1 General Trigonometric Functions Answer Key 11-1 Free Response Homework 1. d dx cos 4 x ( ) = d dx cos x ( ) 4 ⎡ ⎣ ⎤ ⎦ = 4 cos x ( ) 3 ⋅− sin x = −4 cos 3 x sin x 3. d dx sin 2 x + cos 2 x ( ) = d dx 1 () = 0 5. d dx cos 4 2 x ( ) = d dx cos 2 x ( ) 4 ⎡ ⎣ ⎤ ⎦ = 4 cos2 x ( ) 3 ⋅− sin 2 x ⋅ 2 = −8 cos 3 2 x sin 2 x 7. d dx sec x 2 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = sec x 2 3 tan x 2 3 ⋅ 2 3 x = 2 3 x sec x 2 3 tan x 2 3 9. d dx csc x 4 3 ( ) = d dx csc x 4 3 ( ) = − csc x 4 3 cot x 4 3 ⋅ 4 3 x 1 3 = − 4 3 x 1 3 csc x 4 3 cot x 4 3 = − 4 3 x 3 csc x 4 3 cot x 4 3
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General Trigonometric Functions Answer Key · General Trigonometric Functions Answer Key 11-1 Free Response Homework 1. d dx (cos4x)= d dx ... =−4cosx⋅sinx 5. D For f to be continuous
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1
General Trigonometric Functions Answer Key 11-1 Free Response Homework
1. ddx
cos4 x( ) = ddx
cos x( )4⎡⎣ ⎤⎦
= 4 cos x( )3 ⋅−sin x= −4cos3 xsin x
3. ddx
sin2 x + cos2 x( ) = ddx1( )
= 0
5. ddx
cos4 2x( ) = ddx
cos2x( )4⎡⎣ ⎤⎦
= 4 cos2x( )3 ⋅−sin2x ⋅2= −8cos3 2xsin2x
7.
ddx
sec x2
3⎛⎝⎜
⎞⎠⎟= sec x
2
3tan x
2
3⋅ 23x
= 23xsec x
2
3tan x
2
3
9.
ddx
csc x43( ) = ddx
csc x43( )
= −csc x43 cot x
43 ⋅ 43x13
= − 43x13 csc x
43 cot x
43
= − 43
x3 csc x43 cot x43
2
11.
′f x( ) = sec2 x + sec x tan x
′fπ3
⎛⎝⎜
⎞⎠⎟ = sec
2 π3+ secπ
3tanπ
3
= 2( )2 + 2 ⋅ 3
= 4 + 2 3
13.
′f x( ) = 32
+ cos x = 0, DNE, EoaASD
c.v. : x = ± 5π6
,±π
15.
′f x( ) = 32
− sin x = 0, DNE, EoaASD
c.v. : x = π3
, 2π3
,0
17. v t( ) = 2 sin t( )cos t
= 2sin t cos t= sin2t
3
19.
′H t( ) = 4 ⋅−sin π8t − 5( )⎡
⎣⎢⎤⎦⎥⋅ π8
= − π2
sin π8t − 5( )⎡
⎣⎢⎤⎦⎥= 0, DNE, EoaASD
sin π8t − 5( )⎡
⎣⎢⎤⎦⎥= 0
π8t − 5( ) = 0 ± 2πn
π ± 2πn⎧⎨⎩
= 0 ±πn{
t − 5 = 0 ±16n8 ±16n
⎧⎨⎩
= 0 ± 8n{
t = 5 ±16n13±16n
⎧⎨⎩
= 5 ± 8n{
c.v. : t = 5, 13, 0, 20
vt 0 5 13 20← →⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
END + 0 − 0 + END
The weight moves up during the first 5 seconds, then back down until 13 seconds have passed, the back up again until time reaches 20 seconds. 11-1 Multiple Choice Homework
