Now the value 2 is inadmissible, for the cosine of every angle That is, one value of which satisfies the equation is 60o. We shall explain hereafter our reason for writing the word one in italics. (2) To solve the equation 3 sin = 2 cos3 0. 3 sin 0 = 2 (1 − sin3 0); 2 is inadmissible, for the sine of an angle cannot That is, one value of which satisfies the equation is 30°. EXAMPLES.-XXIV. Find a value of which will satisfy the following equations: (11) 3 sin'-cos 0+ (/3+1) (1-2 sin 0) = 0. (12) 3 cos2 0 - sin3 0 + ( √ √/3 + 1)(1 − 2 cos 0) = 0. (13) sec 0. cosec + 2 cot 0 = 4. (15) cosec10 - sec1 0 = 2 cosec3 0 sec3 0. (14) sin + cos 0 = √2. We have already stated (Art. 70) that if the value of a Trigonometrical Ratio be given we cannot fix on one particular angle to which it exclusively belongs. This statement was confirmed by many of the conclusions at which we arrived in Chap. VIII. For instance, since of two values, one of which is 30° and the other 150o. Thus for each given value of any one of the Trigonometrical Ratios there are two angles and two only between 0° and 360° for ' which that Ratio is the same in magnitude and sign. 105. Suppose QE to be the primitive line, and QP the revolving line, and let the angle EQP, less than 360°, be denoted by A. Now suppose QP to make a complete revolution, that is, to start from the position indicated in the diagram and to revolve till it comes back to that position. Then the revolving line will have described an angle 360o + A. Our triangle of reference will then be the same for an angle 360°+A as for an angle A. and Hence, sin (360° + 4) = sin ▲, cos (360° + 4) = cos 4. And the same holds good for the other Ratios. Hence, expressing ourselves for brevity in the symbols of the circular system, if a be the circular measure of an angle for which any one of the trigonometrical ratios has an assigned value, 2+ a will represent an angle for which the value of that particular ratio is the same. Now let the revolving line make a second complete revolution, then it will have described an angle 2π + 2π+ α, or, 4π+a. And so 4+ a will represent an angle for which the value of the above-mentioned ratio will be the same. And, generally, if the revolving line, after having traced out the angle a makes n revolutions, 2n+a will represent an angle for which the value of any particular ratio is the same as it is for a. Now since n may have any integral value from 1 to ∞o there will be an infinite number of angles for which the value of any one of the Trigonometrical ratios is the same as it is for the angle a. Again, if a be the circular measure of an angle traced out by a line revolving in the positive direction, -(2π-a) will be the measure of an angle traced out by the line revolving in the negative direction, for which the triangle of reference will be the same as for the positive angle a. If the line then make n complete revolutions in the negative direction, 2nπ-(2-a) will represent an angle for which the value of any particular ratio is the same as it is for a. We can now explain the way in which general expressions are found for all angles which have a given trigonometrical ratio. 106. To find a general expression for all angles which have a given sine. Let a be an angle whose sine is given, are angles with the same sine. (Art. 105.) Secondly, reckoning in the negative direction, (1) (2) are angles with the same sine, n being any positive integer. Now the angles in (1) and (2) may be arranged thus: 2nπ+a, (2n+1) π - α, · (2n + 2) π + α, − (2n+1) π — all of which, and no others, are included in the formula nπ + (− 1)". α, where n is zero or any positive or negative integer, which is therefore the general expression for all angles which have a given sine. |