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1.
SOLUTIONS Math 30-1 Trigonometry PracοΏ½ce Exam
2. First determine quadrant ππ terminates in. Since π π π π π π is negaοΏ½ve in Quad III and IV, and π‘π‘π‘π‘π π is neg. in II and IV, ππ is in quad IV.
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ππππΒ°
ππ So, ππ = 360Β° β 30Β°
= 330Β° or
11ππ6
So, consider co-terminal angles of 330Β° by adding / subtracting 360Β° (and converting to radians), or use 11ππ
6 and
add/subtract 2ππβ¦
(not one of the choices)
= 330Β° + 360Β° = 690Β°, converts to
ππππππππ
ANSWER: D
This is a reasoning quesοΏ½on, not a calculaοΏ½on quesοΏ½on! Using CAST rule, since πππππ π is negaοΏ½ve in Quad II and III, and π‘π‘π‘π‘π π is neg. in II and IV, ππ is in quad II.
Which means, ππ is between ππ2 and ππ. (1.57 rads and
3.14 as decimals) So, the only possible opοΏ½on there is 2.62 (Note, all answers are presumed to be in radians as no unit is specified!)
First determine quadrant ππ terminates in. Since π π π π π π is negaοΏ½ve in Quad III and IV, and π‘π‘π‘π‘π π is neg. in II and IV, ππ is in quad IV. π½π½
ππ(5,β4)
Next, plot a point ππ in quad IV and sketch angle ππ
π½π½
ππ(5,β4)
ππ
βππ β41
Then make a triangle by connecοΏ½ng ππ to the π₯π₯-axis. Label the sides of the triangle from the coordinates of ππ β¦ use pyth. theorem to get the hyp.
Now use π π π π ππππ = 1ππππππππ
where πππππ π ππ = ππππππβπ¦π¦π¦π¦
β¦.so π π π π ππππ = βπ¦π¦π¦π¦ππππππ
π π π π ππππ =β41
5
β ππ.ππππ
3.
NR #1 Since π π π π π π is negaοΏ½ve in Quad III and IV, and we want the smallest opοΏ½on for ππ draw an angle in st. pos. in quad III.
Given:
π½π½
β1 3
π π π π π π ππ =β13
First find reference angle, letβs call it πΌπΌ (inside the triangle)
πΆπΆ = π π π π π π β1 οΏ½13οΏ½
NOTE: When finding reference angles (by definiοΏ½on less than 90Β° so all trig raοΏ½os are posiοΏ½ve) drop any negaοΏ½ves
(radian mode)
π½π½ β ππ.ππ
ANSWER: B
ANSWER: A
ANSWER: 3.5
4. There are two opοΏ½ons for ππ(β 513
,ππ), it can be drawn in quadrant II or III. (as the π₯π₯-coord is negative) However, it is given that ππππππ is negaοΏ½ve, we know we should draw ππ in quad II. At this point we can pause to consider how fun it is to reason things out like that. (pause for 10 to 15 seconds) Diagram of information given
π½π½ ππ(β5
13,ππ) We can now solve for ππ by either
drawing a triangle and label the hypotenuse 1 (since β unit circle), or by going straight to the unit circle formula:
π₯π₯2 + π¦π¦2 = 1
(β5
13)2 + ππ2 = 1
ππ2 = 1 β25
169
ππ2 =
144169
ππ = Β±οΏ½144169
ππ = Β±
1213
Since quad II, use the + version!