1. B limx→0
1− cos xx2
= 00
=LHlimx→0
sin x2x
= 00
=LHlimx→0
cos x2
= 12
4
3. E y = cos x( )2 − sin x( )2
′y = 2 cos x( ) ⋅sin x − 2 sin x( ) ⋅cos x= −4cos x ⋅sin x
5. D For f to be continuous at x = π : 1. f π( ) = ?
2. limx→π
f x( ) = limx→π
f x( ) = limx→π
tan xsin x
= 00
=LH
limx→π
sec2 xcos x
= 1−1
= −1
3. f π( ) = limx→π
f x( )∴ f π( ) = −1
11-2 Free Response Homework
1. ddx
x3 ⋅sec x( ) = x3 ⋅sec x tan x + sec x ⋅3x2= x2 sec x x tan x + 3( )
3. Dx x2 ⋅sin x + 2x ⋅cos x( )= x2 ⋅cos x + sin x ⋅2x + 2x ⋅−sin x + cos x ⋅2= x2 cos x + 2xsin x − 2xsin x + 2cos x
= x2 cos x + 2xsin x −2xsin x + 2cos x= x2 cos x + 2cos x
= cos x x2 + 2( )
5
5. ′f x( ) = 2 sin2 x ⋅2cos x ⋅−sin x + cos2 x ⋅2sin x ⋅cos x⎡⎣ ⎤⎦
= 2 −2sin3 xcos x + 2cos3 xsin x⎡⎣ ⎤⎦= 4sin xcos x −sin2 x + cos2 x⎡⎣ ⎤⎦= 4sin xcos x cos2 x − sin2 x⎡⎣ ⎤⎦= 2 ⋅2sin xcos x cos2x[ ]= 2sin2x ⋅cos2x= sin 4x
7. dydx
=cos x − 3( )sec2 x − tan x ⋅−sin x
cos x − 3( )2
=cos x − 3( ) ⋅ 1
cos2 x+ sin
2 xcos x
cos x − 3( )2
=
cos x − 3cos2 x
+ sin2 x ⋅cos xcos2 x
cos x − 3( )2
= cos x − 3+ sin2 xcos x
cos2 x cos x − 3( )2
=cos x 1− 3sec x + sin2 x( )cos2 x cos x − 3( )2
=cos x 1− 3sec x + sin2 x( )cos 2 x cos x − 3( )2
= 1− 3sec x + sin2 x
cos x cos x − 3( )2
6
9. ddx
sin x1− cos x
⎛⎝⎜
⎞⎠⎟ =
1− cos x( )cos x − sin x ⋅sin x1− cos x( )2
= cos x − cos2 x − sin2 x
1− cos x( )2
=cos x − cos2 x + sin2 x( )
1− cos x( )2
= cos x −11− cos x( )2
=− 1− cos x( )1− cos x( )2
=− 1− cos x( )1− cos x( ) 2
= −11− cos x
= − 1
2 1− cos x2
⎛⎝⎜
⎞⎠⎟
= − 1
2sin2 12x
= − 12csc2 1
2x
11.
′f x( ) = sec x ⋅sec2 x + tan x ⋅sec x tan x
′fπ4
⎛⎝⎜
⎞⎠⎟ = sec
π4⋅sec2 π
4+ tanπ
4⋅secπ
4tanπ
4
= 2 ⋅ 2( )2 +1⋅ 2 ⋅1
= 2 2 + 2 = 3 2
7
13.
′f x( ) = x ⋅−sin x + cos x + x ⋅cos x + sin x
′fπ4
⎛⎝⎜
⎞⎠⎟ =
π4⋅−sinπ
4+ cosπ
4+ π4⋅cosπ
4+ sinπ
4
= π4⋅− 1
2+ 1
2+ π4⋅ 12+ 1
2
= π4⋅− 1
2+ 1
2+ π4⋅ 12
+ 12
= 22= 2
15.
