4 Graphs of y = tan x, y = cot x, y = sec x, y = csc x
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308
CHAPTER 10 Graphs of the Trigonometric Functions
■ Repeating from above the reciprocal
relationships that are the basis for graphing
y = cot x, y = sec x, and y = csc x, we have
csc x =
1
sin x
cot x =
1
tan x
sec x =
y
y
Period
Period
1
cos x
4
(10.2)
2
4
2
3p
2
p
p
0
−2
p
2
p
3p
2
2p
x
−2
0
−2
y = csc x
−4
p
2p
x
−2
Asymptote
y = sec x
p
2
Fig. 10.25
Asymptote
−4
Fig. 10.26
From these graphs, note that y = tan x and y = cot x have period p and have all
real numbers as their range. The functions y = sec x and y = csc x have period 2p, but
their ranges do not include the real numbers between –1 and 1.
The vertical dashed lines on the graphs are vertical asymptotes (see Section 3.4).
The curves approach these lines but never actually reach them. The values of x for
which the curve has an asymptote are not included in the domain of the function.
As we can see from the graphs, the functions have asymptotes, zeros, and maximum
or minimum values when x is a multiple of p>2, that is to say, at the same key values
which allowed us to sketch the graphs of the sine and cosine functions easily. The characteristics of the trigonometric functions at the key values from 0 to 2p are compared
in Table 10.4. For example, note that when the sine is zero, its reciprocal (the cosecant)
has an asymptote.
Table 10.4
Key
values
0
y = sin x
y = cos x
y = tan x
0
1
0
asymptote
p
2
1
0
p
0
-1
3p
2
-1
0
asymptote
0
1
0
0
2p
Key
values
0
y = csc x
y = sec x
y = cot x
asymptote
1
asymptote
p
2
1
asymptote
0
p
asymptote
3p
2
2p
-1
asymptote
-1
asymptote
0
asymptote
1
asymptote
To sketch functions such as y = a sec x, first sketch y = sec x and then multiply
the y-values by a. Here, a is not an amplitude, since the ranges of these functions are
not limited in the same way they are for the sine and cosine functions.
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10.4 Graphs of y = tan x, y = cot x, y = sec x, y = csc x
y
309
E X A M P L E 1 Sketching the graph of y = a sec x
Sketch the graph of y = 2 sec x.
First, we sketch in y = sec x, shown as the light curve in Fig. 10.27. Then we multiply the y-values of this secant function by 2. Although we can only estimate these values and do this approximately, a reasonable graph can be sketched this way. The
desired curve is shown in in Fig. 10.27.
■
3
p
0
x
2p
−3
E X A M P L E 2 Graph of y = a cot bx
Sketch the graph of y = 0.5 cot 2x.
Since the period of y = cot x is p, the period of y = cot 2x is p>2. Therefore, we
have the following table of key values:
Fig. 10.27
y
2
x
y
1
0
p
4
3p
4
p
2
p
p
Since a = 0.5, the function increases more slowly than y = cot x. We sketch the graph
as shown in Fig. 10.28.
■
x
p
3p
0
p
4
2
4
asymptote 0 asymptote 0 asymptote
−1
Using a graphing calculator, we can display the graphs of these functions more easily and more accurately than by sketching them. By knowing the general shape and
period of the function, the values for the window settings can be determined.
−2
E X A M P L E 3 Calculator graph of y = a csc (bx + c)
Fig. 10.28
View at least two periods of the graph of y = 2 csc12x + p>42 on a graphing
calculator.
Since the period of csc x is 2p, the period of csc12x + p>42 is 2p>2 = p.
Recalling that csc x = (sin x2 -1, the curve will have the same displacement as
p>4
y = sin12x + p>42. Therefore, displacement is - 2 = - p8 . There is some flexibility
in choosing the window settings, and as an example, we choose the following settings:
6
–0.5
6
Xmin = -0.5 1the displacement is -p>8 = -0.42
Xmax = 6 1displacement = -p>8; period = p, -p>8 + 2p = 15p>8 = 5.92
Ymin = -6, Ymax = 6 1there is no curve between y = -2 and y = 22
–6
Fig. 10.29
With y1 = 2(sin12x + p>42 2 -1, and with the calculator in radian mode, Fig. 10.29
shows the calculator view.
■
E XE RC IS ES 1 0 .4
In Exercises 1 and 2, view the graphs on a graphing calculator if the
given changes are made in the indicated examples of this section.
In Exercises 7–14, sketch the graphs of the given functions by use of the
basic curve forms (Figs. 10.23, 10.24, 10.25, and 10.26). See Example 1.
