Calculus: Analytic Geometry and Calculus, with Vectors

440 Trigonometric functions

(8.12), and the number 0 is negative. Even though we could pretend to
be appalled by the idea that numbers have sides, we bow to conventions

Figure 8.13

and agree that the way to find the six fundamental

trigonometric functions of a given angle 0 is to con-

struct "the terminal side of B" as in Figure 8.13, pick

a point P(x,y) on this terminal side, let r be the dis-

tance (positive, of course) from 0 to P, and use the

numbers x, y, r in the usual way.

It is our purpose to present formulas for deriva-

tives of trigonometric functions with derivations to

which no logical objections can be raised. We could undertake

to use the formula .1 = rry0r2 for the area of a circular sector of

radius r which has central angle 0, but it would be immediately

recognized that this formula has not been proved. Even if it be conceded

that we know that the area of a whole circular disk of radius r is rr2, it

is only the docile acceptance of a crafty fraud that would allow us to

"see at once" that the sector "obviously" has area JBr2 because the area

of the sector divided by the area of the circle is "obviously" 0/2r. Noth-

ing is obvious. Even a hazy understanding of the theory of area is

enough to show that nothing can be proved without making substantial

use of precise definitions and completeness of the real-number system

or, what amounts to the same thing, consequences of these things. Since

we do not now wish to be responsible for furnishing a complete treatment

of areas of sectors, we shall base our work on the inequality (7.12). This

inequality, the truth of which should seem thoroughly reasonable to a

person at B who has the choice of walking two paths to D, has a virtue.

It has been proved.

It was indicated in Problem 12 at the end of Section 3.2 that we will be

able to derive the formulas for derivatives of trigonometric functions

when we have proved the two formulas

(8.14) limsin 0= 1, lim1 - cos B= 0

6--0^0 e_ o^0

Like a man who has drilled a hole and filled it with blasting powder to

split a rock, we are ready to produce results. Figure 7.11 can be put

Figure 8.141 into a coordinate system as in Figure

8.141, and we suppose that 0 < 0 < r/2.

In terms of the notation of this figure,

the inequality (7.12) becomes

y<.c y+(r-x).

Dividing by r gives

(8.142) sin 0 <_ 0 < sin 0 + (1 - cos 0).