Volume = l Areatrapezoid( )= l 1
2b1 + b2( )h⎡
⎣⎢⎤⎦⎥
= 12 12
3+ 3+ 2y( ) 4 − y2⎡⎣⎢
⎤⎦⎥
= 12 12
6 + 4sin x( ) 4 − 4sin2 x⎡⎣⎢
⎤⎦⎥
= 12 12
6 + 4sin x( ) ⋅2 ⋅ 1− sin2 x⎡⎣⎢
⎤⎦⎥
= 12 12
6 + 4sin x( ) ⋅ 2 ⋅ 1− sin2 x⎡
⎣⎢
⎤
⎦⎥
= 12 6 + 4sin x( ) cos2 x⎡⎣
⎤⎦
= 12 6 + 4sin x( )cos x⎡⎣ ⎤⎦= 72cos x + 48sin xcos x
Volum ′e = −72sin x + 48 sin x ⋅−sin x + cos x ⋅cos x[ ]= −72sin x − 48sin2 x + 48cos2 x
c.v. : x = 0.439 radians ≡ 25.175!
122
3
2x x
y y
Note:
sin x = y2
y2 + h2 = 4⇒ h = 4 − y2h h
8
17 a.
′x t( ) = 5cos t, ′y t( ) = 2t!v = 5cos t, 2t b.
′′x t( ) = −5sin t, ′′y t( ) = 2!a = −5sin t, 2 c. Speed t=π = 5cosπ( )2 + 2π( )2 = 8.030 19. ′x t( ) = et ⋅cos t + sin t ⋅et
′y t( ) = et ⋅−sin t + cos t ⋅et′x 1( ) = ecos1+ esin1′y 1( ) = −esin1+ ecos1
Min @ x ≈ −1.253 , Max @ x ≈1.253 No sign change @ x = 0
′f x( )
9
3. B ′y = x ⋅−sin x + cos x = 0
⇒ xsin x = cos x⇒ x = cot x
⇒ tan x = 1x
5. E ′h x( ) = f x( ) ⋅ ′g x( ) + g x( ) ⋅ ′f x( ) = f x( ) ′g x( )
⇒ g x( ) ⋅ ′f x( ) = 0⇒ g x( ) = 0 or ′f x( ) = 0We are given that g x( ) > 0, so ′f x( ) = 0.⇒ f x( ) = constantWe are given f 0( ) = 1, ∴ f x( ) = 1.
11-3 Free Response Homework
1.
′y = 1
1− x 2( )2⋅ 2
= 21− 2x2
3.
′y = − 1
1+ 1x
⎛⎝⎜
⎞⎠⎟2 ⋅−x
−2 − 11+ x2
= − 1
1+ 1x2
⋅− 1x2
− 11+ x2
= 1x2 +1
− 11+ x2
= 0
10
5.
′y =x ⋅ 1
x x2 −1− sec−1 x
x2
=
xx x2 −1
− sec−1 x ⋅ x2 −1x2 −1
x2
=
x − sec−1 x x2 −1x2 −1x2
= x − x2 −1sec−1 xx2 x2 −1
7. ′y = −1
2e3x 2e3x( )2 −1⋅2e3x ⋅3
= −1
2e3x 2e3x( )2 −1⋅ 2e3x ⋅3
= −34e6x −1
11
9.
′y = 12x1− x2
⎛⎝⎜
⎞⎠⎟2
+1⋅1− x2( ) ⋅2 − 2x ⋅−2x
1− x2( )2
= 1
4x2
1− x2( )2+1− x2( )21− x2( )2
⋅ 2 − 2x2 + 4x2
1− x2( )2
= 14x2 + 1− x2( )2
1− x2( )2⋅ 2 + 2x
2
1− x2( )2
=1− x2( )2
4x2 + 1− x2( )2⋅2 1+ x2( )1− x2( )2
=1− x2( )2
4x2 +1− 2x2 + x4⋅2 1+ x2( )1− x2( )2
=2 1+ x2( )x4 + 2x2 +1
=2 x2 +1( )
x2 +1( ) x2 +1( )
=2 x2 +1( )
x2 +1( ) x2 +1( )= 2x2 +1
12
11.
