8. = sine (P+Q) 2 sine (P+Q). cos (P+Q_sine (P+Q) sine P-sine 2sine (P-Q). cos (P+Q) ̄sine (P—Q) Now if two fractions be equal to each other, the numerator of the one is to its denominator, as the numerator of the other is to its denominator. The first equation is therefore the same as Prop. XIII. (or Q. 112.) The sixth is the same as Prop. XIV. (or S. 112.) (H) If we suppose B=A, we shall have, from the third equation, (E. 116.) cos2A=rad2 +(rad. cos 2 A); and from the fourth equation of the same article. Sine' Arad (rad. cos 2 A) (I) But when B=A, Q=0 (F. 116.) and cos o° = rad (D. 99.) COS (A+B) have from the first and second equations (D. 115.) tang (A+B)= (sine A. COS B)+(sine B COS A) rad, and since sine a = and sine B COS B. tang B ing these values and dividing the numerator and denominator rad (L. 104.) by substitut tang A+ tang B by cos A. cos B, we obtain tang (A+B)= rad2 - (tang A. tang B) rad', but this has been already obtained by another process (C. 115.) GENERAL PROPERTIES OF SINES, TANGENTS, &c. OF ARCS, IN ARITHMETICAL PROGRESSION. PROPOSITION XIX. (Plate I. Fig. 2.) Let the three arcs BO, BE, and BA, be in arithmetical progression; viz. let Ao be bisected in E, then AEOE is half the difference between the arcs BO and BA and therefore BE is half the sum of the arcs BO and BA (C. 35.) But BO+BABO+ (BO+OE+AE)=BO+ (BO+20E)=2B0 +20E=2BE, that this the sum of the extreme arcs, BO and BA, is equal to double the mean arc BE. Draw the diameter ACL, and join OL; draw Bz parallel to oa, and join xz; also from L, draw Lw perpendicular to ow. Then the following triangles are equiangular and similar, viz. CBR, VBZ, xzv, aop, and oLw. For CBR and VBZ are right angled at R and V, and have the angle at в common. xzv is right angled at v, and the angle at z is equal to the angle at B, for XZB is a right angle, being contained in a semicircle, and VBZ is the complement of VZB to a right angle, and so is xzv. The triangle AOP having the sides AO and OP parallel to the two sides ZB and BV of the triangle VBZ, is equiangular with it; the angle AOL is a right angle, being contained in a semi-circle, therefore woL, the complement of AOP, is equal to the angle OAP: hence the angle OLW is equal to the angle AOP. The following proportions, &c. are naturally derived from these similar triangles; but first we must observe that AG+OD AHWL the sum of the sines of the extreme arcs; for aĦ is parallel to wL, and the angles WLC and HAC being alternate angles are equal (Euclid 29 of I.), therefore the right angled triangles CGA, CYL having one side Ac=LC, and the angles equal, are equal in all respects; hence LY=AG, and wY=OD, consequently LwAG+OD. Now CG+CD is the sum of the cosines of the extreme arcs, but on account of the equality of the triangles. LYC and AGC, YC=CG, therefore Yc+CD=YD= wo, is the sum of the cosines of the extreme arcs. (L) (M) (N) (0) (P) (Q) (R) (S) (T) (U) (W) CB: BRBZ: BV radius: sine of the mean arc::double the sine of the radius: sine of the mean arc::double the cosine of CB: BRAO (=2on): OP (CD-CG) radius: sine mean arc::double the sine of the common difference between the arcs : difference between the cosines, or versed sines of the extreme arcs. CB: BR::OL (=2cn): Lw radius: sine mean arc::double the cosine of the common difference between the arcs: the sum of the sines of the extreme arcs. CB: CR::BZ (=2br): zv radius: cosine of the mean arc::double the sine of the mean arc: sine of double the mean arc. CB: CR::ZX (=2cr): xv radius: cosine mean arc:: double the cosine of the mean arc : versed sine of the supplement of double the mean arc. CB: CRAO (2on): AP (=AG-OD) radius: cosine of mean arc:double the sine of the common difference between the arcs: difference between the sines of the extreme arcs. CB: CR::OL (=2cn): wo (=CG+CD) radius: cosine of the mean arc:: double the cosine of the common difference between the arcs : sum of the cosines of the extreme arcs. BR CRBV: ZV sine of the mean arc: cosine of the mean arc:: versed sine of double the mean arc: sine of double the mean arc. BR: CRZV : xv sine of the mean arc : cosine of the mean arc::sine of double the mean arc : versed sine of the supplement of double