150. In any spherical triangle, the cosine of any angle is equal to the product of the sines of the other two angles into the

or, by (46), whence,

cos a' cos b' cos c' + sin b' sin c' cos A';

cos A cos B cos C — sin B sin С cos a; cos Asin B sin C cos a-cos B cos C. (153)

In like manner may be deduced

cos B sin C sin A cos b. cos C cos A,

(154)

cos Csin A sin B cos c-cos A cos B.

(155)

151. In any spherical triangle, the cotangent of one side into the sine of another side is equal to the cotangent of the angle opposite the first side into the sine of the included angle, plus the cosine of the second side into the cosine of the included angle. By (150) and (152) we have

cos acos b cos c + sin b sin c cos A,

cos c = cos a cos b + sin a sin b cos C;

Substituting these values of cos c and sin e in the first equation,

we obtain

cos a= (cosa cosb+sina sin b cos C) cosb+

or,

sin a sin b cos A sin C

sin A

cosa cosa cos2b+sina sin b cosb cos C+ sin a sinb cot A sin C. Therefore, transposing cos a cos3b, and observing that, by (11), cos a cos a cos2 b = cos a sin2 b,

we have

cos a sin3 b = sin a sin b cot A sin C+ sin a sin b cos b cos C, and dividing the whole by sin a sin b, we obtain

cot a sin b cot A sin C+ cos b cos C.

(156)

152. By interchanging the letters in (156), we obtain
cot a sin ccot A sin B+ cos e cos B,
cot b sin a = cot B sin C+ cos a cos C,
cot b sin ccot B sin A+ cos c cos A,

(157)

(158)

(159)

cot e sin a
cot e sin b

cot C sin B+ cos a cos B,

(160)

cot O sin A+ cos b cos A.

(161)

153. The formula developed in the preceding articles are general, and apply to every case of spherical triangles, but require some transformations to render them more convenient for logarithmic computations.

The formulæ (150), (151), and (152) of Art. 149 are considered the fundamental formulæ of spherical trigonometry, since from them all its other formulæ may be deduced.

154. To express the sine, cosine, and tangent of half an angle of a triangle as functions of the sides.

By means of (150) we have

cos a

cos b cos c

cos A =

sin b sin c

but this formula is not suited to logarithmic computation.

(162)

We then subtract each member of the equation from 1, obtain (Art. 63),

and

Substituting for 1

cos A its value, 2 sin2 A (80), we obtain

cos (b-c)

cos a = 2 sin (a + b −c) sin † (a−b+c), which, substituted in the preceding equation, gives

sin3 A =

sin(a−b+c) sin † (a+b−c)
sin b sin c

(163)

Let, now, s = half the sum of the sides of the triangle; then,

a+b c = 2 (s—c), a − b + c = 2 (s—b). Substituting these values in the last equation, and reducing, we

Adding each member of equation (162) to 1, and observing that 1 + cos A = 2 cos2 ≥ A (81), by means of (65), we have

cos2 A=

sin(a+b+c) sin (b+c-a)

sin b sin c

Introducing s= (a+b+c), and reducing, we have

Again, dividing (164), (165), and (166) by (167), (168), and

(169), respectively, we obtain

155. To express the sine, cosine, and tangent of half a side of a triangle as functions of the angles.

Making (A+B+C)= S, substituting, and reducing, we

cos (A+B+C) cos (B+C — A).
†

sin B sin C

Since S is always greater than 90° and less than 270° (Geom., Prop. X. Bk. IX.), cos S is always negative, and therefore cos S in the numerators of the first and third of the above sets of formulæ is essentially positive.

cos A+ cos B cos C = sin B sin C'cos am sin C cos a sin b, cos B+ cos A cos Csin A sin C cos bm sin C sin a cos b. Adding these equations, factoring, and reducing by (17),

(cos A+ cos B) (1 + cos C) =m sin C sin (a+b). (180)

Dividing (178) by (180), and multiplying by sin C,

Now, by means of (62), (63), and (74), we obtain

sin a + sin b
sin (a+b)