Because the chord line B D is the side of a hexagon, inscribed in a circle, it is the subtense of 60 degrees, and, consequently, equal to the radius CB (corollary to Prop. 15, Book IV., of Euclid); wherefore half the radius BS = BM, will be the sine of 30 degrees FN, which, therefore, is .5000000000. Now, having found the sine of 30 degrees, its cosine may be obtained by Euclid, Book I., Prop. 47: for in the right-angled triangle FNC, the hypothenuse FC is given the radius, or 1.0000000000, and the perpendicular FN half the radius, or .5000000000, to find the base CN the cosine G F; therefore

FC FC - FN

FN

CN .8660254037, or its equal GF; hence the sine of 30 degrees is .5000000000, and its cosine. 8660254037. Again,

In the triangle FNB, the perpendicular FN is given.5000000000, and the base CB-CN=NB. 1339745963, to find the hypothenuse BF: but half the side of a polygon, inscribed in a circle, is equal to the sine of half the circumscribing arc; therefore its half, BT = HO, will be the sine of the arc of 15 degrees: hence FN x FN+ NB x NB = BF.5176380902; the half of which, viz., . 2588190451, is therefore equal to BT, or to its equal HO, the sine of 15 degrees, and from which its cosine HI may be easily obtained; for, in the triangle COH, the hypothenuse CH is given = 1,0000000000, and the perpendicular HO = .2588190451, to find the base CO the cosine HI. Now,

✓CH CHHO× HO CO.9659258263 the cosine HI; hence the sine of 15 degrees is . 2588190451, and its cosine .9659258263. Thus proceeding, the sine of the 12th bisection, viz., 52"44"3":"45"", will be found = .0002556634. And because small arcs are very nearly as their corresponding sines, the measure of 1 minute may be easily deduced from the sine of the small arc, or 12th bisection determined as above; for,

As the arc of 52"44"3""45"" is to an arc of 1 minute, so is the sine of the former to the sine of the latter: that is, as 52"44"3""45"" : 1 :: .0002556634.0002908882; which, therefore, is the sine of 1 minute, the cosine of which is . 9999999577; but this approximates so very closely to the radius, that it may be esteemed as being actually equal to it in all calculations; and hence, that the cosine of 1 is 1.0000000000.

Now, having thus found the sine and cosine of one minute, the sines of every minute in the quadrant may be obtained by the following rule; viz.

As radius is to twice the cosine of 1 minute, so is the sine of a mean arc to the sum of the sines of the two equidistant extremes; from which let either extreme be subtracted, and the remainder will be the sine of the other extreme: as thus,

To find the Sine of the Arc of 2 Minutes.

As radius 1: 2 :: .0002908882 to .0005817764, and .0005817764 -.00000000000005817764; which, therefore, is the sine of the

arc of 2 minutes.

To find the Sine of 3 Minutes.

As radius = 1:2::.0005817764 to .0011635528, and .0011635528. -.0002908882.0008726646; which, therefore, is the sine of an arc

of 3 minutes.

To find the Sine of 4 Minutes.

As radius = 1:2:: .0008726646 to .0017453292, and .0017453292 -.0005817764 = .0011635528; which, therefore, is the sine of the arc of 4 minutes.

To find the Sine of 5 Minutes.

As radius = 1; 2::.0011635528 to .0023271056, and .0023271056 -.008726646 = .0014544407; which, therefore, is the sine of the arc of 5 minutes.

In this manner, the sines may be found to 60 degrees; from which, to the end of the quadrant, they may be obtained by addition only; for the sine of an arc greater than 60 degrees, is equal to the sine of an arc as much less than 60, augmented by the sine of the excess of the given arc above 60 degrees: thus,

All the sines being found to 60 degrees; required the sine of 61 degrees?

Solution.-Sine of 59.8571673, and sine of 19.0174524; their sum.8746197; which, therefore, is the sine of 61 degrees, as required. Again,

All the sines being found to 60 degrees; required the sine of 62 degrees?

Solution.-Sine of 589.8480481, and sine 29.0348995; their sum.8829476; which, therefore, is the sine of 62 degrees, as required.

Now, the natural sines being thus found, the natural versed sines, natural tangents, and natural secants, may be readily deduced therefrom, agreeably to the principles of similar triangles, as demonstrated in Euclid, Book VI., Prop. 4. Thus,

To find the Natural Versed Sine of 30 Degrees NB, in the Diagram.

Since the versed sine of an arc is represented by that part of the diameter which is contained between the sine and the arc; therefore N B is the versed sine of the arc B F, which is the arc of 30 degrees; and since the versed sines are measured upon the diameter, from the extremity B to C continued to the other extremity, the natural versed sines under 90 degrees are expressed by the difference between the radius and the cosine, and those above 90 degrees by the sum of the radius and the sine: hence, the radius CB 1.0000000 the cosine FG, or its equal NC.8660254 = NB, 1339746; which, therefore, is the natural versed sine of 30 degrees.