1. In Example 2, change 0.5 to 5.
7. y = 2 tan x
2. In Example 3, change the sign before p>4.
9. y =
In Exercises 3–6, fill in the following table for each function and plot
the graph from these points.
x
- p2
- p3
- p4
- p6
0
p
6
p
4
p
3
p
2
y
2p
3
3p
4
5p
6
p
1
2
8. y = 3 cot x
10. y =
sec x
4. y = cot x
5. y = sec x
6. y = csc x
csc x
11. y = -8 cot x
12. y = - 0.1 tan x
13. y = -3 csc x
14. y = - 60 sec x
In Exercises 15–24, view at least two cycles of the graphs of the given
functions on a graphing calculator.
15. y = tan 2x
3. y = tan x
3
2
17. y =
1
2
16. y = 2 cot 3x
18. y = 0.4 csc 2x
sec 3x
19. y = 2 cota2x +
p
b
6
20. y = tana3x -
p
b
2
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310
CHAPTER 10 Graphs of the Trigonometric Functions
21. y = 18 csca3x -
p
b
3
23. y = 75 tana0.5x -
22. y = 12 seca2x +
p
b
8
29. A mechanism with two springs is shown in Fig. 10.31, where
point A is restricted to move horizontally. From the law of sines,
we see that b = 1a sin B 2 csc A. Sketch the graph of b as a function of A for a = 4.00 cm and B = p>4.
p
b
4
24. y = 0.5 seca0.2x +
p
b
25
30. A cantilever column of length L will buckle if too large a downward force P is applied d units off centre. The horizontal deflection x (see Fig. 10.32) is x = d1sec1kL2 - 12, where k is a
constant depending on P, and 0 6 kL 6 p>2. For a constant d,
sketch the graph of x as a function of kL.
In Exercises 25 and 26, solve the given problems. In Exercises 27–30,
sketch the appropriate graphs.
25. Write the equation of a secant function with zero displacement, a
period of 4p, and that passes through 10, - 32 .
26. Use a graphing calculator to show that sin x 6 tan x for
0 6 x 6 p>2, although sin x and tan x are nearly equal for the
values near zero.
P
P
27. Near Antarctica, an iceberg with a vertical face 200 m high is
seen from a small boat. At a distance x from the iceberg, the angle
of elevation u of the top of the iceberg can be found from the
equation x = 200 cot u. Sketch x as a function of u.
d
L
28. In a laser experiment, two mirrors move horizontally in equal and
opposite distances from point A. The laser path from and to point
B is shown in Fig. 10.30. From the figure, we see that x = a tan u.
Sketch the graph of x = f1u 2 for a = 5.00 cm.
A
x
x
Fig. 10.32
a
a
b
u
A
B
B
Fig. 10.30
Fig. 10.31
10.5 Applications of the Trigonometric Graphs
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&MFDUSJD$VSSFOU t 'SFRVFODZ
y
When an object moves in a circular path with constant velocity (see Section 8.4), its
projection on a diameter moves with simple harmonic motion. For example, the
shadow of a ball at the end of a string and moving at a constant rate moves with simple
harmonic motion. We now consider this physical concept and some of its applications.
y
E X A M P L E 1 Simple harmonic motion
d
u
t
(R, 0)
In Fig. 10.33, assume that a particle starts at the end of the radius
at 1R, 02 and moves counterclockwise around the circle with
constant angular velocity v. The displacement of the projection
on the y-axis is d and is given by d = R sin u. The displacement
is shown for a few different positions of the end of the radius.
Since u>t = v, or u = vt, we have
Fig. 10.33
d = R sin vt
y(cm)
10
1
p
2
2
3 p
t(s)
(10.3)
as the equation for the displacement of this projection, with
time t as the independent variable.
For the case where R = 10.0 cm and v = 4.00 rad>s, we have
d = 10.0 sin 4.00t
By sketching or viewing the graph of this function, we can
find the displacement d of the projection for a given time t. The
graph is shown in Fig. 10.34.
■
−10
Fig. 10.34
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10.5 Applications of the Trigonometric Graphs
311
In Example 1, note that time is the independent variable. This is motion for which
the object (the end of the projection) remains at the same horizontal position 1x = 02
and moves only vertically according to a sinusoidal function. In the previous sections,
we dealt with functions in which y is a sinusoidal function of the horizontal displacement x. Think of a water wave. At one point of the wave, the motion is only vertical
and sinusoidal with time. At one given time, a picture would indicate a sinusoidal
movement from one horizontal position to the next.
E X A M P L E 2 Simple harmonic motion—application
A windmill is used to pump water. The radius of the blade is 2.5 m, and it is moving
with constant angular velocity. If the vertical displacement of the end of the blade
is timed from the point it is at an angle of 45° 1p>4 rad2 from the horizontal (see
Fig. 10.35(a)), the displacement d is given by
d = 2.5 m
5m
2.
u
d
u
(a)
d = 2.5 sinavt +
If the blade makes an angle of 90° 1p>2 rad2 when t = 0 (see Fig. 10.35(b)), the
displacement d is given by
d = 2.5 sinavt +
(b)
p
b
4
p
b
2
or d = 2.5 cos vt
If timing started at the first maximum for the displacement, the resulting curve for the
displacement would be that of the cosine function.
