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Add together the expanded expressions for sin. (B + A), sin. (BA), that is,

add

to

sin. (BA)

sin. B. cos. A + cos. B. sin. A,

sin. (BA) sin. B. cos. A

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

COS. B. sin. A;

2

CUs. A. sin. B: sin. (B-A).

Thus again, by adding together the expressions for cos(B+A')

and cos. (BA), we have

COS. (B+A) + cos. (B—A) = 2 cos. A . cos. B;

whence, cos. (BA)

2 cos. A. cos. B

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COS. (BA).

Substituting in these expressions for the sine and cosine of

B+ A, the successive values A, 2A, 3A, &c, instead of B; following series will be produced.

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sin. na = 2 cos. A . sin. (~—1)▲ — sin. (n−2)▲ .

Cos. 2A= 2 cos. A. cos. A—cos. 0(=1).

Cos. 3A 2 cos. A. cos. 2A-cos. A.

(xi)

cos. 4A 2 cos. A. cos. 3A-cos. 2A.

cos. NA= 2 cos. A. cos. (n-1) A cos. (12—2)A.

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

*Here we have omitted the powers of a that were necessary to render all the terms homologous, merely that the expressions might be brought in upon the page; but they may easily be supplied, when needed, by the rule in

art. 18.

20.

20. From the equation sin. 2a =

2 sin A. COS A

R

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

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gives R2 sin2 2A = 4R2 Sin2 A 4 sin A.

Here taking sin A for the unknown quantity, we have a quadratic equation, which solved after the usual manner, gives

sin A = ±√÷R2 ± ¿R√/R2 — Sin2 2a.

=

If we make 2A A', then will AA', and consequently the last equation becomes

sin ARRR- sin2 a

or sina√2R2R COS A':

}

(XII.)

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

sin (A+B) sin (A-B)

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2 sin A. cos B

R

; whence,

sin A. COS BR sin (A+B) + R sin (A-B)..(XIII.) Also, taking sín (A—B) from sin (A+B), gives

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-

R

sin B. COS AR sin (A+B) — IR. Sin (A — B). . (XIV.) When AB, both equa. XIII and xiv, become

COS A sin AR sin 2A (XV.)

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22. In like manner, by adding together the primitive expressions for cos (A+B), COS (A-B), there will arise

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

2 cos A. COS B

R

; whence,

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

COS2 AR COS 2A + R2 . . . (XVII.)

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

COS (AB) COS (A+B) =

2 sin A sin B

; whence,

R

sin A.sin B = R. COS (A-E)-R. COS (A+B) (XVIII.) When AB, this formula becomes

sin2 A = R' — R. Cos 2A

(XIX.)

24. Mul

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.

a

Here, since sin2 A sin2 B, is equal to (sin A + sin B) × (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 to n, then the preceding proportion will be converted into sin a : sin (n+1)a− sin na :: sin (n+1)a + sin na : sin (2n+1)a. ... (XXI.)

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 в, obtained in arts. 21, 22, 23, there would arise the following formulæ :

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Dividing the second of these formulæ by the first, there will

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factors of the first member of this equation, are

tan(A-B', and R

R

tan (A+) respectively; so that the equation

manifestly becomes

B')'

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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 formula of the preceding article, as we have already done with the first and second, we shall obtain

tan {(A' + B′) . tan †(▲’— B′) COS B'-COS A'

R2

=

COS ACOS B

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

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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 (AB), there results

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then dividing both numerator and denominator of the second fraction, by cos A. cos B, and recollecting that

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sin

tan

=

we

CUS

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R2 tan A. tan B

tan A tan в.

R2
tan (A+B)
reduction, becomes

which, after a little

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,

28. We might now proceed to deduce expressions for the tangents, cotangents, secants, &c, of multiple arcs, as well as

some

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

sin (54°+a) +'sin (54°— x)—sin (13°+ a) — sín (189—A) sin (90°—▲); sin + sin (36°— a) + sin (72° + a)=sin (36o + a) + sin (72°—▲).

&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, C, be any three arcs or angles, and suppose radius to be unity; then sin A. sin C+ sin B. sin (A+B+C) sin (A+B)

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sin (B+c) = For, by equa. v, sin (A+B+C) = sin A. cos (B+c)+cos A. sin (BC), which, (putting cos B. Cos Csin B. sin c for COS (B+c)), is sin A • cos B. cos csin A. sin B sin c+ COS A. Sin (B+e); 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▲. 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 E) X sin (BC) sin (A + B). sin (B+c). Consequently, by dividing by sin (A+B), we obtain the expression above given.

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=

In a similar manner it may be shown, that

sin (B-C) =

sin A. sin c-
sin B. sin
in (A−B+c)
sin (A-B)

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

sin (B+C+D) =
A like process for five arcs or angles will give

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

sin (B+C+D+E)=?
And for any number, A, B, C, &c, to L,

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

sin A. Sin (c+D+..... L) + sin B. Sin (A + B + C + ...1) sin (A+B)

31. Taking again the three A, B, C, we have
sin (B-C) = sin B. cos csin c. cos: B1

sin (CA) sin c. cos A
sin (AB)

-

sin A COS C,

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sin A. cOS B sin B. cos A.

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

F 2

by

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