EXAMPLES. 1. Find the angle subtended by a man 6 ft. high at a distance of half a 2. What must be the height of a tower, in order that it subtend an angle 1° at a distance of 4000 ft.? = 3.14159265.) 3. Verify the statements made in Art. 11, Note 1. (Take π = 4. The moon's mean angular diameter as observed at the earth is 31' 5", and its actual diameter is about 2160 miles. Find the mean distance of the How many full moons would make a chaplet across the sky ? moon. 5. Taking the earth's equatorial radius as 3963 mi., find the angular semi-diameter of the earth as it would appear if observed from the moon. Compare the relative apparent sizes of the moon as seen from the earth, and the earth as seen from the moon. 6. The semi-diameter of the earth as seen from the sun is very nearly 8.8. (See Art. 11, Note 1.) What is the sun's distance from the earth, the radius of the earth being assumed as 4000 miles? 7. At least how many times farther away than the sun is the nearest fixed star a Centauri, at which the mean distance between the earth and sun (about 92,897,000 miles) subtends an angle something less than 1"? How long, at least, will it take light to come from this star to the earth? 8. Find approximately the distance at which a coin an inch in diameter must be placed so as just to hide the moon, the latter's angular diameter being taken 31′ 5′′. 9. The inclination of a railway to a horizontal plane is 50'. Find how many feet it rises in a mile. 10. Find the angle subtended by a circular target 4 ft. in diameter at a distance of 1000 yd. 11. Find the height of an object whose angle of elevation at a distance of 900 yd. is 1o. 12. Find the angle subtended by a pole 20 ft. high at a distance of a mile. 13. Exs. 5, 6, Art. 34 b. N.B. Questions and exercises suitable for practice and review on the subject-matter of Chapter X. will be found at page 195. CHAPTER XI. GENERAL VALUES. INVERSE TRIGONOMETRIC FUNCTIONS. TRIGONOMETRIC EQUATIONS. 84. General values. Articles 40-43 should be reviewed carefully before this chapter is taken up. It has been seen in these articles that all co-terminal angles have the same trigonometric ratios. It was also pointed out in Art. 43 that two sets of co-terminal angles, each set being infinite in number, correspond to any given ratio. For example, in Art. 42, Ex. 1, Fig. 38, any angle whose terminal line is either OP or OP1 has a sine, ; in Ex. 2, Fig. 39, any angle whose terminal line is either OP or OP has a tangent, —. One of the objects of this chapter is to derive expressions or formulas that will include all angles which have the same sine, cosine, tangent, cotangent, secant, cosecant, respectively. These general expressions are sometimes called general values. The student is advised to deduce, after reading Art. 85, the general values for cosine, tangent, etc., without the help of the book. 85. General expression for all angles which have the same sine. Let s be the given value of the sine. It is required to find an expression that will represent and include every angle whose sine is s. All the angles whose sines are equal to s can be repre sented geometrically, as shown in Art. 42, and indicated in Figs. 84, 85. In Fig. 84, s is positive; in Fig. 85, s is negative. Let XOP be the least positive angle whose sine is s. Let XOP=A; then XOP1 = 180° — A. Every angle whose terminal line is either OP or OP1 has its sine equal to s. Now all angles having OP for a terminal line are obtained by adding all numbers of complete revolutions (positive and negative) to XOP. Hence, these angles are represented by m • 360° + A, i.e. 2 m · 180° + A, (1) in which m is any positive or negative whole number. Similarly, all angles having OP1 for a terminal line are represented by m • 360° + (180° — A), i.e. (2 m + 1) 180° — A. (2) An expression that will include both sets of angles, (1) and (2), will now be obtained. In the expression (1), the coefficient of 180° is even, and the sign of A is positive; in (2), the coefficient of 180° is odd, and the sign of A is negative. Hence, n being any positive or negative whole number, the expression n⚫ 180° + (-1)" A, (3) includes the angles in (1) and (2). This is, accordingly, the general expression for all the angles which have the same sine as A. If radian measure is used, and XOP= α, then (3) takes the form nr + (-1)na. The result may be thus expressed: (4) sin Asin {n 180°+(-1)"A}, sin a=sin {n+(-1)"α}. (5) Since cosec 0 = 1 sin ' the general expression for all angles which have the same cosecant is the same as the general expression for all angles which have the same sine. EXAMPLES. 1. Find an expression to include all angles which have the same sine as 135°. By (3), (4), the expression is n · 180° + (− 1)” 135°, or nw + (− 1)1⁄23TM. 2. Find the general value of the angle whose sine is + least positive angles which have sines equal to + 1 √2 √2 Give the four Hence the general n. 180° + ( − 1)" 45°, i.e. n« +(−1)»TM. To find the four least positive angles, put n = 0, 1, 2, 3, in this expression. 86. General expression for all angles which have the same cosine. Let c be the given value of the cosine. It is required to find an expression to include every angle whose cosine is c. All the angles that have c for a cosine can be represented geometrically, as shown in Figs. 86, 87. In Fig. 86, c is positive; in Fig. 87, c is negative. Let XOP be the least positive angle whose cosine is c, and let XOPA (in degree measure) = a =a (in radian measure). Angle XOP1=- A=a, also has its cosine equal to c. All angles whose terminal line is OP, have cosines equal to c. All these angles are included in n · 360° + A, i.e. 2nπ +a, (1) in which ʼn denotes any positive or negative whole number. Also, all angles whose terminal line is OP1, have cosines equal to c. All these angles are included in n being as before. Both the expressions, (1), (2), are evidently included in n • 360° ± 4, or 2nπ±a, (3) in which n is any positive or negative whole number. Hence (3) is the general expression for all angles which have the same cosine as A or a. The result may be thus expressed: cos A = cos (n • 360° ± A); cos a = cos (2 në ± a). (4) 1 Since seco = the general expression for all angles which have the same secant is the same as the general expression for all angles which have the same cosine. EXAMPLES. 1. What is the general value of the angles which have the cosine, – †? Give the three least positive angles. The least positive angle whose cosine is, value is, by (3), n·360° ± 120°, i.e. 2 nπ ± , is 120°. Hence, the general π. On putting n = 0 and 1, the three least positive angles are found to be 120°, 360° - 120°, or 240°, 360+120°, or 480°. These three angles may also be found by means of a figure. 3. As in Ex. 2 when cos 0 = .99106. 4. As in Ex. 2 when cos 0= .46690. 5. As in Ex. 2 when cos 0=-.72637. 6. As in Ex. 2 when cos 0=-.40141. 87. General expression for all angles which have the same tangent. Let t be the given value of the tangent. It is required to find an expression to include all angles which have the same tangent t. All the angles which have the same tangent t can be represented geometrically as in Figs. 88, 89. In Fig. 88, the tangent t is positive, in Fig. 89, it is negative. |