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sino (

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If then in – 1 is even, writing na = 0 and = $, we have

2λπ
?
φ

ol cos 6 – cos $ = 27-1 P = ="" sin - sin?

2λπ cosh 0 - cos $= 2n-1 p1={(n– 2)

8 = ((sin )

+ sinh? Here put 0 = 0. Thus

2λπ. 1 - cos $ = 2n-1 p1=1(n=2) sin

) Divide each of the above two equations by this last. Thus

φ cos o = 2

1 2

sino (2 + )/n/ φ cosh 8 – cos $ = 2 sin ( 1

2

sin? o/n Here make n infinite. Thus cos 6 - cos = 2

φ

2 cosh 0 cos •

A

sino 0/n) (1

0 sino on/

cos o

sinh )(1+

*)n)... sin? (27 2 *)n).

sino On
'
sinh
?

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2 sing (1-0}{-(**)}{-(**)}... 올

) 7 ${1+(*)} {1+(240}{1+(1.. 4 ()}...

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These products are clearly convergent.

The first formula may be easily remembered by giving to e the values °, 27 + $, 47 +0 &c. which make both sides vanish; and by putting 0 = 0 which gives 1 - cos 0 = 2 sin lo.

524. Many important results may be obtained from the factorisation of cos O and sin 0.

Thus we may equate the coefficients of the several powers of O in the term-expansions of cos 0 and sin 0 with those obtained by multiplying out the factor expressions.

Or we may equate coefficients after taking the logarithms of the two expressions for cos 0 or sin 0. Thus we have, from the two expressions for sin 0,

82 84 3 15

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+
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Equating the coefficients of 6%,
1 1 1 1

72

ad inf.
12 + 22
32 42

6' 525. The method of the following article is of very general application, the process involved being in reality merely to differentiate both sides of an identity.

526. To find series for cot 6 and for tan 6.
Resolving each factor of sin into two factors, we may write

0
A

sin 0 = 0P (1+0),

п

where P denotes the product of all values obtained by giving to r every positive and negative integral value, excluding zero.

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log sin 0 = log 0 + $ log (1+ Now for a write 0+h;

+ ).
log sin (0 + h) = log (6 + h) + £ log (1 + $#%).

O

..

h) =

=

Subtract the identity in 0 from the corresponding identity in 0+h; and expand the remainder in powers of h.

On the left we have log sin (0+ h) – log sin 0 = log (cos h + cot & sin h)

= log (1 + h cot 6 - tk ...)

=h cot 0 }22 - Ž (h cot 6...)? + On the right we have

h h 1 ha log (6 + h) – log 6 = log (1+

72 72+ ... 0

h ( 1

(1+

h 1 ha
2 (r + 0)

+
Hence equating the coefficients of h in the new identity,

10 log (1+0+1) - log (1+)-log16)

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0)2+

1 cot A= a

9

TTT+

where denotes the sum of all values obtained by giving to r every positive and negative integral value including zero.

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In the same way from the identity

0 cos 0 = P(1+

(2r + 1). it follows that

1 – tan 0 =

(2r + 1) 17+ A' where, as above, r has every value including zero. In these expansions of cot 0 and tan 6, the terms having equal and opposite coefficients of a should be added so as to form one term. [Otherwise the series assumes the indeterminate form -0.] Thus

1

20
cot @
A 02 – 7272,

20
tan 0 =

{(2r + 1) {} – 62

+ Σ"

§ 4. CIRCULAR MEASURE IN TERMS OF RATIOS.

527. To express o in terms of tan 0. We have shown in Art. 359 that for all positive integral

values of n,

=

n (n-1) (n - 2)

) n cos”-1 sin 0

COSN—3 A sin3 0 + ... = sin no.

1.2.3 Divide by n cos” 0. Then

(n-1)(n-2) (n-1)(n-2)(n-3)(n-4) tan o tanA +

tan5 0 - ... 1.2.3

1.2.3.4.5 sin no

n cos” 0 If now n is not a positive integer the series on the left will be endless. If moreover tan 0 is arithmetically equal to or less than 1, the series will be convergent as in the binomial theorem.

Now in this case the above equation—as will be shown in Chap. XXI.-still holds, provided that, on the right hand, we make 0 represent the positive or negative acute angle whose tangent = tan 0; and give to cos” O its positive value.

= 0.

n cos”

With this understanding, then, we may make n as small a fraction as we please. Then, in the limit,

sin no sin no 1
= .

no (cos 6)"
Hence from the above equation,

If tan 0 = 1 and 0 is acute; i.e. if lies between - 1 and + ,

tan 0 – } tan3 + } tano 0 - ... = 0.
This is called Gregory's series.
528. The geometrical illustration of Gregory's series.

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Let AOB be any angle not greater than half a right-angle; AB the arc opposite AOB; AT, the tangent at A.

Draw the lines OBT1, T2T2, T,T3, T2T4,... each perpendicular to the preceding to meet AT, and OA alternately. Then

arc AB = AT,- 14T, + AT, - AT,+ ... For, if A0= 1, AB= circular measure of AB= 0 say.

And AT, = tan 0; AT, = tano 0; AT, = tan; and so on, by Euclid VI. 8.

3

3

5

7

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529. To

express o in terms of sin 0. We have shown in Art. 375, that for all odd integral values

of no

n

na – 12 (na – 12) (no – 32)

sin no
sin
sin3 0 +

sin e 3 If now n is not an odd integer the series on the left will be endless. It is always convergent, because sin 6 is arithmetically equal to or less than 1.

Now the above equation—as will be shown in Chap. XXI.still holds provided that, on the right hand, we make 8 represent the positive or negative acute angle whose sine = sin 0.

Thus we may make n as small a fraction as we please. Then in the limit, if 0 is acute,

1 sin0 1.3 sin 0 1.3.5 sin? 0 sin 0 +

= 0. 2 3 2.4 5

6 7

+

+

+

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530. If x is any quantity arithmetically not greater than 1, in Art. 527, we may write tan 6 = x, and in Art. 529, sin 0 = x. Thus

2 – 1 203 + 1.200 – 72c? + E circular measure of the acute angle whose tangent is equal to a. 1 203 1.3 205 1.3.5 2

circular measure of 2.3 2.45 2.4.6 7 the acute angle whose sine is equal to x.

531. Now we have tan (rT+0) = tan 8 where r is any integer. If then is any angle lying between ro - 17 and ra + 17, where r has some assigned integral value,

tan 0 – } tan $ + } tano 0 – ф– т. Again we have cot (rT + - ) = tan 6, where r is any integer. If then ® is any angle lying between ra +11 and ra +17, where r has some assigned integral value,

cot $- {cots $ + } cots $- =rt +31-6. Hence $ may be expanded in powers of tan 6, when $ is in the right or left quadrant, and in powers of cot • when it is in the up or down quadrant.

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