ππ =ππππππππ
ANSWER: C
5. ππππππ (radius)
First find the length of : πππππ π 67Β° =
1.6π΄π΄π΄π΄
π΄π΄π΄π΄ =1.6
πππππ π 67Β°
π¨π¨π¨π¨ β ππ.ππππππ cm
Next find the length of :
(half of π΄π΄π΄π΄ is the opp side β¦) π‘π‘π‘π‘π π 67Β° =
ππππππ1.6
ππππππ = 1.6π‘π‘π‘π‘π π 67Β°
ππππππ β 3.769
ππππππ
π΄π΄π΄π΄ β 2 β 3.769
π¨π¨π¨π¨ β ππ.ππππππ
(radius)
Finally find the arc length: π‘π‘ = ππππ
π‘π‘ = 4.095 β(67Β° β 2)ππ
180Β°
ππππππ β ππ.ππππππ cm ππ in radians
So perimeter is:
2 β 15.84 + 7.54 + 9.58
Third rectangle side, π΄π΄π΄π΄
Rectangle top / bottom
Final βsideβ, the arc length
Perimeter β ππππ.ππ cm
ANSWER: C
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ππ =π‘π‘ππ
ππ =π‘π‘ππ
ππ β8.931.92
ππ β 4.65 So, leash+0.15 β ππ.ππππ m
110Β°ππ180Β°
convert ππ to radians:
re-arrange
π½π½ = ππππππΒ°
ππππππ = ππ.ππππ m
(leash length + 0.15m)
ANSWER: 4.5
ππ
NR #2
(π¦π¦)2 + (2π¦π¦)2 = (βπ¦π¦ππ)2
5π¦π¦2 = (βπ¦π¦ππ)2
βπ¦π¦ππ = β5π¦π¦
π π π π ππππ =β5π¦π¦β2π¦π¦
πππππππ½π½ = ββππππ
hyp
π‘π‘ = ππ ππ = ππ
ANSWER: D
6.
ANSWER: C
π½π½ =ππππππ
ππππππππππ
π·π· =ππππππ
π‘π‘π π πππππ π =5ππ4 β
2ππ3
=15ππ12 β
8ππ12
=ππππππππ
πππ π π‘π‘πππ π π π π π
Orβ¦first convert to degrees:
Converts to 7ππ12
= 95Β° Statement
1 is FALSE
ππ =2ππ3 Principal
angle
ππ =2ππ3 β ππππ
Is co-terminal
ππ = β4ππ3
Statement 2 is TRUE
π¦π¦-coord at 2ππ3
is π π π π π π ππ
π₯π₯-coord at 5ππ4
is πππππ π ππ
Statement 3 is TRUE
y ππππππ
ππππππ
=β32β1
2
x
=β32ββ21
= ββππ
x ππππππ
ππππππ
=ββ2
2
ββ22
y
= ππ
Statement 4 is TRUE
7.
Remember, dogβs extra reach is included here!
adj
Add: 5 + 2 = ππ
8.
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Amplitude, ππ, from ππ to ππ decreases
Graph shifts right, so ππ changes ANSWER: B
Statement 1 is TRUE
Basic sine curve (with no horizontal stretch or shift) has a MIN at π₯π₯ = βππ
2:
So we can see distance between ππ and ππ is greater than 2ππ
Since no horiz. stretch (ππ = 1), period is ππππ.
Statement 2 is TRUE
Median Line
ππ π π
ππ can be visualized as dist. from MAX (or MIN) to median line
π π as dist. from median line to π₯π₯-intercept
Statement 3 is FALSE
For π¦π¦-int, set π₯π₯ = 0:
π¦π¦ = π‘π‘ β π π π π π π (ππ) β ππ
But π π π π π π (0) = 0 , soβ¦
ππ = βπ π Statement 4 is FALSE
For ππ < 1, for exampleβ¦. π¦π¦ = π‘π‘ π π π π π π (2π₯π₯) β ππ
The horiz. str. would be the reciprocal, or here ππ.ππ
For horiz str < 1, all points move closer to the π₯π₯-axis β¦
Statement 5 is TRUE
NR #3
9. First factor ππ for horiz. phase shiοΏ½:
Period is 2ππππ
=ππππ
=2ππ4
Phase shiοΏ½ (to the left, but not relevant)
ANSWER: A
π¦π¦ = 5 π π π π π π [4 οΏ½π₯π₯ +ππ4οΏ½]
NR #4
= ππ.ππ
π¦π¦ = 2.3 π π π π π π (0.1208π₯π₯ β 0.3624) + 5.8
Period is 2ππππ
= ππππ =2ππ
0.1208
MAX is AMPL + Vert. ShiοΏ½
ANSWER: 5281
= π‘π‘ + ππ
= 2.3 + 5.8
10.
ANSWER: D
Domain of π¦π¦ = π‘π‘π‘π‘π π π₯π₯ is π₯π₯ β ππ2
+ π π ππ
That is, where πππππ π π₯π₯ = 0, since π‘π‘π‘π‘π π π₯π₯ = πππ π π π π π πππππππ π
β¦ at the top / botom of the unit circle
So for π¦π¦ = π‘π‘π‘π‘π π 4π₯π₯ Hor. Str. of
14
π₯π₯ β ππππβππ2
+ππππβ π π ππ
Domain is:
π₯π₯ β ππππ +
ππππππ
ANSWER: 125 Note: Mistake on some answer keys
NR #5
Period is ππ2
ππ =2ππππ2
ππ = 2ππ β2ππ
ππ = ππ
med. line
π π = ππ
ANSWER: 41
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12.
