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Measuring Heaven
A Companion Text
for
Precalculus by Dugopolski
&
Precalculus by Ratti & Mcwaters
1
Angles
In this lecture, we review essential geometry and cover some very basic vocabulary. To
start, we recall the Pythagorean Theorem, which states that for any right triangle the sum of the
squares of the lengths of the legs equals the square of the length of the hypotenuse with the
hypotenuse being the side opposite the right angle (the longest side). .
Let
XYZ be a right triangle as shown below.
Y
z
x
Z
y
X
Then, x 2  y 2  z 2 .
In particular, a right triangle with two 45-degree angles has two legs with equal length. Hence,
we acquire the 45º-45º rule below.
x2  y 2  z 2
x2  x2  z 2
2 x2  z 2
z2
2
z
z 2
x
or
2
2
x2 
Let XYZ be a right triangle. Let z represent the length of the hypotenuse.
Let x and y represent the length of the legs. If one of the interior angles
equals 45º, then x  y  z 2 .
Another common rule is the 30º-60º rule stated below.
Let XYZ be a right triangle. Let z represent the length of the hypotenuse.
Let x represent the length of the shortest side, and let y represent the length
of the other leg. If one of the interior angles equals 30º, then the following
equalities
Then, x 2 hold.
y2  z2 .
z
z 3
x  , y  x 3 , y 
2
2
Then, x 2  y 2  z 2 .
2
Now, we turn to some basic vocabulary concerning angles and their measure. A central
angle is an angle with the center of a circle as its vertex. The angle in the figure below is a
central angle.
We denote this angle as AOC or COA or simply O . The measure of a central angle equals
the measure of its intercepted arc. In other words, if a central angle cuts away N of a circle
then the angle measures N . Thus, the degree measure of O equals the degree measure of
AC .
An angle in standard position has its initial side on the positive x-axis and its vertex at the
origin. PQR pictured below is in standard position.
A quadrantal angle is an angle in standard position with its terminal side on the x or y-axis.
Coterminal angles have a common terminal side when in standard position.
We commonly measure angles in degrees and radians. We have discussed degree measure
above when we defined a central angle, but the box below gives the formal definition of the
degree measure of an angle.
The degree measure of an angle is the number of degrees in the intercepted arc
of a circle centered at its vertex. The degree measure is positive if the rotation is
counterclockwise and negative if the rotation is clockwise.
Using this definition, we can more formally define coterminal angles as below.
Let angles A and B be coterminal. Then,  k 
 m  B   m  A  k  360
3
For example, consider two angles, one with degree-measure 549 and one with degreemeasure 171 . Let m  A  549 and m  B   171 , then solve for k in the equation below.
171  549  k  360
171  549  k  360
720 k  360

360
360
2k
Since k is an integer, then the two angles are coterminal in standard position. If k had not been
an integer, then the angles would not be coterminal in standard position.
The degree is a unit of angle measure used in practical applications such as surveying and
navigation. Radian measure is another way to measure angles more common to scientific work
and advanced mathematics. The radian measure of a central angle equals the ratio of the arc
length intercepted by the angle to the radius of the circle. Formally, we define radian measure
below.
The radian measure of an angle in standard position equals the length of
the intercepted arc on any circle centered at the origin divided by the
circle’s radius. Let θ denote the radian measure of an angle in standard
position. Let s denote the intercepted arc of a circle centered at the origin
with radius r. Then,
s
θ .
r
If the rotation is counterclockwise the radian measure is positive;
otherwise, the radian measure is negative.
On the unit circle, the radian measure of an angle in standard position equals the length of the
intercepted arc (the unit circle is a circle centered at the origin with a radius of one unit). In the
figure below, the radian measure of angle  is s.
4
Since the circumference of a circle is 2 r , the circumference of the unit circle is 2 . Hence, if
we rotate the terminal side 180 (one-half of a full revolution), the measure of the resulting angle
is  . We use this equality for conversion from degrees to radians or from radians to degrees.
For example, to convert m  36 to radians, we solve the proportion below.
m


36 180
m

5
 0.6283
Thus, an angle measuring 36 degrees measures  5 radians, which is approximately 0.6283
radians.
Radian measure can be used to calculate the length of an arc intercepted by a central angle.
Let  represent the radian measure of a central angle inscribed in a circle with radius r. Let s
represent the length of the arc intercepted by the angle. The radian measure of the central angle
represents a fraction of the circumference of the circle, namely,  2 . Multiplying this fraction
by the full circumference gives the arc length as below.

 2 r
2
s  r
s
Hence, we have the Arc Length Theorem stated below.
Arc Length Theorem: Let s represent the length
of an arc intercepted by a central angle of 
radians on a circle of radius r. Then, s   r .
Using the theorem below, we can calculate the distance from point P to the North Pole given the
latitude of point P. The latitude or parallel of a point on the earth’s surface in the northern
hemisphere equals the complement of the measure of the central angle subtended by the arc
extending from the point to the North Pole. In other words, if point P in the diagram below has
latitude of 60º, then  measures 30º.
 60th parallel

60
5
If the radius of the earth is 3,950 miles, then how far is point P from the North Pole? To answer,
convert the angle measure to radians and use the Arc Length Theorem as below.
s  r
s

6
 3,950 miles  2, 068 miles
Example Exercise #1
Show that angles measuring 15 and 1,065  are coterminal.
If angles A and B are coterminal, then there exists an integer k such that we have the equation
below.
m  B   m  A  k 360
Substituting our angle measures, we solve for k and show that it is an integer.
1, 065  15  k  360
1, 080  k  360
1, 080
k
360
3k
Since k  , the angles measuring 15 and 1,065  are coterminal.
Example Exercise #2
Convert 12 to radian measure.
Recall that   180 . Write and solve the proportion.
x


12 180
x  12 

180
x  12 

180
15
x

15
 0.21
6
2nd edition:
4th edition:
Suggested Homework in Dugopolski
Section 5.1: #1-3 odd, #23-27 odd, #71-91 odd, #93-99 odd, #107
Section 5.1: #1-3 odd, #23-27 odd, #55-79 odd, #91-101 odd
Suggested Homework in Ratti & McWaters
Section 5.1: #7-13 odd, #27-45 odd, #61-67 odd
Application Exercise
If a point is in motion on a circle of radius r through an angle of  radians in
time t, then its linear velocity is v  s t where s is the arc length and its angular
velocity is    t . Use these definitions as well as the Arc Length Theorem to prove
the theorem below:
If v is the linear velocity of a point on a circle of radius r and  is its angular
velocity then v  r .
Homework Problems
#1 What is the radian measure of a central angle that “cuts away” an arc 15 centimeters long in a circle
with a radius of 4 centimeters?
#2 What is the radian measure of a central angle intersecting an arc 2 inches long in a circle whose radius
measures 5 inches?
#3 What is the radian measure of a central angle intersecting an arc 5 midfraqs long in a circle whose
radius equals 5 midfraqs?

#4 What is the radian measure of an angle in standard form that intersects the unit circle at  12 ,
3
2
?
#5 What is the radian measure of an angle whose degree measure is 135 ?
#6 Let β be an angle in standard position that measures 225 . What is the least positive angle in
standard position that is coterminal with β ?
#7 Let γ be an angle in standard position that measures 740 . What is the least positive angle in standard
position that is coterminal with γ ?
#8 Let α be an angle in standard position that measures 210 . Find three angles in standard position that
are coterminal with α ?
#9 A circle has a radius of 24 inches. What is the length of the arc “cut away” by a central angle of 210 ?
#10 Queen’s Port (latitude 45 ) is due north of King’s Landing (latitude 39 ). What is the distance
between the two cities?
7
Trigonometric Functions Defined
From algebra, we are familiar with the idea of a function, a set of ordered pairs  x, y 
generated by a rule or mapping that pairs every distinct x-value with one and only one y-value.
We call the x-values inputs or domain values and the y-values outputs or range values. For
instance, if f  2   8 , then f is a rule such that the input 2 is mapped to the output 8, creating
the ordered pair (2,8).
In trigonometry, we study functions whose inputs are often thought of as the measures of
interior angles of a right triangle and whose outputs are the ratios of side lengths of the triangle.
To generate an example, let’s consider the right triangle below.
B

c
a

C
b
A
In the diagram of ABC , the length of the hypotenuse equals c. A measures  while B
measures  . The length of leg opposite A equals a while the length of the leg opposite B
measures b. Using the measure of A as an input, we have six different ratios to discuss as
listed below.
a c b c a
b
, , , , ,
c a c b b
a
To discuss these ratios, we will define six trigonometric functions listed below respectively.
sine, cosecant, cosine, secant, tangent,  cotangent
We said that the inputs of these functions are usually thought of as interior angles of a right
triangle. Since the interior angles of a right angle are restricted to measures on the interval
0,  2 (where zero represents the degenerate case), this way of thinking turns out to be too
restrictive. To eliminate this restriction, we first define reference angles as below.
If  is a nonquadrantal angle in standard position, then the reference angle for 
is the positive acute angle  ' formed by the terminal side of  and the x-axis.
In the diagram below, we have the obtuse angle  and its acute reference angle  ' .
'

8
Consider angle  below. In degrees, this angle measures 405 . The reference angle, then is the
acute angle formed between the positive x-axis and the terminal side of  , which measures 45
because 405  360  45 .
m     405 
m   '  45 
9
4

4
We are now ready to define the six trigonometric functions without restricting the inputs.
We start with sine.
If  is an angle in standard position and  x, y  is the point of
intersection of the terminal side and the circle centered at the origin with
radius r , then sine maps the measure of  to y r . We conflate the
designation of the angle with its actual measurement and denote the sine
function as sin    y r .
At first glance, this definition seems to abandon the idea of inputs representing the measure of
interior angles of right angles, that is, until we recall the concept of a reference angle. Recall
angle  from our diagram above whose measure was 405 degrees. Note that the intersection of
the terminal side of  with the circle centered at the origin with radius r is by definition the same
intersection of the terminal side of  ' with the circle. Hence, sine maps this obtuse measure to
the same output as the corresponding reference angle, which is by definition a positive acute
angle and therefore can be imagined as the interior angle of a right triangle as sketched below.
y
 x, y 
r
'
x
9
Let  be an angle in standard position. We denote the cosine function as cos   ,
tangent as tan   , cosecant as csc   , secant as sec   , and cotangent as cot   . We define
these functions as below.
If  is an angle in standard position and  x, y  is the point of
intersection of the terminal side and the circle centered at the origin with
radius r , then we define the six trigonometric functions as follows.
sin    y r
cos    x r
tan    y x
csc    r y
sec    r x
cot    x y
Now that we have the trigonometric functions defined, we will evaluate the functions for
various real number inputs. Let’s consider sine and cosine for the input value of 45-degrees or
 4 radians. The terminal side for a 45-degree angle in standard position lies on the line y  x .
We need the intersection of this line with x 2  y 2  r 2 , which describes the circle centered at the
origin with radius r. Using substitution, we find the intersection below.
x2  y 2  r 2
x2  y2  r 2
x2  x2  r 2
y2  y2  r 2
2x2  r 2
2 y2  r 2
r2
2
r
x
2
y2 
x2 
r2
2
r
y
2

Since r represents the radius of the circle and is therefore positive, the points  r

and r
r
2, r

2  and   r
2,  r
2, r
2

2 are extraneous solutions while the actual points of intersection are
2,  r

2 . Remember that the terminal side of the 45-degree angle

in standard position lies on the line y  x only in the first quadrant. Therefore, r
2, r

2 is
the solution of interest for our purposes. Using the definition, we can now evaluate sine and
cosine at 45 degrees as follows.
10
sin  45   sin  4  
y r 2
r
r 1
1


r 
 
r
r
2
2 r
2
cos  45   cos  4  
x r 2
r
r 1
1


r 
 
r
r
2
2 r
2
What about the other point of intersection between y  x and x 2  y 2  r 2 ? Consider the angle
in standard position measuring 225  or 5 4 . The terminal side of this angle lies on y  x in

the third quadrant so that its intersection with the circle is the point  r
2,  r

