MA 108-001 Test 2 Version B Solutions
For the graphs of y = A sin(ωx − φ) + B or y = A cos(ωx − φ) + B, where ω > 0:
• The period is given by the formula
2π
ω
• The phase shift is given by the formula
φ
.
ω
The period of a graph of the form A cot(ωx) + B or A tan(ωx) + B is given by
π
ω
Problem 1(12 points)
π
Graph one fundamental period of the function y = −2 sin
x + π . Label the x and
2
y coordinates of the five key points with exact values.
Amplitude: 2, Vertical Shift: None. Period :
2
2π
π = 2π π = 4.
2
π
π
x + π = −2 sin
x − (−π) . So our phase shift is negative. We have
Now −2 sin
2
2
π
2
φ = − π = −π = −2
π
2
2π
Without the phase shift our fundamental interval would be 0,
= [0, 4]. But we
ω
have a phase shift of −2, so our fundamental interval is
φ φ 2π
, +
= [−2, 2]
ω ω
ω
The total length of [−2, 2] is four units, so when we divide [−2, 2] into four subintervals
of equal length, each subinterval will have width 1:
[−2, −1],
[−1, 0],
[0, 1],
[1, 2]
So the x-coordinates for our five key points are -2, -1, 0, 1 and 2.
π
x
y(x) = −2 sin
x+π
2
π
−2
−2 sin
(−2) + π = −2 sin(0) = −2(0) = 0
2
π
π (−1) + π = −2 sin
−1 −2 sin
= −2(1) = −2
2
2
π
0
−2 sin
(0) + π = −2 sin (π) = −2(0) = 0
2
π
3π
1 −2 sin
(1) + π = −2 sin
= −2(−1) = 2
2
2
π
2
−2 sin
(2) + π = −2 sin (2π) = −2(0) = 0
2
So the key points you must plot are
(−2, 0), (−1, −2), (0, 0), (1, 2), (2, 0)
For full credit you must label the ordered pairs and connect them with an upside down
sine curve of amplitude 2, as shown in the picture.
Problem 2 (12 points) Graph one fundamental period of the function y = 2 cot(4x)+1.
Label the asymptotes and x, y coordinates of the three key points with exact values.
This exact question appeared on the practice exam, and I did graph this in class on
Monday, but I’ll write out the solution again here. The fundamental period for the
regular un-transformed cot x graph is [0,h π], and
since the period of our new function
π
π
πi h πi
is
= , we’ll graph over the interval 0,
= 0, . Divide this new interval into
ω
4
ω
4
four subintervals of equal length:
h
0,
π i h π π i π 3π
3π π
,
,
,
,
,
,
16
16 8
8 16
16 4
The regular cot x function has asymptotes at x = 0 and x = π. Ours has asymptotes
π
at either end of our fundamental interval, namely x = 0 and x = . Then we need to
4
find the y-values for the three key points:
x
π
16
π
8
3π
16
y(x) = 2 cot(4x) + 1
π
π 2 cot(4(
)) + 1 = 2 cot
+ 1 = 2(1) + 1 = 3
16
4
π 2 cot(4(
)) + 1 = 2 cot (0) + 1 = 2(0) + 1 = 1
8
3π
3π
2 cot(4(
)) + 1 = 2 cot
+ 1 = 2(−1) + 1 = −1
16
4
Plot the three points, label with x and y coordinates, and label the two asymptotes.
Connect the three points with a cotangent-shaped curve that lives between the asymptotes, as shown in the picture.
Other comments: the phrase Amplitude = 2 does not mean anything here. Cotangent
and tangent graphs don’t have amplitude; they have vertical stretch.
Problem 3(6 points)
Solve the equation 4 sin−1 (3x) = −π for x. Your answer should be in exact form.
First
4 sin−1 (3x) = −π ⇐⇒ sin−1 (3x) = −
π
4
Then
sin−1 (3x) = −
π
π
1
⇐⇒ sin(sin−1 (3x)) = sin −
⇐⇒ 3x = − √
4
4
2
So finally divide both sides by 3 to obtain
1
1
1
−√
x=
=− √
3
2
3 2
Problem 4 (12 points)
(a) State the domain and range of sin−1 (x), cos−1 (x) and tan−1 (x).
Part (a) appears in the course summary for 2/13-2/15 (and the book 8.1)
(b) State the domain and range of sec−1 (x), csc−1 (x), and cot−1 (x).
Part (b) appears in the course summary for 2/17 (and the book 8.2)
Problem 5(8 points)
Rewrite the expression tan(cos−1 (u)) as an algebraic expression in the variable u.
Now cos−1 (u) is an angle θ in [0, π] such cos θ = u. We can draw a reference triangle
containing θ such that u is the value of the side adjacent to θ and 1 is the value of
the hypoteneuse. We need an expression for the side b that is opposite θ. By the
Pythagorean Theorem
p
u2 + b2 = 12 ⇐⇒ b2 = 1 − u2 ⇐⇒ b = ± 1 − u2 .
Now θ ∈ [0, π] means√the terminal side of θ lives is in the top half of the unit circle and
b is positive. So b = 1 − u2 . Finally we know that
√
b
1 − u2
tan θ = =
a
u
Problem 6 (30 points) Find the exact value of each expression
(a) tan(tan−1 (100))
π π
The expression tan−1 (−1) represents an angle θ ∈ − ,
such that tan θ = 100.
2 2
We know that such an angle exists because the domain of tan−1 x is (−∞, ∞).
