13
MAC 1147 Exam #5 Review Instructions: Exam #5 will consist of 10 questions plus a bonus problem. Some questions will have multiple parts and others will not. Some questions will be multiple choice and some will be free response. For the free response questions, be sure to show as much work as possible in order to demonstrate that you know what you are doing. The point value for each question is listed after each question. Notice that all four parts of question 10 are multiple choice but you will get partial credit based on your work. So it is possible to get the correct answer and not receive full credit and it is possible to get the wrong answer and still receive some credit. The bonus problem will be worth 10 extra credit points. Attempting the bonus problem can only help you; it can't hurt you. A scientific calculator may be used but no graphing calculators or calculators on any device (cell phone, iPod, etc.) which can be used for any other purpose. The exam will be similar to this review, although the numbers and functions may be different so the steps and details (and hence the answers) may work out different. But the ideas and concepts will be the same. Additionally, you will be allowed to use the Trigonometric Identities and Unit Circle which I have posted on my website. Neither will be provided on the test, so if you wish to use either or both , you must bring them with you. The Trigonometric Identities and the Unit Circle may not be shared nor can they be used if they have any wr iting on them.

MAC 1147 Exam #5 Review - Department of Mathematicsdscheib/teaching/summer2011/mac1147_files/mac114… · MAC 1147 Exam #5 Review . Instructions: Exam #5 will consist of 10 questions

  • Upload
    others

  • View
    9

  • Download
    0

Embed Size (px)

Citation preview

Page 1: MAC 1147 Exam #5 Review - Department of Mathematicsdscheib/teaching/summer2011/mac1147_files/mac114… · MAC 1147 Exam #5 Review . Instructions: Exam #5 will consist of 10 questions

MAC 1147 Exam 5 Review

Instructions Exam 5 will consist of 10 questions plus a bonus problem Some questions will have multiple parts and others will not Some questions will be multiple choice and some will be free response For the free response questions be sure to show as much work as possible in order to demonstrate that you know what you are doing The point value for each question is listed after each question Notice that all four parts of question 10 are multiple choice but you will get partial credit based on your work So it is possible to get the correct answer and not receive full credit and it is possible to get the wrong answer and still receive some credit The bonus problem will be worth 10 extra credit points Attempting the bonus problem can only help you it cant hurt you A scientific calculator may be used but no graphing calculators or calculators on any device (cell phone iPod etc) which can be used for any other purpose The exam will be similar to this review although the numbers and functions may be different so the steps and details (and hence the answers) may work out different But the ideas and concepts will be the same Additionally you will be allowed to use the Trigonometric Identities and Unit Circle which I have posted on my website Neither will be provided on the test so if you wish to use either or both you must bring them with you The Trigonometric Identities and the Unit Circle may not be shared nor can they be used if they have any writing on them

~I (~ tl) _ - -Jfs 3 -~l~

IT -+~~-r--

----shy

(1) Find the exact value of each expression (4 points each)

(i) sin-1(sin 4)

gt-1 (b) 2 (c) 4 (d) 1 (e) None of the above

~ _ L][) - -lL l ~ - 1

][ 3

(ii) sin (sin-1 13)

(a) 03 (b) -13 (c) 13 Undefined (e) None of the above

~ ~ not ~ - ( g

(iii) tan (COS-l~)

(a) ~-9 (b)-fi7 (d) VJ7 (e) None of the above

2-~N1fJ e - (0)_1 T~ jr

1- 1- - A l 0 tl -

Cl- t-4(-=- ~ - 4---I

(2) Write the trigonometric expression as an algebraic expression in u (4 points)

csc (tan-1 u) ----rshy

vu2+1 (c) vu2-1( )a u 2 +1 u 2 -1

A2 -t -=(1

c=- JAL- ---shy

(3) Establish the identity (10 points)

1 - sinx cos x =-shy

cos x 1 + sinx

( l ~ f)c) (l+il gtc)

Co~ x bull (l+-SlhX)

-

vr~ nVftUAW t ~~ 7 I+trgtlt

--

(4) Find the exact value of each expression (5 points each)

(i) tan 1950 = ~ ( ~(o -doOo)

- -~ 13So + ~ (00deg

1shy ~ I~S middot ~~Olgt

s3 - +- 3 (- ~ ~) ~ - ~+J3

- (- 1)( i) G+- 4)1 ~~

(ii) cos 311 8

- +

- +shy-

=+~- ~ - -~ -

J t-~ 2

(5) Find the exact value of tan(a - (3) under the given conditions (4 points) It vj r of

5 7r 15 7r cos ex = - - - lt ex lt 7r sm (3 = - - lt (3 lt 7r

13~ 17~_

(a) -io (b) i ~ (c) to) e ~l None of the ~f ~ ( above

QvJ1t ~f CWrJ II

If) ~ ~r -If~

~

kJ - ~r -J (-- 1pound)s - ~

(-l--~o~r

(-~~ Jf)(~b) -1L- lzu- Gt I~o )(10)

(6) Establish the identity (10 points)

sin (a - (3) cos (a + (3) = sin a cos a - sin (3 cos (3

----------------

(7) Find the exact value of each expression (4 points)

(i) cos (tan-1 ~ - sin-1 ~)

~T (a) 2~ (b) 1 (c) 2~ (e) None of the

above

rJ-t- CWJII ~

(ii) tan [2 cos- l (-~)J

(a) 2i (b) -2~ (c) - 2l (d) -~ ~None of the ryabove

2~ (tod-~)) _ El)(I~) _ (1- Z) ((lt0)

_ ~l (CraquoS-1 (~raquo) ~

(8) Establish the identity (10 points)

4cos (4x) = cos X - 6middotsm2 x cos2 x + sin4 x

(9) (i) Express the product as a sum containing only sines or cosines (4 points)

sin (30) cos (40)

GY j[sin (78) - sin 81 (b) sincos(1202)

(c) Hcos(70) -cosO] (d) ~[cos(7e)+sinel

(ii) Express the sum or difference as a product of sines andor cosines (4 points)

cos (60) + cos (40)

