This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Trigonometry I
 Answers  Trigonometry I Diploma Practice Exam  ANSWERS 1 www.puremath30.com Formulas
These are the formulas for Trig I you will be given on your diploma.
a = rθ
sin θ
cos θ
1
sec θ =
cos θ
tan θ = cos θ
sin θ
1
csc θ =
sin θ
cot θ = Answer Sheet 1. A NR 2) 0.33 19. B 29. D 2. C NR 3) 0.28 20. D 30. C NR 1) 2.83 11. C 21. C 31. D 3. C 12. B 22. D 32. A 4. B 13. C 23. D 5. A 14. A 24. D 6. A NR 4) 111 NR 5) 14.7 7. C 15. D 25. B 8. C 16. C 26. B 9. B 17. B 27. C 10. D 18. B 28. D Copyright © Barry Mabillard, 2006
Trigonometry I Diploma Practice Exam  ANSWERS 2 www.puremath30.com
www.puremath30.com 1. Draw the following graphs
in your graphing calculator: f (θ ) = y1 = sin θ
g (θ ) = y2 = sin 2θ − 2
The transformations applied to the original
graph are a horizontal stretch by a
1
factor of , and a shift down by two units.
2
As can be seen on the graph, the new range for g (θ ) is −3 ≤ y ≤ −1 .
The answer is A. 2. First state all the information you know:
• The length of arc is 8952 km. • The angle is 55˚. However, we never use degrees in the arc length formula, so this
must be converted to radians. 550 • π 1800 = 0.9599 rad Now use the arc length formula a = rθ to solve for the radius.
a = rθ
a
8952
= 9325.97 km
r= =
θ 0.9599
To find the height of the satellite above the surface of the planet, take the value you just
found (height of satellite from the centre of the planet) and subtract the radius of the
planet.
9325.97 km – 5000 km = 4325.97 km
The answer is C. Trigonometry I Diploma Practice Exam  ANSWERS 3 www.puremath30.com ⎛π
⎞
NR #1) When a question gives a point, such as ⎜ , −2 ⎟ , these values can be put in for x and y.
⎝2
⎠
In the context of this question, x and y are called θ and f (θ ) . π⎞
⎛
f (θ ) = a cos ⎜ θ − ⎟ − 4
4⎠
⎝
⎛π π ⎞
Insert the given point.
−2 = a cos ⎜ − ⎟ − 4
⎝2 4⎠
⎛π ⎞
−2 = a cos ⎜ ⎟ − 4
⎝4⎠
⎛π ⎞
2 = a cos ⎜ ⎟
⎝4⎠
2=a 2
2 ππ
 24 = π
4 , or thinking in degrees, 900 − 450 = 450 Bring the 4 over to the other side
2
⎛π ⎞
Since cos ⎜ ⎟ =
2
⎝4⎠ 4=a 2
a = 2.83 The value of a, to the nearest hundredth, is 2.83 3. Look at the graph on the right. Notice
that the midline has been shifted three units
down, and the amplitude is k.
That means: Minimum = 3  k
Maximum = 3 + k
The range of the graph is
−3 − k ≤ f (θ ) ≤ −3 + k
The answer is C. Trigonometry I Diploma Practice Exam  ANSWERS 4 www.puremath30.com π⎞
⎛
4. Draw the graph of y = cos ⎜ θ + ⎟
2⎠
⎝
This graph is identical to the graph
of sin θ , reflected in the xaxis.
The answer is B. 5. At the yintercept, θ = 0. π⎞
⎛
f (θ ) = −3cos ⎜ kθ + ⎟ − b
2⎠
⎝
π⎞
⎛
= − 3cos ⎜ k (0) + ⎟ − b
2⎠
⎝
⎛π ⎞
= − 3cos ⎜ ⎟ − b
⎝2⎠
⎡
⎤
⎛π ⎞
= − 3 ( 0 ) − b ⎢since cos ⎜ ⎟ = 0⎥
⎝2⎠
⎣
⎦
= −b
The answer is A. 6. By inspection, the midline is at 7,
and the amplitude is 4. To determine the correct answer, it is
necessary to check each of the possible
answers to see which one actually
matches the graph provided.
The only equation which is a
correct match is
f (θ ) = −4sin (θ − 300 ) − 7
The answer is A. Trigonometry I Diploma Practice Exam  ANSWERS 5 www.puremath30.com 7. a and d will be the same. The bvalue will be the same. (see below for proof)
3600 3600
=
=1
P
3600
2π 2π
=
=1
Radian Mode: b =
P 2π Degree Mode: b = Thus, the only value which will change is the cvalue, which changes from 30˚ to π
6 = 0.5236 The answer is C. 8. At point A, the yvalue of both graphs is equal to m. Therefore, f ( k ) + g ( k ) = m + m = 2m
The answer is C.
csc θ < 0 9. First determine the quadrant the
angle is found in. Based on the diagrams to the right,
the angle is in Quadrant IV.
