Hence nм NO::AH: AP; but nм: no::EK: EI per similar triangles; therefore AH: AP::EK: EI; or sin AB + sin OB: sin AB-sin OB:: tan (232) Hence, if the two arcs BA by A and B, we shall have (BA+BO): tan (BA-BO). and Bо be represented sin A+ sin B sin A-sin B:: tan sin A+ sin B tan (A+B) or, = sin A-sin B tan (A—B)' (A+B): tan (A—B). PROPOSITION XIV. (Plate I. Fig. 2.) (233) The sum of the cosines of two arcs is to their difference, as the cotangent of half the sum of those arcs is to the tangent of half their difference. Let BA and BO be the two arcs, as in the preceding proposition, BE their half sum, and OE their half difference. Draw no and cs parallel to AG, and nit to BC; also produce CA to T, and na, ET to r, s. Then because AO is bisected in n, GD (=PO) will be bisected in Q, viz. GQ = QD. Hence 2cQ=CD+CG the sum of the cosines, and 2GQ=CD-CG the difference of the cosines. Now nt (co) ni (=GQ): : nr : na by similar triangles, or 2cQ: 2GQ::nr: na But nr: na:: ES: ET therefore 2cQ: 2GQ:: ES: ET, where Es is the tangent of the complement of the arc BE, or its cotangent, and ET EI the tangent of the arc OE. ..COS BO+COS BA: COS BO-COS BA :: cot (BA+BO): tan (BA-BO). (234) Let the two arcs BA and Bo be represented by A and B, COS B+COS A cot (A+B) then PROPOSITION XV. (235) The sum of the tangents of two arcs is to their difference, as the sine of the sum of the arcs is to the sine of their difference. Let BE and OE be the two arcs, the construction remaining as in Props. XIII. and XIV. Because AO is bisected in n, IT will be bisected in E, so that TK is equal to the sum of the tangents EI and EK, and IK to their difference. Since the arc AO is bisected in E, the arc BA is equal to the sum of the arcs BE and OE, and Bo to their difference; also AG and OD are the sines of the arcs BA and BO. The lines TK and AM being parallel, we have (236) If the two arcs BE and OE be represented by a and B sin (A+B) tan B sin (A-B) (237) The difference between the rectangle of the sines and the rectangle of the cosines of two arcs, is equal to the rectangle of the radius and the cosine of the sum of these arcs. Let BO and OA be the two arcs, OD and an their sines, CD and cn their cosines; then AG will be the sine of the sum of the arcs, and CG will be the cosine. Draw ni parallel to CB, and no parallel to OD; then the triangles COD, cnq, and ani, are equiangular and similar; for the ▲ Anc= ▲ inq, are each right angles, and if the common Linc be taken away, there will remain the ▲ ani = ▲ cnq. Then if the arc Bo be represented by a, and the arc ao by B, we shall have co: cn:: CD: cq, viz. rad: cos B:: CÙS A: CQ= : COS A. COS B rad CO: OD: AN : ni = GQ, viz. rad sin A:: sin B: GQ = sin A. sin B rad But CQ-GQ=CG (the cosine of the arc BA)=cos (A+B) sin A sin B rad = COS (A+B) that is cos a. cOS B-sin A. sin в rad. cos (A+B). (238) The same construction remaining, we shall have co: cn::OD: nq, viz. rad : cos B::sin A : NQ= co: CD:: an: ai, viz. rad: cos a:: sin B: ai= sin A. COS B rad COS A. sin B rad But no+AiAG (the sine of the arc BA)=sin (A+B) sin A. cos B COS A. sin B (239) The sum of the rectangle of the sines and the rectangle of the cosines of two arcs, is equal to the rectangle of the radius and the cosine of the difference of these arcs. Let BO and OA be the two arcs, make OR=AO, then the chord AR will be bisected in n, and BR will be the difference between the arcs BO and AO; through R draw RP parallel to GB, and RH parallel to AG, then will AG be the sine of the sum of the arcs BO and AO, and RH the sine of their difference. By similar triangles an: AR:: Ai: AP; but an is the half of AR, therefore ai is the half of AP, that is, ai=ip. Now CQ+GQ=CQ+QH=CH=COS (A-B); and from the investigation of the preceding proposition, we have CQ+GQ= COS A. COS B sin A. sin B + rad cos (AB), cos A. cos B+sin a. sin в=rad. cos (A—B). (240) Again, no—ai =Gi—pi=PG=RH=Sin (A—B); and it may be shown, as in art. (238) that rad rad sin A . COS B-COS A sin Brad. sin (A—B). PROPOSITION XVIII. (Plate I. Fig. 3.) (241) The square of the radius is to the square of the radius diminished by the rectangle of the tangents of two arcs, as the tangents of the sum of these arcs is to the sum of their tangents. The same construction remaining, let BO and AO be the Draw the tangent вt, and produce co, CA to meet it in T, t; through a draw AI perpendicular to ct, and through T draw ET parallel to AI: when the arc ao is greater than the arc BO, the line ET will fall within the circle, but that will not affect the investigation. arcs. The triangles CAI and CET are equiangular and similar; also the triangles tвc and tET are equiangular and similar; for the triangles CBT and tET are right angled at B and E, and have the angle at t common. Hence tr (tB-TB): tE (=tc-Ec): tc: tb, or tв2-tB. TB=tc2-tc. EC, .. tc. Ec tc2-tв2 + tB. TB=CB2+tB. TB. By similar tr: ET:: tc: CB triangles JET AI :: EC: AC=CB Hence tT: AI:: tc. EC: CB2 or tB-TB : AI :: CB2+tB. TB: CB2 .. tB. CB2-TB. CB2=AI. CB2 + AI . tB. TB, hence CB2: CB2-AI. TB::tB: AI+TB. (242) The square of the radius is to the square of the radius increased by the rectangle of the tangents of two arcs, as the tangent of the difference of these arcs is to the difference of their tangents. Let BA and Bo be the two arcs; then the last analogy but one in the investigation of preceding article is the property here enunciated. (243) If two arcs be represented by a and b, rad 2 tan Atan B tan (A-B) and rad2+tan A tan B tan A- - tan B (244) From the propositions already given, a great variety of formulæ for the sums and differences of arcs may be deduced, some of the most important of which will here be pointed out. By Props. xvI. and XVII. it is shown, (245) By adding the first and third equations together, and dividing by 2, sin A. cos B sin (A + B) + sin (A-B) rad (5) By subtracting the third equation from the first, and dividing by 2, By adding the second and fourth equations together, and dividing by 2, By subtracting the second equation from the fourth, and dividing by 2, (246) If in these formulæ there be substituted P for A+B, and Q for A-B, then A= (P+Q), B= (P-Q), and we shall obtain 2 COS Q-COS P= sin (P+Q). sin † (P—Q) . . (12) rad sin P+ sin Q_sin † (P+Q) . cos † (P—Q) __ tan § (P+Q) =. sin P-sin Q cos (P+Q). sin (P-Q) tan 3 (P—Q) sin(P-Q) sin P-sin Q_sin COS PCOS Q cos (P-Q) = (13) rad (16) rad (P+Q) = (17) rad (P-Q) __cot (P-Q) tan 1⁄2 (P+Q) (P+Q) cos (P+Q) (P-Q) cos (P-Q) (20) sin P-sin Q COS Q- -COS P COS P+ cos Q cot (P+Q) cot = (P+Q) __ sin § (P+Q) sin (P+Q) ___ 2 sin 1 (P+Q). cos cos2 Arad2 + rad. cos 2 A |