■
Fig. 10.35
Practice Exercise
1. If the windmill blade in Example 2 starts
at an angle of 135°, what equation gives
its displacement in terms of the cosine
function?
Other examples of simple harmonic motion are (1) the movement of a pendulum bob
through its arc (a very close approximation to simple harmonic motion), (2) the motion of
an object “bobbing” in water, (3) the movement of the end of a vibrating rod (which we
hear as sound), and (4) the displacement of a weight moving up and down on a spring. Other
phenomena that give rise to equations like those for simple harmonic motion are found in the
fields of optics, sound, and electricity. The equations for such phenomena have the same
mathematical form because they result from vibratory movement or motion in a circle.
E X A M P L E 3 Alternating current
A very important use of the trigonometric curves arises in the study of alternating current, which is caused by the motion of a wire passing through a magnetic field. If the
wire is moving in a circular path, with angular velocity v, the current i in the wire at
time t is given by an equation of the form
i = Im sin1 vt + a 2
i(A)
6
1
120
0
1
40
1
60
1
30
t(s)
−6
Fig. 10.36
where Im is the maximum current attainable and a is the phase angle.
The current may be represented by a sinusoidal wave. Given that Im = 6.00 A,
v = 120p rad>s, and a = p>6, we have the equation
i = 6.00 sin 1 120pt +
p
6
2
1
From this equation, note that the amplitude is 6.00 A, the period is 60
s, and the dis1
placement is - 720 s. From these values, we draw the graph as shown in Fig. 10.36.
Since the current takes on both positive and negative values, we conclude that it moves
alternately in one direction and then the other.
■
It is a common practice to express the rate of rotation in terms of the frequency f,
the number of cycles per second, rather than directly in terms of the angular velocity v,
the number of radians per second. The unit for frequency is the hertz (Hz), and
1 Hz = 1 cycle>s. Since there are 2p rad in one cycle, we have
■ The unit Hertz is named for the German
physicist Heinrich Hertz (1857–1894).
v = 2pf
(10.4)
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312
CHAPTER 10 Graphs of the Trigonometric Functions
It is the frequency f that is referred to in electric current, on radio stations, for
musical tones, and so on.
E X A M P L E 4 Frequency—hertz
For the electric current in Example 3, v = 120p rad>s. The corresponding frequency f is
f =
120p
= 60 Hz
2p
This means that 120p rad>s corresponds to 60 cycles>s. This is the standard frequency
used for alternating current.
■
E XE R C I SES 1 0 .5
In Exercises 1 and 2, answer the given questions about the indicated
examples of this section.
1. In Example 1, what is the equation relating d and t if the end of
the radius starts at 10, R 2?
2. In Example 2, if the blade starts at an angle of -45°, what is the
equation relating d and t as (a) a sine function? (b) A cosine
function?
A graphing calculator may be used in the following exercises.
In Exercises 3 and 4, sketch two cycles of the curve of the projection
of Example 1 as a function of time for the given values.
3. R = 2.40 cm, v = 2.00 rad>s
4. R = 1.80 m, f = 0.250 Hz
In Exercises 5 and 6, a point on a cam is 8.30 cm from the centre of
rotation. The cam is rotating with a constant angular velocity, and the
vertical displacement d = 8.30 cm for t = 0 s. See Fig. 10.37. Sketch
two cycles of d as a function of t for the given values.
5. f = 3.20 Hz
8.30 cm
6. v = 3.20 rad>s
Centre
Fig. 10.37
In Exercises 7 and 8, a satellite is orbiting the earth such that its
displacement D north of the equator (or south if D 6 0) is given by
D = A sin1vt + a 2. Sketch two cycles of D as a function of t for the
given values.
7. A = 500 km, v = 3.60 rad>h, a = 0
8. A = 850 km, f = 1.6 * 10
-4
Hz, a = p>3
In Exercises 9 and 10, for an alternating-current circuit in which the
voltage V is given by V = E cos 1vt + a 2, sketch two cycles of the
voltage as a function of time for the given values.
9. E = 170 V, f = 60.0 Hz, a = - p>3
10. E = 80 V, v = 377 rad>s, a = p>2
In Exercises 11 and 12, refer to the wave in the string described in
Exercise 37 of Section 10.3. For a point on the string, the displacement
x
t
y is given by y = A sin 2pa - b. We see that each point on the
T
l
string moves with simple harmonic motion. Sketch two cycles of y as a
function of t for the given values.
11. A = 3.20 cm, T = 0.050 s, l = 40.0 cm, x = 5.00 cm
12. A = 0.750 cm, T = 0.250 s, l = 24.0 cm, x = 20.0 cm
In Exercises 13 and 14, the air pressure within a plastic container
changes above and below the external atmospheric pressure by
p = p0 sin 2pft. Sketch two cycles of p as a function of t for the
given values.