2 . Hence,
we can evaluate any one of the trigonometric functions at 5 4 . For example, we will evaluate
tangent.
y r 2
tan  225   tan  5 4   
1
x r 2
For most inputs, the trigonometric functions are more difficult to evaluate exactly, and
we will rely on calculators to attain approximations. Nevertheless, as we see from the above
discussion, calculating the exact value of a trigonometric function when the input is some
multiple of  4 is not difficult. The same can be said for multiples of  6 or 30 . Consider an
angle of 210  , which is coterminal with the angle measuring 150 . The reference angle is
always positive, so the reference angle measures 30 as shown below.
  210
 '  30
r
From the intersection of the terminal side of   210 and the circle centered at the origin with
radius r, we drop a line segment perpendicular to the negative x-axis forming a right triangle as
pictured above. From the 30º-60º-90º Rule, we know that the leg of the right triangle opposite
the 30 angle has a length equal to one-half the length of the radius. Hence, the y-coordinate of
the intersection point must be r 2 (the negative sign appears because the y-coordinate is below
zero). Similarly, the other leg is
2 times longer. Hence, the x-coordinate of the intersection
11
 r r 2
point is r 2 2 . Using the intersection point   , 
 , we can evaluate any of the six
2
2


trigonometric functions at 210  . We evaluate secant below.
sec  210   sec  30   sec  6  
 2
r
r
 r

 r      r      2 .
x r 2
 2
 r 
Alert readers have probably noticed that each time we evaluate a trigonometric function
the simplification process always involves dividing r into itself. Since r is an arbitrary radius, it
eases the calculations to assign the radius to equal one. We call a circle with r  1 a unit circle
since its radius measures one unit. Since our definition of the six trigonometric functions relies
on the intersection of the terminal side of an angle with a circle centered at the origin with any
radius, we can simply use a radius of one. Hence, we can define the trigonometric functions
equivalently as below.
If  is an angle in standard position and  x, y  is the point of
intersection of the terminal side and the unit circle centered at the origin,
then we define the six trigonometric functions as follows.
sin    y
cos    x
tan    y x
csc    1 y
sec    1 x
cot    x y
Note that x  cos  and y  sin  by definition. Substituting into the definition of tangent, we
obtain a fundamental identity.
tan    y x
tan    sin  cos 
In this way, we obtain the Fundamental Identities stated below.
Fundamental Identities:
sin 
, cos   0
cos 
cos 
cot  
, sin   0
sin 
tan  
12
Similarly, we also obtain the Reciprocal Identities stated here.
Reciprocal Identities:
1
sin 
1
sin  
csc 
1
sec  
cos 
1
cos  
sec 
1
cot  
tan 
1
tan  
cot 
csc  
, sin   0
, csc  0
, cos   0
, sec  0
, tan   0
, cot   0
13
Example Exercise #1
Find the reference angle for an angle whose radian measure is
13
.
4
8
12
is a full rotation in the clockwise direction. Moreover, 3 or 
4
4
13
equals 1.5 rotations in the clockwise direction. Hence, 
equals 1.75 rotations in the
4
13
.
clockwise direction, and RQP shown below has a radian measure of 
4
Recall that 2 or 
T
The reference angle equals the acute angle between the terminal side and the x-axis and is always
given a positive measure regardless of the direction of rotation. In this case, the reference angle
equals TQP . Its measure equals  4 .
Example Exercise #2
 13
Find the exact value of cos  
 4

.

We evaluate cos  4  , but remember that the angle is actually in quadrant two where cosine is
negative. We note that if the terminal side intersects the unit circle, then cos    x . Dropping a
perpendicular from the intersection of the terminal side of the angle and the unit circle, we create
a right triangle with a 45  interior angle. Using the Isosceles Right Triangle rule, we note that
since the hypotenuse equals 1 unit (because the radius is one), then x  1 2 units. Hence,
cos  4   1
2 and cos  13 4   1
2 . If we rationalize, we have the answer below.
 13
cos  
 4
2


2

14
Suggested Homework in Dugopolski
Section 5.2: #29-39 odd, #49-59 odd, #89
Section 5.4: #3-27 odd
Suggested Homework in Ratti & McWaters
Section 5.3: #7-43 odd, #55-63 odd
Application Exercise
Sine and cosine appear in functions that model the motion of a spring. If a
weight is at rest while hanging from a spring, then the spring is in a position of
equilibrium. On a vertical number line, zero represents the equilibrium point. If the
weight is set in motion with an initial velocity v0 from location x0 , then the function
below gives the vertical location at time t in relation to the point of equilibrium.
x t  
v0

sin t   x0 cos t 
The letter  represents a constant that depends on the stiffness of the spring and the
amount of weight on the spring. Positive values of x indicate that the weight is below
equilibrium while negative values of x indicate that the weight is above equilibrium.
Since upward velocity is negative and locations below equilibrium are
positive, use v0  3, x0  2 cm, and    to locate the spring after 0.75 seconds.
Homework Problems
Find the reference angle measures for the following angle measures.
#1 210
#2  43π
#3
5π
3
Find the exact values of the following expressions.
#4 sin   76π 
#5 cos  225
#6 tan  73π 
#7 cot  112π 
#8 sec  23π 
#9 csc  23π 
Find the exact values of sec  θ  , sin  θ  , csc  θ  , tan  θ  , and cot  θ  given the following.
#10 The terminal side of θ is in quadrant II, and cos  θ    135 .
15
Bonus Exercise
Use the 30º-60º-90º Rule and the 45º-45º-90º Rule from Geometry, to label a
unit circle like the one pictured below with all the appropriate points of intersection
for the terminal sides of angles in standard position measuring 0º, 30º, 45º, 60º, 90º,
120º, 135º, 150º, 180º, 210º, 225º, 240º, 270º, 300º, 315º, & 330º.
r 1
Please do not try to label this particularly tiny unit circle but instead draw your own
larger circle centered at the origin and assign the radius to equal one or use the circle
on the following page. Show your work in the form of all the appropriate triangles
formed by dropping perpendiculars from the points of intersection to the x-axis.
16
r 1
17
Unit Circle Winding Function
In this lecture, we want to construct a correspondence between the real number line and
the unit circle, x 2  y 2  1. We will visualize a vertical number line with its origin at the point
1, 0 
on the unit circle.
x2  y 2  1
Our goal is to map every number t on this number line to a point P on the unit circle in such a
way that the arc length from 1, 0  to P on the circle represents the number’s absolute value
(allowing for multiple revolutions). This type of mapping would map the number  to the point
 1, 0  because the arc length along the unit circle from 1, 0  to  1, 0  equals  . Similarly,
we want to map 3 also to  1, 0  since the arc length from 1, 0  to  1, 0  equals 3
allowing for one complete revolution. Our mapping is actually wrapping or “winding” numbers
from the number line around the unit circle. Indeed, we will call it a winding function and define
it formally below.
Let t be any real number. Let W be a function that maps t to  x, y  . We define
W as

1, 0  if t  0
W t   

 x, y  if t  0
where  x, y  is on the circle x 2  y 2  1 such that the arc length along the graph
of x 2  y 2  1 from 1, 0  to  x, y  equals t with the revolution from 1, 0  to
 x, y  being counterclockwise if t  0 and clockwise if t  0 (allowing for
complete and multiple revolutions).
18
In other words, W maps any real number to a point on the unit circle so that Evaluating
 7 
W 
 requires finding the point P on the unit circle so that the arc length from 1, 0  to P
 3 
7
equals 
.
3
Obviously, P lies somewhere in the fourth quadrant. Dropping a perpendicular from P to
the x-axis creates a right triangle with the radius of the circle serving as the hypotenuse. The
reference angle for  7 3 is an interior angle of the triangle. Since the reference angle for
 7 3 is  3 , we can employ the 30º-60º-90º Rule, to find the coordinates of point P, namely,
1
3
3
 7   1
 .
 , 
 . Hence, W  
   , 
3
2
2
2
2






The winding function produces points on the unit circle, so it produces outputs of cosine
and sine and leads to the following theorem.
Let t be any real number and let W  t    x, y  be the winding function.
Then, we have the following.
cost  x and sin t  y
In other words, W  t    x, y    cos t ,sin t  .
This theorem is really an alternate definition of the cosine and sine functions, but it leads us
directly to the Pythagorean Identity. Recall that if W  t    x, y  , then  x, y  is on the circle
x 2  y 2  1. Since W  t    x, y    cos t ,sin t  , we can substitute cost and sin t for x and y
respectively to obtain  cos t    sin t   1 . It is customary to write  sin t  as sin 2  t  , so we
have the Pythagorean Identity below.
2
2
2
The Pythagorean Identity is cos2  t   sin 2  t   1 .
From the Pythagorean Identity, we see that cos t  1 and sin t  1 . Hence, the range of both
functions equals the interval  1,1 .
19
Example Exercise
 9
Let W  t  be the winding function defined in the lecture. Evaluate W 
 4

.

8
equals a full rotation of a circle. Obviously, an angle with a measure of
4
9 4 lies in the first quadrant, forming a reference angle of  4 .
Recall that 2 or
We will call the intersection of the terminal side of this angle with the unit circle point P.
P
Dropping a perpendicular from point P to the x-axis creates a right triangle with the radius of the
circle serving as the hypotenuse. The reference angle for 9 4 is an interior angle of the
triangle. Since the reference angle for 9 4 is  4 and since the degree measure of  4 is
45 , we can employ the Isosceles Right Triangle Rule, to find the coordinates of point P,
 2 2
 9   2 2 
namely, 
,
,
 . Hence, W 
.

 4   2 2 
 2 2 
20
Suggested Homework in Dugopolski
The Dugopolski text does not have any comparable problems.
Suggested Homework in Ratti & McWaters
The Ratti & McWaters text does not have any comparable problems.
Application Exercise
A rational point on a plane curve is a point on the curve with
rational coordinates. For example,  1, 0  is a rational point on the circle
with equation x 2  y 2  1.
 1  t 2 2t 
,
Let t be a natural number. Let Q : t  
. Then, Q is a
2
2 
 1 t 1 t 
mapping from the natural numbers to the rational points on the circle
x 2  y 2  1, excluding the point  1, 0  . Use Q to find a rational point on
the unit circle that is not on the x-axis or y-axis.
Homework Problems
Let W  t  be the winding function defined in the lecture and evaluate the following expressions.
 
#1 W  
2
 3 
#2 W  
 2 
 7 
#3 W 

 6 
 11 
#4 W 

 6 
#5 W  
 3 
#6 W  
 4 
 5 
#7 W  
 4 
 4 
#8 W 

 3 
 2 
#9 W 

 3 
 5 
#10 W  
 6 
21
Right Triangle Trigonometry
In the previous lecture, we said that the inputs of the trigonometric functions are usually
thought of as interior angles of a right triangle, but we decided to define the functions in a less
restrictive way. For instance, we said if  x, y  is any point other than the origin on the terminal
side of an angle  in standard position and r  x 2  y 2 (i.e., r is the radius of a circle centered
at the origin), then sin   y r . This definition allows  to take any real value, which was our
goal, but in its own way this definition can be restrictive since it ties the trigonometric functions
to a coordinate system.
 x, y 
r
y

x
In this lecture, we will take away the coordinate system and regard the values of the
trigonometric functions simply as ratios of the lengths of the sides of a right triangle.
hypotenuse
leg opposite angle 

leg adjacent to angle 
Hence, using the right triangle above, we have the following theorem.
If  is an acute interior angle of a right triangle, then
opposite
hypotenuse
adjacent
cos  
hypotenuse
opposite
tan  
adjacent
sin  
Since cosecant, secant, and cotangent are the reciprocals of sine, cosine, and tangent
respectively, we can use the acute angle of a right triangle to calculate the values of all six
trigonometric functions.
22
Using right triangle trigonometry, we can find the height of an object without actually
measuring the object. Consider the water tower below. Suppose a surveyor stands 25 meters to
the side of the tower and measures the angle of elevation  to the top of the tower at 43 .
  43
25 meters
In this situation, calculating the height of the tower is a simple task as follows.
opposite
adjacent
height
tan  43  
25 meters
25 meters  tan  43   height
tan  43  
Using a table of trigonometric values (i.e., our calculator in degree mode), we see that
tan  43  0.932515 , so we conclude below.
25 meters  tan  43   height
25 meters  0.932515  height
23.3 meters  height
23
Example Exercise
To get home from school, Tameka can follow the roads seven miles south then
even farther due east, or she can take a shortcut following the old railroad
tracks. How many miles does Tameka walk if she takes the shortcut?
28°
Using a table of values for sine, i.e., our calculator in degree mode, we see that
sin  28  0.46947 . Using the right triangle theorem for the trigonometric functions, we let s
represent the length of the shortcut and solve as below.
opposite side length
hypotenuse length
7 miles
sin  28  
shortcut length
7 miles
0.46947 
s
s  0.46947  7 miles
sin   
s  0.46947 7 miles

