300
HISTORY OF
TRACT li?'
complements by the 3d; then the sines of the halves of these,and of their complements, by the same theorems; and so on,alternately, of the halves and complements, till they arrivedat an arc which is nearly equal to its sine. Thus, beginningwith the above arc of 12 degrees, and its sine, the halves wereobtained as follows:
the halves.33° . '
ie 30
8 15
■27 45
Sines.
5446390
2840153
1434926
4656145
Comps.
57
73 30
81 45
62 15
8386706
9583197
9396514
8849876
Halves.
‘28 30
14 15
36 45
4771588
2461533
5983246
Comps.
61 30
75 43
53 15
8783171
9692309
8012538
Half.
30 45
5112931
Comp.
59 15
8594064
The halves.
Sines.
The comp.
Sines.
of these.
6° '
1045285
48°
7431448
3
523360
69
9335804
1 30
261769
79 30
9832549
45
130896
84 45
9958049
79 ( >37<1A
The Comp.
68 15
9288095
of these.
45 45
7163019
84
9945218
87
9986295
The halves
88 30
9996573
of these.
89 15
9999143
24
4067366
34 ^0
5664062
The halves
17 15
2965416
of these.
39 45
6394390
42
6691306
23 15
3947439
21
3583699.
10 30
1822355
The comp.
5 15
915016
66
9135455
43 30
6883545
55 30
8241262
21 45
3705574
72 45
9550199
44 15
6977905
50 15
7688418
66 45
9187912
The sines of small arcs are then deduced in this manner.From the sine of 45', above determined, arc found the halves,which will be thus: ,
45' 0" - . _ _ 130896
22 30 - - _ . 65449,4
11 15. 32724,8
Now these last two sines being evidently in the same ratio astheir arcs, the sines of all the less single minutes will be foundby single proportion. So the 45th part of the sine of