13. p0 = 280 kPa, f = 2.30 Hz
14. p0 = 45.0 kPa, f = 0.450 Hz
In Exercises 15–22, sketch the required curves.
15. The angular displacement u of a certain pendulum bob in
terms of its initial displacement u0 is u = u0 cos vt. If
v = 2.00 rad>s, and u0 = p>30 rad, draw two cycles for the
resulting equation.
16. A study found that, when breathing normally, the increase in volume V (in L) of air in a person’s lungs as a function of the time t
(in s) is V = 0.30 sin 0.50pt. Sketch two cycles.
17. Sketch two cycles of the radio signal V = 0.014 cos 12pft +
p>42 (V in volts, f in hertz, and t in seconds) for a station broadcasting with f = 950 kHz (“95” on the AM radio dial).
18. Sketch two cycles of the acoustical intensity I of the sound wave
for which I = A cos 12pft - a 2, given that t is in seconds,
A = 0.027 W>cm2, f = 240 Hz, and a = 0.80.
19. The rotating beacon of a parked police car is 12 m from a straight
wall. (a) Sketch the graph of the length L of the light beam, where
L = 12 sec pt, for 0 … t … 2.0 s. (b) Which part(s) of the graph
show meaningful values? Explain.
20. The motion of a piston of a car engine approximates simple harmonic motion. Given that the stroke (twice the amplitude) is
0.100 m, the engine runs at 2800 r>min, and the piston starts at
the middle of its stroke, find the equation for the displacement d
as a function of t. Sketch two cycles.
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10.6 Composite Trigonometric Curves
21. The paddle wheel of the S. S. Beaver, Canada’s first steamship,
had a radius of 1.98 m and rotated at 30 r/min when moving at top
speed. Find the equation of motion of the vertical displacement
(from the centre of the wheel) y of the end of a paddle as a function of the time t (in s) if the paddle was initially horizontal.
Sketch two cycles.
313
22. The sinusoidal electromagnetic wave emitted by an antenna in a
cellular phone system has a frequency of 7.5 * 109 Hz and an
amplitude of 0.045 V>m. Find the equation representing the wave
if it starts at the origin. Sketch two cycles.
Answer to Practice Exercise
1. d = 2.5 cos 1vt + p>42
10.6 Composite Trigonometric Curves
Many applications involve functions that in themselves are a combination of two or
more simpler functions. In this section, we discuss methods by which the curve of such
a function can be found by combining values from the simpler functions.
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&RVBUJPOT t -JTTBKPVT'JHVSFT
E X A M P L E 1 Function that is a sum of simpler functions
y
Sketch the graph of y = 2 + sin 2x.
This function is the sum of the simpler functions y1 = 2 and y2 = sin 2x. We may
find values for y by adding 2 to each important value of y2 = sin 2x.
For y2 = sin 2x, the amplitude is 1, and the period is 2p>2 = p. Therefore, we
obtain the values in the following table and sketch the graph in Fig. 10.38.
3
2
1
0
−1
p
_
4
p
_
2
3p
__
p
x
4
Fig. 10.38
x
0
p
4
p
2
3p
4
p
sin 2x
2 + sin 2x
0
2
1
3
0
2
-1
1
0
2
Note that this is a vertical shift of 2 units of the graph of y = sin 2x, in the same way as
discussed on page 106.
■
"%%*5*0/0'03%*/"5&4
Another way to sketch the resulting graph is to first sketch the two simpler curves and
then add the y-values graphically. This method is called addition of ordinates and is
illustrated in the following example.
E X A M P L E 2 Addition of ordinates
y
y = 2 cos x + sin 2x
2
y = 2 cos x
y = sin 2x
0
2p
p
A
C
−2
B
Fig. 10.39
3p
Sketch the graph of y = 2 cos x + sin 2x.
On the same set of coordinate axes, we sketch the curves y = 2 cos x and y = sin 2x.
These are shown as dashed and solid light curves in Fig. 10.39. For various values of x,
we determine the distance above or below the x-axis of each curve and add these distances, noting that those above the axis are positive and those below the axis are
negative. We thereby graphically add the y-values of these curves to get points on
the resulting curve, shown in colour in Fig. 10.39.
At A, add the two lengths (shown side-by-side for clarity) to get the length for y.
At B, both lengths are negative, and the value for y is the sum of these negative values. At C, one is positive and the other negative, and we must subtract the lower
length from the upper one to get the length for y.
x
We combine these lengths for enough x-values to get a good curve. Some points
are easily found. Where one curve crosses the x-axis, its value is zero, and the resulting curve has its point on the other curve. Here, where sin 2x is zero, the points for
the resulting curve lie on the curve of 2 cos x.
We should also add values where each curve is at its maximum or its minimum.
Extra care should be taken for those values of x for which one curve is positive and
the other is negative.
