Math 115 Spring 11
Written Homework 13 Solutions
1. Suppose θ is an angle in standard position whose terminal side falls in the third quadrant
and 0 ≤ θ ≤ 2π.
(a) If the radian measure of θ is an integer, what is its value?
Solution: The angles in the third quadrant are between the quadrantal angles π = 3.14159 . . .
and
3π
2
= 4.7123 . . .. Thus, the only integer value between these is 4. So, θ = 4.
(b) If the terminal side of the angle had fallen in any other quadrant, could you still
answer the question with a single value? Explain.
Solution: If the terminal side had fallen in quadrant I, we would be able to answer this
question because those angles fall between 0 and
π
2
= 1.5707 . . .. Hence, in quadrant I, θ
would be equal to 1. In quadrants II and IV, there would be two possible values for θ; 2 and
3 in II and 5 and 6 in IV.
2.
(a) In a circle of radius r, an arc of length 10 is subtended by an angle of 3 radians.
What is the exact radius of the circle?
Solution: The radian was defined in such a way that the arc length, s, is equal to the radius
times the angle, when the angle is in radians. i.e.
s = rθ.
Here we have that s = 10 and θ = 3. Thus,
10 = r · 3 ↔ r =
10
.
3
(b) In a circle of radius r = 6, what is the exact length of the arc subtended by an angle
of 210◦ ?
Solution: Before we can apply the arc length formula, we must have an angle in radians.
Thus, here we must convert the angle from degree measure into radian measure.
210◦
Now we have r = 6 and θ =
π 210π
21π
7π
=
=
=
.
◦
180
180
18
6
7π
.
6
s = rθ = 6
7π
6
= 7π.
The length of the are subtended by the angle 210◦ on the circle is 7π.
3. For each angle θ given below, state the quadrant in which it lies, sketch θ in standard
position, and find the reference angle, θr .
37π
(a) θ =
6
π
37π
= 6π + . 6π is 3 full revolutions. Thus, the terminal side of θ will end
6
6
π
up in the first quadrant and the reference angle will be θr = .
6
Solution: θ =
(b) θ = −
11π
4
11π
3π
= −2π −
. −2π is 1 full revolution clockwise. Thus, the terminal
4
4
π
side of θ will end up in the third quadrant and the reference angle will be θr = .
4
Solution: θ = −
(c) θ = 5.5
Solution: θ = 5.5 is between the quadrantal angles
3π
2
= 4.7123 . . . and 2π = 6.2831 . . . .
Thus, the terminal side of θ will end up in the fourth quadrant and the reference angle will
be θr = 2π − 5.5 = 0.7831 . . .
4. Suppose θ is an angle in standard position whose terminal side intersects the unit circle
1
at the point (x, y). If y = , what are the possible values of cos θ? If you know that cos θ is
3
positive, in which quadrant does the terminal side of θ lie?
Solution: The equation of the unit circle is x2 +y 2 = 1. If we plug y = − 13 into the equation,
we get two possible values of x and thus two possible values of cos θ:
x2 + y 2 = 1
1
x2 + ( ) 2 = 1
3
1
x2 +
= 1
9
8
x2 =
9√
x = ±
√
The possible values are cos θ = −
8
3
8
3
√
and cos θ =
8
.
3
If we know that cos θ is positive,
then the angle must lie in quadrant I because both the x and the y coordinate of the corresponding point on the unit circle are positive.
5. Find the exact
value of each of the following.
9π
(a) cos
2
9π
8π + π
π
π
=
= 4π + . Thus the angle is coterminal to . Thus,
2
2
2
2
π 9π
cos
= cos
=0
2
2
4π
(b) sin
3
Solution:
4π
Solution: The terminal side of
ends up in quadrant 3 so it’s sine value will be negative.
3
π
The reference angle here is . Using either a special right triangle or the unit circle, we find
3
that
√
π 4π
3
sin
= − sin
=−
3
3
2
11π
(c) tan
6
11π
ends up in quadrant 4 so it’s tangent value will be
π6
negative. The reference angle is . Using either a special right triangle or the unit circle,
6
we find that
1
π sin π6
11π
1
2
√ = −√
tan
=−
= − tan
=
−
π
3
6
6
cos 6
3
2
5π
(d) csc −
4
Solution: The terminal side of
5π
π
will fall in quadrant 2 with a reference angle of θr = . Using
4
4
either a special right triangle or the unit circle, we find that
π √
5π
1
1
=
=
=
2
csc −
= csc
√1
4
4
sin π4
2
48π
(e) sec
3
Solution: The angle −
48π
= 16π is a quadrantal angle that is coterminal to 2π. Thus,
3
48π
1
1
1
sec
= sec(16π) =
=
= =1
3
cos 16π
cos 2π
1
Solution: The angle
6. As we did with the sine function in lecture, state the properties of the cosine function.
Solution: The domain is x ∈ R.
