than 10, but the logarithmic tangents, secants, cotangents, and cosecants admit of all possible values. ON THE SIGNS OF TRIGONOMETRICAL QUAN TITIES. When geometrical quantities that are measured from a given point or line are considered analytically, they are considered as positive or negative, and are accordingly designated as + or -, according as they lie on the same side or on opposite sides of the same point or line. Thus in fig 2, the sines are estimated from the diameter AA', and in the semicircle A EA' they are considered as +; but, as in the other semicircle A E A they fall on the other side of the diameter AA', they are then considered as -. The cosines, being estimated from the centre C, are considered as in the first quadrant A E; but as in the second quadrant EA', and the third A E', they fall on the other side of the centre C they are then considered as — ; and again in the fourth quadrant E'A they become +, as in the sin. first quadrant. As tan., tan. is + in the COS. same sine, and the cosecant and sine have the same sign. PROP. I.-The chord of 60° and the tangent of 45° are each equal to the radius; the sine of 30°, the versed sine of 60°, and the cosine of 60°; are each equal to half the radius; and the secant of 60° is double the radius. Let D, fig. 4, be the centre of a circle, AB an arc of 60°, and AC an arc of 45°; join A B, B D, and from B draw BE perpendicular to A D, and from D draw DG H perpendicular to A B, and it will bisect both the chord A B and the arc AC B. Draw the tangent AG meeting the secants DBG and DCF in G and F. Now, as the angle BDA is 60°, the sum of the two equal angles D A B and D BA is 120°; therefore each of these angles is 60°, and consequently the triangle A B D is equilateral: whence A B, the chord of 60° is equal to the radius. And as, in the equilateral triangle ABD, the perpendicular BE bisects A D, and the perpendicular DG H bisects both the chord A B and the arc AHC B, D E the cosine of 60°, E A the versed sine of 60°, and B C the sine of 30°, are each equal to half the radius. And by similar triangles DE: DB:: DA: DG; whence, as DB=2 DE, DG 2 DA, or the secant of 60° is double the radius. angle AFD must also be half a right angle; and therefore A D and AF are equal, or the tangent of 45° is equal to the radius. PROP. II.--The secant of an arc is equal to the sine of its tangent and the tangent of half its complement. Let A B, fig. 5, be any arc, A D its tangent, and CD its secant; produce A D till DE is equal to DC; join C E, and draw DH a perpendicular on C E. Then as the right angled triangles CAE and DHE have the common angle E and the right angles CAE and DHE equal; the remaining angle ACE in the one is equal to the remaining angle EDH in the other. But the triangles EDC being isosceles, the angle D is bisected by the perpendicular DH; there fore the angle ACE is equal to half ADC, or to half the complement, of AC D. Hence A D the tangent of AC D, and AE the tangent of half its complement, are equal to C D its se cant. PROP. III.-If A represent the greater and B the less of two arcs, it is proposed to investigate the relations between the sines and cosines of those arcs and the sines and cosines of their sums. Let K, fig. 6, be the centre of the circle, CE the greater arc A, and EF or ED the less arc B; join F, D, and K, E, which will bisect. F D in H and cut it at right angles. Draw FL, HM, EN, and DO perpendicular to KC and GH, ID parallel to KC. Then the triangles F G H and HID are obviously identical, GF and HI being equal; and GH, ID, LM, and MO are also equal. If from the right angles FHK and G H M the common part GHK be taken, the angles FHG and KH M will remain equal; and as the angles at G and M are also equal, being right angles, the angles HKM and HFG are equal, and consequently the triangle FG H, which is identical with the triangle HID, is similar to KH M, which again is evidently similar to KEN. Now EN is the sine, and K N the cosine of the greater arc A, FH the sine and K II the cosine of the less arc B; F L the sine and KL the cosine of the sum of the arcs A+ B; DO the sine and KO the cosine of the difference of the arcs A-B. And LFHM + CF; DO= HM- HI HM-GF; KL = KM - GH, and KO = KM+ID = KM+GH. = - From the properties of similar triangles we have KE:EN:: KH: HM; or rad. sin. A :: cos. B: HM; whence if radius be unity HM : sin. A. cos. B. And K E: KN:: FH: F G; or rad. cos. A :: sin. B: F G; whence FG cos. A sin. B. And K E K N:: KH KM; or rad. : cos. A:: cos. B: K M; whence K M cos. A cos. B. Lastly, K E EN:: FH HG; or rad. : sin. A: sin. B: HG, whence HG = sin. A sin. B. + cos. A sin. B. Again as the angle DAF is a right angle, and the angle ADF is half a right angle, the (I.) Hence FL, sin. A + B = sin. A cos. B DO, sin. A B sin. A· cos. B · cos. A cos. B sin. A cos. A K L, cos. A + B By taking the sum and the difference of the first two of these four equations, and also of the last two, we have cos. A + cos. A B + В. cos. A cos. A+B On the four equations which we have now obtained it may be remarked that A is half the sum of B, and that B is half the difference of A + B and A if therefore inA+B and A B; stead of A+B we were to put A', and instead of A- B we were to put B', we must put for A A' + B' , and for B A'B' and the equations would then stand in this form: A' + B' • COS. A'B' sin. A' sin. B' 2 cos. sin. 2 A'B' cos. A'oos. B' 2 cos. . 