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Cos. + cos. B

2 cos. (A+B).cos. (AB)....[7].

2

cos. B - cos. A = 2 sin. (A+B). sin. (A-B)......[8].

2

COR. 5. Divide [5] by [6], and

sin. A + sin. B

2

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A-B

sin. A-sin. B-cos. (A+B) sin. (4-8)

similarly,

2

2

= tan. (A+B).cot. (A-B),

or,

cos. A+cos. B

cos. B-cos. A

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1

2

.......[].

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2

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2

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which, in a particular case, that is, when B = 0, becomes

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From this last expression, we may express, cos. A in terms of the tan. 4, &c.; for, since

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1 + cos. A

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*** COR. 6. The forms for the sine and cosine of A + B being obtained, it is easy to deduce from them, the sine or cosine of

Μπ

n

any arc, such as + A, and nearly with as much convenience as by reference to the Table given in page 16, thus, by......[1],

sin. (ᅲ + A) = sin..cos. A + cos. π. sin. A,

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and, in like manner, other instances may be reduced ***.

We will now proceed to deduce expressions for the tangents of the sums and differences of arcs: and, the first step will be the solution of the following Problem, which, indeed, is little else than a corollary from the preceding results :

PROBLEM 4. It is required to express the tangent of the sum and difference of two arcs in terms of the tangents of the simple arcs.

Since (p. 8.) the tangent of an arc is equal to the sine divided by the cosine, if the arc be A + B,

tan. (A + B) =

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

sin. A.cos. B±cos. A. sin. B cos. A. cos. Bsin. A.sin. B

Now, since the object is to obtain an expression involving tan. A, tan. B, we must divide both numerator and denominator of the above fraction by cos. A. cos. B, an operation which will not change its real value; beginning then with the numerator

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(rad. = 1) [9].

consequently, tan. (A + B) = tan. A ± tan. B

1+tan. A. tan. B

This formula may be used for determining the tangents of such arcs as 90° ± A, 180° ± A, &c. exactly, as in Cor. 6. p. 34, we shewed the formulæ for sin. (A + B), &c. might be used in

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COR. 3. Let t, t', t", &c. be the tangents of the arcs A, B,

C, &c. then by formula [9], considering A + B as one arc,

tan. (A + B + C) =

But, tan. (A + B) =

tan. (A + B) + tan. C ........ [9] 1 - tan. (A + B) tan. C

t+t' 1-tt

;

therefore the numerator of the fraction [q] equals

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and the denominator, of the same fraction, equals

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If A + B + C = ", (which is the case when A, B, C, are

the three angles of a triangle), since tan. " = 0,

t + t' + t" - tt'ť" = 0, or
t + t' + t" = tt't",

which is the theorem given in the Phil. Trans. 1808. p. 122.

But the theorem has an origin much more remote; for, the above formula for tan. (A + B + C) and similar formulæ for the tangents of A + B + C + D, &c. were given as far back as the year 1722, by John Bernoulli, and are inserted in the Leipsic Acts for that year, p. 361, and in the second volume of his Works, at p. 526.

The formulæ for the tangents of the sums of any arcs A, B, C, &c. are symmetrical in their composition, and their law is easily defined: suppose, the symbols S3 (tt), S, (tt't"), &c. be made to represent, respectively,

tí' + tť" + t't",

tťť" + tť l'" + tt" t'" + t't" t", &c.

,

then, tan. (A + B + C) = S3(t) - S, (til")
1 - S3 (t t')
S1 (t) - S1 (tt't")
1-S(tt) + S (tt't"t")'

tan. (A + B + C + D) =

,

tan. (A + B + C + D + E) = Ss (t)-S, (tt't") + Ss (tt'...t") 1 - Ss(tt) + Ss (tt'ť"t")'

&c.

These formulæ are easily shewn to be true on the principle, that, if the formula for the tangent of n arcs be true, that for (n + 1) arcs must be true also: the latter inference being made by means of the form [9], p. 35.

If, instead of a radius = 1, we would introduce a radius = r into the preceding formulæ, we must avail ourselves of the rule laid down in p. 19. Thus the formula [1], p. 27, becomes r. sin. (A + B) = sin. A x cos. B + cos. A x sin. B, the formula [c], p. 31, is the same, whether the radius be 1

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but, the formula [f], p. 33, remains the same, whether the radius be equal 1 or r.

*** We will now subjoin a few additional formulæ for the sines and cosines, &c. of the sums and differences of arcs, the investigation of which the Student, by pursuing a track similar to what has been already proceeded on, will easily discover,

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