36. With regard to the tangents of more than two arcs, the following property (the only one we shall here deduce) is a very curious one, which has not yet been inserted in works of Trigonometry, though it has been long known to mathe. maticians. Let the three arcs A, B, C, together make up the whole circumference, O : then, since tan (A + B) = sum of the tangents % any three arcs which together con. stitute a circle, multiplied by the square of the radius, is equal to the product of those tangents. . . . (XXX.) Since both arcs in the second and fourth quadrants have their tangents considered negative, the above property will apply to arcs any way trisecting a semicircle ; and it will therefore apply to the angles of a plane triangle, which are, together, measured by arcs constituting a semicircle. So that, if radius be considered as unity, we shall find, that the sum of the tangents of the three angles of any plane triangle, is equal to the continued product of those tangents. (XXXI.) 37. Having thus given the chief properties of the sines, tangents, &c. of arcs, their sines, products, and powers, we shall merely subjoin investigations of theorems for the 2d and 3d cases in the solutions of plane triangles. Thus, with respect to the second case, where two sides and their included angles are given : By equ. Iv, a b : : sin A : sin B. By compos. and division. but, eq. xxii, tan #(A + B): tan (A - B); : sin A + sin B : sin A — sin B ; whence, ex equal. a+b : a-b :: tan (A+B) : tan #(A-R). . . . (XXXII.) Agreeing with the result of the geometrical investigation at pa. 387, vol. i. 38. If, instead of having the two sides a, b, given, we know their logarithms, as frequently happens in geodesic operations, tan (A - B) may be readily determined without first finding the number corresponding to the logs. of a and b. For if a and b were considered as the sides of a right-angled triangle, in which p denotes the angle opposite the side a, then would tan p = +. Now, since a is supposed greater than b, this angle will be greater than half a right angle, or it will be measured by an arc greater than } of the circumfer And, from the preceding article, a—b tan #(A–B) tan (A-R) a-Fi, Tū, iaTV, *Toots. consequently, tan (A–B) = cot c. tan (p-HO). . . . (XXXIII.) From this equation we have the following practical rule : Subtract the less from the greater of the given logs, the remainder will be the log tan of an angle : from this angle / take 45 degrees, and to the log tan of the remainder add the log cotan of half the given angle ; the sum will be the log tan of half the difference of the other two angles of the plane triangle. 39. The remaining case is that in which the three sides of the triangle are known, and for which indeed we have already obtained expressions for the angle in arts. 6 and 8. But, as neither of these is best suited for logarithmic computation, (however well fitted they are for instruments of investigation), another may be deduced thus: In the equation - - - b2+ c2—a? for cos. A, (given equation it), viz. cos A = +; substitute, instead of cos A, its value 1 – 2 sin” #A, change the signs of all the terms, transpose the 1, and divide by 2, a2-k2 – c.2 + 2bc a” —(b — c)3 H-- = —so-. Here, the numerator of the second member being the product of the two factors (a+b – c.) and (a – b + c), the equa#(a + h-c). (4–h-H c) tion will become sin’ A = −. But, since we shall have sin”; A = 40. The student will find it advantageous to collect into one place all those formulae which relate to the computation of sines, tangents, &c.”; and, in another place, those which are of use in the solutions of plane triangles: the former of these are equations v, viii, Ix, x, x1, x, zi, xII, XIII, xiv, xv, xvi., xvii, xviii, xix., xx, xxii, Trii, xxii.1, xxiv, xxvii.; the latter are equa. II, III, IV, vii, xxxii, xxxiii, xxxiv. "What is here given being only a brief sketch of an inexhaustible subject ; the reader who wishes to pursue it further is referred to the copions Introduction to our Mathematical Tables, and the treatises on Trigonometry, by Emerson. Gregory, Bonnycastle, Woodhouse, Lard. ner, and many other modern writers on the same subject, where he will find his curiosity richly gratified. To exemplify the use of some of these formulae, the foskwing exercises are subjoined. EXERCISEs. Er. 1. Find the sines and tangents of 15', 30°, 45.60°; and 75°: and show how from thence to find the sines and tangents of several of their submultiples. First, with regard to the arc of 45°, the sine and cosine are manifestly equal; or they form the perpendicular and base of a right-angled triangle whose hypothenuse is equal to the assumed radius. Thus, if radius be R, the sine and cosine of 45° will each be = v Ro-R V =Ry/2. If R be equal to l, as is the ease with the tables in use, then sin 45° = cos 45° = }v 2 = .707106S sin Secondly, for the sines of 60 and of 30°: since each angle in an equilateral triangle contains 60°, if a perpendicular be demitted from any one angle of such a triangle on the opposite side, considered as a base, that perpendicular will be the sine of 60°, and the half base the sine of 30°, the side of the triangle being the assumed radius. Thus, if it be R, we shall have 4R for the sine of 30°, and v (R*— R*) = }Ry3, for the sine of 60°. When R = H, these become sin 30° = 5 . . . . . . sin 60° = cos 30 = -8660254. Hence, tan 30° =#3–3; - v 3 = .5773503. tan 60° = *** = v 3 = . . . . 1732050s. equation solved by any of the approximating rules for such equations, will give z = -1045285, which is the sine of 6°. Next, to find the sine of 2°, we have again, from equation x, sin 3A = 3 sin A — 4 sin” A : that is, if a be put for sin 2°, 3r — 4r" = 1045285. This cubic solved, gives a = -0348995 = sin 29. Then, if s = sin 19, we shall, from the second of the equations marked x, have 2s v(1 — s”) = 034S995; whence s is found = -0174524 = sin 19. Had the expression for the sines of bisected arcs been applied successively from sin 15°, to sin 7°30', sin 3°45', sin 1°52}", sin 56}', &c. a different series of values might have been obtained : or, if we had proceeded from the quinqui. section of 45°, to the trisection of 9°, the bisection of 39, and so on, a different series still would have been found. But what has been done above is sufficient to illustrate this method. The next example will exhibit a very simple and compendious way of ascending from the sines of smaller to those of larger arcs. Cor. By the same method, knowing the sines of 5°, 10°, and 15°, the sines of 20°, 25°, 35°, 55°, 65°, &c. may be found, each by a single proportion. And the sines of 1°, 9°, and 10°, will lead to those of 199,299, 399, &c. So that the |