(a) 2cos(50)sinO (b) - 2 cos (50) cos 0

(c) 2 sin (50) sin 0 2 cos (50) cos e

(10) (i) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2cos2 0 - 1 = 0

rr5rr ~(b) E 3rr 5rr 7rr (c) E 7rr()a 33 ~4444 44

(d) rr 2rr 4rr 5rr ( e) None of the above3333

or

(ii) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2 cos (20) = v3 fr)( ) rr llrr 13rr 23rrvJ 12 12 12 12

(b) 3 2 rr (c) llrr

6 6

(d) ~ (e) None of the above

j (1 I

L 1T Il

(iii) Solve the equation on the interval 0 x lt 27f (6 points - 3 points for the answer and 3 points for the steps)

2 cos2 X + 3 sin x - 3 = 0

7r 7r 57r 77r ( ) (b) 77r 117r (c) 7r 57r a 6266 2 6 6 6 6

7r 7r 27r(d) ~None of the above323

JCb~l)( +- 3tin )c - 3 =--0 l-

I-~ttt X S( )lt -=--

~ (_Sl-)() -t3~ X - ~ -=- D X =-~ ()

~ _ ~~t +S~ X - 3 - 0 X -= Jt2x~ Sli ( )

-= lC I [X fO fa

(iv) Solve the equation Give a general formula for all the solutions (6 points shy3 points for the answer and 3 points for the steps)

esc (~) = 2~ ~11=7f+6n7f 11= 27f+6n7f (b) 11=~+2n7f 11=2+2n7f

(c) 11=~+2n7f B=~~+2n7f (d) B = ~ + 6n7f B = 5 + 6n7f

(e) None of the above

1) ~)~ ~ ~ In~(~ ~ (~~) =0f+ Jlf)()

e~ 1+ ~Vlf (1 nr- ~1

-----------------

Bonus Write the trigonometric expression as an algebraic expression containing u and v (10 points)

1 1b -- -4

b~W 5 ( ~ J15 1i

SivSlA - -4iIf) =~~J105 ( ~- lIT) - togt (141 S ( (-bI- )

- fu) (- I -([F4)( ~ ~ Jt1)L+ I ) r~ t )

-

Page 2: MAC 1147 Exam #5 Review - Department of Mathematicsdscheib/teaching/summer2011/mac1147_files/mac114… · MAC 1147 Exam #5 Review . Instructions: Exam #5 will consist of 10 questions

~I (~ tl) _ - -Jfs 3 -~l~

IT -+~~-r--

----shy

(1) Find the exact value of each expression (4 points each)

(i) sin-1(sin 4)

gt-1 (b) 2 (c) 4 (d) 1 (e) None of the above

~ _ L][) - -lL l ~ - 1

][ 3

(ii) sin (sin-1 13)

(a) 03 (b) -13 (c) 13 Undefined (e) None of the above

~ ~ not ~ - ( g

(iii) tan (COS-l~)

(a) ~-9 (b)-fi7 (d) VJ7 (e) None of the above

2-~N1fJ e - (0)_1 T~ jr

1- 1- - A l 0 tl -

Cl- t-4(-=- ~ - 4---I

(2) Write the trigonometric expression as an algebraic expression in u (4 points)

csc (tan-1 u) ----rshy

vu2+1 (c) vu2-1( )a u 2 +1 u 2 -1

A2 -t -=(1

c=- JAL- ---shy

(3) Establish the identity (10 points)

1 - sinx cos x =-shy

cos x 1 + sinx

( l ~ f)c) (l+il gtc)

Co~ x bull (l+-SlhX)

-

vr~ nVftUAW t ~~ 7 I+trgtlt

--

(4) Find the exact value of each expression (5 points each)

(i) tan 1950 = ~ ( ~(o -doOo)

- -~ 13So + ~ (00deg

1shy ~ I~S middot ~~Olgt

s3 - +- 3 (- ~ ~) ~ - ~+J3

- (- 1)( i) G+- 4)1 ~~

(ii) cos 311 8

- +

- +shy-

=+~- ~ - -~ -

J t-~ 2

(5) Find the exact value of tan(a - (3) under the given conditions (4 points) It vj r of

5 7r 15 7r cos ex = - - - lt ex lt 7r sm (3 = - - lt (3 lt 7r

13~ 17~_

(a) -io (b) i ~ (c) to) e ~l None of the ~f ~ ( above

QvJ1t ~f CWrJ II

If) ~ ~r -If~

~

kJ - ~r -J (-- 1pound)s - ~

(-l--~o~r

(-~~ Jf)(~b) -1L- lzu- Gt I~o )(10)

(6) Establish the identity (10 points)

sin (a - (3) cos (a + (3) = sin a cos a - sin (3 cos (3

----------------

(7) Find the exact value of each expression (4 points)

(i) cos (tan-1 ~ - sin-1 ~)

~T (a) 2~ (b) 1 (c) 2~ (e) None of the

above

rJ-t- CWJII ~

(ii) tan [2 cos- l (-~)J

(a) 2i (b) -2~ (c) - 2l (d) -~ ~None of the ryabove

2~ (tod-~)) _ El)(I~) _ (1- Z) ((lt0)

_ ~l (CraquoS-1 (~raquo) ~

(8) Establish the identity (10 points)

4cos (4x) = cos X - 6middotsm2 x cos2 x + sin4 x

(9) (i) Express the product as a sum containing only sines or cosines (4 points)

sin (30) cos (40)

GY j[sin (78) - sin 81 (b) sincos(1202)

(c) Hcos(70) -cosO] (d) ~[cos(7e)+sinel

(ii) Express the sum or difference as a product of sines andor cosines (4 points)

cos (60) + cos (40)

(a) 2cos(50)sinO (b) - 2 cos (50) cos 0

(c) 2 sin (50) sin 0 2 cos (50) cos e

(10) (i) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2cos2 0 - 1 = 0

rr5rr ~(b) E 3rr 5rr 7rr (c) E 7rr()a 33 ~4444 44

(d) rr 2rr 4rr 5rr ( e) None of the above3333

or

(ii) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2 cos (20) = v3 fr)( ) rr llrr 13rr 23rrvJ 12 12 12 12

