PreCalc B Second Semester Final Exam Review

I.    Linear  velocity,  angular  velocity,  and  arc  length.  

1A.θ = 40π t = 1min

ω = θt=

20 2π( )1min

i1min60sec

= 2π3

rad/sec

 

21B. 303

20 cm/sec

V r πω

π

⎛ ⎞= = ⎜ ⎟⎝ ⎠=

 

1C. 0 cm/sec  

2A.V = 5m20sec

= 14

m/sec

V = rω = 14

m/sec = 2ω

2B. ω = 18

rad/sec

 

( )

( )( )2 2 2

3. 6 2 12feet

1 1 6 2 36ft2 2

S r

A r

θ

θ

= = =

= = =  

 II.    Trigonometry  

1A. 2 13

131B. − 13

31C. −3

21D. 13

2  

2. 22

, 22

⎛

⎝⎜

⎞

⎠⎟  

3A. 2

23B. −1 3C. 2 3

33D. 2

   

4. A : none P : 3π

D :°except− 5π2

+ 3πk R :°    

                             

5. A : 2 P : 4D :° R : −1.5,2.5[ ]                    

6. A : none P : 2π

D :°except 3π2

+πk R : (−∞,−2]∪[2,∞)  

       

 

 

 

 

 

 

 

7. y = 19cosπ32

x +10( )⎛⎝⎜

⎞⎠⎟+ 31

8. y = 23cosπ9

x − 2( )⎛⎝⎜

⎞⎠⎟−1

 

 

 

9A. cos2 x2

⎛⎝⎜

⎞⎠⎟− sin2 x

2⎛⎝⎜

⎞⎠⎟

= cos x

9B.2sin xcos x

1+ 2cos2 x −1= sin x

cos x

= tan x

9C. tan(2(50)) = tan100o

9D. sin(2(35)) = sin70o

 

 

9E. tan2θ +1

= sec2 x  

9F. 2sinθ cosθ

= sin2θ  

10A. 2 13

⎛⎝⎜

⎞⎠⎟2 23

⎛⎝⎜

⎞⎠⎟= 4 2

9

10B.1− 2 13

⎛⎝⎜

⎞⎠⎟2

= 1− 2 19

⎛⎝⎜

⎞⎠⎟ =

79

10C.1− 2 2

3

1+ 2 23

= 3− 2 23+ 2 2

   

 

11A. tan−1 −1( )= − π

4  

11B. arccos − 2

2

⎛

⎝⎜

⎞

⎠⎟ =

3π4  

11C. tan sin−1 5

13⎛⎝⎜

⎞⎠⎟= 5

12  

12. cos 1502

⎛⎝⎜

⎞⎠⎟ =

1+ − 32

⎛⎝⎜

⎞⎠⎟

2

= 2 − 34

= 2 − 32

   

13A.1− sin2β = 1− 2 13

⎛⎝⎜

⎞⎠⎟2

= 79

13B. 1− cosα2

=1− 122

= 12

13C. 1− tan2α

2 tanα=1− − 3

1⎛⎝⎜

⎞⎠⎟

2

2 − 3( ) = 33

13D.1− 2 2

3

1+ 2 23

= 3− 2 23+ 2 2

   

14.    Proof  answers  may  vary    

15A.sec2θ = 4; cosθ = ±1/ 2

θ = 60o ,120o ,240o ,300o  

15B.tan2θ +5tanθ + 6 = 0= (tanθ + 3)(tanθ + 2)

θ = 108.43o ,116.57o ,288.44o ,296.57o  

15C.2sin2θ − cosθ −1= 0

θ = 60o ,180o ,300o  

16A. sin2 x − sin x − 6 = 0

sin x − 3( ) sin x + 2( ) = 0

sin x = 3 or sin x = −2

NoSolution

 

16B. 2cot2 x + csc2 x − 2 = 0

2cot2 x +1+ cot2 x − 2 = 0

3cot2 x = 1 cot2 x = 13

tan2 x = 3 tan x = ± 3

x = π3

,2π3

,4π3

,5π3

 

16C. 2cos2 x − sin x −1= 0

2 1− sin2 x( )− sin x −1= 0

2− 2sin2 x − sin x −1= 0

2sin x −1( ) sin x +1( ) = 0

sin x = 12

orsin x = 1

x = π6

,5π6

,3π2

 

17D. sin2θ = 22

θ = π8

, 3π8

, 9π8

,11π8

 

17E. 0, π2

, 3π2

, π

17F. tan(3θ + π4

) = 1

θ = 0,π3

, 2π3

,π , 4π3

,5π3

 

17G. sec2θ + tanθ = 0

tan2θ + tanθ +1= 0

tanθ = −1± i 32

θ = no solution

 

   

17H. sinθ + cosθ = 1

θ = 0,π2

 

     III.    Polar  Graphing  

( )

21. 4 25 295tan 111.82

29,111.8

r r

θ θ

= + =

= → =−

o

o

 

5 5 22. cos 5cos4 25 5 2sin 5sin4 2

5 2 5 2,2 2

x r

y r

πθ

πθ

= = = −

= = = −

⎛ ⎞− −⎜ ⎟⎜ ⎟⎝ ⎠

 

83. 3,7π⎛ ⎞−⎜ ⎟⎝ ⎠

 

 