Now draw in the triangle and
use Pythagoras to solve for
the hypotenuse. From the triangle, sin θ = −4
5 The answer is A. Trigonometry I Diploma Practice Exam  ANSWERS 6 www.puremath30.com 10. • First use the information from cos A = 30
, 0 < θ < 900 to solve for the angle A.
2 Based on the unit circle, A = 30˚
• Then use B = 600 + A to
determine the value of B.
B = 600 + A • = 600 + 300
= 900 Finally, evaluate secB
1
sec B =
cos B
1
=
cos 900
1
=
0
=undefined The answer is D. NR #2) The length of one complete period is 12 units.
The graph starts at π, ends at 7π, so the length is 6π b= 2π
P 2π
6π
1
=
3
= The answer is 0.33
NR #3)First determine the quadrant
the angle can be found in. Then use Pythagoras to
complete the triangle.
4
5
3
cos θ = −
5 sin θ = − sin 2 θ − cos 2 θ
2 ⎛ 4⎞ ⎛ 3⎞
⎜− ⎟ −⎜− ⎟
⎝ 5⎠ ⎝ 5⎠
= 0.28 The answer is 0.28
Trigonometry I Diploma Practice Exam  ANSWERS 7 www.puremath30.com 2 11. The correct answer is C, since the new graph has a phase shift of c units to the right.
This will move all the θ  intercepts by the same amount. (Remember: In a trig graph, the
xintercepts are called θ  intercepts)
The answer is C.
12. The only graph which has no effect on the yintercept is y3 = cos 3θ , since the
horizontal stretch will not transform points on the yaxis.
The answer is B.
13. •
•
•
• max − min 112 − 28 84
= 42
2
2
2
2π 2π
bvalue: The period is 2π, so b =
=
=1
P 2π amplitude: cvalue: For cosine, it’s = = π 2
max + min 112 + 28 140
dvalue:
= 70
=
=
2
2
2 π⎞
⎛
The equation is y = 42 cos ⎜ θ − ⎟ + 70
2⎠
⎝
The answer is C.
14. First convert to degrees to make the calculation simpler: 16π 1800
•
= 3200
π
9 The reference angle is 360˚  320˚ = 40˚
Now reconvert to radians: 400 • π
180 0 = 2π
9 The answer is A.
NR #4)
Use the arc length formula to determine the length of the curved portion of the sidewalk.
a = rθ π a = 30 • 1.693 (Don’t forget to convert to radians! 97 0 •
= 1.693 )
1800
a = 51 m
Add up all the lengths: 30 m + 51 m + 30 m = 111 m
The answer is 111. Trigonometry I Diploma Practice Exam  ANSWERS 8 www.puremath30.com 15.
3π
< θ < 2π tells you that the angle is in quadrant IV
2 Use Pythagoras to determine the remaining
side of the triangle
a 2 + b2 = c2
4 2 + b 2 = 52
b 2 = 25 − 16
b2 = 9
b=3
4
3
The answer is D.
Thus, cot θ = − 16. The amplitude of each graph being quadrupled means all the yvalues are being
multiplied by 4. ⎛π 2 ⎞
⎛ 5π − 2 ⎞
⎟
⎜,
⎟,⎜ ,
⎜8
⎜8 2 ⎟
2⎟
⎝
⎠
⎝
⎠
⎛π
⎛ 5π
− 2⎞
2⎞
⎜ , 4•
⎟ , ⎜ , 4•
⎟
⎜8
⎜8
2⎟
2⎟
⎝
⎠
⎝
⎠
⎛π
⎞
⎛ 5π
⎞
⎜ , 2 2 ⎟ , ⎜ , −2 2 ⎟
⎝8
⎠
⎝8
⎠
The answer is C.
17. The values can be represented in a triangle as follows.
Solve for the hypotenuse using Pythagoras. −6 k
3
=−
10k
5
8k 4
cos θ =
=
10k 5
3
−6 k
tan θ =
=−
8k
4 sin θ = a 2 + b2 = c 2
(−6k ) 2 + (8k ) 2 = c 2
36k 2 + 64k 2 = c 2
100k 2 = c 2
10k = c The answer is B. Trigonometry I Diploma Practice Exam  ANSWERS 9 www.puremath30.com 18. 13π
π
is equivalent to
on the unit circle since they are coterminal angles.