To find the Natural Tangent of 60 degrees = BQ, in the Diagram.

As the cosine C M is to the sine D M, so is the radius CB to the tangent BQ: that is,

As CM.5000000: DM.8660254 :: CB 1, 0000000 : BQ = 1.7320508; which is the natural tangent of 60 degrees.

To find the Natural Secant of 60 Degrees C Q, in the Diagram.

As the cosine CM is to the radius CD, so is the radius C B to the secant CQ: that is,

As CM .5000000: CD 1.0000000 :: CB 1.0000000 : CQ = 2. 0000000; which is the natural secant of 60 degrees. Hence, the manner of computing the natural co-tangent AP, the natural co-secant CP, and the natural co-versed sine EA, will be obvious. The versed sine supplement of an arc is represented by the difference between the versed sine of that arc and the diameter or twice the radius: thus, the versed sine supplement of the arc BF is expressed by the difference between twice the radius C B, and the versed sine NB; viz., twice C B = 2.0000000 NB .1339746 1.8660254; which, therefore, is the natural versed sine supplement of the arc B F or the arc of 30 degrees, and so of any other.

Now, the natural sines, versed sines, tangents, and secants, found as above, being principally decimal numbers, on account of the radius being assumed at unity or 1; therefore, in order to render these numbers all affirmative, they are to be multiplied by ten thousand millions respectively; and then the common logs. corresponding thereto will be the logarithmic sines, versed sines, tangents, and secants, which are generally given in the different mathematical Tables under these denominations.

Of the Table.

In this Table, the natural versed sinès are given to every tenth second of the semicircle; the corresponding arcs being arranged at the top, in numerical order, from 0 to 180 degrees. The natural versed sines supplement are given to the same extent; but their corresponding arcs are placed at the bottom of the Table, and numbered from the right hand towards the left, or contrary to the order of the versed sines. The natural co-versed sines begin at the end of the first quadrant, or of the 90th degree of the versed sines; the arcs corresponding to which are given at the bottom of the page and numbered, like the versed sines supplement, towards the left hand from 0 to 90 degrees, and then continued at top of the page from 90 to 180 degrees, towards the right hand, until they terminate at the 90th degree of the versed sines, where they first began. The natural sines begin where the co-versed sines end; viz., at the end of the first quadrant, or 90th degree of the versed sines, with which they increase by equal increments; the arcs corresponding to those are placed at the top of the page to every tenth second of the quadrant, the 90th degree of which terminates with the 180th of the versed sines. The natural cosines begin with the versed sines supplement; the arcs corresponding to which are given at the bottom of the page, being numbered, like the latter, contrary to the order of the versed sines and natural sines, to every tenth second from 0 to 90 degrees, or to the end of the first quadrant of the versed sines, thus ending where the co-versed sines begin.

Note. In the general use of this Table, it is to be remarked, that the natural versed sine supplement, natural co-versed sine under 90 degrees, or natural cosine, of a given degree, is found in the same page with the next less degree in the column marked O" at top, it being the first number in that column; that answering to a given degree and minute is found on the same line with the next less minute in the column marked 60" at the bottom of the page; and that corresponding to an arch expressed in degrees, minutes, and seconds, is obtained by deducting the proportional part, at bottom of the page, from the natural versed sine supplement, natural coversed sine under 90 degrees, or natural cosine of the given degree, minute, and less tenth second.

PROBLEMS TO ILLUSTRATE THE USE OF THE TABLE.

PROBLEM.

To find the Natural Versed Sine, Natural Versed Sine Supplement, Natural Co-versed Sine, Natural Sine, and Natural Cosine, of any given Arch, expressed in Degrees, Minutes, and Seconds.

RULE.

Enter the Table, and find the natural versed sine, versed sine supplement, co-versed sine, natural sine, or natural cosine, answering to the given degree, minute, and next less tenth second; to which add the proportional part answering to the odd seconds, found at the bottom of the page, if a natural versed sine, co-versed sine above 90°, or natural sine be wanted; but subtract the proportional part, if a versed sine supplement, co-versed sine under 90%, or natural cosine, be required: and the sum, or remainder, will be the natural versed sine, natural sine, natural versed sine supplement, co-versed sine, or natural cosine, of the given arch.

Example 1.

Required the natural versed sine, versed sine supplement, co-versed sine, natural sine, and natural cosine, answering to 42:12:36"?

Given Arch 42:12:36% Natural versed sine = 259313