ππ
ππ
ππ.ππ
ππ
Median line
ππ =πππ‘π‘π₯π₯ βπππ π π π
2
Amplitude is 30 Note: you can also use formula
ππππππππππ Distance (from π¦π¦-axis) of point where curve upswing intersects median line.
ππππππππππ = ππ.ππππππ
For πππππ π π π π π determine π₯π₯-coords of max / minsβ¦
20 rotaοΏ½ons in 60 secβ¦.
1 rotaοΏ½on in 60 20
= ππ ππππππ (this is the period)
(1 min)
Halfway between ππ (where min is) and ππ.ππ ππ (where max is) ANSWER: A
13.
Two numbers must mult. to β2
Solve by factoring: 2πππππ π 2ππ + 3πππππ π ππ β 2 = 0
(2πππππ π ππ )(πππππ π ππ ) = 0
(2πππππ π ππ β 1)(πππππ π ππ + 2) = 0
ANSWER: A
Now set each factor to zero β¦ 2πππππ π ππ β 1 = 0
πππππ π ππ + 2 = 0
πππππ π ππ = 12
or πππππ π ππ = β2
15. Factor π π π π ππ2π₯π₯ β π π π π πππ₯π₯ β 2 = 0
(π π π π πππ₯π₯ + 1)(π π π π πππ₯π₯ β 2) = 0
Set each factor to zero:
Since ππππππππ is reciprocal of : πππππ π π₯π₯ = β1 or πππππ π π₯π₯ = 1
2
Find where on unit circle the ππ-coord is β1 ππππ 1/2
120Β°
ππππ
ππππππ
ππ
120Β°
120Β°
Note that all three soluοΏ½ons
are 120Β°, or 2ππ3
, apart
π½π½ =ππππ
+ππππππππ
; π π β πΌπΌ
first soluοΏ½on diff between sols
π π π π πππ₯π₯ = β1 or π π π π πππ₯π₯ = 2
14. trig term:
First isolate 3πππ π ππ2ππ β 4 = 0
3πππ π ππ2ππ = 4
Sq. root both sides
πππ π ππ2ππ =43
πππ π ππππ = Β±2β3
πππ π ππππ = Β±οΏ½43
Since οΏ½ππππ
= βππβππ
Since πππππππ½π½ is reciprocal of πππππππ½π½ π π π π π π ππ = Β±
β32
Find where on unit circle the ππ-coord is Β± β32
π½π½ =ππππ
,ππππππ
,ππππππ
,ππππππ
ANSWER: C
2β1 = π π π π π π π₯π₯
ππππππ2 οΏ½π π π π π π π₯π₯πππππ π π₯π₯ β πππππ π π₯π₯οΏ½ = β1
π π π π π π π₯π₯ =12
ππππππ2(π‘π‘π‘π‘π π π₯π₯) + ππππππ2(πππππ π π₯π₯) = β1 First isolate and use log laws on LS
ππππππ2(π‘π‘π‘π‘π π π₯π₯ β πππππ π π₯π₯) = β1 Use trig idenοΏ½οΏ½es to simplify
Convert to log form
ππππππ2(π π π π π π π₯π₯) = β1
Find where on unit circle the ππ-coord is 1
2
or ππ =ππππ
ππππππ
ππ
16.
ANSWER: B
ππ3
(12
,βππππ
)
5ππ3
(12
,ββππππ
)
5ππ3
(β12
,ββππππ
)
2ππ3
(β12
,βππππ
)
ANSWER: C
= π π π π π π ππ πππππ π ππ β πππππ π ππ π π π π π π ππ
Sketch the angle ππ and draw a triangle to determine trig raοΏ½os:
π½π½
ππ(3,β5)
ππ
βππ βππππ
By pyth. theorem
Now, for π π π π π π (ππ β ππ) β¦
= (0) ( 3β34
) β (β1) ( β5β34
)
From unit circle From diagram
πππππ π = ππππππβπ¦π¦π¦π¦
= β ππβππππ
ANSWER: D
17.
πππ π πππ π ππππ =2ππππ
11.
