PROP. I. THEOR. The rectangle under the radius and the sine of the sum of two arcs, is equal to the sum of the rectangles under their alternate sines and cosines. Let A and B denote two arcs, of which A is the greater; then, R.sin(A+B)=sinA.cos B+cos A.sinB. For it is evident that AC will represent the sum of the arcs AB and BC; make BC' equal to BC, and join OB and CC', and draw HFH' parallel, and CE, FG, BD, and HC'E' perpendicular, to the radius OA. B H F O E GD E' A The triangles COF and C'OF, having the side CO equal to C'O, OF common, and the contained angles FOC and FOC' measured by the equal arcs BC and BC', are equal; wherefore OF bisects CC' at right angles. But the triangles OBD and OFG being similar, OB: BD:: OF: FG, or HE, and consequently OB.HE= BD.OF. The triangles OBD and CFH are likewise similar, for the right angle CFO being equal to HFG, if HFO be taken from both, the remaining angle CFH is equal to OFG or OBD; whence OB OD:: CF: CH, and OB.CH=OD.CF. Wherefore OB.HE+OB.CH, or OB.CE=BD.OF+ OD.CF. But BD and OD are the sine and cosine of the arc AB, CF and OF the sine and cosine of BC, and CE is the sine of the compound arc AC. Consequently, R sin AC sin AB cos BC+cos AB sin BC. Cor. 1. Hence, likewise, the rectangle under the radius and the sine of the difference of two arcs, is equal to the difference of the rectangles under their alternate sines and cosines; or R sinAC'sin AB cosBC-cos AB sinBC. Cor. 2. If the two arcs A and B be equal, it is obvious that R sin2A=sinA 2cos A. Cor. 3. Let the arc A contain 45°; then R sin(450B)=sin45°(cosBsinB)= √†R2 (cos B sinB) or R sin(45° B)=R√(cos B±sinB). Cor. 4. Let 2A=C, and, by the second corollary, PROP. II. THEOR. The rectangle under the radius and the cosine of the sum of two arcs, is equal to the difference of the rectangles under their respective cosines and sines. Let A and B denote two arcs, of which A is the greater; then R cos(A+B)=cos A cos B-sinA sinB. For, in the preceding figure, the triangles OBD and OFG being similar, OB: OD :: OF: OG, and OB.OG= OD.OF, and the triangles OBD and CFH being likewise similar, OB: BD:: CF: FH, or GE, and consequently OB.GE=BD.CF. Wherefore OB.OG OB.GE= OB.OE=OD.OF-BD.CF; that is, R cosAC=cos AB cosBC-sinAB sin BC. Cor. 1. Hence, likewise, the rectangle under the radius and the cosine of the difference of two arcs is equal to the sum of the rectangles under their respective cosines and sines; or R.cos AC' = cos AB cosBC+sinAB sin BC. Cor. 2. If A and B represent two equal arcs, it will follow, that R.cos2A=cos A-sin A=(cosA+sin A)(cos A-sinA); or, since cos A2 — R2—sinA2, = R cos2A=R2sinA=2cos A*-R2. Cor. 3. Since, sinA2=+R(R—cos2A), and sin B2 = R(R-cos2B); therefore sin A-sinB = R(cos2 B-cos2A). Cor. 4. Let the arc A be equal to 45°, and Cor. 5. Let 2A=C, and by the second corollary, PROP. III. THEOR. Of the equidifferent arcs, the rectangle under the radius and the sum of the sines of the extremes, is equal to twice the rectangle under the cosine of the common difference and the sine of the mean arc. Let A-B, A, and A+B represent three arcs increasing by the difference B; then R(sin(A+B)+sin(A—B))=2cosB sinA. 1 20F.BD; that is, R(sinAC+sinAC')=2cos BC sinAB. Cor. 1. Hence, likewise, of three equidifferent arcs, the rectangle under the radius and the difference of the sines of the extremes, is equal to twice the rectangle under the sine of the common difference and the cosine of the mean arc; or R(sin(A+B)—sin(A—B))=2sinB cosA. Cor. 2. Hence R(cos(A-B)+cos(A+B))=2cosB cos A, and R(cos(A-B)—cos(A+B))=2sinB sinA. For OB OD:: OF: OG::20F: 20G or OE'+OE, and OB(OE'+OE)=20F.OD; that is, R(cos AC'+cosAC)=2cos BC cosAB. Again, OB BD CF: FH 2CF: 2FH, or : :: : : OE'-OE, and OB(OE'—OE)=2CF.BD; that is, R(cosAC-cosAC)=2sinBC sinAB. Cor. 3. Let the radius be expressed by unit, and the arcs B and A, denoted by a and na; then collectively 2sin a.cos na sin(n+1)a-sin(n-1)a, Cor. 4. Since versB=R-cos B, it follows that and consequently R sin(A+B)=2R sin A—R sin(A—B)— 2versB sinA, or R(sin(A+B)—sinA)=R(sinA—sin(A–B)) -2versB sinA. In the same way, it may be shown that R(cos(A-B)—coșA) = R(cosA—cos(A+B))—2versB cosA. Cor. 5. If the mean arc contain 60°, then R(sin(60°+B) -sin(60°-B))=2sinB cos60°, or sinB 2sin30°. But twice the sine of 30° being (cor. 1. def.) equal to the chord of 60° or the radius, it is evident that sin(60° + B)— sin(60°-B)=sinB, or sin(60°+B)=sin(60°—B)+sinB. Cor. 6. Produce CE to the circumference, join C'I meeting the production of FG in K, and join OK. Since FK is parallel to CI and bisects CC', it likewise bisects IC'; and hence OK is perpendicular to KC', which is, therefore, the sine of half the arc IAC', or of half the sum of the arcs AC and AC', as CF is the sine of half their difference. But (II.21.El.)IC-CC"2=IC.2C'E', or C’K2-CF2 =CE.C'E'; consequently sin2 AB-sin2 BC=sinAC sinAC', or, employing the general notation, |