(b) 3 2 rr (c) llrr

6 6

(d) ~ (e) None of the above

j (1 I

L 1T Il

(iii) Solve the equation on the interval 0 x lt 27f (6 points - 3 points for the answer and 3 points for the steps)

2 cos2 X + 3 sin x - 3 = 0

7r 7r 57r 77r ( ) (b) 77r 117r (c) 7r 57r a 6266 2 6 6 6 6

7r 7r 27r(d) ~None of the above323

JCb~l)( +- 3tin )c - 3 =--0 l-

I-~ttt X S( )lt -=--

~ (_Sl-)() -t3~ X - ~ -=- D X =-~ ()

~ _ ~~t +S~ X - 3 - 0 X -= Jt2x~ Sli ( )

-= lC I [X fO fa

(iv) Solve the equation Give a general formula for all the solutions (6 points shy3 points for the answer and 3 points for the steps)

esc (~) = 2~ ~11=7f+6n7f 11= 27f+6n7f (b) 11=~+2n7f 11=2+2n7f

(c) 11=~+2n7f B=~~+2n7f (d) B = ~ + 6n7f B = 5 + 6n7f

(e) None of the above

1) ~)~ ~ ~ In~(~ ~ (~~) =0f+ Jlf)()

e~ 1+ ~Vlf (1 nr- ~1

-----------------

Bonus Write the trigonometric expression as an algebraic expression containing u and v (10 points)

1 1b -- -4

b~W 5 ( ~ J15 1i

SivSlA - -4iIf) =~~J105 ( ~- lIT) - togt (141 S ( (-bI- )

- fu) (- I -([F4)( ~ ~ Jt1)L+ I ) r~ t )

-

Page 3: MAC 1147 Exam #5 Review - Department of Mathematicsdscheib/teaching/summer2011/mac1147_files/mac114… · MAC 1147 Exam #5 Review . Instructions: Exam #5 will consist of 10 questions

(2) Write the trigonometric expression as an algebraic expression in u (4 points)

csc (tan-1 u) ----rshy

vu2+1 (c) vu2-1( )a u 2 +1 u 2 -1

A2 -t -=(1

c=- JAL- ---shy

(3) Establish the identity (10 points)

1 - sinx cos x =-shy

cos x 1 + sinx

( l ~ f)c) (l+il gtc)

Co~ x bull (l+-SlhX)

-

vr~ nVftUAW t ~~ 7 I+trgtlt

--

(4) Find the exact value of each expression (5 points each)

(i) tan 1950 = ~ ( ~(o -doOo)

- -~ 13So + ~ (00deg

1shy ~ I~S middot ~~Olgt

s3 - +- 3 (- ~ ~) ~ - ~+J3

- (- 1)( i) G+- 4)1 ~~

(ii) cos 311 8

- +

- +shy-

=+~- ~ - -~ -

J t-~ 2

(5) Find the exact value of tan(a - (3) under the given conditions (4 points) It vj r of

5 7r 15 7r cos ex = - - - lt ex lt 7r sm (3 = - - lt (3 lt 7r

13~ 17~_

(a) -io (b) i ~ (c) to) e ~l None of the ~f ~ ( above

QvJ1t ~f CWrJ II

If) ~ ~r -If~

~

kJ - ~r -J (-- 1pound)s - ~

(-l--~o~r

(-~~ Jf)(~b) -1L- lzu- Gt I~o )(10)

(6) Establish the identity (10 points)

sin (a - (3) cos (a + (3) = sin a cos a - sin (3 cos (3

----------------

(7) Find the exact value of each expression (4 points)

(i) cos (tan-1 ~ - sin-1 ~)

~T (a) 2~ (b) 1 (c) 2~ (e) None of the

above

rJ-t- CWJII ~

(ii) tan [2 cos- l (-~)J

(a) 2i (b) -2~ (c) - 2l (d) -~ ~None of the ryabove

2~ (tod-~)) _ El)(I~) _ (1- Z) ((lt0)

_ ~l (CraquoS-1 (~raquo) ~

(8) Establish the identity (10 points)

4cos (4x) = cos X - 6middotsm2 x cos2 x + sin4 x

(9) (i) Express the product as a sum containing only sines or cosines (4 points)

sin (30) cos (40)

GY j[sin (78) - sin 81 (b) sincos(1202)

(c) Hcos(70) -cosO] (d) ~[cos(7e)+sinel

(ii) Express the sum or difference as a product of sines andor cosines (4 points)

cos (60) + cos (40)

(a) 2cos(50)sinO (b) - 2 cos (50) cos 0

(c) 2 sin (50) sin 0 2 cos (50) cos e

(10) (i) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2cos2 0 - 1 = 0

rr5rr ~(b) E 3rr 5rr 7rr (c) E 7rr()a 33 ~4444 44

(d) rr 2rr 4rr 5rr ( e) None of the above3333

or

(ii) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2 cos (20) = v3 fr)( ) rr llrr 13rr 23rrvJ 12 12 12 12

(b) 3 2 rr (c) llrr

6 6

(d) ~ (e) None of the above

j (1 I

L 1T Il

(iii) Solve the equation on the interval 0 x lt 27f (6 points - 3 points for the answer and 3 points for the steps)

2 cos2 X + 3 sin x - 3 = 0

7r 7r 57r 77r ( ) (b) 77r 117r (c) 7r 57r a 6266 2 6 6 6 6

7r 7r 27r(d) ~None of the above323

JCb~l)( +- 3tin )c - 3 =--0 l-

I-~ttt X S( )lt -=--

~ (_Sl-)() -t3~ X - ~ -=- D X =-~ ()

~ _ ~~t +S~ X - 3 - 0 X -= Jt2x~ Sli ( )

-= lC I [X fO fa

(iv) Solve the equation Give a general formula for all the solutions (6 points shy3 points for the answer and 3 points for the steps)

esc (~) = 2~ ~11=7f+6n7f 11= 27f+6n7f (b) 11=~+2n7f 11=2+2n7f

(c) 11=~+2n7f B=~~+2n7f (d) B = ~ + 6n7f B = 5 + 6n7f

(e) None of the above

1) ~)~ ~ ~ In~(~ ~ (~~) =0f+ Jlf)()

e~ 1+ ~Vlf (1 nr- ~1

-----------------

Bonus Write the trigonometric expression as an algebraic expression containing u and v (10 points)

