therefore, if tables be constructed, exhibiting in numbers the sines, tangents, secants, and versed sines of certain angles to a given radius, they will exhibit the ratios of the sines, tangents, &c. of the same angles to any radius whatsoever. In such tables, which are called Trigonometrical Tables, the radius is either supposed 1, or some in the series 10, 100, 1000, &c. The use and construction of these tables are about to be explained. 8. The difference between any angle and a right angle, or between any arc and a quadrant, is called the Complement of that angle, or of that arc. Thus, if BH be perpendicular to AB, the angle CBH is the complement of the angle ABC, and the arc HC the complement of AC; also, the complement of the obtuse angle FBC is the angle HBC, its excess above a right angle; and the complement of the arc FC is HC. F H K L B D 9. The sine, tangent, or secant of the complement of any angle is called the Cosine, Cotangent, or Cosecant of that angle. Thus, let CL or DB, which is equal to CL, be the sine of the angle CBH; HK the tangent, and BK the secant of the same angle: CL or BD is the cosine, HK the cotangent, and BK the cosecant of the angle ABC. COR. 1. The radius is a mean proportional between the tangent and the cotangent of any angle ABC; that is, tan. ABC × cot. ABC=R2. For, since HK, BA are parallel, the angles HBK, ABC are equal, and KHB, BAE are right angles; therefore the triangles BAE, KHB are similar, and therefore AE is to AB, as BH or BA to HK. COR. 2. The radius is a mean proportional between the cosine and secant of any angle ABC; or cos. ABC X sec. ABC=R2. Since CD, AE are parallel, BD is to BC or BA, as BA to BE. PROP. I. In a right angled plane triangle, as the hypotenuse to either of the sides, so the radius to the sine of the angle opposite to that side; and as either of the sides is to the other side, so is the radius to the tangent of the angle opposite to that side. Let ABC be a right angled plane triangle, of which BC is the hypotenuse. From the centre C, with any radius CD, describe the arc DE; draw DF at right angles to CE, and from E draw EG touching the circle in E, and meeting CB in G; DF is the sine, and EG the tangent of the arc DE, or of the angle C. Also, because EG touches the circle in E, CEG is a right angle, and therefore equal to the angle BAC; and since the angle at C is common The two triangles DFC, BAC, are equiangular, because the angles DFC, BAC are right angles, and the angle at C is common. Therefore, CB: BA :: CD: DF; but CD is the radius, and DF the sine of the angle C, (Def. 4.); therefore CB: BAR sin. C. B D to the triangles CBA, CGE, these triangles are equiangular, wherefore CA: AB :: CE : EG; but CE is the radius, and EG the tangent of the angle C; therefore, CA: AB :: R: tan. C. COR. 1. As the radius to the secant of the angle C, so is the side adjacent to that angle to the hypotenuse. For CG is the secant of the angle G (def. 7.), and the triangles CGE, CBA being equiangular, CA : CB :: CE: CG, that is, CA: CB:: R: sec. C. COR. 2. If the analogies in this proposition, and in the above corollary be arithmetically expressed, making the radius = 1, they give sin. C = AB ; tan. C = Also, since sin. C=cos. B, because B BC AB sec. C = BC COR. 3. In every triangle, if a perpendicular be drawn from any of the angles on the opposite side, the segments of that side are to one another as the tangents of SCHOLIUM. D The proposition, just demonstrated, is most easily remembered, by stating it thus: If in a right angled triangle the hypotenuse be made the radius, the sides become the sines of the opposite angles; and if one of the sides be made the radius, the other side becomes the tangent of the opposite angle, and the hypotenuse the secant of it. PROP. II. THEOR. The sides of a plane triangle are to one another as the sines of the opposite angles. From A any angle in the triangle ABC, let AD be drawn perpendicular to BC. And because the triangle ABD is right angled at D, AB AD::R: sin. B; and for the same reason, AC AD: R: sin. C, and inversely, AD: AC sin. CR; therefore, ex æquo inversely, AB : AC sin. C sin. B. In the same manner it may be demonstrated, that AB : BC sin. C: sin. A. : B PROP. III. THEOR. A 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 to the tangent of half their difference. Let AB, AC be two arcs of a circle ABCD; let E be the centre, and AEG the diameter which passes through