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If in these expressions, 1—so be substituted for co, 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.
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+sinE : sin A-sinb
For, let AC and AC be the sum aad difference of the arcs AB and BC, or BC'; draw the perpendiculars co 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 : HL; and consequently CE +CE : CE—C'E' : : HL-H HI, : 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+sinAC': sinAC—
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 : HL, or BL–BH ; that is, resuming the general notation, sin(A+B) : sin(A–B) :: tan A+ tanb : tan A–tanb.
Cor. 2. Let the greater arc be equal to a quadrant; and R+sin B : R—sin B :: tan(45°-i-#B) : tan(45°—#B), or cot(45°-i-;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+sin B : cosB : : R : tan(45°—#B), and R—sin B : cosB, or cosB : R+sin B : : R : tan(45°-- B).
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 al’CS.
Let A and B denote any two arcs; then, R*—tan.A.tan B : R* :: tan A+tan B : 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 perpendicular to OD and OG making the F 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 ! : 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=OA”, and hence GH, AD = OA* – 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