1 1b -- -4

b~W 5 ( ~ J15 1i

SivSlA - -4iIf) =~~J105 ( ~- lIT) - togt (141 S ( (-bI- )

- fu) (- I -([F4)( ~ ~ Jt1)L+ I ) r~ t )

-

Page 4: MAC 1147 Exam #5 Review - Department of Mathematicsdscheib/teaching/summer2011/mac1147_files/mac114… · MAC 1147 Exam #5 Review . Instructions: Exam #5 will consist of 10 questions

(3) Establish the identity (10 points)

1 - sinx cos x =-shy

cos x 1 + sinx

( l ~ f)c) (l+il gtc)

Co~ x bull (l+-SlhX)

-

vr~ nVftUAW t ~~ 7 I+trgtlt

--

(4) Find the exact value of each expression (5 points each)

(i) tan 1950 = ~ ( ~(o -doOo)

- -~ 13So + ~ (00deg

1shy ~ I~S middot ~~Olgt

s3 - +- 3 (- ~ ~) ~ - ~+J3

- (- 1)( i) G+- 4)1 ~~

(ii) cos 311 8

- +

- +shy-

=+~- ~ - -~ -

J t-~ 2

(5) Find the exact value of tan(a - (3) under the given conditions (4 points) It vj r of

5 7r 15 7r cos ex = - - - lt ex lt 7r sm (3 = - - lt (3 lt 7r

13~ 17~_

(a) -io (b) i ~ (c) to) e ~l None of the ~f ~ ( above

QvJ1t ~f CWrJ II

If) ~ ~r -If~

~

kJ - ~r -J (-- 1pound)s - ~

(-l--~o~r

(-~~ Jf)(~b) -1L- lzu- Gt I~o )(10)

(6) Establish the identity (10 points)

sin (a - (3) cos (a + (3) = sin a cos a - sin (3 cos (3

----------------

(7) Find the exact value of each expression (4 points)

(i) cos (tan-1 ~ - sin-1 ~)

~T (a) 2~ (b) 1 (c) 2~ (e) None of the

above

rJ-t- CWJII ~

(ii) tan [2 cos- l (-~)J

(a) 2i (b) -2~ (c) - 2l (d) -~ ~None of the ryabove

2~ (tod-~)) _ El)(I~) _ (1- Z) ((lt0)

_ ~l (CraquoS-1 (~raquo) ~

(8) Establish the identity (10 points)

4cos (4x) = cos X - 6middotsm2 x cos2 x + sin4 x

(9) (i) Express the product as a sum containing only sines or cosines (4 points)

sin (30) cos (40)

GY j[sin (78) - sin 81 (b) sincos(1202)

(c) Hcos(70) -cosO] (d) ~[cos(7e)+sinel

(ii) Express the sum or difference as a product of sines andor cosines (4 points)

cos (60) + cos (40)

(a) 2cos(50)sinO (b) - 2 cos (50) cos 0

(c) 2 sin (50) sin 0 2 cos (50) cos e

(10) (i) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2cos2 0 - 1 = 0

rr5rr ~(b) E 3rr 5rr 7rr (c) E 7rr()a 33 ~4444 44

(d) rr 2rr 4rr 5rr ( e) None of the above3333

or

(ii) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2 cos (20) = v3 fr)( ) rr llrr 13rr 23rrvJ 12 12 12 12

(b) 3 2 rr (c) llrr

6 6

(d) ~ (e) None of the above

j (1 I

L 1T Il

(iii) Solve the equation on the interval 0 x lt 27f (6 points - 3 points for the answer and 3 points for the steps)

2 cos2 X + 3 sin x - 3 = 0

7r 7r 57r 77r ( ) (b) 77r 117r (c) 7r 57r a 6266 2 6 6 6 6

7r 7r 27r(d) ~None of the above323

JCb~l)( +- 3tin )c - 3 =--0 l-

I-~ttt X S( )lt -=--

~ (_Sl-)() -t3~ X - ~ -=- D X =-~ ()

~ _ ~~t +S~ X - 3 - 0 X -= Jt2x~ Sli ( )

-= lC I [X fO fa

(iv) Solve the equation Give a general formula for all the solutions (6 points shy3 points for the answer and 3 points for the steps)

esc (~) = 2~ ~11=7f+6n7f 11= 27f+6n7f (b) 11=~+2n7f 11=2+2n7f

(c) 11=~+2n7f B=~~+2n7f (d) B = ~ + 6n7f B = 5 + 6n7f

(e) None of the above

1) ~)~ ~ ~ In~(~ ~ (~~) =0f+ Jlf)()

e~ 1+ ~Vlf (1 nr- ~1

-----------------

Bonus Write the trigonometric expression as an algebraic expression containing u and v (10 points)

1 1b -- -4

b~W 5 ( ~ J15 1i

SivSlA - -4iIf) =~~J105 ( ~- lIT) - togt (141 S ( (-bI- )

- fu) (- I -([F4)( ~ ~ Jt1)L+ I ) r~ t )

-

Page 5: MAC 1147 Exam #5 Review - Department of Mathematicsdscheib/teaching/summer2011/mac1147_files/mac114… · MAC 1147 Exam #5 Review . Instructions: Exam #5 will consist of 10 questions

--

(4) Find the exact value of each expression (5 points each)

(i) tan 1950 = ~ ( ~(o -doOo)

- -~ 13So + ~ (00deg

1shy ~ I~S middot ~~Olgt

s3 - +- 3 (- ~ ~) ~ - ~+J3

- (- 1)( i) G+- 4)1 ~~

(ii) cos 311 8

- +

- +shy-

=+~- ~ - -~ -

J t-~ 2

(5) Find the exact value of tan(a - (3) under the given conditions (4 points) It vj r of

5 7r 15 7r cos ex = - - - lt ex lt 7r sm (3 = - - lt (3 lt 7r

13~ 17~_

(a) -io (b) i ~ (c) to) e ~l None of the ~f ~ ( above

QvJ1t ~f CWrJ II

If) ~ ~r -If~

~

kJ - ~r -J (-- 1pound)s - ~

(-l--~o~r

(-~~ Jf)(~b) -1L- lzu- Gt I~o )(10)

