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A Trigonometrical Canon is a Table, which, beginning from one second or one minute, orderly expresses the lengths that every sine, tangent, and secant have, in respect of the radius, which is supposed unity; and is conceived to be divided into 10000000 or more decimal parts. And so the sine, tangent, or secant of an arc, may be had by help of this table; and, contrariwise, a sine, tangent, or secant being given, we may find the arc it expresses. Take notice, that in the following tract, R signifies the radius, S a sine, Cos. a cosine, T a tangent, and Cot. a cotangent; also ACq signifies the square of the right line AC; and the marks or characters, +, -, =, ;, : ;, and ✔, are severally used to signify addition, subtraction, equality, proportionality, and the extraction of the square root. Again, when a line is drawn over the sum or difference of two quantities, then that sum or difference is to be considered as one quantity.

Constructions of the Trigonometrical Canon.


THE two sides of any right angled triangle being
given, the other side is also given.

For (by 47. 1.) ACq=ABq+BCq, and ACq-BCq= Fig. 28.
ABq, and interchangeably ACq-ABq=BCq. Whence,
by the extraction of the square root, there is given AC=
✔ABq+BCq; and AB✔ ACq-BCq; and BC=


THE sine DE of the arc BD, and the radius CD, Fig. 29. being given, to find the cosine DF.


The radius CD, and the sine DE, being given in the right angled triangle CDE, there will be given (by the last Prop.) ✓CDq-DEq= (CE=) DF.


Fig. 29. THE sine DE of any arc DB being given, to find DM or BM, the sine of half the arc.

DE being given, CE (by the last Prop.) will be given, and accordingly EB, which is the difference between the cosine and radius. Therefore DE, EB, being given, in the right angled triangle DBE, there will be given DB, whose half DM is the sine of the arc DL the are BD.


Fig. 29. THE sine BM of the arc BL being given, to find the sine of double that arc.

The sine BM being given, there will be given (by Prop. 2.) the cosine CM. But the triangles CBM, DBE, are equiangular, because the angles at E and M are right angles, and the angle at B common: Wherefore (by 4. 6.) we have CB: CM:: (BD, or) 2 BM: DE. Whence, since the three first terms of this analogy are given, the fourth also, which is the sine of the are DB, will be known.

COR. Hence CB : 2 CM :: BD : 2 DE; that is, the radius is to double the cosine of one half of the arc DB, as the subtense of the arc DB is to the subtense of double that Also CB : 2 CM :: (2 BM: 2 DE : :) BM : DE:: CB: CM. Wherefore the sine of an arc, and the sine of its double being given, the cosine of the arc itself is given.



Fig. 30. THE sines of two arcs, BD, FD, being given, to find FI, the sine of the sum, as likewise EL, the sine of their difference.

Let the radius CD be drawn, and then CO is the cosine of the arc FD, which accordingly is given, and draw OP through O parallel to DK; also let OM, GE, be drawn parallel to CB. Then because the triangles CDK, COP,

CHI, FOH, FOM, are equiangular; in the first place, CD:DK::CO: OP, which, consequently, is known. Also we have CD: CK:: FO: FM; and so, likewise, this will be known. But because FO=EO, then will FM=MG= = ON; and so OP+FM=FI=sine of the sum of the arcs: And OP-FM: that is, OP-ON-EL-sine of the difference of the arcs: which were to be found.

COR. Because the differences of the arcs BE, BD, BF, are equal, the arc BD is an arithmetical mean between the ares BE, BF.


THE same Things being supposed, the radius is to double the cosine of the mean arc, as the sine of the difference is to the difference of the sines of the extremes.

For we have CD: CK :: FO: FM; whence by doubling Fig. 30. the consequents, CD: 2CK :: FO: (2 FM, or) to FG, which is the difference of the sineș EL, FI. Q. E. D.

COR. If the arc BD be 60 degrees, the difference of the sines FI, EL, will be equal to the sine FO of the difference. For, in this case, CK is the sine of 30 degrees; the double whereof is equal to the radius (by 15. 4.); and so, since CD-2 CK, we shall have FO FG. And, conse quently, if the two arcs BE, BF, are equidistant from the are of 60 degrees, the difference of the sines will be equal to the sine of the difference FD.

COR. 2. Hence, if the sines of all arcs distant from one another by a given interval, be given, from the beginning of a quadrant to 60 degrees, the other sines may be found by one addition only. For the sine of 61 degrees the sine of 59 degrees + the sine of 1 degree; and the sine of 62 degrees the sine of 58 degrees + the sine of 2 degrees. Also, the sine of 63 degrees the sine of 57 degrees + the sine of 3 degrees, and so on.

COR, 3. If the sines of all arcs, from the beginning of a quadrant, to any part of a quadrant, distant from each other by a given interval, be given, thence we may find the sines of all arcs to the double of that part. For example: let all the sines to 15 degrees be given; then, by the preceding analogy, all the sines to 30 degrees may be found, For the

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