Section 3.3 Derivatives of Trigonometric Functions
190 ¤ CHAPTER 3 DIFFERENTIATION RULES
63. For () = 2, 0() = 2 + (2) = (2 + 2). Similarly, we have
00() = (2 + 4+ 2)
000() = (2 + 6+ 6)
(4)() = (2 + 8+ 12)
(5)() = (2 + 10+ 20)
It appears that the coefficient of in the quadratic term increases by 2 with each differentiation. The pattern for the
constant terms seems to be 0 = 1 · 0, 2 = 2 · 1, 6 = 3 · 2, 12 = 4 · 3, 20 = 5 · 4. So a reasonable guess is that
()() = [2 + 2 + (− 1)].
Proof: Let be the statement that ()() = [2 + 2 + (− 1)].
1. 1 is true because 0() = (2 + 2).
2. Assume that is true; that is, ()() = [2 + 2+ ( − 1)]. Then
(+1)() =
()()
= (2 + 2) + [2 + 2+ ( − 1)]
= [2 + (2 + 2)+ (2 + )] = [2 + 2( + 1) + ( + 1)]
This shows that +1 is true.
3. Therefore, by mathematical induction, is true for all ; that is, ()() = [2 + 2+ (− 1)] for every
positive integer .
64. (a)
1
()
=
() ·
(1)− 1 ·
[()]
[()]2[Quotient Rule] =
() · 0− 1 · 0()
[()]2=
0− 0()
[()]2= − 0()
[()]2
(b)
1
3 + 22 − 1
= −
3 + 22 − 1
0(3 + 22 − 1)
2= − 32 + 4
(3 + 22 − 1)2
(c)
(−) =
1
= − ()0
()2[by the Reciprocal Rule] = −−1
2= −−1−2 = −−−1
3.3 Derivatives of Trigonometric Functions
1. () = 2 sinPR⇒ 0() = 2 cos+ (sin)(2) = 2 cos+ 2 sin
2. () = cos+ 2 tan ⇒ 0() = (− sin) + (cos)(1) + 2 sec2 = cos− sin + 2 sec2
3. () = cos ⇒ 0() = (− sin) + (cos) = (cos− sin)
4. = 2 sec− csc ⇒ 0 = 2(sec tan)− (− csc cot) = 2 sec tan+ csc cot
5. () = 3 cos ⇒ 0() = 3(− sin ) + (cos ) · 32 = 32 cos − 3 sin or 2(3 cos − sin )
6. () = 4 sec + tan ⇒ 0() = 4 sec tan + sec2
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SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS ¤ 191
7. () = csc + cot ⇒ 0() = − csc cot + (− csc2 ) + (cot ) = − csc cot + (cot − csc2 )
8. = (cos+ ) ⇒ 0 = (− sin+ ) + (cos+ ) = (cos− sin+ + )
9. =
2− tan⇒ 0 =
(2− tan)(1)− (− sec2 )
(2− tan)2=
2− tan+ sec2
(2− tan)2
10. = sin cos ⇒ 0 = sin (− sin ) + cos (cos ) = cos2 − sin2 [or cos 2]
11. () =sin
1 + cos ⇒
0() =(1 + cos ) cos − (sin )(− sin )
(1 + cos )2=
cos + cos2 + sin2
(1 + cos )2=
cos + 1
(1 + cos )2=
1
1 + cos
12. =cos
1− sin⇒
0 =(1− sin)(− sin)− cos(− cos)
(1− sin)2=− sin+ sin2 + cos2
(1− sin)2=− sin+ 1
(1− sin)2=
1
1− sin
13. = sin
1 + ⇒
0 =(1 + )( cos + sin )− sin (1)
(1 + )2=
cos + sin + 2 cos + sin − sin
(1 + )2=
(2 + ) cos + sin
(1 + )2
14. =sin
1 + tan ⇒
0 =(1 + tan ) cos − (sin ) sec2
(1 + tan )2=
cos + sin − sin
cos2 (1 + tan )2
=cos + sin − tan sec
(1 + tan )2
15. Using Exercise 3.2.61(a), () = cos sin ⇒ 0() = 1 cos sin + (− sin ) sin + cos (cos ) = cos sin − sin2 + cos2
= sin cos + (cos2 − sin2 ) = 12
sin 2 + cos 2 [using double-angle formulas]
16. Using Exercise 3.2.61(a), () = cot ⇒ 0() = 1 cot + cot + (− csc2 ) = (cot + cot − csc2 )
17.
