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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
tan
82 44 06

A2n
0
3
63

4"-1
A3 A5

A2n+1 .. tan 0 > +

3 15 4" 1 and à fortiori tan 6 > sum of any number of these terms.

1>

+

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+

15

+

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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.

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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.,

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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.,

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(*

An
-
n

;

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On

are in ascending order of

n

n

Since (sin"."(tan

D)(9). ... dividing by (sin ); () (100 %)

1, (sin d",

n

n

0 sec

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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

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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)
Thus (sec 6)* +0" = [(1 + tan 6) cote]tano 0+e".
21

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.
If m= 2, the limit = elæ?.
If m> 2, the limit =

an
The particular case

when n is infinite, may be treated directly as follows:

- cosec?

- In sin?

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(cos".

n

n

(cos 9)* = [(1 - sino 9)

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=

n

n

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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
2
n (n − 1) (n 2) (n − 3)

cos”–4 a sino a
4

a

a

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tano a

1

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na

n4

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n

1

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Now let n increase indefinitely while 0 remains constant and finite so that a decreases indefinitely.

a

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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.

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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.

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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.

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Hence each of the above brackets is positive; .. sin 0 <0; but > 0 103 ; but <0-383 + 12705; and so on.

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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.

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