(6) Establish the identity (10 points)

sin (a - (3) cos (a + (3) = sin a cos a - sin (3 cos (3

----------------

(7) Find the exact value of each expression (4 points)

(i) cos (tan-1 ~ - sin-1 ~)

~T (a) 2~ (b) 1 (c) 2~ (e) None of the

above

rJ-t- CWJII ~

(ii) tan [2 cos- l (-~)J

(a) 2i (b) -2~ (c) - 2l (d) -~ ~None of the ryabove

2~ (tod-~)) _ El)(I~) _ (1- Z) ((lt0)

_ ~l (CraquoS-1 (~raquo) ~

(8) Establish the identity (10 points)

4cos (4x) = cos X - 6middotsm2 x cos2 x + sin4 x

(9) (i) Express the product as a sum containing only sines or cosines (4 points)

sin (30) cos (40)

GY j[sin (78) - sin 81 (b) sincos(1202)

(c) Hcos(70) -cosO] (d) ~[cos(7e)+sinel

(ii) Express the sum or difference as a product of sines andor cosines (4 points)

cos (60) + cos (40)

(a) 2cos(50)sinO (b) - 2 cos (50) cos 0

(c) 2 sin (50) sin 0 2 cos (50) cos e

(10) (i) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2cos2 0 - 1 = 0

rr5rr ~(b) E 3rr 5rr 7rr (c) E 7rr()a 33 ~4444 44

(d) rr 2rr 4rr 5rr ( e) None of the above3333

or

(ii) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2 cos (20) = v3 fr)( ) rr llrr 13rr 23rrvJ 12 12 12 12

(b) 3 2 rr (c) llrr

6 6

(d) ~ (e) None of the above

j (1 I

L 1T Il

(iii) Solve the equation on the interval 0 x lt 27f (6 points - 3 points for the answer and 3 points for the steps)

2 cos2 X + 3 sin x - 3 = 0

7r 7r 57r 77r ( ) (b) 77r 117r (c) 7r 57r a 6266 2 6 6 6 6

7r 7r 27r(d) ~None of the above323

JCb~l)( +- 3tin )c - 3 =--0 l-

I-~ttt X S( )lt -=--

~ (_Sl-)() -t3~ X - ~ -=- D X =-~ ()

~ _ ~~t +S~ X - 3 - 0 X -= Jt2x~ Sli ( )

-= lC I [X fO fa

(iv) Solve the equation Give a general formula for all the solutions (6 points shy3 points for the answer and 3 points for the steps)

esc (~) = 2~ ~11=7f+6n7f 11= 27f+6n7f (b) 11=~+2n7f 11=2+2n7f

(c) 11=~+2n7f B=~~+2n7f (d) B = ~ + 6n7f B = 5 + 6n7f

(e) None of the above

1) ~)~ ~ ~ In~(~ ~ (~~) =0f+ Jlf)()

e~ 1+ ~Vlf (1 nr- ~1

-----------------

Bonus Write the trigonometric expression as an algebraic expression containing u and v (10 points)

1 1b -- -4

b~W 5 ( ~ J15 1i

SivSlA - -4iIf) =~~J105 ( ~- lIT) - togt (141 S ( (-bI- )

- fu) (- I -([F4)( ~ ~ Jt1)L+ I ) r~ t )

-

Page 6: MAC 1147 Exam #5 Review - Department of Mathematicsdscheib/teaching/summer2011/mac1147_files/mac114… · MAC 1147 Exam #5 Review . Instructions: Exam #5 will consist of 10 questions

(5) Find the exact value of tan(a - (3) under the given conditions (4 points) It vj r of

5 7r 15 7r cos ex = - - - lt ex lt 7r sm (3 = - - lt (3 lt 7r

13~ 17~_

(a) -io (b) i ~ (c) to) e ~l None of the ~f ~ ( above

QvJ1t ~f CWrJ II

If) ~ ~r -If~

~

kJ - ~r -J (-- 1pound)s - ~

(-l--~o~r

(-~~ Jf)(~b) -1L- lzu- Gt I~o )(10)

(6) Establish the identity (10 points)

sin (a - (3) cos (a + (3) = sin a cos a - sin (3 cos (3

----------------

(7) Find the exact value of each expression (4 points)

(i) cos (tan-1 ~ - sin-1 ~)

~T (a) 2~ (b) 1 (c) 2~ (e) None of the

above

rJ-t- CWJII ~

(ii) tan [2 cos- l (-~)J

(a) 2i (b) -2~ (c) - 2l (d) -~ ~None of the ryabove

2~ (tod-~)) _ El)(I~) _ (1- Z) ((lt0)

_ ~l (CraquoS-1 (~raquo) ~

(8) Establish the identity (10 points)

4cos (4x) = cos X - 6middotsm2 x cos2 x + sin4 x

(9) (i) Express the product as a sum containing only sines or cosines (4 points)

sin (30) cos (40)

GY j[sin (78) - sin 81 (b) sincos(1202)

(c) Hcos(70) -cosO] (d) ~[cos(7e)+sinel

(ii) Express the sum or difference as a product of sines andor cosines (4 points)

cos (60) + cos (40)

(a) 2cos(50)sinO (b) - 2 cos (50) cos 0

(c) 2 sin (50) sin 0 2 cos (50) cos e

(10) (i) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2cos2 0 - 1 = 0

rr5rr ~(b) E 3rr 5rr 7rr (c) E 7rr()a 33 ~4444 44

(d) rr 2rr 4rr 5rr ( e) None of the above3333

or

(ii) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2 cos (20) = v3 fr)( ) rr llrr 13rr 23rrvJ 12 12 12 12

(b) 3 2 rr (c) llrr

6 6

(d) ~ (e) None of the above

j (1 I

L 1T Il

(iii) Solve the equation on the interval 0 x lt 27f (6 points - 3 points for the answer and 3 points for the steps)

