Earth's as 8000, compare the apparent magnitudes of the Sun and Earth, as seen from each other. MISCELLANEOUS EXAMPLES. 133 75. If, in any triangle, a2=b2+bc+c2, shew that A=120~. 76. The sides of a triangular field are 330 yds, 275 yds, 275 yds. Find its area. 77. In any triangle shew that sin (A-B)_(a2-b2) sinA si (B-C) (b2 —c2) sinC' 78. Solve the triangle ABC, given A=30~, A C=3a/3, BC=3. 79. Given L cot 72~ 15'=9.5052819, L cot 72~ 16'=9.5048538, find L cot 72~ 15' 35", and L tan 17~ 44' 20"' 80. In any triangle, shew that the area may be expressed by a sinB sin C c bsinB+ c sin C' 81. In a triangle, right-angled at C, given A=54~, c=108ft solve the triangle. 82. Shew that, in any triangle, sinA+sin C cosC-cosA=cot B. cosA+ cosC sinA-sinC 83. Given log2=.3010300, find the logs of 1.28,.00128, and (T-, )3. 84. Find the area of a parallelogram, whose adjacent sides are 25 ft and 30ft, and included angle 54~. $5. Solve the triangle ABC in which AB= 100ft, BC=50ft, and ABC=60~. 86. Given AB=125 yds, A C=80 yds, and A=60~: find the area of the triangle. 87. Find the logs of 12.5,.00132, 2.16, 149, from the logs given in (137). 88. The ascent of a railroad incline is 1 in 472: find the angle of inclination in minutes and seconds. 89. Find the value of V7, having given the mantissa for 7, 13204, 13205, to be 8450980, 1207055, 1207384, respectively. 90. Prove the truth of the following Rule in Surveying: Take 8, of the square of the distance in chains, and it will give the correction for curvature in inches. 91. The perimeter of an equilateral field is 20 chains; find its area. 92. The angles of a triangle are as 1: 2; 7; compare the greatest and least sides. 93. Shew that cot-' 2-cot' 4=cot' 3+cot-' 5 + cot-' 7 +cot' 21. 94. Shew that a foot will subtend an angle of 39", nearly, at a distance of a mile. 95 If A+B+ C-900, shew that cos2(A+ B)+-cos'(C+ B)-1 134 MISCELLANEOUS EXAMPLES. 96. The perimeter of an isosceles triangle is 50 yards, and the vertical angle is 120~: determine the area. 97. A field is in the form of a right-angled triangle, whose hypothenuse (AB) is 200 feet, and one angle (A) 30~. How much longer would it take to run along the sides AC, CB, than to go straight from A to B, at the rate of 4 miles an hour? 98. Shew that 1 +cos (A-B) +cos (B-C) +cos(C-A) =4 cos (A-B) cos - (B-C) cos (C-A). 99. The angle of elevation of a tower is 60~, to an observer at a certain distance from its base, and 100feet further off, the angle of elevation is 45~. Find its height, allowing 6 feet for the observer's height. 100. On a horizontal line AB, between two towers AC and BD, I observed the angle of elevation DAB=22~ 56'; then in the same line I measured AE, 96 feet, and at E found the elevation of AC to be 49~ 50'. Going on as far as I could towards B, viz. to the point F, I there found the elevations of AC and BD to be 36~ 48' and 44~ 15', respectively. What were the heights of the towers?