Really we should return to the Sines: if for which, it will have been wished to make a whole Table of Sines, for thousandths of a Degree; For a given Chord of 60 : 0': Degrees the Chord for 20:0': should be sought by Trisection, which by five-fold multiplication shall give the 100: 0': Chord . Then by Bisection the squares of the Chords 50: 0': ˝25: 0': ˝12: 30': ˝6: 15': ˝being sought. These Squares will give the 50: 0': | 25: 0': | 12: 30': | 6: 15': Chords themselves. And these Chords shall give by Triplication the Chords 18: 45': | 56: 15' | 37: 30': | 75: 0' | 112: 30' | .
And by 5-fold multiplication the Chords 31: 15' | 62: 30': | 125: 0': | 93: 45'.
And by 7-fold multiplication the Chords 43: 45': | 87: 30': | .
For it should become by multiplication [of 6: 15':] by 11. 13. 17. 19. to be giving the Chords 68: 45': | 81: 15': | 106: 15': | 118: 45': |
Of these Chords the halves namely the Sines of 3 1/8 , 6 2/8 , 9 3/8 , 12 1/2 Degrees, etc. have been placed below here.
From these few given, by dividing the Quadrant into 144 equal parts, of these I have cared to give their own individual Sine. First by the quinquisection rule being propounded [on] page 38 [not a valid reference, as this refers to Chapter 12, whereas the rule is in Chapter 5], thus I shall have filled out all the Sines for the first hundred parts, as far as the Sine of 62: 30': Degrees. Then the rest of the 44 Sines as far as the end of the Quadrant, being found* by the final section of the following chapter. For being given the Sines 375 and 5625, of these the sum being equal to the Sine 6375. As we see here. By the same method the rest may be computed. [For sin(60 + x) - sin(60 - x) = sin(x)]
6. If the sum of two Arcs being equal to 60:0': Degrees, the sum of the Sines of the given Arcs will be equal to the Sine of the Arc composed from 60 Degrees, and from the other given Arc.
For the Triangles ACE,NCO, NXG, are right angled and similar: and the Angles CAE, CNO of 30:0': and therefore OC, EC equal and EC the Difference of the Sines AF, CG. [For, if the angle BXA is x, then the angle AXF = 60 - x; while the angle CXF is 60 + x. OC = R sin x, while EC = 2.OC.sin 30 = OC. ]
Hence : if two Arcs keep the same [angular] distance from 60 Degrees, and the third being equal to the same distance: by being given two of the Sines, it is permitted to find the Sine of the third. For if the Sines of the smaller Arcs will have been given, the sum of these will be the Sine of the largest Arc. And by taking away the Sine of the other small Arc from the Sine of the third there will remain the Sine of the remaining small [Arc].
The Canon of Sines for Dividing the Quadrant into 144 equal parts.
If we shall have divided the Quadrant into 144 equal parts, of these individual [angles] being set out these Sines above being described, from which the whole Canon of Sines being found for the smallest part, being resolved by quinquisection alone, of which examples follow nearby.
1. Page 45 [here the next 4 page for all of these], with the individual Degrees being divided into eight parts. As the total fourth part will have 720 equal small parts.
2. Page 46. The degree cut in 40 parts. The fourth part in 3600.
3. Page 47. The degree cut in 200 parts. The fourth part in 18000.
4. Page 48. The degree cut in 40 parts. The fourth part in 90000.
Section 3. The Sines being given [on] Page 43.
The remaining Sines being placed between those you see here following the method being expounded on page 38.
Section 4. The Sines being given [on] Page 45.
As before the rest which here have been computed being placed between.
Section 5. The Sines being given [on] Page 46.
As before the rest being placed between.
Section 6. The Sines being given [on] Page 47.
As before the rest being placed between.
And by this most convenient means the Canon of Sines will be able to be prepared to one thousandth of a Degree. Because if we are content to make hundredths of Degrees, other Sines being taken from the Canon which gives the Sines for 144 parts of the Quadrant. And so with the Quadrant being divided into 72 equal parts, we will be able by the same method of quinquisection to increase the number of Sines first to 360 parts, then for 1800, finally for 9000. For the individual Degrees should have 100 Sines computed. There is a desire for me to show the Sines, Tangents, and Secants with Logarithms.