cumference of the circle may differ by a line less than any given line, or, which is the same, may be nearly to one another in the ratio of equality. Therefore their like parts will also be nearly in the ratio of equality, so that the side of the polygon will be to the arch which it subtends nearly in the ratio of equality; and therefore, half the side of the polygon to half the arch subtended by it, that is to say, the sine of any very smail arch will be to the arch itself, nearly in the ratio of equality. Therefore, if two arches are both very small, the first will be to the second as the sine of the first to the sine of the second. Hence, from the sine of 52". 44"". 03''''. 45,, being found, the sine of 1' becomes known; for, as 52". 44"". 03""". 45,. to 1, so is the sine of the former arch to the sine of the latter. Thus the sine of 1' is found 0.0002908882. 4. The sine 1' being thus found, the sines of 2', of 3', or of any number of minutes, found by the following proposition. THEOREM. Let AB, AC, AD be three such arches, that BC the difference of the first and second is equal to CD the difference of the second and third; the radius is to the cosine of the common difference BC as the sine of AC, the middle arch, to half the sum of the sines of AB and AD, the extreme arches. D Draw CE to the centre: let BF, CG, and DH perpendicular to AE, be the sines of the arches AB, AC, AD. Join BD, and let it meet CE in I; draw IK perpendicular to AE, also BL and IM perpendicular to DH. Then, because the arch BD is bisected in C, EC is at right angles to BD, and bisects it in I; also BI is the sine, and El the cosine of BC or CD. And, since BD is bisected in I, and IM is parallel to BL, (2. 6.), LD is also bisected in M. Now BF is equal to HL, therefore, BF+DH-DH+HL =DL+2LH=2LM+2LH=2MH or 2KI; and therefore IK is half the sum of BF and DH. But because the B C M triangles CGE, IKE are equiangular, A F GK CE: EI:: CG: IK, and it has been HE shewn that EI=cos BC, and IK = (BF+DH); therefore R: cos BC: sin AC: (sin AB÷sin AD). Q. E. D. COR. Hence, if the point B coincide with A, R: cos BC:: sin BC: sin BD, that is, the radius is to the cosine of any arch, as the sine of the arch is to half the sine of twice the arch; or if any arch=A, sin 2A=sin Axcos A, or sin 2A 2 sin Axcos A Therefore also, sin 2'-2' sin l'xcos l'; so that from the sine and cosine of one minute the sine of 2' is found. H h Again, 1',2',3' being three such arches that the difference between the first and second is the same as between the second and third, R: cos 1':: sin 2: (sin l'+sin 3'), or sin l'+sin 3'=2 cos 1'+sin 2', and taking sin l' from both, sin 3-2 cos 1'xsin 2'-sin 1. In like manner, sin 4'-2' cos l'xsin 3'-sin 2', sin 5'-2' cos l'xsin 4'-sin 3', sin 6'2' cos l'xsin 5'-sin 4', &c. Thus a table containing the sines for every minute of the quadrant may be computed; and as the multiplier, cos 1' remains always the same, the calculation is easy. For computing the sines of arches that differ by more than 1', the method is the same. Let A, A+B, A+2B be three such arches, then, by this theorem, R: cos B:: sin (A+B): (sin A+sin (A+ 2B)); and therefore making the radius 1, sin A+sin (A+2B)=2 cos Bxsin (A+B), or sin (A+2B)=2 cos Bxsin (A+B)-sin A. = sin A cos A' By means of these theorems, a table of the sines, and consequently also of the cosines, of arches of any number of degrees and minutes, from 0 to 90, may be constructed. Then, because tan A= the table of tangents is computed by dividing the sine of any arch by the cosine of the same arch. When the tangents have been found in this manner as far as 45°, the tangents for the other half of the quadrant may be found more easily by another rule. For the tangent of an arch above 45° being the co-tangent of an arch as much under 45°; and the radius being a mean proportional between the tangent and co-tangent of any arch, (1 Cor. def. 9.), it follows, if the difference between any arch and 45° be called D, that tan (45° –D): 1 :: 1: tan (45°+D), so that tan (45°+D)=; 1 tan (45°-D) Lastly, the secants are calculated from (Cor. 2. def. 9.) where it is shewn that the radius is a mean proportional between the cosine and the secant of any arch, so that if A De any arch, sec A= 1 cos A The versed sines are found by substracting the cosines from the radius. 5. The preceding Theorem is one of four, which, when arithmetically expressed, are frequently used in the application of trigonometry to the solution of problems. 