that when the trigonometrical line is + in the fourth quadrant, the corresponding trigonometrical line of the negative arc has the same sign as that of a positive arc of the same magnitude, and when the trigonometrical line is --- in the fourth quadrant, a contrary sign. The truth of this assertion may be seen, by trying negative arcs of various magnitudes upon the diagram, laying them off downwards from the right point of the circle, and observing in which quadrant their extremities fall. They will be found in every case to give results agreeable to the rule just stated. THE COTANGENT AND COSECANT. = since = 0. = R. 28. The cotangent of 0° is equal to the tangent of 90° (art. 23) and is therefore co. The cotangent of 900 is equal to the tangent of 0° and is 0. The cotangent of 180° is equal to the tangent of 90° — 180o = the tangent of — 90° ° 90° is a negative arc, and terminates at the bottom of the circle, or the 270° point. The tangent of 270° the tangent of 900 270o = the tangent of — 180° = 0. When the tangent has its least value which is 0, the cotangent has its greatest which is co, and vice versa. 29. The cosecant of 0o = the secant of 900 The cosecant of 900 the secant of 00 The cosecant of 1800 the secant of 900 180o = 0. The cosecant of 2700 the secant of 1800 = 0. When the secant has its least value which is R, the cosecant has its greatest, which is co, and vice versa. The cotangent and cosecant have their greatest values together and their least values together, viz., that of the one 0, of the other R, at the top and bottom of the circle, and both o at the right and left points. 30. With regard to the signs of the cotangent and cosecant in the different quadrants, they will be most conveniently discovered from the analytical expressions for these lines which we shall presently have. We add here, however, which so far as the cotangent and cosecant are concerned must be for a moment taken for granted, that the six trigonometrical lines may be arranged in three pairs, each pair having always the same algebraic sign. We have seen that the secant and cosine go together in this way; so do also the cosecant and sine ; and so do the tangent and cotangent. The positive sines and cosecants are separated from the negative, by the horizontal diameter ; the positive cosines and secants from the negative, by the vertical diameter; and the tangent and cotangent are together + and alternately in the successive quadrants. 31. The following algebraic notation is employed for the six trigonometrical lines. Let a be the algebraic expression for the number of degrees in any arc, then the trigonometrical lines of the arc a will be expressed thus; sin a, tan a, sec a, cos a, cot a, cosec a. Cot a tan a = r2 is read, the cotangent of the arc a multiplied by the tangent of the same arc is equal to the square of the radius of the circle in which these trigonometrical lines are supposed to be drawn. Cot a and tan a are expressions for straight lines, and the equation above expresses that the rectangle formed by the tangent and cotangent of an arc is equivalent to the square formed upon the radius. . The two members of the above equation contain the same number of dimensions, and are therefore homogeneous. This ought to be the case in all trigonometrical equations; because a line cannot be equal to a surface, nor either of these to a solid. Sometimes in analytical investigations R is supposed to be equal to 1; R’ and R’ would also be equal to 1. Whether this 1 is a unit of length, of surface, or of solidity, must be determined by what is required to preserve the homogeneity of the equation. 32. The tangent, secant, cotangent and cosecant may be expressed in terms of the sine and cosine. a = The values of the four former in terms of the two latter are derived geometrically as follows: Call the arc AM D E in the diagram a, T M then DM = 900 complement of a, DE = cota and ce = cosec a. A P In the similar triangles comand CAT, since homologous sides are proportional we have CP : PM :: CA : AT or cos a : sin a ::R: tan a : whence multiplying the means and dividing by the first term, we obtain the last R X sin a tan a = COS a that is the tangent of any arc is equal to radius multiplied by the sine divided by the cosine of the same arc. If R be made equal to 1, then sin tan = COS 34. In the triangles CMP and CED, which have their sides respectively parallel, and are therefore similar, we have the proportion* COS cot sin 35. The same triangles give also the proportion 36. In the expressions for the tangent and cotangent which we have here derived, it will be observed that we have the quotient of the sine and cosine, and that therefore when the sine and cosine have contrary signs, the tangent and cotangent will be negative. This occurs in the second and fourth quadrants. It appears hence, that the cotangent changes its sign always with the tangent. Also that both the tangent and cotangent of an arc are equal to those of its supplement with contrary signs. From the expressions for the secant and cosecant, it appears that the former must always have the same sign as the cosine, and the latter the same as the sine. The formula derived in the last four articles should be committed to memory. * The homologous sides are those which are parallel. (Geom. B. 4, Prop. 21, Schol.) 37. Multiplying the expression for the tangent given in art. 32 by that of the cotangent in art. 34, we have tan a cot a = R? whence, cot =: tan 38. We are now prepared to find formulæ for the solution of right angled plane triangles in all cases, and plane triangles in general in a few particular ones. The remaining cases of triangles in general will require further preliminary matter. DERIVATION OF FORMULÆ FOR THE SOLUTION OF RIGHT ANGLED PLANE TRIANGLES, A MP Let ABC be any M right angled triangle. B. With C as a centre describe, with any radius at pleasure, the M arc mn terminating at the sides of the angle. This arc will be the C P N P N measure of the angle c. Draw MP perpendicular to CN. will be the sine of the arc Mn because it is drawn from one extremity m of the arc perpendicular to the diameter which passes through the other extremity n, MP is also the sine of the angle c. CM is the radius of the circle to which the arc mn belongs. The two triangles CMP and CBA are equiangular and similar, and give the proportion. CM : MP :: CB: BA or R: sin c::CB : BA. Had an arc been described with B as a centre in a similar manner we should have had r : sin b :: BC : from a |