(csc) =
1
sin
=
(sin)(0)− 1(cos)
sin2 =− cos
sin2 = − 1
sin· cos
sin= − csc cot
18.
(sec) =
1
cos
=
(cos)(0)− 1(− sin)
cos2 =
sin
cos2 =
1
cos· sin
cos= sec tan
19.
(cot) =
cos
sin
=
(sin)(− sin)− (cos)(cos)
sin2 = −sin2 + cos2
sin2 = − 1
sin2 = − csc2
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SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS ¤ 193
(b)
Note that 0 = 0 where has a minimum and 00 = 0 where 0 has a
minimum. Also note that 0 is negative when is decreasing and 00 is
negative when 0 is decreasing.
29. () = sin ⇒ 0() = (cos ) + (sin ) · 1 = cos + sin ⇒
00() = (− sin ) + (cos ) · 1 + cos = − sin + 2cos
30. () = sec ⇒ 0() = sec tan ⇒ 00() = (sec ) sec2 + (tan ) sec tan = sec3 + sec tan2 , so
004
=√
23
+√
2(1)2 = 2√
2 +√
2 = 3√
2.
31. (a) () =tan− 1
sec⇒
0() =sec(sec2 )− (tan− 1)(sec tan)
(sec)2=
sec(sec2 − tan2 + tan)
sec 2 =
1 + tan
sec
(b) () =tan− 1
sec=
sin
cos− 1
1
cos
=
sin− cos
cos1
cos
= sin− cos ⇒ 0() = cos− (− sin) = cos + sin
(c) From part (a), 0() =1 + tan
sec=
1
sec+
tan
sec= cos+ sin, which is the expression for 0() in part (b).
32. (a) () = () sin ⇒ 0() = () cos + sin · 0(), so
0(3) = (
3) cos
3+ sin
3· 0(
3) = 4 · 1
2+√
32· (−2) = 2−√3
(b) () =cos
()⇒ 0() =
() · (− sin)− cos · 0()
[()]2
, so
0(3) =
(3) · (− sin
3)− cos
3· 0(
3)
3
2 =4−√
32
− 1
2
(−2)
42=−2√
3 + 1
16=
1− 2√
3
16
33. () = + 2 sin has a horizontal tangent when 0() = 0 ⇔ 1 + 2 cos = 0 ⇔ cos = − 12⇔
= 23
+ 2 or 43
+ 2, where is an integer. Note that 43and 2
3are ±
3units from . This allows us to write the
solutions in the more compact equivalent form (2+ 1) ± 3, an integer.
34. () = cos has a horizontal tangent when 0() = 0. 0() = (− sin) + (cos) = (cos− sin).
0() = 0 ⇔ cos− sin = 0 ⇔ cos = sin ⇔ tan = 1 ⇔ = 4
+ , an integer.
35. (a) () = 8 sin ⇒ () = 0() = 8 cos ⇒ () = 00() = −8 sin
(b) The mass at time = 23has position
23
= 8 sin 2
3= 8
√3
2
= 4√
3, velocity
23
= 8cos 2
3= 8
− 12
= −4,
and acceleration
23
= −8 sin 2
3= −8
√3
2
= −4
√3. Since
23
0, the particle is moving to the left.
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194 ¤ CHAPTER 3 DIFFERENTIATION RULES
36. (a) () = 2 cos + 3 sin ⇒ () = −2 sin + 3cos ⇒() = −2 cos − 3 sin
(b)
(c) = 0 ⇒ 2 ≈ 255. So the mass passes through the equilibrium
position for the first time when ≈ 255 s.
(d) = 0 ⇒ 1 ≈ 098, (1) ≈ 361 cm. So the mass travels
a maximum of about 36 cm (upward and downward) from its equilibrium position.
(e) The speed || is greatest when = 0, that is, when = 2 + , a positive integer.
