Page images

tion, if the establishment of the French metrical system had not required the calculation of new tables corresponding to the decimal division of the circle.

To give some idea of the methods proper to be followed in the construction of tables, suppose we are required to calculate the sines of all the arcs from minute to minute, which lie between 1 minute and 10000 minutes or 100 degrees. We shall make the radius 1, the arc of one minute a; and in the first place we shall have to find the sine and cosine of the arc a with a great degree of accuracy.

[ocr errors]


The radius being 1, the semicircumference or the arc of 200 degrees, we already know is 3.14159 26535 897932; dividing this number by 20000, we have the arc of 1 minute or a=0.00015 70796 32679 48966, a value which is correct to the twentieth decimal. When an arc is very small, its sine may be considered as almost equal to the arc: hence we have very nearly sin a = 0.00015 70796 32679 48966. But this value is wrong from the thirteenth decimal, which is only the tenth significant figure. To obtain a more correct value, the simplest method is to have recourse to the

formulas in Art. 36, by which, making




n 10000'


the first two or three terms of each series, we shall immediately have

sin a 0.00015 70796 32033 525563, cos a = 0.99999 99876 62994 52400 5253, values, which are correct to the twentieth decimal for the sine, and to the twenty-fourth for the cosine.

* XXXVIII. Knowing the sine and the cosine belonging to the arc of one minute designated by a, in order to obtain from it successively the sines of all the arcs which are multiples of a, we shall, in the formulas of Art. 22, make p=a+a, q=x—1. By this substitution, R being always 1, the first and the third formulas will give

sin (x+a)=2 cos a sin x

[ocr errors]
[ocr errors]

sin (x — a)

cos (x

cos (x+a) = 2 cos a cos x From these formulas it appears, that if a series of arcs are in arithmetical progression, the difference being a, their sines will form a recurring series, whose scale of relation will be 2 cos a, -1; in other words, two consecutive sines A and B being calculated, the following sine C will be found by multiplying B by 2 cos a, A by -1, and adding the two products, which will give C = 2 B cos a A. The cosines of these arcs will


in like manner form a recurring series, whose scale of relation will be 2 cos a, -1; hence we shall have in succession,

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

* XXXIX. All that now remains is to execute the operations here indicated, by substituting the values of sin a and cos a. If our tables of sines are to be constructed with ten decimal places, it will be enongh to take the values of sin a and cos a, carried to 16 places, namely,

sin a

0.00015 70796 320335,

cos a 0.99999 99876 629945.

But as cos a differs very little from unity, there is a method of abbreviation, of which it will be right to take advantage. Put k=2(-cos a)= 0.00000 00246 740110; we shall have 2 cos a=2-k, which will give

[ocr errors]
[ocr errors]

·sin (x — a) k sin x,
cos (x-a)-k cos x.

sin (x+a) — sin x = sin x cos (x+a)To obtain the term sin (x+a), we have only to increase the preceding term sin x by the difference sin (x+a) - sin x, which will in every case be very small. Now by the formula, this difference is equal to a similar difference already calculated, namely, sin x― sin (x-a) minus the product of sin x by the constant number k. The finding of this product is thus the only operation of any length connected with the deducing of a sine from the two preceding ones. We may also observe, first, that this product need not be carried beyond the sixteenth place of decimals, which gives only a small number of ciphers to calculate; secondly, that these multiplications may be greatly abridged by finding beforehand the products of the constant number 246740110 by 1, 2, 3, &c. to 9; by which means, we shall immediately obtain the partial products resulting from the different ciphers of the multiplier sin x, and nothing will remain for us but to add these products together, rejecting all the decimals beyond the sixteenth.

The same process will require to be followed in calculating the cosines; and after carrying the series of both these quantities to 50°, the table will be complete.

* XL. It is requisite, we say once more, to calculate sines with 16 decimals, that is, with five or six decimals beyond what we

mean to retain in the tables, in order to make sure that no errors, though augmented in the course of 5000 operations, may influence the tenth decimal in the concluding results. When the calculation is finished, the superfluous decimals may be struck off, and only ten retained in the table.

It is evident enough, that in executing so many calculations, we should endeavour to verify our results as often as possible. In the example, which we have exhibited, of a table calculated from minute to minute, it would be requisite to calculate beforehand the sines and cosines from degree to degree, which would afford a very useful verification at the end of every 100 terms. Now, for calculating the sines from degree to degree, we have the formulas and values which follow:

sin(x+1)—sin x-sin x-sin (x-1°)—h sin x,
cos(x+1°)—cos x cos x-cos (x—1°)—h cos x,

sin 1° 0.01570 73173 11820 676,

cos 1°=0.99987 66324 81660 599,

h=2(1-cos 1°)=0.00024 67350 36678 802.

