Download 1 sec 1 tan sin

568536 - Mathematics
Guide
2
A
3
Work : (example)


1.
sin 2 x tan 2 x  1
1  cos 2 x

sec 2 x  1
cos 2 x
= tan2x
2.
sin 2 x sec 2 x
sin 2 x

cos 2 x
tan 2 x
= tan2x
3.
sin 2 x sec 2 x
2
 tan x
2
tan x
= tan2x
4.
sin2x  sec2x
= tan2x
5.
sin2x 
6.
1
cos 2 x
= tan2x
tan2x
= tan2x
4
Work : (example)
1.
2 sin t cos t cos 2 t  sin 2 t sec t
=

sin t
cos t
cos 2 t  sin 2 t
=
cos t
2.
2 cos t 
3.
2 cos 2 t  cos 2 t  sin 2 t
=
cos t
4.
cos 2 t  sin 2 t
=
cos t
5.
1
=
cos t
6.
sec t =
5
A
6
A
7
Example of an appropriate solution
1.
cosec A (cosec A + cot A) =
1
1  cos A
2.
1  1
cos A 
1
+

=
sin A  sin A sin A  1  cos A
3.
1  1 + cos A 
1

=
sin A  sin A  1  cos A
4.
1 + cos A
1
=
2
1  cos A
sin A
5.
1  cos A
1
=
2
1  cos A 1  cos A
6.
1 + cos A
1
=
(1 + cos A) (1  cos A) 1  cos A
7.
1
1
=
1  cos A 1  cos A
8
C
9
B
10
A
11
Work :
Example 1 :
Example 2 :
sin t sec t
 sec2 t  -1
cot t
sin t sec t
 sec2 t  -1
cot t
sin t tan t
 sec2 t  -1
cos t
sin t
tan t  tan t  sec2t = -1
sin 2 t
1

 -1
2
cos t cos2 t
sin t
1
1


 -1
cos t cos t cos2 t
tan2t  (tan2t + 1) = -1
-1 = -1
-
cos2 t
 -1
cos2 t
-1 = -1
12
The rule of this function is
f x   - 300 sin

x  500
4
f x   - 300 sin

x  4  500
4
f x   - 300 cos

x  2  500
4
or
or
13
B
14
C
15
D
16
The exact values of x are -,
- 
, ,
3 3
17
Example of an appropriate solution
Rabbit population is of the form
R(x) = a sin b(x  h) + k
For simplicity
h=0
k = 1000 (vertical shift)
a = 200
3
of the period is given by the 12 years
4



3 2
•
= 12  b =
and R(x) = 200 sin x + 1000
4
8
8
b
Likewise, f(x) is of the form
F(x) = a cos b(x  h) + k
h = 0, a = 75, k = 175, b =
F(x) = 75 cos

8

x + 175
8
In 2004
x = 14 and R(14)  F(14)  630.545
Answer:
The difference between the rabbit and the fox populations is 631.
Accept 630 or 631.
Note:
18
A
Students who have determined the rule for one of the populations have shown they have a partial
understanding of the problem.
19
Example of an appropriate solution
The minimum height of the pendulum is 150 cm.
The maximum height of the pendulum is 150 + (31  31 sin 45) = 159.08 cm.
y  a cos b x  h   k
159.08  150
2
 4.54
a
2
0.875
16

7
b
159.08  150
2
 154.54
k 
h  0 if a is negative
 16t 
f t   - 4.54 cos 
  154.54
 7 
1 hour  3600 seconds
f 3600  156 cm
Answer:
To the nearest cm, the pendulum will be 156 cm above the ground after 1 hour.
Note:
Students who use an appropriate method in order to determine parameters a, b and k have shown
they have a partial understanding of the problem.
Students’ answers may vary considerably, depending on the rounding they applied in the course of
their work.
20
Example of an appropriate solution
Rule of the function
t: time in seconds
H(t)
22
H(t): height in metres
12
H t   a sin bx  h  k
2
7.5
22  2
2
 10
a 
H t   10 sin bx  7.5  12
1
period  15, P  30 seconds
2
2
b
2
b 
30
30 