the mean arc. BR: CR::PO (=CD—CG): AP (=AG—QD) sine of the mean arc : cosine of the mean arc:: difference between the cosines (=difference between the versed sines) of the extreme arcs: the difference between the sines of the extreme arcs. (X) BR: CR::LW (AG+OD): WO (=CG+CD) sine of the mean arc: cosine of the mean arc::the sum of the sines of the extreme arcs: the sum of the -cosines of the extreme arcs. (Y) By comparing the above proportions, various other proportions may be formed. Thus BZ: BV:: ZX: Zv by comzv paring (L. 119. with M.) that is double the sine of the mean arc versed sine of double the mean arc::double the cosine of the mean arc sine of double the mean arc, &c. for the rest. (Z) By what has been already shewn, two arcs may be compared, and their sums and differences formed into a proportion. The arcs being in arithmetical progression, the mean arc BE is half the sum of the extreme arcs Bо and BA. Therefore, in all the above proportions, and their variations, whereever the words mean arc occur, read half the sum of the arcs. Also the sine on, of the arc EO, the common difference between the arcs, is the sine of half the difference between the arcs BO and BA; hence in the above proportions for common difference between the arcs, read half the difference between the arcs. Thus for example (N. 119.) will run thus; radius: sine of half the sum of the arcs::double the sine of half the difference of the arcs: difference between the cosines of the arcs. (A) Hence it follows that the difference between the cosines of two arcs x radius=2 × sine of half the difference x sine of half the sum of these arcs. But this has been obtained in another manner, see the fourth equation, F. 116. Indeed from this proposition, the greater part of the formulæ noted in Prop. XVIII. may be deduced, and it will be no bad exercise for the young student to trace out the coincidence, PROPOSITION Xx. (Plate I. Fig. 2.) (B) If three arcs be in arithmetical progression, the sum of the sines of the extreme arcs, is to their difference, as the tangent of the mean arc is to the tangent of the common difference between the arcs. The same construction remaining as in the preceding proposition, AG+OD sum of the sines of the extreme arcs BA and BO, and AP AG-PG-AG-OD-the difference of the sines of the extreme arcs. But AG+ODAH, because GHDF=OD =PG. By Prop. XIII. AH: AP::EK : EI. Now EK is the tangent of the mean arc BE, and EI is the tangent of EO, the common difference between the arcs. Hence the proposition is evident. Q. E. D. (C) Viz. If the extreme arcs be represented by A and B, of which a is greater than B, then sine A+ sine B_tang (A+B) sine A-sine B tang PROPOSITION XXI. (A-B) (See Z. 120.) (Plate I. Fig. 2.) (D) If three arcs be in arithmetical progression, radius is to double the cosine of the middle arc, as the sine of the common difference of the arcs, is to the difference between the sines of the extreme arcs. Let the three arcs be BO, BE, and BA, as in the proceding propositions, and OEAE the common difference between the arcs. Then An is the sine of AE, and AP difference between the sines of BA and Bо (B. 120.) The triangles Ain, cno, are equiangular and similar, for inc and ing are right angles; if therefore inc be taken away from both, there will remain the cng; but the triangle cno is equiangular with the triangle CEM. Therefore CE cm::an: ai; but aiip, for Ani= an: ao::ai: ap, and since an=ao, ai=tap. Hence CE cm::An: iP, and doubling the consequents, Α Q.E.D. (E) Viz. If A and B represent the extreme arcs, a being the greater. PROPOSITION XXII. (Plate I. Fig. 2.) (F) If three arcs be in arithmetical progression, the rectangle of radius, and the sum of the sines of the extreme arcs, is equal to the rectangle of double the cosine of the common difference of the arcs, and the sine of the mean arcs. The same construction remaining as in the foregoing propo But AG+ODAH (B. 120.)=2ng. For PG GH (B. 120.) and Ai=ip (D. 121.); now Ai+GH= iP+PG no, therefore Ai+iP+PG+GHAH=2nQ. (G) Let the extreme arcs be represented by A and B, then the mean arc will be (A+B), and the common difference of the arcs (A-B). Hence, rad. (sine A+ sine B) = |