nated by the characters: ,'," : thus the expression 16° 6' 15" represents an arc, or an angle, of 16 degrees, 6 minutes, and 15 seconds. III. The complement of an angle, or of an arc, is what remains after taking that angle or that arc from 90°. Thus the complement of 25° 40' is equal to 90°—25° 40'=64° 20'; and the complement of 12° 4' 32" is equal to 90°-12° 4' 32"=770 55' 28". In general, A being any angle or any arc, 90°—A is the complement of that angle or arc. If any arc or angle be added to its complement, the sum will be 90°. Whence it is evident that if the angle or arc is greater than 90°, its complement will be negative. Thus, the complement of 160° 34' 10" is -70° 34' 10". In this case, the complement, taken positively, would be a quantity, which being subtracted from the given angle or arc, the remainder would be equal to 90°. The two acute angles of a right-angled triangle, are together equal to a right angle ; they are, therefore, complements of each other. IV. The supplement of an angle, or of an arc, is what remains after taking that angle or arc from 180°. Thus A being any angle or arc, 180°—A is its supplement. În any triangle, either angle is the supplement of the sum of the two others, since the three together make 180°. If any arc or angle be added to its supplement, the sum will be 180°. Hence if an arc or angle be greater than 180°, its supplement will be negative. Thus, the supplement of 200° is —20°. The supplement of any angle of a triangle, or indeed of the sum of either two angles, is always positive. The secant of an arc is the line drawn from the centre of the circle through one extremity of the arc and limited by the tangent drawn through the other extremity. Thus CT is the secant of the arc AM, or of the angle ACM. The versed sine of an arc, is the part of the diameter intercepted between one extremity of the arc and the foot of the sine. Thus, AP is the versed sine of the arc AM, or the angle ACM. These four lines MP, AT, CT, AP, are dependent upon the arc AM, and are always determined by it and the radius ; they are thus designated : MP=sin AM, or sin ACM, AP-ver-sin AM, or ver-sin ACM. VI. Having taken the arc AD equal to a quadrant, from the points M and D draw the lines MQ, DS, perpendicular to the radius CD, the one terminated by that radius, the other terminated by the radius CM produced; the lines MQ, DS, and CS, will, in like manner, be the sine, tangent, and secant of the arc MD, the complement of AM. For the sake of brevity, they are called the cosine, cotangent, and cosecant, of the arc AM, and are thus designated : MQ=cos AM, or cos ACM, CS=cosec AM, or cosec ACM. cos A=sin (90°-A), cosec A= sec (90°—A). The triangle MQC is, by construction, equal to the triangle CPM; consequently CP=MQ: hence in the right-angled triangle CMP, whose hypothenuse is equal to the radius, the two sides MP, CP are the sine and cosine of the arc AM: hence, the cosine of an arc is equal to that part of the radius intercepted between the centre and foot of the sine. The triangles CAT, CDS, are similar to the equal triangles CPM, CQM ; hence they are similar to each other. From these principles, we shall very soon deduce the different relations which exist between the lines now defined : before doing so, however, we must examine the changes which those lines undergo, when the arc to which they relate increases from zero to 180°. The angle ACD is called the first quadrant ; the angle DCB, the second quadrant; the angle BCE, the third quadrant ; and the angle ECA, the fourth quadrant. = P VII. Suppose one extrem s S ity of the arc remains fixed in A, while the other extremity, M Q MT marked M, runs successively throughout the whole extent of the semicircumference, B Ρ Α. from A to B in the direction P A ADB. When the point M is at A, or when the arc AM is zero, N R the three points T, M, P, are confounded with the point A; whence it appears that the sine and tangent of an arc zero, are zero, and the cosine and secant of this same arc, are each equal to the radius. Hence if R represents the radius of the circle, we have sin 0=0, tang 0=0, cos (=R, sec 0=R. VIII. As the point N advances towards D, the sine increases, and so likewise does the tangent and the secant; but the cosine, the cotangent, and the cosecant, diminish. When the point M is at the middle of AD, or when the arc AM is 45°, in which case it is equal to its complement MD, the sine MP is equal to the cosine MQ or CP; and the triangle CMP, having become isosceles, gives the proportion MP : CM ::1: V2, R =cOS 450 =!RV 2. In