James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 3.3 Derivatives of Trigonometric Functions 191
Let’s try to confirm our guess that if f sxd − sin x, then f 9sxd − cos x. From the definition
of a derivative, we have
We have used the addition formula for
sine. See Appendix D.
f 9sxd − lim
h l 0
f sx 1 hd 2 f sxd
h
sinsx 1 hd 2 sin x
− lim
h l 0 h
− lim
h l 0
sin x cos h 1 cos x sin h 2 sin x
h
− lim
h l 0F
sin x cos h 2 sin x
h
− lim x
h l 0Fsin S cos h 2 1
h
1
cos x sin h
h
D 1 cos x S sin h
h
G
DG
cos h 2 1
sin h
1 − lim sin x ? lim 1 lim cos x ? lim
h l 0 h l 0 h
h l 0 h l 0 h
Two of these four limits are easy to evaluate. Since we regard x as a constant when computing
a limit as h l 0, we have
lim sin x − sin x and lim cos x − cos x
h l 0 h l 0
The limit of ssin hdyh is not so obvious. In Example 2.2.3 we made the guess, on the basis
of numerical and graphical evidence, that
sin
2 lim − 1
l 0
1
B
D
E
We now use a geometric argument to prove Equation 2. Assume first that lies between
0 and y2. Figure 2(a) shows a sector of a circle with center O, central angle , and
radius 1. BC is drawn perpendicular to OA. By the definition of radian measure, we have
arc AB − . Also | BC | − | OB | sin − sin . From the diagram we see that
| BC | , | AB |
, arc AB
O
¨
O
(a)
C
B
A
E
A
Therefore sin , so
sin
Let the tangent lines at A and B intersect at E. You can see from Figure 2(b) that the
cir cumference of a circle is smaller than the length of a circumscribed polygon, and so
arc AB , | AE | 1 | EB | . Thus
− arc AB , | AE | 1 | EB |
, | AE | 1 | ED |
, 1
− | AD | − | OA | tan
(b)
− tan
FIGURE 2
(In Appendix F the inequality < tan is proved directly from the definition of the
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