is equal to that of the same function of 0. Whether the signs agree or not will depend on the angle and the function in question. For example: To construct the second figure, lay off the angles 0, 90° - 0, 90° + 0, 270° – 0, and 270° + 0, with OX for the initial line. On their terminal lines take P(x, y), P1(x1, Y1), P3(x, y) and P4(4, P P P2(X2, Y2), 4), so that = 24. Since P is symmetrical to P with respect to the bisector of the first quadrant (why?), X1 У and Ут = X. P2, P3, and P4 are symmetrical to P1 with respect to the y-axis, the origin, and the x-axis respec In tively. Hence the numerical values of x1, x2, x3, and x4 are equal to y, and those of y1, y2, Yз, and y1 are equal to x. spection of the definitions on page 166 shows that if x and y are interchanged, each function is replaced by its co-function. Hence the numerical values of any function of one of the angles 90° ± 0 or 270° ± 0 is equal to that of the co-function of 0. ± Ø) whence In this survey we have taken tions of are positive. Hence f(n90°) will equal = f(0), or co-f(0), according as f(n90° 0) is positive or negative, which can be determined readily in a given case. Ө To illustrate the use of the rule, express cos (270° - 0) in terms of a function of 0. Since 270° = 3.90°, and 3 is odd, we will obtain the co-function, sin 0. If 0 is acute, 270° - 0 is in the third quadrant, in which the cosine is negative (see graph). Hence cos (270° — 0) sin 0. (7) Writing down a formula by the rule does not constitute a proof. The proof of (7), for example, which is identical with part of (6), is given above. The graphical significance of any one of the formulas is easily found. Consider, for example, formulas (1) and (4), and construct the graph of sin 0. Let OA Let OA = 0, and construct OB = 180° – 0, OC 180° + 0, OD = 360° 0. Then by (1), the ordinates at A and B are equal, and by (4), the ordinates at C and D are equal nu merically to that at A, but are opposite in sign. If increases from 0° to 90°, it follows -1 that the graph consists FIG. 114. of four congruent parts. Periodicity of tan 0. From (5) it follows at once that the period of tan is 180°, or π. The following formulas are stated for purposes of reference: If 360° are subtracted from the angle on the left in (11), (12), 69. Application to the Use of Tables. A positive angle less than 360° may be put in the form 180° - 0, 180° +0, or 360° -0, (1) where 0 is acute, according as the angle lies in the second, third or fourth quadrant. Hence the functions of such an angle may be expressed in terms of the functions of an acute angle 0, and the latter may be found from the tables. The functions of a positive angle greater than 360° may be found by using the periodicity of the function, and then proceeding as above; for example, sin 985° = sin (2·360° + 265°) = sin 265° = sin 265° = sin (180° + 85°) The functions of a negative angle are found by first using either the relations The functions of a positive acute angle less than 360° may also be found by putting the angle in one of the forms 90° + 0, 270° — 0, or 270° + 0, (2) where O is acute. The form (1) is somewhat less confusing because the application of the rule in Section 68 does not involve a change to the co-function. The examples following show how to find all the angles for which a given function has a given value. The solutions depend upon the fact that the numerical value of a function is the same for the three angles (1) as for the acute angle 0. EXAMPLE 1. Find all values of for which sin ✪ 0.4332. First find the positive values less than 360°. From the tables, one value is 0 = 25°.67. The graph of sin 0 shows that a line parallel to the 0-axis and 0.4332 unit above it cuts the graph in two points, one in the first quadrant, corresponding to the value 0 25°.67 found from the tables, and one in the second quadrant. Formula (1) Section 68, shows that the second value is 0 180° -25°.67 154°.33. All values of for which sin 0 = 0.4332 are given by Ө 0 = 25°.67 + n360° and 0 = 154°.33 + n360°. EXAMPLE 2. Find 0 if cos 0 0.5. If we neglect the negative sign and seek an acute angle whose cosine is 0.5, we know it to be 60° (table, page 161). The graph of cos 0 shows that a line parallel to the 0-axis and 0.5 unit below it cuts the graph in two points, one in the second and one in the third quadrant. Then formulas (2) page 192, and (9), page 195, show that the values of corresponding to these points are 0 = 180° - 60° = 120°, and 180° + 60° = 240°. All values of are then, by the periodicity of cos 0, Ө 120° + n360° and 0 = 240° + n360°. If we notice that one of the angles of the second set is, for n = 1, - 120°, which may also be obtained from = 120° by the relation cos (- 0) = cos 0, all the angles may be expressed by the single equation 0 ± 120° + n360°. EXAMPLE 3. Find all the values of ◊ if tan 0 = −√3. Neglecting the negative sign, we recognize that one value of ◊ is 60° (table, page 161). A line below the x-axis cuts the graph of tan 0 in two points, one in the second quadrant, and one in the fourth. Hence, by formulas (3) page 192, and (13), page 195, the required values of less than 360° are 180° - 60° = 120° and 360° - 60° 300°. Then all the required values are 0 = 120° + n360° and 0 = 300° + n360°. If we use the fact that the period of tan is 180°, noticing that 300° = 120° + 180°, all these angles may be expressed by the single equation Ꮎ 0 = 120° + n180°. EXERCISES 1. Prove formulas (8) (13), Section 68, for ◊ acute. 2. State, prove, and give the graphical significance of the formula for 3. Prove the six formulas for the functions of the angle: 4. By means of the proper formulas and the table on page 161 find (a) sin (— 60°), cos 300°, cos 240°, tan 315°. (b) sin 330°, cos (— 120°), cot 210°, sec 150°. (c) cos (- 135°), tan 120°, tan (– 150°), sin 225°. 5. By means of the formulas for the sine, cosine, and tangent, and the reciprocal relations, derive the formulas for the cotangent, secant and cosecant of (a) — 0; (b) 90° – 0; (c) 180° – 0; (d) 180° + 0; (e) 360° − 0. 6. Find all the functions of 142°.30; of 118°.17. 7. Find all positive values of less than 360° for which sin (90° – 0) (b) cos (0 – 180°). (c) tan (0 (c) tan (0-270°). 10. Construct a table of values of 0 and sin 0 for values of 0 taken every 10° from 0° to 360°, expressing 0 in radians decimally instead of in terms of π (see Tables, page 32), and giving the values of 0 and sin @ to two decimal places. Construct the graph as accurately as possible from this table of values. 11. As in the preceding exercise, construct a table of values and draw the graph of (a) cos 0, (b) tan 0, (c) cot 0, (d) sec. 0, (e) csc 0. |