2 cos(AB). sine (A+B) and if rad1, sine a+sine в= 2 cos(A-B). sine (A+B). Consequently. If the sine of the mean of three equidifferent arcs be multiplied by twice the cosine of the common difference, and the sine of either of the extreme arcs be deducted from the product, the remainder will be the sine of the other extreme arc; the radius being 1. OF THE SINES, COSINES, TANGENTS, &c. OF THE MULTIPLES OF ARCS." (H) Sine (A+) B = (sine A. cos B)+(sine B. cos A) rad Let B-2A, and rad=1, then we shall have (D.115.) (O. 108.) and cos 2A-2 cos2 A-I (2 cos2A-1)=sine 2A. =(sine 2A. cos A)-sine A, and substituting this in the first equation. Sine 3A (sine 2a. cos a)—sine a+ (sine 2a. cos a). And by making B=3A, and following the same method, Sine 4A (2 cos A. sine 3A)-sine 2A. Hence we obtain (1) + Sine A=sine A Sine 2A=2 cos A. sine A Sine 3A (2 cos A. sine 2A)-sine a OR, (K) By substituting the value of cos A1-sine A, &c. as in the single arcs (M. 104.), sine 2A found above, &c. Sine A sine a. Sine 2A=2 sine A1-sine A * See Euler's Introductio ad Analysin Infinitorum; Traité de Trigonométrie, par M. Cagnoli; Crakelt's translation of Mauduit's Trigonometry; Baron Masere's Trigonometry; Emerson, &c. + Cagnoli, page 27. Legendre, page 358. Crakelt's translation of Mauduit, page 15. &c: (L) Sine 3A-3 sine A-4 sine3 A. Sine 4A (4 sine A-8 sines A)1-sine A Sine 6A (6 sinea—32 sine3 a +32 sine3 A)/1—sine 2A, OR, Sine A1-COS2 A Sine 2A-2 cos A/1-cos2 A Sine 3A (4 cos2 A-1)/1-COS2 A Sine 4A=(8 cos3 A-4 cos A)/1—cos2 A Sine 5A (16 cos4 A-12 cos2 A+1)/1-cos2 A Sine 6A (32 cos5A-32 cos3A+6 cosa)/1—cos2a,&c. (cos B.COS A)-(sine A.Sine B) (M) Cosine (A+B)= rad (D.115.) Now if B-2A, and radius=1 as before, we shall obtain (1— cos 2a).cosa. (P.108.)=cos A-(COSA. cos 2A) therefore Cos 3A=(cos 2A. COSA)-COS A+ (COS A.cos 2A)=(2 cos A.cos 2A) - Cos A. And by making B=3A, and pursuing the same method, we shall find cos 4A-2 cos A.COS 3A-cos 2A. (N)*Cos A=cos A Cos 2A=2 cos A.COS A-1 Cos 3A 2 cos A.cos 2A-COS A Cos 4A 2 cos A.cos 3A-COS 2A Cos 5A=2 cos A.cos 4A-cos 3A Cos 6A 2 cos A.cos 5a-cos 4A, &c. OR, (0) By substituting the value of cos A1-sine2A, cos 2A, found above, &c. +Cos A1- sine2 a Cos 2A-1-2 sine2 A Cos 3A (1-4 sine2 A)√1-sine2 A Cos 4A 8 sinet A 8 sine2 A+1 - Cos 5A (16 sinet A-12 sine2 A+1)/1-sine A (P) Cos A=COS A OR, * Cagnoli, page 27. Legendre, page 358. Cos 2A-2 cos2 A-1 Côs 3A 4 cos3 A-3 cOS A Cos 4A 8 cos1 A-8 cos2 a+1 Cos 5A 16 cos5 A-20 cos3 A+5 COS A Cos 6A 32 coso A-48 cos1 A+18 cos2 A—1, &c. Also if &c. 1 be substituted for cos a (N. 123.) the secant of sec A the multiple of any arc may be obtained. OF THE SINES AND COSINES OF THE POWERS OF ARCS.+ (S) Cos 2A=1–2 sine2 A (Q. 123.) * Cagnoli, page 29. Crakelt's translation of Mauduit, page 34. Therefore 2 sine® A-1-COS 2 A Sine 3 A3 sine ▲-4 sine3 a (K. 123.) Cos 4A 8 sine* A-8 sine* A+1 (O. 123.) The law of continuation is obvious, for the odd powers are expressed in terms of the sines, and the even powers in terms of the cosines of the multiples of A; and the signs are alternately + and -. The numerators of the co-efficients (reckoning from the right hand towards the left), are the co-efficients of a binomial whose power is the same as that of the sine of A; except in the even powers, where the term in which a is not found, has the numerator of its co-efficient only one half of the corresponding co-efficient of the binomial, and the denominators are 2 involved to the power of the sine-1. (T) To deduce formulæ for the successive powers of the cosine of any arc, we must apply P. 124. in the same manner as above. The law of continuation is the same as S. 125, only all the terms here are positive. (U) The sine and cosine of any arc, or the sine and cosine of any multiple of that arc, may also be derived by substituting the imaginary quantities -1 and--1, successively for z, in the exponential expression e2=1+ Where e is the number whose hyperbolic logarithm is 1.* The addition of the two new equations, obtained by substitution, if divided by 2, will give a series expressing the sine; and the subtraction, if divided by 2√−1, will give a series for the cosine of any arc. THE DETERMINATION OF THE VALUE OF THE SINE AND OF THE COSINE, &c. OF ANY ARC, IN TERMS OF THAT ARC. PROPOSITION XXIII. (Plate IV. Fig. 1.) To determine the increment of an arc, in terms of the increment of the sine, tangent, secant, &c.; and thence to deduce several useful formula. Let Am be any arc; Pm its sine, cp its cosine, AT its tangent, CT its secant, &c. Take the arc mo indefinitely small, draw onv parallel to pm, and mn parallel to Ac; also, from the centre c, with the radius CT, describe the arc ST. Then mo is the increment of the arc am; mn is the decrement of the cosine, or the increment of the versed sine; on is the increment of the sine; Tt the increment of the tangent, and st the increment of the secant. The triangles mno, and cpm, are equiangular and similar ; for the arc mo, being extremely small, may be considered as a straight line. Likewise the triangles CAT and Tst are equiangular and similar, for the lines ct and CT are supposed nearly to coincide, so that Ts may be considered as a straight line, the Tst a right angle, and the str=/ CTA. Lastly, the sectors com and CST are similar. Hence we deduce the following proportions: (W) cm: CP::mo: on, hence mo=(cm.on)÷CP (X) cm CT::mo: Ts, or, Ts=(cT.mo)÷cm (Y) CA=cm: CT::TS: Tt, hence mo= (cm2.Tt)÷CT2 (Z) CA AT::TS: st, hence mo=(cm2.st÷AT.CT (A) Pm cm::mn: mo, hence mo=(cm.mn)÷Pm (B) Now it is shewn by writers on fluxions, that the limiting ratio of the cotemporary increments or decrements of any two quantities, will be the ratio of the fluxions of those quan * Éléments de Géométrie par A. M. Legendre, 6th edit. page 354. Vince's Trigonometry, pro. 27. and 28. page 79. and 80. 2d edit. |