If π΄π΄ is at π₯π₯ = 0.00125β¦ then period is double, ππ.ππππππππ ππ
Use
formula to solve for ππ: 0.0025 π π =
2ππ2ππππ
1ππ = 0.0025 π π
ππ =1
0.0025 π π
ππ = 400 π»π»π»π»
ANSWER: D
NR #6 First re-write in degrees: πππππ π (105Β°) Next find any two standard unit circle angles that add (or subtract) to 105Β°
Such as: πππππ π (45Β° + 60Β°) Note: There are many opοΏ½ons, another is πππππ π (135Β° β 30Β°)
ππππ π π (πΌπΌ + π½π½) = πππππ π πΌπΌ πππππ π π½π½ β π π π π π π πΌπΌ π π π π π π π½π½
= πππππ π 45Β°πππππ π 60Β° β π π π π π π 45Β°π π π π π π 60Β° Finally refer to unit circle and simplify
= οΏ½β22οΏ½ οΏ½1
2οΏ½ β οΏ½β2
2οΏ½ οΏ½β3
2οΏ½
= β2
4β β6
4
=βππ β βππ
ππ
ANSWER: 264
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18. Where on the unit circle is the π₯π₯-coord 0?
πππππ π π₯π₯ β 0 π π π π π π π₯π₯πππππ π π₯π₯
1 + π π π π π π π₯π₯
π‘π‘π‘π‘π π π₯π₯1 + π π π π π π π₯π₯
Simplifies to β¦
πππππ π π₯π₯ β 0 So, NPVs β¦
1 + π π π π π π π₯π₯ β 0
1 + π π π π π π π₯π₯ β 0 π π π π π π π₯π₯ β β1 Where on the unit circle is the π¦π¦-coord equal to β1 ? (already covered above!)
ππππ
ππππππ
ANSWER: C
So, NPV at the top / botom of the unit circle
π₯π₯ β ππ2 then every ππ
Which we write as:
ππ β ππππ + ππππ
where π π β πΌπΌ
NR #7 Simplify: π‘π‘π‘π‘π π π₯π₯ +πππππ π π₯π₯
1 + π π π π π π π₯π₯
=π π π π π π π₯π₯πππππ π π₯π₯ +
πππππ π π₯π₯1 + π π π π π π π₯π₯
=π π π π π π π₯π₯(1 + π π π π π π π₯π₯)πππππ π π₯π₯(1 + π π π π π π π₯π₯) +
πππππ π π₯π₯(πππππ π π₯π₯)1 + π π π π π π π₯π₯(πππππ π π₯π₯)
=π π π π π π π₯π₯ + π π π π π π 2π₯π₯ + πππππ π 2π₯π₯
πππππ π π₯π₯(1 + π π π π π π π₯π₯)
Pyth. identity, this is β1β
=π π π π π π π₯π₯ + 1
πππππ π π₯π₯(1 + π π π π π π π₯π₯)
=1
πππππ π π₯π₯
= ππππππππ Code: ππ ANSWER: 41
Simplify: πππ π πππ₯π₯π π π π π π π₯π₯
+πππππ‘π‘π₯π₯π‘π‘π‘π‘π π π₯π₯
=1
π π π π π π π₯π₯π π π π π π π₯π₯ +
πππππ π π₯π₯π π π π π π π₯π₯π π π π π π π₯π₯πππππ π π₯π₯
=1
π π π π π π π₯π₯ β1
π π π π π π π₯π₯ +πππππ π π₯π₯π π π π π π π₯π₯ β
πππππ π π₯π₯π π π π π π π₯π₯
=1
π π π π π π 2π₯π₯ +πππππ π 2π₯π₯π π π π π π 2π₯π₯
=1 + πππππ π 2π₯π₯π π π π π π 2π₯π₯
=π π π π π π 2π₯π₯π π π π π π 2π₯π₯ = ππ Code: ππ
πππππ‘π‘ππ = β45
Start with appropriate addiοΏ½on / subtracοΏ½on formula:
π π π π π π (πΌπΌ + π½π½) = π π π π π π πΌπΌ πππππ π π½π½ + πππππ π πΌπΌ π π π π π π π½π½
= π π π π π π ππ3πππππ π ππ + πππππ π ππ
3π π π π π π ππ
Use unit circle for this and πππππ π ππ
3
Draw ππ using info given π½π½
ππ( β4,5)
5
β4
β41
In Quad II since adj side is neg and π π π π π π ππ > 0 Use pyth. theorem for hyp., (β4)2 + 52 = βπ¦π¦ππ2
To sketch ππ, use: adj
opp
To sketch ππ, use:
πππππ π ππ =β4β41
π π π π π π ππ =5β41
soβ¦
= οΏ½β32οΏ½οΏ½
β4β41
οΏ½+ οΏ½12οΏ½ οΏ½
5β41
οΏ½
=β4β32β41
+5
2β41
=β4β3 + 5
2β41
ANSWER: B
19.