6
6 ⎛π ⎞
Evaluate the value of −3 tan ⎜ ⎟ using the unit circle.
⎝6⎠
⎛π ⎞
−3 tan ⎜ ⎟
⎝6⎠
⎛π ⎞
sin ⎜ ⎟
⎝6⎠
−3
⎛π ⎞
cos ⎜ ⎟
⎝6⎠
1
Now rationalize the denominator.
2
−3 •
3
3
3
−
•
2
3
3
12
−3 3
−3 • •
2
3
3
3
−3
−
3
The answer is B.
19. Draw the graph of w(t ) = cos3 t − sin(t − 3) + 8 using technology in radian mode. Using the minimum/maximum features of the
calculator, the lowest point of a cycle occurs
at 6.7 cm, and the highest point at 9.3 cm. The answer is B. Trigonometry I Diploma Practice Exam  ANSWERS 10 www.puremath30.com 20. Draw in the asymptotes as shown in
the diagram to the right. The first asymptote occurs at π , and every other asymptote
2
is a multiple of π units away. The answer is D. 21. To answer this question, determine which of the angles does not reduce to 150˚. A. 930˚ + 360˚ + 360˚ + 360˚ = 30˚
17π
= 510˚ 510˚  360˚ = 150˚
B.
6
23π
C.
= 690˚ 690˚  360˚ = 330˚
6
D. 3.67 rad = 210˚ 210˚ + 360˚ = 150˚
23π
will give an angle that isn’t 30˚ or 150˚
6
The answer is C. Only 22. • avalue = 15 − 3
2 = 12
2 = 6m The period is 40 s. •
•
• 2π
2π π
=
=
Period 40 20
cvalue: None
15 + 3 18
dvalue:
=
= 9m
2
2
bvalue: The equation is h(t ) = −6 cos π t + 9 . Use a negative since the cosine graph starts at the
20
bottom of the ride instead of the top. The answer is D. Trigonometry I Diploma Practice Exam  ANSWERS 11 www.puremath30.com 1350
23. Position B, which is 135˚ away from the starting position, is
= 0.375 revolutions
3600
Multiply this by the period for one complete revolution to find the time. 40 s × 0.375 = 15 s.
The answer is D. 24. Graph the following: y1 = −6 cos π
20 t +9 y2 = 13 Determine the time a rider spends over 13 m for one cycle, then multiply by 3 to get the
complete time for the entire ride.
The total time is 10.7 × 3 = 32 seconds.
The answer is D. 25. When the Ferris Wheel rotates counterclockwise, there will be no change to the
shape of the graph, since the height with respect to time will be identical. Look for the
answer which will result in exactly the same graph. The answer which gives the same graph is y = f (t − 40) , since moving the cosine shape
exactly one full period to the right will have everything realign.
The answer is B. 26. When the ride speeds up, the only thing to be affected is the period. Since the period
is related to the bvalue, the bvalue will change.
The answer is B. Trigonometry I Diploma Practice Exam  ANSWERS 12 www.puremath30.com 27. The graph is f (θ ) = sin 4θ is shown to the right.
There are 12 θ  intercepts between 0 ≤ θ < 3π
(Note that you don’t include the one at 3π)
So, there are 12 asymptotes.
The answer is C. 28. • 26 − 20 = 6 = 3 cm
2
2
The period is 2 seconds.
2π
2π
• bvalue:
=
=π
Period
2
• cvalue: None
20 + 26
46
• dvalue:
=
= 23 cm
2
2 avalue = The equation is h(t ) = 3cos π t + 23 .
The answer is D.
29. First isolate sin θ 2 sin θ = 3
sin θ = 3
2 Solving from the unit circle, θ = π 2π ,
33
Each solution repeats for every coterminal angle, so the general solutions are
π
2π
θ = ± 2nπ ,
± 2nπ
3
3
The answer is D. Trigonometry I Diploma Practice Exam  ANSWERS 13 www.puremath30.com 30. From the equation, determine the min/max positions, and the length of a cycle. y = 20.1sin 2π
(t  265) + 6.2
300 Minimum: 6.2 – 20.1 = 13.9
Maximum: 6.2 + 20.1 = 26.3 b= 2π
2π
=
Period 300 Period = 300 A window setting that allows us to see all these points is x: [0, 600, 100], y: [15, 30, 5]
The answer is C. π⎞
⎛
31. First factor the 3 out of the brackets: g (θ ) = sin [3θ − π ] = sin 3 ⎜ θ − ⎟
3⎠
⎝
This tells you there is a horizontal stretch by 1
π
, then a phase shift
units right.
