and add the results) and determine the part of the circle it describes from Find the velocity and acceleration (direction and magnitude) at these times

t

==

0 to t = 1.

10. A water resevoir is in the form of the figure generated by revolving the curve 12y = x2 about the y-axis. How much does it contain if the water is 10 feet deep at the center? If water is entering at the rate of 10 cubic feet per second when the depth at the center is 12 feet, how rapidly is the surface rising?

CHAPTER VIII

PROPERTIES OF TRIGONOMETRIC FUNCTIONS

LOGARITHMIC SOLUTION OF TRIANGLES, CASES III AND IV

116. Introduction. In this Chapter we shall derive formulas which express properties of the trigonometric functions analogous to certain properties of algebraic, exponential, and logarithmic functions with which we are already familiar. These properties enable us to change the form of expressions involving trigonometric functions, a process of great importance in those parts of more advanced mathematics where these functions appear. These properties, together with the formulas for the functions of n90° (page 192), are used more, perhaps, than the solution of triangles, except in such fields as surveying. As an application of these formulas we shall obtain formulas for the solution of triangles, which are adapted to the use of logarithmic tables, for those cases in which we have hitherto used the law of cosines.

In the sections immediately following we shall consider the interdependence of the trigonometric functions of an angle 0, with applications to the solution of equations and the proof of identities.

117. Fundamental Trigonometric Relations. As an immediate consequence of the definitions of the trigonometric functions we obtained the reciprocal relations

in which sin20 denotes the square of sin 0, or (sin 0)2, and cos2 denotes (cos )2.

To prove (5), we have, for any value of 0,

(4)

(5)

EXAMPLE. Express each of the trigonometric functions in terms of sin 0.

±√1 − sin2 0.

From (5) we obtain

The sign of the radical must be chosen from a knowledge of the quadrant in which lies.

These results may also be obtained from a figure by means of the definitions of the functions (see Exercises 9 and 10, page 170). The given x function, sin 0, may be regarded as a fraction with unit denominator. Describe a circle

about the origin with radius r = 1, and draw the line y = sin 0, regarding

O as constant. Let the line cut the circle in P and P'.

Then the angles

XOP and XOP' are such that their sines are equal to

these angles are values of 0.

From the right triangles, using the Pythagorean Theorem, we get

Then the results obtained above may be written down from the figure. In this figure it is assumed that sin 0 is positive. If sin 0 is negative, the angles would lie in the third and fourth quadrants.

This example shows that it would be entirely possible to get along with only one of the trigonometric functions, but frequently it would be inconvenient.

The formulas (1)-(5) are identities (Definition, page 134). There can not be more than five independent equations connecting the six functions, for if there were six, it would be possible to solve them for the six functions, which would then be constants. Hence any other identity involving functions of a single angle must be a consequence of these five, and may be proved by them.

Three other identities are of sufficient importance to be listed with the fundamental formulas. They are:

The first two may be proved as follows, the proof of (8) being analogous to that of (7).

The formulas in this and the following sections should be memorized.

118. Trigonometric Equations. No specific rules for the solution of equations involving trigonometric functions can be given, but the following remarks may be of service.

Express the equation in terms of a single function by means of the formulas in the preceding section, and solve for this function. The values of the angle may then be found from the tables.

If, when all the terms are written on the left of the equality sign, the left-hand member can be factored, equate each factor to zero, and solve the resulting equations. For any solution makes the product of the factors zero, and hence it must make one of the factors zero. Conversely, any solution of one of the resulting equations makes one of the factors zero, and hence also the product.

All the other values of x may be obtained by adding 2nπ to each of these four solutions.

To check the results, we substitute each of the values of x in the given equation. This gives: