Page images
PDF
EPUB
[ocr errors][merged small][merged small][merged small][merged small][merged small][merged small]

eos B sin (A+B) + sin (AB) (art. 21), is equal to sin A- - sin (A — B) But, since sin a-sin B'=2 cos (A'+B′) ·

2(1-COS B)

sin (A-B), by art. 25, it follows, that sin A-sin (A-B) = 2 cos (AB) sin; besides which, we have 1 — COS B= 2 sin B. Consequently the preceding expression becomes ssin a + sin (A+B) + sin (A + 2B) + sin (a+3b)+&c. COS (AB)

ad infinitum ===

2 sin B

[ocr errors]

(XXVI.)

35. To find the sum of n + 1 terms of this series we have simply to consider that the sum of the terms past the (n+1)th, that is, the sum of sin [a + (n+1) B] + sin [A+(n+2)B] + sin [▲ + (n+3)в] + &c. ad infinitum, is, by the preceding theorem, = COS [A+ (n+1)B] Deducting this, therefore, from

2 sin B

the former expression, there will remain, sin a + sin (A+B) + sin (A + 2B) + sin (A + 3B) + . . .

sin (A+B) =

COS(AB)-COS [A+ (n+1)R} ____sin (A+ \nB). sin }(n+1)B

2 sin B

sin B

[ocr errors]

(XXVII.)

By like means it will be found, that the sums of the cosines of arcs or angles in arithmetical progression, will be COS A+COS (A + B) + COS (1 + 2B) + cos (1+3B) + &c. sin (AB) ad infinitum =

2 sin B

Also,

(XXVIII.)

cos a + cos (A + B) +cos (a + 2b) + cos (a+3b+) .

[merged small][merged small][ocr errors]

(A+B) sin (n+1)в
sin B.

[blocks in formation]

36. With regard to the tangents of more than two arcs, the following property (the only one we shall here deduce) is a very curious one, which has not yet been inserted in works of Trigonometry, though it has been long known to mathe. maticians. Let the three arcs A, B, C, together make up the whole circumference, O: then, since tan (A + B) = (by equa. xxIII), we have R3 X (tan A+tan B+ tan c) = R3× [tan A+ tan в-tan (A+B)] R2X (tan A+

R3 (tan Atan x)

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small]

sum of the tangents of any three arcs which together con stitute a circle, multiplied by the square of the radius, is equal to the product of those tangents.... (XXX.)

Since both arcs in the second and fourth quadrants have their tangents considered negative, the above property will apply to arcs any way trisecting a semicircle; and it will therefore apply to the angles of a plane triangle, which are, together, measured by arcs constituting a semicircle. So that, if radius be considered as unity, we shall find, that the sum of the tangents of the three angles of any plane triangle, is equal to the continued product of those tangents. (XXXI.)

37. Having thus given the chief properties of the sines, tangents, &c. of arcs, their sines, products, and powers, we shall merely subjoin investigations of theorems for the 2d and 3d cases in the solutions of plane triangles. Thus, with respect to the second case, where two sides and their included angles are given :

By equ. Iv, ab :: sin a: sin B.

By compos.

and division.

a+ba-b:: sin a+sin B: sin Asin B ;

but, eq. XXII, tan (A+B): tan (A — B) : : sin a + sin в : sin A-sin B; whence, ex equal. a+b: a-b:: tan (A+B): tan (A-B). ... (XXXII.)

Agreeing with the result of the geometrical investigation at pa. 387, vol. i.

38. If, instead of having the two sides a, b, given, we know their logarithms, as frequently happens in geodesic operations, tan (AB) may be readily determined without first finding the number corresponding to the logs. of a and b. For if a and b were considered as the sides of a right-angled triangle, in which o denotes the angle opposite the side a, then would tan =

Now, since a is supposed greater

than b, this angle will be greater than half a right angle, or it will be measured by an arc greater than of the circumfer

[ocr errors]

tantan

ence, or than O. Then, because tan (p—}0)= 1+tan & tan

and because tan = R = 1, we have

a-b

a+b

tan (p-1) = (-1) ÷ (1 + 7 ) = = +6·

And, from the preceding article,

a-b tan (A—B) tan (A-R)

a+b tsu &a+l).

=

cot c

: consequently,

tan (A-B) cot c. tan (p-O). . . . (XXXIII.)

From this equation we have the following practical_rule: Subtract the less from the greater of the given logs, the remainder will be the log tan of an angle: from this angle

take 45 degrees, and to the log tan of the remainder add the log cotan of half the given angle; the sum will be the log tan of half the difference of the other two angles of the plane triangle.

39. The remaining case is that in which the three sides of the triangle are known, and for which indeed we have already obtained expressions for the angle in arts. 6 and 8. But, as neither of these is best suited for logarithmic computation, (however well fitted they are for instruments of investigation), another may be deduced thus: In the equation b2+ c2― a2 if we , 26c

for cos. A, (given equation II), viz. cos a =—

JA,

substitute, instead of cos A, its value 12 sin2 A, change the signs of all the terms, transpose the 1, and divide by 2, we shall have sin2 A =

a2 — b2 — c2 + 2bc a2

46c

=

-(bc)3
4bc

Here, the numerator of the second member being the product of the two factors (a+b-c) and (a−b+c), the equa.

tion will become sin3 A =

[ocr errors]

400

But, since (a+b−c)=(a+b+c)—c, and (a−b+c) = (a+b+c)~ b; if we put sa+b+c, and extract the square root, there will result,

[merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

These expressions, besides their convenience for logarith. mic computation, have the further advantage of being perfectly free from ambiguity, because the half of any angle of a plane triangle will always be less than a right angle.

