H K к B D (sinus) of the arc, from one end of it, perpending upon the diameter which' meets the other end, -as CD. The greatest sine is that of a quadrant or 90 degrees; which is equal to the radius. 6. The tangent is a straight line touching the arc, parallel to the sine, and intercepted by the two sides which contain the angle, as AE. The tangent of 45 degrees is equal to the radius. 7. The secant is one of the sides which contain the angle, produced to meet the tangent, through the point in which the sine meets the arc, -as BE. 8. The cosine, cotangent, and cosecant, are the sine, tangent and secant, of the complement of the arc or angle. 9. The sine, tangent and secant of an arc, or angle, are also the sine, tangent and secant of its supplement. 10. The subsine, or versed sine, is a segment of the diameter intercepted between the sine and tangent, -as AD. 11. The subtense*, or hypothenuse, is the side opposite to a right angle. Note. It appears from the diagram, that any one of the lines, BA, BC, BH, &c., is a radius; that CD or BL is a sine; CL or BD a cosine; AE a tangent; HK a cotangent; BE a secant, and BK A cosecant. Moreover, if the arc AC, F and its complements CH were equal, the tangent, cotangent and radius, would also be equal; as also the sine and cosine. These lines form several equiangular triangles, whose sides are therefore proportionals, viz: 1. Cosine is to sine as radius is to tangent: Ex. BD: DC :: BA: AE. 2. Tangent is to radius as radius is to cotangent: Ex. AE : AB :: BH : HK. 3. Cosine is to radius as radius is to secant: Ex. BD: BC :: BA: BE. 4. Sine is to radius as radius is to cosecant: Ex. CD: CB :: BH : BK. 5. Sine is to cosine as radius is to cotangent: Ex. BL:LC :: BH: HK. 12. The sine, subsine,* tangent and secant of an arc, taken as the measure of an angle, is to the sine, subsine, tangent and secant of any other arc measuring the same angle, as the radius of the first arc is to the radius of the other. E A Given the arcs AC and MN, as measures of the angle ABC ; and let CD be the sine, DA the subsine, AE the tangent, and BE the secant, of the arc AC. Also, let NO be the sine, OM the subsine, MP the tangent, and BP the secant, of the arc MN. The first lines are to the second, each to each, as the radius BC is to the radius BN. B * The terms subsine and subtense are more convenient than versed sine and hypothenuse. 03 Cor. If numerical tables of sines, subsines, tangents and secants, of certain angles to a given radius, be constructed, they will show the ratios of the sines, &c., of the same angles to any radius whatever. Such are called Trigonometrical Tables; in which the radius is represented as some term of the decimal series, 1, 10, 100, 1000, &c. Propositions. 1. Th. In a right-angled triangle, the subtense of the right angle is to either side, as radius is to the sine of the angle opposite to that side ; and either of the sides is to the other, as radius is to the tangent of the angle opposite that other. Given the triangle ABC, right angled at A; the subtense BC, and the sides CA, AB. From the centre C, with any radius CD, describe the arc DE; draw DF at right angles to CE, and EG parallel to it, touching the arc in E, and meeting CB in G. F EA Then DF is the sine, EG the tangent, and CG the secant of the angle C, or of the arc DE. The three triangles DFC, GEC, BAC, are equiangular: because the angles at F, E, A are right angles; and the angle Č is common to all. Therefore, 1. CB : BA:: CD: DF;—but CD is the radius, and DF is the sine of the angle C (def. 5, trig.); Therefore CB : BA :: R: sine of the angle C. 2. CA : AB :: CE: EG;-but CE is the radius, and EG is the tangent of the angle C (def. 6, trig.); Therefore CA : AB :: R : tangent of the anglo C. Wherefore in a right angled triangle, &c. Q. E. D. Cor. 1. Radius is to the secant as the side adjacent to the angle C is to the subtense. For CD, or CE: CG :: CA : CB. Cor. 2. Let Radius =1; then the preceding analogies are as follows: Sine of the angle C=BA divided by CB. Cor. 3. If, in the triangle ABC, a perpendicular be drawn to the base BC; then (making AD radius), AD: DC ::R: tan. CAD, and AD: DB::R: tan. BAD. Cor. 4. Since, in a right angled triangle, the two acute angles are equal to the right angle, one of them is the complement of the other (def. 3 trig.), therefore any position in a proportion, which the sine, tangent, or secant of one of them takes, may be given to the cosine, cotangent, or cosecant of the other. 