2 If in these last expressions the arc B' were to be considered as evanescent, we should have 1+cos. A'= 2 cos.2 whence sin A+ B➡ 2 sin. 1+cos. A = 2 2 sin, A+B 2 sin. A cos. B sin. A B 2 cos. A sin. B + sin. A—B’ and recollecting that it has been shown in proposition I. that sin. 30° and cos. 60° are each half radius; if we consider A in the first of these equations as 30°, and in the second as 60°, the formule will become PROP. IV. A representing the greater, and B the less of two arcs, it is proposed to investigate the relation between the tangents of those arcs and the tangents of their sum and their difference. If t, t', t", &c., be the tangents of the arcs A, B, C, &c., then, considering A + B as one arc, tan. A+B+ tan. C we have tan. A + B + C = But tan. A+B= t + ť (III.) Hence tan. A + B + C = And if A + B + C = 180°, which is the case when A, B, and C are the angles of a plane triangle; then, since tan. 180° 0, the theorem gives t+t+t' tt' t'; (IV) or when radius is unity, the sum of the tangents of three arcs, which together make 180°, is equal to the continued product of those tangents. PROP. V. From the sine and cosine of any given arc A, it is proposed to investigate expressions for the sines and cosines of its multiples. In formula 1, proposition III., we have sin. A + B = sin. A. cos. B + cos. A. sin. B. (I.) sin. 2 A = 2 sin. A. cos. A. sin. 3 A sin. 4 A sin. A. cos. 2 A+ cos. A . sin. 2 A. sin. A. cos. 3 A + cos. A. sin. 3 A, &c. Again, in the same proposition, we have cos. A + B cos. A. cos. B sin. A . sin. B, and substituting for B, A, 2 A, 3 A, &c., in succession, we have (II.) cos. 2 A = cos. A cos. 3 A cos. 4. A cos. A. cos. 2 A— Again, by formula 8, proposition III, sin. A sin. A+B 2 sin. A. cos. B. If in this formula B-1. A, it becomes (III.) sin. n A = 2 cos. A. sin. n - 1. A sin. n — 2. A. By adding together the third and fourth expressions, formula 2, proposition III., and transposing, we have cos. A+B 2 cos. A. cos. B A class of very elegant and curious formulæ may be deduced from the last class by means of a particular substitution. 2 1 (VII.) Make 2 cos. A = z+; then 2 cos. 2 A= 2 . (2 cos, 3A = 32 + &c. and a like substitution for 2 cos. 3 A, 2 cos. 4 A, &c., will give 2 cos. 3 A = &3 + 2 cos. 4 A = 21 + formula required for the solution of the different cases of right angled plane triangles :-Let ABC (fig. 7,) be a plane triangle, right-angled at B; round A as a centre with any radius, A D, describe the arc DG; and from G and D draw GF and D E perpendicular to A B. Then G F will be the sine of the angle A, AF its cosine, DE its tangent, and AE its secant; and, as the angle C is the complement of A, the sine, tangent, and secant, of one of these angles will be the cosine, coangent, and cosecant respectively, of the other. Now, by the properties of right angled triangles A C√ A B2 + BC2, AB = √(A C + B C). (A C — BC), and BC =√(AC+ A'B). (A C — A B). A C. sin. A Again, by similar triangles A C: CB::AG: GF, that is:: rad. : sin. A; hence A C. sin. A BC rad; therefore BC= A C cos. C. rad. BC. sec. C rad. and A C B C rad. sin. A BC cosec A = rad. AC cos. A And similarly A B= rad. rad. n- 1 2 n R-4 x &c. #4-2 + n2 Again, A B: BC:: AD: DE, that is:: rad. : by collecting into pairs tan. A; therefore A B tan. A =BC rad.; hence n=3, 22. cos. 3A n=4, 23 cos. 'Acos. 4 A+ 4 ros. 2 A +3 n5, 2. cos. A cos. 5 A + 5 cos. 3 A+ 10 cos. A, &c. To obtain a general form for sin. "A; in the expression for cos. "A, substitute q A, (q signifying a right angle); then 2 5-1 cos. "q-A cos. n q-n A + n. cos. n — 2 (g—A) + n. PROP. VIII.—It is proposed to investigate the formulæ required in the solution of the different cases of oblique-angled plane triangles. Formula 1-A B. sin. A=CB. sin. C (figs. 8 and 9). For let BD, in each figure, be a perpendicular from B on the base A C, or AC produced. Then, by the last proposition, in the right-angled triangle CBD CB. sin C = BD rad., and A B sin. BAD=BD. rad.; therefore A B. sin. BAD C B · sin. C; and, sin. BAD (fig. 8): = sin. BAC; whence generally A B. sin. A B C sin. C. Hence C B: BA:: sin. A: sin. C. Formula 2.-In the same figures AD:DC:: tan. A BD : tan. DBC, or :: cot. A: cot. C. For (prop. VII.) BD. tan. A BD AD rad., and BD.tan. DBC = C D, rad. and therefore A D· rad. CD rad.:: BD: tan. ABD: BD . tan. DBC'; hence AD: CD:: tan ABD: tan. DBC; or cot. A cot. C. Formula 3.-In any triangle a A, BC (fig. 10), in which A B and C represent the angles, and a, b, and c, the sides opposite those A+ B A--B. angles, a+ba-b:: tan. 2 2 : tan. For formula 1 of this proposition a: b:: sin. A: sin. B; therefore a+ba-b:: sin. A + sin. B : sin. A — sin. B:: (formula 1, prop. III.) tan. A+B A-B 2 : tan. 2 Formula 4.-In any triangle, as ABC (fig. 9), if B D be a perpendicular, let fall from the vertex upon the longest side, then AC: AB+ BC:: AB-BC: AD-DC. For |