Let, therefore, S,nan's + ns3 + ns5, &c.; where s denotes the sine of the arc a, and n, n, n, &c. the successive odd orders of the functions of n. It was shown (Prop. 3. Trig.) that S,n+1.a + S,n — 1.a = 2Cos,a × S,na; whence, by substitution, (n+1+n—1')s + (n+1 + n−1)s3 + (n+1 + n−1)s3 +&c. = 2.√✓ ( 1—s1) S,na = ( 2—s2—¡s*, &c.) (ns + n's3+ns3, &c.) =2n's +(2n−n')s3 + (2n—n—¿n')s5, &c. Now, equating corresponding terms, and rejecting the powers of s, we obtain these general results: It remains to discover hence the several orders of the functions of n. 1. The equation 2n'≤ 2n' contains a mere identical proposition; but other considerations indicate that n must always denote the first term, or that the first function of n is n itself. 2. The equation n+1 + n−1 = 2n n fixes the conditions of the third function of n, which, from the nature of the relation, is obviously imperfect, and wants the second term. Put, therefore, n"" an3 + ßn; and, by substitution, 2an3 + 6xn+28a2an3 + 2ẞn-n. Equating now the corresponding terms, and 6a1, or a = but therefore B. Whence n = — UN a+Bo, and · In3 + !n = — n. 3. Again, in the third equation, n+1 + n−1 = 2n−n—in', substitute nan + ßn3 + 2n, and this compound expression will arise for the conditions of the fifth order of the function of n: 2an3 + (20α +28) n3 + (10α+6,3+2y)n=2an3 + (2ß+ž)n3 (2y-1-4) n. Equate the corresponding terms, and 10x +63 + 2y = 2y-1-4, and B=-=-= Whence, resuming all the terms, S,nans-n. 2.3 n. n2—I n3—9 §3 _ &c. in which the law of continuation is suffi 2.3 4.5 ciently apparent. From the expression for the sine of a multiple arc, may be deduced the series for the sine of any arc, in terms of the arc itself, A and conversely. Let na=A, and therefore a== n ; if n be sup posed indefinitely great, then a must be indefinitely small, and consequently in a ratio of equality to s. Whence, substituting A A for na, and for s in the general expression, there results, But n being indefinitely great, the composite fractions 222 -I &c. are each in effect equal to unit, which forms their ex Again, putting a=A and s=S, suppose n to be indefinitely small, and S,nananA; whence, by substitution, But, if n vanish from all the terms, the series will pass 2.3 I 4.5 into a By a similar investigation, the series for the cosine of an arc is likewise found. A Ασ 2.3.4.5.6 +, &c. These series' are very commodious for the calculation of sines, since they converge with sufficient rapidity when the arc is not a large portion of the quadrant. Though the method explained in the text is on the whole much simpler, yet as the errors of computation are thereby unavoidably accumulated, it would be proper at intervals to calculate certain of the sines by an independent process. The series' now given furnish also various modes for the rectification of the circle. Thus, assuming an arc equal to the radius, cosine .500000. Wherefore (Pr. 2. T.) the sine of the difference of these two arcs is .866025 x .540302.841471 x .500000 = .04718, and consequently, by the series, that interval itself is .0472. Hence the length of the arc of 60° is 1.0472, and the circumference of a circle which has unit for its radius is 3×1.0472=3.1416; an approximation extremely commodious. Note XXXII.-Page 419. The series for the tangent in terms of the arc, is easily derived, by the theory of functions, from the expression of the tangent of 2t the double arc. Since T, 2a = │ =2t2t3+ 2ts + &c. ` I Put ta+Aa3 + Bas + &c. and, by substitution, T,2a = = 2a + (2A + 2) a3 + 2a + 8Aa3 + 32 Ba3 + &c (2B + 6A + 2)as +, &c. Equating, therefore, the corresponding terms, we obtain, 8A = 2A + 2, or A = †, and 32B2B + 6A + 2, or 30B = 4, and B. Whence, in general, T,a= a + ža3 + †a3, &c. Again, revert this series The last series affords the most expeditious mode for the rectification of the circle. Assume an arc a, whose tangent t is one. fifth part of the radius, and (Sch. Prop.6.T.) T,4a== 4t-- 413 120 119 ; consequently (Prop. 5. Trig.) T,4a-45° I -6t2+t+ 239 .004,184,100,418. Wherefore, computing the terms of the series, a=.197,395,559,850, and 4α = .789,582,239,400. In like manner, we find 4a 45°.004,184,076,000, and hence the diffe rence between these values, or .785,398,1634 exhibits the length of the octant; which number, multiplied by 4, gives 3,1415926536 for the circumference of a circle whose diameter is 1. 2 I Note XXXIII.—Page 421. This proposition would also furnish a simple quadrature of the circle. The sine of a semiarc being equal to half the chord, it follows that the ratio of an arc to its chord is compounded of the successive ratios of the radius to the cosines of the continued bisections of half that arc. Assuming therefore the arc of 60°, whose chord is equal to the radius, the logarithm of the ratio of the circumference of a circle to its diameter will be thus computed: exceeds only by 3 in the last place the logarithm of 3,141592654. As the successive terms come to form very nearly a progression that descends by quotients of 4, the third of the last one is, for the reason stated in page 422, considered as equal to the result of the continued addition. Note XXXIV.-Page 425. AN elegant mode of forming the approximate sines corresponding to any division of the quadrant, may be derived from the same |