Add together the expanded expressions for sin. (в + A),

sin. (B-A), that is,

add

to

sin. (B+A) = sin. B. cos. A + cos. B. sin. A, sin. (B-A) = sin. B. cos. A there results sin. (B+A) + sin. (B-A) = 2 whence,

sin. (B+A) = 2 cos. A. sin. B

Thus again, by adding together the expressions and cos. (B-A), we have

cos. B. sin. A; cos. A. sin. B: sin. (в - л). for cos(B+A')

COS. (B+A) + Cos. (B-A) = 2 cos. A. COS. B; whence, cos. (B+A) = 2 COS. A. COS. B - Cos. (B-A). Substituting in these expressions for the sine and cosine of B + A, the successive values A, 2A, 3A, &c, instead of B; the

following series will be produced.

sin. 2A = 2 cos. A. sin. A.

sin. 3A = 2 cos. A. sin. 2A - sin. A.

sin. 4A = 2 cos. A. sin. 3A

sin. 2A.

sin. Na = 2 cos. A. sin. (n-1)A-sin. (n-2)A.

COS. 2A = 2 COS. A. COS, A-Cos. O (=1).

COS. 3A

2 Cos. A.COS. 2A-COS. A.

Cos. 4A = 2 cos. A. Cos. 3A-COS. 2A.

COS. NA = 2 cos. A.cos. (n-1) cos. (n-2)A.

(2.)

(χι.)

Several other expressions for the sines and cosines of multiple arcs, might readily be found: but the above are the most useful and commodious.

2 sin A.coS A

R

20. From the equation sin. 2A = easy, when the sine of an arc is known, to half. For, substituting for cos A its value 2 sin A(R2-Sin? A) there will arise sin 2A =

R

, it will be find that of its (R2 - sin2A), This squared

gives R2 sin2 2A = 4R2 sin2 A 4 sin4 A. Here taking sin a for the unknown quantity, we have a quadratic equation, which solved after the usual manner, gives

2

sin A = ± R2±R/R2- sin2 2A.

If we make 2A = A', then will A = A', and consequently

the last equation becomes

by putting cos a' for its value ✔ R2-- sin2 A', multiplying the quantities under the radical by 4, and dividing the whole second number by 2. Both these expressions for the sine of half an arc or angle will be of use to us as we proceed.

21. If the values of sin (A+B) and sine (A - B), given by

equa. v, be added together, there will result

R

; whence,

sin A. cos B = R Sin (A+B) + R Sin (A - B).. (XIII.)

Also, taking sin (A-B) from sin (A+B), gives

sin (A+B)

2

sin (A-B) = SinB.cos A; whence,

R

sin B. COS AR sin(A+B) - R. sin (A - B).. (XIV.)

When A = B, both equa. XIII and xiv, become

cos a. sin a = R sin 2A (XV.)

22. In like manner, by adding together the primitive ex

pressions for cos (A+B), cos (A - B), there will arise

COS (A+B) + cos (A - B) =

R

COS A. COS BR.Cos(A+B)++R.COS (A-B) (XVI.) And here, when A = B, recollecting that when the arc is nothing the cosine is equal to radius, we shall have

COS AR.COS 2A + R (XVII.)

...

23. Deducting cos (A + B) from cos (A - B), there will

remain

COS (A-B) - COS (A+B) =

2 sin A sin B

; whence,

R

sin A.sinn = R. COS (A - B)-R.COS (A+B) (XVIII.)

When A = B, this formula becomes

sin2 A = R - R. COS 2A... (XIX.).

i

24. Multiplying together the expressions for sin (A + B) and sin (A - B), equa. v, and reducing, there results sin (A + B). sin (A - B) = sin2 A sin2 B.

And, in like manner, multiplying together the values of cos (A + B) and cos (A - B), there is produced

COS (A+B).cos (A - B) = cos2 A

cos2 B.

Here, since sin2 A sin2 B, is equal to (sin A + sin B) х (sin A sin B), that is, to the rectangle of the sum and difference of the sines; it follows, that the first of these equations converted into an analogy, becomes

sin (A - B): sin A-sin B :: sin A + sin B: sin (A + B) (XX.) That is to say, the sine of the difference of any two ares or angles, is to the difference of their sines, as the sum of those sines is to the sine of their sum.

...

