PROPOSITION III. PROBLEM. The Sine DE of any Arc D B being giveu, to find DM or B M, the Sine of Half the Arc. DE being given, CE (by the laft Prop.) will be given, and accordingly EB, which is the Difference between the Cofine and Radius. Therefore DE, E B, being given, in the Right-angled Triangle DBE, there will be given D B, whofe Half D M is the Sine of the Arc D L= the Arc B D. PROPOSITION IV. PROBLEM. The Sine B M of the Arc B L being given, to find the Sine of double that Arc. T HE Sine B M being given, there will be given (by Prop. 2.) the Cofine C M. But the Triangles CBM, DBE, are equiangular, because the Angles at E and M are Right Angles, and the Angle at B common: Wherefore (by 4. 6.) we have CB: CM:: (B D, or) 2 B M: DE. Whence, fince the three first Terms of this Analogy are given, the fourth also, which is the Sine of the Arc DB, will be known. Coroll. Hence CB: 2 CM:: BD: 2 DE; that is, the Radius is to double the Cofine of one Half of the Arc D B, as the Subtenfe of the Arc D B is to the Subtenfe of double that Arc. Alfo CB: 2 CM :: (2 BM: 2 DE: :) BM : DE::CB: CM. Wherefore the Sine of an Arc, and the Sine of its Double, being given, the Cofine of the Arc itfelf is given. PRO PROPOSITION V. PROBLEM. The Sines of two Arcs, B D, F D, being given, to find F I, the Sine of the Sum, as likewife EL, the Sine of their Difference. LET the Radius CD be drawn, and then CO is the Cofine of the Arc FD, which accordingly is given, and draw OP thro' O parallel to D K; alfo let OM, GE, be drawn parallel to CB: Then, because the Triangles CDK, COP, CHI, FOH, FOM, are equiangular; in the firft Place CD: DK:: COOP, which, confequently, is known. Alfo, we have CD: CK::FO: FM; and fo, likewise, this shall be known. But becaufe F O EO, then will FM MG=ON; and fo OP+FM FI=Sine of the Sum of the Arcs: And O P-F M; that is, OP-ON-EL Sine of the Difference of the Arcs; which were to be found. Coroll. Because the Differences of the Arcs BE, BD, BF, are equal, the Arc B D shall be an Arithmetical Mean between the Arcs BE, B F. PROPOSITION VI. The fame Things being fuppofed, the Radius is to double the Cofine of the mean Arc, as the Sine of the Difference is to the Difference of the Sines of the Extremes. FOR OR we have CD; CK:: FO: FM; whence, by doubling the Confequents, CD: 2 CK:: FO: (2 FM, or) to FG, which is the Difference of the Sines EL, FI. W. W. D. Coroll. If the Arc BD be 60 Degrees, the Difference of the Sines FI, EL, fhall be equal to the Sine FO, of the Difference. For, in this Cafe, CK is the Sine of 30 Degrees; the double whereof is equal to the Radius; and fo, fince C D=2 C K, we fhall have FQ FO F G. And, confequently if the two Arcs BE, Coroll. 2. Hence, if the Sines of all Arcs diftant from one another by a given Interval, be given, from the Beginning of a Quadrant to 60 Degrees, the other Sines may be found by one Addition only. For the Sire of 61 Degrees the Sine of 59 Degrees + the Sine of 1 Degree; and the Sine of 62 Degrees the Sine of 58 Degrees + the Sine of 2 Degrees: Alfo, the Sine of 63 Degrees the Sine of 57 Degrees the Sine of 3 Degrees, and fo on. = Coroll. 