-Given, nat. cot44~ 15'=1.0265287; nat. cot 22~ 56'=2.3634946. LOGARI11HMS OF SELECTED NUMBERS, ETC. 135 Logarithms of Selected Numbers. I No. 10066 10067 10067 10228 10229 10354 10355 10985 10986 11373 11374 12345 12346 14037 14038 14161 14162 15203 15204 15388 15389 17720 17721 23456 23457 24119 Log. 0028569 0029001 0097907 0098332 0151082 0151501 0408001 0408396 0558750 0559132 0914911 0915263 1472743 1473052 1510939 1511246 1819293 1819579 1871822 1872104 2484637 2484882 3702540 3702725 3823593 I No. I 24120 24623 24624 26507 26508 36831 36832 38831 38832 38852 38853 39712 39713 39982 39983 41277 41278 63455 63456 63830 63831 76543 76544 93970 1 93971 _. Log. 3823773 3913410 3913586 4233606 4233770 5662135 5662253 5891786 5891898 5894134 5894246 5989218 5989327 6018645 6018754 6157081 6157186 8024658 8024727 8050248 8050316 8839055 8839112 9729892 97299:i8 Logarithms of Selected Sines and Tangents. Log. Log. tanll0 0' 9.2886523 sin35~ 46' 9.7667739 tanl8 0 9.5117760 sin37 0 9.7794630 sin21 19 9.5605310 sin44 0 9.8417713 sin22 0 9.5735754 sin44 15 9.8437250 tan22 0 9.6064096 tan49 50 10.0736222 sin22 55 9.5903869 tan53 12 10.1260429 sin22 56 9.5906856 sin53 30 9.9051787 tan25 10 9.6719628 sin62 14 9.9468707 sin27 45 9.6680265 sin62 15 9.9469372 sin27 46 9.6682665 sin64 40 9.9560886 tan27 45 9.7210893 sin66 0 9.9607302 tan27 46 9.7213958 sin70 0 9.9729858 sin31 10 9.7139349 tan71 30 10.4754801 tan35 0 9.8452268 sin81 15 9 9949158 tan?5 44 98.570039 sin83 10 9 9969044). _. l. t 136 ) EXAMINATION PAPERS. Paper t. I. What is meant by the circular measure of an angle? 2. Define the tangent of an angle; and express it in terms of the sine. 3. Find the sine of 30~, and the secant of 45~. 4. The length of an arc of 35~, in a circle whose radius is 20 inches, is 3 of an arc of 25~ in another circle; find the circumference of the latter circle. 5. Prove that sin(A-B)=sinAcosB-cosAsinB; and thence deduce the value of sin(A+B) in trigonometrical functions of A and B. 6. Shew that in any plane triangle, the perpendicular from A to _a 2 sin C+ c sinB b+c 7. Eliminate 0 between the equations sin0-cosf'-m, sin20=n. 8. Construct an angle whose cosecant is 34, and find the corresponding numbers for the other trigonometrical ratios. 9. What values of x will satisfy the equation cos (a - x) = cos? cos a Paper XZ. 1. Define the secant of an angle. Trace the changes, in magnitude and algebraic sign, of the secant of an angle, as the angle increases from zero to four right angles. 2. The sine of an angle x, between 90~ and 180~, is |; find tan x, sec x, sin 2x. cos 2x, tan 2x. 3. In a plane triangle, ABC, the sum of A and B is 102 grades, and the circular measure of C exceeds that of A by -,7r; by how many degrees does B exceed A? 4. Shew that the sum of the angles of any polygon of n sides =(n-2)180~, and hence that each angle of a regular pentagon is 108~. EXAMINATION PAPERS. 