with the second member of equation (m), we perceive that equation (m) is readily reduced to Considering (b+c) as one are, and then making application of equation (18), Plane Trigonometry, we have, The second member of this equation gives the value of the cosine when the radius is unity. To a greater radius, the cosine would be greater; and in just the same proportion as the radius increases, all the trigonometrical lines increase; therefore, to adapt the above equation to our tables where the radius is R, we must write R in the second member, as a factor; and if we put it under the radical sign, we must write R2. For the other angles we shall have precisely similar equations: To deduce from formulæ (S), formulæ 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.' A1-cos.A. Substituting the value of cos. A, taken from formula (S), and we have, 2sin.'A=1 But, cos.(bsc) cos.a-cos.b cos.c sin.b sin.c (sin.b sin.c+cos.b cos.c) cos.a (0) = Sec. I, Plane Trig.). sin.b sin.c sin.b sin.c+cos.b cos.c, (Eq. 10, Considering (bsc) as a single arc, and applying equation 18, Sec. I, Plane Trig., we have Dividing equation (o') by 2, and making these substi tutions, we have —b=S―b. sin.24 = sin.(S—c) sin.(S—b) 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 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; 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°-a), But, (2) 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 formulæ (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.A = sin.B sin. C cos.a - cos.B cos. C sin. A sin. C cos.b cos.A cos. C (V) sin. A sin.B cos.c cos.A cos.B. From these we can find formula 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. Acos. B cos. C + sin.B sin. C But, cos.A+cos.(B-C) 1 + cos.a = 2cos.'a; hence, 2cos.1 a = cos.A + cos.(B — C') sin. B sin.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=28; then A+B-C-28-20, A+ C―B=2S—2B, (A + B—C) S― C, and †(4 +C—B) = S-B; whence To find the sin.a in terms of the functions of the angles, we must subtract each member of eq. (3) from 1, by which we get But, 1-cos.a 2sin.'a; hence we have, 2sin.'Ja= (sin.B sin. C-cos. B cos. C)-cos.A Operating upon this in a manner analogous to that which cos.a was found, we get, |