0.46947
0.46947
s  14.9 miles
24
Suggested Homework
Section 5.6: #9-13 odd, #23, #24, #41, #43, #51
Suggested Homework in Ratti & McWaters
Section 5.2: #7-17 odd, #39-43 odd, #53-59 odd
Application Exercise
In meteorology, the vertical distance from the ground to the base of the clouds
is called the ceiling. To measure the ceiling, weather watchers directed a spotlight
vertically overhead. An observer made the measurements shown in the figure below.
What is the ceiling?
Homework Problems
The kite flyer shown to the right has let out 800 feet of string from a roll
of 1,000 feet, and she is holding the string 4 feet off the ground.
#1 If θ  41 , how high is the kite from the ground?
#2 If θ  54 , how high is the kite from the ground?
#3 How much additional string has the kite flyer released if the kite
is 700 feet directly above the ground and θ  56 ?
#4 How much additional string has the kite flyer released if the kite
is 854 feet directly above the ground and θ  63 ?
25
Trigonometric Functions
In the last few lectures, we defined the trigonometric functions. In this lecture, we will
discuss their graphs. We will limit this introductory discussion to the graphs without
transformation. We begin with the function y  sin x . Note the subtle change in variable
designation. Previously, x and y were variables used to calculate the output of a function sin 
where  was the input. Now x represents the input and y represents the output. Or, when we let
y  f  x  , then f  x  represents the output. Generally, when x represents the input of a function
then x represents any real number (unless the function places some restriction on x). This is the
case with y  sin x . The input of the sine function is the measure of an angle in standard
position, but using radians or even degrees, the measure of an angle in standard position can
equal any real number, rational or irrational, negative or nonnegative. Hence, the domain of sine
equals , and we can assign x or t or any variable to represent domain values. Similarly, we
can assign y or f  x  to represent range values.
Recall our original definition of the sine function. Consider this definition alongside the
figure below.
If  is an angle in standard position and  x, y  is the point of intersection of the
terminal side and the circle centered at the origin with radius r , then sine maps the
measure of  to y r . We conflate the designation of the angle with its actual
measurement and denote the sine function as sin    y r .
y
 x, y 
r
'
x
The diagram demonstrates the basic inequality y  r . When y  r , the output t, y r , equals a
proper fraction. When y  r , then the output, y r , equals 1. Since the radius is always
positive, when y is negative y r is negative and when y is positive y r is positive. Hence, the
26
range of sine equals the interval  1,1 while the domain equals the interval  ,   as
discussed above.
Now recall our simplified definition using the unit circle. Consider this definition
alongside the figure below.
If  is an angle in standard position and  x, y  is the point of intersection of the
terminal side and the unit circle centered at the origin, then sin    y .
The value of sin  at the multiples of 30º and 45º reveal a repetitive pattern to the function
especially when the reader reflects that all these same values will be repeated as the angle
measures extend beyond 2 . For instance, sine has the same value at 13 6 as it does at  6 .
This repetition creates what we call a periodic function, one that repeats itself after a given
interval called its period. We define a periodic function formally below.
A function f  x  is periodic if  a number P  f  x   f  x  kP  for
k  . The least period equals the smallest positive constant P.
In the case of sine, we see that sin  x   sin  x  2k  , so sine is a periodic function with a least
period equal to 2 . One complete least period of the sine function is called a cycle. The interval
27
0, 2  is called the fundamental cycle for f  x   sin  x  . The image below shows the graph of
f  x   sin  x  over the fundamental cycle using points from the table.
sin  x 
x
0
0
2 2  0.7
1
 4
 2
3 4

5 4
3 2
7 4
2 2
0
 2 2  0.7
1
2
 2 2
0
The wave pattern repeats continuously in both directions. We call this graph or any translation or
dilation of this graph a sine wave or a sinusoidal wave. We call the magnitude of the wave’s
oscillation its amplitude. We define the amplitude formally below.
Amplitude equals the maximum displacement from a zero position.
Accordingly, the amplitude of a sine wave equals the absolute value of half
the difference between the maximum and minimum y-values on the curve.
The other five trigonometric functions also represent periodic functions. Note that sine,
cosine, secant, and cosecant all have a least period of 2 while tangent and cotangent have a
least period of  . Only sine and cosine have real number amplitudes.
We leave graphing the remaining trigonometric functions as an exercise, but before we
do, we comment on a characteristic of some functions that aid in graphing them, namely,
asymptotes, defined in layman’s terms below.
When the graph of a function mimics a relation, we call the mimicry
asymptotic behavior, and we call the mimicked relation an asymptote. If
the asymptote is linear, we call it a linear asymptote.
For example, consider the reciprocal function y  1 x . Since the reciprocal of large positive
numbers greater than one are small proper fractions, the reciprocal function takes on small values
as the x-values increase. Indeed, the larger the x-value, the smaller the y-value; hence, as the xvalues approach infinity, the y-values approach zero. For large enough x-values, the function
y  1 x mimics the horizontal line y  0 . Hence, y  0 is an asymptote of the reciprocal
function. In particular, we call y  0 a horizontal linear asymptote of y  1 x . Similarly, the
reciprocals of proper fractions are numbers greater than one. The closer a positive number is to
28
zero, the further its reciprocal is from zero; therefore, as x-values approach zero, the reciprocal
function approaches infinity. Hence, x  0 , which describes a vertical line through the origin,
serves as an asymptote of the reciprocal function. In particular, we call x  0 a vertical linear
asymptote of y  1 x .
The graphs of tangent, cotangent, secant, and cosecant all have domain restrictions
(which is obvious from their definitions), and they all exhibit asymptotic behavior near these
domain restrictions. For example, cosecant is undefined at x   since it is the reciprocal of sine
which equals zero at x   . Since sine decreases to zero as x approaches  from the right, then
its reciprocal, cosecant, must grow larger and larger as x approaches  from the right. Hence,
the vertical line x   serves as an asymptote for the graph of cosecant.
29
Example Exercise
Graph y  tan x
We start by generating a table of sine and cosine values because we recall that tangent
equals the ratio of sine to cosine (Fundamental Identity). From this table, we can find tangent’s
values.
tan x
x
cos x sin x sin x
cos x
0
undefined
1
1
 2
0
1
 3
 3
3  3 2

2
12
2
1
1
1
 4
1 2

2
2
1 2
 6
0
 6
 4
3
2
1
3
2
1
2
 3
1
2
 2
0
1
2
1 2
3 2
0
1
12
3 2
1
2
1
1
1
2
0

3
2
1
2
2
3 2
12
1
0

1
3
0
1
3
1
3
undefined
If we extended our table, it would become obvious that tangent begins to repeat itself, so we see
that tangent has a period equal to  . We are using    2,  2 as the fundamental cycle.
Plotting the ordered pairs from the table and suspecting vertical asymptotes where tangent is
undefined, we generate the graph below.
30
Suggested Homework in Dugopolski
Graph each of the six trigonometric functions over their fundamental cycle. Compare the results
with the graphs that appear in the Function Gallery: Trigonometric Functions on page xxii.
Suggested Homework in Ratti & McWaters
The Ratti & McWaters text does not have appropriate practice problems.
Application Exercise
The frequency F of a sinusoidal wave with least period P is defined by
2
1
. The least period of y  sin x is 2 . The least period of y  sin ax is
.
a
P
What is the frequency of y  sin ax ?
F
Homework Problems
#1 Graph f  x   sin x over   ,   .
#2 Graph g  x   csc x over   ,   .
#3 Graph y  cos x over  0, 2  .
#4 Graph y  sec x over  0, 2  .
#5 Graph T  x   tan x over  0, π  .
#6 Graph C  x   cot x over  0, π  .
#7 Graph F  x   sin x over  0,   .
#8 Graph G  x   csc x over  0,   .
31
Dilations and Reflections of Trigonometric Functions
In the last lecture, we graphed the trigonometric functions. In this lecture, we will
discuss dilations and reflections of these graphs. We start by defining dilation.
Let y  f  x  and let A  0 , then we call y  A  f  x  a vertical
dilation of f  x  and A represents the scale of dilation.
Let y  f  x  and let a  0 , then we call y  f  ax  a horizontal
dilation of f  x  and a represents the scale of dilation.
The graphs below demonstrate the effect of a vertical dilation on a sine wave by a scale of two.
y  2sin x
y  sin x
Note that the vertical dilation does not affect periodicity but has a dramatic impact on the
amplitude of the wave. This demonstrates the following theorem.
The amplitude of y  A sin  x  or of y  A cos  x  equals A .
As just noted, a vertical dilation affects the amplitude of the graph. Horizontal dilations
affect the periodicity of periodic functions as demonstrated by the hand-drawn graphs below.
32
The graph of y  sin  2 x  completes its full cycle twice as often as the graph of y  sin  x  ,
which demonstrates the following theorem.
Let P be the least period of a periodic function y  f  x  .
Then,
P
is the least period of y  f  ax  .
a
Alert readers will note that we defined the scale of dilation to be a positive number. We
do that because a negative scale of dilation creates a mirror image called reflection.
Let y  f  x  , then the graph of y   f  x  is a reflection of f  x  over the
x-axis and the graph of y  f   x  is a reflection of f  x  over the y-axis.
The hand-drawn graph below demonstrates a basic reflection over the x-axis.
The topic of reflection leads us to even and odd functions defined below.
If f  x   f   x  , then we call f  x  an even function. If  f  x   f   x  ,
then we call f  x  an odd function.
The graphs of the trigonometric functions illustrate the following identities.
Even Identities:
cos  x   cos   x  and sec  x   sec   x 
Odd Identities:
 sin  x   sin   x  and  csc  x   csc   x 
 tan  x   tan   x  and  cot  x   cot   x 
33
Suggested Homework in Dugopolski
Section 5.3: #1, #11, #27, #15, #57, #63
Section 5.4: #61, #69
Suggested Homework in Ratti & McWaters
Section 5.4: #9, #11, #15, #17
Section 5.5: #13, #15, #17, #21, #23
Application Exercise
Scientists use the same types of terms to describe ocean waves that describe
sine waves. The wave period equals the time between crests. Assume that the graph
of y  sin  10 x  where x represents seconds elapsed models an ocean wave, and
identify the wave period.
Homework Problems
#1 Graph y  sin  2 x  .
#2 Graph y  2sin  x  .
#3 Graph y   cos  x  .
#4 Graph y  tan   x  .
#5 Graph y  2cos  4 x  .
#6 Graph y  sin  3x  .
#7 Graph y  csc  2 x  .
#8 Graph y  sec  4 x  .
#9 Graph y  14 tan x .
#10 Graph y  6sin  πx  .
34
Translations of Trigonometric Functions
In the last lecture, we dilated and reflected the graphs of trigonometric functions. In this
lecture, we will discuss other type of transformations called translations. While dilations change
the shape of a graph, translations simply move the graph. Accordingly, we call translations shifts
because they shift the graph to a new location. We will concern ourselves with two types of
translations, vertical and horizontal. A vertical translation simply moves the graph up or down
some set number of units.
Let y  f  x  , then we call y  f  x   C a vertical translation of f  x  and C
represents the number of units each point on the graph is translated vertically. If
C  0 , then the points of y  f  x   C lie C -units above their corresponding
points on y  f  x  . If C  0 , then the points of y  f  x   C lie C -units
below their corresponding points on y  f  x  .
The hand-drawn graphs below show how a vertical shift affects a parent graph.
Horizontal translations shift the graph left or right. We call horizontal translations phase shifts.
Let y  f  x  be a periodic function, then we call y  f  x  c  a horizontal
translation of f  x  , and we call c the phase shift. The phase shift represents
the number of units each point on the graph is translated horizontally. If c  0 ,
then the points of y  f  x  c  lie c -units to the right of their corresponding
points on y  f  x  . If c  0 , then the points of y  f  x  c  lie c -units to
the left of their corresponding points on y  f  x  .
The hand-drawn graph below shows a translation to the right. The phase shift equals  4 .
35
Graphing any single transformation is straightforward, but generating a graph with more than
one transformation is a little trickier. Consider the function y  2sin  3x   2  . We start by
finding the phase shift as below.
y  2sin  3 x   2 
 
 
y  2sin 3  x   
6 
 
     
y  2sin 3  x      
   6  
Normally, y  sin  x  has a fundamental cycle  0, 2  . The dilation y  sin 3x has a least
period of 2 3 , so it has a fundamental cycle of  0, 2 3 . Note that y  sin x has x-intercepts
at x  0 (the beginning point of the fundamental cycle), x   (the halfway point), and x  2
(the endpoint). Accordingly, the dilation y  sin 3x has x-intercepts at x  0 (its beginning
point), x   3 (its half-way point), and x  2 3 (its end-point). However, since
 
 
y  sin 3  x    has a phase shift equal to  6 , the x-intercepts each move  6 -units to the
6 
 
 
 
right. Hence, the x-intercepts of y  sin 3  x    are x   6 , x   2 (because
6 
 
 
 