(b) State the range of f (x) := cos x.
Solution: The range is f (x) ∈ [−1, 1].
(c) Determine all values of x where f (x) is zero.
Solution: The zeros occur when x =
π
+ kπ where k is any integer.
2
(d) Determine the y-intercept for the graph y = cos x.
Solution: The y-intercept occurs at (0, 1).
8. For each of the following, use the basic sine and cosine graphs and your knowledge of
transformations to sketch the graph of the function through one full period. Identify the
amplitude and period.
(a) f (x) := 2.5 sin
2π
x
5
Solution: Start with one period of the basic sine function.
y
3
2
1
0
Π
2
Π
3Π
2
x
2Π
-1
-2
-3
Then determine where the argument
2π
x moves the zeros of the basic sine function. The
5
zeros of the shown period occur when
2π
2π
2π
x = 0, x = π, and
x = 2π.
5
5
5
Solving these three equations, we see the zeros are translated to
5
x = 0, x = , and x = 5.
2
This corresponds to the graph y = sin 2π
x
5
y
3
2
1
0
Π
2
5
2
Π
3Π
5
2
x
2Π
-1
-2
-3
We note that the period P of f (x) is P = 5 − 0 = 5.
Now applying the vertical stretch caused by A = 2.5 yields the final graph y = 2.5 sin
2π
x
5
y
3
2
1
0
Π
2
5
2
Π
3Π
5
2
x
2Π
-1
-2
-3
By definition, the amplitude of f (x) is |A| = 2.5.
(b) f (x) := −10 sin
hπ
4
(x + 4)
i
Solution: Again start with one period of the basic sine function.
y
10
5
0
Π
2
Π
3Π
2
x
2Π
-5
-10
π
(x + 4) moves the zeros of the basic sine function.
4
The zeros of the shown period occur when
Then determine where the argument
π
π
π
(x + 4) = 0, (x + 4) = π, and (x + 4) = 2π.
4
4
4
Solving these three equations, we see the zeros are translated to
x = −4, x = 0, and x = −4.
This corresponds to the graph y = sin
π
4
(x + 4)
y
10
5
Π
2
-4
Π 4
3Π
2
x
2Π
-5
-10
We note that the period P of f (x) is P = 4 − (−4) = 8.
Here A = −10 is less than 0. We will apply this in two-steps. By definition, the amplitude
of f (x) is |A| = 10. We first apply the the stretch factor of 10. This corresponds to the
graph y = 10 sin π4 (x + 4)
y
10
5
Π
2
-4
Π 4
3Π
2
x
2Π
-5
-10
Now recall the negative sign reflects the previous graph about the x-axis. This yields the
final graph y = −10 sin π4 (x + 4) .
y
10
5
Π
2
-4
-5
-10
Π 4
3Π
2
x
2Π
π
+3
(c) h(x) := 6 cos x −
4
Solution: Starting with one period of the basic cosine function. Note, that for ease of
graphing, I choose to work with three consecutive zeros (as we do with the sine function).
This makes graphing significantly easier as it is easier to keep track of the max and minimum
values. (They are always located halfway between the zeros.)
y
10
5
- Π2
Π
2
Π
x
3Π
2
-5
Then determine where the argument x −
π
moves the zeros of the basic cosine function.
4
The zeros of the shown period occur when
x−
π
π
π
π
π
3π
= − , x − = , and x − =
.
4
2
4
2
4
2
Solving these three equations, we see the zeros are translated to
π
3π
7π
x = − ,x =
, and x =
.
4
4
4
This corresponds to the graph y = cos x − π4 .
y
10
5
- Π4
Π
4
3Π
4
5Π
4
7Π
4
x
-5
-10
7π π − −
= 2π.
4
4
Now applying the vertical stretch cause by A = 6 yields the graph y = 6 cos x − π4 .
We note that the period P of f (x) is P =
y
10
5
- Π4
Π
4
3Π
4
5Π
4
7Π
4
x
-5
-10
By definition, the amplitude of f (x) is |A| = 6.
Lastly, we apply the vertical shift caused by D = 3. This results in the final graph
y = 6 cos x − π4 + 3.
y
10
5
- Π4
Π
4
3Π
4
5Π
4
7Π
4
x
-5
-10
Remember - there is more than correct answer for each of these graphing questions. ANY
complete cycle is correct. Some of you may have used a different method on part (c) and
ended up with the following graph:
y
10
5
0
-5
-10
Π
4
3Π
4
5Π
4
7Π
4
9Π
4
x
9. Find an equation that produces the given graph in each of the following forms.
(a) y = A sin[B(t + C)] + D
(b) y = A cos[B(t + C)] + D
y
12
10
8
6
4
2
0
90
180
270
360
x
Solution: Part (a)
Solution: As always, start with one period of the basic sine function.
y
6
4
2
0
Π
x
2Π
-2
-4
-6
On the given graph, we see the maximums of our function occur at x = 45 and x = 405.