If A and B be to each other as n + 1 ton, then the preceding proportion will be converted into sin A: sin (n+1)Asin na :: sin (n+1) + sin na: sin (2n+1)A. (ΧΧΙ.) These two proportions are highly useful in computing a table of sines; as will be shown in the practical examples at the end of this chapter,

25. Let us suppose A + B = A', and A-B = B'; then the half sum and the half difference of these equations will give respectively A =(A'+B'), and B=(A'-B'). Putting these values of a and B, in the expressions of sin A. cos B, sin B. cos A, cos A. cos B, sin A. sin B, obtained in arts. 21, 22, 23, there would arise the following formulæ :

sin(A'+ B').cos (A'B') =

sin (A'B'). cos(A+B) =

Cos(A+B). COS(A-B') =
sin(A+B).sin (A'B') =

R(sin A'+ sin B ́),

R(sin A' - sin B'),

R(COS A'+ COS B'),
R(COS B' - COS A').

Dividing the second of these formule by the first, there will be had

sin(A'B'). COS(A'+ B') sin(A'-B') COS(A'+ B') sin A'-sin B'

factors of the first member of this equation, are tan(A'B', and

R

R

tan(A+B) respectively; so that the equation

manifestly becomes tan(A'B')

sin A'siu B'

tan (A' + B') sin A' + sin B

This equation is readily converted into a very useful proportion, viz, The sum of the sines of two arcs or angles, is to their difference, as the tangent of half the sum of those arcs

or angles, is to the tangent of half their difference.

VOL. III.

F

26. Operat

26. Operating with the third and fourth formulæ of the preceding article, as we have already done with the first and

second, we shall obtain

Making B = 0, in one or other of these expressions, there

These theorems will find their application in some of the

investigations of spherical trigonometry.

27. Once more, dividing the expression for sin (A + B) by that for cos (A + B), there results

Sin (AB) sin A, cos B± sin B.cOS A
COS (AB)

=

COS A COS B SIN A sin B

:

then dividing both numerator and denominator of the second

fraction, by cos A. cos B, and recollecting that

shall thus obtain

sin

tan

we

COS

R

28. We might now proceed to deduce expressions for the

tangents, cotangents, secants, &c, of multiple arcs, as well as

some of the usual formulæ of verification in the construction of tables, such as

sio (54°+A) + sin (540-A)-sin (180+A)-sin (18°-A)-sin (90°-4); sin+sin (36°- A) + sin (72° + A)=sin (36°+A) + sin (72°-A).

&c. &c.

But, as these enquiries would extend this chapter to too great a length, we shall pass them by; and merely investigate a few properties where more than two arcs or angles are concerned, and which may be of use in some subsequent parts of this volume.'

29. Let A, B, Cc, be any three arcs or angles, and suppose radius to be unity; then

sin (B+c) =

sin a. sin c + sin B. sin (A+B+C)

sin (A+B)

For, by equa. v, sin (A+B+C) = sin A. cos (B+C) + COS A. sin (B+c), which, (putting cos B. cos c - sin B. sin c for cos (B + c)), is = sin A. cos B. cos c-sin A. sin B. sinc+ cos A. sin (B+c); and, multiplying by sin B, and adding sin A. sin c, there results sin A. sin c+ sin B. sin (A+B+C)

sin A.Cos B. cos C. sin B + sin A. sin c. cos2 B + Cos A. sin B. sin (B+c) = sin A. COS B. (sin B. cos c + cos B. sin c) + cos A. sin B. sin (B + c) = sin A. cos B. sin (B + c) + cos A. sin A sin (B+C) = (sin A. cos B + cos A. sin B) x sin (B + c) = sin (A + B). sin (B+C). Consequently, by dividing by sin (A + B), we obtain the expression above given.

.

In a similar manner it may be shown, that sina.sincsin B. sin (A-B+C) sin (A-B)

sin (B-C) =

30. If A, B, C, D, represent four arcs or angles, then writing c + D for c in the preceding investigation, there will result,

sin (B+C+D) =

sin A. sin (c+D) + sin B. sin (A+B+C+D)

sin (A+B)

A like process for five arcs or angles will give

sina.sin (C+D+E) + sin B. sin(A+B+C+D+E)

sin a. sin (c + D +... L) + sin B. sin (A+B+C+...L)

sin (B+C+D+E)-SIA. Sin

sin (A+B)

And for any number, A, B, C, &c, to L,

sin (B+C+....L)=

sin (c-A) = sin c. cos A

sin (A - B) = sin A. cos B

Multiplying the first of these equations by sin a, the second

C, we have
sin c. cos Bi
sin A. cos c,

sin B. cos A.