In this section we will determine the derivatives of the basic trigonometric functionssine and cosine. Angles are measured in radians, unless stated otherwise.
The Derivatives of Sine and Cosine at 0
Let’s begin by dierentiating sine and cosine at 0. By the denition of the derivative, d
dxsin (x)¯¯
¯¯x=0= lim
h0
sin (h)sin (0)
h = lim
h0
sin (h) h . Let’s set
F(h) =sin (h) h .
122 CHAPTER 2. THE DERIVATIVE Graphically, F(h) is the slope of the secant line that passes through the points (0,0) and (h,sin (h)), as illustrated in Figure 1.
Π Π x
1 1 y
h sinh
Figure 1: The secant line through(0,0)and(h,sin (h))
Figure 2 shows the graph ofF, as produced by a graphing utility. The picture indicates that
h0limF(h)= lim
k0
sin (h) h =1.
2Π Π Π 2Π h 1
y
Figure 2: y= sin (h) h
Table 1 displays sin (h)/h (rounded to 10 signicant digits) and 1sin(h)/h (rounded to 2 signicant digits) for h = 10k, where k = 1,2,3,4,5. The numbers in Table 1 denitely support the claim that limh0sin(h)/h= 1(the last row displaysF(105)as1.0, rounded to 10 signicant digits).
h sin(h)/h 1sin(h)/h 101 .9983341665 1.7×103 102 .9999833334 1.7×105 103 .9999998333 1.7×107 104 .9999999983 1.7×109 105 1.0 1.7×1011
Table 1 The above graphs and numbers indicate that
d
dxsin (x)¯¯
¯¯x=0= 1.
By the denition of the derivative, d
dxcos (x)¯¯
¯¯x=0= lim
h0
cos (h)cos (0)
h = lim
h0
cos (h)1
h .
Let’s set
G(h) = cos (h)1
h .
Graphically, G(h) is the slope of the secant line that passes through the points (0,1) and (h,cos (h)), as illustrated in Figure 3.
x
1 y
h, cosh
Π Π 2
2
0, 1
h
Figure 3: The secant line through(0,1)and(h,cos (h))
Figure 4 displays the graph ofG, as produced by a graphing utility. The picture indicates that
h0limG(h) = lim
h0
cos(h)1
h = 0.
2Π 2Π h
1 1 y
Figure 4: y= cos (h)1 h
Table 2 displays (cos(h)1)/h (rounded to 6 signicant digits) for h = 10k, where k = 1,2,3,4,5. The numbers support the claim thatlimh0(cos(h)1)/h= 0.
h cos (h)1 101 4.995 83h ×102 102 4.999 96×103 103 5×104 104 5×105 105 5.0×106
Table 2 The above graphs and numbers indicate that
d
dxcos (x)¯¯
¯¯x=0= 0.
124 CHAPTER 2. THE DERIVATIVE Proposition 1 The derivative of sine at 0 is 1, and the derivative of cosine at 0 is 0:
d
dxsin(x)¯¯
¯¯{=0= lim
k0
sin(h) h = 1, and
d
dxcos(x)¯¯
¯¯{=0
= lim
k0
cos(h)1
h = 0
You can nd the proof of Proposition 1 at the end of this section.
Example 1 Determine the tangent line to the graph of sine at(0,sin (0)).
Solution
By Proposition 1,
d
dxsin (x)¯¯
¯¯x=0
= 1.
Therefore, the tangent line to the graph of sine at(0,sin (0)) is the graph of the equation y= sin (0) +
d
dxsin (x)¯¯
¯¯x=0ả
(x0) =x.
Figure 5 shows the graph of sine and the liney=x.¤
Π Π x
1
1 y
Π Π 2
2
Figure 5
Example 2 Determine the tangent line to the graph of cosine at(0,cos (0)).
Solution By Proposition1,
d
dxcos (x)¯¯
¯¯x=0= 0.
Therefore, the tangent line to the graph of cosine at(0,cos (0))is the graph of y= cos (0) +
d
dxcos (x)¯¯
¯¯x=0ả
(x0) = 1.
Thus, the horizontal liney= 1is tangent to the graph of cosine at(0,1), as illustrated in Figure 6. the tangent line to the graph of sine at(/3,sin (/3)). ¤
Π Π x
1 1 y
y1
Π Π 2
2
Figure 6
The Derivative Functions Corresponding to Sine and Cosine
The formulas of the derivatives of sine and cosine are elegant and easy to remember:
Theorem 1 We have d
dxsin(x) = cos(x)and d
dxcos(x) =sin(x) for each real numberx.