 3   6  3 6   2 ), and x  5 6 . Finally, y  2sin 3  x    vertically dilates
6 
 
 
 
y  sin 3  x    so that the amplitude changes from 1 to 2. A vertical dilation does not affect
6 
 
x-intercepts (since zero times any number is still zero), which is why it is always convenient to
plot the x-intercepts first.
36
In general, it is wise to adhere to the following procedure when generating a graph with
multiple transformations. The reader will note that the procedure effectively follows the order of
operations.
Let y  f  x  be a trigonometric function with least period P. To graph
y  Af  B  x  C    D , we complete the following steps.
 P
1. Sketch one cycle of y  f  Bx  over its fundamental cycle, 0,  .
 B
2. Reflect the cycle over the y-axis if B  0 .
3. Translate the cycle C units to the right if C  0 or C units to the left if C  0 .
4. Change the amplitude of the cycle by a factor of A .
5. If A  0 , reflect the curve over the x-axis.
6. Translate the graph D units upward if D  0 or downward if D  0 .
Suggested Homework in Dugopolski, 2nd Edition
Section 5.3:
Section 5.4:
#17, #19, #21, #25, #65
#59, #63, #65
Suggested Homework in Ratti & McWaters
Section 5.4: #19, #21, #25, #33, #45, #49
Section 5.5: #9, #11, #27, #41
Application Exercise
Many functions can be approximated to any degree of accuracy by polynomial
functions called Taylor polynomials. The Taylor polynomials that approximate the
sine wave function are given below.
yx
x3
y  x ,
3!
x3
y  x 
3!
x3
y  x 
3!
x5
,
5!
x5 x 7
 ,
5! 7!
Demonstrate the approximation of y  sin x by the third-degree Taylor polynomial
x3
by sketching both functions over the interval   ,   . Keep in mind that
3!
3!  3  2 1  6 .
y  x
Homework Problems
#1 Graph y  1  sin x .
#2 Graph y  1  cos x .
#3 Graph y  sin  2 x  π  .
#4 Graph y  cos  12 x  π4  .
#5 Graph y  csc  x  π2  .
#6 Graph y  sec  x  π4  .
#7 Graph y  sin  πx  π  .
#8 Graph y  cos  x   2 .
#9 Graph y  sin  3x  π2  .
#10 Graph y  tan  π2  x  .
38
Inverse Trigonometric Functions
Any one-to-one function has an inverse. A function is one-to-one if no two ordered pairs
in the function have the same second component. In other words, one-to-one functions do not
have any repeated values. Consequently, y  sin x is not one-to-one. A quick test for one-toone correspondence is the horizontal-line test. If all the horizontal lines of the Cartesian plane
intersect the function only once, the function is one-to-one and has an inverse. A quick glance at
the trigonometric functions shows that all six fail the horizontal line test. Hence, the
trigonometric functions are only invertible when we restrict their domain over an interval so that
the resulting graph will pass the horizontal-line test.
If a function is a one-to-one function over a given interval, then its inverse is the function
formed by reversing all the ordered pairs.
Suppose f  x  is a one-to-one function over a given interval I. Then,
reversing all the ordered pairs of f  x  creates a function we call the
inverse of f  x  , denoted f 1  x  .
In other words, if a function is a one-to-one function, then its inverse is the function formed by
reversing all the ordered pairs. For example, for the function f :  1, 4  ,  2,3 ,  5,0  , the
inverse is f 1 :  4, 1 ,  3, 2  ,  0,5 . Note the importance of the original function being one-toone: if the original function is not one-to-one, then reversing the ordered pairs would yield a
relation that is not a function. Graphically, the points of f 1 are reflections of the points of f
across the line y = x, as shown below.
(0,5)
(–1,4)
(2, 3)
(3, 2)
(5,0)
(4,–1)
Every one-to-one function has an inverse, and the graph of the inverse of any one-to-one
function f can be found by reflecting the graph of f across the line y  x .
39
We can define inverse functions for all six trigonometric functions by restricting their
domain. However, we will concern ourselves only with inverses of sine, cosine, and tangent.
We denote the inverse of sine as sin 1  x  or arcsin  x  .
y  sin 1  x  or y  arcsin x means sin y  x and 

2
 y

2
.
Similarly, we denote the inverse of tangent as cos1  x  or arccos  x  .
y  cos1  x  or y  arccos x means sin y  x and 0  y   .
Likewise, we denote the inverse of tangent as tan 1  x  or arctan  x  .
y  tan 1  x  or y  arctan x means tan y  x and 

2
 y

2
.
40
In the definitions above, the range restrictions result from arbitrary restrictions on the domain of
  
sine, cosine, and tangent respectively. It is usual to restrict the domain of sine to   ,  so
 2 2
that the function will be one-to-one and have an inverse. Other restrictions could be chosen.
  
The usual restriction for cosine is  0,   while the usual restriction for tangent is   ,  .
 2 2
Recall that we often think of the input values of a trigonometric function as angle
measures and their outputs as the ratios of side lengths (although we went through a great deal of
trouble to realize the inputs can be any real number for sine and cosine). Since inverse functions
reverse the ordered pairs, it is helpful to think of the inputs as ratios of side lengths and the
outputs as angle measures in inverse trigonometric functions. For example, consider a few
ordered pairs of y  sin x .
x
y
0
0
 6
 4
12
1
2
 3
3 2
 2
1
Since the inverse function reverses the ordered pairs, we have the table below for y  arcsin x .
x
y
0
12
0
 6
1
2
 4
3 2
 3
1
 2
Note how the convenient x-values for y  sin x are common angle measures, but for the inverse
function, the common angle measures are y-values.
41
Example Exercise 1
Find the exact value of arcsin


3 2 .
We start by restricting the domain of sine so that it is a one-to-one function. If we restrict
  
the domain of sine to   ,  , then its graph passes the horizontal line test, which means sine
 2 2
  
is one-to-one over   ,  and has an inverse.
 2 2
Now, we recall that arcsine is the function that produces the set of ordered pairs that are
interchanges between x and y-values of the sine function. (Recall also that we think about the xvalues of sine as angle measures despite the fact that the inputs are really just real numbers. This
means that the outputs of arcsine are numbers that we can think of as angle measures.) For
arcsin

the value

3 2 then, we are looking for the angle measure from   2 and  2 where sine takes
3 2 , which is  3 .
 3 
arcsin 
 
 2  3
Example Exercise 2
Find the exact value of cot 1 1 .
We start by restricting the domain of tangent so that it is a one-to-one function. If we
restrict the domain of cotangent  0,   , then its graph passes the horizontal line test, which
means cotangent is one-to-one over  0,   and has an inverse.
Now, we recall that inverse cotangent is the function that produces the set of ordered
pairs that are interchanges between x and y-values of the cotangent function. (Recall also that we
think about the x-values of sine as angle measures despite the fact that the inputs are really just
real numbers. This means that the outputs of arcsine are numbers that we can think of as angle
measures.) For cot 1 1 then, we are looking for the angle measure from 0 and  where
cotangent equals one. Remember that cotangent equals the ratio or cosine to sine, so for
cotangent to equal one, cosine and sine must be equal. These two functions equal one another at
 4 . Hence, cotangent equals one at  4 . Hence, inverse cotangent equals  4 at one.
cot 1 1 

4
42
Example Exercise 3
Solve the triangle in the figure below. In other words, find the angle measures
for alpha and beta and the length for c.

c
15

10
We use the Pythagorean Theorem to find the hypotenuse.
152  10 2  c 2
225  100  c 2
325  c 2
325  c 2
13  25  c
5 13  c
Now, we turn to the right triangle theorem for the trigonometric functions, which gives us the
opposite side length
10
relationship, sin   
. Thus, we have sin   
, and we solve using
hypotenuse length
5 13
a table of arcsine values (i.e., our calculator) as below.
10
5 13
sin     0.5547001962 
sin   
  sin 1  0.5547001962 
  33.7
We find  using the fact that the interior angles of a triangle are supplementary.
90      180
90  33.7    180
123.7    180
  180  123.7
  56.3
43
Suggested Homework in Dugopolski
2nd Edition: Section 5.5:
4th Edition: Section 5.6:
#1-11 odd, #17-27 odd, #33, #41, #65, #67, #71
#25-31 odd
Suggested Homework in Ratti & McWaters
Section 5.6: #9-43 odd, #55-59 odd
Application Exercise
The function x 
v0

sin t   x0 cos t  gives the position x at time t for a
weight in motion on a vertical spring where v0 is the initial velocity, x 0 is the initial
position, and  is a constant relative to the particular spring.
At time t  0 , a 1-kilogram block is positioned so that the spring compresses
to a position 1 meter below the zero position. When released from this position, the
block moves upward with an initial velocity of 2 meters per second. Assume   1 .
Hence, x  2sin  t   cos  t  gives the position of the block in relation to the zero
position after t seconds. When is the weight at the zero position? In other words,
what are the solutions to the equation below?
2sin t  cos t  0
Hint: we can simplify the equation as follows.
2sin t  cos t  0
2sin t  cos t
sin t
1
cos t
1
tan t 
2
2
Homework Problems
 .
#1 Evaluate sin 1  1 .
#2 Evaluate arccos 
#3 Evaluate arcsin  12  .
#4 Evaluate arctan  1 .
 3 .
#6 Given that cos  12  
#5 Evaluate tan 1
3
2
2 3
2
, evaluate cos 1

2 3
2
.
44
Basic Identities
A large part of studying trigonometry involves studying the numerous interrelationships
among the trigonometric functions. With this lecture, we begin verifying trigonometric
identities. This particular lecture will list identities already mentioned in previous lectures then
demonstrate how to prove new relationships using these basic identities. We start by listing
previously mentioned identities.
Quotient Identities:
sin 
, cos   0
cos 
cos 
, sin   0
cot  
sin 
tan  
Reciprocal Identities:
1
sin 
1
sin  
csc 
1
sec  
cos 
1
cos  
sec 
1
cot  
tan 
1
tan  
cot 
csc  
, sin   0
, csc  0
, cos   0
, sec  0
, tan   0
, cot   0
Odd Identities:
 sin  x   sin   x 
 csc  x   csc   x 
 tan  x   tan   x 
 cot  x   cot   x 
Even Identities:
cos  x   cos   x 
sec  x   sec   x 
45
We have also mentioned the Pythagorean Identity, namely, sin 2   cos2   1 . There are two
more identities commonly referred to as Pythagorean Identities. The box below gives all three.
Pythagorean Identities:
sin 2   cos2   1
1  tan 2   sec2 
1  cot 2   csc2 
Deriving the second two Pythagorean identities is quite simple based on the first. For
instance, consider the first Pythagorean identity. If we divide through by cos2  we arrive at the
second Pythagorean identity as shown below.
sin 2   cos 2   1
sin 2  cos 2 
1


2
2
cos  cos  cos 2 
Of course, dividing by a quantity introduces the difficulty of worrying about whether or not the
divisor-quantity equals zero, but in this case tangent and secant are not defined if cos   0 , so
we simply divide for all values of  except where cos   0 . To conclude, we employ the
fundamental and reciprocal identities as below.
sin 2  cos 2 
1


2
2
cos  cos  cos 2 
tan 2   1  sec2 
We leave the proof for 1  cot 2   csc2  to the reader.
The main point of this lecture is to demonstrate how to use the basic identities to verify
other identities. For example, we will show that csc  cos cot   sin  . First, we apply the
reciprocal identity for cosecant as below.
csc   cos  cot   sin 
1
 cos  cot   sin 
sin 
Second, we apply the fundamental identity for cotangent.
1
 cos  cot   sin 
sin 
1
cos 
 cos 
 sin 
sin 
sin 
Next, we add the resulting fractions as follows.
46
1
cos 
 cos 
 sin 
sin 
sin 
1
cos 2 

 sin 
sin  sin 
1  cos 2 
 sin 
sin 
Noting that sin 2   cos2   1 implies sin 2   1  cos2  , we conclude as below.
1  cos 2 
 sin 
sin 
sin 2 
 sin 
sin 
sin   sin 
 sin 
sin 
Since all the steps performed on the left can be reversed, we conclude csc  cos cot   sin  .
1
For a second example, we will verify that 1  sin x 1  sin x  
. We start by
1  tan 2 x
using the algebraic identity,  a  b  a  b   a 2  b2 as below.
1  sin x 1  sin x  
1  sin 2 x 
1
1  tan 2 x
1
1  tan 2 x
Since the Pythagorean identity sin 2   cos2   1 implies sin 2   1  cos2  , we can substitute
next as below.
1
1  sin 2 x 
1  tan 2 x
1
cos 2 x 
1  tan 2 x
Applying the reciprocal identity followed by the Pythagorean identity 1  tan 2   sec2  , we
conclude as follows.
47
1
1  tan 2 x
1
1

2
sec x 1  tan 2 x
1
1

2
1  tan x 1  tan 2 x
cos 2 x 
Verifying a non-identity is a simple matter of finding a counter-example. For instance,
consider the equation 2sin  cos  sin   0 . This equation holds if   0 , but to show that the
equation is not an identity, we simply let    2 as follows.
?
2sin  cos   sin   0
?
2sin  2  cos  2   sin  2   0
?
2 1  0  1  0
1  0
As implied above, an identity can usually be verified several ways. However, the
suggestions below are helpful.
Suggestions for Verifying Identities
1. Work with only one side of the equation at a time. It is usually easier to begin with
the more complicated side and then simplifying using the remaining suggestions.
2. Make substitutions using known identities. Often it is helpful to rewrite one side in
terms of sine or cosine.
3. Perform indicated algebraic operations such as adding or subtracting rational
expressions or multiplying polynomials.
4. Consider reversing algebraic operations such as factoring polynomials and
decomposing rational expressions.
5. Keep checking the result against the other side of the identity.
48
Example Exercise 1
Verify
1
1  sin 
.