37. From the diagram we can see that sin = 6 ⇔ = 6 sin . We want to find the rate
of change of with respect to ; that is, . Taking the derivative of the above
expression, = 6(cos ). So when = 3,
= 6cos 3
= 6
12
= 3 mrad.
38. (a) =
sin + cos ⇒
=
( sin + cos )(0)− ( cos − sin )
( sin + cos )2
=(sin − cos )
( sin + cos )2
(b)
= 0 ⇒ (sin − cos ) = 0 ⇒ sin = cos ⇒ tan = ⇒ = tan−1
(c) From the graph of =06(20)(98)
06 sin + cos for 0 ≤ ≤ 1, we see that
= 0 ⇒ ≈ 054. Checking this with part (b) and = 06, we
calculate = tan−1 06 ≈ 054. So the value from the graph is consistent
with the value in part (b).
39. lim→0
sin 5
3= lim
→0
5
3
sin 5
5
=
5
3lim→0
sin 5
5=
5
3lim→0
sin
[ = 5] =
5
3· 1 =
5
3
40. lim→0
sin
sin= lim
→0
sin
·
sin· 1
= lim
→0
sin
· lim→0
sin · 1
[ = ]
= 1 · lim→0
1
sin
· 1
= 1 · 1 · 1
=
1
41. lim→0
tan 6
sin 2= lim
→0
sin 6
· 1
cos 6·
sin 2
= lim
→0
6 sin 6
6· lim→0
1
cos 6· lim→0
2
2 sin 2
= 6 lim→0
sin 6
6· lim→0
1
cos 6· 1
2lim→0
2
sin 2= 6(1) · 1
1· 1
2(1) = 3
42. lim→0
cos − 1
sin = lim
→0
cos − 1
sin
=lim→0
cos − 1
lim→0
sin
=0
1= 0
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
194 ¤ CHAPTER 3 DIFFERENTIATION RULES
36. (a) () = 2 cos + 3 sin ⇒ () = −2 sin + 3cos ⇒() = −2 cos − 3 sin
(b)
(c) = 0 ⇒ 2 ≈ 255. So the mass passes through the equilibrium
position for the first time when ≈ 255 s.
(d) = 0 ⇒ 1 ≈ 098, (1) ≈ 361 cm. So the mass travels
a maximum of about 36 cm (upward and downward) from its equilibrium position.
(e) The speed || is greatest when = 0, that is, when = 2 + , a positive integer.
37. From the diagram we can see that sin = 6 ⇔ = 6 sin . We want to find the rate
of change of with respect to ; that is, . Taking the derivative of the above
expression, = 6(cos ). So when = 3,
= 6cos 3
= 6
12
= 3 mrad.
38. (a) =
sin + cos ⇒
=
( sin + cos )(0)− ( cos − sin )
( sin + cos )2
=(sin − cos )
( sin + cos )2
(b)
= 0 ⇒ (sin − cos ) = 0 ⇒ sin = cos ⇒ tan = ⇒ = tan−1
(c) From the graph of =06(20)(98)
06 sin + cos for 0 ≤ ≤ 1, we see that
= 0 ⇒ ≈ 054. Checking this with part (b) and = 06, we
calculate = tan−1 06 ≈ 054. So the value from the graph is consistent
with the value in part (b).
39. lim→0
sin 5
3= lim
→0
5
3
sin 5
5
=
5
3lim→0
sin 5
5=
5
3lim→0
sin
[ = 5] =
5
3· 1 =
5
3
40. lim→0
sin
sin= lim
→0
sin
·
sin· 1
= lim
→0
sin
· lim→0
sin · 1
[ = ]