■
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314
CHAPTER 10 Graphs of the Trigonometric Functions
We have seen how a fairly complex curve can be sketched graphically. It is expected
that a graphing calculator (or computer grapher) will generally be used to view most
graphs, particularly ones that are difficult to sketch. A graphing calculator can display
such curves much more easily, and with much greater accuracy. Information about the
amplitude, period, and displacement is useful in choosing window settings. For these
graphs, it is important that the calculator is in radian mode.
4
–1
7
–2
E X A M P L E 3 Graphing calculator addition of ordinates
(a)
y
x
2
y=
2
y=
0
p
x
2
− cos x
x
2p
y = −cos x
(b)
−2
Use a graphing calculator to display the graph of y = 2x - cos x.
Here, we note that the curve is a combination of the straight line y = x>2 and the
trigonometric curve y = cos x. There are several good choices for the window settings,
depending on how much of the curve is to be viewed. To see a little more than one
period of cos x, we can make the following choices:
Xmin =
Xmax =
Ymin =
Ymax =
-1 (to start to left of y-axis)
7 (period of cos x is 2p = 6.3)
-2 (line passes through 10, 02; amplitude of y = cos x is 1)
4 (slope of line is 1>2)
See Fig. 10.40(a). The graphs of y = x>2 - cos x, y = x>2, and y = -cos x are
shown in Fig. 10.40(b).
We show y = -cos x rather than y = cos x because for addition of ordinates, it is
easier to add graphic values than to subtract them. Here we can add y1 = x>2 and
y2 = -cos x in order to get y = x>2 - cos x.
■
Fig. 10.40
E X A M P L E 4 Graphing calculator addition of ordinates
View the graph of y = cos px - 2 sin 2x on a graphing calculator.
The combination of y = cos px and y = 2 sin 2x leads to the following choices for
the window settings:
3
–1
7
–3
Fig. 10.41
Xmin = -1 (to start to the left of the y-axis)
Xmax = 7 (the periods are 2 and p; this shows at least two periods of each)
Ymin = -3, Ymax = 3 (the sum of the amplitudes is 3)
There are many possible choices for Xmin and Xmax to get a good view of the graph
on a calculator. However, since the sum of the amplitudes is 3, note that the curve cannot be below y = -3 or above y = 3.
The graphing calculator view is shown in Fig. 10.41.
This graph can be constructed by using addition of ordinates, although it is difficult
to do so very accurately.
■
-*44"+064'*(63&4
An important application of trigonometric curves is made when they are added at right
angles. The methods for doing this are shown in the following examples.
E X A M P L E 5 Graphing parametric equations
Plot the graph for which the values of x and y are given by the equations y = sin 2pt
and x = 2 cos pt. Equations given in this form, x and y in terms of a third variable, are
called parametric equations.
Since both x and y are in terms of t, by assuming values of t, we find corresponding
values of x and y and use these values to plot the graph. Since the periods of sin 2pt and
2 cos pt are t = 1 and t = 2, respectively, we will use values of t = 0, 1>4, 1>2, 3>4, 1,
and so on. These give us convenient values of 0, p>4, p>2, 3p>4, p, and so on to use in
the table. We plot the points in Fig. 10.42.
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315
10.6 Composite Trigonometric Curves
y
2
6
10
1
3
5
9
1
7
2
Ϫ2
Ϫ1
4
8
Fig. 10.42
■ Lissajous figures are named for the French
physicist Jules Lissajous (1822–1880).
■ See the chapter introduction.
x
1
4
1
2
3
4
t
0
x
y
Point number
2 1.4 0 -1.4
0 1
0 -1
1 2
3
4
1
-2
0
5
5
4
3
2
7
4
2
9
4
-1.4
1
6
0
0
7
1.4
-1
8
2
0
9
1.4
1
10
■
Since x and y are trigonometric functions of a third variable t, and since the
x-axis is at right angles to the y-axis, values of x and y obtained in this way
result in a combination of two trigonometric curves at right angles. Figures
obtained in this way are called Lissajous figures. Note that the Lissajous figure
in Fig. 10.42 is not a function since there are two values of y for each value of x
(except x = -2, 0, 2) in the domain.
In practice, Lissajous figures can be displayed on an oscilloscope by applying an
electric signal between a pair of horizontal plates and another signal between a pair of
vertical plates. These signals are then seen on the screen of the oscilloscope. This type
of screen (a cathode-ray tube) is similar to that used on most older TV sets.
E X A M P L E 6 Graphing Lissajous figures
If a circle is placed on the x-axis and another on the y-axis, we
may represent the coordinates 1x, y2 for the curve of Example 5
by the lengths of the projections (see Example 1 of Section 10.5)
of a point moving around each circle. A careful study of Fig. 10.43
will clarify this. We note that the radius of the circle giving the
x-values is 2 and that the radius of the circle giving the y-values is
1. This is due to the way in which x and y are defined. Also, due to
these definitions, the point revolves around the y-circle twice as
fast as the corresponding point around the x-circle.
y
x = 2 cos pt
3
4
2
5
2
1
2
y = sin 2pt
1
3
5
7
9
4
■ See the chapter introduction.
