math analysis chapter 5 jeopardy

Jeopardy Chapter 5

LCCHS

Spring 2013

Author: R. Hunsperger

Edited By: J. Teague

Math Analysis

Trig

Game 1

RULES: •  Work only with those in your group! Support your group members!

•  As a group, keep score of your points for accuracy; we will keep score as well.

•  This is a review game! We will work out solutions as requested after each question is completed.

•  Write your group’s final answer on your whiteboard and let us know when your group is done with each question.

•  Alternate who writes on the whiteboard each question.

•  Do not use your notes during any part of Jeopardy; this is a game that tests your current knowledge and readiness!

•  Identity sheets are allowed.

•  IMPORTANT: Everyone must write their own solution to EVERY question, then collaborate as a group. We will be collecting each person’s scratch work

at the end and it will be worth 10 points.

•  During Final Jeopardy, you may only wager up to your current point value.

Game Board

300

Exact Values Triangles Verifying

Identities

Half and Double Angles

Equations

400

500

200

100

300

400

500

200

100

300

400

500

200

100

300

400

500

200

100

300

400

500

200

100

Final Jeopardy

Exact Values, 100

Answer:

Find the exact value of:

cos65°cos5°+ sin65°sin5°

cos(65°− 5°) = cos60° = 12

Exact Values, 200

Answer:

sin80°cos50°− cos80°sin50°Find the exact value of:

sin(80°− 50°) = sin30° = 12

Exact Values, 300

Answer:

sin105°Find the exact value of:

sin105°= sin(135°−30°)= sin135°cos30°− cos135°sin30°

=22×32

#

$%

&

'(− −

22×12

#

$%

&

'(

=6 + 24

Using the fact that

105° =135°−30°

Exact Values, 400

Answer:

sin2αUsing the fact that

sinα = 45,α


lies in quadrant II.

2524

53

542

cossin22sin

−=

⎟⎠

⎞⎜⎝

⎛ −⎟⎠

⎞⎜⎝

⎛=

= αα

α

-3

5 4

α

Exact Values, 500

Answer:

tan 5π12


tan 5π12

= tan 3π12

+2π12

!

"#

$

%&=

tan π4+ tan π

61− tan π

4tan π

6

=1+ 3

3

1− 1⋅ 33

!

"#

$

%&

=1+ 3

3

1− 33

=1+ 3

3

1− 33

⋅1+ 3

3

1+ 33

=1+ 2 3

3+13

1− 13

=

33+2 33

+13

23

=4+ 2 33

⋅32=4+ 2 32

= 2+ 3

Triangles, 100

Find

Answer:

θcos given that 53sin =θ

where θ lies in quadrant II.

3 5

-4 θ

54cos −==

hypotenuseadjacent

θ

Triangles, 200

Answer:

Find cos α −β( ) given that

tanα = 43

,α in quadrant III, tanβ = 512

,β in quadrant I.

α -4

12

5 β

13

-3

5

cos α −β( ) = cosα cosβ + sinα sinβ

= −35

"

#$

%

&'1213"

#$

%

&'+ −

45

"

#$

%

&'513"

#$

%

&'

= −3665

"

#$

%

&'+ −

2065

"

#$

%

&'

= −5665

Triangles, 300

Answer:

Find tan(α +β) given that

sinα = −13

,α in quadrant III, cosβ = −13

,β in quadrant III.

-1 -√8 3 β

-1 -√8

3

tan α +β( ) = tanα + tanβ1− tanα tanβ

=

18

"

#$

%

&'+

81

"

#$

%

&'

1− 18

"

#$

%

&'⋅

81

"

#$

%

&'

=

18

"

#$

%

&'+

81

"

#$

%

&'

1−1

=

18

"

#$

%

&'+

81

"

#$

%

&'

0undefined

α

Triangles, 400

Answer:

Find sin2α and cos2α given that

tanα = −3,α in quadrant II.

-1

√10

α 3

sin2α = 2sinα cosα = 2 310

!

"#

$

%&⋅ −

110

!

"#

$

%&= −

610

= −35

cos2α =1− 2sin2α =1− 2 ⋅ 310

#

$%

&

'(2

=1− 2 ⋅ 910

=1− 1810

= −810

= −45

Triangles, 500

Answer:

Find cos β

2 given that

cotβ = −3,β in quadrant IV.