3
3 The answer is D. 32. The graph of f (θ ) = cot 4θ is shown to the right. The first asymptote occurs along the yaxis,
and every other asymptote is a multiple
of 45˚ away.
The domain is x ∈ R, x ≠ ± nπ
4 The answer is A. Trigonometry I Diploma Practice Exam  ANSWERS 14 www.puremath30.com Written Response #1
Equation of Sunrise
10.18 − 2.57 7.61
=
= 3.805
2
2
The period is 365 days.
2π
2π
• bvalue:
=
Period 365
• cvalue: 10 days left, since the
maximum point is on Dec. 21
10.18 + 2.57 12.75
=
= 6.375
• dvalue:
2
2
2π
T ( x ) = 3.805cos
( x + 10 ) + 6.375
365 • avalue = Equation of Sunset
22.75 − 15.00 7.75
=
= 3.875
2
2
The period is 365 days.
2π
2π
• bvalue:
=
Period 365
• cvalue: 172 days right, since the maximum
point is on June. 21
22.75 + 15.00 37.75
=
= 18.875
• dvalue:
2
2
2π
T ( x ) = −3.875cos
( x − 172 ) + 18.875
365 • avalue = Use a negative since the cosine graph is upsidedown The following transformations are required to
transform f ( x ) = cos x into the graph representing
the sunset time in Yellowknife:
Vertical stretch by a factor of 3.875
Reflection in the xaxis.
365
2π
(Remember to use the reciprocal for a horizontal
stretch)
Horizontal stretch by a factor of Translation of 172 units right, and 18.875 units up. 2π
y1 = 3.805cos
( x + 10 ) + 6.375
Graph
365
y2 = 4 Determine the intersection points, which occur
on days 120 and 225.
Yellowknife experiences a sunrise earlier than 4
AM for 105 days each year. Feb. 15 is the 46 th day of the year
2π
Graph y = 3.805cos
( x + 10 ) + 6.375
365
2nd Trace Value: x = 46. Sunrise = 8.54
2π
Graph y = 3.875cos
( x − 172 ) + 18.875
365
2nd Trace Value: x = 46. Sunset = 16.69 Length of daylight = 16.69 – 8.54 = 8.15 hours.
(Can also be written as 8 hours, 9 minutes.)
Trigonometry I Diploma Practice Exam  ANSWERS
15
www.puremath30.com Written Response #2
•
• If it takes 1 second for the tire to go around twice, that means the time for one
spin is 0.5 seconds. The period is 0.5 s.
Parameter
a
b
c
d Value
42
12.57
0
42 • h(t ) = −42sin12.57t + 42 • Contact between the wheel and ground occurs when the height, h ( t ) , is zero. Solve the equation: 0 = −42sin12.57t + 42 by graphing to obtain the tintercepts.
The first contact occurs at 0.125 seconds, and every future contact occurs one period
later. The general formula is: Time of contact = 0.125 + 0.5n
The fifth contact will be 0.125 + 0.5(4) = 2.125 s • avalue: 39 instead of 42
Period: Still 0.5 seconds
bvalue: No change, since no change in period
cvalue: No change
dvalue: Midline is at 39 instead of 42. Trigonometry I Diploma Practice Exam  ANSWERS 16 www.puremath30.com Written Response #3
a  value 1
1
2
0
0
4π
x ≠ ±2nπ
y ≤ −1, y ≥ 1
None
None b  value
Phase Shift
Vertical Displacement
Period
Domain
Range
xintercepts
yintercepts
Asymptotes
(general equation) •
• x = ±2nπ 1
The equation of the graph is g (θ ) = sin θ
2
The asymptotes in f (θ ) occur at the θ  intercepts of g (θ ) 1 ⎛ 10π ⎞
⎛ 10π ⎞
f⎜
⎟ = csc ⎜
⎟
3⎠
2⎝ 3 ⎠
⎝
⎛ 10π ⎞
= csc ⎜
⎟
⎝6⎠
⎛ 5π ⎞
= csc ⎜ ⎟
⎝3⎠ 1
⎛ 5π ⎞
sin ⎜
⎟
⎝3⎠
1
=
3
−
2
2
= 1× −
3
2
=−
3
= Trigonometry I Diploma Practice Exam  ANSWERS Now rationalize
the denominator:
=−
=− 17 2
3
×
3
3
23
3 www.puremath30.com ...
View
Full
Document
 Spring '09
 JACK
 Calculus, Trigonometry, Formulas

Click to edit the document details