Section 3.3 Derivatives of Trigonometric Functions 46. Find the limit. lim

Section 3.3 Derivatives of Trigonometric Functions

46. Find the limit. lim

x→0 sin x sin πx. Solution:

194 ¤ CHAPTER 3 DIFFERENTIATION RULES

36. (a) () = 2 cos  + 3 sin  ⇒ () = −2 sin  + 3 cos  ⇒

() = −2 cos  − 3 sin 

(b)

(c)  = 0 ⇒ ²≈ 255. So the mass passes through the equilibrium position for the first time when  ≈ 255 s.

(d)  = 0 ⇒ ¹≈ 098, (¹) ≈ 361 cm. So the mass travels

a maximum of about 36 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,^_ = 6 cos^₃ = 6₁

2

= 3mrad.

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  = 06(20)(98)

06 sin  + cos  for 0 ≤  ≤ 1, we see that



 = 0 ⇒  ≈ 054. Checking this with part (b) and  = 06, we calculate  = tan⁻¹06 ≈ 054. 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 3 lim

→0

sin 5

5 = 5 3 lim

→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

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62. Find the given derivative by finding the first few derivatives and observing the pattern that occurs.

d³⁵

dx³⁵(x sin x).

Solution:

SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS ¤ 195

43. lim

→0

sin 3

5³− 4 = lim

→0

sin 3

3 · 3

5²− 4



= lim

→0

sin 3

3 · lim_

→0

3

5²− 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_→05 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

→0csc  sin(sin ) = lim

→0

sin(sin ) sin  = lim

→0

sin 

 [As  → 0,  = sin  → 0.] = 1

47. lim

→0

cos  − 1 2² = lim

→0

cos  − 1

2² ·cos  + 1 cos  + 1 = lim

→0

cos² − 1

2²(cos  + 1) = lim

→0

− sin² 2²(cos  + 1)

= −1 2 lim

→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

 + 2 lim

→1

sin( − 1)

 − 1 = ¹₃ · 1 = ¹3

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 have ⁹⁹

⁹⁹(sin ) = ³

³(sin ) = − cos .

52. Let () =  sin  and () = sin , so () = (). Then ⁰() = () + ⁰(),

⁰⁰() = ⁰() + ⁰() + ⁰⁰() = 2⁰() + ⁰⁰(),

⁰⁰⁰() = 2⁰⁰() + ⁰⁰() + ⁰⁰⁰() = 3⁰⁰() + 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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66. A semicircle with diameter P Q sits on an isosceles triangle P QR to form a region shaped like a two-dimensional ice-cream cone, as shown in the figure. If A(θ) is the area of the semicircle and B(θ) is the area of the triangle, find

lim

θ→0⁺ A(θ) B(θ).

SeCtion3.4 The Chain Rule 197

55. Differentiate each trigonometric identity to obtain a new (or familiar) identity.

(a) tan x − sin x

cos x (b) sec x − 1

cos x (c) sin x 1 cos x − 1 1 cot x

csc x

56. A semicircle with diameter PQ sits on an isosceles triangle PQR to form a region shaped like a two-dimensional ice-cream cone, as shown in the figure. If Asd is the area of the semicircle and Bsd is the area of the triangle, find

llim0¹

Asd Bsd

P Q

R B(¨) A(¨)

¨

10 cm 10 cm

57. The figure shows a circular arc of length s and a chord of length d, both subtended by a central angle . Find

llim0¹

s d

d

¨ s

58. Let fsxd − x s1 2 cos 2x.

(a) Graph f . What type of discontinuity does it appear to have at 0?

(b) Calculate the left and right limits of f at 0. Do these values confirm your answer to part (a)?

; If the rope makes an angle  with the plane, then the

magnitude of the force is

F − W

sin  1 cos 

where  is a constant called the coefficient of friction.

(a) Find the rate of change of F with respect to .

(b) When is this rate of change equal to 0?