8
7
6
6
9
8
6
7
3
1
5
9
4
8
Fig. 10.43
x
■ On an oscilloscope, the curve would result when two electric signals are
plotted against each other, one in x and the other in y. The first would have
■
twice the amplitude and half the frequency of the other.
Most graphing calculators can be used to display a curve
defined by parametric equations. It is necessary to use the mode
feature and select parametric equations. Use the manual for the
calculator, as there are some differences as to how this is done on
various calculators.
E X A M P L E 7 Graphing calculator Lissajous figures
Use a graphing calculator to display the graph defined by the parametric equations
x = 2 cos pt and y = sin 2pt. These are the same equations as those used in Examples
5 and 6.
First, select the parametric equation option from the mode feature and enter the parametric equations x1T = 2 cos pt and y1T = sin 2pt. Then make the following window
settings:
1
–2
2
–1
Fig. 10.44
Tmin = 0 (standard default settings, and the usual choice)
Tmax = 2 (the periods are 2 and 1; the longer period is 2)
Tstep = .1047 (standard default setting; curve is smoother with 0.01)
Xmin = -2, Xmax = 2 (smallest and largest possible values of x), Xscl = 1
Ymin = -1, Ymax = 1 (smallest and largest possible values of y), Yscl = 0.5
The calculator graph is shown in Fig. 10.44.
■
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CHAPTER 10 Graphs of the Trigonometric Functions
E XE R C I SES 1 0 .6
In Exercises 1–8, sketch the curves of the given functions by addition
of ordinates.
1. y = 1 + sin x
3. y =
1
3x
5. y =
1 2
10 x
2. y = 3 - 2 cos x
+ sin 2x
4. y = x - sin x
6. y = 14x2 + cos 3x
- sin px
7. y = sin x + cos x
8. y = sin x + sin 2x
In Exercises 9–20, display the graphs of the given functions on a
graphing calculator.
9. y = x3 + 10 sin 2x
10. y =
11. y = sin x - 1.5 sin 2x
1
x2 + 1
- cos px
12. y = cos 3x - 3 sin x
1
2
13. y = 20 cos 2x + 30 sin x
14. y =
15. y = 2 sin x - cos 1.5x
16. y = 8 sin 0.5x - 12 sin x
sin 4x + cos 2x
17. y = sin px - cos 2x
18. y = 2 cos 4x - cos ax 19. y = 2 sina2x -
p
b
4
p
p
b + cos a2x + b
6
3
In Exercises 21–24, plot the Lissajous figures.
23. x = 2 cos 2pt, y = sin pt
22. x = 2 cos t, y = cos 1t + 42
24. x = cos at +
p
b, y = sin 2t
4
In Exercises 25–32, use a graphing calculator to display the Lissajous
figures.
25. x = cos pat +
1
b, y = 2 sin pt
6
26. x = sin2 pt, y = cos pt
27. x = 2 cos 3t, y = cos 2t
28. x = 2 sin pt, y = 3 sin 3pt
29. x = sin1t + 12, y = sin 5t
30. x = 5 cos t, y = 3 sin 5t
31. x = 2 cos pt, y = 3 sin 1 2pt -
p
4
32. x = 1.5 cos 3pt, y = 0.5 cos 5pt
38. The world’s highest tides occur in the Bay of Fundy on the
Atlantic coast of Canada. One day in August 2012, the first low
tide at Minas Basin, Nova Scotia, occurred at 6:21 a.m. The water
level at low tide was 1.8 m; later, at high tide, it was 14.9 m. The
next low tide occurred at 6:45 p.m. Assuming a sinusoidal type of
function, find the height of the water h (in m) as a function of time
t (in hours since midnight). Sketch the graph for one day.
39. The electric current i (in mA) in a certain circuit is given by
i = 0.32 + 0.50 sin t - 0.20 cos 2t, where t is in milliseconds.
Sketch two cycles of i as a function of t.
40. The available solar energy depends on the amount of sunlight,
and the available time in a day for sunlight depends on the time of
the year. An approximate correction factor (in min) to standard
1
1
time is C = 10 sin 29
1n - 802 - 7.5 cos 58
1n - 802, where n
is the number of the day of the year. Sketch C as a function of n.
41. Two signals are seen on an oscilloscope as being at right angles. The
equations for the displacements of these signals are x = 4 cos pt
and y = 2 sin 3pt. Sketch the figure that appears on the oscilloscope.
20. y = 3 cos 2px + sin p2 x
21. x = 3 sin t, y = 2 sin t
37. A normal person with a pulse rate of 60 beats>min has a blood
pressure of “120 over 80.” This means the pressure is oscillating
between a high (systolic) of 120 mm of mercury (shown as mmHg)
and a low (diastolic) of 80 mmHg. Assuming a sinusoidal type of
function, find the pressure p as a function of the time t if the initial
pressure is 120 mmHg. Sketch the graph for the first 5 s.
