IV. RELATIONS BETWEEN THE TRIGONOMETRICAL FUNCTIONS OF THE SIDES AND ANGLES OF A SPHERICAL TRIANGLE. 37. To express the cosine of an angle of a triangle in terms of sines and cosines of the sides. B E Let ABC be a spherical triangle, O the centre of the sphere. Let the tangent at A to the arc AC meet OC produced at E, and let the tangent at A to the arc AB meet OB produced at D; join ED. Thus the angle EAD is the angle A of the spherical triangle, and the angle EOD measures the side a. From the triangles ADE and ODE we have DE AD2 + AE2-2AD. AE cos A, = DE2 = OD2 + OE2 - 20D. OE cos a; also the angles OAD and OAE are right angles, so that OD2 = OA2 + AD2 and OE2 = ОA2 + AE2. we have Hence by subtraction 0 = 20A2 + 2AD. AE cos A-20D. OE cos a; 38. We have supposed, in the construction of the preceding article, that the sides which contain the angle A are less than quadrants, for we have assumed that the tangents at A meet OB and OC respectively produced. We must now shew that the formula obtained is true when these sides are not less than quadrants. This we shall do by special examination of the cases in which one side or each side is greater than a quadrant or equal to a quadrant. (1) Suppose only one of the sides greater than a quadrant, for example, AB. Produce BA and BC to meet at B'; and put AB'=c', CB'= d. B B Then we have from the triangle ABC, by what has been already proved, cos a = cos b cos c + sin b sin c cos A. (2) Suppose both the sides which contain the angle A to be greater than quadrants. Produce AB and AC to meet at A'; put A'B=c, A'C=6'; then from the triangle ABC, as before, A B COS α = cos b' cos c + sin b' sin c' cos A'; but bb, c'π-c, A'A; thus T. S. T. cos a = cos b cos c + sin b sin c cos A. (3) Suppose that one of the sides which contain the angle A is a quadrant, for example, AB; on AC, produced if necessary, A A take AD equal to a quadrant and draw BD. If BD is a quadrant π B is a pole of AC (Art. 11); in this case a = and A 2 π 2 2 as c= Thus the formula to be verified reduces to the identity 0=0. If BD be not a quadrant, the triangle BDC gives cos a = cos CD cos BD + sin CD sin BD cos CDB, and this is what the formula in Art. 37 becomes when c = π (4) Suppose that both the sides which contain the angle A are quadrants. The formula then becomes cos a = cos A; and this is obviously true, for A is now the pole of BC, and thus A =α. Thus the formula in Art. 37 is proved to be universally true. 39. The formula in Art. 37 may be applied to express the cosine of any angle of a triangle in terms of sines and cosines of the sides; thus we have the three formulæ, cos a = cos b cos c + sin b sin c cos A, These may be considered as the fundamental equations of Spherical Trigonometry; we shall proceed to deduce various formulæ from them. 40. To express the sine of an angle of a spherical triangle in terms of the sides. √(1-cos3a - cos2 b - cos3 c + 2 cosa cos bcos c) sin b sin c The radical on the right-hand side must be taken with the positive sign, because sin b, sin c, and sin A are all positive. 41. From the value of sin A in the preceding article it follows that for each of these is equal to the same expression, namely, √(1-cos2a - cos3b - cos3c + 2 cosa cosb cos c) Thus the sines of the angles of a spherical triangle are proportional to the sines of the opposite sides. We will give an independent proof of this proposition in the following article. 42. The sines of the angles of a spherical triangle are proportional to the sines of the opposite sides. Let ABC be a spherical triangle, O the centre of the sphere. Take any point P in OA, draw PD perpendicular to the plane BOC, and from D draw DE, DF perpendicular to OB, OC respectively; join PE, PF, OD. Since PD is perpendicular to the plane BOC, it makes right angles with every straight line meeting it in that plane; hence PE2 = PD2 + DE2 = PO2 — OD2 + DE2 = PO2 – OE2; thus PEO is a right angle. Therefore PE=OP sin POE=OP sin c; and PD = PE sin PED = PE sin B = OP sin c sin B. Similarly, PD = OP sin b sin C; therefore therefore OP sin c sin B = OP sin b sin C'; sin B sin b sin C sin c The figure supposes b, c, B, and C each less than a right angle; it will be found on examination that the proof will hold when the figure is modified to meet any case which can occur. If, for instance, B alone is greater than a right angle, the point D will fall beyond OB instead of between OB and OC; then PED will be the supplement of B, and thus sin PED is still equal to sin B. 43. To shew that cot a sin b = cot A sin C+ cos b cos C. We have |