Proof
The above expressions follow from Proposition 1 on the derivatives of sine and cosine at 0, with the help of the addition formulas for these functions.
Let’s begin with sine. For anyxRand incrementh6= 0, sin(x+h)sin(x)
h = sin(x) cos(h) + cos(x) sin(h)sin(x) h
= sin(x)
cos(h)1 h
ả
+ cos(x)
sin(h) h
ả . Therefore,
d
dxsin (x) = lim
h0
sin(x+h)sin(x) h
= lim
h0
sin(x)
cos(h)1 h
ả
+ cos(x)
sin(h) h
ảả
= sin(x) lim
h0
cos(h)1
h + cos(x) lim
h0
sin(h) h
= sin(x)(0) + cos(x)(1)
= cos(x).
We have made use of the sum and constant multiple rules for limits (as far as the limit process is concerned,sin(x)and cos(x)are constants, sincexis kept xed), and Proposition 1.
The expression for the derivative of cosine is derived in a similar manner. We make use of the addition formula for cosine:
cos(x+h)cos(x)
h = cos(x) cos(h)sin(x) sin(h)cos(x) h
= cos(x)
cos(h)1 h
ả
sin(x)
sin(h) h
ả .
126 CHAPTER 2. THE DERIVATIVE Therefore,
d
dxcos (x) = lim
h0
cos(x)
cos(h)1 h
ả
sin(x)
sin(h) h
ảả
= cos(x) lim
h0
cos(h)1
h sin(x) lim
h0
sin(h) h
= cos(x) (0)sin(x) (1)
=sin(x).
¥
Example 3 Letf(x) = sin (x). Determine the tangent line to the graph of sine at(/3,sin (/3)).
Solution We have
f
³ 3
´= sin³ 3
´= 3
2 , and
f0
³ 3
´= d
dxsin (x)¯¯
¯¯x=/3
= cos (x)|x=/3= 1 2.
Therefore, the tangent line to the graph of sine at(/3,sin (/3))is the graph of the equation y=f
³ 3
´+f0
³ 3
´ ³
x
3
´= 3
2 +1 2
³
x
3
´ . Figure 7 shows this tangent line and the graph of sine. ¤
Π Π x
1 1 y
Π 3
Π3, 32
Figure 7
The Proof of Proposition 1
In order to prove Proposition 1 we will establish the following inequalities: Ifh6= 0and/2<
h < /2, a)
0< sin (h) h <1, b)
0<1cos (h)< h2 2, c)
sin (h)
h >cos (h).
Assume that0 < h < /2. With reference to Figure 9, the area of triangle AOP is less than the area of the circular sector determined by the points A, Oand P. The area of the triangle AOP is
1
2×base×height= 1
2(1) (sin(h)) =1
2sin (h).
1
1 1
O sinh h
cosh A P
Q
Figure 9
The area of a sector of a disk of radiusrthat corresponds to the angleh(in radians) is 1
2hr2.
Therefore, the area of the circular sector determined byA,O andP ish/2. Thus, 0< 1
2sin(h)< 1
2h0<sin(h)< h0< sin (h) h <1 Now assume that/2< h <0. Since sine is an odd function,
sin (h)
h = sin (h)
h = sin (h) h . Since0<h < /2,
0< sin (h) h <1.
Therefore,
0< sin (h) h <1
ifh6= 0and/2< h < /2. Thus, we have established inequality a).
In order to establish inequality b), we will use the identity cos (h) = 12 sin2
h 2
ả . Indeed, by the addition formula for cosine,
cos(h) = cos h
2+ h 2
ả
= cos2 h
2
ả sin2
h 2
ả
=
1sin2 h
2
ảả
sin2 h
2
ả
= 12 sin2 h
2
ả .
128 CHAPTER 2. THE DERIVATIVE Assume thath6= 0and/2< h < /2. Since
0< sin (h/2) h/2 <1, we have
sin2(h/2) h2/4 <1, so that
sin2(h/2)< h2 4. Therefore,
cos (h) = 12 sin2 h
2
ả
>12 h2
4
ả
= 1h2 2 . Thus,
h2
2 >1cos (h).