Hence, one period occurs between x = 45 and x = 405. Hence P = 405 − 45 = 360. Using
2π
the fact that P =
we determine that
B
2π
π
= 360 or B =
.
B
180
π
The graph of one period of y = sin( 180
x) is:
y
6
4
2
0
45 90
180
270
360
x
-2
-4
-6
By definition, the amplitude is the distance between the average value and the maximum
max value − min value
.
(or minimum) value. Here, it is easier to use the fact that the amplitude is
2
From inspection we see that the max value is 11 and the min value is 3. Hence, the amplitude
11 − 3
is
= 4. We will define A = 4.
2
π
x) is
The graph of y = 4 sin( 180
y
6
4
2
0
45 90
180
270
360
x
-2
-4
-6
Recall that the vertical shift D is determined by either (max value − amplitude) or (min
π
value + amplitude). In either case, we see that D = 7. The graph of y = 4 sin( 180
x) + 7 is
y
12
10
8
6
4
2
0
0 45 90
180
270
360
x
The last thing we need to now accomplish is the horizontal shift require to match up
with the given graph. Notice that on our most recent graph, the maximum is located
at (90, 11). We see that we can use a shift backwards (to the left) to (45, 11). As in
our solution to problem 6, we see that the transformation f (x) := x + 45 will accomplish
π
this. Hence, a function whose graph matches the given picture is y = 4 sin( 180
f (x)) + 7 =
π
4 sin 180 (x + 45) + 7
y
12
10
8
6
4
2
0
0 45 90
180
270
360
x
Solution: Part (b)
The discussion here is identical to the solution to part (a), until it comes time to apply
π
the horizontal shift. Again, A = 4, B = π/180, and D = 7. The graph of y = 4 cos( 180
x) + 7
is
y
12
10
8
6
4
2
0
0 45 90
180
270
360
x
The last thing we need to now accomplish is the horizontal shift require to match up
with the given graph. Notice that on our most recent graph, the maximum is located at
(0, 1). We see that we can use a shift forwards (to the right) to (45, 11). Here g(x) :=
x − 45 will accomplish this. Hence, a function whose graph matches the given picture is
π
π
y = 4 cos( 180
g(x)) + 7 = 4 cos 180
(x − 45) + 7
y
12
10
8
6
4
2
0
0 45 90
180
270
360
x
10.
(a) Find a function, f (x) so that the graphs of y = cos x and y = sin(f (x)) are identical.
Solution: Consider the graph of the basic sine function and the basic cosine function.
y
y
3
3
2
2
1
1
Π 7Π 5Π 3Π
- 92- 2 2 2
Π 3Π 5Π 7Π 9Π
2 2 2 2 2
x
Π 7Π 5Π 3Π
- 92- 2 2 2
-1
-1
-2
-2
-3
-3
Figure 1: y = sin x
Π 3Π 5Π 7Π 9Π
2 2 2 2 2
Figure 2: y = cos x
Geometrically, these are the same (infinitely long) waves, but they are out of phase. We
can use a horizontal shift to move the cosine wave so that it appears identical to the sine
wave.
x
To match up the curves, we need to focus on any single point and move it. I choose
to work a point on y = sin x where a maximum occurs: (π/2, 1). If I want to pull this
point back to the coincide with (0, 1) on the cosine curve, we simply need argument function
f (x) of sin(f (x)) to cause a horizontal shift to the left of π/2. Using our knowledge of the
π
transformations of graphs, we know that f (x) := x + will cause this shift.
2
y
3
2
1
x
Π 3Π 5Π 7Π 9Π
2 2 2 2 2
Π 7Π 5Π 3Π
- 92- 2 2 2
-1
-2
-3
Figure 3: y = sin x +
π
2
Notice that plotting the curve y = cos x and y = sin x +
π
2
on the same graph, we see
that they are the same curve.
y
3
2
1
Π 7Π 5Π 3Π
- 92- 2 2 2
Π 3Π 5Π 7Π 9Π
2 2 2 2 2
x
-1
-2
-3
Figure 4: y = sin (f (x)) (dashed red curve) and y = cos x (blue curve)
(b) Is there only one correct answer to this question? Explain your answer.
Solution: There are an infinite number or answers to this question. We can start by choosing any of the infinite number of maximums on the graph on y = sin x. The maximum
π
values of the basic sine curve have the form
+ 2πk, 1 for any integer value of k. Then
2
for any specific maximum, there is an associated integer K. For fixed K, the function
i
h
π
+ 2πK will have a graph identical to the graph of y = cos x.
f (x) := sin x +
2
12. Evaluate
each
expression.
