College trigonometry by William Le Roy Hart

By William Le Roy Hart

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Following facts: It ahhreviates the I. There is no tangent fQr 90°. FUNDAMENTAL points (x, y) from the table are plotted in Figure 53, and lie on the line l, which is the graph of 3x II. The absolute value of tan 8 becomes greater than any specified number, however large, for all angles 8 sufficiently near 90°. - 5y = 15. ILLUSTRATION tan 89° = 57; 1. From trigonometric tables it is found that tan 89° 59' = 3438; tan 89° 59' 59" = 206,265. 66. Variation of eot (), see (), ese (J Since cot 8 = Iitan 8, the angle 8 has no cotangent when tan 8 = 0, because 1/0 has no meaning.

Repeat the proof of identities (1), Section 59, by use of a figure in which (J is taken as an angle between 180° and 270°. Express the function sin (- 25°). 2. cas (- 128°). 6. csc (- 148°). 10. cas (- a). 14. sin 439°. 18. If f(x) = 3x2 - X - 5, find the value of each symbol. 22. f(3). 23. f( - 1). 24. f(i). 25. ). If G(y) = 2y 26. G(a). 30. If F(x, y) - 73 3y2,find an expression for each symbol. ~7. G(2b). 28. G(y + 4). 29. G(tx). T- y2,find F(3, 2); F( - 1, 4). 60. Identities for functions of (0 + 180°) and (0 - 180°) sin (0 + 180°) = - sin 0; csc (0 + 180°) = - csc 0; (I) cos (0 + 180°) = - cos 0; see (0 + 180°) = - see 0; (2) tan (0 + 180°) = tan 0; cot (0 + 180°) = cot O.

The bearing of OB is 840° W. To describe the bearing of a line, we first write the letter N (or 8), then the acute angle between the given line and ON (or 08), and finally write E or W to show on which side the given direction falls. N K I I L-IJ; D w B S FIG. 38 EXAMPLE1. 3 miles east of O. OF RIGHT TRIANGLES 61 SOLUTION. 1. 3. 3 OF OF ; 2. In f::,OFK, tan a = cos a or OK = -. 0. 0 miles from 0 in the direction N 68° 39' E. OD and OF are, respectively, the east and north projections, or components, of OK.

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