I. Then let KOF= A, then IOF= A + 90°. .. (1) becomes cos (A + 90°) = − sin A (2) becomes sin (A + 90°) = + cos A II. Let IOF -- A, then KOF = A − 90°. 327. .. (1) becomes sin (A - 90°) = cos A (2) becomes cos (4 − 90°) = + sin A................ .(3), ..(4). .(5), .(6). These formulæ may be applied to find the ratios of A 180°, A 270° &c., in terms of ratios of A. Thus sin (4+180°) i.e. sin {(A + 90°) +90°} = cos (4 +90°) by (4) - sin A by (3), cos (4 + 180°) i.e. cos {(4 + 90°) + 90°} = − sin (A + 90°) by (3) = cos A by (4), and so on. 328. In the next articles we give the usual proofs of the formulæ for the sine and cosine of the sum and difference of two angles, confining ourselves to positive acute angles. 329. To find the sine, cosine, and tangent of the sum of two angles, in terms of the ratios of the angles. Let XOY=A, and YOZ = B, then XOZ = A + B. In OZ, the final line of the compound angle A + B, take any point P. From P, draw PM, PQ perpendiculars upon OX, OY-the initial and final lines of A-respectively. From Q, draw QN, QR perpendiculars upon OX, MP respectively. Then 4 QPR 90° - 4 RQP = LOQR = L XOQ = A. = [Or: RP, QP being perpendiculars upon OX, OY contain the same angle as OX, OY.] = [introducing required hyps. under the numerators] = sin NOQ. cos QOP + cos QPR . sin QOP = sin A. cos B + cos A. sin B. ON OQ RQ QP OQ OP QP' OP = cos NOQ. cos QOP - sin QPR . sin QOP = cos A. cos B-sin A. sin B. Since the triangles NOQ, RPQ are similar, 330. To find the sine, cosine, and tangent of the difference of two angles in terms of the ratios of the angles. Let XOY = A, and ZOY = B. Then XOZ = A − B. In OZ, the final line of the compound angle AB, take any point P. From P draw PM, PQ perpendiculars upon OX, OY—the initial and final lines of A-respectively. From tively. draw QN, QR perpendiculars upon OX, MP respec Then 4 QPR = 90° RQP = 4 RQY = 4XOQ=A. [Or: RP, QP being perpendiculars upon OX, OY contain the same angle as OX, OY.] = [introducing required hyps. under the numerators] = [introducing required hyps. under the numerators] = cos NOQ. cos POQ + sin QPR. sin POQ Since the triangles NOQ, RPQ are similar, 331. We may combine the above two figures into one, if we dash the letters Z, P, M, R in the second figure. Thus In OY take any point Q; through Q draw PQP' perpendicular to OY, cutting OZ, OZ' in P, P'. and = Draw QN, PM, P'M' perpendiculars on OX; RQR' perpendicular to QN, cutting PM, P'M' in R, R'. Then OP = OP', RQ = QR', RP = P'R'. Also LQP'R' = QPR = 90° - L RQP = ¿OQR=A, sin (4 + B) + sin (4 – B)=sin XOZ + sin XOZ sin (A + B) – sin (4 – B) = sin XOZ − sin XOZ (NQ+RP)-(NQ-P'R') 2. RP 2. RP = QP ОР cos (AB) + cos (A + B) = cos XOZ' + cos XOZ= = _ = (ON + QR')+(ON-RQ) 2.ON 2.ON OQ OP cos (AB) - cos (A + B) = cos XOZ' – cos XOZ = = 2 sin A sin B. 332. Taking the angle XOZ to be P, and XOZ' to be Q in the above figure, from Art. 116, we see that and XOY= (XOZ+XOZ) = (P+Q) YOZ = (XOZ- XOZ') = } (P – Q). Hence the above formulæ take the form sin P + sin Q = 2 sin † (P + Q) cos sin P - sin Q = 2 cos |