is the sum of the arcs BO and BA, and AO is their difference; BE is the half sum, and OE the half difference; but EK and EI are the tangents of the arcs BE and OE. PH being bisected in G, ON will be bisected in м; now since AO is bisected in n, and on in м, nм will be the half of AN, therefore AN AO::NM: no and AN AO::AH: AP per similar triangles. Hence nм: no::AH: AP but nм: no::EK: EI per similar triangles; therefore AH: AP::EK: EI viz. sine AB+ sine OB : sine AB-sine OB:: (Q) Hence, if the two arcs BA and Bo be represented by a and B, we shall have, sine a+sine B: sine a-sine B::tang (A+B): fang (A—B), or, sine A+ sine B_tang (A+B) (A-B) PROPOSITION XIV. (Plate I. Fig. 2.) (R) The sum of the cosines of two arcs, is to their difference; as the co-tangent 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 parallel to AG, then because ao is bisected in n, GD (PO) will be bisected in g, viz. GQ=QD. Hence 2cQ=CD+CG the sum of the cosines and 2GQ CD-CG the difference of the cosines. nt (=cq): ni (=GQ)::nr : na or 2cq: 260::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, is the tangent of the arc OE. Consequently CD+CG: CD-CG::cot arc BE tang arc OE. Q. E. D. (S) Let the two arcs BA and Bo be represented by A and B, COS A+COS B cot (A+B) then COS B-COS A PROPOSITION XV. = (T) 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 dif ference. The same construction remaining as in Props. XIII. and xiv. Because AO is bisected in n, IT will be bisected in E, Wherefore TK is equal to the sum of the tangents, And since the arc AO is bisected in E, the arc BA is equal to the sum of the arcs BE and OE, and Bo is equal to their difference; and because of the parallel lines TK, AM; TK: IK::AM: Oм, and DO, therefore TK IK: AG: OD. (U) If the two arcs BE and OE be represented by a and в then tang A + tang B__sine (A+B) tang A sine (B-A) (W) 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 сB, and no parallel to OD; then the triangles COD, cno, and ani, are equiangular and similar. The Ancing, being each of them a right angle, and if the common inc be taken away, there will remain the ▲ ani = 2 cng. Then if the arc Bo be represented by ▲, and the arc ao by B, we shall have .co: cn::CD co, viz. rad: cos B::COS A: CQ= : CO: OD:: An ni÷GQ, viz. rad sine A:: sine sine A. sine B rad COS B.COS A & rad B: GO= But co-GO-CG the cosine of the arc BA,COS (A+B) Q.E.D. therefore =COS (A+B) that is (cos A. COS B)-(sine A. sine B)=cos (A+B). rad (X) The same construction remaining, we shall have COS B. Sine A co: cn::OD: no, viz. rad : cos B¦¦sine a: no⇒ I rad COSA.sine B co CD::an: ai, viz. rad : cos a sine B: Ai = rad But no+Ai AG the sine of the arc BA,=sine (A+B). therefore rad + rad = sine (A+B) that is (cos B. sine A) + (cos A. sine B) = sine (A+B). rad. PROPOSITION XVII. (Plate I. Fig. 3.) (Y) 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 between these arcs. Let BO and OA be the two arcs, make ORAO, 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 Az is the half of AP, that is ai=ip. co+ni=cQ+GQ=CQ+OH=CH. But by the preceding proCOS B.COS A sine A. sine B position, co+GQ= rad + therefore #cos (A→B), hence (COS B.COS A)+(sine A. sine B)=COS (A—B). rad. Q.E.D. (Z) no-Ai=Gi_P=PG=RH. But no-Ai= COS B sine A COS A. sine B ; therefore (cos B. Sine A)-(cos A. sineB) = sine (A—B). rad. PROPOSITION XVIII. (Plate I. Fig. 3.) (A) The square of the radius, is to the square of the radius diminished by the rectangle of the tangents of two arcs; as the tangent of the sum of these arcs, is to the sum of their tangents. The same construction remaining, let BO and Ao be the arcs. Draw the tangent вt and co and CA to meet it in T and 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 conclusion of the problem. The triangles CAI and CET are equiangular and similar; also the triangles tbc 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. 11 tr (=tB-TB): te (=tc-Ec)::tc: tB. or, tc × EC=tc2 — tв2+tв X TB or tB-TB: AI :: CB2+tB X TB: CB2 therefore tв X CB2-TB X CB2=AI X CB2 + AI × TB X TB Q. E. D. or, tв x CB2-AI X tв × TB≈AI+TB X CB hence CB2: CB2-AI X TB::tB: A1+BT. (B) 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 between these arcs, is to the difference between their tangents. This is proved in the last proportion but one, by inversion; in the above investigation. (C) If the two arcs BO and AO be represented by A and B, tang (A+B) rad-(tang A. tang B) tang A+tang B rad2 rad2+(tang A. tang B) tang (A-B) tang A-tang B. (D) 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 shewn, (E) By adding the first and third equations together, and dividing by 2. 1. Sine A. cos B = sine (A+B)+ sine (A−B)* * Éléments de Géométrie, par A. M. Legendre, sixième edition, p. 348. et seq. By subtracting the third equation from the first, and dividing by 2. 2. COS A. sine B rad = sine (A+B)-sine (A—B) By adding the second and third equations together, and dividing by 2. By subtracting the second equation from the third, and dividing by 2. 4. sinė A. sine B rad = 1⁄2 COS (A—B) — COS (A+B) ́ P (F) If in the formulæ (E. 115.) there be substituted p for A+B, and Q for A-B, then A= (P+Q), B=} (P−2), and we •sine (P-Q). cos (P+Q) 2 cos (P+2).cos (P-2) rad 1. we shall obtain by division and reduction. sine P+sine sine (P+Q).cos (P-Q)_ tang (P+Q) 'sine P-sine Qcos (P+Q).sine sine P+sine (P+Q) tang (P+Q) 2. = (P+Q) rad sine P+ sine Q cos 3. = cos Q+cos Psine Q sine (P-2)_tang (P-Q) 4. 5. 6. 7. = cot (P+Q) cot cos Q-cos Psine COS Q COS P sine (P+Q) cot / (P+Q) (P+Q). cos (P-Q)_cot (P+Q) sine (P + 2) 2 sine (P+2).cos (P+Q) cos (P+Q) sine P+sine Q2 sine (P+Q). cos (P-Q) cos (P-Q) |