- 1 < s < 1, is satisfied by THEOREM I. The equation sin x = s, where exactly two angles between 0° and 360°.* Every angle congruent to either of these is a solution, and conversely, every solution of the equation is congruent to one or the other of these angles. y B a A FIG. 69 -X The following construction will indicate the method of solution of such an equation for any particular value of s. Draw a pair of coördinate axes and a unit circle whose center is at the origin. On the y-axis lay off OB = s (above O if s > 0, below if s< 0); draw through B a line parallel to the This line cuts the circle in two and only two points, C and D Draw the radii OC and OD. Then the positive angles x-axis. α = AOC, B = AOD are two angles (and the only two angles between 0° and 360°) such that sin α = sin ẞ= 8. Every angle congruent to either of these has the same sine. Therefore every such angle is a solution of the given equation. Conversely, if any angle y is a solution, when placed upon the axes, its terminal side must fall either upon OC or OD since no other radius meets the circle at height s. Hence y must be congruent to a or to ß. THEOREM II. The equation cos x = c, where 1<c<1, has exactly two solutions between 0° and 360°. Every angle congruent to either of these two is a solution, and conversely, every solution is congruent to one or the other of them. C y B X To see this, draw a unit circle as for theorem I and lay off on the x-axis, OB = c (to the right if c> 0, to the left if c < 0); through B draw a line parallel to the y-axis; this line meets the circle in two and only two points, C and D. The positive angles α = AOC, B= AOD are two angles (and the only angles between 0° and 360°), which satisfy the equation cos x = c. The student will easily see, as in theorem I, that all angles congruent to either of these are solutions, and that every solution is congruent to a or to ß. FIG. 70 THEOREM III. The equation tan xt, where t is any real number, has exactly two solutions between 0 and 360°. Every angle congruent to either of these two is a solution, and conversely every solution is congruent to one or the other of these two angles. *In this chapter, by "between 0° and 360°" we shall mean 0° included and 360° excluded; i.e. "x between 0° and 360°" means 0°< x < 360°. To see this, draw a unit circle as in the previous theorems and draw the tangent TAT'; on this tangent lay off AB = t (upward if t>0, downward if t<0); through B draw a diameter; this diameter meets the circle in two and only two points C and D. The positive angles α = AOC, B = AOD are the only solutions of the equation tan x = t between 0 and 360°. The student may complete the demonstration as in the previous theorems. THEOREM IV. The equation ctn x = 0 is equivalent* to cos x = 0. The equations ctn x = c, sec x = c, csc x = c, where c 0, are equivalent, sin x = The proof, which is respectively, to tan x = 1 с 80. Use of the Graph. A second method for solving the equation sin x =s is as follows: Plot the graph of sin x, on the y-axis lay off OA = s (above if s> 0, below if s<0), and draw through O a line parallel to the x-axis. If −1<s<1, this line will cut the curve in points P1, Q1, P2, Q2, etc., and the projections of these points on the x-axis determine the angles α1, B1, α2, B2, etc., which are solutions of the equation sin x = s. It is obvious that aα2, B1 B2, etc. This method has an advantage over that shown on p. 94, in that it shows graphically more than the two solutions which lie between 0° and 360°; the former method has, however, the advantage that it requires very much less time to make the construction accurately. This method can clearly be applied to any of the trigonometric functions. *Two equations are equivalent when every solution of either is a solution of the other also. [NOTE. The equations (k) and (1) are not included under theorems I and II, but the student can readily solve them by a similar method.] 2. 2 sin2 x + sin x = = 1. [HINT. Solve this quadratic for sin x and apply theorem I.] 3. (a) 2 sin2 x − 5 sin x + 2 = 0. 4. (a) tan x = 1. (d) tan x = 2.6. (b) 4 cos2 0 + 8 cos 0 = 5. (b) tan x=- 1/2. (e) tan x = = 5.3. (c) tan x = 2. (f) tan x = 0. (c) tan2 0 = 6–4√2. (b) 3 tan2 x - 4√3 tan x + 3 = 0. (b) ctnx.73. (e) 2 ctn2 x 3.1. (b) cscx = 15. (b) secx 3 sec x + 2 = 0. 5 csc x + 2 = 0. (c) ctn x=- 1.31. 