Abbildungen der Seite

C1 perpendicular to CE. Of this assumed arc AB, the complement is BC, and the supplement | ] BCF; the sine is BD, the cosine C BG or OD, the versed sine AD, G the coversed sine CG, and the supplementary versed sine FD ; the

[ocr errors]

tangent of AB is AH, and its co- O A. tangent CI; and the secant of the same arc is OH, and its cosecant OI.


Several obvious consequences flow from these definitions : * \

1. Since the diameter which bisects an arc bisects also the chord at right angles, it follows that half the chord of any arc is equal to the sine of half that arc.

[ocr errors]

2. In the right-angled-triangle ODB, BD”--OD*= OB"; and hence the squares of the sine and cosine of an arc are together equal to the square of the radius.

3. The triangle ODB being evidently similar to OAH, OD: DB : OA : AH; that is, the cosine of an arc is to the sine, as the radius to the tangent.

4. From the similar triangles ODB and OAH, OD: OB :: OA : OH ; wherefore the radius is a mean proportional

, between the cosine and the secant of an arc.

5. Since BD*= AD.FD, it is evident that the sine of an arc is a mean proportional between the versed sine and the


supplementary versed sine, or between the sum and difference of the radius and the cosine.

6. Hence also the chord of an arc is a mean proportional between the versed sine and the diameter; for AB’= AD.A.F.

7. The triangles OAH and ICO being similar, AH; OA :: OC: CI; and hence the radius is a mean proportional between the tangent of an arc and its cotangent.

8. Since OD* = BG*=CG.C.E., it follows that the cosine of an arc is a mean proportional between the sum and the difference of the radius and the sine.

The circumference of the circle is commonly divided into 360 equal parts, called degrees, each of them being subdivided into 60 minutes, and these again being each distinguished into 60 seconds. It very seldom is required to carry this subdivision any farther. Degrees, minutes, seconds, or thirds, are conveniently noted by these marks,

o / In ///

Thus, 23° 27' 48" 42", signifies 23 degrees, 27 minutes, 43 seconds, and 42 thirds.

Scholium. To discern more clearly the connection of the lines derived from the circle, it will be proper to trace their successive values, while the corresponding arc is supposed to increase. Let the arc AB', on the opposite side, be made equal to AB, draw the diameter FOA, extend the diameters b/OB and boB', join BB' and bb', and at A apply the double tangent HAH'. It is evident that BE=be, or that the sine of the arc AB is equal to the sine of its supplement ABb. But B'E and b'e, or the sines of ABFb' and ABFö'B' which lie on the opposite side of the diameter, are likewise equal to BE; that is, the inverted sine of an arc is equal to the sine of that arc or of its supplement, augmented, each by a semicircumference. The arc AB, and its defect ABFB" from a whole circumference, have both the same cosine OE; and the supplemental arc ABb, and its defect from a whole circumference, have likewise the same cosine, although with an inverted position. AH and OH are respectively the tangent and secant not only of AB, but of the arc ABöFb', which is compounded of the original arc and a semicircumference; and the similar lines AH' and OH', on the opposite side, are at once the tangent and secant of the supplementary arc ABb, and of ABöFö'B', likewise compounded of that arc and a semicircumference. As the prolonged diameter b/OBH, therefore, turns about the centre, the sine and tangent both increase, till the arc attains 90°, when the sine becomes equal to the radius, and the tangent vanishes into unlimited extent. Between 90° and 180°, the sine again diminishes, and the tangent, re-appearing in the opposite direction, likewise contracts by successive diminutions. In the third quadrant, the sine emerges with a contrary position, and increases till it becomes equal to the radius; while the tangent, resuming its first position, stretches out till it vanishes away. Between 270° and 360°, the opposite sine again contracts, and the tangent, re-appearing on the same side, shrinks also by degrees to a point. In the first and fourth quadrants, the cosine lies on the same side of the centre, while the secant stretches from it in the direction of the extremity of the arc; but, in the second and third quadrants, the cosine shifts to the opposite side, and the secant shoots from the centre in a direction opposite to the termination of the arc.


The same phases are thus repeated at each succeeding revolution. Hence, if m denote any integral number, the sine of an arc a is equal to the sine of the arc (2m—1) 180°–a, and to opposite sines of (2m-1) 180°-Ha and of 2m. 180°–a the cosine and secant of an arc a are equal to the cosine and secant of 2m. 180°–a, and to the opposite cosines and secants of (2m—1) 180°–a and of (2m—1) 180°--a ; and the tangent or cotangent of an arca is equal to the tangent or cotangent of the arc (2m—1) 180°-Fa, and to the opposite tangents or cotangents of the arcs (2m—1) 180—a and 2m. 180—a.

An arc may, by a simple extension of analogy, be conceived to comprehend innumerable other arcs. Thus, the arc AB, in fact, represents all the arcs which have their origin at A and their termination at B ; it therefore includes not only the small arc AB, but that arc as augmented by successive revolutions, or the repeated addition of entire circumferences. Hence the sine or tangent of an arc a are the same with the sine or tangent of any arc m.360°-Ha.


The rectangle under the radius and the sine of the sum of two arcs, is equal to the sum of the rectangles under their alternate sines and cosines.

Let A and B denote two arcs, of which A is the greater; then, R.sin(A+B)=sinA.cosB+cos A.sinb.

For it is evident that AC will represent the sum of the arcs AB and BC; make BC equal to BC, and join OB and CC, and draw HFH' parallel, and CE, FG, BD, and HC'E' perpendicular, to the radius OA.

The triangles COF and COF, having the side CO equal to C/O, OF common, and the contained angles FOC and FOC' measured by the equal arcs BC and BC, are equal; wherefore —É OF bisects CC at right angles. But a H}{N-II the triangles OBD and OFG being N a similar, OB: BD :: OF : FG, or HE, and consequently OB.HE= / BD.OF. The triangles OBD and CFH are likewise similar, for the right angle CFObeing equal to HFG, if HFObe taken from both, the remaining angle CFH is equal to OFG or OBD; whence OB : OD: : CF : CH, and OB.CH=OD.CF. . Wherefore OB.HE+OB.CH, or OB.CE–BD.OF-H OD.CF. But BD and OD are the sine and cosine of the arc AB, CF and OF the sine and cosine of BC, and CE is the sine of the compound arc AC. Consequently, R sinAC=sinAB cosBC+cos/AB sinBC.

[ocr errors]
« ZurückWeiter »