1 + π π π π π π π΄π΄ β (πππππ π 2π΄π΄ β π π π π π π 2π΄π΄)πππππ π π΄π΄ + 2π π π π π π π΄π΄πππππ π π΄π΄
Step 1
=1 β πππππ π 2π΄π΄ + π π π π π π π΄π΄ + π π π π π π 2π΄π΄
πππππ π π΄π΄ + 2π π π π π π π΄π΄πππππ π π΄π΄
=π π π π π π 2π΄π΄ + π π π π π π π΄π΄ + π π π π π π 2π΄π΄πππππ π π΄π΄ + 2π π π π π π π΄π΄πππππ π π΄π΄
=
2π π π π π π 2π΄π΄ + π π π π π π π΄π΄πππππ π π΄π΄ + 2π π π π π π π΄π΄πππππ π π΄π΄
=π π π π π π π΄π΄(2π π π π π π π΄π΄ + 1)πππππ π π΄π΄(1 + 2π π π π π π π΄π΄)
= πππππππ¨π¨
Correct steps:
ANSWER: A
20. The mistake
is here! π½π½ 3
β4
5
Sketch:
By pyth. theorem
π‘π‘π‘π‘π π ππ = β34 So,
Formula sheet:
=2 οΏ½β3
4οΏ½
1 β οΏ½β34οΏ½
2
=β6
41 β 9
16
=β3
27
16
= β32β
167
= βππππππ
ANSWER: 247
NR #8
Writen #1
First bullet P(β5, 1)
ππ
βππ
ππ βππππ
(βπ¦π¦ππ)2 = (β5)2 + (1)2 (βπ¦π¦ππ)2 = 26
By pyth. theorem:
π½π½πΉπΉ
To find π½π½, first find reference angle π½π½πΉπΉ (inside the triangle)
πππ π = π‘π‘π‘π‘π π β1(1/5) or π π π π π π β1(1/β26) or πππππ π β1(5/β26)
ππ β 180Β° β 11Β° π½π½ β ππππππΒ°
so β¦
Q(β3,β4)
π½π½
βππ ππ
π·π·πΉπΉ By pyth. theorem:
βππ
To find π·π·, first find reference angle π·π·πΉπΉ (inside the triangle)
π½π½π π = π‘π‘π‘π‘π π β1(4/3) or π π π π π π β1(4 5β ) or πππππ π β1(3/5)
ππ β 180Β° + 53Β° π½π½ β ππππππΒ°
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Second bullet
Refer to your formula sheet: opp π π π π π π ππ =
1β26
hyp
adj πππππ π ππ =
β5β26
hyp
similarlyβ¦
π π π π π π π½π½ =β45
πππππ π π½π½ =β35
Weβll need all this: (leave exact)
π π π π π π (ππ + π½π½) = οΏ½1β26
οΏ½ οΏ½β35οΏ½+ οΏ½
β5β26
οΏ½ οΏ½β45οΏ½
Soβ¦
=β3
5β26+
205β26
=ππππ
ππβππππ
Note: There was a mistake in some answer keys!
WR #1
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Writen #2
First bullet 2οΏ½ππ β πππππππππ½π½οΏ½ β πππππ π π₯π₯ β 1 = 0
2 β 2πππππ π 2ππ β πππππ π π₯π₯ β 1 = 0 0 = 2πππππ π 2ππ + πππππ π π₯π₯ β 1
Use πππππππππ½π½ + πππππππππ½π½ = ππ, which re-arranges to πππππππππ½π½ = ππ β πππππππππ½π½ , to re-write equation in terms of πππππ π ππ only
πππππππππππ½π½ + ππππππππ β ππ = ππ
2nd bullet (2πππππ π ππ )(πππππ π ππ ) = 0
Factor
Find two #s that mult to β1, and expression will expand back out to 2πππππ π 2ππ + πππππ π π₯π₯ β 1
(2πππππ π ππ β 1)(πππππ π ππ + 1) = 0= 0
2πππππ π ππ β 1 = 0
Set each factor to zero πππππ π ππ + 1 = 0
or
ππππ
ππππππ
For GENERAL SOLUTION, consider all three primary solutions together
ππ
120Β°
120Β°
120Β°
Theyβre all ππππππ
(that is, 120Β°) apart
π½π½ =ππππ +
ππππππ ππ
π π β πΌπΌ
GENERAL SOLUTION:
πππππ π ππ = 1/2
ππππ
ππππππ
Where on the unit circle is the π₯π₯-coord Β½ ?