3. If the Sines of all Arcs, from the Beginning of a Quadrant to any Part of a Quadrant, diftant from each other by a given Interval, be given, thence we may find the Sines of all Arcs to the Double of that Part. For Example, Let all the Sines to 15 Degrees be given; then, by the precedent Analogy, all the Sines to 30 Degrees may be found: For the Radius is to double the Cofine of 15 Degrees, as the Sine of 1 Degree is to the Difference of the Sines of 14 Degrees, and 16 Degrees: So, alfo, is the Sine of 3 Degrees to the Difference between the Sines of 12 and 18 Degrees; and fo on continually, until you come to the Sine of 30 Degrees. After the fame manner, as the Radius is to double the Cofine of 30 Degrees, or to double the Sine of 60 Degrees, fo is the Sine of 1 Degree to the Difference of the Sines of 29 and 31 Degrees:: Sine 2 Degrees to the Difference of the Sines of 28 and 32 Degrees :: Sine 3 Degrees, to the Difference of the Sines of 27 and 33 Degrees. But, in this Cafe, the Radius is to double the Cofine of 30 Degrees, as I to 3. See FIG. of the DEFINITIONS. Let BD be an Arc of 30 Degrees: Rad. Tan. Cofine Sine Then, as CB: BG::FD: DE. DO=CB; ergo DE=1; CD: 2 CE: 1:2√√3. 2. E. D. And, And, accordingly, if the Sines of the Distances from the Arc of 30 Degrees, be multiplied by ✔ 3, the Differences of the Sines will be had. So, likewife, may the Sines of the Minutes in the Beginning of the Quadrant be found, by having the Sine, and Cofines of one and two Minutes given. For, as the Radius is to double the Cofine of 2':: Sine 1' Difference of the Sines of 1' and 3:: Sine 2: Difference of the Sines of o' and 4'; that is, to the Sine of 4'. And fo, the Sines of the four first Minutes being given, we may thereby find the Sines of the others to 8', and from thence to 16, and so on. PROPOSITION VII. THEOREM. In fmall Arcs, the Sines and Tangents of the fame Arcs are nearly to one another, in a Ratio of Equality. FOR, becaufe the Triangles CED, CB G, are equiangular, CE:CB: ED: C G. But as the Point E approaches B, E B will vanish in respect of the Arc BD; whence CE will become nearly equal to C B, and fo ED will be alfo nearly equal to BG. If I EB be less than the Part of the Radius, Coroll. Since any Arc is lefs than the Tangent, and greater than its Sine, and the Sine and Tangent of a very finall Arc are nearly equal; it follows, that the Arc fhall be nearly equal to its Sine: And fo, in very fmall Arcs, it hall be, as Arc is to Arc, fo is Sine to Sine. PRO PROPOSITION VIII. To find the Sine of the Arc of one Minute. THE Side of a Hexagon infcribed in a Circle, that is, the Subtenfe of 60 Degrees, is equal to the Radius (by Coroll. 15th of the 4th); and fo the Half of the Radius fhall be the Sine of the Arc of 30 Degrees. Wherefore the Sine of the Arc of 30 Degrees being given, the Sine of the Arc of 15 Degrees may be found (by Prop. 3.) Alfo the Sine of the Arc of 15 Degrees being given (by the fame Prop.) we may have the Sine of 7 Degrees 30 Minutes: So, likewife, can we find the Sine of the Half of this viz. 3 Degrees 45'; and so on, until twelve Bifections being made, we come to an Arc of 523, 443, 034, 45', whofe Cofine is neatly equal to the Radius; in which Cale (as is manifeft from Prop. 7.) Arcs are proportional to their Sines: And fo, as the Are of 522, 443, 04, 453, is to an Arc of one Minute, fo fhall the Sine before found be to the Sine of an Arc of one Minute, which therefore will be given. And when the Sine of one Minute is found, then (by Prop. 2. and 4.) the Sine and Cofine of two Minutes will be had. PROPOSITION IX. THEOREM. If the Angle B A C, being in the Periphery of a Circle, be bifected by the Right Line AD, and if AC be produced until D E=AD meets it in E; then fhall CE=A B. IN N the quadrilateral Figure ABDC (by 22. 3.) the Angles B and DCA are equal to two Right Angles DCE+DC A (by 13. 1.): Whence the Angle B DCE. But, likewife, the Angle E-DAC (by 5. 1.)D A B, and D C D B: Wherefore the Triangles B AD and CED are congruous, and fo C E is equal to A B. W. W. D. PRQ |