tan   sec
cos
Multiply the left side by a propitious choice of “1” to obtain the following.
1
tan   sec  1  sin 


tan   sec  tan   sec 
cos 
1 tan   sec   1  sin 

tan 2   sec 2 
cos 
By the Pythagorean Identity, we have tan 2   sec2   1 ; hence, we have
1 tan   sec  1  sin 

sec 2   1  sec 2 
cos 
1 tan   sec  1  sin 

sec 2   sec 2   1
cos 
1  tan   sec  1  sin 

cos 
1
1  sin 
tan   sec 
.
cos 
Next, we substitute a Fundamental Identity and a Reciprocal Identity as below.
sin 
1
1  sin 


cos cos
cos
Adding the fractions yields the identity.
sin   1 1  sin 

cos
cos
49
Example Exercise 2
Verify cot  x   sin  x   cos2  x   sec  x   0.
By Fundamental Identity, we have the following.
cos  x 
 sin  x   cos 2  x   sec  x   0
sin  x 
Next, we reduce as below.
cos  x 
sin  x 
 sin  x   cos 2  x   sec  x   0
cos  x   cos2  x   sec  x   0
By Reciprocal Identity and reduction, we conclude as below.
cos  x   cos 2  x  
1
0
cos  x 
cos  x   cos 2  x  
1
cos  x 
0
cos  x   cos  x   0
00
50
Suggested Homework in Dugopolski
Section 6.2: #55-87 odd
Suggested Homework in Ratti & McWaters
Section 6.1: #29-69 odd
Application Exercise
A projectile is fired with initial velocity of v0 . The projectile can be pictured
as being fired from the origin into the first quadrant, making an angle  with the
positive x-axis as shown in the figure below.
y

x
If there is no significant air resistance, then at time t the coordinates of the projectile
16
are x  v0t cos  and y  16t 2  v0t sin  . Show that y   2 sec2  x 2  x tan  .
v0
Homework Problems
Verify the following identities.
#1 sin x sec x  tan x
#2 sin x cos x  tan x  cot x   1
#3  sec   1 sec   1  tan 2 
#4 csc  sin   cos  cot 
#5
tan  x  sin  x 
 1  cos  x 
sec  x   1
1  tan 2 x
tan 2 x
cos   sin 
#9 1  tan  
cos 
tan x  cot x
#11 sec2 x  csc2 x 
sin x cos x
#7 csc2 x 
#6 2cot x 
#8
csc x cos x

sec x sin x
cot x sec x

 sec2 x csc x
cos x cot x
#10 sec4 x  sec2 x  tan 4 x  tan 2 x
#12 sec  2 tan  
cot   2cos 
csc  sin 
51
Sum and Difference Identities
In the last lecture, we learned how to use basic identities to verify other interrelationships
between the trigonometric functions. In this lecture, we will append some additional identities
called the Sum and Difference Identities to our list of basic facts that we can use to verify further
interrelationships.
We want to establish an identity for the sine of the sum of two angles  and  .
Consider AB in the diagram below. Let AB revolve to point C and sweep out angle CAB ,
measuring  , then revolve further to point D and sweep out angle CAD , measuring  , as
shown.
B
From point D, we draw DE perpendicular to AB . Then, sin      ED DA , and
cos      AE DA .
Next, we draw DF perpendicular to AC , FG perpendicular to AB , and FH
perpendicular to ED . This makes AC a transversal crossing the parallels AB and HF .
Hence, HFA and CAB are alternate interior angles and congruent. Moreover, HDF
and HFA are also congruent because they are both complementary to HFD . We have
CAB  HFA and HDF  HFA . By the transitive property, CAB  HDF . Thus,
m  HDF    .
Now we note ED  GF  HD , and we substitute as below.
sin     
ED GF  HD GF HD



DA
DA
DA DA
52
Multiplying by a judicious form of 1, we obtain the following.
GF HD

DA DA
GF AF HD FD
sin     



DA AF DA FD
GF AF HD FD
sin     



AF DA FD DA
sin     
Now, we note the following equalities by definition, sin    GF AF , cos     AF DA ,
cos    HD FD , and sin     FD DA . We conclude by substitution.
GF AF HD FD



AF DA FD DA
sin      sin    cos     cos    sin   
sin     
Hence, we have the following identity.
Sine of a Sum Identity:
sin      sin  cos   cos  sin 
The remaining Sum and Difference Identities we will state without proof.
Sum and Difference Identities:
sin      sin  cos   cos  sin 
sin      sin  cos   cos  sin 
cos      cos  cos   sin  sin 
cos      cos  cos   sin  sin 
tan     
tan   tan 
1  tan  tan 
tan     
tan   tan 
1  tan  tan 
53
Example Exercise 1


Verify cos  x    cos x tan x .
2

By the Difference Identity of Cosine, we obtain the following.


cos  x    cos x tan x
2

 
 
cos  x  cos    sin  x  sin    cos x tan x
2
2
 
 
Evaluating cos   and sin   yields
2
2
cos  x   0  sin  x  1  cos x tan x
sin  x   cos x tan x.
Multiplying by a fortuitous form of “1” yields
cos x
 sin x  cos x tan x
cos x
sin x
cos x 
 cos x tan x.
cos x
The Quotient Identity completes the verification.
cos x  tan x  cos x tan x
54
Example Exercise 2
Verify
cos  x  y  1  tan x tan y
.

sin  x  y  tan x  tan y
By the Quotient Identity, we have
sin x sin y
cos  x  y 
cos x cos y
.

sin x sin y
sin  x  y 

cos x cos y
1
Adding the fractions in the denominator, we obtain
sin x sin y
cos  x  y 
cos x cos y

sin  x  y  sin x  cos y sin y  cos x

cos x  cos y cos y  cos x
sin x sin y
1
cos  x  y 
cos x cos y

.
sin  x  y  sin x  cos y  sin y  cos x
cos x  cos y
1
Division, yields
cos  x  y 

sin x  sin y 
cos x  cos y
 1 

sin  x  y   cos x  cos y  sin x  cos y  sin y  cos x
cos  x  y 
cos x  cos y
sin x  sin y


.
sin  x  y  sin x  cos y  sin y  cos x sin x  cos y  sin y  cos x
Adding the fractions, we obtain
cos  x  y 
sin  x  y 

cos x  cos y  sin x  sin y
.
sin x  cos y  sin y  cos x
The Cosine of a Difference Identity and the Sine of a Sum Identity complete the verification.
cos  x  y 
sin  x  y 

cos  x  y 
sin  x  y 
55
Suggested Homework in Dugopolski
Section 6.3: #11-17 odd, #79-97 odd
Suggested Homework in Ratti & McWaters
Section 6.2: #27-31 odd, #63-69 odd
Application Exercise
In the electromagnetic wave theory of light, scientists use the following equation
where k is the index of refraction, sin i sin r .
 k cos r  cos i 
E "  E 
 k cos r  cos i 
Show that the above equation is equivalent to the equation below.
 sin  i  r  
E "  E 

 sin  i  r  
Homework Problems
Use identities to find the exact value of the following expressions.
#1 sin  75 
#2 cos  12π 
#3 sin  1112π 
#4 cos  912π  cos  512π   sin  912π  sin  512π 
Verify the following identities.
#5 cos  α  180   cos α
#6 sin x  cos  32π  x 
#7 tan  x   tan  π  x 
#8 sin  360  x    sin x
#9 cos x cos y  12 cos  x  y   cos  x  y 
#10 tan x  tan  x  3π 
#11 cos 2 x  1  2sin 2 x
#12 sin 2 x  2sin x cos x
56
Double-Angle and Half-Angle Identities
In the last lecture, we showed sin      sin  cos   cos  sin  . In this lecture, we
will acquire the so-called Double-Angle Identity. We let     x and substitute into the Sine of
a Sum Identity as below.
sin      sin  cos   cos  sin 
sin  x  x   sin x cos x  cos x sin x
sin  2 x   2sin x cos x
Hence, we have the following identity.
sin  2   2sin  cos 
For cosine, we do the same as below.
cos      cos  cos   sin  sin 
cos  x  x   cos x cos x  sin x sin x
cos  2 x   cos 2 x  sin 2 x
Using the Pythagorean Identity, we attain another common identity for cos 2x as below.
cos  2 x   cos 2 x  sin 2 x
cos  2 x   cos 2 x  1  cos 2 x 
cos  2 x   cos 2 x  1  cos 2 x
cos  2 x   2 cos 2 x  1
If we replace cos2 x using the Pythagorean Identity instead of sin 2 x , we attain sill another
common identity for cos 2x .
cos  2 x   cos 2 x  sin 2 x
cos  2 x   1  sin 2 x   sin 2 x
cos  2 x   1  sin 2 x  sin 2 x
cos  2 x   1  2sin 2 x
So, we have three Double-Angle Identities for cosine stated below.
cos  2   cos2   sin 2   2cos2   1  1  2sin 2 
57
Similarly, we can attain a Double-Angle Identity for tangent.
tan   tan 
1  tan  tan 
tan x  tan x
tan  x  x  
1  tan x tan x
2 tan x
tan  2 x  
1  tan 2 x
tan     
Thus, we have the Double-Angle Identities stated here.
Double-Angle Identities:
sin  2   2sin  cos 
cos  2   cos2   sin 2   2cos2   1  1  2sin 2 
tan 2 
2 tan 
1  tan 2 
We can use the Double-Angle Identities to acquire the following Half-Angle Identities.
x
For instance, we have cos 2  1  2sin 2  . Let   . Then, we have the following.
2
x
cos x  1  2sin 2  
2
 x
2sin 2    1  cos x
2
 x  1  cos x
sin 2   
2
2
1  cos x
 x
sin 2   
2
2
1  cos x
 x
sin    
2
2
58
x
x
lies. If
is an angle in the first or second
2
2
1  cos x
1  cos x
 x
 x
quadrants, then sin   
; otherwise, sin    
.
2
2
2
2
Similarly, we may obtain the formulas in the box below.
The sign depends on the quadrant wherein
Half-Angle Identities:
1  cos 
 
sin    
2
2
1  cos 
 
cos    
2
2
1  cos 
sin 
1  cos 
 
tan    


1  cos  1  cos 
sin 
2
59
Example Exercise 1
Verify sin  2s   2sin s  sin  s   2  .
By the Sine of a Difference Identity, we have the equality below.
sin  2s   2sin s  sin s  cos  2   cos s  sin  2 
Evaluating sine and cosine at  2 , we obtain the following.
sin  2s   2sin s    cos s 
By the commutative property of multiplication, we arrive at the Double Angle Identity for Sine.
sin  2s   2  1 sin s  cos s
sin  2s   2sin s  cos s
Example Exercise 2
Verify
cos 2 x
 sec2 x  2 tan 2 x .
2
cos x
First, we apply the Double-Angle Identity and decompose the resulting fractions as follows.
cos 2 x  sin 2 x
 sec2 x  2 tan 2 x
2
cos x
2
cos x sin 2 x

 sec2 x  2 tan 2 x
2
2
cos x cos x
Second, we apply the Pythagorean Identity and decompose yet again as below.
1  sin 2 x sin 2 x

 sec2 x  2 tan 2 x
2
2
cos x
cos x
1
sin 2 x sin 2 x


 sec2 x  2 tan 2 x
cos 2 x cos 2 x cos 2 x
Finally, we apply the Reciprocal Identity and Fundamental Identity, and we add.
sec2 x  tan 2 x  tan 2 x  sec2 x  2 tan 2 x
sec2 x  2 tan 2 x  sec2 x  2 tan 2 x
60
Suggested Homework in Dugopolski
Section 6.4: #53-65 odd
Suggested Homework in Ratti & McWaters
Section 6.3: #13-29 odd, #43-47 odd, #63
Application Exercise
Physicists often approximate the acceleration due to gravity, g , as 9.8 meters
per second squared. However, g varies slightly with latitude. If  is the measure of
latitude in degrees, then the following formula approximates g in meters per second
squared.
g  9.78049 1  0.005288sin 2   0.000006sin 2 2 
Write an expression for g in terms of sin  only.
Homework Problems
Use double-angle identities to find the exact values of the following expressions.
#1 2sin  12π  cos  12π 
#2 2cos2  512π   1
Use half-angle identities to find the exact values of the following expressions.
#3 cos  π8 
#4 tan  78π 
Verify the following identities.
1  cos 2 x
#5 cot x 
sin 2 x
2
#7 1  sin 2 x   sin x  cos x 
#9
1  cos 2 x
 tan x
sin 2 x
sin 2 x
1  cos 2 x
#8 cos 2 λ  cos4 λ  sin 4 λ
#6 cot x 
#10
2cos 2 x
 cot x  tan x
sin 2 x
60
Product to Sum Identities
In this lecture, we will acquire a new identities from the sum and difference identity.
We add sin (α + β ) = sin α cos β + cos α sin β to sin (α − β ) = sin α cos β − cos α sin β as
below.
sin (α + β ) + sin (α − β ) = sin α cos β + cos α sin β + sin α cos β − cos α sin β
sin (α + β ) + sin (α − β ) = sin α cos β + sin α cos β + cos α sin β − cos α sin β
sin (α + β ) + sin (α − β ) = 2sin α cos β
sin (α + β ) + sin (α − β )
= sin α cos β
2
Similarly, we produce the Product-to-Sum Identities below.
Product-to-Sum Identities:
sin α cos β =
sin (α + β ) + sin (α − β )
2
sin α sin β =
cos (α − β ) − cos (α + β )
2
cos α sin β =
sin (α + β ) − sin (α − β )
2
cos α cos β =
cos (α − β ) + cos (α + β )
2
Next, we let α + β = x and α − β = y . Adding these equations, we get x + y = 2α .
x+ y
x− y
Subtracting the two equations, gives us x − y = 2β . Thus, we have α =
,
and β =
2
2
sin (α + β ) + sin (α − β )
as below.
which we substitute into sin α cos β =
2
sin (α + β ) + sin (α − β )
2
⎛ x+ y x− y⎞
⎛ x+ y x− y⎞
sin ⎜
+
+ sin ⎜
−
⎟
⎟
2
2 ⎠
2
2 ⎠
⎛ x+ y⎞
⎛ x− y⎞
⎝
⎝
sin ⎜
⎟ cos ⎜
⎟=
2
⎝ 2 ⎠
⎝ 2 ⎠
sin α cos β =
⎛ x+ y⎞
⎛ x− y⎞
2sin ⎜
⎟ cos ⎜
⎟ = sin ( x ) + sin ( y )
⎝ 2 ⎠
⎝ 2 ⎠
61
Similarly, it is possible to obtain the following Sum-to-Product Identities.
Sum-to-Product Identities:
⎛ x+ y⎞
⎛ x− y⎞
sin x + sin y = 2sin ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
⎛ x+ y⎞ ⎛ x− y⎞
sin x − sin y = 2 cos ⎜
⎟ sin ⎜
⎟
⎝ 2 ⎠ ⎝ 2 ⎠
⎛ x+ y⎞
⎛ x− y⎞
cos x + cos y = 2 cos ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
⎛ x+ y⎞ ⎛ x− y⎞
cos x − cos y = −2sin ⎜
⎟ sin ⎜
⎟
⎝ 2 ⎠ ⎝ 2 ⎠
62
Suggested Homework in Dugopolski
Section 6.5: #51-59 odd
Suggested Homework in Ratti & McWaters
Section 6.4: #51-53 odd
Review Exercises on page 624: #5-11
Application Exercise
Use the Sum and Difference Identities from the previous lecture to derive one
sin (α + β ) + sin (α − β )
of the Product-to-Sum Identities other than sin α cos β =
.
2
61
Conditional Identities
Consider the equation, 2sin x 1  0 . We call equations like this conditional identities
because they are true only on the condition that x takes certain values. In this lecture, we learn to
solve for these conditional x-values.
Let’s consider the given equation, 2sin x 1  0 . Isolating the trigonometric expression,
we obtain, sin x  1 2 ; hence, x   6 and x  5 6 if we restrict x to values less than 2 .
Abdicating this restriction, the solutions reoccur every cycle of sine’s period be it in the positive
or negative direction. Consequently, we can say, 2sin x 1  0 if x   6  2k and
x  5 6  2k where k is an integer.
To understand our general condition above, let’s look at the solution x   6 . It is easy
to see that the identity is true again if x  13 6 (note that  6 and 13 6 are co-terminal
angles). Consider the expression x   6  2k where k = 1:
x
x
x
x

