84. Some relations between inverse functions derived from the formulas for double angles, half angles, and the addition formulas.

The general method applicable to this class of problems will be illustrated by a few examples.

1. To express cos (2 sec-1 u) in terms of u. cos (2 sec-1 u) = 2 (cos sec-1 u)2

1 Art. 71, Eq. 4.

2. To express tan (1 cos-1 u) in terms of u.

u2

3. To express sin (sin-1 u + cos-1 v) in terms of u and v.

85.

sin‍1 u + cos−1 v = sin−1 (uv ± √1 —u2 √1 — v2).

EXAMPLES

Find the value of each of the following:

1. sin-1V3.

3. tan-11.

2. cos1(√3).

4. tan cot-14.

Express the following in terms of u and v:

Find x in terms of a and b.

34. tan-1x= sin-1a+sin-1b.

35. cos-1x sec-1a - sec-1 b.

36. sin1x=2 cos1a + cos1b.

CHAPTER VIII

OBLIQUE TRIANGLE

86. In the present chapter we develop the formulas by means of which a triangle may be completely solved when any three independent parts are given.

87. Law of sines. In a plane triangle any two sides are to each other as the sines of the opposite angles.

Let a, b, c be the sides of a triangle and a, ß, y the angles opposite these sides, respectively.

From the vertex of y draw h perpendicular to the side c, or c produced.

Dividing the first equation by the second, we have

In a similar manner, dropping a perpendicular from the vertex of a, it is seen that

88. Law of tangents. The tangent of half the difference of two angles is to the tangent of half their sum, as the difference of the corresponding opposite sides is to their sum.

From the law of sines we have

89. Cyclic interchange of letters. Each formula pertaining to the oblique triangle gives rise to two other formulas of the same type by a cyclic interchange of

letters.

β

a

α

с

A cyclic interchange of letters may be accomplished by arranging the letters around the circumference of a circle as in the figure, and then replacing each. letter by the next in order as indicated by the arrows. Thus by this cyclic interchange of letters formula (1) of Art. 88 gives rise to formula (2); likewise formula (2) gives rise to formula (3).

90. Law of cosines. The square of any side of a plane triangle is equal to the sum of the squares of the other sides minus twice the product of those sides into the cosine of the included angle.

Substituting values, this equation becomes

a2 = (b sin α)2 + (c − b cos α)2

91. To find the sine of half an angle of a plane triangle in terms of the sides of the triangle.

COS α =

(2)