137 5. If the earth's radius is 4000 miles, shew that the radius of the circle of latitude 30~ is about 3464 miles. 6. Having given cos(A + B) = cosA cosB-sinA sinB, deduce c 1 -tan2A cos2A — - +tan2A 7. The base of a pyramid is a square, and the sides are equilateral triangles; find the angle at which a side is inclined to the base. 8. In any plane triangle, shew that sinB=- sinA; and point out a what is called the ambiguous case in the application of that formula to the solution of a triangle. 9. A person attempts to swim directly across a stream of given breadth. If he swim n times as far as he would have done had there been no current, where will he reach the opposite side, and what angle does his course make with it? Paper III. 1. Describe the sine and cosine of an angle in relation (i) to those of its supplement, and (ii) to those of its complement. 2. The hypotenuse of a triangle ABC, right-angled at C, is 20, and the base, AC, is 16; find the numerical values of sinA, cosA, and cotA. 3. What are the three angles of the triangle ABC, if | of A, g of B, and i of C, are in arithmetical progression, and A is - of C? 4. Given sin30~=-, and sin45~=cos45~=V/2; find the values of tan30~, sin75~, and cos22~ 30'. 5. Find the cotangent, cosecant, and versed sine of an arc. 6. If sin(AO-BO)=cos(Ao+-B~), what is the value of B? 7. Prove sin(A+B)=sinA cosB+cosA sinB. 8. In a right-angled triangle, given the sum of the hypotenuse and perpendicular equal to the square of the base, and the value of the sine of the angle between the hypotenuse and base =.97561; find each side. 9. From the top of a hill I observed two consecutive mile-stones on a horizontal road, running directly from the base. The angles of depression of the mile-stones were 45~ and 30~; find the height of the hill. EXAMINATION PAPERS. Paper IV. 1. Define the sine and cosine of an angle. Trace the values of these ratios as the angle increases from zero to 360 degrees. 2. How often is an arc of 63 66' 19-" contained in half the circumference? 8. Shew that the greater the number of sides of a regular polygon, the greater is the magnitude of each of its angles. 4. Determine the radius of the inscribed circle of a regular heptagon whose side is a. 1 - sinA 5. Shew that tan (45~ — A) = 1 sA cosA 6. Shew that, if tan A=b, then // a/ a+ 2 os- a a' 'a-b4 a+-b Vcos- A 7. In the triangle ABC, the side a, and the angles A, B, are given: find the area of the triangle in terms of these data. 8. In any triangle ABC, prove that a=b cosC+c cosB; and thence deduce the formula for the cosine of an angle of a triangle in terms of a, b, c. 9. In order to determine the distance between two points A, B, I took a station C, 500 yards from A and 300 yards from B, and observed the angle ACB, 126~ 30'. Given the natural cosine of 53~ 30'=.5948228; find AB. Paper V. 1. Shew that the complement of P -7q 100 grades =. 