2 cos2 X + 3 sin x - 3 = 0

7r 7r 57r 77r ( ) (b) 77r 117r (c) 7r 57r a 6266 2 6 6 6 6

7r 7r 27r(d) ~None of the above323

JCb~l)( +- 3tin )c - 3 =--0 l-

I-~ttt X S( )lt -=--

~ (_Sl-)() -t3~ X - ~ -=- D X =-~ ()

~ _ ~~t +S~ X - 3 - 0 X -= Jt2x~ Sli ( )

-= lC I [X fO fa

(iv) Solve the equation Give a general formula for all the solutions (6 points shy3 points for the answer and 3 points for the steps)

esc (~) = 2~ ~11=7f+6n7f 11= 27f+6n7f (b) 11=~+2n7f 11=2+2n7f

(c) 11=~+2n7f B=~~+2n7f (d) B = ~ + 6n7f B = 5 + 6n7f

(e) None of the above

1) ~)~ ~ ~ In~(~ ~ (~~) =0f+ Jlf)()

e~ 1+ ~Vlf (1 nr- ~1

-----------------

Bonus Write the trigonometric expression as an algebraic expression containing u and v (10 points)

1 1b -- -4

b~W 5 ( ~ J15 1i

SivSlA - -4iIf) =~~J105 ( ~- lIT) - togt (141 S ( (-bI- )

- fu) (- I -([F4)( ~ ~ Jt1)L+ I ) r~ t )

-

Page 7: MAC 1147 Exam #5 Review - Department of Mathematicsdscheib/teaching/summer2011/mac1147_files/mac114… · MAC 1147 Exam #5 Review . Instructions: Exam #5 will consist of 10 questions

(6) Establish the identity (10 points)

sin (a - (3) cos (a + (3) = sin a cos a - sin (3 cos (3

----------------

(7) Find the exact value of each expression (4 points)

(i) cos (tan-1 ~ - sin-1 ~)

~T (a) 2~ (b) 1 (c) 2~ (e) None of the

above

rJ-t- CWJII ~

(ii) tan [2 cos- l (-~)J

(a) 2i (b) -2~ (c) - 2l (d) -~ ~None of the ryabove

2~ (tod-~)) _ El)(I~) _ (1- Z) ((lt0)

_ ~l (CraquoS-1 (~raquo) ~

(8) Establish the identity (10 points)

4cos (4x) = cos X - 6middotsm2 x cos2 x + sin4 x

(9) (i) Express the product as a sum containing only sines or cosines (4 points)

sin (30) cos (40)

GY j[sin (78) - sin 81 (b) sincos(1202)

(c) Hcos(70) -cosO] (d) ~[cos(7e)+sinel

(ii) Express the sum or difference as a product of sines andor cosines (4 points)

cos (60) + cos (40)

(a) 2cos(50)sinO (b) - 2 cos (50) cos 0

(c) 2 sin (50) sin 0 2 cos (50) cos e

(10) (i) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2cos2 0 - 1 = 0

rr5rr ~(b) E 3rr 5rr 7rr (c) E 7rr()a 33 ~4444 44

(d) rr 2rr 4rr 5rr ( e) None of the above3333

or

(ii) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2 cos (20) = v3 fr)( ) rr llrr 13rr 23rrvJ 12 12 12 12

(b) 3 2 rr (c) llrr

6 6

(d) ~ (e) None of the above

j (1 I

L 1T Il

(iii) Solve the equation on the interval 0 x lt 27f (6 points - 3 points for the answer and 3 points for the steps)

2 cos2 X + 3 sin x - 3 = 0

7r 7r 57r 77r ( ) (b) 77r 117r (c) 7r 57r a 6266 2 6 6 6 6

7r 7r 27r(d) ~None of the above323

JCb~l)( +- 3tin )c - 3 =--0 l-

I-~ttt X S( )lt -=--

~ (_Sl-)() -t3~ X - ~ -=- D X =-~ ()

~ _ ~~t +S~ X - 3 - 0 X -= Jt2x~ Sli ( )

-= lC I [X fO fa

(iv) Solve the equation Give a general formula for all the solutions (6 points shy3 points for the answer and 3 points for the steps)

esc (~) = 2~ ~11=7f+6n7f 11= 27f+6n7f (b) 11=~+2n7f 11=2+2n7f

(c) 11=~+2n7f B=~~+2n7f (d) B = ~ + 6n7f B = 5 + 6n7f

(e) None of the above

1) ~)~ ~ ~ In~(~ ~ (~~) =0f+ Jlf)()

e~ 1+ ~Vlf (1 nr- ~1

-----------------

Bonus Write the trigonometric expression as an algebraic expression containing u and v (10 points)

1 1b -- -4

b~W 5 ( ~ J15 1i

SivSlA - -4iIf) =~~J105 ( ~- lIT) - togt (141 S ( (-bI- )

- fu) (- I -([F4)( ~ ~ Jt1)L+ I ) r~ t )

-

Page 8: MAC 1147 Exam #5 Review - Department of Mathematicsdscheib/teaching/summer2011/mac1147_files/mac114… · MAC 1147 Exam #5 Review . Instructions: Exam #5 will consist of 10 questions

----------------

(7) Find the exact value of each expression (4 points)

(i) cos (tan-1 ~ - sin-1 ~)

~T (a) 2~ (b) 1 (c) 2~ (e) None of the

above

rJ-t- CWJII ~

(ii) tan [2 cos- l (-~)J

(a) 2i (b) -2~ (c) - 2l (d) -~ ~None of the ryabove

2~ (tod-~)) _ El)(I~) _ (1- Z) ((lt0)

_ ~l (CraquoS-1 (~raquo) ~

(8) Establish the identity (10 points)

4cos (4x) = cos X - 6middotsm2 x cos2 x + sin4 x

(9) (i) Express the product as a sum containing only sines or cosines (4 points)

sin (30) cos (40)

GY j[sin (78) - sin 81 (b) sincos(1202)

(c) Hcos(70) -cosO] (d) ~[cos(7e)+sinel

(ii) Express the sum or difference as a product of sines andor cosines (4 points)

cos (60) + cos (40)

