69. Application to the Use of Tables. A positive angle less than 360° may be put in the form

180° +0,

or 360° -0,

(1)

180° - 0, where 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 (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 is acute. The form (1) is somewhat less confusing because the application of the rule in Section C3 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.

=

0.4332.

EXAMPLE 1. Find all values of 0 for which sin 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 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.

=

=

=

for which sin 0 = 0.4332 are given by

= 25°.67 + n360° and 0 = 154°.33 + n360°.

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 = 180° - 60° = 120°, and 0 = 180° + 60° = 240°. All values of 0 are then, by the periodicity of cos 0,

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

EXAMPLE 3. Find all the values of 0 if tan 0 = −√3.

Neglecting the negative sign, we recognize that one value of is €0° (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

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

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 0 less than 360° for which

(a) cos 0 (b) sin 0 (c) sec 0

=

=

0.6460; cos 0 = 0.5348; tan = 2.638. 1.4788; csc 0 = 4.865; cot 0 = 33.96.

9. Express the following as functions of 0.

=

1.4770.

sin (0 – 90°) = sin [− (90° – 0)] = − sin (90° – 0)

(b) cos (0 - 180°). (c) tan (0 – 270°).

(d) sin (0 - 180°).

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 0 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 (d) sec. 0, (e) csc 0.

(a) cos 0,

(b) tan 0,

(c) cot 0,

70. Inclination and Slope of a Straight Line. The function tan 0 enables us to express precisely the relation between the slope and the direction of a line (see

page 52).

The upper, right-hand angle (the northeasterly angle) which a line makes with the x-axis is called the inclination of the line.

Parallel lines have the same inclination, and conversely (why?).

на

FIG. 115.

Consider any line through the origin. Its slope m is the ratio of the difference of the ordinates of any two of its points to the difference of the abscissas. For these points take the origin and any point P(x, y) on the line, above the x-axis. then have, in either figure,

We

Hence the slope of the line through the origin is the tangent of the inclination. As parallel lines have the same slope and the same inclination, the result holds for any line. We thus

Theorem 1. The slope of a line is the tangent of the inclination, i.e., m = tan 0.

In most of the applications of the linear equation y = mx + b, the slope is of greater importance than the inclination, because the slope is the rate of change of y with respect to x.

The chief importance of the theorem above is in geometry. For from the slope the direction of the line may be found, provided the same unit has been used on the x and y-axes.

(see

Let a line start in a horizontal position and rotate about one of its points through half a revolution. Its inclination will increase from 0° to 180°. The variation of tan graph, page 176) shows that the slope of the line, m = tan 0, will increase from zero through all positive values, and become infinite as the inclination approaches 90°, when the line becomes vertical. As the inclination increases from 90° to 180°, the slope is negative, and its numerical value decreases to

zero.

EXAMPLE 1. Find the angle formed by the lines and l' whose equations are 3x 4y+ 120 and 2x + y - 8 = 0.

-2

-1 0

FIG. 117.

From the tables and the rule in Section

69, the inclinations of the lines are

0 = 36°.87; 0′ = 180° – 63°.43 = 116°.57.

If a is the angle between the lines, then a = 0' – 0 (why?). Hence

If two lines are perpendicular, then, using the notation in Example 1, a = 90°, so that '90° +0. Hence

That is, the slope of one is the negative reciprocal of the slope of the other. The converse may be proved by retracing the steps in the reverse order. Hence we have

Theorem 2. Two lines are perpendicular if and only if the slope of one is the negative reciprocal of the slope of the other.