15
H t   10 sin

15
x  7.5  12
15
t
Height of Tom’s seat after 20 seconds
H 20  10 sin

15
 17 metres
20  7.5  12
Answer:
Tom’s seat is 17 m above the ground 20 seconds after the Ferris wheel begins to turn.
Note:
H t   - 10 cos

15
x  12 and H t   - 10 sin

15
x  22.5  12 are examples of other
acceptable rules.
Note:
Students who use an appropriate method in order to determine parameters a and b have shown
they have a partial understanding of the problem.
21
Example of an appropriate solution

2 cos 2 x  3 sin x  3  0

2 1  sin 2 x  3 sin x  3  0
2  2 sin x  3 sin x  3  0
2
- 2 sin 2 x  3 sin x  1  0
2 sin 2 x  3 sin x  1  0
2 sin
x  1sin x  1  0
2 sin x  1  0
sin x  1 = 0
1
2
sin x = 1
sin x 
Answer: The exact values of x are x 
22
C


or x  .
6
2
23
Example of an appropriate solution
Construction of the function
p(t) = a cos (bt) + k
where
k
18.2  9.2
 13.7
2
a
18.2  9.2
 4.5
2
In order to have the function increase from the initial value, parameter a must be negative.
Therefore, a = -4.5 .
Since the period is 12.5 hours, we calculate
b
2
 0.16
12.5
Therefore, the equation is
p(t) = -4.5 cos (0.16t) + 13.7
or


pt   4.5 sin  0.16t    13.7
2

Solving for t when p(t) = 14.5 m.
14.5  - 4.5 cos 0.16t   13.7
0.8  - 4.5 cos 0.16t 
0.18  cos 0.16t 
cos 1 - 0.18
0.16
t1  3.5 hours
or

t 
t 2  - 3.5
-3.5 + 12.5 = 9
The function reaches a height of 14.5 m at 3.5 hours and again at 9 hours.
Therefore,
t = 9  3.5 = 5.5 hours
Answer:
The oil tanker has 5.5 hours to dock safely between two consecutive low tides.
24
Example of an appropriate solution
Let
t: time, in seconds, that has past since 13:00
f(t): height of the jet, in metres
The rule of correspondence
f(t) = a sin b(t  h) + k
5 1
2
2
2
p
b
2
60 
b
2
b 
60
f(t)
(m)
a
b 
7
6
5
4
3
2
1
15

30
45
60
75
90
105
120 135
150
t(s)
30
f  x   2 sin

30
t
 h  k
Translation (h, k) = (15, 3)
f(t) = 2 sin

(t  15) + 3
30
Height at 13 h 12 min 40 s
Since the function has a period of one minute, the jet will be at the same height in 40 seconds.
f 40  2 sin
 3.09
Answer:

30
40  15  3
At 13 hours 12 minutes and 40 seconds, the water jet will be at a height of 4 m.
25
Example of an acceptable solution
Temperature of 15 C
 
x  + 2 = 15
 13 
20 sin 
 
x  = 13
 13 
 20 sin 
   13
x =
 13  20
 sin 
 
x  = 0.65
 13 
 sin 
Period = p =
2
2
p=
 p = 26

b
13
sin  = 0.65  1  0.71 and 2  2.43
x
13
x
13
= 0.71  x1  2.93
= 2.43  x2  10.07
x1  2.93
x2  10.07
Given that the period is 26 weeks
x3 = 2.93 + 26 = 28.93
x4 = 10.07 + 26 = 36.07
Intervals in which the temperature is greater than 15 C
x 2  x1 =
7.14
x 4  x3 =
7.14
Answer:
During the course of the year, the average temperature is higher than 15 C for 14.28 weeks.
26
Example of an appropriate method
Solving the equation for i(t) = 9
2 
 t
6 sin 