Ifh6= 0and /2< h < /2, thencos (h)<1. Therefore, 1cos (h)>0. Thus, 0<1cos (h)< h2
2. We have established inequality b).
Now we will establish inequality c). Assume that 0< h < /2. With reference to Figure 10, the lineAT is tangential to the unit circle atA.
1
1 1
O
h sinh cosh A
P
Q T
Figure 10 Since
length ofAT
length ofOA = length ofAT
1 = tan(h), the area of the triangleAOT is
1
2×base×height= 1
2(1) (tan(h)).
As in part a), the area of the circular sectorAOP ish/2. The area of the circular sectorAOP is less than the area of the triangleAOT. Therefore,
1 2h <1
2tan(h).
Thus,
tan (h)> hif0< h <
2, i.e.,
sin(h)
cos(h) > hif 0< h <
2. Therefore,
sin(h)
h >cos(h)if0< h <
2. If/2< h <0, we have0<(h)< /2, so that
sin(h)
(h) >cos (h).
Since cosine is an even function and sine is an odd function, we obtain sin(h)
h >cos(h)sin(h)
h >cos(h).
Therefore,
sin(h)
h >cos(h)if
2 < h <
2 andh6= 0.
Thus, we have established inequality c).
Now we will show that Proposition 1 follows from inequalities a), b) and c). We restricth so thath6= 0and/2< h < /2.
By inequalities a) and c),
cos (h)< sin (h) h <1 Figure 11 illustrates the above inequality.
Π2
Π 2
h
1 1 y
y1
ysinh ycosh h
Figure 11
We see that the graph of y = sin (h)/h is squeezed between the graphs of y = cos (h) and y= 1. Since cosine is continuous at 0,limh0cos (h) = cos (0) = 1. Of course, limh01 = 1.
Therefore,
h0lim sin (h)
h = 1 as well, by the Squeeze Theorem.
Now we will show that
h0lim
cos (h)1
h = 0.
130 CHAPTER 2. THE DERIVATIVE By inequality b),
0<1cos (h)< h2 2 Sinceh2=|h|2,
1cos (h)
|h| < |h| 2 . Since1cos (h)>0,|cos (h)1|= 1cos (h). Therefore,
¯¯¯¯cos (h)1 h
¯¯¯¯< |h| 2
Since ¯¯
¯¯cos (h)1 h
¯¯¯¯< |h| 2
|h|
2 < cos (h)1 h < |h|
2 , the graph of
y= cos (h)1 h
on the interval[/2, /2]is squeezed between the linesy= |h|/2andy=|h|/2, as illustrated in Figure 12.
1 1 h
1 y
ycosh1 h y h
y h 2
2
Figure 12: |h|
2 < cos (h)1 h < |h|
2 Since
h0lim
|h| 2
ả
= lim
h0
|h| 2 = 0, we have
h0lim
cos (h)1
h = 0,
as well, again by the Squeeze Theorem. ¥
Problems
In problems 1 - 4, determine the limit.
1. lim
x0
sin (4x) x 2. lim
x0
sin (6x) sin (3x)
3. limx0cos (2x)1 4x 4. lim
x0
cos (x)1 3 sin (x)
In problems 5 - 12,
a) Determine the expression forf0(x), b) Evaluatef0(a).
5. f(x) = 2 sin (x), a= 3 6. f(x) = 4 cos (x), a= 6
7. f(x) = 3 sin (x)4 cos (x), a= 4 8. f(x) = 4
x+ 3 sin (x), a= 3
9. f(x) = 4 cos (x) 2 x, a=
10. f(x) =x23x+ 7 + cos (x), a= 1 11. f(x) = 2 sin (x)3 cos (x), a=
2 12. f(x) = 4 sin (x) + 2 cos (x)3x4, a= 0
In problems 13 and 14,
a) Determine the tangent line to the graph off at(a, f(a))(the point-slope form with basepoint awill do),
b)[C]Make use of your graphing utility to plot the graph off and the tangent line at(a, f(a)).
Zoom in towards(a, f(a))until you are unable to distinguish between the graph off and the tangent line. Do the pictures reinforce the identication of the slope of the graph offat(a, f(a)) with the slope of the tangent line at that point?
13.
f(x) = sin (x), a=/6.
14.
f(x) = cos (x), a=/4.