3 ctn x + 1 = 0. (c) secx = : 10.57. (c) csc x 7.4. = (b) sec2 x = √2 sec x. If a trigo 81. Reduction of Equations to Standard Form. nometric equation contains more than one of the trigonometric functions, all but one can usually be eliminated; the resulting equation may then be solved algebraically for the function which remains, as in Ex. 2, above; the solutions may then be found by the methods of § 79. Example 1. Solve the equation cos2 t sin2 t = sin t. In this equation cost may be replaced by its equal 1-sin2t; the equation then becomes a quadratic in sint, viz.: 2 sin2 t + sint 1 0. This equation is equivalent to the given one; i.e. every solution of either is a solution of the other. The solutions may now be found by factoring : Hence we have either sin t + 1 = 0, whence sint = −1, and t≈ 270°; or else, 2 sint 1 = 0, whence sin t = 1/2, and t≈ 30° or t≈ 150°. There are no other solutions. The process of solving equations consists chiefly in replacing one equation by another (or by a set of others) which is equiv alent, and which is more easily solved; in carrying out the necessary transformations, use may be made of any identities previously proved.* Certain operations (for example, squaring both sides of an equation) yield a new equation which, though not equivalent to the original, has all the solutions of the original equation, and perhaps other solutions. If such a transformation is employed, every solution of the transformed equation must be tested by substitution in the given equation. Example 2. Solve the equation cos x √3 sin x 10. Substituting √1 - cos2 x for sin x, transposing the radical, squaring both sides, and collecting terms, we obtain 4 cos2 x + 2 cos x 20. The solutions of this equation are all included in the set x ≈ 60°, x ≈ 180°, ≈≈ 300°. By substitution in the given equation we see that it is not satisfied by any of the values x 300°; hence these are not solutions of the given equation. Similarly, it is found that all the values x 60° and x 180°, do satisfy the given equation and these are therefore the values required. EXERCISES XXXII. — SOLUTION OF TRIGONOMETRIC EQUATIONS Solve completely the following equations: * Attention is called to the fact that we need for this process only the following rules of algebra: (1) The following changes in an equation lead to an equivalent equation: (a) transposition of terms, (b) multiplication (or division) of all the terms by the same constant 0. [Changing the signs of all the terms is the same as multiplying by — 1.] (c) substituting for one expression, another identically equal to it. (2) If an equation is of the form A = 0 (i.e. has all of its terms on the lefthand side), and if A can be factored into B x C so that the equation can be written BC= 0, then the equation A = 0 is equivalent to the pair B = 0, C = 0. In (1) (c), and (2) make sure that the expressions used are defined throughout the range of values of the unknown in which we are interested. H 82. Special Methods of Solution. An equation of the form a sin x + b cos xc, may be solved by the following device: h=Va2 + b2 B a FIG. 73 a or ẞ, of the a b Construct a right triangle whose sides are a and b and compute the hypotenuse h=√a2+ b2, as in § 64, p. 79. Divide the given equation through by h and replace a/h and b/h by the appropriate function of one of the acute angles, triangle. Then solve the equation as in § 79. Example 1. Solve the equation 3 sin x 4 cos x 2. Here the hypotenuse of the auxiliary triangle is 5 and the equation reduces to (3/5) sin x-(4/5) cosx = 2/5, or sin a sin x-cos α cos x =2/5, whence cos (a + x)= −2/5, where a 36° 52'. By theorem II, the complete solution of this equation is x + α ≈ 113° 35′ or x + α ≈ 246° 25'; i.e. x 76° 43' or x 209° 33'. If we had employed the angle ẞ = 53° 08′ instead of a, we should have obtained sin x cos ẞ cos x sin ß = 2/5. Show that this equation leads to the values of x found above. Many equations can be solved by the following principles: (1) If sin a sin ß, then aß, or a 180°-ẞ; and conversely. (2) If cos a = cos ß, then aß, or a≈- ß; and conversely. (3) If tan α=tan ß, then « ≈ ß, or a ≈ 180°+ß; and conversely, provided either a or ẞ has a tangent. Example 2. Solve the equation sin 50 = sin 2 0. By (1) either 50 ≈ 2 0 ; i.e. 3 0 = ± n ⋅ 360°, and hence 0 = ±n · 120°; or, 50 180°-20; i.e. 70 180°n 360° and = = (1 ± 2n) 180°/7. Example 3. Solve the equation cos (5 0/6) = cos (0/3). = = ±2 n. 360°; or 50/60/3, whence (6/7)n · 360° = ±n · (308° 34' 17'' 1/7). Example 4. Solve the equation tan 7x = tan 3 x. By (3) either 7 x 3 x, whence 4 x = n · 360°, that is, x = n ⋅ 90° ; or 7x3x + 180°, whence 4 x = 180° + n ⋅ 360° = (2 n + 1)180°, that is, X = (2n+1)45°. All these results may therefore be written x = n • 45°. Of these values, those which make x an odd multiple of 90° must of course be excluded; and the only solutions are x = n · · 45°, provided n is any integer except twice an odd number. Example 5. Solve the equation csc 2 x = csc 5 x. Any solution of the equation sin 2 x = sin 5 x, except those that make sin 2 x=sin 5x=0 are solutions of the given equation. Hence x = ±n · 120° are solutions, except when n is a multiple of 3; and x = (2 m±1)180°/7 are solutions, except when 2 m + 1 is a multiple of 7. See Example 1. |