2 Th. The sides of a plane triangle are to each other as the sines of the angles opposite to them. In right angled triangles this proposition is obvious: for if the subtense be made radius, the sides are the sines of their opposite angles ; and the radius is the sine of 90°, that is, of a right angle (a). But in any oblique angled triangle, as ABC, any two sides, AC, BC, will have to each other the ratio of the sines of ABC, BAC, which are opposite to BL them. From the angles A, B, draw AD, BE, perpendicular to BC, AC, produced if required : then, if AB be made radius, AD will become the sine of the angle B; and BE will become the sine of the angle A. Since the triangles CDA, CEB, have right angles at D, E, and the angle C common, they are equiangular (6); therefore AC : AD:: BČ: BE, and alternatelv AC: BC:: AD: BE (c): that is, the sides are as the sines of the angles opposite to them. Wherefore, the sides of a plane triangle, &c. Q. E. D. Recite (a) def. 5, P. T. (6) p. 32, 1, and p. 4, 6; (c) p. 16, 5. Cor. Of two sides and two angles opposite to them, in a plane triangle, any three being given, the fourth is also given. Lemma 2. Of two unequal magnitudes, the greater equals half their sum more half their difference, the less equals half their sum less half their difference. Given two magnitudes AB, BC, of which AB is the greater. А, Е ов с And since AC=AB+BC, the sum; make AE=BC: therefore AB-AE=EB, the difference of the two magnitudes. Bisect AC in D: then since AD=CD, and AE=CB; taking these equals from the former, the remainders ED, BD are equal; and EB, the difference is also bisected in D. It is therefore evident that AB=AD+DB, and BC=AD-DB; that is, the greater equals half the sum + half the difference, and the less equals half the sum half the difference, as stated. Otherwise : let s represent the sum, and d the difference. sd). But f(std)=s+id; and 1(s-dl= d. arcs. 3 Th. The sum of the sines of any two arcs of a circle, is to the difference of their sines, as the tangent of half the sum of the arcs is to the tangent of half their difference. Let AB, AC be two arcs of a circle ABCD; E its centre ; AEG the diameter which pas B through A: it is to be proved that, the sin. AC+sin. AB : sin. AC—sin. AB::tan. I(AC+ А AB): tan. $(AC-AB). Draw BF parallel to AG, meeting the circle in F; draw the sines BH, CL perpendicular to AE ; produce CL to D; join BE, CE, CF, DE, DF. Since the perpendicular EL bisects CD and the arc CAD, DL equals CL, the sine of AC; the arcs AC, AD are also equal; and KL equals BH, the sine of AB. Therefore. DK is the sum, and KC the difference of the sines; and DAB is the sum, and BC the difference of the Now, in the triangle DFC, the perpendicular FK being made radius, DK: KC :: tan. DFK : tan. ČFK. But DK is the sum, and KC the difference of the sines : also DFK, CFK are halves of the central angles DEB, CEB, which are measured by the arcs DAB, BC, the sum and difference of the arcs AC, AB. Therefore the sin. AC+sin. AB : sin. AC—sin. AB : : tan. I(AC+ AB): tan. &(AC-AB). Q. E. D. 4 Th. In any triangle, the sum of two sides is to their difference, as the tangent of half the sum of the angles at the base is to the tangent of half their difference. Given the triangle ABC, the side AB being greater than AC; AB+AC: AB-AC: : tan. (ACB+ABC): tan. (ACB-ABC). With A as centre, and radius AB, describe the circle BEDF, meeting AC produced in E, F, and BC in D. Join EB, BF, AD; draw FG parallel to BC, meeting EB produced in G. Then, because of the equal radii, EC=AB+ AC, and CF=AB-AC: and since EBF is an angle in a semicircle, FB is perpendicular to EG; therefore EB, BG are the tangents of the angles BFE, BFG, or DBF, the alternate. Moreover, the angles DBF, BFE are halves of the central angles DAF, EAB; also the exterior EAB=ACB+ABC, and the exterior ACB=CDA+CAD: but, because of the equal radii, ABD=ADB; therefore CAD=ACB-ABC. Therefore EB is the tangent of half ACB+ABC, and BG is the tangent of half ACB-ABC. And because, in the triangle EFG, BC is drawn parallel to FG the base EC: CF :: EB : BG; that is, the sum of two sides is to their difference, as the tangent of half the sum of the angles at the base is to the tangent of half their difference. Therefore, in any triangle, &o. Q. E. D. 5 Th. The base of a triangle is to the sum of both the sides, as the difference of the sides is to the difference of the segments of the base, made by a perpendicular falling upon it from the vertex, or to the sum of the segments, when the base must be produced to meet the perpendicular. Given the triangle ABC, and AD, a perpendicular drawn from A to BČ: and let AC be greater than AB. With A, as centre, and AC radius, describe a circle, meeting AB produced, in E, F, and CB in G. E Then, because the chords CG, EF, in a circle, cut one another, the rectangles EBXBF, GBXBC are equal (a); therefore BC: BF:: BE : BG. But BC is the base, BF=AC+AB, BE=AC-AB, and BG=CD-DB, or CD+DB, if the perpendicular fall on the base produced. Wherefore, the base, &c. Q. E. D. Recite (a) p. 35 of b. 3. D AN B В 6 Th. In any triangle, the sum of two rectangles contained by the sides, is to the difference of the squares of those sides and the square of the base, as radius is to the cosine of the angle included by the two sides. Given the triangle ABC; in which AC is the base, and AB, BC the sides: let AB be the radius; then, 2ABXBC: AB+BC2-AC?::R: cos. B. Draw AD perpendicular to BC; then 2BCXBD=AB2+BC2-AC2 (a). But BCXBA:BCXBD::BA: BD (b) B Now 2BCXBD is the difference between AB%+BC” and AC2. BA: BD::R: cos. B, BD=BAXcos. B (a); and 2BCXBAX cos. B=2BCXBD (6); therefore when B is acute, 2BCXBAXcos. B=BC2+BA2-AC (C). Add to both AC2, then AC +2BCXBAX cos. B=BC2+BA%. Take from both 2BCXBAXcos. B; then AC?=BC2+BA-2BCXBAXcos. B. Wherefore, AC=V(BC2+BA?—2BCXBAXcos. B); and the same result will appear by a similar process when the angle B is obtuse (c). Recite (a) p. 16, 6; (b) p. 16, 6, and 15, 5; (c) p. 13, 12, of b. 2. |