AD. Draw BD cutting CE in I. Draw IK perp. to AE, and IM, BL perp. to DH. Because the arc BD is bisected in C, EC will be perp. to the chord BD, and therefore will bisect it in I (3.3); therefore BI will be the sine, and EI the cosine of BC or CD. Since BD is bisected in I, and IM is parallel to BL, LD is also bisected in M (2. 6). Now BF=HL, therefore BF+DH= DH+HL=DL+2LH=2 LM+2 LH=2MH=2IK; therefore IK (BF+ DH) = half the sum of the sines of the extreme arcs AB, AD. By similar triangles CGE, IKE, CE : EI :: CG : IK, that is, R: cos. BC :: s. AC : 1 (s. AB+s. AD). 114. Cor. 1. Hence R x (s. AB+s. AD) =2 s. AC x cos. BC, that is, the rectangle under radius and the sum of the sines of two arcs is equal to twice the rectangle under the sine of half the sum and the cosine of half the difference of those arcs. 115. Cor. 2. If the point B coincide with A, then R: cos. BC :: s. BC: s. BD, that is, radius is to the cosine of any arc, as the sine of the arc is to half the sine of twice the arc. Let A denote any arc or angle, then sine A x cos. A= Rx sine 2 A. If R=1, then s. 2 A=s. A x cos. A, or s. 2 A2 s. A x cos. A. 116. The preceding theorem is one of four which, when arithmetically expressed as follows, are frequently used in the application of trigonometry to the solution of the more difficult problems in the higher parts of mathematics. If in the figure the arc ACA, BC= B, and radius EC= 1, then AD=A+ B, and AB=A-B; therefore by the proposition, 1 cos. B:: s. A: s. (A + B) + 1 6. (A — B), therefore s. A x cos. B = s. (A+B) + 1⁄2 s. (A — B). 117. Because BF, IK, DH are parallel, the straight lines BD, FH are cut proportionally (2.6); therefore FH, the difference of the straight lines FE, HE is bisected in K; therefore, as was shown in the proposition, KE is half the sum of FE and HE, that is, of the cosines of the arcs AB and AD. But, by the similar triangles EGC, EKI, EC: EI :: EG : EK, that is, R: cos. BC :: cos. AC: cos. AD + cos. AB, or, 1: cos. B:: cos. A : 1⁄2 cos. (A + B) + 1⁄2 cos. (A — B), that is, the radius: cos. of the common difference :: cos. of the middle arc half the sum of the cosines of the extreme arcs. Hence cos. A x cos. B = cos. (A + B) + cos. (A—B). 118. Again, the triangles IDM, CEG, having the angles DMI, EGC right, and the angles DIM, EIK or ECG, equal, because the angle EIM is the complement of each, are equiangular; therefore EC: CG:: DI IM. Now IM is half thedifference between the cosines FE and EH; therefore R: s. AC: s. BC: or, 1 s. A :: s. B: cos. (A — B) — 1⁄2 cos. (A+B), therefore s. A x s. B= cos. (A-B) cos. (A+B). 1⁄2 119. In the same triangles ECG, DIM, EC: EG::ID: DM. Now DM is half the difference of the sines DH and BF; therefore R cos. AC :: s. BC: s. AD—1 s. AB, or, 1 : cos. A :: s. B : 1 s. (A + B) — 1 s. (A therefore cos. A × s. B ≈ 1 s. (A+B) -B), 120. Hence if A, B denote any two arcs, and the radius = 1, then I. S. A x cos. B = 1 s. (A + B) + 1⁄2 s. (A —B). 121. Cor. If A = 30°, then s. A=R (45); therefore theorem I becomes cos. B=s. (30°+B) + s. (30-B), therefore s. (30°+ B)=cos. B-s. (30°-B). If A60°, then cos. AR (45); therefore theorem IV becomes s. Bs. (60° + B)—s. (60°-B), therefore s. (60° + B) = s. (60° — B)+s. B. 122. From these four theorems other four are deduced. By adding the first and fourth theorems, s. Ax cos. B+ cos. A x s. B=s. (A+B). By subtracting the fourth theorem from the first, s. A x cos. B cos. A x s. B =S. (A-B). By adding the second and third theorems, cos. A x cos. B+s. A x s. B = cos. (A — B). By subtracting the third theorem from the second, cos. A x cos. B—s. A x s. B=cos. (A+B). The last four theorems are presented in one view, as follows. I. S. (A+B)=s. A x cos. B+ cos. A xs. B. II. S. (A-B)= s. A x cos. B -cos. A x s. B. III. Cos. (A+B) = cos. A x cos. B. -s. Axs. B. 1 123. Since s. A x cos. B = s. (A+B)+1 s. (A—B)(120), if A+B=S, and A-BD, then A = (S+ D) (57), and B= (S-D); therefore, s. (S+D) x cos. (SD) = 1 s. S+ 1 s. D. Now, because S and D denote any two arcs, to preserve the former notation, they may be called