6

6

6

 2 k
 2 1 
 2

6
13
x
6
12
6
In the example above, we isolated the trigonometric expression. Another strategy
requires factoring. Consider the conditional identity: cot x cos2 x  2cot x . To solve for the
conditional values of x, we will set one side equal to zero and factor as below.
cot x cos 2 x  2 cot x
cot x cos 2 x  2 cot x  0
cot x  cos 2  2   0
We have a product equal to zero; hence, one of the factors must equal zero for the condition to be
true. Setting each factor equal to zero, we obtain solutions.
cot x  0
x

2
cos 2  2  0
 k
cos x   2
Note that one of the factors could not equal zero (  2 falls outside the range of cos x ).
Conditional identities can take the form of a quadratic as below.
2sin 2 x 1  sin x
62
In which case, we apply strategies similar to a quadratic equation, i.e., we factor if possible and
otherwise rely on the quadratic formula.
2sin 2 x  1  sin x
2sin 2 x  sin x  1  0
 2sin x  1 sin x  1  0
2sin x  1  0, sin x  1  0
sin x  1 2,
sin x  1
7

 2 k , x   2 k 
6
2
11
x
 2 k
6
x
Sometimes using non-conditional trigonometric identities help solve a conditional
identity as below.
2sin 2 x  3cos x  3
2sin 2 x  3cos x  3
2 1  cos 2 x   3cos x  3
2  2 cos 2 x  3cos x  3  0
2 cos 2 x  3cos x  1  0
2 cos 2 x  3cos x  1  0
 2 cos x  1 cos x  1  0
2 cos x  1  0,
cos x  1  0
cos x  1 2,
cos x  1

 2k , x  2 k
3
5
x
 2 k
3
x
Sometimes squaring both sides offers an avenue towards a solution, but we must
remember that squaring both sides of an equation can introduce extraneous roots.
cos x  1  sin x
 cos x  1
2
  sin x 
2
cos 2 x  2 cos x  1  sin 2 x
cos 2 x  2 cos x  1  1  cos 2 x
2 cos 2 x  2 cos x  0
 cos x  1  2 cos x  0
cos x  1 or
x 
cos x  0
or x 
 3
2
,
2
Checking the solutions shows that 3 2 is extraneous, so the solutions are x    2 k and
x   2  2 k .
Some equations involve multiple angles. Start by solving for the multiple angle.
63
2 cos  3 x   1  0
2 cos  3 x   1
cos  3 x  
1
2
Concentrating on the argument, 3x, we see:
3x 

3
 2 k and 3x 
5
 2 k
3
Dividing by the multiple, we obtain the solutions.
x

9

2 k
5 2 k
and x 
.

3
9
3
All the previous examples involve common angles (or some simple fraction of a common
angle). The solutions in the following conditional identity are not nice, convenient angles.
sec 2 x  2 tan x  4
1  tan 2 x  2 tan x  4
tan 2 x  2 tan x  3  0
 tan x  3 tan x  1  0
tan x  3  0,
tan x  1  0
tan x  3,
tan x  1
Here we come to an impasse, until we realize that we can undo a trigonometric function by its
inverse as demonstrated below.
arctan  tan x   arctan  3
arctan  tan x   arctan  1
x  arctan  3
x  arctan  1
x  arctan  3   k
x  arctan  1   k
Use a calculator to acquire approximations for the x-values above.
64
Suggested Homework in Dugopolski
Section 6.6: #1-17 odd, #31-53 odd, #65, #69, #73, #75, #77, #79, #81, #85, #91, #93
Suggested Homework in Ratti & McWaters
Section 6.5: #7-25 odd, #39-45 odd, #55-69 odd
Section 6.6: #7-19 odd
Application Exercise
The distance d traveled by a projectile fired at an angle  is related to the initial
velocity v0 (in feet per second) by the equation v02 sin 2  32d . If the muzzle
velocity for a rifle equals 230 feet per second, then at what angle would it have to be
aimed for the bullet to travel 826.5625 feet?
Homework Problems
Find all the solutions on the interval 0  x  2π to the conditional identities.
#1 2sin  x   1  2
#2 2sin 2 x  sin x  0
#3 tan  2 x    3
#4 8cos  x  π4   6  10
#5 1  sin x  3 cos x
#6 cos 2 x  cos x  0
Find all the solutions to the conditional identities.
#7 cos θ  2sin θ
#8 sin 2θ  sin θ
#9 2sin 2 θ  9sin θ  5
#10 2sin 2 θ  3sin θ  1  0
65
Law of Sines
The Law of Sines says that the ratio of the sine of an angle and the length of the side opposite
the angle is the same for each angle of a triangle. For proof, we consider the case of an acute triangle
here and leave the case of the obtuse triangle as an exercise.
Consider ABC below.
B
c
a
A
C
b
We begin by constructing an altitude h1 from vertex B to side b as shown.
B
c
h1
A
a
C
b
Now, we note sin A  h1 c , so h1  c sin A . Likewise, sin C  h1 a , so h1  a sin C . Hence, we have
c sin A  a sin C , which implies the following.
sin A sin C

a
c
Constructing another altitude leads to the proof’s conclusion. Hence, we have the Law of Sines stated
below.
The Law of Sines states that for any
c
A
b
ABC acute, right, or obtuse, then
B
a
C
sin A sin B sin C


a
b
c
66
We can also write the Law of Sines as below.
a
b
c


sin A sin B sin C
We use the Law of Sines to solve triangles, that is, to solve for some unknown measure of a
triangle be it an angle or side.
The ASA or AAS Cases
The Law of Sines will solve a triangle in the case where two angle measures and the length of
the included side are known (ASA case). If given two angle measures and an un-included side length
(AAS case), then subtract the sum of the two angles from 180º to find the third angle measure to
change the case to the ASA case. For instance, imagine that A measures 28º, B measures 66º, and
the side opposite C measures 8.2 units. Since the three angles of a triangle are supplementary, we
know C measures 86 . By the Law of Sines, we have the following.
8.2
b

sin 86 sin 66
8.2
b  sin 66
sin 86
b  7.5
Again, by Law of Sines, we finish below.
7.5
a

sin 66 sin 28
7.5
a  sin 28
sin 66
a  3.9
Hence, the three angles measure 28, 66, and 86 degrees, and the three sides measure 8.2, 7.5, and 3.9
units in length.
The SSA or Ambiguous Case
If only one angle is given, then two sides are needed. If the given angle is opposite from one of
the two given sides, we have what is called the ambiguous case since more than one triangle (or no
triangle) is possible. Assume that the measure of A is known as well as sides b and a . Then, the
diagram below demonstrates the ambiguity.
C
b
a
C
a
h
The boxes
below
summarize
A
B'
B the situation.
A
b
C
b
ha
B
A
a
h
B
67
The Ambiguous Case with a Given Obtuse Angle:
Suppose A is an obtuse angle whose measure is known. Suppose side a is
opposite A and side b is adjacent to A and that the lengths of a and b are
known. Then,
1. If a  b , then no triangle is formed.
2. If a  b , exactly one triangle is formed.
If the given angle is acute, solve for the height of the triangle and use the summary below.
The Ambiguous Case with a Given Acute Angle:
Suppose A is an acute angle whose measure is known. Suppose side a is opposite
A and side b is adjacent to A and that the lengths of a and b are known. Let h
represent the height of the proposed triangle. Then,
1.
2.
3.
4.
If
If
If
If
a  b , then exactly one triangle is formed.
a  h , then no triangle can be formed.
a  h , then exactly one right triangle is formed.
h  a  b , then exactly two triangles are formed.
68
Suggested Homework in Dugopolski
Section 7.1: #1-17 odd
Suggested Homework in Ratti & McWaters
Section 7.1: #7-23 odd
Section 7.2: #9, #11, #15, and #27-33 odd
Application Exercise
The formulas below calculate the area A of
B
c
a
C
b
A
ABC .
1
A  bc sin A
2
1
A  ac sin B
2
1
A  ab sin C
2
Show that a formula for the area K of parallelogram
ABCD is K  bd sin A .
c
D
C
b
a
A
d
B
Homework Problems
#1 How far is the plane represented in the diagram from the Hazard station?
#2 Consider a triangle such that a  12 , b  9 , and A  143 . Solve the triangle (or triangles).
#3 Consider a triangle such that b  5 , c  10 , and B  72 . Solve the triangle (or triangles).
#4 Consider a triangle such that a  6.2 , b  7.5 , and A  39.4 . Solve the triangle (or triangles).
69
Law of Cosines
In the previous lecture, we used the Law of Sines to solve triangles given two angle
measures (AAS and ASA cases) or two side measures and the un-included angle measures (the
ambiguous SSA case). In this lecture, we use the Law of Cosines stated below to solve triangles
given three side measures (SSS case) or given two side measures with the included angle’s
measure (SAS case).
The Law of Cosines states that given any oblique triangle ABC or any
right triangle ABC where c is the hypotenuse, then we have the following
equality.
c2  a2  b2  2ab cos C
B
c
A
a
b
C
For proof, we consider a general triangle. Without loss of generality, we have one vertex
on the origin and another on the x-axis.
Let the vertex at  0, 0  be C (and let C denote the vertex, the interior angle at the vertex, and
the angle measure). Let A be on the x-axis. Note that A   b, 0  and B   a  cos C, a  sin C  .
Calculating the distance between A and B using the distance formula, we have the following.
70
d
 x2  x1    y2  y1 
c
 a  cos C  b    a  sin C  0 
c
 a  cos C  b    a  sin C 
2
2
2
2
2
2
Next, we square both sides as below.
c
2


 a  cos C  b    a  sin C 
2
c 2   a  cos C  b    a  sin C 
2
2

2
2
c 2   a  cos C  b  a  cos C  b   a 2 sin 2 C
c 2  a 2 cos 2 C  ab cos C  ab cos C  b 2  a 2 sin 2 C
c 2  a 2 cos 2 C  2ab cos C  b 2  a 2 sin 2 C
Finally, we conclude using the commutative property of addition and the Pythagorean Identity as
follows.
c 2  a 2 cos 2 C  2ab cos C  b 2  a 2 sin 2 C
c 2  a 2 cos 2 C  a 2 sin 2 C  b 2  2ab cos C
c 2  a 2  cos 2 C  sin 2 C   b 2  2ab cos C
c 2  a 2  b 2  2ab cos C
We can use the Law of Cosines to solve triangles where the Law of Sines is insufficient
such as case where two side lengths and the included angle measure are given (SAS case) and
the case where three side lengths are known (SSS case).
Technical Note for the SAS case: If the given angle is acute and a
solution reverts to the Law of Sines after finding the third side with
the Law of Cosines, then solve for the angle opposite the smallest
side. If the triangle has an obtuse angle, then it will be opposite the
largest side of the triangle. Thus, the angle opposite the smaller of
the two given sides cannot be obtuse. This is important because
there are two solutions between 0˚ and 180˚ to the equation
sin   y , that is, the acute angle   sin 1  y  and the obtuse
angle  '  180  sin 1  y  .
Technical Note for the SSS Case: If a solution reverts to the Law
of Sines after finding an initial angle using the Law of Cosines,
then find the largest angle first. Since a triangle can have at most
one obtuse angle, solving for the largest angle first eliminates any
71
further candidates for obtuse angles, and no SSA-style ambiguities
arise.
If three side lengths can possibly form a triangle, then the sum of any two of the lengths
must be greater than the length of the third side according to the Triangle Inequality below.
Triangle Inequality: For all real numbers a and b , we have a  b  a  b .
Consider a triangle with side lengths a  48 , b  36 , and c  31 . The sum of any two of
these lengths will exceed the third; hence, a triangle is possible. Since a is the longest side, we
will solve for the measure of angle A in order to solve for the largest angle first. To solve for the
measure of A , we simply substitute the known quantities into the Law of Cosines then solve as
follows.
a 2  b 2  c 2  2bc cos A
 48
2
  36    31  2  36  31 cos A
2
2
2,304  1, 296  961  2, 232 cos A
2,304  2, 257  2, 232 cos A
47  2, 232 cos A
47
 cos A
2, 232
0.0210573477  cos A
cos 1  0.0210573477   A
91.2  A
Since the largest angle is obtuse, we preclude the possibility of two triangles. Now, we can
employ either the Law of Sines or the Law of Cosines to find a second angle. We will do the
former.
sin  91.2  sin C