42. In the study of optics, light is said to be elliptically polarized if
certain optic vibrations are out of phase. These may be represented by Lissajous figures. Determine the Lissajous figure for
two light waves given by w1 = sin vt and w2 = sin 1 vt + p4 2 .
43. In checking electric circuit elements, a square wave such as that
shown in Fig. 10.45 may be displayed on an oscilloscope. Display
px
4
3px
4
sina
b on a graphthe graph of y = 1 + sina b +
p
4
3p
4
ing calculator, and compare it with Fig. 10.45. This equation gives
the first three terms of a Fourier series. As more terms of the
series are added, the approximation to a square wave is better.
y
2
In Exercises 33–44, sketch the appropriate curves. A graphing
calculator may be used.
33. An object oscillating on a spring has a displacement (in m) given
by y = 0.4 sin 4t + 0.3 cos 4t, where t is the time (in s). Sketch
the graph.
34. The voltage V in a certain electric circuit is given by V = 50
sin 50pt + 80 sin 60pt, where t is the time (in s). Sketch the graph.
35. An analysis of the temperature records for Montreal indicates that the
average daily temperature T (in °C) during the year is approximately
T = 6 - 15 cos 3 p6 1x - 0.52 4 , where x is measured in months
(x = 0.5 is Jan. 15, etc.). Sketch the graph of T vs. x for one year.
36. An analysis of data shows that the mean density d (in mg>cm3) of
a calcium compound in the bones of women is given by
d = 139.3 + 48.6 sin10.0674x - 0.2102, where x represents
the ages of women 120 … x … 80 years 2. (A woman is considered to be osteoporotic if d 6 115 mg>cm3.) Sketch the graph.
0
−4
−8
4
8
12
x
Fig. 10.45
44. Another type of display on an oscilloscope may be a sawtooth
wave such as that shown in Fig. 10.46. Display the graph of
3px
1
y = 1 - p82 1 cos px
2 + 9 cos 2 2 on a graphing calculator and
compare it with Fig. 10.46. See Exercise 43.
y
2
−4
−2
0
Fig. 10.46
2
4
x
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317
Review Exercises
E QU A T IONS
CH A PTER 1 0
For the graphs of y = a sin 1 bx + c2
and y = a cos 1 bx + c2
Amplitude = ͉ a ͉
2p
b
Period =
y
c
2p
b
a
c
0
−b
(10.1)
y
2p
b
a
c
b
Displacement = -
−b +
2p
b
x
For each
a > 0, b > 0
−a
y = a sin (bx + c), c > 0
c
c
0
−b +
−b
2p
b
x
−a
y = a sin (bx + c), c < 0
(a)
(b)
1
sin x
1
cos x
1
tan x
Reciprocal relationships
csc x =
Simple harmonic motion
d = R sin vt
(10.3)
Angular velocity and frequency
v = 2pf
(10.4)
In Exercises 1–28, sketch the curves of the given trigonometric functions.
1. y = 23 sin x
2. y = -4 sin x
3. y = -2 cos x
4. y = 2.3 cos 1 -x2
5. y = 2 sin 3x
6. y = 4.5 sin 12x
7. y = 0.4 cos 4x
8. y = 24 cos 6x
1
3x
10. y = 3 sin1 -0.5x2
11. y = sin px
12. y = 36 sin 4px
13. y = 5 cos 1 px
2 2
14. y = -cos 6px
15. y = - 0.5 sin 1 - px
6 2
18. y = 3 sina
19. y = -2 cos 14x + p2
x
p
20. y = 0.8 cos a - b
6
2
17. y = 2 sina3x -
p
b
2
21. y = - sinapx +
23. y = 8 cos a4px 25. y = 0.3 tan 0.5x
27. y = - 13 csc x
cot x =
(10.2)
RE V IEW EXERCISES
CH A PTER 1 0
9. y = 3 cos
sec x =
p
b
6
p
b
2
16. y = 8 sin p4 x
p
x
+ b
2
2
22. y = 250 sin13px - p2
24. y = 3 cos 12px + p2
26. y = 14 sec x
28. y = -5 cot px
In Exercises 29–32, sketch the curves of the given functions by addition of ordinates.
29. y = 2 +
1
2
sin 2x
31. y = sin 2x + 3 cos x
30. y = 12x - cos 13x
32. y = sin 3x + 2 cos 2x
In Exercises 33–40, display the curves of the given functions on a
graphing calculator.
33. y = 2 sin x - cos 2x
34. y = 10 sin 3x - 20 cos x
p
35. y = cos ax + b - 0.4 sin 2x
4
36. y = 2 cos px + cos 12px - p2
37. y =
sin x
x
38. y = 1x sin 0.5x
39. y = sin2 x + cos2 x
1sin2 x = 1sin x2 2 2
What conclusion can be drawn from the graph?
p
p
b - cos ax - b + 1
4
4
What conclusion can be drawn from the graph?
