Math Analysis Chapter 5 Jeopardy
Transcript of Math Analysis Chapter 5 Jeopardy
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Jeopardy Chapter 5
LCCHS
Spring 2013
Author: R. Hunsperger
Edited By: J. Teague
Math Analysis
Trig
Game 1
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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.
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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
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Exact Values, 100
Answer:
Find the exact value of:
cos65°cos5°+ sin65°sin5°
cos(65°− 5°) = cos60° = 12
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Exact Values, 200
Answer:
sin80°cos50°− cos80°sin50°Find the exact value of:
sin(80°− 50°) = sin30° = 12
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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°
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Exact Values, 400
Answer:
sin2αUsing the fact that
sinα = 45,α
Find the exact value of:
lies in quadrant II.
2524
53
542
cossin22sin
−=
⎟⎠
⎞⎜⎝
⎛ −⎟⎠
⎞⎜⎝
⎛=
= αα
α
-3
5 4
α
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Exact Values, 500
Answer:
tan 5π12
Find the exact value of:
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
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Triangles, 100
Find
Answer:
θcos given that 53sin =θ
where θ lies in quadrant II.
3 5
-4 θ
54cos −==
hypotenuseadjacent
θ
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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
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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
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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
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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
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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 .
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Verifying Identities, 200
Answer: d
Verify the identity and choose what the R.H.S. might look like:
xxdc
xbxxa
cossin1 . x tan1 .
2sec . cos1
tan .
2 ++
++
?sin1cos
=− xx
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Verifying Identities, 300
Answer: a
Verify the identity and choose what the R.H.S. might look like:
θθθθ
θθθθ2222
2222
cotec . cossc .cosin . cscsec .−+
−+
sdccsba
?)cot(tan 2 =+ θθ
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Verifying Identities, 400
Answer: d
Verify the identity and choose what the R.H.S. might look like:
θθ
θθ
tan2 . 1cot .2cos . 1tan2 .
2
2
dcba
+
+
?sin12sin2 =
− θθ
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Verifying Identities, 500
Answer: c
?)cos1(2
tan =+ xx
Verify the identity and choose what the R.H.S. might look like:
xdxcxbxa
2
2
sin . sin .cos . cos .
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Half and Double Angles, 100
Find the exact value of
Answer:
°−° 15sin15cos 22
23
30cos)15(2cos15sin15cos 22
=
°=
°=
°−°
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Half and Double Angles, 200
Answer:
Find the exact value of
125tan1125tan2
2 π
π
−
33
65tan
1252tan
125tan1125tan2
2
−=
=⎟⎠
⎞⎜⎝
⎛=−
πππ
π
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Half and Double Angles, 300
Answer:
Find the exact value of °5.22sin
222
422
2222
2221
245cos1
245sin5.22sin
−=
−=
−
=−
=
°−+=
°=°
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Half and Double Angles, 400
Find the exact value of
12tan π
322312
21231
6sin
6cos1
26tan
12tan −=⎟
⎟⎠
⎞⎜⎜⎝
⎛−=
−=
−=
⎟⎠
⎞⎜⎝
⎛
=π
πππ
Answer:
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Half and Double Angles, 500
Answer: c
Verify the identity and choose what the R.H.S. might look like:
?2sin2cos1
=+
xx
xdxcxbxa
sin . cot .sec . csc .
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Equations, 100
Find all solutions:
sin x = 22
sin x = 22
⇒ x = π4+ 2nπ
and
⇒ x = 3π4+ 2nπ
Answer:
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Equations, 200
Answer:
Find all solutions:
3 tan x −1= 0
ππ nx
x
x
x
+=⇒
==⇒
=⇒
=−
6
33
31tan
1tan3
01tan3
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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
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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
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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
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Final Jeopardy
Final Jeopardy
Category:
Solving Trigonometric Equations
Please wager now!
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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