180~. P+q p+q 2. What is the numerical value of the tangent of an angle, when that of its cosine is.98? 3. One of two regular polygons has three sides more than the other, and the angle of one exceeds that of the other by 27~. What number of sides has each? 4. Prove sin A = + /(2-2 V1-sin — 2. 5. Investigate a general expression for the area of a triangle in terms of its sides. 6. Determine the radius of the circumscribed circle of a triangle. 7. If I be the length in miles of an arc of a great circle of the Earth, and d the depression in feet of one extremity of it below a tangent to the other; shew that d = 12 nearly. 8. The angles of a triangle are in arithmetical progression, and EXAMINAT1 ON PAPERS. ]39 the sines of twice the angles are in harmonic progression; find the angles. 9. From a station, A, I found the angle of elevation of P, the top of a pyramid to be 250 10' 24"; but being prevented from approaching to or receding from the pyramid in a direct line, I measured in the horizontal plane in which the pyramid was situated a distance AB=300 feet, and observed the angles BAP and ABP to be 68~ 10' and 74~ 35', respectively. Find the height of the pyramid. -Given Lsin74~ 35'= 9.9840852; Lsin37~ 15'=9.7819664; Lsin25~ 10'=9.6286472; Lsin25~ 11 =9.6289160; log6.774=.8308452; log6.775=.8309093. Paper VI. 1. The least angle of a triangle is X of the greatest, and the remaining one is 4~ less than the greatest; find each of the angles. 2. Determine the area of a regular polygon circumscribing a circle, and thence deduce the area of a circle in terms of its diameter. 3. Shew that the perimeters of an equilateral triangle, a square, and a regular hexagon, each including the same area, are as V27, t/16, /12, respectively. 4. Prove the formula sinA-sinB=2 cos2(A+B) sin-(A-B); and thence find two angles of which the sum is 120~ and the difference of the sines - the sine of 8~. 5. Prove sin870-sin59~ —(sin930~-sin61~)=sinl~. 6. The difference of the natural sines of 58~ 20' and 44~ 31' is.15. Verify in this instance the formula in Quest. 5, having given Leos of 51~ 25=9.7949425, and of 51~26'=9.7947841; Lsin of 6~ 54'=9.0796762, and of 6~ 55'=9.0807189; also, log2=.30103, and log3=.4771213. 7. Prove the formula si --- = tan sinA-sinB tan -(A-B) 8. Given O=the circular measure of an arc of n degrees. Prove that, when 0 is indefinitely diminished, the limiting value of n is unity; but that, when n is indefinitely diminished, the. sinn", 0 limiting value of sinn is not unity. n 14b 1EXAMINATION PAPERS. 