(c) If W − 50 lb and  − 0.6, draw the graph of F as a function of  and use it to locate the value of  for which dFyd − 0. Is the value consistent with your answer to part (b)?

39–50 Find the limit.

39. lim

xl0

sin 5x

3x 40. lim

xl0

sin x sin x 41. lim

tl0

tan 6t

sin 2t 42. lim

l0

cos  2 1 sin  43. lim

xl0

sin 3x

5x³24x 44. lim

xl0

sin 3x sin 5x x² 45. lim

l0

sin 

 1tan  46. lim

xl0 csc x sinssin xd 47. lim

 l0

cos  2 1

2² 48. lim

xl0

sinsx²d x 49. lim

x ly4

1 2 tan x

sin x 2 cos x 50. lim

xl1

sinsx 2 1d x²1x 22

51–52 Find the given derivative by finding the first few deriva- tives and observing the pattern that occurs.

51. d⁹⁹

dx⁹⁹ssin xd 52. d³⁵

dx³⁵sx sin xd

53. Find constants A and B such that the function

y − A sin x 1 B cos x satisfies the differential equation y99 1 y9 22y − sin x.

54. (a) Evaluate lim

xl` x sin 1 x. (b) Evaluate lim

xl0 x sin 1 x.

(c) Illustrate parts (a) and (b) by graphing y − x sins1yxd.

;

;

Suppose you are asked to differentiate the function

Fsxd −

sx

²

1 1

The differentiation formulas you learned in the previous sections of this chapter do not enable you to calculate F9 sxd.

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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Solution:

196 ¤ CHAPTER 3 DIFFERENTIATION RULES

53.  =  sin  +  cos  ⇒ ⁰=  cos  −  sin  ⇒ ⁰⁰= − sin  −  cos . Substituting these expressions for , ⁰, and ⁰⁰into the given differential equation ⁰⁰+ ⁰− 2 = sin  gives us

(− sin  −  cos ) + ( cos  −  sin ) − 2( sin  +  cos ) = sin  ⇔

−3 sin  −  sin  +  cos  − 3 cos  = sin  ⇔ (−3 − ) sin  + ( − 3) cos  = 1 sin , so we must have

−3 −  = 1 and  − 3 = 0 (since 0 is the coefficient of cos  on the right side). Solving for  and , we add the first equation to three times the second to get  = −10¹ and  = −10³.

54. (a) Let  = 1

. Then as  → ∞,  → 0⁺, and lim

→∞ sin1

 = lim

→0⁺

1

sin  = lim

→0

sin 

 = 1.

(b) Since −1 ≤ sin (1) ≤ 1, we have (as illustrated in the figure)

− || ≤  sin (1) ≤ ||. We know that lim_

→0(||) = 0 and

lim→0(− ||) = 0; so by the Squeeze Theorem, lim_→0 sin (1) = 0.

(c)

55. (a) 

tan  = 



sin 

cos  ⇒ sec² = cos  cos  − sin  (− sin )

cos² = cos² + sin²

cos² . So sec² = 1 cos². (b) 

sec  = 



1

cos  ⇒ sec  tan  = (cos )(0) − 1(− sin )

cos² . So sec  tan  = sin  cos². (c) 

(sin  + cos ) = 



1 + cot 

csc  ⇒

cos  − sin  = csc  (− csc²) − (1 + cot )(− csc  cot )

csc² = csc  [− csc² + (1 + cot ) cot ]

csc²

= −csc² + cot² + cot 

csc  = −1 + cot  csc  So cos  − sin  = cot  − 1

csc  .

56. We get the following formulas for  and  in terms of :

sin 2 = 

10 ⇒  = 10 sin

2 and cos 2 = 

10 ⇒  = 10 cos 2 Now () = ¹₂²and () = ¹₂(2) = . So

lim

→0⁺

()

() = lim

→0⁺ 1 2²

 = ¹₂ lim

→0⁺



= ¹₂ lim

→0⁺

10 sin(2) 10 cos(2)

= ¹₂ lim

→0⁺tan(2) = 0

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