Cese wears crating for any three angles whatFe, aur say be tree angles of any triangle. sin c; then adding together the equausarmed, and reducing; there will result, ----+ sin c.sin (4—B)=0, sinc-4)+cos c. sin (A—B)=0. SIT 4-3 3-4 US -S# 7 པ་ 24, =sia A-siin B, sian3 C, sin D, -2014-S 1-4=0. Ting III smiar manner with sin (1-B), cos (A+B), 1. dere at length be obtained SIT 1-3 -2 3-C.sin. B-c)+ &c... 140 = sin2 A. gether, the second memist, and of consequence we shall ... form an arithmetical Nove - 0, c850 +8=0} (XXV.) "rion, however, only shows the s +(x+2)+sin(a+ s sis is a recurring series, w GRAN Jominter is 1-2.cos B preceded in this article. The result therefore is, that the sum of the tangents of any three arcs which together constitute 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 angle are given: By equa Iv, By compos and division ab sin A: sin B. a+b: a- b sin A + sin B: sin A → :: sin B. 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 (AB). . . . (XXXIÏ.) Agreeing with the result of the geometrical investigation, at pa. 10, vol. ii. 38. If, instead of having the two sides a, b, given, we know their logarithms, as frequently happens in geodesic operations, tan (AB) 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 denotes the angle opposite the side a, then would tan. 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 ence, or than O. Then, because tan (-)=1+ tan tan o tan 4-tan 10, and because tan OR = 1, we have a tan (? — § 0) = ( − 1 ) ÷ (1 + † ) = And, from the preceding article, a-b tan (A-B) tan (A-B) = a+b tan (A + B) == tan (A-B) cot c. tan (90) ... (XXXIII.) From 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 angles 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 for cos A, (given equation 11), viz, cos a = if we substitute, instead of cos A, its value 12 sin2 A, change the signs of all the terms, transpose the 1, and divide by 2, a2 — b2 — c2 + 2bc a2—(b-c)1 we shall have sin2 A = 4bc b2+ c2 — a2 4bc 2bc Here, the numerator of the second member being the product of the two factors (a+b-c) and (a-b+c), the equa4(a+b-c). (a−b+c) tion will become sin2 = 41c But, since (a+b−c)=(a+b+c)−c, and 4(a−b+c)=4(a+b+c)−b; if we put sa 4b+c, and extract the square root, there will result, These expressions, besides their convenience for logarithmic computation, have the further advantage of being perfectly free from ambiguity, because the half of any angle of a plane triangle will always be less than a right angle. 40. The student will find it advantageous to collect into one place all those formule 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 * 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 copious Introduction to our Mathematical Tables, and the comprehensive treatises on Trigonometry, by Emerson and many other modern writers on the same subject, where he will find his curiosity richly gratified. these these are equations v, VIII, IX, X, XI, x, xi, xf, XIII, XIV, XV, XVI, XVII, XVIII, XIX, xx, xxII, xxй, XXIII, XXIV, XXVII; the latter are equa. II, III, IV, VII, XXXII, XXXIII, XXXIV. To exemplify the use, of some of these formulæ, the following 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 RR=R2, If R be equal to 1, as is the case with the tables in use, then sin 45° cos 45° = 4√√/2 = ·7071068. sin COS tan 45° = == = Cotangent 45o. COS 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 R for the sine of 30°, and R-R2R√3, for the sine of 60°. When R = 1, these become Thirdly, for the sines of 15° and 75°, the former arc is the half of 30°, and the latter is the compliment of that half arc. Hence, substituting 1 for R and √3, for cos A, in the expression sina = ± ± √2R2 ± 2R COS A it becomes sin 15° 플 2√32588190. (equa. XII) Hence, sin 75° = cos 15°/1-(2-√3) = {√2 + √3 = |