5π
(a) arcsin sin −
6
5π
5π
Solution: Begin by evaluating, sin −
ends up in
. The terminal side of the angle −
6
6
π
quadrant 3 and has a reference angle of . Using a special right triangle or the unit circle,
6
5π
1
we find that sin −
=− .
6
2
5π
1
Thus, arcsin sin −
= arcsin − . By the definition of arcsine, we know that
6
2
π
1
π
1
−→ sin θ = −
AND − ≤ θ ≤
θ = arcsin −
2
2
2
2
1
π
Thus, arcsin −
=− .
2
6
Be careful: the inverse sine and sine functions do not ”undo” each other here because
5π
−
is not within the range of the inverse sine function.
6
5
(b) cos arccos −
12
5
5
5
Solution: cos arccos −
= −
since −
is within the domain of the arccosine
12
12
12
5
function. This could also be solved by drawing a right triangle. Let θ = arccos −
.
12
5
This means θ is an angle in the interval [0, π] whose cosine is − . Thus, θ is in quadrant
12
5
II. Drawing a right triangle in quadrant II, it should be clear that cos θ = − .
12
h π i
(c) cos−1 sin −
3
π
π
Solution: Begin by evaluating, sin − . The angle − is in quadrant IV. Using a special
3
3 √
h π i
π
3
= −
. Thus, cos−1 sin −
=
right triangle or the unit circle, we find sin −
3
2
3
√ !
3
cos−1 −
. By the definition of arccosine, we know that
2
θ = cos−1
Thus, cos
−1
h
√ !
3
−
−→ cos θ =
2
√ !
3
−
AND0 ≤ θ ≤ π
2
π i 5π
sin −
=
.
3
6
1
(d) sec arcsin
2
1
1
π
π
1
π
Solution: θ = arcsin
−→ sin θ = AND − ≤ θ ≤ . So arcsin
= and
2
2
2
2
2
6
π 1
1
2
1
√ = √
sec arcsin
=
= sec
π =
3
2
6
cos 6
3
2
−1
(e) tan cos
11
−
61
11
11
AND0 ≤ θ ≤ π. Thus, θ is in quadrant II.
Solution: θ = cos
−→ cos θ = −
−
61
61
Drawing a right triangle in quadrant II, you can use the Pythagorean Theorem to show that
−1
the opposite side of the triangle is 60. Thus,
11
60
−1
= tan θ = −
tan cos
−
61
11
13. Explain why the following expression arccos sec
3π
4
is not defined.
3π
Solution: Begin by evaluating sec
:
4
√
3π
1
1
sec
=
2
3π = √1 =
4
cos 4
2
But
√
2 is not in the domain of the inverse cosine function because it is outside of the interval
[−1, 1].
14. Use a right triangleto simplify
each function. State the domain of each function.
x
+
3
(a) f (x) := cos sin−1
4
x+3
Solution: Let θ = sin
. By the definition of arcsine, we know that sin θ =
4
Sketch this angle within a right triangle:
−1
Use the Pythagorean Theorem to find the value of b:
b2 + (x + 3)2 = 42
b2 + (x + 3)2 = 16
b2 = 16 − (x + 3)2
p
b =
16 − (x + 3)2
p
16 − (x + 3)2
x
+
3
.
Thus, f (x) := cos sin−1
= cos θ =
4
4
The domain of this function is found by considering the domain of sin−1 x:
Recall that for sin−1 (x), −1 ≤ x ≤ 1. Here, this requires
−1 ≤
x+3
≤ 1.
4
−4 ≤ x + 3 ≤ 4.
−7 ≤ x ≤ 1.
The domain of f (x) is [−7, 1].
x+3
4
.
(b) g(x) := sin(arccos(4x)) + cos(arcsin(5x))
Solution:
Let θ = arccos 4x and let β = arcsin(5x). The right triangles associated with these
equalities are:
As in (a), use the Pythagorean Theorem to find the third side of each triangle:
a=
Thus,
√
sin(θ) + cos(β) =
√
1 − 16x2 and b =
1 − 16x2
+
1
√
√
1 − 25x2 .
√
1 − 25x2 √
= 1 − 16x2 + 1 − 25x2 .
1
For the domain of g, we consider the domain of the arcsine and arccosine function individually. Here, we require
−1 ≤ 4x ≤ 1 and − 1 ≤ 5x ≤ 1.
Equivalently, the conditions
1
1
1
1
− ≤ x ≤ and − ≤ x ≤
4
4
5
5
must both be satisfied. Combining these restrictions, we see the domain must be the smaller
of the two intervals. Hence, the domain of g is − 15 ≤ x ≤ 15 .