69
3 
 12
2 
 t
6 sin 

3
3 
 12
2  1
 t
sin 


3  2
 12

2  
1
 t
arc sin  sin 

   arc sin
3 
2
 12

t
12
t
12
t
12



2 

3
6

6

4
6
- 3
6
t = -6
or
t
12
t
12
t
12

2
5

3
6

5
4

6
6


6
t=2
Period of the function
2
2

 24

b
12
(24 seconds elapse between each sound signal)
Number of seconds that have elapsed since the start of the experiment
2
26
50
74
98
122
Answers not within the
given interval
-6
Answer:
18
42
66
90
114
During the 120 seconds, the apparatus emitted 10 sound signals
Trigonometry
2
The population growth P of the native people of Northern Quebec is represented by the graph of the
sinusoidal function with equation
Pt   400 sin

t  4  600
4
where t is the number of years elapsed since 1984.
Which one of the graphs below corresponds to this situation?
A)
C)
E(t)
1 000
600
600
200
200
2
B)
4
6
8 t
D)
E(t)
1 000
600
600
200
200
4
6
8 t
Prove that,


2
4
6
8 t
2
4
6
8 t
E(t)
1 000
2
3
E(t)
1 000
sin 2 x tan 2 x  1 1  cos 2 x

 tan 2 x
2
2
sec x  1
cos x
4
Peter and Lucy are studying the repeated up and down movement of a piston in a gasoline engine.
Peter claims that the piston moves according to the rule :
d 
2 sin t cos t cos 2 t  sin 2 t

sin t
cos t
Lucy believes that the piston moves according to the rule :
d = sec t
Prove that Peter and Lucy are both correct by showing that
2 sin t cos t cos 2 t  sin 2 t

 sec t
sin t
cos t
5
A sinusoidal function is shown below.
f(x)
6
5
4
3
2
1

6
-1
2
3
-2
Which of the following rules define this function?
A)


f x   4 sin 2 x    2
6

C)


f x   4 sin  x    2
6

B)


f x   4 sin 2 x    2
6

D)


f x   4 sin  x    2
6

7
6
x
6
Given the function f x   - 100 sin

x  4  100
4
Which of the following is true?
7
8
9
A)
6 is a zero of the function f on the interval [0, 8].
B)
The function f is decreasing on the interval [4, 8].
C)
Dom f =  and range f = [-100, 100].
D)
The period of the function f is 8 and the phase shift is -4.
Prove the following trigonometric identity:
cosec A (cosec A + cot A) =
1
1  cos A
For all values of A (for which A is defined), the expression tan A + cot A is equal to
A)
sin A cos A.
C)
sec A cosec A.
B)
sec A cos A.
D)
sin A cosec A.
The expression sec A 
tan A
is equivalent to
cosec A
A)
2  cos2 A
cos A
C)
sin A
B)
cos A
D)
(1  sin A) sin A
cos A
10
The equation of a sinusoidal function is f(t) = 2 sin
t
+ 4 where t  [0, 4[.
2
On which interval is this function increasing?
11
12
A)
[0, ]  [3, 4[
B)
  3   5 7 
 2 , 2    2 , 2 
C)
  3   5 7   9 11  13 15 
 4 , 4    4 , 4    4 , 4    4 , 4 
D)
[2, 6]
Prove the following identity :
sin t sec t
 sec2 t = - 1.
cot t
Many factors influence the deer population in a given habitat: climate, hunting, predators, etc.
The following graph shows the evolution of a population of deer as a function of time.
f(x)
Number
of deer
800
500
200
2
6
10
14
18
Time (years)
Write the rule that can be used to represent this function.
x
13
14
15
16
What is the domain of the tangent function?

where k  Z}
2
A)
{x   | x = (2k + 1)
B)
{ x   | x  (2k + 1)
C)
{ x   | x  k  where k  Z}
D)


where k  Z}
2
In which of the following intervals does the sine function decrease?
A)
-  
 2 , 2 
C)
  3 
 2 , 2 
B)
  5 
 6 , 6 
D)
], 2[
The amplitude of function f(x) = a sinx ALWAYS has a value of ...
A)
2
C)
a
B)

