EXERCISES In the following exercises, express the given function as a function of an acute angle. Since 270° = 3 × 90° is an odd multiple of 90°, take the cosine of the acute angle 20°. Since sin 290° is negative, the minus sign precedes cos 20°. -sin 70°. Here the multiple of 90° Again, sin 290° = sin (360° — 70°) = is even, therefore the same function, that is, the sine of the acute angle 70°, is taken. The sign is determined as before. These results do not differ, for -cos 20° = -sin 70°, the angles being complementary and the functions co-functions. It is to be noted that any angle that is not a multiple of lies between two consecutive multiples of, and either multiple may be used in the reduction. 2. tan 165° = tan (90° +75°) = -cot 75°. Since 90° is an odd multiple of 90° we take cot 75°, and since tan 165° is negative a minus sign is prefixed to cot 75°. = - cos 825° = cos (2 × 360° + 105°) = cos 105° = cos (90° + 15°) = −sin 15°. 5. cot (-1115°) -cot 1115° =-cot (12 x 90° +35°) = -cot 35°. 6. By the use of the table of natural functions, find correct to four decimal places the sine, cosine, tangent, and cotangent of the following angles: -115°, 153°, 212°, 265°, 295°. 7. Find sine, cosine, tangent, and cotangent of 120°, 135°, 150°, 210°, 225°, 300°, and 330° by expressing them in terms of functions of 30°, 45°, or 60°. Compare the results with the table of values given in Art. 16. sin (0) cos (} π + 0) tan (+0) 8. Simplify sin ( . Verify Exercises 9 to 15. Ans. -1. tan 180° tan of 0. 52. Proof of the reduction formulas of Art. 51 for any value Although the reduction formulas of Art. 51 were proved for acute values only of 0, they are true in general. The proof can readily be carried Y These results are seen to agree with the results previously determined for the functions of π + where 0 is acute. In a similar manner, the formulas for functions of ± 0, π0, etc., for any value of may be proved. П 53. Given the function to find the angle. In Art. 12 it was proved that for any angle there is but one value for each of the trigonometric functions. The converse of this proposition is not true, however. It will now be shown that for every function of an angle 0, there are any number of angles, both positive and negative, which satisfy the relation expressed. If is restricted to positive values not greater than 360°, it will be shown that there are, in general, two values of 0. Since angles with the same terminal side have the same values as functions, it will then follow that an integral multiple of 360°, or 2 π radians, added to or subtracted from these two values will give other angles having the same values as functions. П In the following articles only positive values of less than 360° will be considered. The general expressions for such angles may be obtained by adding multiples of 2 π to each angle. 54. Values for all angles that have a given sine or cosecant. Given sin = c, where c is a positive or negative number not greater than 1 nor less than -1. (1) Suppose that c is positive. Draw a unit circle with its center at the origin, Fig. 65. Draw RN | X'X at a distance c above it. RN intersects the circle at P1 and P2. Draw OP1 and OP2 forming the angles 01 and 02, and draw M1P1 and M2PLX'X. From the geometry of the figure 0201, that is, two angles less than 360° having the same positive sine are supplementary. +. It is to be noted that, if c = 1, 0, (2) Suppose that c is negative. 02, and = 0. A similar discussion for the case when c is a negative number, as in Fig. 66, shows that sin 03 04 = π + φ. In general, then, the terminal sides of angles that have the same sine differ by the same angular magnitude from the axis of ordinates. That is, from π, ¦ π, § π, еtс. = Thus, sin 30° sin 150°, because 90° - 30° = 150°-90°, that is, both 30° and 150° differ from 90° by the angle 60°. sin 20° = sin 220° sin 160° because 90° - 20° = sin 320° because 270° - 220° = 160° - 90°. = 320° - 270°. Example. Given sin = -0.25882, find all values of ◊ < 360°. ✪ Solution. = From the tables we find that when sin 0 = +0.25882, Ө 0 15°. But when sin = -0.25882, 0 is in the third and fourth quadrants. Now all angles having equal sines differ by the same angle from 90°, 270°, etc., p = 90° — 15° = 75°. = 345° or 195°. A similar discussion shows that the rule for determining all angles that have a given cosecant is the same as that for all angles which have a given sine. 55. Values for all angles having the same cosine or secant.— Given cos = c, where c is a positive or negative number not greater than 1 nor less than -1. Draw a unit circle as in Fig. 67. Draw RN || Y'Y, at the distance c from it and cutting the circle at P1 and P4. Draw OP1 and OP4. Then cos 01 OM1 = cos 04. = A similar discussion applied to Fig. 68 shows that cos 02 = cos 03. But P4OH = HOP1 and P2OM2 = ZM2OP3, therefore, in general, the terminal sides of angles that have the same cosine differ by the same angular magnitude from the axis of abscissas. That is, from 0, π, 2π, etc. Thus, cos 50° = cos (360° - 50°) = cos 310°. cos 160° = cos (180° + 20°) = cos 200°. = = Example. Given cos 0 = 0.57358, find all values of <360°. 0 Solution. From the tables we find that 55° when cos 0 0.57358. Since the cosine is positive, is in the first and fourth quadrants. Therefore, by the rule, 0 = 0° 55° 55° or 360° - 55° = 305°. + = A similar discussion shows that the rule of this article applies when the angle is determined from the secant. 56. Values for all angles that have the same tangent or cotan = gent. Given tan 0 c, where c is any number positive or neg- If c is negative, similarly tan 02 = HP2 tan (+92). Therefore, in general, all angles that have the same tangent differ by integral multiples of π. IR Y H P3 P2 IN FIG. 69. Thus, tan 30° = tan (180° + 30°) =tan 210°, tan 160° = tan (180° + 160°) = tan 340°. Example. Given tan = -0.70021, find all values of <360°. Solution. From the tables we find when tan = 0.70021 that = = -0.70021 the angle is 35°. But when tan 0 and fourth quadrants. Now since tan and tan in the second are equal in absolute value but opposite in sign, angle in the second quad rant is the supplement of . by the rule, = 145° + 180° = 325°. = A similar discussion shows that the rule of this article applies when the angle is determined from the cotangent. 57. Method by corresponding angles. The rules of Arts. 54, 55, and 56 for determining angles from their trigonometric functions may be stated in one general rule as follows: RULE. First find the acute angle which corresponds to the absolute value of the given function. The remaining, or corresponding, angles which have the same trigonometric function in absolute value are # ± 4 and 2 п 4. From these the angles can be chosen which satisfy the given function. Example 1. Given sin = 1; find <360°. 0 Solution. First find = sin-1 = 30°. By the rule, the remaining angles which have their sine equal to in absolute value are 180° - 30° = 150°, |