48
32
sin  91.2 
32 
 sin C
48
0.6665204556  sin C
sin 1  0.6665204556   C
41.8  C
To find the final angle, we can simply subtract the measures of angles A and C from 180 degrees.
Interestingly, we can use the Law of Cosines to produce the Full Pythagorean Theorem
stated below.
72
Full Pythagorean Theorem: Consider ABC . If C is a right angle, then
c2  a 2  b2 . If C is obtuse, then c2  a 2  b2 . If C is acute, then c2  a 2  b2 .
This extension of the Pythagorean Theorem seems obvious when we consider the Law of
Cosines, c2  a2  b2  2ab cos C , and that if C is acute, cosC is positive, but if C is
obtuse, cosC is negative.
73
Suggested Homework in Dugopolski
Section 7.2:
#1-13 odd
Section 7.3:
#19-27 odd
Suggested Homework in Ratti & McWaters
Application Exercise
An engineer wants to position three pipes at the vertices of a triangle, as
shown in the figure below. If the pipes, A, B, and C have radii 2 inches, 3
inches, and 4 inches, respectively, then what are the measures of the angles of
triangle ABC.
Homework Problems
#1 How far is the distance across the pond represented by BC ?
#2 Consider a triangle such that b  13 , c  6 , and A  36 . Solve the triangle.
#3 Consider a triangle such that a  7 , b  19 , and c  15 . Solve the triangle.
1
1  cos x

#4 Verify the identity
.
csc x  cot x
sin x
74
The Conic Sections
The conic sections include a family of curves generated when planes intersect a
cone. These curves include the circle, ellipse, parabola, and hyperbola. The online
encyclopedia MathWorld illustrates the conic sections with the graphic below.
Weisstein, Eric W. "Conic Section." MathWorld. Wolfram Web Resource. http://mathworld.wolfram.com/ConicSection.html
Parabolas
The intersection of a cone and a plane parallel to one of the straight lines that
generate the cone is a parabola. More precisely, a parabola is the set of points in a plane
such that the distance of each point in the set from a fixed point called the focus equals
the perpendicular distance from the point to a fixed line called the directrix. Since the set
of points equidistant from a fixed point and a fixed line defines a parabola, the distance
formula will give the standard equation of a parabola. The focus lies on the parabola's
axis of symmetry, and the directrix is perpendicular to the parabola's axis of symmetry.
The vertex is the point of the parabola that lies on the axis of symmetry. The vertex is an
equal distance from the focus and the nearest point on the directrix according to the
definition of a parabola. Let the vertex be the point (h,k). Assume the parabola has a
vertical axis of symmetry. Let c be the distance from the vertex to the focus. The focus
has the coordinates  h, k  c  . The equation y  k  c describes the directrix. To get the
equation of the set of points that are the same distance from the line y = k–f and the point
 h, k  c  , choose any point P with coordinates  x, y  such that the distance from P to the
focus equals the length of a perpendicular line segment stretching from P to y  k  c .
This line segment intersects y  k  c at a point D with the coordinates  x, k  f  . Since
the distance from P to the focus must equal the distance from P to D, the distance
formula gives an equation of a parabola.
 x  h   y   k  c 
2
2

Simplify and square both sides of the equation.
 x  x    y  (k  c ) 
2
2
75
 x  h   y   k  c 
2


 x  h   y   k  c 
2
 x  h   y   k  c 
2
Solve for y.
2
2
 y  (k  c ) 

2
2
 
 
2
 y  (k  c ) 
  y  (k  c ) 
2


2
2
 x  h    y   k  c     y  (k  c ) 
2
2
2
 x  h  y2  2 y k  c  k  c  y2  2 y k  c   k  c 
2
 x  h   2 yk  2 yc  k 2  2kc  c 2  2 yk  2 yc  k 2  2kc  c 2
2
 x  h   4kc  4 yc
2
x  h

y
k
2
2
2
4c
y
Let
1
2
  x  h  k
4c
1
 a to obtain the standard form of a parabola with a vertical axis of symmetry.
4c
1
2
y    x  h  k
4c
y  a  x  h  k
2
Accordingly, the equation in standard form of a parabola with vertex  h, k  and
the vertical line x  h as the axis of symmetry appears in the box below.
Let  x, y  be a point on a parabola with vertex  h, k  and the vertical
line x  h as the axis of symmetry, then
y  a  x  h  k
2
where  h, k  is the vertex and where a  c  1 4 and c represents the
distance from the vertex to the focus.
The equation in standard form of a parabola with vertex  h, k  and the horizontal line
y  k as the axis of symmetry appears in the box on the following page.
76
Let  x, y  be a point on a parabola with vertex  h, k  and the
horizontal line y  k as the axis of symmetry, then
x  a y  k  h
2
where
 h, k 
is the vertex and where a  c  1 4 and c represents the
distance from the vertex to the focus.
To graph a parabola, it is convenient to know its vertex. Consider the equation in
quadratic form, y = –2x2 – 4x + 6. The graph of a quadratic (or second-degree
polynomial) is a parabola. The parabola has a y-intercept at (0, 6). The x-intercepts
occur where y = 0, so solving the equation –2x2 – 4x + 6 = 0 gives the x-intercepts.
 2x2  4x  6  0
x2  2x  3  0
( x  1)( x  3)  0
x  1  0, x  3  0
x  1, x  3
(1,0) and ( 3,0)
The vertex, however, is not self-evident since the equation appears in quadratic form.
Completing the square can transform the parabola, however, to standard form. The goal
is to obtain an equivalent equation of the form y = a(x – h)2 + k for y = –2x2 – 4x + 6.
Write the equation.
y  2 x 2  4 x  6
Ignore the constant and factor out the leading term from the binomial.
y  2 x 2  4 x  6
y  2x 2  2 x   6
Use the binomial within the parenthesis to compose a perfect square trinomial in x.
Recall that (a ± b)2 = a2 ± 2ab + b2. In a perfect square trinomial, the linear term equals
twice the product of the two terms of the binomial. To complete a square, therefore, the
constant term must be the square of half the linear term. To complete the square, add the
square of half the linear term, [½(2)]2, which equals 1. Adding one within the parenthesis
actually adds negative two to the right side of the equation, so add negative two to the left
side of the equation.
y  2x 2  2 x   6
y  2  2x 2  2 x  1  6
Factor the perfect square trinomial to a squared binomial.
77
y  2  2x  1x  1  6
y  2  2x  1  6
2
y  2x  1  8
2
Now the equation is in standard form, and the vertex can be identified as (–1, 8).
The focus has the coordinates  h, k  c  , and the directrix has the equation
y  k  c . Since a  c  1 4 , c  1 8 . Thus, the focus is the point  1, 63 8 and the
equation y  65 8 describes the directrix.
78
Suggested Homework in Dugopolski
Section 10.1: #19, #27-35 odd, #51-59 odd
Suggested Homework in Ratti & McWaters
Section 10.2: #7, #27, #31, #41-45 odd
Application Exercise
Eavesdroppers use a microphone placed at the focus of a parabolic shield as
shown below in order to “pull in voices” from far away.
microphone at focus
vertex
sound waves
1 2
y trace out the parabola of the shield. How far from the
12
vertex of the shield should the microphone appear to maximize it’s effectiveness?
Let the equation x  
Homework Problems
2
2
Write each equation in y  a  x  h   k or x  a  y  k   h form. Identify the vertex,
focus, and directrix of each parabola.
#1 y  x 2  8x  3
#3 y  2 x2  12 x  5
#5 x   12 y 2  y  4
#2 x  y 2  y  6
#4 y  x 2  2 x  5
#6 y   18 x 2  5
Verify the identities below.
1
1  sin x

#7
1  sin x cos 2 x
#8 1  csc x sec x 
cos   x   csc   x 
cos x
79
Circles
The intersection of a cone and a plane perpendicular to the axis of the cone is a circle. A
circle is a set of points in a plane, each of which is equidistant from a fixed point called the
center. The distance between the points of the circle and the center is the radius. Since the set of
points equidistant from a fixed point defines a circle, the distance formula will give the standard
equation of a circle. Suppose that a circle has center (h,k) and radius r > 0. If (x,y) is any point
on the circle, then we can use the distance formula to write ( x  h) 2  ( y  k ) 2  r . Squaring
both sides of this equation gives the standard equation of a circle:
x  h 2   y  k 2  r 2
This equation can be expanded as shown here: x 2  2 xh  h 2  y 2  2 yk  k 2  r 2 . This
expansion is called the general form of the equation of a circle when –2h equals c, –2k equals d,
and k2 – r2 equals e, so that the general form is x2 + y2 + cx + dy + e = 0 where c, d, and e are real
numbers.
To graph a circle, affix an endpoint of a line segment whose length equals the radius at
the center then rotate the segment about 360o. Consider a circle described by the equation
x2 + y2 = 16. The circle has a radius of 4 and a center at (0,0). The graph appears below.
The points that compose the circle satisfy the equation x2 + y2 = 16, and each point is 4 units
from the center (0,0).
If the radius increases, the circle widens. Consider the circle described by x2 + y2 = 49
with a radius of 7 and a center at (0,0) as shown on the following page.
80
Similarly, if the radius decreases, the circle tightens. Consider the circle described by x2 + y2 = 4
with a radius of 2 and a center at (0,0). The graph is below.
If the center moves, the entire circle moves, creating a translation of the circle with center at the
origin.
81
Recall the standard equation of a circle, (x – h)2 + (y – k)2 = r2 where (h,k) is the center
and r is the radius. Consider the original example, x2 + y2 = 16, which has its center at (0,0). If
the center moves from (0,0) to (–3,2), the equation changes from x2 + y2 = 16 to
(x + 3)2 + (y – 2)2 = 16, and the graph moves as shown on the next page.
To graph a circle, it is convenient to know its center and radius. Consider the equation in
general form, x2 + y2 – 6y + 4y – 3 = 0. The center is not self-evident when the equation is
written in general form. This equation can be written in the standard form following a process
called completing the square. The goal is to obtain an equivalent equation of the form
(x – h)2 + (y – k)2 = r2 for x2 + y2 – 6y + 4y – 3 = 0. Write the equation with the constant on the
right.
x 2  y 2  6x  4 y  3  0
x 2  y 2  6x  4 y  3
Use the commutative property to gather like variables.
x 2  y 2  6x  4 y  3
x 2  6x  y 2  4 y  3
Compose a perfect square trinomial in both x and y. Creating perfect square trinomials is called
completing the square. Recall that (a ± b)2 = a2 ± 2ab + b2. The middle term is frequently a
linear term, and the last term is frequently a constant. In a perfect square trinomial, the linear
term equals twice the product of the two terms of the binomial. To complete a square, therefore,
the constant term must be the square of half the linear term. To complete the square in x, add the
square of half the linear term, [½(–6)]2. To complete the square in y, add the square of half the
linear term, [½(4)]2.
x2  6 x  y 2  4 y  3
x 2  6 x   12 6  y 2  4 y   12 4  3   12 6   12 4
2
Simplify, creating two perfect square trinomials.
2
2
2
82
x 2  6 x   3  y 2  4 y   2   3   3    2 
2
2
2
2
x2  6x  9  y 2  4 y  4  3  9  4
x 2  6 x  9  y 2  4 y  4  16
Factor the perfect square trinomials to squared binomials.
x 2  6 x  9  y 2  4 y  4  16
x  3x  3   y  2 y  2  16
x  32   y  22  16
Now the equation is in standard form, and the center can be identified as (3, –2) and the radius as
four.
The circle is a special form of an ellipse, which is a conic section defined as the set of
points in a plane such that the sum of their distances from two fixed points called the foci is
constant.
Ellipses
The box below contains the equation of an ellipse.
An ellipse centered at  h, k  with horizontal major axis of length 2a and
vertical minor axis of length 2b has the equation below.
x  h 2   y  k 2
a2
b2
1
Similarly, an ellipse centered at  h, k  with vertical major axis of length 2a
and horizontal minor axis of length 2b has the equation below.
 y  k 2  x  h 2
a2
b2
1
An ellipse has a pair of foci that lie on the major axis. The foci of ellipses described by
the equations in the above box lie c units from the center  h, k  . The equation c2  a 2  b2
yields c.
To graph an ellipse it is convenient to know the length of the major and minor axes and to
know its center. It is also important to find the foci since the definition of the ellipse depends
upon the focal points. Consider the ellipse described by the equation y 2 16  x2 9  1 The
equation shows that h and k equal zero, so the center occurs at the origin. The equation also
shows that a2 = 16 and b2 = 9; therefore, a = 4 and b = 3. The major axis is vertical and it
83
extends from (0,–4) to (0,4). The minor axis is horizontal and it extends from (–3,0) to (3,0).
The equation c2  a 2  b2 obtains the distance from the center to the foci as shown here.
c 2  a 2  b2
c 2  16  9

c 7
 

Consequently, the foci occur at 0, 7 and 0, 7 . The graph of the ellipse is below.
Let’s assume we need to identify the foci of an ellipse with the equation below.
 x  2
16
2