this same case, the triangle CAT becomes isosceles and equal to the triangle CDS ; whence the tangent of 45° and its cotangent, are each equal to the radius, and consequently we have tang 45°=cut 15°=R. IX. The arc AM continuing to increase, the sine increases till M arrives at D; at which point the sine is equal to the radius, and the cosine is zero. Hence we have sin 90°=R, cos 90o=0; and it may be observed, that these values are a consequence of the values already found for the sine and cosine of the arc zero ; because the complement of 90° being zero, we have sin 90°=cos 0°-R, and cos 90°=sin 0°=0. or As to the tangent, it increases very rapidly as the point M approaches D; and finally when this point reaches D, the tangent properly exists no longer, because the lines AT, CD, being parallel, cannot meet. This is expressed by saying that the tangent of 90° is infinite ; and we write tang 90°=0 The complement of 90° being zero, we have tang 0=cot 90° and cot 0=tang 90°. Hence cot 90°=0, and cot (=0. X. The point. M continuing to advance from D towards B, the sines diminish and the cosines increase. Thus M'P' is the sine of the arc AM', and M'Q, or CP' its cosine. But the arc M'B is the supplement of AM', since AM' +M'B is equal to a semicircumference; besides, if M'M is drawn parallel to AB, the arcs AM, BM', which are included between parallels, will evidently be equal, and likewise the perpendiculars or sines MP, MP. Hence, the sine of an arc or of an angle is equal to the sine of the supplement of that arc or angle. The arc or angle A has for its supplement 1800-A: hence generally, we have sin A=sin (1800—A.) The same property might also be expressed by the equation sin (90°+B)=sin (90°—B), B being the arc DM or its equal DM'. XI. The same arcs AM, AM', which are supplements of each other, and which have equal sines, have also equal cosines CP, CP'; but it must be observed, that these cosines lie in different directions. The line CP which is the cosine of the arc AM, has the origin of its value at the centre C, and is estimated in the direction from C towards A; while CP', the cosine of AM' has also the origin of its value at C, but is estimated in a contrary direction, from C towards B. Some notation must obviously be adopted to distinguish the one of such equal lines from the other; and that they may both be expressed analytically, and in the same general formula, it is necessary to consider all lines which are estimated in one direction as positive, and those which are estimated in the contrary direction as negative. If, therefore, the cosines which are estimated from C towards A be considered as positive, those estimated from C towards B, must be regarded as negative. Hence, generally, we shall have, cos A=~cos (180°-A) that is, the cosine of an arc or angle is equal to the cosine of its supplement taken negatively. The necessity of changing the algebraic sign to correspond T А. V with the change of direction S' S MT P P ver-sin AM=R-cos AM. R E ver-sin AM-R--Cos AM, cos 180°=-R. For all arcs, such as ADBN', which terminate in the third quadrant, the cosine is estimated from C towards B, and is consequently negative. At E the cosine becomes zero, and for all arcs which terminate in the fourth quadrant the cosines are estimated from C towards A, and are consequently positive. The sines of all the arcs which terminate in the first and second quadrants, are estimated above the diameter BA, while the sines of those arcs which terminate in the third and fourth quadrants are estimated below it. Hence, considering the former as positive, we must regard the latter as negative. + XII. Let us now see what sign is to be given to the tangent of an arc. The tangent of the arc AM falls above the line BA, and we have already regarded the lines estimated in the direction AT as positive: therefore the tangents of all arcs which terminate in the first quadrant will be positive. But the tangent of the arc AM', greater than 90°, is determined by the intersection of the two lines M'C and AT. These lines, however, do not meet in the direction AT ; but they meet in the opposite direction AV. But since the tangents estimated in the direction AT are positive, those estimated in the direction AV must be negative: therefore, the tangents of all arcs which terminate in the second quadrant will be negative. When the point M' reaches the point B the tangent AV will become equal to zero: that is, tang 180°=0. When the point M' passes the point B, and comes into the position N', the tangent of the arc ADN'will be the line AT: |