(a) 2cos(50)sinO (b) - 2 cos (50) cos 0

(c) 2 sin (50) sin 0 2 cos (50) cos e

(10) (i) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2cos2 0 - 1 = 0

rr5rr ~(b) E 3rr 5rr 7rr (c) E 7rr()a 33 ~4444 44

(d) rr 2rr 4rr 5rr ( e) None of the above3333

or

(ii) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2 cos (20) = v3 fr)( ) rr llrr 13rr 23rrvJ 12 12 12 12

(b) 3 2 rr (c) llrr

6 6

(d) ~ (e) None of the above

j (1 I

L 1T Il

(iii) Solve the equation on the interval 0 x lt 27f (6 points - 3 points for the answer and 3 points for the steps)

2 cos2 X + 3 sin x - 3 = 0

7r 7r 57r 77r ( ) (b) 77r 117r (c) 7r 57r a 6266 2 6 6 6 6

7r 7r 27r(d) ~None of the above323

JCb~l)( +- 3tin )c - 3 =--0 l-

I-~ttt X S( )lt -=--

~ (_Sl-)() -t3~ X - ~ -=- D X =-~ ()

~ _ ~~t +S~ X - 3 - 0 X -= Jt2x~ Sli ( )

-= lC I [X fO fa

(iv) Solve the equation Give a general formula for all the solutions (6 points shy3 points for the answer and 3 points for the steps)

esc (~) = 2~ ~11=7f+6n7f 11= 27f+6n7f (b) 11=~+2n7f 11=2+2n7f

(c) 11=~+2n7f B=~~+2n7f (d) B = ~ + 6n7f B = 5 + 6n7f

(e) None of the above

1) ~)~ ~ ~ In~(~ ~ (~~) =0f+ Jlf)()

e~ 1+ ~Vlf (1 nr- ~1

-----------------

Bonus Write the trigonometric expression as an algebraic expression containing u and v (10 points)

1 1b -- -4

b~W 5 ( ~ J15 1i

SivSlA - -4iIf) =~~J105 ( ~- lIT) - togt (141 S ( (-bI- )

- fu) (- I -([F4)( ~ ~ Jt1)L+ I ) r~ t )

-

Page 9: MAC 1147 Exam #5 Review - Department of Mathematicsdscheib/teaching/summer2011/mac1147_files/mac114… · MAC 1147 Exam #5 Review . Instructions: Exam #5 will consist of 10 questions

(8) Establish the identity (10 points)

4cos (4x) = cos X - 6middotsm2 x cos2 x + sin4 x

(9) (i) Express the product as a sum containing only sines or cosines (4 points)

sin (30) cos (40)

GY j[sin (78) - sin 81 (b) sincos(1202)

(c) Hcos(70) -cosO] (d) ~[cos(7e)+sinel

(ii) Express the sum or difference as a product of sines andor cosines (4 points)

cos (60) + cos (40)

(a) 2cos(50)sinO (b) - 2 cos (50) cos 0

(c) 2 sin (50) sin 0 2 cos (50) cos e

(10) (i) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2cos2 0 - 1 = 0

rr5rr ~(b) E 3rr 5rr 7rr (c) E 7rr()a 33 ~4444 44

(d) rr 2rr 4rr 5rr ( e) None of the above3333

or

(ii) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2 cos (20) = v3 fr)( ) rr llrr 13rr 23rrvJ 12 12 12 12

(b) 3 2 rr (c) llrr

6 6

(d) ~ (e) None of the above

j (1 I

L 1T Il

(iii) Solve the equation on the interval 0 x lt 27f (6 points - 3 points for the answer and 3 points for the steps)

2 cos2 X + 3 sin x - 3 = 0

7r 7r 57r 77r ( ) (b) 77r 117r (c) 7r 57r a 6266 2 6 6 6 6

7r 7r 27r(d) ~None of the above323

JCb~l)( +- 3tin )c - 3 =--0 l-

I-~ttt X S( )lt -=--

~ (_Sl-)() -t3~ X - ~ -=- D X =-~ ()

~ _ ~~t +S~ X - 3 - 0 X -= Jt2x~ Sli ( )

-= lC I [X fO fa

(iv) Solve the equation Give a general formula for all the solutions (6 points shy3 points for the answer and 3 points for the steps)

esc (~) = 2~ ~11=7f+6n7f 11= 27f+6n7f (b) 11=~+2n7f 11=2+2n7f

(c) 11=~+2n7f B=~~+2n7f (d) B = ~ + 6n7f B = 5 + 6n7f

(e) None of the above

1) ~)~ ~ ~ In~(~ ~ (~~) =0f+ Jlf)()

e~ 1+ ~Vlf (1 nr- ~1

-----------------

Bonus Write the trigonometric expression as an algebraic expression containing u and v (10 points)

1 1b -- -4

b~W 5 ( ~ J15 1i

SivSlA - -4iIf) =~~J105 ( ~- lIT) - togt (141 S ( (-bI- )

- fu) (- I -([F4)( ~ ~ Jt1)L+ I ) r~ t )

-

Page 10: MAC 1147 Exam #5 Review - Department of Mathematicsdscheib/teaching/summer2011/mac1147_files/mac114… · MAC 1147 Exam #5 Review . Instructions: Exam #5 will consist of 10 questions

(9) (i) Express the product as a sum containing only sines or cosines (4 points)

sin (30) cos (40)

GY j[sin (78) - sin 81 (b) sincos(1202)

(c) Hcos(70) -cosO] (d) ~[cos(7e)+sinel

(ii) Express the sum or difference as a product of sines andor cosines (4 points)

cos (60) + cos (40)

(a) 2cos(50)sinO (b) - 2 cos (50) cos 0

(c) 2 sin (50) sin 0 2 cos (50) cos e

(10) (i) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2cos2 0 - 1 = 0

rr5rr ~(b) E 3rr 5rr 7rr (c) E 7rr()a 33 ~4444 44

(d) rr 2rr 4rr 5rr ( e) None of the above3333

or

(ii) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2 cos (20) = v3 fr)( ) rr llrr 13rr 23rrvJ 12 12 12 12

(b) 3 2 rr (c) llrr

6 6

(d) ~ (e) None of the above

j (1 I

L 1T Il

(iii) Solve the equation on the interval 0 x lt 27f (6 points - 3 points for the answer and 3 points for the steps)