Speed t=4 = tan−1 45
⎛⎝⎜
⎞⎠⎟
2
+ ln17( )2 = 2.912
′′x t( ) = 1t
t +1⎛⎝⎜
⎞⎠⎟
2
+1⋅t +1( ) ⋅1− t ⋅1
t +1( )2 , ′′y t( ) = 1t 2 +1
⋅2t
′′x t( ) = 1t 2
t +1( )2 +1⋅ 1t +1( )2 , ′′y t( ) = 2t
t 2 +1
′′x t( ) = 1t 2
t +1( )2 +t +1( )2
t +1( )2
⋅ 1t +1( )2 , ′′y t( ) = 2t
t 2 +1
′′x t( ) = 1t 2 + t +1( )2
t +1( )2
⋅ 1t +1( )2 , ′′y t( ) = 2t
t 2 +1
′′x t( ) = t +1( )2
t 2 + t +1( )2 ⋅1
t +1( )2 , ′′y t( ) = 2tt 2 +1
′′x t( ) =t +1( )2
t 2 + t +1( )2 ⋅1
t +1( )2, ′′y t( ) = 2t
t 2 +1
′′x t( ) = 1t 2 + t +1( )2 , ′′y t( ) = 2t
t 2 +1
′′x 4( ) = 142 + 4 +1( )2 , ′′y t( ) = 2 ⋅4
42 +1
a t( ) = 841
, 817
11-3 Multiple Choice Homework
1. B Graph y = arcsin x and y = 2arccos x . Use the intersection menu item – x = 0.866 ≈ 0.9
11. No fractions with a denominator of zero, no radicals, and no logarithms ∴ Domain: x ∈ −π , π[ ] h=0 EB: None POEs: None VAs: None Axis Points: sin x = 0
x = 0 ± 2πnπ ± 2πn
⎧⎨⎩
= 0 ±πn{
⇒ x = −π , 0, π−π , 0( ), 0, 0( ), π , 0( )
Extreme Points:
−π/2 π/2 π 3π/2 2π
−6
−5
−4
−3
−2
−1
1
2
3
4
5
6
x
y
19
′y = ex ⋅cos x + sin x ⋅ex ⋅= 0, DNE , EoaASDex cos x + sin x( ) = 0
ex = 0
ln ex( ) = ln0
DNE or
cos x + sin x = 0sin x = −cos xsin xcos x
= −cos xcos x
tan x = −1
x =− π
4± 2πn
3π4
± 2πn= π
4±πn⎧
⎨⎩
⎧
⎨⎪⎪
⎩⎪⎪
c.v. : x = π4
, 3π4
, −π , π
−π , 0( ), π , 0( ),
− π4
, − 0.322⎛⎝⎜
⎞⎠⎟ , 3π
4, 7.460⎛
⎝⎜⎞⎠⎟
Range: y∈ 0, 1[ ]
11-4 Multiple Choice Homework
1. C
′f x( ) = 2sin x ⋅cos x > x = ′g x( )
−π −π/2 π/2 π 3π/2
−8
−7
−6
−5
−4
−3
−2
−1
1
2
3
4
5
6
7
8
x
y
20
3. E I. limx→0
sin xx
= 00
=LHlimx→0
cos x1
= 1≠ ∞
Since the limit is finite (not infinite) we have a POE rather than a VA at x = 0 . II. limx→±∞
sin xx
= 0
Since the limit is 0 (which is finite) we have an HA of y = 0 . III. sin xx
= 0
⇒ sin x = 0
⇒ x = 0 ± 2πnπ ± 2πn
⎧⎨⎩
= 0 ±πn{
We have an infinite number of zeros. Statements II and III are true. 5. B dydx
= 2cos 2x( ) ⋅−sin 2x( ) ⋅2= −4cos2xsin2x
336
General Trigonometric Functions Practice Test Answer Key
Multiple Choice – Calculator Allowed
1. E ′f x( ) = cos e− x( ) ⋅e− x ⋅ −1( )
= −e− x cos e− x( )
2. B ′y = 1− sin x′y x=0 = 1− sin0 = 1
y −1= 1 x − 0( )⇒ y = x +1
3. B
Graph ′f x( ) = cos2 xx
− 15
.
The sign of the derivative changes 3 times – i.e. from + to –, then – to +, and finally + to – again.
