0 (1 + ) + Increasing indefinitely the number of these inequalities, adding them, and observing that the term on the right involving A2n is A2 n 1 1 02n 1 A21 4" 42n 4" '1-1/4" 4"-1' tan 0/2 and that the limit of on the left is 1, we have 0/2 A2n 4"-1 A2n+1 .. tan 0 > + 3 15 4" 1 and à fortiori tan 6 > sum of any number of these terms. 1> + + + + 15 + + + +... 497. The results in these last three articles are nearer approximations to the values of the ratios than the corresponding results in Arts. 485, 486, 487. Since sec = 1 when n is indefinitely large, we have here n to evaluate a limit of the form 1"—the index being a function of the base. See Art. 468. Now, if x is any quantity greater than 1, the following three quantities are in ascending order of magnitude, viz., For if the last quantity is expanded by the binomial theorem, 0 its first two terms are 1 + tano and the remaining terms are + n all positive. Raising each to the power in the following three are in ascending order, viz., (* An ; On are in ascending order of n n Since (sin"."(tan D)(9). ... dividing by (sin ); () (100 %) 1, (sin d", n n 0 sec n magnitude, 61 0/n ( sinn 0/ are in ascending order of magnitude, On tan 0/n are in ascending order of magnitude. and dividing by (tan)"; (cosa)", (.). 1 Hence, by the last article, the limits of (sinon)" and of (can'on)" and of their reciprocals are each unity. n is e. n m 500. The above limits—and others of the form 1°—may be found from the known theorem that the limit of (1 + (1+2) 0. Hence, when A= (), the expression in the square brackets becomes equal to e, and the limit of (sec 6) **+g is as follows: If m <2, the limit = e = 1. an when n is infinite, may be treated directly as follows: - cosec? - In sin? (cos". n n (cos 9)* = [(1 - sino 9) = n n n-2 COS 501. To expand cos 8 and sin 0 in powers of e. We have n (n − 1) Cos na = cos” a a sino a cos”–4 a sino a a a tano a 1 | 4 na n4 n 1 Now let n increase indefinitely while 0 remains constant and finite so that a decreases indefinitely. a reaches its limit 1 independently of the number of them, therefore the limit of their product is 1. Similarly since the diminution of a in is inde (tama)" (tama) pendent of the index r, therefore the limit of is 1. Thus A2 04 46 cos 6 =1 + 2 4 16 Exactly in the same manner, we may prove that A3 05 A7 sin A=0 13 5 17 + t... 502. The student should observe that the expansions of cos 6 and sin 0 in powers of Q are obtained from those of cos no and sin no in powers of cos Q and sin 0 in the same way that the expansion of eix is obtained from the binomial theorem. And hence that The terms of cos 0 and of sin @ are taken alternately but with alternate change of sign from the expansion of e. 503. The above expansions of sin 0 and cos 0 hold for all finite values of 0 whatever. For the ratio of any term to the 02 preceding is - : which, when n is large enough, is clearly n(n - 1) less than a quantity less than 1. Now if we increase 0 by 21, the values of cos 0 and of sin 0 are unaltered; hence the values of these series are unaltered by adding 27 to 0. 504. All the theorems with respect to the sine and cosine that have been investigated from the definition by means of an angie, may be proved from the above series. A few cases may be worked out. 505. Since every index in sin 0 is odd, .. sin (-0) = - sin 0. Since every index in cos 0 is even, .. cos (- 6) = cos 0. 506. If @ measures an acute angle, 02 < 472; ., à fortiori A< 6. Hence each of the above brackets is positive; .. sin 0 <0; but > 0 – 103 ; but <0-383 + 12705; and so on. Hence, since the series beginning at any term has the same sign as that term, the difference between sin @ and any number of its first terms is less than the first term omitted. Similar propositions hold with respect to the cosine. |