If in these expressions, I-s be substituted for c2, in the sines of the odd multiples of a, and in the cosines of the even multiples,-the sines and cosines of such multiple arcs will be represented merely by the powers of the sine a. Sin 3a 3s-4s3. (3.) Sin 5a = 5s—20s3 +16s. Sin 7a7s-56s3+112s5-6457. (4.) Cos 4a = + 1 Cos 6a+ 1-18s2+48s4—32s*. &c. &c. &c. If the terms of the first table be repeatedly multiplied by 2s, and those of the second by 2c, observing the substitutions of cor. 2, there will result expressions for the sines and cosines. Thus, 2sin a2=2s.s= cos 2a + 1, 4 sin a3 - 2s.cos 2a+2s =— sin 3a + sin a + 2s = -sin 3a3s, and 8 sin a4=-2s.sin 3a+2s.3s+ cos 4a -cos 2a-3cos 2a+3=cos 4a-4 cos 2a+3. Again, 2 cos a2 =2c.c = cos 2a + 1,4 cos a3 = 2c.cos 2a+2c=cos 3a + cos a + 2c=cos 3a + 3 cosa, and 8 cos a* = 2c.cos 3a + 2c.3cosa cos 4a+cos 2a+3cos 2a+3=cos4a+4cos 2a+3. In this manner, the following tables are formed. The sum of the sines of two arcs is to their dif ference, as the tangent of half the sum of those arcs to the tangent of half the difference. If A and B denote two arcs; sinA+sinB: sin A-sinB A-B :: tan A+B 2 : tan 2 For, let AC and AC' be the sum aad difference of the H B O E F E'A KL arcs AB and BC, or BC'; draw the perpendiculars CE, and C'E', extend the chord CC', and apply at B the parallel tangent HBL, meeting in K and L the diameter produced, and draw OCH, OFB and OC'H'. Because CE is parallel to C'E', and CK to HL, CE: C'E':: CK: C'K (VI. 2. El.) HL: H'L; and consequently CE+C'E' : CE-C'E' HL+H'L: HL-HL', that is, 2BL: 2BH, or BL: BH. But CE and C'E' are the sines of the arcs AC and AC', and BL and BH are the tangents of AB and BC, or of half the sum and half the difference of those arcs. Wherefore sin AC+ sin AC': sinACAC+AC AC-AC 2 sinAC' tan 2 : tan Cor. 1. The sines of the sum and difference of two arcs are proportional to the sum and difference of their tangents. For CE C'E':: HL, or BL+BH: H'L, or BL-BH; that is, resuming the general notation, sin(A+B) : sin(A–B)::tanA+tanB:tanA—tanB. : Cor. 2. Let the greater arc be equal to a quadrant; and R+sinB: R-sinB :: tan(45° +B) : tan(45°—÷B), or cot (45°+B). But, the radius being a mean proportional between the tangent and cotangent of any arc, and the cosine of an arc being a mean proportional between the sum and difference of the radius and the sine, it follows that R+sinB: cosB:: R: tan(45°—B), and R-sin B: cosB, or cosB: R+sinB :: R: tan(45° +{B). Or, if instead of B, there be substituted its complement, these analogies will become R+cos B: sinB:: R: tan B, and R-cosB: sinB: R: cot B. Cor. 3. Since cosB: R:: R-sinB : tan(45°-B), and cosB: R:: R+sinB: tan(45° +B), therefore (VI. 19. El.) cosB: R 2R : tan(45°-B)+tan(45°+÷B); that is, supposing B to be the complement of 2C, sin2C: 2R:: R: tanC+cotC. But (Prop. 1. cor. 1.) R.sin2C=2cosC. sinC, and consequently cosC.sinC: R:: R: tanC+cotC. Cor. 4. Since (4 cor. def.) cosB: R::R: secB, and (3. cor. def.) cos: sin B:: R: tanB, therefore cos B: R+sinB :: R: tanB+secB, and consequently (2. cor. def.) tan(45°+÷B)=tanB+sec B.-This also appears clearly from the figure, on supposing OH'H'L', or the angle LOH' equal to OLH', and consequently the arc AC' equal to the complement of AB. PROP. V. THEOR. As the difference of the square of the radius and the rectangle under the tangents of two arcs, is to the square of the radius,-so is the sum of their tangents, to the tangent of the sum of the arcs. Let A and B denote any two arcs; then, R*—tanA.tanB : R2 :: tanA+tanВ : tan(A+B.) In reference to a diagram, let AB and BC be the two arcs, AD and BE their tangents, and AF consequently. the tangent of their sum HC. From the centre O, draw to meet the extension of this tangent, draw OH perpen dicular to OD and OG making the angle AOG equal to BOC; and from D draw DI parallel to BE, or (I. 23. El.) OH. The triangle AOG is evidently equal (I. 20. El.) to BOE, and therefore AG equal to BE. Because the parallels BE and DI (VI. 2. El.) cut the diverging lines OD and OI, BE or AG : DI :: OB or OA: OD; but the right angled triangle DOH being (VI. 15. El.) divided by the perpendicular OA into similar triangles, OA: OD: : AH: OH, and consequently AG: DI::AH: OH, or by alternation AG AH :: : E B H DI OH. Again, since the parallels DI and OH are intercepted by the diverging lines FH and FO, (VI. 2.) DI: OH :: FD: FH; wherefore AG AH:: FD: FH, and (V. 10. El.) GH: AH :: DH: FH:: (V. 19. 1. cor. El.) DG: AF. Consequently (V. 25. cor. 2. El.) GH.AD : AH.AD::DG: AF; but (VI. 15. cor. El.) AH.AD=OA2, and hence GH,AD OA- AD.AG; AD.AG; wherefore OA-AD.BE: OA:: DG AF. Now OA is the radius, AD and BE the tangents of the arcs AB and BC, DG their sum, and AF the tangent of the compound arc AC; consequently the proposition is manifest. Cor. 1. Hence it follows, by changing the position of the figure-That as the sum of the square of the radius, and R |