40. y = sinax +
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318
CHAPTER 10 Graphs of the Trigonometric Functions
In Exercises 41–44, give the specific form of the indicated equation
by evaluating a, b, and c through an inspection of the given curve.
42. y = a cos 1bx + c2
(Fig. 10.47)
41. y = a sin1bx + c2
(Fig. 10.47)
43. y = a cos 1bx + c2
(Fig. 10.48)
44. y = a sin1bx + c2
(Fig. 10.48)
y
p
3p
__
x
4
0
1
9
x
Fig. 10.48
Fig. 10.47
In Exercises 45–48, display the Lissajous figures on a graphing
calculator.
45. x = -cos 2pt, y = 2 sin pt
p
b, y = sin t
6
p
47. x = 2 cos a2pt + b, y = cos pt
4
48. x = cos at -
p
p
b, y = cos a2t + b
6
3
In Exercises 49–60, solve the given problems.
49. Display the function y = 2 ͉ sin 0.2px ͉ - ͉ cos 0.4px ͉ on a
graphing calculator.
50. Display the function y = 0.2 ͉ tan 2x ͉ on a graphing calculator.
51. Show that cos 1 x +
52. Show that tan 1 x -
v0
u
Fig. 10.49
Ϫ1
−2
46. x = sinat +
a function of u for v0 = 1000 m>s
and g = 9.8 m>s2. See Fig. 10.49.
v 20 sin 2u
. Sketch R as
g
R
1
0
61. The range R of a rocket is given by R =
y
2
− _4
In Exercises 61–82, sketch the appropriate curves. A graphing calculator may be used.
2 = sin 1 p4 - x 2 on a graphing calculator.
2 = - tan 1 p3 - x 2 on a graphing calculator.
p
4
p
3
62. The blade of a saber saw moves vertically up and down at 18
strokes per second. The vertical displacement y (in cm) is given
by y = 1.2 sin 36pt, where t is in seconds. Sketch at least two
cycles of the graph of y vs. t.
63. The velocity v (in cm>s) of a piston in a certain engine is given
by v = vD cos vt, where v is the angular velocity of the crankshaft in radians per second and t is the time in seconds. Sketch
the graph of v vs. t if the engine is at 3000 r>min and D = 3.6 cm.
64. A light wave for the colour yellow can be represented by the
equation y = A sin 3.4 * 1015 t. With A as a constant, sketch
two cycles of y as a function of t (in s).
65. The electric current i (in A) in a circuit in which there is a fullwave rectifier is i = 10 ͉ sin 120pt ͉. Sketch the graph of
i = f1t2 for 0 … t … 0.05 s. What is the period of the current?
66. A circular disc suspended by a thin wire attached to the centre of
one of its flat faces is twisted through an angle u. Torsion in the
wire tends to turn the disc back in the opposite direction
(thus, the name torsion pendulum is given to this device). The
angular displacement u (in rad) as a function of time t (in s) is
u = u0 cos 1vt + a 2, where u0 is the maximum angular displacement, v is a constant that depends on the properties of the disc and
wire, and a is the phase angle. Sketch the graph of u vs. t if
u0 = 0.100 rad, v = 2.50 rad>s, and a = p>4. See Fig. 10.50.
53. What is the period of the function y = 2 cos 0.5x + sin 3x?
54. What is the period of the function y = sin px + 3 sin 0.25px?
55. Find the function and graph it if it is of the form y = a sin x and
passes through 15p>2, 32.
56. Find the function and graph it if it is of the form y = a cos x and
passes through 14p, - 32.
57. Find the function and graph it if it is of the form y = 3 cos bx
and passes through 1p>3, - 32 and b has the smallest possible
positive value.
58. Find the function and graph it if it is of the form y = 3 sin bx
and passes through 1p>3, 02 and b has the smallest possible
positive value.
59. Find the function and graph it for a function of the form
y = 3 sin1px + c2 that passes through 1 -0.25, 02 and for
which c has the smallest possible positive value.
60. Write the equation of the cosecant function with zero displacement, a period of 2, and that passes through 10.5, 42.
u
Fig. 10.50
67. The vertical displacement y of a point at the end of a propeller
blade of a small boat is y = 14.0 sin 40.0pt. Sketch two cycles
of y (in cm) as a function of t (in s).
68. In optics, two waves are said to interfere destructively if, when
they both pass through a medium, the amplitude of the resulting
wave is zero. Sketch the graph of y = sin x + cos 1x + p>22
and explain whether or not it would represent destructive interference of two waves.
69. The vertical displacement y (in dm) of a buoy floating in water
is given by y = 3.0 cos 0.2t + 1.0 sin 0.4t, where t is in seconds. Sketch the graph of y as a function of t for the first 40 s.