D)
|a|
Given the trigonometric equation:
2 cos2x + cos x = 1, x  [, ]
What are the exact values of x that satisfy this equation?
17
The following diagram (not drawn to scale) represents a predator-prey situation of the population of rabbits
and foxes in a region of Mt.Tremblant Park, both of which follow a sinusoidal model.
Population
rabbits
1000
800
250
foxes
175
4
Years after
1990
Consider t = 0 to be 1990. Initially there are 1000 rabbits and 250 foxes. The graph shows the rabbit population
increasing for 4 years then decreasing to its minimum population of 800 in 2002. Over the same period of time
the fox population starting from its maximum decreases, reaches its minimum, then increases to a population
of 175 in 2002. If the model continues as shown, what is the difference between the rabbit and fox
populations in 2004?
18
A hyperbola and a trigonometric function are drawn on the same Cartesian plane. The equation of the
hyperbola is
x2
y2

 1.
25 144
The foci of the hyperbola are directly below two of the maxima of the
trigonometric function. Which of the following is an equation of the
trigonometric function?
A)


y  28.8 cos  x  13
 13

C)


y  23.3 cos  x  11
 11

B)
 2
x  13
y  28.8 cos 
 13

D)
 2
x  11
y  23.3 cos 
 11

19
A cuckoo clock uses a pendulum to keep time. The
movement of the pendulum can be described by a
sinusoidal function.
The length of the pendulum is 31 cm. At its lowest
point, the pendulum is 1.5 m from the ground.
31 cm
t3
t2
t1
The pendulum starts its movement at t1. The interior
angle between t2 and t3 is 90 and it takes the
pendulum 0.875 second to go from t2 to t3.
What is the height of the pendulum relative to the
ground after 1 hour? Round your answer to the
nearest centimetre.
1.5 m
20
Tom’s ride on the Ferris wheel at La Ronde can be described by the graph of the sinusoidal function shown
below. The graph represents the height of Tom’s seat above the ground, in metres, as a function of the time, t,
in seconds. The distance between the minimum and maximum heights of Tom’s seat is 20 metres. Tom’s seat
reaches its first maximum height 15 seconds after the Ferris wheel begins to turn.
H(t)
22
2
15
t
How many metres above ground is Tom’s seat 20 seconds after the Ferris wheel begins to turn?(Assume Tom’s
seat started from rest at the bottom of the Ferris wheel.)
21
22
Given


2 cos2 x + 3 sin x  3 = 0, x  0,

. What are the exact values of x?
2 
In the sinusoidal function below, x represents time in seconds.


f x   5 sin  x  1  3
2

Which of the following correctly describes the properties associated with the sinusoidal function?
A)
Maximum Value: 5
C)
Minimum Value: 2
Minimum Value: -2
 
 sec
2
Period: 4 sec
Period: 
B)
Maximum Value: 8
Maximum Value: 8
D)
Maximum Value: 5
Minimum Value: 3
Minimum Value: 3
Period: 5 sec
Period: (2 ) sec
23
The depth of water at the port of St. Marie-Elise varies according to the tides. A sinusoidal function can be
used to predict water depth.
At low tide, the depth of the water is 9.2 m.
At high tide, it is 18.2 m.
The time between two low tides is 12 hours and 30 minutes. In order for an oil tanker to dock safely, the
depth of water must be at least 14.5 m.
For how many hours can an oil tanker dock safely between two consecutive low tides?
24
A fountain in a shopping centre has a single jet of water. The height of the jet of water varies according to a
sinusoidal function. Joel notes that, in exactly one minute, the jet goes from a minimum height of 1 m to a
maximum height of 5 m and back to 1 m.
At 13:00, the jet of water is at a height of 1 m.
What will be the height of the jet of water, to the nearest tenth of a metre, when the clock reads 13:12:40? (13
hours, 12 minutes, 40 seconds)
25
The average temperature of a fictitious town is given by the equation
 
x + 2
 13 
(x) = 20 sin 
where x represents the number of weeks since the beginning of the year.
During the course of the year, for how many weeks was the average temperature higher than 15 C?.
26
During an experiment, the intensity i(t) of the electric current of a device as a function of time t elapsed since
the beginning of the experiment is given by:
 t 2 

  6 where t is expressed in seconds.
3 
 12
i(t) = 6 sin 
The device emits a sound signal each time the current’s intensity is equal to 9.
The experiment lasts 120 seconds.
How many sound signals does the device emit during the experiment?