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Perform the same operations on the other two formulas cos A+B=cos A cos B-sin A sin B, and cos (A-B=cos A cos B +sin A sin B, and we shall deduce

Cos A cos B= cos(A+B) + {cos (A-B)

Sin A sin B=cos (A+B) - cos (A+B) These four last formulas are useful when we wish to transform products of sines into simple sines. The four following forms enable us to substitute for the sums or differences of sines, the product of other sines, in order that the logarithmic calculus may apply to them.

+ 17. Let A+B=P, A-B-Q, we shall have A=

PEQ, and

COS

COS

COS

Cos Q-Cos P=2 sin P+Q

P-Q B=

Therefore from the preceding article
2

P+Q P-Q
Sin P+sin sin-

2

2

P-Q P+Q
Sin P-sin Q=2 sin

2

2
P+Q

P-Q
Cos P+cos Q=2 cos

2

2
P-Q
sin
2

2 18. Suppose that in the two first of the foregoing formula P= 90°, and in the two last that Q=0; and we shall have 1 +sin Q=2 sin (45° + } Q) cos (45°-Q)=2 sin* (45° + {Q) 1-sin Q=2 sin (45o

_Q) cos (45° +10=2 sin? (45°-IQ = 2 cos? (45° +1 Q)= coversed sine of Q

1+cos P-2 cos? £ P

1-cos P=2 sin? IPversed sine of P. 19. Divide the forinulas of No. 17, one by another, and we shall have

P+Q

P-Q
sin
Sin P +sin Q

2

2
tang

Х
Х

2
Sin P- sin Q

sin 2

2 P+Q

tang P-Q

2 cot

P-Q
2
tang,

2
And in like manner,

P-Q Sin P+sin Q

= tang

2 Cos P+cos Q.

2

cos P +cos Q

P-Q.
Sin P+sin Q

sin P-sin Q. P+Q

cot

; Cos Q-cos P

2 cos P+cos Q Cos P + cos Q

P+Q

P_Q
Cos Q-cos P

2

COS

P+Q

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+Q

P+Q

con P;

= tang

P+Q; sin P-sin Q

= cot

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Pg;

=cot

x cot

2

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=

2

cos B

+

cos B

20. In like manner dividing one by another, some of the for. mula of No. 18, we find 1 +sin Q. sin (45° +} Q)_sin (45° +} Q) 1-sin Q

-tang? (45° +*Q? sin (45° - {cos* (45° + Q) 1+cos P

cos P

=cot | P 1-cos P sin? 5 P 1+sin P

sin? (45° +}Q) 1+cos P

cos' } P 1-sin Q_ coversin Q_sin (45°— Q)

? '1-cos Q versin Qsin' IQ

21. Resuming the values of sin (A+B), sin (A-B, cos (A+B), cos (A-B), we deduce from them

sin A cos B sin B cos A Sin '(A+B)_sin A cos B+sin B cos A

+

sin A sin B sin A sin B Sin (A-B) sin A cos B-sin B cos A

sin A cos B sin B cos A

sin A sin B sin A sin B cos A

1

1

t. --sin B

sin A_cot B+cot A tang B'tang A
cos A cot B-cot A 1

1 sin B sin A

tang B tang A tang A +tang B tang A-tang B

+ sin (A+B)_ sin A cos B+sin B cos A _ sin B cos (A-B) cos A cos B+sin A sin B cos A cos B

4.2

sin A sin B cot B +cot A tang A +tang B

1+cot B cot A 1+tang A tang B sin (A—B) sin cos B-sin B cos A cot B-cot A cos (A + B) cos cos B-sin A sin Becot B cot A-1 tang A-tang B 1-tang A tang B COS (A+B) cos A cos B-sin A sin B

cot B_tang A cos (A-B)cos A cos B+sin A sin B cot B+tang A

1-tang A tang B_ cot A-tang B 1+tang A tang Bocot A +tang B sin (A+B)

tang A +tang B cot A +cot B =tang (A+B= cos (A+B)

1-tang A tang B cot A cot B-1

1 Therefore, cot (A+B)=;

1-tang A tang B

tang(A + B)tang A +tang B cot A cot B-1 cot A +cot B

cos B

cos A sin A

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; or tang C_2 tang { C

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we have

sin A'

sin (A-B)

tang A-tang B cot B-cot A == Cos (A-B) = tang (A-B) =

1+tang A tang Bocot B cot A +1 Therefore cot (A-B)=

1+tang A tang B cot B cot A +1

tang A--tang B cot B-cot A 22. Let A=45°, and we shall have

1+tang B_cot B+1 tang (45° +B)=

; 1-tang Bocot B-1

B

1-tang B tang (45°-B=

=cot (45° +B)=

cot B-1 1+tang B

cot B+1 If we make A=B=C we shall have

2 tang A tang 2 A= 1-tango A

1-tang? IC

And cot 2 A_1-tang®

Acot A–tang A, therefore

2 tang A cot C=cot C-4 tang C and cot C=2 cot C+tang 1 c.