πππππ π ππ = β1
ππ
Where on the unit circle is the π₯π₯-coord β1 ?
π½π½ =ππππ ,ππ,
ππππππ
PRIMARY SOLUTIONS:
Writen #3
First bullet
Proceed on leοΏ½ side:
=1
π π π π π π π₯π₯ β πππππ π π₯π₯π π π π π π π₯π₯πππππ π π₯π₯ + πππππ π π₯π₯
π π π π π π π₯π₯
=ππππππππππππππππ
ππππππππ β π π π π π π π₯π₯ππππππππ β π π π π π π π₯π₯ + ππππππππ β πππππ π π₯π₯
ππππππππ β πππππ π π₯π₯
=πππππ π π₯π₯π π π π π π π₯π₯
π π π π π π 2π₯π₯ + πππππ π 2π₯π₯πππππ π π₯π₯π π π π π π π₯π₯
= ππππππππππ =πππππ π π₯π₯π π π π π π π₯π₯ β
πππππ π π₯π₯π π π π π π π₯π₯1
Write each expression in terms of π π π π π π and πππππ π
Re-write bottom expressions with a common denom.
Pyth. identity, this is β1β !
Invert and multiply
2nd bullet =
1π π π π π π π₯π₯ β πππππ π π₯π₯π π π π π π π₯π₯πππππ π π₯π₯ + πππππ π π₯π₯
π π π π π π π₯π₯
Go back to first step, restriction at any expression in the denominator. (Canβt divide by 0)
π π π π π π π₯π₯ β 0 πππππ π π₯π₯ β 0
βwhere on the unit circle is the π₯π₯ or π¦π¦ coord equal to 0 ?β
ππ
ππ,ππππ
ππππ
ππππππ
ππ β ππππ ππ
π π β πΌπΌ
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Writen #4
First bullet
ππ =14.4 + 9.6
2
ππ =πππ‘π‘π₯π₯ + πππ π π π
2
π π = ππππ
Median line ("π π " value)
("ππ" value) Dist. from median line to max or min, can also use:
π‘π‘ =πππ‘π‘π₯π₯ βπππ π π π
2
π‘π‘ =14.4 β 9.6
2 ππ = ππ.ππ
For ππ, determine the period:
This shows half of the period (dist from max to min)
Half of Period = 183 days (356 β 173)
Period = 366 days Actually, this was given!
ππ =ππππππππ
ππ =2ππ
366 Finally for ππππππππ , determine the horiz. dist of the max from the π¦π¦-axis
Basic cos curve βstartsβ at the MAX Here the max has shifted, 173 to the right
ππ = ππππππ π―π― = ππ.ππ ππππππ οΏ½
ππππππππ (ππ β ππππππ)οΏ½ + ππππ
2nd bullet
For π₯π₯-max, graph over 1 period
πππ π ππ =2ππ
0.524
β ππππ
For π¦π¦-max, use π΄π΄π¨π¨π΄π΄ = ππ + π π = 5.1 + 23.9
= ππππ
Graph π¦π¦1 = 5.1 π π π π π π οΏ½0.524(π₯π₯ β 2.75)οΏ½ + 23.9 π¦π¦2 = 26
And find the INTERSECTS
So, total # months above 26β can be found by subtractingβ¦.
= 7.94 β 3.56
ππ.ππ months
Third bullet
ππ would be higher, as the range of Calgary temperatures (between min and max) would be greater π π would be lower, as the median temperature for Calgary (represented by ππ) would be lower
Alsoβ¦. (not needed in your answer) ππ would be unchanged, as the period for each city would be the same (12 months). Similarly, ππ would be essentially unchanged, as the number of months after which the min / max temperature occurs would be approximately the same as both cities are in the northern hemisphere.
ALL DONE!
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