9. Two objects, P and Q, are observed from stations A, B; AP, AQ, make angles a, f, with the line BA produced; BP, BQ, make angles a', S', with the same line. Shew that the *t *1 2., -D Ar\ ' - sin a' sin I' sin (I-as) area of the triangle PA Q is AB2 x in sin- ' sin (/-a) 2 sin (a - a') sin (3 - ')' Paper VIZ. 1. Prove that the area of a circle whose radius is R is 7rR2; and find the area of a circle in which an arc of 75~ measures 18 inches. 2. In any plane triangle, prove that tanA + tanB +tanC =tanA tanB tanC: and thence deduce tan60~= /3. 3. Having given cos(A.-B)=sinA sinB+cosA cosB, deduce the formulae cos2A=-1-2 sin2A, tan(A+B)= tanA tanB 1-tanA tanB 4. Determine the radius of the inscribed circle of a triangle. 5. Eliminate 0 from the equations tan30 sec3e Xadse2O, y=atan"8 6. The interior angles of a rectilineal figure are successively as the numbers 2, 5, 6, 8, 3, and 12; find the least and the greatest. 7, If two plumb-lines suspended from points at a given distance apart on the Earth's surface are inclined at an angle of n", and at n" when at an elevation h; shew that the Earth's nh radius = -'- very nearly. 8. Determine the numerical value of cos30~; and, assuming the Earth's radius to be 4000 miles, find at what rate per hour a point on the circle of latitude 36~ is carried by the Earth's rotation on her axis. 9. Two stations, A, B, are chosen, and four points, P, Q, R, S, are observed; and it is found that the angles which the directions of these points, as seen from A, make with BA produced are 60~, 150~, 240~, 300~; the angles which the directions as seen from B make with BA are 30~, 120~, 270O, 330~. Shew that the area inclosed by PQRS is AB2 x V/3. ANSWERS TO THE EXAMPLES. 1. 1. 2. 3 5. 8 740 22'58"/;-52 15' 29'; 88~ 58' 56".8. 340 44' 40"; 2070 17' 7". 80~ 50'51"; 9~.1525, 800.8475. 4. 570 12', 32~48f. 600 '2". 6. 72~, 54~, 540. 7 70~, 50~, 60~. 80~, 40~, 60~. 9. 30~, 60~, 90~. 10. 81-A3, 40~-0, 67 A0 2 1. 1156; 72 75\; 99'67'99". 2. 1989 97' 97"; 1109 99' 96"; —25 10' 1". 3. 33~ 26' 8", 56~ 33' 52", 62~ 84' 93". 4. g 11' 11", 3".08; 54', 32".4. 5. 108~, 120~. 7. 1: 3. ' 8. 54~, 81~, 108~, 135~, 162~. 10. 7 and 5. 11. 144~, 108~. 12. 16 or 9. 3. 1..157 in. 2. 68~ 45' 18t'. 3. 5~, ~T. 4. 5ft. 9 in 5. 7958 miles. 6. 15 miles. 7. 1: 15: 360. 8. 1136 miles. 9. 69.8 miles; 25143 miles. 10. 171 miles. 8. 1. 8 0, t r. 2. 0 o-6. 3. 5' 7 -7,?-, T0. 4. ~; 19~05' 55"'. 5. 5043'46/". 6. ~a'ru. 7. 47~ 44'47". 8. 88j~; a..4.. 9. 016297; 56' 1I'!. 10. 45~ s. 1. 100:81. 2. 4;214051'33"/,229010'59'. 3. 39.24yds 4. 2.57 sq.ft. 5. 9: 16. 6. 25.77 sq.ft.; 773 cub.ft. 7. 31 sq. in.. i. 9. r'( I+). 10. 720 53/ 33" 2i /(sec2-1) 1 2/(cosec2-1) cot sec V /(l+cot2)' cosec ' v/(1 +cot2) sin 1 cos 1 3 v(1-sin2)' V (cosec2-1)' 4 1-cos2)' /C(sec2 —)' 1/(1+cotL) sec 5.(1 cot) sec ) 6. V(2 vers-vers2). cot,(seC"2..). 7. i,, 11, 1I, It 8. 56, '. 9. V5, v"/5. 10. ~v15, {. 1.42 ANSWERS TO THE EXAMPLES. 7. 1. 31983'10". 3. 73:64. 4. 