40. The student will find it advantageous to collect into one place all those formula which relate to the computation of sines, tangents, &c.*; and, in another place, those which are of use in the solutions of plane triangles: the former of these are equations v, VIII, IX, X, XI, x, xi, XII, XIII, XIV, XV, XVI, XVII, XVIII, XIX, XX, XXII, xxii, XXIII, XXIV, XXVII; the latter are equa. II, III, IV, VII, XXXII, XXXIII,

XXXIV.

*What is here given being only a brief sketch of an inexhaustible sub. ject the reader who wishes to pursue it further is referred to the copious Introduction to our, Mathematical Tables, and the treatises on Trigonometry, by Emerson, Gregory, Bonnycastle, Woodhouse, Lardner, and many other modern writers on the same subject, where be will and his curiosity richly gratified.

To exemplify the use of some of these formulæ, the fol lowing exercises are subjoined.

EXERCISES.

Er. 1. Find the sines and tangents of 15%, 30°, 45°, 60°; and 75°: and show how from thence to find the sines and tangents of several of their submultiples.

First, with regard to the arc of 45°, the sine and cosine are manifestly equal; or they form the perpendicular and base of a right-angled triangle whose hypothenuse is equal to the assumed radius. Thus, if radius be R, the sine and cosine of 45° will each be R R R2. If R be equal to 1, as is the case with the tables in use, then

sin 45°

tan 45°=

cos 45° = 27071068

[blocks in formation]

Secondly, for the sines of 60° and of 30°: since each angle in an equilateral triangle contains 60°, if a perpendicular be demitted from any one angle of such a triangle on the opposite side, considered as a base, that perpendicular will be the sine of 60°, and the half base the sine of 30°, the side of the triangle being the assumed radius. Thus, if it be R, we shall have R for the sine of 30°, and (R-R3) =R/3, for the sine of 60°.

[ocr errors]

When R = I, these become

5.....

sin 30° = Hence, tan 30°

sin 60°

[blocks in formation]

cos 30 = 8660254.

tan 60° =

Consequently, tan 60°

√3 =

√3 = ·5773503.

[ocr errors]

3 tan 30°.

1.7320508.

Thirdly, for the sines of 15° and 75°, the former are is the half of 300, and the latter is the complement of that half arc. Hence, substituting 1 for R and 3 for cos A, in the expression sin = ± √ √ (2R2+2R cos A)... (equa. xII), it becomes sin 15° = }√ (2−√/3) ·2588190. Hence, sin 75°cos15°[1-(2−√/3)]=√(2·+ √/3)=

[blocks in formation]

Now, from the easily be found. equation x, sin 51

sin .2588190

COS -9659258

3.7320508.

[ocr errors]

=2679492.

sine of 30°, those of 6o, 2o, and 1o, may For, if 5 = 30°, we shall have, from = 5 sin a 20 sin3 A 16 sin3 A: or, if 20x3 + 5x = 5. This

sin Ar, this will become 16r5

[ocr errors]

equation solved by any of the approximating rules for such equations, will give x = 1045285, which is the sine of 6°. Next, to find the sine of 2°, we have again, from equation x, sin 31 = 3 sin a 4 sin3 A: that is, if x be put for sin 2°, 3x 4x3 =1045285. This cubic solved, gives

[ocr errors]

0348995: = sin 2°.

[ocr errors]

Then, if s=sin 1°, we shall, from the second of the equations marked x, have 2s ✓(1 − s2) =.0348995; whence s is

found0174524 = sin 1°.

Had the expression for the sines of bisected arcs been applied successively from sin 15°, to sin 7°30', sin 3°45', sin 1°52', sin 561', &c. a different series of values might have been obtained: or, if we had proceeded from the quinquisection of 45°, to the trisection of 9°, the bisection of 3°, and so on, a different series still would have been found. But what has been done above is sufficient to illustrate this method. The next example will exhibit a very simple and compendious way of ascending from the sines of smaller to those of larger arcs.

[ocr errors]

Ex. Given the sine of 1°, to find the sine of 20, and then the sines of 3o, 4o, 5o, 6o, 70, 80, 90, and 10°, each by a single proportion.

Here, taking first the expression for the sine of a double arc, equa. x, we have sin 20-2sin1°√(1-sin31°)='0348995. Then it follows from the rule in equa. xx, that

sin 20

sin 10 sin 20-sin 1° :: sin 2°+sin 1° : sin 3°0523360 sin 3°-sin 1° :: sin 30+ sin 1° sin 4°0697565 sin 4°-sin 1° :: sin 4°+sin 1° : sin 5°.0871557

sin 30

:

:

sin 40 sin 5°-sin 1° :: sin 50+sin 1° : sin 6o•1045285 sin 50 sin 6°-sin 1° :: sin 60+sin 1° : sin 7°1218693 sin 60 sin 70-sin 1° :: sin 70+sin 1° : sin 80=1391731 sin 7 sin 8

:

:

-sin 1°

sin 80: sin 9°

sin 8°+sin 1°: sin 9°1564375 sin 9°+sin 1° : sin10° 1736482

[blocks in formation]

To check and verify operations like these, the proportions should be changed at certain stages. Thus,

sin 10 sin 3°-sin 2° :: sin 30+ sin 2: sin 5°, sin 10 sin 4°-sin 3°:: sin 40+ sin 3°: sin 7°, sin 40 sin 7°-sin 3°:: sin 70+ sin 3°: sin 10°. The coincidence of the results of these operations with the analogous results in the preceding, will manifestly establish the correctness of both.

Cor. By the same method, knowing the sines of 5°, 10°, and 15, the sines of 20°, 25°, 35°, 55°, 65°, &c. may be found, each by a single proportion. And the sines of 1o, 9°, and 10°, will lead to those of 19°, 29°, 39°, &c. So that the

« PreviousContinue »