A and B, which will also express any arcs whatever. Thus, s. (A + B) x cos. (AB) = s. A+ s. B, In the same manner is deduced 2 cos. (A + B) × cos. } (A — B)= cos. B+ cos. A, from theor. II, art. 120, 2 s. 1 (A+B) × s. 1 (A-B)= cos. B—cos. A, from th. III, 2 cos. (A+B) xs. (A-B) = s. As. B, from th. IV. In all these theorems the arc B is supposed less than A. 124. Theorems of the same kind with respect to the tangents of arcs may be deduced from the preceding theorems. Because the tangent of any arc is equal to the sine cosine s. (A+B) (41), tan. (A + B) = cos. (A+B) But s. (A+B)=s. A x cos. B+ cos. A x s. B, and cos. (A+B) = cos. A x cos. B- s. A xs. B; s. A x cos. B+ cos. A x s. B cos. A x cos. B-s. A x s. B Divide both the numerator and denominator of this fraction by cos. A x cos. B, and it becomes s. A s. B + cos. A by substitution. 125. By art. 123, s. A+s. B=2 s. 1⁄2 (A+B) × cos. 1⁄2 (A—B), and s. A- s. B=2 cos. (A + B) × s. 1⁄2 (A — B), s. A + s. B s. (A+B) x cos. (A — B) s. A therefore tan. (A+B) 1 tan. — (A — B) ' s. B = cos. 1 (A + B) × s. 1 (A — B) = Hence s. A + s. B: s. As. B :: tan. 1 (A + B) : tan. (AB), that is, the sum of the sines of any two arcs : difference of the sines :: tangent of half the sum of the arcs : tangent of half their difference. If A= 90°, and B = 2x, then R+s. 2x: R s. 2 x :: tan. (45°+x) : tan. (45° — x). cos. A = cos. (A+B) x cos. (A-B)_cot. (A + B). s. (A + B) × s. (A — B) - tan. (AB) Hence cos. B+ cos. A : cos. B -cos. A :: cot. (A + B) : tan. (A — B). (A+B) × cos. 1⁄2 (A — B) 126. By art. 122 sine (A+B): s. (AB) :: s. A x cos. B+ cos. A x s. B: s.. A x cos. B the last two terms by cos. A x cos. B) S. B cos. A xs. B:: (by dividing : cos. A cos. B cos. A :: tan. A+ tan. B: cos. B :: cot. B+cot. A cot. B -cot. A. Again, cos. (A+B) : cos. (A — B) :: cos. A x cos. B→ s. Ax s. B: cos. A x cos. B+s. Axs. B:: (by dividing by 127. In all the preceding theorems the radius R is supposed equal to unity, because in this manner the propositions are most concisely expressed, and are also most readily applied to trigonometrical calculation. But if it be required to enunciate any of them geometrically, the multiplier R, which has disappeared by being supposed = 1, must be restored; and it will always be evident from inspection in what terms the multiplier R is wanted. Thus, in theorem I, art. 120, 2 s. A x cos. B=s. (A+B)+s. (A-B) is a true proposition, taken arithmetically; but if taken geometrically it is absurd, unless we supply the radius as a multiplier of the terms on the right hand of the sign of equality. It then becomes 2 s. A x cos. B =RXs. (A+B) + Rxs. (A-B), or, twice the rectangle under the sine of A and the cos. of B is equal to the rectangle under radius and the sum of the sines of A+B and A-B. In general, the number of linear multipliers, that is, of lines whose numerical values are multiplied together, must be the same in every term, otherwise we shall compare unlike magnitudes with one another, which is absurd. The propositions in this section are useful in many of the higher branches of the mathematics, and are the foundation of what is called the Arithmetic of Sines. Almost every branch of the mathematical sciences has been simplified and extended by the application of this calculus; but it has chiefly facilitated the investigations in physical astronomy. III. Construction of Trigonometrical Tables.* 128. In all the calculations performed by the preceding rules, tables of sines, tangents, &c. are necessarily employed. A trigonometrical canon is a table exhibiting the length of the sine, tangent, &c. to every degree and minute of the quadrant, from 1' to 90°, the radius being supposed equal to unity, and to be divided into 10000000 or more decimal parts. In constructing the tables the first thing to be done, is to compute the sine and cosine of an arc of 1', or the sine and cosine of the least arc in the tables. *This subject is treated fully and clearly in Horsley's Mathematics, vol. III, p. 120; also in Legendre's Elements de 'Geometrie, p. 358. |