2: 1; 76~23' 40", 42r44' 13S 9 2 6. + 1 7. /5 4' 8..2 or; 1 or 3 a2+b2 ab 10. or -. 11.a; _b' 12. A= — 45~, B= + 30~. 8. 1. - /3, 2; 1V2, -1; -/3, -2. 2. -4V3, -, /V3 3. A/3, 12; 1,. 4. l1; 2(2+ /2). 5. +cos 100, -cos 200, -cot 300, +cot 400, -sec400, -sec300. 9. 1. -sinO, +cos0, -sinO, -cos0, +tan0, +cot0. 2. +cosO, +sinQ, ~sinO, +cos, -cot 0. 3. 0. 4. +cose; Fcot0. 6. +coseca. 10. 1. (4n-1) 7r. 2. (4n+l1)-7r. 3. (6n~2) Jr. 4. (6n+1) 7r. 5. (3n+1) 5r. 6. 26ft; 15 ft. 7. 156ft. 8. 214ft. 9. 86.4 yds. 115.2yds.64.8 yds. 10. 51ft. 11. 79ft. 12. 36ft; 63ft. cotA cot B 1:L. 5. 2 sinA cosA, 2 cos2A-1. 6. cot cotA secA secB 7. 1* /{ (sec2A-1 ) (sec2B-] )} cosecA cosecB s. 8. (cosec2B-1)+ /V(cosec.2A- 1) 1l 1..(cos4A-cos6A), I(cosA-cos5A), I(coskA-cosiA); sin2 3A-sin2 2A, sin2 A-sin2'A, sin2 ]A-sin2 "A. 2. I(oA+ sA) (cos 3A+cosA), — (c os3A+cosA), (os +cosA) cos2 7A-sin2 5A, cos2 A —sin2 'A, cos2 A — sin2 IA. 8. ~ (sin 5A- sinA), I (sin 3A-sin 2A), I(sin -A-sin IA); cos (450+~A) —sin2(45~ —A), cos2(450+A)-sin2(450~-A) sin2(450 —A)-sin2(450- A). 4 {cos (A+B-C)-cos (A+B+C)}, f {cos(A —B+ C) +cos (A —B —C), I (sin 2A+ sin 2B), sin2 (A+ B+ - - sin (A+-B-C), cOs2(A-B+ C)-sin2(A-B — C),cos2(450-B)-sin2(450-A) ANSWERS TO THE EXAMPLES. 143 15.. 2 sin 4A cosA; -2 sinA sinIA, 2. 2 sin (A-B) cosB; 2 sin (45 —A +- B) cos (45~-1B). 3. 2 sin IA cos (kA+B); 4 sin -A cos -A. 4. 2 sin 3A sin A; 2 cos (A- C) cos (A-2B+C). 6. sin 2 (A-B) sin 2B; -sin AA sin IA. 6. -2 sin I (A+-3B) sin - (3A-B); 2 sin 3A cos}A. 7. 2sin (A+B+2C) sin (A-B); sin(A-B) sin(A-3B). 8. 4 sin(A-~B) sin B; 2cos45~cos(45~-A) or v2 cos(45~-A). 9. - cos 2 (A+B); cos 2A cos 2B. 10. 2 cos (45~-B) cos (45~-A); -2 sin 45~ sin(45~ —A), or — 2 sin (45~ — A). 2 1/2- 4+3-1 17. 9. V/(2+v/2). 10. V/(2-/2). 11. a2V2-_3-l 12. /(2V2-/3-1) 13. 2/2-V3-1 2 V2 2, /2 14. i /{2+V/(2+V/2)}. 15. /{2+ /V(2- /2)1. 16. V/(2+ /2) 18. 1. 9: 14. 3. 2182~ miles. 6. 141 miles. 7. 205 miles. 8. 4000 miles. 9. 8448 miles. 10. 7.7 miles, 13.4 miles. 19. 1..4166114; 24~ 37' 19'/. 2..3712598; 68~ 12'37", 3. 1.0003394; 45~ 0 24". 4. 1.7317213; 30~ 0'43O', 5. 1.9994125; 59~ 59' 4". 6. 1.1545810; 60~ 0/30". 21. 1. 21 sq. yds. 2. 1559 sq. yds. 3. 415 sq. yds. 4. 750. 5. 3A 2R 18P I 1- yds. 6. 57 yds. 7. 901 sq. yds. 8. ~101 14s. 1id. 9. B=60~; a=2.625ft, b=4.5465ft; 10. 8660 sq. yds. area= 6 sq.ft. 11. B=60~, C=90; b = 17.32. 12. 90~; 0, %. a c 2a-e 22. 1. R-1aV3; r=1a^3. 2. R= /4a; r — /a 7. u r(ia2 ) a.+c V'~ a+3+y 6. 3:4. +r1r+r2r, 7. r= - 8. 6 in., 19 in., 18 in., 3.6 in, * 1 2 a r' 1' - 3 ' 2'a3 144 ANSWERS TO THE EXAMPLES. 26. 1. 2A 2R 20P. 2. 94ft. 3. 1426 ft. 4. 260ft. 5. 2A lR 9P. 6. 324yds. 7. 513 yds. 8. 72ft. 9. 100 yds, 125 yds; 3750 sq. yds. 10. 1099yds. 11. 173 yds, 100 yds, 200yds; 1A3R6P. 12. 150ft. 13. 185 yds. 14. 30ft. 15. 11.6 miles, 16.4 miles. 16. 14A 2R 27P 3, sq. yds. 17. 