To deduce from formule (S), formule for the sines of the half of each of the angles of a spherical triangle, we proceed as follows: From Eq. 35, Sec. I, Plane Trig., we have 2sin. A = 1- cos.A. Substituting the value of cos. A, taken from formule (S), and we have, But, cos. (bs c) = sin.b sin.c + cos.b cos.c, (Eq. 10, Sec. I, Plane Trig.). This equation reduces equation (0) to Considering (bsc) as a single arc, and applying equa tion 18, Sec. I, Plane Trig., we have 2sin. (+-) sin. (+-) (0) a+b-c a+b+c 2 sin.b sin.c 2 • - c = S-c, if we put S= 2 Dividing equation (o') by 2, and making these substi tutions, we have sin. A = sin.(S - c) sin. (S - 6) sin.b sin.c when radius is unity. The above equations are now adapted to our tables. We shall show the application of these formulæ, and those in group (T), hereafter. PROPOSITION VIII. The cosine of any of the angles of a spherical triangle, is squal to the product of the sines of the other two angles multiplied by the cosine of the included side, minus the product of the cosines of these other two angles. Let ABC be a spherical triangle, and A'B'C' its supplemental or polar triangle, the angles of the first being denoted by A, B, and C, and the sides opposite these angles by a, b, c, respectively; A', B', C'', a', b', c', denoting the angles and corresponding sides of the second. A' C' a' a A C B B' By Prop. 6, Sec. I, we have the following relations between the sides and angles of these two triangles. A'= 180° — a, B' = 180° — b, C' = 180° c; a' = 180° - А, в' = 180° - В, с' = 180° - С. The first of formulæ (S), Prop. 7, when applied to the polar triangle, gives cos.a' cos.b' cos.c' + sin.b' sin.c' cos.A' (1) which, by substituting the values of a', b', c', and A', becomes cos.(180°-A)=cos. (180° - B) cos. (180° - C) + sin.(180° - B) sin.(180° — C) cos.(180° -α), (2) But, cos.(180°-A) = - cos. A, etc., sin. (180°-B) =sin.B, etc.; and placing these values for their equals in eq. (2), and changing the signs of both members of the resulting equation, we get cos.A = sin.B sin. C cos.a - cos. B cos. C, which agrees with the enunciation. By treating the other two of formule (S), Prop. 7, in the same manner, we should obtain similar values for the cosines of the other two angles of the triangle ABC; or we may get them more easily by a simple permutation of the letters A, B, C, a, etc. Hence, we have the three equations cos.B cos.C cos. A = sin.B sin. C cos.a From these we can find formulæ to express the sine or the cosine of one half of the side of a spherical triangle, in terms of the functions of its angles; thus: Add 1 to each member of eq. (3), and we have 1 + cos.a cos. A + cos. B cos. C + sin. B sin.C sin.B siu.C and since cos. A + cos.(B - C) = 2cos.(A+B-C)cos. (A+C-B) (Eq. 17, Sec. I, Plane Trig.), we have 2cos.2 a = 2cos.(A + B - C)cos.(A + C-B) sin.B sin.C Make A+B+C=2S; then A+B-C = 2S-2C, A+C-B = 2S-2B, (A + B - C) = S - C, and (A +C-B) = S-B; whence To find the sin.za in terms of the functions of the angles, we must subtract each member of eq. (3) from 1, by which we get 1-cos.a=1 cos. A + cos.B cos.C But, 1-cos.a= 2sin.a; hence we have, 2sin.2a = (sin.B sin.C-cos. B cos.C) -cos. A sin.B sin.C Operating upon this in a manner analogous to that h which cos.ła was found, we get, If the first equation in (W) be divided by the first in If the value of cos.c, expressed in the third equation of group (S), Prop. 7, be substituted for cos.c, in the second member of the first equation of the same group, we have, cos.a = cos.a cos.26 + sin.a sin.b cos.b cos. C+ sin.b sin.c cos.A; which, by writing for cos.b its equal, 1-sin.b, becomes, cos.a=cos.a-cos.a sin. 26+sin.a sin.b cos.b cos. C+sin.b sin.c cos. A. Or, 0=-cos.a sin.2b+sin.a sin.b cos.b cos. Ctsin.b sin.c cos.A. Dividing through by sin.b, and transposing, we find, cos. A sin.c = cos.a sin.b-sin.a cos.b cos. C; cos.a sin.b-sin.a cos.b cos. C hence, cos. A = sin.c (1) By substituting the value of cos.c, in the second of the equations of group (S), Prop. 7; or, merely writing B for A, and interchanging b and a, in the above value, for cos. A, we obtain, cos.b sin.a-sin.b cos.a cos. C. cos.B= sin.. (2) |