2 cos2 X + 3 sin x - 3 = 0

7r 7r 57r 77r ( ) (b) 77r 117r (c) 7r 57r a 6266 2 6 6 6 6

7r 7r 27r(d) ~None of the above323

JCb~l)( +- 3tin )c - 3 =--0 l-

I-~ttt X S( )lt -=--

~ (_Sl-)() -t3~ X - ~ -=- D X =-~ ()

~ _ ~~t +S~ X - 3 - 0 X -= Jt2x~ Sli ( )

-= lC I [X fO fa

(iv) Solve the equation Give a general formula for all the solutions (6 points shy3 points for the answer and 3 points for the steps)

esc (~) = 2~ ~11=7f+6n7f 11= 27f+6n7f (b) 11=~+2n7f 11=2+2n7f

(c) 11=~+2n7f B=~~+2n7f (d) B = ~ + 6n7f B = 5 + 6n7f

(e) None of the above

1) ~)~ ~ ~ In~(~ ~ (~~) =0f+ Jlf)()

e~ 1+ ~Vlf (1 nr- ~1

-----------------

Bonus Write the trigonometric expression as an algebraic expression containing u and v (10 points)

1 1b -- -4

b~W 5 ( ~ J15 1i

SivSlA - -4iIf) =~~J105 ( ~- lIT) - togt (141 S ( (-bI- )

- fu) (- I -([F4)( ~ ~ Jt1)L+ I ) r~ t )

-

Page 11: MAC 1147 Exam #5 Review - Department of Mathematicsdscheib/teaching/summer2011/mac1147_files/mac114… · MAC 1147 Exam #5 Review . Instructions: Exam #5 will consist of 10 questions

(10) (i) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2cos2 0 - 1 = 0

rr5rr ~(b) E 3rr 5rr 7rr (c) E 7rr()a 33 ~4444 44

(d) rr 2rr 4rr 5rr ( e) None of the above3333

or

(ii) Solve the equation on the interval 0 0 lt 21f (6 points - 3 points for the answer and 3 points for the steps)

2 cos (20) = v3 fr)( ) rr llrr 13rr 23rrvJ 12 12 12 12

(b) 3 2 rr (c) llrr

6 6

(d) ~ (e) None of the above

j (1 I

L 1T Il

(iii) Solve the equation on the interval 0 x lt 27f (6 points - 3 points for the answer and 3 points for the steps)

2 cos2 X + 3 sin x - 3 = 0

7r 7r 57r 77r ( ) (b) 77r 117r (c) 7r 57r a 6266 2 6 6 6 6

7r 7r 27r(d) ~None of the above323

JCb~l)( +- 3tin )c - 3 =--0 l-

I-~ttt X S( )lt -=--

~ (_Sl-)() -t3~ X - ~ -=- D X =-~ ()

~ _ ~~t +S~ X - 3 - 0 X -= Jt2x~ Sli ( )

-= lC I [X fO fa

(iv) Solve the equation Give a general formula for all the solutions (6 points shy3 points for the answer and 3 points for the steps)

esc (~) = 2~ ~11=7f+6n7f 11= 27f+6n7f (b) 11=~+2n7f 11=2+2n7f

(c) 11=~+2n7f B=~~+2n7f (d) B = ~ + 6n7f B = 5 + 6n7f

(e) None of the above

1) ~)~ ~ ~ In~(~ ~ (~~) =0f+ Jlf)()

e~ 1+ ~Vlf (1 nr- ~1

-----------------

Bonus Write the trigonometric expression as an algebraic expression containing u and v (10 points)

1 1b -- -4

b~W 5 ( ~ J15 1i

SivSlA - -4iIf) =~~J105 ( ~- lIT) - togt (141 S ( (-bI- )

- fu) (- I -([F4)( ~ ~ Jt1)L+ I ) r~ t )

-

Page 12: MAC 1147 Exam #5 Review - Department of Mathematicsdscheib/teaching/summer2011/mac1147_files/mac114… · MAC 1147 Exam #5 Review . Instructions: Exam #5 will consist of 10 questions

(iii) Solve the equation on the interval 0 x lt 27f (6 points - 3 points for the answer and 3 points for the steps)

2 cos2 X + 3 sin x - 3 = 0

7r 7r 57r 77r ( ) (b) 77r 117r (c) 7r 57r a 6266 2 6 6 6 6

7r 7r 27r(d) ~None of the above323

JCb~l)( +- 3tin )c - 3 =--0 l-

I-~ttt X S( )lt -=--

~ (_Sl-)() -t3~ X - ~ -=- D X =-~ ()

~ _ ~~t +S~ X - 3 - 0 X -= Jt2x~ Sli ( )

-= lC I [X fO fa

(iv) Solve the equation Give a general formula for all the solutions (6 points shy3 points for the answer and 3 points for the steps)

esc (~) = 2~ ~11=7f+6n7f 11= 27f+6n7f (b) 11=~+2n7f 11=2+2n7f

(c) 11=~+2n7f B=~~+2n7f (d) B = ~ + 6n7f B = 5 + 6n7f

(e) None of the above

1) ~)~ ~ ~ In~(~ ~ (~~) =0f+ Jlf)()

e~ 1+ ~Vlf (1 nr- ~1

-----------------

Bonus Write the trigonometric expression as an algebraic expression containing u and v (10 points)

1 1b -- -4

b~W 5 ( ~ J15 1i

SivSlA - -4iIf) =~~J105 ( ~- lIT) - togt (141 S ( (-bI- )

- fu) (- I -([F4)( ~ ~ Jt1)L+ I ) r~ t )

-

Page 13: MAC 1147 Exam #5 Review - Department of Mathematicsdscheib/teaching/summer2011/mac1147_files/mac114… · MAC 1147 Exam #5 Review . Instructions: Exam #5 will consist of 10 questions

-----------------

Bonus Write the trigonometric expression as an algebraic expression containing u and v (10 points)

1 1b -- -4

b~W 5 ( ~ J15 1i

SivSlA - -4iIf) =~~J105 ( ~- lIT) - togt (141 S ( (-bI- )

- fu) (- I -([F4)( ~ ~ Jt1)L+ I ) r~ t )

-