4 miles. 18. 23P 9 sq. yds. 4ft. 19. asina, acosa 20. 101. 21. 721ft. 22. 6.34 miles. 23. 243 yds. 24. 130 53/52". 27. 1..9030900,.9542426, 1.0791813. 2. 1.3010300, 1.3979400, 1.7781513. 3. 1.5228787, 1.3979400, 1.6020600. 4. 2.4771213, 2.5228787, 3.5185140. 5..2552726, 2.1461280,.1583626. 6. 2.7958800, 4.9172146,.0211893. 7. 1.0253059,.6232493,.6434527. 8. 2.0453230,.0880456, 0.416462. 9. 1.5000000, 1.4336766,.0527875. 10. 3.4647060, 1.8957576, 1.5355396. 11. 1.8035760. 12. 1.8149306. a8. 1. 1.5527807, 1.5530013, 3.5528198, X D. Pro..5530092. - 2. 3571.693,.3573465,.00035709, 217 217 i.3572892. 3. 1.3010504,.0200087. 4 65 4. 6.4913649, 4.4913652, 1.4913722. 6 10 5. 4.498527. 6..5738614. a 8 174 29. 1. 9.2961738; 11~ 24' 7". 2. 9.6988970, 9.698805! 3. 10.0001685; 9.9998526. 4. 9.6991340; 30~ 0' 3f 5. 9.9375063, 9.9374881, 10.0624997. 6. 9.9998315, 10.0001263, 44059 31". ( 145 ) ANSWERS TO MISCELLANEOUS EXAMPLES. 1. 600, 80~, 40~. 2. 61~, 31~, 88~. 3. 72~, 54~, 54~. 5. 1382 yds. 6. 94~0=105=. 7. 4; L 2ol 8 84~, 36~, 60~. 9. 32~ 42' 15"; 136' 33' 80". 10. 1000~. 11. 9 and 10. 12. 88~ 25' 45". 13. -.', 7r, 7r, Tr. 14. r; 4~ 46' 29"; 5g 30' 52". 15. 120~, 450, 15~. 16. 12. 18. 114~ 35/ 30"/, 320 42' 15", 32~ 42/ 15". 19. 4 3 20. 1I. 21. 330 yds. 22. 2-1- (nearly)=120~ 19' 16". 23. 44. 24. a2+b, 2ab. 26. 174~ 1 52 — 19336' 78". a2+b2' a2-fb2 28. ]. 30. +3. +i,. 31. nl-n+p m + mn-p 32. 35.4 yds, nearly. 33. 1140 35 30. 34. +cos20~, -cos40~, -tan30~, +tanl00, -secl0~, -sec30~. 35. {6n+(-1)"}7r. 36. m2+-n2=2. 37. 988. 38. 8. 39. (6n+-1)l 7r. 40. (m2 —71)2=8(m2+ 72). 42. +cosec;. 43. (6n+1)l7r, (6n+_2)7Tr. 44. 8660 sq. yds 61. 328ft. 64. 17". 65. 27 miles. 68..4998740; 29~ 59/ 7" 70. Sun's diameter= 879,645 miles=880,000 miles, nearly. 72. 20,806ft. 74. 12100: 1. 76. 7.A. 78. B=60~, C7=90~, AB=6; or B=120~, C=30, AB=3. 79. 9.5050322, 9.5049965. 81. B=36~, a=87.4ft, b=63.5ft. 83..1072100, 3.1072100, 7.6783700. 84. 6064 sq. ft. 85. AC=86.6ft, A=30~, C=90~. 86. 3R 23P 4~ sq. yds. 87. 1.0969100, 3.1205740,.3344539, 1.1529674. 88. 7' 17". 89. 1.320469. 91. 1A 3R28p, nearly. 92. v/5+1: /5 —1 96. 77~ sq. yds. 97. 12 /" nearly. 99. 243 ft. 100 Each 113'7 ft (1.) ( 146 ) ANSWERS TO EXAMINATION PAPERS. PAPER I. 4. 22 ft. nearly. 7. m2 + n =1. 8. sin =.28. cos =.96; &c. 9. xa-, or 180 + a. PAPER II. 2. tan xS -1, sec x= -12, sin 2x= 24; &c. 3. 41~ 24'. 7. cos-11V3. 9. Vn2 -1 times the river's breadth - distance lower down; sin-'1. fn PAPER III. 2..6,.8, 1. 6. Arbitrary. 9. (/3 + l) mi. PAPER 1V. 2. 3.1416 times. 9. 720y. PAPER V. 2..203. 8. Each 60~. PAPER VI. 1. 46~, 65~, 69~. PAPER VII. 1. 1.44 sq. yd. 6. 40, 2400. 3. 36~, 480, 96~. 8. 41, 40, 9. a2 sin B. 7 si a sin(A + B). 2 Tsinl A 3. 8,5. 9. 203~ ft. nearly. 4. 68~, 520. 5. X5y5 (y5-X5) = a 8. -(V5 + 1): 847 mi. 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