So tan(tan−1 (100)) = tan(θ) = 100.
−1
(b) sin
1
−
2
h π πi
1
The expression sin
−
represents an angle θ ∈ − ,
such that sin θ =
2
2 2
π π
1
1
and so the reference angle for θ must be
.
− . We know that sin
=
2
6
2
6
π
Since sin θ < 0, θ must be in Quadrant IV. So θ = −
6
13π
(c) cos−1 cos
7
−1
13π
The key to this question is to recognize is the find the angle θ =
cos
7
13π
13π
in the interval [0, π] such that cos(θ) = cos
. In particular, θ and
will
7
7
have the
reference angle. We are looking for an angle, not the exact value
same
13π
(which we can’t calculate anyways).
of cos
7
14π
13π
Note that 2π =
and so the terminal side of
is in Quadrant IV. Therefore
7
7
13π
cos
> 0 and cos θ > 0 must also hold. Recall our choices for where θ is
7
are Quadrants I and II. So θ is in Quadrant I. The angle in Quadrant I with the
π
13π
is .
same reference angle as
7
7
cos−1
−1 1
(d) sec sin
3
−1
The expression sin
h π πi
1
represents an angle θ ∈ − ,
(Quadrants I and IV)
3
2 2
1
such that sin θ = . Furthermore, since sin θ > 0, we know that the terminal side
3
of θ is in Quadrant I. Draw a reference triangle for θ. We know that b (the side
across from θ) is labeled with value 1, and the hypoteneuse is labeled 3. We want
r
sec θ = so we need to calculate a. By the Pythagorean Theorem
a
√
a2 + 12 = 32 ⇐⇒ a2 = 9 − 1 ⇐⇒ a = ± 8.
r
3
Since θ is in Quadrant I, we know that a > 0 so sec θ = = √
a
8
4π
(e) cos−1 sin
3
4π
4π
π
We know what sin
is because
has reference angle and the terminal side
3
3
3
4π
of
is in Quadrant III. So
3
√
4π
π
3
sin
= − sin = −
3
3
2
So we know that θ = cos−1
√
cos θ = −
√ !
3
. Equivalently θ is an angle in [0, π] such that
−
2
3
2 .
So θ is in Quadrant II, where the cosine function is negative, and
π
5π
has reference angle . The unique angle satisfying these two conditions is
.
6
6
(f) csc(tan−1 (−2))
π π
b
First, tan−1 (−2) is an angle θ in θ ∈ − ,
such that tan θ = −2 = . Since
2 2
a
tan θ < 0, θ must be in Quadrant IV. Draw a reference triangle for θ. Since θ is
−2
b
. This tells us to label the
in Quadrant IV, a > 0, and we writetan θ = =
a
1
side opposite θ with -2 and thepside adjacent to
√ θ with 1. By the Pythagorean
Theorem, the hypoteneuse r = 12 + (−2)2 = 5. So
√
r
5
−1
csc(tan (−2)) = csc θ = =
b
−2
Problem 7(12 points)
The plankton population, measured millions of specimens per liter, varies periodically
with the seasons in Apalachee Bay. In 2011, the maximum population of 10 million
specimens per liter was recorded September 7, the 250th day of the year, and the
minimum population of 2 million specimens per liter was recorded March 8, the 67th
day of the year. Note: a decimal approximation for the phase shift φ is acceptable.
(a) (9 points) Find a function of the form y = A sin(ωx − φ) + B that models the
population in units of millions of plankton per liter (same units as the problem
statement uses) as a function of time.
10 − 2
10 + 2
We have A =
= 4, B =
= 6. The period T is 365 days. We use
2
2
the period to solve for ω:
2π
2π
=
.
ω=
T
365
To find the phase shift, we use the fact that the maximum population value occurs
at x = 250. Then we set
π
2π
2π
π
=
(250) − φ ⇐⇒ φ =
(250) − ≈ 2.73
2
365
365
2
So our function is
y = 4 sin
2π
x − 2.73 + 6
365
(b) Use your function to estimate the population density per liter on April 9, the
100th day of the year.
For (b) we just need to evaluate our function at x = 100.
y(100) = 4 sin
2π
(100) − 2.73 + 6 ≈ 4 sin (−1.01) + 6 = 4(−0.85) + 6 = 2.6
365
So, we estimate there are about 2.6 million plankton per liter on April 9.
Problem 8 (8 points) For (a)−(d) write the Roman numeral for the graph that matches
the equation.
(a)
1
csc (2x − π)
2
(c)
− sec(3x)
I
II
(b)
(d)
I
II
III
IV
2 sec(3x)
csc(2x) + 1
III
IV
1
and the only graph with this amplitude is I.
2
Then (b) has amplitude 2 and the only graph with amplitude 2 is III. To sort out the
other two, we will calculate values for the two functions. For c, note that − sec(3(0)) =
1
1
−
= − = −1. However, for (d), there is an asymptote at x = 0 because if
cos(3(0))
1
Explanation: First, (a) has amplitude
we try to plug in x = 0 this is what happens:
csc(2(0)) + 1 =
1
1
=
+1
sin(2(0))
sin(0)
which is not defined. Since graph II has y-value of -1 at x = 0, and graph IV has an
asymptote at x = 0, we choose graph II for (c) and graph IV for (d). If you’re still not
convinced, observe that graph IV has a vertical shift of 1, and the only equation with
any vertical shift is IV.