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.

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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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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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