A; sin. AC+sin. AB : sin. AC -sin. AB :: tan. (AC+AB): tan. (AC-AB). Draw BF parallel to AG, meeting the circle again in F. Draw BH and CL perpendicular to AE, and they will be the sines of the arcs AB and AC; produce CL till it meet the circle again in D; join DF, FC, DE, EB, EC, DB. E B Now, since EL from the centre is perpendicular to CD, it bisects the line CD in L and the arc CAD in A: DL is therefore equal to LB, or to the sinę of the arc AC; and BH or LK being the sine of AB, DK is the sum of the sines of the arcs AC and AB, and CK is the difference of their sines; DAB also is the sum of the arcs AC and AB, because AD is equal to AC, and BC is their difference. Now, in the triangle DFC, because FK is perpendicular to DC, (3. cor. 1.), DK : KC tan. DFK tan. CFK; but tan. DFK=tan. arc. BD, because E L /H. the angle DFK (20. 3.) is the half of DEB, and therefore measured by half the arc DB. For the same reason, tan. CFK=tan. arc. BC; and consequently, DK: KC:: tan. arc. BD: tan. arc. BC. But DK is the sum of the sines of the arcs AB and AC; and KC is the difference of the sines; also BD is the sum of the arcs AB and AC, and BC the difference of those arcs COR. 1. Because EL is the cosine of AC, and EH of AB, FK is the sum of these cosines, and KB their difference; for FK=1FB+EL=EH +EL, and KB=LH = EH-EL. Now, FK KB:: tan. FDK : tan. BDK; and tan. DFK=cotan. FDK, because DFK is the complement of FDK; therefore, FK : KB :: cotan. DFK : tan. BDK, that is, FK KB: cotan. arc. DB: tan. arc. BC. The sum of the cosines of two arcs is therefore to the difference of the same cosines as the cotangent of half the sum of the arcs to the tangent of half their difference. COR. 2. In the right angled triangle FKD, FK: KD :: R: tan. DFK; Now FK=cos. AB+cos. AC, KD= sin. AB+sin. AC, and tan. DFK= tan. (AB+AC), therefore cos. AB+cos. AC: sin. AB+sin. AC:: R: tan. (AB+AC). In the same manner, by help of the triangle FKC, it may be shewn that cos. AB+cos. AC: sin. AC-sin. AB: R: tan. (AC—AB). COR. 3. If the two arcs AB and AC be together equal to 90°, the tangent of half their sum, that is, of 450, is equal to the radius. And the arc BC being the excess of DC above DB, or above 90°, the half of the arc BC will be equal to the excess of the half of DC above the half of DB, that is, to the excess of AC above 45°; therefore, when the sum of two arcs is 90°, the sum of the sines of those arcs is to their difference as the radius to the tangent of the difference between either of them and 45°. PROP. IV. THEOR. The sum of any two sides of a triangle is to their difference, as the tangent of half the sum of the angles opposite to those sides, to the tangent of half their difference. Let ABC be any plane triangle ; CA+AB: CA-AB:: tan. (B+C): tan. (B-C). For (2.) CA: AB:: sin. B: sin. C; and therefore (E. 5.) CA+AB: CA-AB:: sin. B+sin. C: sin. B-sin. C. Otherwise, without the 3d. Let ABC be a triangle; the sum of AB and AC any two sides, is to the difference of AB and AC as the tangent of half the sum of the angles ACB and ABC, to the tangent of half their difference. About the centre A with the radius AB, the greater of the two sides, describe a circle meeting BC produced in D, and AC produced in E and F. Join DA, EB, FB; and draw FG parallel to CB, meeting EB in G. Because the exterior angle EAB is equal to the two interior ABC, ACB (25. 1.): and the angle EFB, at the circumference is equal to half the angle EAB at the centre (20.3.); therefore EFB is half the sum of the angles opposite to the sides AB and AC. Again, the exterior angle ACB is equal to the two interior CAD, ADC, and therefore CAD is the difference of the angles ACB, ADC, that is, of ACB, ABC, for ABC is equal to ADC. Wherefore also DBF, which is the half of CAD, or BFG, which is equal to DBF, is half the difference of the angles opposite to the sides AB, AC. Now because the angle FBE, in a semicircle is a right angle, BE is the tangent of the angle EFB, and BG the tangent of the angle BFG to the radius FB; and BE is therefore to BG as the tangent of half the sum of the angles ACB, ABC to the tangent of half their difference. Also CE is the sum of the sides of the triangle ABC, and CF their difference; and because BC is parallel to FG, CE: CF:: BE: BG, (2. 6.) that is, the sum of the two sides of the triangle ABC is to their difference as the tangent of half the sum of the angles opposite to those sides to the tangent of half their difference. |