1mo, If in the last Theorem, the arch AC-A, the arch BC-B, and the radius EC=1, then AD=A+B, and AB=A-B; and by what has just been demonstrated, 1: cos B:: sin A: sin (A+B)+ sin (A−B), sin Axcos B sin (A+B) + (A-B). = 2do, Because BF, IK, DH are parallel, the straight lines BD and FH are cut proportionally, and therefore FH, the difference of the straight lines FE and HE, is bisected in K; and therefore, as was shown in the last Theorem, KE is half the sum of FE and HE, that is, of the cosines of the arches AB and AD. But because of the similar triangles EGC, EKI, EC : EI :: GE: EK; now, GE is the cosine of AC, therefore, R: cos BC:: cos AC: cos AD+ cos AB, or 1: cos B:: cos A: cos (A+B) + cos (A−B); and therefore, cos Ax cos B1⁄2 cos (A+B) + cos (A-B); 3tio, Again, the triangles IDM, CEG are equiangular for the angles KIM, EID are equal, being each of them right angles, and therefore, taking away the angle EIM, the angle DIM is equal to the angle EIK, that is, to the angles ECG; and the angles DMI, CGE are also equal, being both right angles, and therefore the triangles IDM, CGE have the sides about their equal angles proportionals, and consequently, EC: CG:: DI: IM; now, IM is half the difference of the cosines FE and EH, therefore, R: sin AC sin BC: or 1 sin A sin B: cos AB- cos AD, sin Axsin B= cos (A-B) - cos (A+B). 4to, Lastly, in the same triangles ECG, DIM, EC: EG:: ID: DM; now, DM is half the difference of the sines DH and BF, therefore, R: cos AC: sin BC : 1⁄2 sin AD- sin AB, or 1: cos A :: sin B : & sin (A+B)—1⁄2 sin (A+B), and therefore, cos Axsin B sin (A+B) – 1⁄2 sin (A —B). 6. If therefore A and B be any two arches whatsoever, the radius being supposed 1; From these four Theorems are also deduced other four. For adding the first and fourth together, sin Axcos B+cos Axsin B-sin (A+B). Also, by taking the fourth from the first sin Axcos B-cos Axsin B-sin (A-B). Again, adding the second and third, cos Axcos B+sin Axsin B-cos (A-B.); And, lastly, substracting the third from the second, cos Axcos B-sin Axsin B=cos (A+B). 7. Again, since by the first of the above theorems, sinAxcos B-sin(A+B)+sin(A-B),if A+B=S,andA-B=D, then (Lem. 2.) A= and B= S+D = S-D ; wherefore sin S+D 2 S-D sin S+ sin D. But as S and D may be any arches what 2 ever, to preserve the former notation, they may be called A and B, which also express any arches whatever: thus, sinA+BxcosA-B 2 A+B A-B 2 2 sin Xcas 2 2 · sin A+ sin B, or sin A+sin B. In the same manner, from Theor. 2 is derived, 2 cosA+B A-B Xsin =cos B-cos A; and from the 4th, 2 In all these Theorems, the arch B is supposed less than A. 8. Theorems of the same kind with respect to the tangents of arches may be deduced from the preceding. Because the tangent of any arch is equal to the sine of the arch divided by its cosine, tan (A+B)= sin (A+B). But it has just been shewn, that cos (A+B) sin (A+B)=sin Axcos B+cos Axsin B, and that cos (A+B)=cos Axcos B-sin Axsin B; therefore tan (A+B) sin Axcos B+cos Axsin B cos Axcos B-sin Axsin B' , and dividing both the numerator and denominator of this fraction by cos Ax cos B, tan (A+B)= In like manner, tan (A—B)= -tan A+tan B I tan Axtan B tan A tan B 1+tan Axtan B' 9. If the theorem demonstrated in Prop. 3. be expressed in the same manner with those above, it gives And by Cor. 2, to the same proposition, sin A+ sin B_tan (A+B), or since R is here supposed cos A+cos B R 10. In all the preceding theorems, R, the radius is supposed=1, because in this way the propositions are most concisely expressed, and are also most readily applied to trigonometrical calculation. But if it be required to enunciate any of them geometrically, the multiplier R, which has disappeared, by being made=1, must be restored, and it will always be evident from inspection in what terms this multiplier is wanting. Thus, Theor. 1, 2 sin A x'cos B=sin (A+B)+ sin (AB), is a true proposition, taken arithmetically; but taken geometrically, is absurd, unless we supply the radius as a multiplier of the terms on the right hand of the sine of equality. It then becomes 2 sin Axcos B=R (sin (A+B)+ sin (A−B)); or twice the rectangle under the sine of A, and the cosine of B equal to the rectangle under the radius, and the sum of the sines of A+B and A-B. In general, the number of linear multipliers, that is, of lines whose numerical values are multiplied together, must be the same in every term, otherwise we will compare unlike magnitudes with one another. The propositions in this section are useful in many of the higher branches of the Mathematics, and are the foundation of what is called the Arithmetic of sines. |