1 - cos C

(1 But (12) tang C=V 1 + cos C

-cos C)?_1

-COS C 1- cosC

sin

C 23. Since sec A=

1
-and cosec As-

1
COS A

1

I sec (A+B)=

cos A cos B cos A cos B-sin A sin Basin A sin B

17

cos A cos B sec A sec B

cosec A cosec B 1-tang A tang Bocot A cot B-1 2

sec A sec B sec (A-B)

1+tang A tang B

cosec A cosec B cosec (A+B)=

cot B +cot A

cosec A cosec B cosec (A-B)=

cot B-cot A 24. Let A=B, and we shall have

cosec? A
cosec 2 A.
1 + cot? A cot A +tang

A
2 cot A
2 cot A

2
Therefore cosec A=
cot A+tang. A

But cotA=2 cotA +

2 tang Aby (22) therefore cosec A=cot A+tang ;A cotA+tang A

=cot · 1-cot A, by writing for tang į A its 2 value, cot} A_2 cot A. We have also sec 2 A=

sec? A

1+tango A

1 tang” đ 1tang A (1+tang A)^_2 tang A 1+tang A_2 tang A l-tang* A 1.-tang* A1_tang A 1_tang* A

ܪ

2

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cos A

But
1+tang A
=tang (45° + A,) and

2 tang A

tang 2 A 1-tang A

–tang A Therefore sec 2 A=tang (45° +A)- tang 2 A; and tec A=tang (45+ A-tang A=cot (45°-A)-tang A 1

1 Since sec AS

and cosec A

; we have sec A

sin A tang A cosec A; and substituting all the values of cosec A found above, we shall have sec A=tang A(cst A+tang | A)=tang

į A ) A (cot A +tang A)=1+tang A tang ! Atang A cot Acot A)=tang A cot A-1_tang A =· =

- 1 These formulas may be varied in an infinity of ways by adding, subtracting, dividing them, &c. But it is useless to dwell longer upon so easy a matter.

See the Introduction to the Analysis of lufinities, by EULER.)

tang 5 A

CALCULATION OF THE TABLES OF SINES, BY

SERIES.

The same thing has occurred with respect to 'Tables of Sines as had before taken place with the Tables of Logarithms. The first calculators had already completed their labours, when means were found to simplify them. These means are not the less ingenious on this account, as we may judge from the method proposed by John Bernoulli, in the second volume of his works.

We shall give the analysis of it.

25. If we turn back to the values of tang (A+B) we shall deduce tang (A+B+C)=

tang (A+B) +tang C

1 –tang C tang (A+B) Let then a, b, c be the respective tangents of the arcs A, B, C; and we shall have (by writing for tang (A + B) its value) tang (A+B+C)=

a+b+c-abc

1-ab-ac-bc Similarly, if a, b, c, d, are the respective tangents of the arcs A, B, C, D, we shall have

+-AB+ tang (A+B+C+D)=2+b+c+d—abc-abdacd_bod

1-ab-ac-ad-bc-bd-cd + abcd Whence in general if there be any number of arcs A, B, C, D, &c.; then calling s the sum of their tangents, s'i the products of them two by two, sili their products three by three, we shall have tang (A+B+C+D+&c.) =

- g" ts' -8"* + &c.

-g" +8i-8" + &c. Suppose for a moment that the arcs A, B, C, &c. are all equal , then if we call the number of them ni, and tang A the tangent of

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n. –

any one of them, we shall have (Page 111)s=" n=tangi A, ...

=
-tang' A S”-n. 1-1. n-2. n-3

tang* A,

2

n. n-1. n–2.

2. 3

2. 3. 4

&c.

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+ &c.

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+ &c.

We have therefore in general Tang nA= n. no-l. n--2,

n. n-j. n.-2. n--3. n--4, n tang AtangA+

tango A&c. 2. 3

2. 3. 4. 5 n. n-1

n, n-1. n-2. n-3 1tango A+

tang4 A-&c. 2

2. 3. 4. or, writing for tang A its value sin A

.

the above fraction becomes n Sin A n. n--1. n-2 sin? A

2 3 cos3 A n. n-1 sin? A 1

2 cosA or, multiplying both numerator and denominator by Cos" A, we have finally Tang nA= a cos"-? A sin A

n. nl. n-2

(coso sin' A + &c.

2. 3 n. n-1 cosa Acogn- sin? A+ n. n-1. n--2. n-3

cost sin* A,&c. 2

2. 3. 4 Let N be the numerator of this last quantity, and D its denomia nator; then by actually performing the calculation we shall find N+D2 = cos2n A +n cogan-2 A sin? A++

n. n-1

cos

2 ...... sin A=(cos? A +sin? A)"=1

But since on the one hand N+D2=1, and on the other N

sin? n A Da=tangon A=

; it is clear that N=sin n A, and that

cos? n A D=cos n A. We have therefore in general

n. n-1, nQ sin a A=n cos - A sin A

cos-3 A sind A+

2. 3 n, 0-1. n2. n-3. n-4

cos"

5-6 A sin' A-&c. and 2. 3. 4. 5

n. n-1 cos n A=cos A

cos?-? A sin? A+

n. n-1. n–2. n-3 2

2. 3. 4 cog. A sin 4 A-&c.

Suppose now that the arc A is infinitely small, so that n must be infinite, in order that the arc n A may be of a finite magnitude a, we shall have

1st. Sin ASA, because an infinitely small arc does not sensibly differ from its sine ;

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a

a A

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