(E) (F) BZ: CZ::GF: bg sine of an arc : its cosine::sine double arc: versed sine supplement of double arc. BG GF GF bG versed sine double arc : sine double arc:: sine double arc versed sine of the supplement of the double arc. (G) By comparing these proportions, with the proportions Prop. XI. and casting out the equal terms, various other portions may be formed. Thus (W. 103.) it is said pro radius: sine::secant: tangent, and (X. 106.) we have radius sine::double sine: versed sine double arc, therefore secant tangent::double the sine: versed sine double the arc, &c. for others. (H) COROL. I. The rectangle of the radius and the sine of any arc, is equal to double the rectangle of the sine and cosine of half that arc. For (A. 106) CB: cz::2bz: gf Therefore CB x gf=2bz x cz. (I) COROL. II. The rectangle of radius and half the versed sine of an arc, is equal to the square of the sine of half that arc. For (X 106.) CB BZ:: BF And CB BZ::BF BG That is CB BZ::BZ: BG BG Therefore CBX BG=BZ2; now CB is the radius, BG the versed sine of BF, and Bz is the sine of Bi, which is half the arc BF. (K) COROL. III. The rectangle of radius and half the versed sine of the supplement of an arc, is equal to the square of the cosine of half that arc. For (Z. 106) CB: cz::bF: bG CB: cz::bF: bG CB: cz::cz: 1bG 1bG=cz2. Therefore CB x bG-cz. Where CB is the radius, be the versed sine of the supplement of the arc BF, and cz the cosine of half the arc BF-cosine of the arc Bi. (L) COROL. IV. The diameter of the circle is to the versed sine of any arc, as the square of the radius is to the square of the sine of half that arc. For, (I. 107.) CB × BG=BZ2, multiply by the radius CB, then CB2 × BG=CB X BZ2 therefore CB that is 2CB BG:: CB2: BZ2 BG:: CB2: BZ2 (M) COROL. V. The rectangle of the sine of any arc, and of the cotangent of its half, is equal to double the square of the co sine of half that arc. And the same is true, writing tangent and sine, for cotangent and cosine. For, if any arc, sine of = COS A. rad A = (L.104.) and cos A cot A tang A And rad rad: cos A:: sine 2A= cos A::2 sine A: sine 2A; (A. 106.), hence 2 cos A. rad cot A 2 cos2 A cot A sine 2A; therefore ; or sine 2A. cot A=2cos A, that is A Again, by substituting the value of the cosine of a in the second term of the first proportion, we shall obtain sine 2A. tang A=2 sine A, that is sine A. tang A-2 sine2 A. (N) Using the same notation as in K. 104. the following formulæ may be easily obtained, by simple algebraical reductions.* (0) I. Sine 2A= 2 rad. sine A sec A 2 rad2 cot A+ tang A 2 cos A. sine A 2 sine2 A rad 2 rad. cos A CoSec A 2 rad2. cot a rad+cot A (P) II. Cos 2A = 2 cos2 A-rad2 rad tang A 2 cos2 A cot A 2 rad. tang A_2 rad. tang A rad+tang A rad 2 rad2. cot a cosec2 A sec2 A rad2-tang2 A .rad sec2 A cot A+tang A (Q) III. Tang 2A=rad-tang2 A sec A sec A rad. 2 rad2 cot A-tang ▲ . 2 rad. cos A. sine A 2 rad. cos A. sine A * Emerson's Trigonometry, 2d edit. Prop. II. Scholium. (X) The coversed sine may be expressed in terms of the rest by substituting its value in any of the above forms. Covers 2 Arad sine 2 A rad 2 cos A. sine A rad 2cosec A- 2sineA rad= • rad CoSec A 2 rad-vers A 2 rad 2 (U) VII. Vers 2A= 2 sine2 A 2 rad2 2 cos2 A rad 2vers ▲ . (2rad-vers a) ___ 2rad.tang2 a 2 rad3 rad+tang A (Y) Also if A be substituted for a in each of the foregoing expressions, we shall have, (H) By finding the values of sine a, cosa, tang a, &c. from the most convenient of the foregoing equations, the sine, cosine, &c. of the half arc will be obtained in terms of the sine, cosine, &c. of the whole arc, by easy algebraic reductions. sec A-rad (I) I. Sine & A=rad\/rad-cos A rad rad2 + (rad. cos a) 2 rad 2 sec A =√rad2 + (rad. sine a) + 1⁄2 √rad2 — (rad . sine a) = 2 And in the same manner the versed sines, coversed sines, chords, &c. of the half arcs may be found. GENERAL PROPERTIES OF SINEs, tangents, &c. OF THE SUMS, AND OF THE DIFFERENCES OF ARCS. PROPOSITION XIII. (Plate I. Fig. 2.) (P) The sum of the sines of two arcs is to their difference, as the tangent of half the sum of those arcs is to the tangent of half their difference. Let BA and Bo be the two arcs; draw the diameter BX, and OD and AG perpendicular to it. Produce OD to meet the circumference in F, and draw FHN parallel to the diameter; join AO and produce it to N, draw cnE perpendicular to Ao, and at E draw EIK perpendicular to cne, meeting co and CB (produced) in I and K. Because of is perpendicular to BC it is bisected in D (Euclid 3 of III.): hence PH is bisected in G; therefore AH is the sum of the sines AG and OD, and AP their difference. The arcs AO and of are bisected in E and Bв (Euclid 30 of III.), therefore AF |