= 1 · lim→0
1
sin
· 1
= 1 · 1 · 1
=
1
41. lim→0
tan 6
sin 2= lim
→0
sin 6
· 1
cos 6·
sin 2
= lim
→0
6 sin 6
6· lim→0
1
cos 6· lim→0
2
2 sin 2
= 6 lim→0
sin 6
6· lim→0
1
cos 6· 1
2lim→0
2
sin 2= 6(1) · 1
1· 1
2(1) = 3
42. lim→0
cos − 1
sin = lim
→0
cos − 1
sin
=lim→0
cos − 1
lim→0
sin
=0
1= 0
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS ¤ 195
43. lim→0
sin 3
53 − 4= lim
→0
sin 3
3· 3
52 − 4
= lim
→0
sin 3
3· lim→0
3
52 − 4= 1 ·
3
−4
= −3
4
44. lim→0
sin 3 sin 5
2= lim
→0
3 sin 3
3· 5 sin 5
5
= lim
→0
3 sin 3
3· lim→0
5 sin 5
5
= 3 lim→0
sin 3
3· 5 lim
→0
sin 5
5= 3(1) · 5(1) = 15
45. Divide numerator and denominator by . (sin also works.)
lim→0
sin
+ tan = lim
→0
sin
1 +sin
· 1
cos
=lim→0
sin
1 + lim→0
sin
lim→0
1
cos
=1
1 + 1 · 1 =1
2
46. lim→0
csc sin(sin) = lim→0
sin(sin)
sin= lim
→0
sin
[As → 0, = sin→ 0.] = 1
47. lim→0
cos − 1
22= lim
→0
cos − 1
22· cos + 1
cos + 1= lim
→0
cos2 − 1
22(cos + 1)= lim
→0
− sin2
22(cos + 1)
= −1
2lim→0
sin
· sin
· 1
cos + 1= −1
2lim→0
sin
· lim→0
sin
· lim→0
1
cos + 1
= −1
2· 1 · 1 · 1
1 + 1= −1
4
48. lim→0
sin(2)
= lim
→0
· sin(2)
·
= lim→0
· lim→0
sin(2)
2= 0 · lim
→0+
sin
where = 2
= 0 · 1 = 0
49. lim→4
1− tan
sin− cos= lim
→4
1− sin
cos
· cos
(sin− cos) · cos = lim→4
cos− sin
(sin− cos) cos= lim
→4
−1
cos=
−1
1√
2= −√2
50. lim→1
sin(− 1)
2 + − 2= lim
→1
sin(− 1)
( + 2)(− 1)= lim
→1
1
+ 2lim→1
sin(− 1)
− 1= 1
3· 1 = 1
3
51.
(sin) = cos ⇒ 2
2(sin) = − sin ⇒ 3
3(sin) = − cos ⇒ 4
4(sin) = sin.
The derivatives of sin occur in a cycle of four. Since 99 = 4(24) + 3, we have99
99(sin) =
3
3(sin) = − cos.
52. Let () = sin and () = sin, so () = (). Then 0() = () + 0(),
00() = 0() + 0() + 00() = 20() + 00(),
000() = 200() + 00() + 000() = 300() + 000() · · · ()() = (−1)() + ()().
Since 34 = 4(8) + 2, we have (34)() = (2)() =2
2(sin) = − sin and (35)() = − cos.
Thus,35
35( sin) = 35(34)() + (35)() = −35 sin− cos.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS ¤ 195
43. lim→0
sin 3
53 − 4= lim
→0
sin 3
3· 3
52 − 4
= lim
→0
sin 3
3· lim→0
3
52 − 4= 1 ·
3
−4
= −3
4
44. lim→0
sin 3 sin 5
2= lim
→0
3 sin 3
3· 5 sin 5
5
= lim
→0
3 sin 3
3· lim→0
5 sin 5
5
= 3 lim→0
sin 3
3· 5 lim
→0
sin 5
5= 3(1) · 5(1) = 15
45. Divide numerator and denominator by . (sin also works.)
lim→0
sin
+ tan = lim
→0
sin
1 +sin
· 1
cos
=lim→0
sin
1 + lim→0
sin
lim→0
1
cos
=1
1 + 1 · 1 =1
2
46. lim→0
csc sin(sin) = lim→0
sin(sin)
sin= lim
→0
sin
[As → 0, = sin→ 0.] = 1
47. lim→0
cos − 1
22= lim
→0
cos − 1
22· cos + 1
cos + 1= lim
→0
cos2 − 1
22(cos + 1)= lim
→0
− sin2
22(cos + 1)
= −1
2lim→0
sin
· sin
· 1
cos + 1= −1
2lim→0
sin
· lim→0
sin
· lim→0
1
cos + 1
= −1
2· 1 · 1 · 1
1 + 1= −1
4
48. lim→0
sin(2)
= lim
→0
· sin(2)
·
= lim→0
· lim→0
sin(2)
2= 0 · lim
→0+
sin
where = 2
= 0 · 1 = 0
49. lim→4
1− tan
sin− cos= lim
→4
1− sin
cos
· cos
(sin− cos) · cos = lim→4
cos− sin
(sin− cos) cos= lim
→4
−1
cos=
−1
1√
2= −√2
50. lim→1
sin(− 1)
2 + − 2= lim
→1
sin(− 1)
( + 2)(− 1)= lim
→1
1
+ 2lim→1
sin(− 1)
− 1= 1
3· 1 = 1
3
51.
