chord of part of the semi-periphery to what accuracy you please: then, by proceeding as above (for finding the sine of 15′), the sine of 1' will also be obtained to a very great degree of exactness.

PROP. IV.

To show the manner of constructing the trigonometrical

canon.

First, find the sine of an arch of one minute, by the preceding Prop. and then its co-sine by Prop. I. which let be denoted by C; then (by Theor. I. p. 17.) we shall have sine O' = sine 2'.

2C x sine 1'

And thus are the sines of 6', 7', 8', &c. successively derived from each other.

The sines of every degree and minute, up to 60°, being thus found; those of above 60° will be had by addition only (from Theor. 2. p. 18.); then, the sines being all known, the tangents and secants will likewise become known, by Prop. 1.

Note. If the sine of every 5th minute, only, be computed according to the foregoing method, the sines of all the intermediate arches may be had from thence, by barely taking the proportional parts of the differences, and that so near as

to give the first six places true in each number; which is sufficiently exact for all common purposes.

SCHOLIUM.

Although what has been hitherto laid down for constructing the trigonometrical canon, is abundantly sufficient for that purpose, and is also very easily demonstrated; yet, as the first sine, from whence the rest are all derived, must be carried on to a great number of places, to render the numerous deductions from it but tolerably exact (because in every operation the error is multiplied), I shall here subjoin a different method, which will be found to have the advantage, not only in that, but in many other respects.

First, then, from the co-sine of 15°, which is 2+3* =,965925826, &c.f, and the sine of 18°, which is = - 1,309017, &c. (equal to half the side of a decagon inscribed in the circle), let the co-sine of 3°, the difference

*The side of a hexagon being equal to the radius of its circumscribing circle (15. 4.), it follows that the sine of 30° (radius being unity) = 1, and its co-sine (VT — }) = √3; whence the versed sine of 150° 2+1/3 2

(1 co-s. of 30°)

12+√3

; consequently the co sine of 15° =

(prop. 1, p. 54) = {√2+√3.

+ It appears (by 11. 2. and 10. 4.) that if the side of an isosceles triangle,

each of whose angles at the base is double the third angle, be

= 1, the base

条 ; and if from the vertex of such a triangle a perpendicular be

demitted on the base, the semibase is manifestly the sine of half the vertical angle, that is, the sine of 18°

between 18° and 15° be found*; from which the co-sine of 45' will be had, by two bisections only: whence the sines of all the arches in the progression 1° 36', 2° 15′, 3° 00′, 3° 45', &c. may be determined (by Theor. 1. p. 17), and that to any assigned degree of exactness.

The sines of all the terms of the progression 45', 1° 30', 2° 15', &c. up to 60°, being thus derived, the next thing is to find, by the help of these, the sines of all the intermediate arches, to every single minute.

This, if you desire no more than the 4 or 5 first places of each (which is exact enough where nothing less than degrees and minutes is regarded), may be effected by barely taking the proportional parts of the differences.

But if a greater degree of accuracy be insisted on, and you would have a table carried on to 7 or 8 places, each number (which is sufficient to give the value of an angle to seconds, and even to thirds, in most cases), then the operation may be as follows:

1o. Multiply the sum of the sines of any two adjacent terms of the progression 45', 1° 30', 2° 15′, 3 00′, 3° 45', &c. (betwixt which you would find all the intermediate sines) by the fraction ,0000000423, for a first product; and this, again, by 22, for a second product; to which last let

of the difference of the two proposed sines (or extremes) be added, and the sum will be the excess of the first of the intermediate sines above the lesser extreme.

Note. The co-sine of the difference of two arches (suppos ng radius unity) is found by adding the product of their sines to that of their co-sines; as is hereafter demonstrated.

2o. From this excess let the first product be continually subtracted; that is, first, from the excess itself; then from the remainder; then from the last remainder, and so on 44. times.

3°. To the lesser extreme add the forementioned excess; and to the sum add the first remainder; to this sum add the next remainder, and so on continually: then the several sums thus arising will respectively exhibit the sines of all the intermediate arches, to every single minute, exclusive of the last; which, if the work be right, will agree with the greater extreme itself, and therefore will be of use in proving the operation.

But to illustrate the matter more clearly, let it be proposed to find the sines of all the intermediate arches between 3° 00′ and 3° 45′ to every single minute, those of the ́extremes being given, from the foregoing method, equal to ,05233595 and ,06540312 respectively. Here, the sum of the sines of the extremes being multiplied by ,0000000423, the first product will be ,00000000498, &c. or ,0000000050, nearly (which is sufficiently exact for the present purpose); and this, again, multiplied by 22, gives ,00000011 for a 2d product; which added to,0002903815, part of the difference of the two given extremes, will be ,0002904915, the excess of the sine of 3° 01' above that of 3° 00'. From whence, by proceeding according to the 2d and 3d rules, the sines of all the other intermediate arches are had, by addition and subtraction only. See the operation.

4415 10th. rem. ,0552406400 sine 3° 10′

Again, as a second example, let it be required to find the sines of all the arches, to every minute, between 59° ‍15′ and 60° 00'; those of the two extremes being first found, by the preceding method. In this case, the two extremes being ,85940641 and ,86602540, their sum will be 1,72543, &c. and their difference=,00661899; whereof the former, multiplied by ,0000000423 (see the rule), gives ,00000007298,