4. 2 r 2 cos2θ( )− 2 r 2 sin2θ( ) = 5r sinθ

2r 2 cos2θ − 2r 2 sin2θ = 5r sinθ  

5. r 2 = r − 2r sinθ

x2 + y2 = x2 + y2 − 2y  

 

6A.    symmetry:    x-­‐axis,  y-­‐axis ,2π⎛ ⎞

⎜ ⎟⎝ ⎠  pole  

 

6B.    symmetry:    y-­‐axis  2π⎛ ⎞

⎜ ⎟⎝ ⎠  

 

6C.    symmetry:    y-­‐axis  2π⎛ ⎞

⎜ ⎟⎝ ⎠  

 6D.    symmetry:    x-­‐axis,  y-­‐axis   ,

2π⎛ ⎞

⎜ ⎟⎝ ⎠  pole  

 

   IV.  Law  of  Sines  and  Cosines  

1A. 51o 1B. 18.59 cm

1C. 14.45 cm

2A. 12.64 cm 2B. 50.91o

2C. 81.09o

3A. 27.45 in 3B. 29.71 in

3C. 28o

4A. b1=51.60m b2 =5.135

4B. B1=153.45o B2 =2.55o

4C. C1=14.55o C2 =165.45o

5A. A = 15(7)(5)(3) = 1575 = 15 7 ft2

5B. A = 12

sin38o 6( ) 4( ) = 7.39mi2

 

 V.  Conics    

1A. x2 + 4x + 4( )− 3 y2 − 2y +1( ) = −13+ 4− 3

x + 2( )2− 3 y −1( )2

= −12

y −1( )2

4−

x + 2( )2

12= 1

C : −2,1( )V : −2,3( ) −2,−1( )F : −2,5( ) −2,−3( )A : y −1( ) = ± 3

3x + 2( )

 

1B. 4 x2 − 4x( ) + 9 y2 + 6y( ) = −61

4 x2 − 4x + 4( ) + 9 y2 + 6y + 9( ) = −61+16+81

4 x − 2( )2+ 9 y + 3( )2

= 36

x − 2( )2

9+

y + 3( )2

4= 1

C : 2,−3( )V : −1,−3( ) 5,−3( )F : 2 ± 5,−3( )Major length : 6Minor length : 4

 

1C. 4 x2 − 2x +1( )− y2 + 6y + 9( ) = 6+ 4− 9

4 x −1( )2− y + 3( )2

= 1

x −1( )2

14

−y + 3( )2

1= 1

C : 1,−3( )V : 3

2,−3

⎛⎝⎜

⎞⎠⎟

12

,− 3⎛⎝⎜

⎞⎠⎟

F : 1± 52

,−3⎛

⎝⎜

⎞

⎠⎟

A : y + 3( ) = ±2 x −1( )

 

2. x2 +8x +16( )− 4y = −8+16

x + 4( )2− 4y = 8

x + 4( )2= 4y +8

x + 4( )2= 4 y + 2( )

V : −4,−2( )F : −4,−1( )D : y = −3

 

3. a = 5 c = 325− b2 = 9 b2 = 16

x − 2( )2

25+

y − 4( )2

16= 1

 

4.V : 5,−4( )x − h( )2

= −4a y − k( )x −5( )2

= −8 y + 4( )  

( )( ) ( )2 2

5. : 4,6

4 61

5 9

C

x y− −+ =

 

( ) ( )

2 2

2 2

6. 4 9 5

4 11

4 5

b b

x y

+ = =

− +− =

 

( )

2

2

2

7. 430 4 20900 80 11.25

45

5, 0.56 ht=19.44ft

10, 2.22 ht=17.78ft

20, 8.89 ht=11.11ft

x ayaa a

x y

x y

x y

x y

== −= − = −

= −

= = − →

= = − →

= = − →

 

 VI.    Parametrics  

( )

( ) ( ) ( )( ) ( )

( ) 4 2 cos 12

( ) 4 2 cos15

( ) 4 2 2 cos 24 4 2 2

2( ) 4 2 2 cos 4 2 215

x t t

or x t t

y t t

or y t t

π

π

=

⎛ ⎞− − = ⎜ ⎟⎝ ⎠

= − + −

⎛ ⎞− − = − + −⎜ ⎟⎝ ⎠

 

x(t)  

 y(t)                      

 VII.  Sequences  and  Series  

1A. 3, 7, 11, 15, 19; Arithmetic

1B. −143

, −133

, − 4, −113

,−103

; Arithmetic

1C. 23

, 34

, 45

, 56

, 67

; Neither

2A. an = 32−5 n−1( )2B. a n = 11 3( )n−1

3. −333 4. 3.45×1011

5.166 = 30+ 4 n−1( ) n = 35

6A. S25 =252

5+ 77( ) = 1025

S5 =52

5+17( ) = 55 1025−55= 970

6B. S23 =3 1− 323( )

1− 3= 1.41×1011

6C. 1339280

7. 8748

8. 164

= 2 12

⎛⎝⎜

⎞⎠⎟

n−11

128= 1

2⎛⎝⎜

⎞⎠⎟

n−1

ln 1128

⎛⎝⎜

⎞⎠⎟= n−1( ) ln 1

2⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

7 = n−1 n = 89. a100 = 2+ 99 3( ) = 299

S100 =100

22+ 299( ) = 15050

10. S15 =13

1− −2( )15

1− −2( )⎛

⎝⎜⎜

⎞

⎠⎟⎟= 3641