-1 √10

β 3 cos β

2= −

1+ cosβ2

= −1+ 3

102

= −1+ 3 10

102

−

1010

+3 10102

= −10+3 10

20= −

10+3 102 5

= −10+3 102 5

⋅55= −

50+15 1010

Verifying Identities, 100

Answer: b

Verify the identity and choose what the R.H.S. might look like:

?tansincos =+ xxx

xdxcxbxa

sin . cot .sec . csc .


Answer: d


xxdc

xbxxa

cossin1 . x tan1 .

2sec . cos1

tan .

2 ++

++

?sin1cos

=− xx


Answer: a


θθθθ

θθθθ2222

2222

cotec . cossc .cosin . cscsec .−+

−+

sdccsba

?)cot(tan 2 =+ θθ


Answer: d


θθ

θθ

tan2 . 1cot .2cos . 1tan2 .

2

2

dcba

+

+

?sin12sin2 =

− θθ


Answer: c

?)cos1(2

tan =+ xx


xdxcxbxa

2

2

sin . sin .cos . cos .

Half and Double Angles, 100

Find the exact value of

Answer:

°−° 15sin15cos 22

23

30cos)15(2cos15sin15cos 22

=

°=

°=

°−°


Answer:


125tan1125tan2

2 π

π

−

33

65tan

1252tan

125tan1125tan2

2

−=

=⎟⎠

⎞⎜⎝

⎛=−

πππ

π


Answer:

Find the exact value of °5.22sin

222

422

2222

2221

245cos1

245sin5.22sin

−=

−=

−

=−

=

°−+=

°=°



12tan π

322312

21231

6sin

6cos1

26tan

12tan −=⎟

⎟⎠

⎞⎜⎜⎝

⎛−=

−=

−=

⎟⎠

⎞⎜⎝

⎛

=π

πππ

Answer:


Answer: c


?2sin2cos1

=+

xx

xdxcxbxa

sin . cot .sec . csc .

Equations, 100

Find all solutions:

sin x = 22

sin x = 22

⇒ x = π4+ 2nπ

and

⇒ x = 3π4+ 2nπ

Answer:

Equations, 200

Answer:

Find all solutions:

3 tan x −1= 0

ππ nx

x

x

x

+=⇒

==⇒

=⇒

=−

6

33

31tan

1tan3

01tan3

Equations, 300

Answer: tan x = 2cos x tan x

0cos21

0tan0)cos21(tan

0tancos2tantancos2tan

=−⇒

=⇒

=−⇒

=−⇒=

andx

xxxxxxxx

Solve on the interval [0,2π):

π,00tan

=⇒

=

xx

35,

3

21cos

0cos21

ππ=⇒

=⇒

=−

x

x

x

Equations, 400

Answer:

Solve on the interval [0,2π): 4sin2 x =1

4sin2 x =1

⇒ sin2 x = 14

⇒ sin x = ± 12

65,

6

21sin

ππ=⇒

=

x

x

611,

6721sin

ππ=⇒

−=

x

x

Equations, 500

Answer:

Solve on the interval [0,2π):

cos2x = 22

cos2x = 22

⇒ 2x = π4+ 2nπ and 2x = 7π

4+ 2nπ

2x = π4+ 2nπ ⇒ x = π

8+ nπ

n = 0 : x = π8

, n =1: x = 9π8

2x = 7π4+ 2nπ ⇒ x = 7π

8+ nπ

n = 0 : x = 7π8

, n =1: x = 15π8

Final Jeopardy

Final Jeopardy

Category:

Solving Trigonometric Equations

Please wager now!

Final Jeopardy

Solve the equation on the interval [0,2π):

2sin3 x − sin2 x − 2sin x +1= 02sin3 x − sin2 x − 2sin x +1= 0⇒ sin2 x(2sin x −1)− (2sin x −1) = 0⇒ (2sin x −1)(sin2 x −1) = 0

Answer:

2sin x −1= 0⇒ sin x = 12

⇒ x = π6, 5π6

sin2 x −1= 0⇒ sin2 x =1⇒ sin x = ±1sin x = +1

⇒ x = π2

sin x = −1

⇒ x = 3π2

math analysis chapter 5 jeopardy

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