on the horizontal axis. The point b, thus determined, has as coördinates (15°, sin 15°). In the same way locate c, (30°, sin 30°); d, (45°, sin 45°); e, (60°, sin 60°); etc. Through these points draw a curve. With the sine curve thus constructed, one can determine the value of the sine of an angle approximately by measurement. For example, find the sin 51°. By measurement the ordinate for sin 51° is 23.3 spaces. Since the unit is 30 spaces, sin 51° 23.3 30 = From the table of natural functions sin 51° = = 0.7766. 0.77715. A comparison of the results for a number of angles will give an idea of the accuracy of the graph. Exercise. Measure the ordinates for the angles given in the following table, compute the sines, and tabulate the results. Find the sines of the same angles from the Tables and tabulate. Compare the results. 1. Plot y = cos e by both of the methods employed in plotting sin 0. 2. Plot y 3. Plot y = = tan 0 and y = cot on the same set of axes. sint, y = sin t, and y = sin 2 t on the same set of axes. Compare the curves as to the maximum and minimum values, and as to their periods. 4. Plot y = sint, y = 2 sint on the same set of axes. Compare the curves as directed in Exercise 3. sin x, and y2 = cos x, using the same units and Y1+y2 may be computed by means of the dividers. 61. Inverse functions. We have seen that sin-1t means the angle whose sine is t. In Art. 58, it was shown that the sine function varied from 1 to +1. Then the equation = sin-1t has real solutions when and only when t is not less than -1 nor greater than +1. In the same way it can be shown that @ = cos ̄1 t has a solution when and only when t is not less than -1, nor greater than +1. = Since tan and cot @ can have any value from ∞ to +∞ the equation = tan-1t and cot-1t have solutions for all values of t. The two expressions sint and = sin-1t both express the same thing, namely, that is an angle whose sine is equal to t. In the first expression t is a function of @ and in the second function of t. k 'y' y-sin-1x FIG. 73. is a In sin 0 = t, there is but one value of t for every value of 0. sin is then said to be a single valued function of 0. In 0 = sin-1t for every value of t, there are an indefinite number of values of 0, as was seen in Art. 50. sin-1t is then said to be a multiple valued function of t. In many mathematical operations where sin-1x enters, it is often desirable and, indeed, necessary to consider a portion of the curve, Fig. 73, for which there will be but one value of У for every value of x. A glance at the figure will show that for the portion AOC of the curve, the function is single valued. That is, for every value of x between and including -1 and +1, y takes values between and including andπ. Definition. The values of sin-1 and between and including - 1π for each value of x, are called the principal values of sin-1x. To represent the principal value of the function the s is often written a capital, thus, Sin-1 x. The other functions are denoted in a similar manner. EXERCISES In the following exercises, find the numerical values of the given expressions, using the principal values of the angles.* x = 5, then substitute x = 2, and subtract the second result from the first. Determine the result correct to the third decimal. Ans. 11.891. 2, and subtract the second result from the first. 63. Relation between sin 0, 0, and tan 0, for small angles. — Draw angle BOE = 0, Fig. 74. With O as a center and OB = 1 as radius, describe the arc BD. Draw DAL to OB and BE tangent to the arc at B. Then sin But A OBDOB X AD, sector OBDOB2 0, where is in radians, see Art. 5, and A OBE OB. BE. Then OB. AD < 1⁄2 OB2 · 0 < 1⁄2 OB · BE. Dividing by and substituting OB = 1, AD = sin 0, and BE = tan 0, Dividing [12] by sin and remembering that tan sin 0 cos 0 ' Now as approaches 0 as a limit sec @ approaches 1 as a limit, written lim In sec many applications of anti-functions, as in the calculus, they enter into the expressions for areas, volumes, etc., and the angles must be expressed in radians. Then since sin is always less than a quantity which approaches 1 as a limit, and at the same time is greater than 1, we have By computing the following table the student will find the theorem verified for several angles. These results show that for small angles, sin 0 and tan 0 may be replaced by in radians and the results will be approximately correct. For a smaller angle the agreement would be still closer. EXERCISES 1. A tower is 125 ft. high. The angle of elevation of the top of the tower, from a point in the same horizontal plane as the base, is 1°. tance from the point of observation to the tower. Find the dis 2. A railway track has a 2% grade for a certain distance. Find the inclination of the track to the horizontal. Solution. The per cent of a grade is the ratio of the number of feet rise per mile to the number of feet in a mile. mile would be 2% of 5280 = 105.6 ft. Then for a 2% grade the rise per 3. A railway track rises 80 feet to the mile. Find the angle of inclina 4. A certain plane is inclined to the horizontal at an angle of 45'. Find the per cent of the grade of a railway track constructed on this plane. Ans. 1.309%. 64. Many problems arise which involve triangles, two sides of which are practically equal, and each very great compared with the third side, as shown in Fig. 77. The long sides may be taken as the radii of an arc, of which the third side BC is the chord. Since the limit of the ratio of the arc to its chord is unity as the angle approaches zero, chord BC may be replaced by the arc BC. arc BC if the chord BC is sub stituted for the arc, the results will be accurate enough for many problems.* *In solving exercises in which radian measure is changed to degrees or vice versa, it is convenient to use Table V. |