Example. Given tan 30 cot 2 0, find 0. = To solve equations of this type, formulæ (19) and (20) are used. Example. Solve the equation sin 5 x- sin 3x + sin x=0. sin 5 x — sin 3 x = 2 cos 1 (5 x + 3 x) sin † (5 x − 3 x) 67. Simultaneous equations involving trigonometric functions can in many cases be solved. Example 1. Given y = 1 — cos x, y = 1 + sin x, find x Example 2. Given r cos (-) = a, rcos (0-3)-a, find r and 6. We have 68. Equations Involving Inverse Trigonometric Functions may, in general, be solved by transforming to other, equivalent equations involving the direct functions. The method of solution is illustrated by the following examples. Example 1. Solve the equation 2 tan-1 x = cot-1 x. or That is, Example 2. Solve the equation cos-1x + sin-1 2 x = 0. We have sin (cos-1x+sin-12x) = 0, sin (cos-1x) • cos (sin-12 x) + cos (cos ̄1 x) · sin (sin−1 2 x) = 0. That is, ± √1 − x2 . √1 − 4 x2 + x ⋅ 2 x = 0. Whence, (1 - x2)(1 − 4 x2) = 4 x4, 5 x2 = 1, x = ± 1. √5 A second method of solving example 2 is as follows: In √5 every case the values must be checked by substitution in the original equation. It is often convenient, in dealing with inverse functions, to assume the angle whose function is given and to construct a figure to show the values of the remaining functions. Thus, in example 1, we wish to find cot (tan-1 x). Let tan1x = a and construct the angle a, Fig. 30, with ordinate equal tox and abscissa equal to ±1. The distance is then √1+x2, and all the functions are readily found. Thus, cot (tan-1 2) = 1. Similarly, in example 2, we wish x) to find sin (cos1) and cos (sin-12x). By figure 31 we see assuming cos ̄1xa and sin-12 x = b, that sin (cos ̄1x) V1-22 and cos (sin-12x)= ±√1 − 4x2. |