672
4. D
′E B( ) = 2 B +1( ) ⋅ dBdt
= 2 B +1( ) ⋅20cos t10
⎛⎝⎜
⎞⎠⎟ ⋅
110
= 2 20sin t10
⎛⎝⎜
⎞⎠⎟ + 50 +1⎡
⎣⎢⎤⎦⎥⋅20cos t
10⎛⎝⎜
⎞⎠⎟ ⋅
110
′E 100( ) = 4 20sin 10010
⎛⎝⎜
⎞⎠⎟ + 51⎛
⎝⎜⎞⎠⎟
cos 10010
⎛⎝⎜
⎞⎠⎟ = −134.396 dollars/day
Free Response – Calculator Allowed
1. No fractions with a denominator of zero, no radicals, and no logarithms ∴ Domain: x ∈ −2π , 2π[ ]
′y = 32
− sin x = 0, DNE , EoaASD
sin x = 32
x =
π3± 2πn
2π3
± 2πn
⎧
⎨⎪⎪
⎩⎪⎪
c.v. : x = − 5π3
, − 4π3
, π3
, 2π3
, − 2π , 2π
′yx −2π − 5π
3 4π
3 π
3 2π
3 2π
← →⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ END + 0 − 0 + 0 − 0 + END
Extreme Points:
−2π , − 4.441( ), 2π , 6.441( ), π
3 , 1.407⎛⎝⎜
⎞⎠⎟, 2π
3 , 1.314⎛⎝⎜
⎞⎠⎟
,
−5π
3 , − 4.034⎛⎝⎜
⎞⎠⎟, − 4π
3 , − 4.128⎛⎝⎜
⎞⎠⎟
673
2. No fractions with a denominator of zero, no radicals, but there is a logarithm –
tan xx 0 π
2 π 3π
2 2π
← →⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯0 + DNE − 0 + DNE − 0
∴ Domain: x ∈ 0, π2
⎛⎝⎜
⎞⎠⎟ ∪ π , 3π
2⎛⎝⎜
⎞⎠⎟
′y = 1
tan x⋅sec2 x
= 1sin xcos x
= 0 , DNE, EoaASD
If ′y = DNE , then
sin x = 0
x = 0 ± 2πnπ ± 2πn
= 0 ±πn{⎧⎨⎩
or
cos x = 0
x =
π2± 2πn
− π2± 2πn
= ± π2± 2πn⎧
⎨⎩
⎧
⎨⎪⎪
⎩⎪⎪
c.v. : x = 0 , π , 2π , π2
, 3π2
No critical values No extreme points
674
Multiple Choice – No Calculator Allowed
5. E ′f x( ) = sec2 2x( ) ⋅2
′fπ6
⎛⎝⎜
⎞⎠⎟ = sec
2 π3
⎛⎝⎜
⎞⎠⎟ ⋅2 = 8
6. D ddx
sin xy( ) = x⎡⎣ ⎤⎦
cos xy( ) x ⋅ dydx
+ y ⋅1⎡⎣⎢
⎤⎦⎥= 1
xcos xy( ) dydx
+ ycos xy( ) = 1
xcos xy( ) dydx
= 1− ycos xy( )dydx
=1− ycos xy( )xcos xy( )
7. A y x=0 = arctan0 = 0
′y = 1
1+ x3
⎛⎝⎜
⎞⎠⎟2 ⋅13
′yx=0 = 11+ 0( )2
⋅ 13= 13
Point: 0, 0( )
Slope: 13
y − 0 = 13x − 0( )
y = 13x⇔ 3y = x⇔ x − 3y = 0
675
8. B
′f x( ) = limh→0
f x + h( )− f x( )h
= limh→0
cos x + h( )− cos xh
⇒ f x( ) = cos x∴ ′f x( ) = −sin x
OR
′f x( ) = limh→0
cos x + h( )− cos xh
= 00
=LHlimh→0
−sin x + h( ) ⋅1− 01
= limh→0
−sin x + h( )1
= −sin x
Note: When using L’ Hopital’s Rule, you must take the derivative with respect to h.