(sin) = cos ⇒ 2
2(sin) = − sin ⇒ 3
3(sin) = − cos ⇒ 4
4(sin) = sin.
The derivatives of sin occur in a cycle of four. Since 99 = 4(24) + 3, we have99
99(sin) =
3
3(sin) = − cos.
52. Let () = sin and () = sin, so () = (). Then 0() = () + 0(),
00() = 0() + 0() + 00() = 20() + 00(),
000() = 200() + 00() + 000() = 300() + 000() · · · ()() = (−1)() + ()().
Since 34 = 4(8) + 2, we have (34)() = (2)() =2
2(sin) = − sin and (35)() = − cos.
Thus,35
35( sin) = 35(34)() + (35)() = −35 sin− cos.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS ¤ 195
43. lim→0
sin 3
53 − 4= lim
→0
sin 3
3· 3
52 − 4
= lim
→0
sin 3
3· lim→0
3
52 − 4= 1 ·
3
−4
= −3
4
44. lim→0
sin 3 sin 5
2= lim
→0
3 sin 3
3· 5 sin 5
5
= lim
→0
3 sin 3
3· lim→0
5 sin 5
5
= 3 lim→0
sin 3
3· 5 lim
→0
sin 5
5= 3(1) · 5(1) = 15
45. Divide numerator and denominator by . (sin also works.)
lim→0
sin
+ tan = lim
→0
sin
1 +sin
· 1
cos
=lim→0
sin
1 + lim→0
sin
lim→0
1
cos
=1
1 + 1 · 1 =1
2
46. lim→0
csc sin(sin) = lim→0
sin(sin)
sin= lim
→0
sin
[As → 0, = sin→ 0.] = 1
47. lim→0
cos − 1
22= lim
→0
cos − 1
22· cos + 1
cos + 1= lim
→0
cos2 − 1
22(cos + 1)= lim
→0
− sin2
22(cos + 1)
= −1
2lim→0
sin
· sin
· 1
cos + 1= −1
2lim→0
sin
· lim→0
sin
· lim→0
1
cos + 1
= −1
2· 1 · 1 · 1
1 + 1= −1
4
48. lim→0
sin(2)
= lim
→0
· sin(2)
·
= lim→0
· lim→0
sin(2)
2= 0 · lim
→0+
sin
where = 2
= 0 · 1 = 0
49. lim→4
1− tan
sin− cos= lim
→4
1− sin
cos
· cos
(sin− cos) · cos = lim→4
cos− sin
(sin− cos) cos= lim
→4
−1
cos=
−1
1√
2= −√2
50. lim→1
sin(− 1)
2 + − 2= lim
→1
sin(− 1)
( + 2)(− 1)= lim
→1
1
+ 2lim→1
sin(− 1)
− 1= 1
3· 1 = 1
3
51.
(sin) = cos ⇒ 2
2(sin) = − sin ⇒ 3
3(sin) = − cos ⇒ 4
4(sin) = sin.
The derivatives of sin occur in a cycle of four. Since 99 = 4(24) + 3, we have99
99(sin) =
3
3(sin) = − cos.
52. Let () = sin and () = sin, so () = (). Then 0() = () + 0(),
00() = 0() + 0() + 00() = 20() + 00(),
000() = 200() + 00() + 000() = 300() + 000() · · · ()() = (−1)() + ()().
Since 34 = 4(8) + 2, we have (34)() = (2)() =2
2(sin) = − sin and (35)() = − cos.
Thus,35
35( sin) = 35(34)() + (35)() = −35 sin− cos.
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