The sines, thus calculated from degree to degree, will themselves be verified at every tenth term, by the already known values of sin 10°, sin 20°, &c. And finally, when the whole table is constructed, it may farther be verified in as many places as we choose, by the equation


*XLI. The sines, as they result from the calculations here explained, are expressed in parts of the radius, and called natu ral sines: but in practice it has been found, that great advantages result from using the logarithms of those sines instead of the sines themselves; and accordingly our tables for the most part do not exhibit the natural sines at all, but merely their logarithms. It is easy to conceive how, when the sines are calculated, their logarithms may be found; but as the supposition of radius=1 would make the logarithms of all the sines negative, it has been thought proper to take radius =10000000000, in other words, to multiply by 10000000000 all the sines found on the supposition of radius=1. By this arrangement, the radius or the sine of 100°, which is often met with in calculation, has 10 units for its logarithm; and scarcely any angle can occur in practice so small as to have a negative logarithm for its


The logarithmic sines being found, the logarithms of the tangents are easily deduced from them, by a series of operations in simple subtraction: for, having tang x=

Ꭱ sin Ꮖ cos

we must

As to the

also have log. tang x=10+log. sin x-log, cos x. logarithms of the secants, they may be found in a manner still

more simple, by help of the equation sec x=


It is because these deficiencies may be so easily supplied, that in the tables, no logarithms are inserted but those of the sines and tangents.

We might now explain the species of interpolation usually employed for finding the logarithmic sines or tangents of such arcs as contain fractions of a minute; or for finding the arc which corresponds to a given logarithmic sine or tangent, when the latter lies between two numbers given in the table. But for these details it is better to consult the explanations, with which the tables are always accompanied.


XLII. In all right-angled triangles, the radius is to the sine of one of the acute angles, as the hypotenuse is to the side opposite this angle.

Let ABC be the proposed triangle right-angled at A: from the point C as a centre, with a radius CD equal to the radius of the tables, describe the arc DE, which will measure the angle C; on CD let fall the perpendicular EF, which will be the sine of the




angle C. The triangles CBA, CEF are similar, and give the proportion CE: EF :: CB: BA; hence

R sin C: BC: BA.

XLIII. In all right-angled triangles, radius is to the tangent of one of the acute angles as the side lying adjacent to this angle is to the side lying opposite.

Having described the arc DE (see the last figure), as in the preceding Article, draw DG perpendicular to CD; it will be the tangent of the angle C. From the similar triangles CDG, CAB, we shall have the proportion CD: DG: CA: AB; hence

R: tang C: CA : AB.

XLIV. In any rectilineal triangle, the sines of the angles are as the opposite sides.

Let ABC be the proposed triangle; AD the perpendicular, let fall from the vertex A on the opposite side BC: there may be two


First. If the perpendicular falls within the triangle ABC, the right-angled triangles ABD, ACD (Art. 42.) will give

[blocks in formation]


In these two proportions, the extremes are equal; hence with the means we shall have

sin C: sin B::AB: AC.

Secondly. If the perpendicular falls without the triangle ABC, (see the fig. in the next page) the right-angled triangles ABD, ACD will still give the proportions

R: sin ABD :: AB : AD,

R: sin C :: AC: AD;

from which we derive sin C: sin ABD:: AB: AC. But the angle ABD is the supplement of ABC or B; hence sin ABD= sin B; hence we again have

sin C: sin B::AB: AC.

XLV. In all rectilineal triangles, the cosine of one angle is to radius, as the sum of the squares of the sides which contain that angle, minus the square of the third side is to twice the rectangle of the former two sides; in other words, we have

cos B: R: AB+BC2-AC2: 2AB. BC, or

[blocks in formation]

From the vertex A, let AD be again drawn perpendicular to the side BC.

First. If this perpendicular falls within the triangle (see the preceding figure) we shall have AC2¬AB2+BC2—2BC × BD AB2+BC2 AC

(Prop. 31. II.); hence BD:


But in the right

angled triangle ABD, we have R : sin BAD :: AB : BD; also the angle BAD being the complement of B, we have

« PreviousContinue »