First, we rewrite the equation in standard form.
 y  2
25
2
1
84
 x   2    y  2
2
42
52
2
1
Since a > b and 5  4 , a cursory examination of the given equation reveals that it describes an
ellipse with a vertical major axis. The center of the ellipse is  2, 2  since h  2 and k  2 .
The major axis lies on the line x  2 .
The foci of an ellipse are always c units from the center on the major axis where
2
2
c  a  b2 and a  b . Substituting for a and b, we obtain the following.
c 2  25  16
c2  9
c  3
Since the major axis is vertical, add 3 to k, the y-value of the center, to find the foci. The foci
are  2, 1 and  2,5 .
The box below defines the eccentricity e of an ellipse.
Let e represent the eccentricity of an ellipse whose foci
are c units from the center and whose major axis equals
2a units in length. Then, we define eccentricity as
below.
ec a.
Since 0  c  a , we have 0  e  1 . For an ellipse that appears circular, the foci are close to the
center. Hence c is small compared to a, so the eccentricity of a nearly circular ellipse is near
zero. Elongated ellipses have eccentricity near one.
85
Suggested Homework in Dugopolski
Section 10.2: #25-33 odd, #37, : #47-63 odd
Suggested Homework in Ratti & McWaters
Section 10.3: #7, #25, #29, #33, #45, #51
Application Exercise
For a satellite in an elliptical orbit around a planet, perigee, P, is defined to be its
closest distance to the planet and apogee, A, is defined to be its greatest distance from
the planet.
Let e represent the eccentricity of an ellipse. Show that e 
A P
.
A P
Homework Problems
#1 Write x  4 x  y  6 y  9 in standard form. Identify the center and radius and graph the
circle in the Cartesian Plane.
2
#2 Consider
2
 x  2 2
16

y 1
4
2
 1 . Graph the ellipse and identify its foci.
#3 Consider 9 x2  18x  4 y 2  16 y  11 . Identify the center and foci.
#4 Consider 4 x2  16 x  y 2  6 y  21 . Sketch the graph and label the foci.
#5 Calculate the eccentricity of the ellipse described by
#6 What is the eccentricity of a circle?
#7 Verify the identity
csc x sin x
sin x
 cot 2 x .
x2
5

y2
9
 1.
86
Hyperbolas
A hyperbola has a center  h, k  and two important axes, the conjugate and transverse
axes. The transverse axis connects the vertices of the hyperbola. The conjugate axis is
perpendicular to the transverse axis. If the distance from the center to one of the vertices is a and
the distance from the center to one of the foci is c, the length of the conjugate axis is said to have
a length of 2b where c2  a 2  b2 . The transverse and conjugate axes form a rectangle with
dimensions a × b. This rectangle, the fundamental rectangle, has corner points that share
common x-coordinates with the vertices if the transverse axis is horizontal and common ycoordinates with the vertices if the transverse axis is vertical.
2a
2b
The following box contains the equation of a hyperbola.
A hyperbola centered at  h, k  with horizontal transverse axis of
length 2a and vertical conjugate axis of length 2b has the equation below.
x  h 2   y  k 2  1
a2
b2
This hyperbola has foci at  h  c, k  where c2  a 2  b2 .
Similarly, a hyperbola centered at  h, k  with vertical transverse
axis of length 2a and horizontal conjugate axis of length 2b has the
equation below.
 y  k 2  x  h2
a2
b2
1
This hyperbola has foci at  h, k  c  where c2  a 2  b2 .
87
Hyperbola approach a pair of asymptotes. For hyperbola with a horizontal transverse axis the
b
points on the hyperbola approach the lines y  k  x  h  as |x| approaches infinity. These
a
lines are asymptotes of the hyperbola and extended diagonals of the fundamental rectangle. If
the transverse axis is vertical the asymptotes will have the same equations with reciprocal slopes
a
y  k  x  h  .
b
The diagram below summarizes the important aspects of a hyperbola.
2a
2b
2b
2a
To graph a hyperbola it is convenient to know the dimensions, a×b, of the fundamental
rectangle and to know its center because the asymptotes pass through the vertices of the
fundamental rectangle and the center. It is also important to find the foci since the definition of
the hyperbola depends upon the focal points. Consider the hyperbola given by the equation
x  12   y  22  1 . The transverse axis is horizontal. The center is (1,–2). The vertices are
4
9
a-units away from the center along the transverse axis, so they occur at (3,–2) and (–1,–2). The
foci are c-units away from the center along the transverse axis. Recall that c2 = a2 + b2, so c2 =
13 and c = 13 . The foci, therefore, occur at (1+ 13 ,–2) and (1– 13 ,–2). The asymptotes are
the lines y  2  3 x  1 and y  2  3 x  1 . The graph appears on the following page.
2
2
88
 x  1
4
2

 y  2
9
2
1
89
Suggested Homework in Dugopolski
Section 10.3: #19-27 odd, #35, #47
Suggested Homework in Ratti & McWaters
Section 10.4: #15, #27, #31, #49, #65
Application Exercise
A cooling tower for a nuclear power plant has a hyperbolic cross section as shown
in the diagram below.

48 11 feet
 0, 0 
100 feet
240 feet
The diameter of the tower at the top and bottom is 240 feet while the diameter in the
middle is 200 feet. The height of the tower equals 48  11 feet. Find the equation of
the hyperbola, using the coordinate system shown in the diagram.
Homework Problems
#1 Find the equations of the asymptotes for the hyperbola described by 4 y 2  9 x2  36 .
#2 Graph the hyperbola
y2
25
x
 144
 1.
2
#3 Determine the center, foci, and asymptotes of the hyperbola given by x 2  y 2  1 .
#4 Graph the hyperbola
 x  2 2
9

y 1
4
2
 1.
#5 Let 9 x  16 y  18x  64 y  199 . Find the center, vertices, foci and asymptotes of the conic.
2
2
#6 Verify the identity sin x  3cos2 x  sin 2 x   sin  3x  .
90
Vectors
We use the term vector to refer to a list of numbers. If the list includes just two numbers,
an ordered pair, we call the list a vector in 2 . In this course, we will deal exclusively with
vectors in 2 and commonly call them two-space vectors or just simply vectors.
Let z1  z2 be real numbers. Then, u , v , w , z , OA , and 0 below are all examples of
vectors and accepted notations for vectors.
3
u    , v  1, 4 , w   0.2,0.3 , z  z1 , z2 , OA   2,3 , and 0  0, 0
 2 
Two 2 vectors are equal if and only if their corresponding entries are equal. Hence, if
z  v from the examples above, then z1  1 and z2  4 . Given two 2 vectors like u and v ,
their sum is the vector u  v whose entries are the sums of the corresponding entries of u and
v . Let u  3, 2 and v  1, 4 . Then, u  v  3  1, 2  4  2, 2 . Given a vector u and
a real number c, then cu is a scalar multiple of u obtained by multiplying each entry in u by c.
For instance, if u  3, 2 and c  7 ,then cu  7 3, 2  7  3 ,7  2   21, 14 .
In summary then, we have the following definitions.
A two-space vector is an ordered pair of real numbers. If
both numbers are zero, we call the vector the zero vector.
Let u  u1 , u2 and v  v1 , v2 be two-space vectors and let k be any
real number. Then, ku  ku1 , ku2 , u  v  u1  v1 , u2  v2 , and
u  v  u1  v1 , u2  v2 . We call ku a scalar multiple of u . We call u  v
the sum (or the resultant) of the vectors u and v . We call u  v the
difference of the vectors u and v .
Geometric Representation of Vectors
Consider a rectangular coordinate system in the plane. Since every point in the plane is
determined by an ordered pair, we can identify a geometric point  a, b  with a vector a, b .
Hence, we may regard 2 as the set of all vectors with two real number entries just as we regard
2
as the set of all points in the plane.
The geometric visualization of a nonzero vector is a directed line segment from the origin
to the point. We aid this visualization by including an arrow at the terminal end of the segment
directing our attention to the point that is the vector. It turns out, that for some applications, we
can visualize this directed line segment as free of its standard position in the plane and just pick
91
it up and place it elsewhere in the plane. Nevertheless, we will adhere to directed line segments
in standard position exemplified by vector OA below.
We call the angle that the vector makes with the x-axis its direction. We call the length
of the vector its magnitude or norm. We denote magnitude with absolute value or double
absolute value. Hence, OA or OA refers to the magnitude of the vector OA , which we obtain
from the Pythagorean Theorem.
Let u  u1 , u2 be a two-space vector. We call u  u12  u22 the magnitude of u .
The sum of two vectors has a useful geometric representation summarized by the
Parallelogram Rule stated below.
Let u  u1 , u2 and v  v1 , v2 be two-space vectors. Then, the resultant
u  v corresponds to the fourth vertex of the parallelogram whose other
vertices are u , v , and 0 .
The following figure demonstrates the rule in action.
92
Considering the Parallelogram Rule, we can view any nonzero vector as the sum or
resultant of two special vectors called the horizontal component and the vertical component. If
u is a vector with magnitude r and direction  , we denote the horizontal component of u as u x
and the vertical component of u as u y . The figure below provides an illustration.
ux
u
uy
Using the figure above and the trigonometric ratios, we obtain the following equations.
cos  
sin  
ux
r
uy
r
 u x  r cos 
 u y  r sin 
If the direction of u is negative, then we can write the following.
u x  r cos 
u y  r sin 
Above, we defined the sum and difference of two vectors. We also defined scalar
multiplication. Intuitive readers may wonder about vector multiplication. Actually, there is more
than one definition for multiplying two vectors. We are interested in the dot product, defined
below.
The dot product of two nonzero vectors v  v1 , v2 and w  w1 , w2 is the
number
v w  v1w1  v2 w2  v w cos 
where  is the angle between v and w , 0     . If either v or w is 0 ,
we define v w  0 .
93
The following figure illustrates the dot product.
If the angle between v and w is 90 , then the dot product of v and w will be zero because
cos90  0 . Since any nonzero vector has a positive magnitude, the only way the dot product of
two nonzero vectors can equal zero is if the angle between the two vectors is a right angle.
Hence, we have the theorem below.
Two nonzero vectors u and v are perpendicular if and only if u  v  0 .
Linear Combinations
We call any vector with a magnitude of one unit a unit vector. Two special unit vectors
are the vectors i  1, 0 and j  0,1 . We can represent any vector u  u1 , u2 using i and j
as below.
u  u1 , u2  u1 1,0  u2 0,1  u1i  u2 j
We call the form u1i  u2 j a linear combination.
Given vectors v1 , v 2 ,
given by
, v n and given scalars c1 , c2 ,
z  c1v1  c1v 2 
is called a linear combination of v1 , v 2 ,
, cn , the vector z
 cn v n
, v n with scalars c1 , c2 ,
, cn .
94
Applications of Vectors
We can use vectors to represents a force because a force has both magnitude and
direction. If several forces are acting on an object, the resultant force experienced by the object
is the vector sum of these forces.
Consider a 100-lb weight hanging from a horizontal surface by two wires as shown in the
diagram below.
50
32
T1
T2
100 lb
We can find the magnitude of the tensions T1 and T2 in both wires. We first express T1 and T2
in terms of their horizontal and vertical components as below.
T1   T1 cos  50  i  T1 sin  50  j
T2  T2 cos  32  i  T2 sin  32  j
The resultant T1  T2 of the tensions counterbalances the weight w , so we have the following
equalities below.
T1  T2  w  100 j
 T
1
cos  50  i  T1 sin  50  j   T2 cos  32  i  T2 sin  32  j  100 j
 T1 cos  50  i  T2 cos  32  i  T1 sin  50  j  T2 sin  32  j  100 j
 T
1
cos  50   T2 cos  32   i   T1 sin  50   T2 sin  32   j  0i  100 j
Thus, we have the system of equations below.
 T1 cos  50   T2 cos  32   0
T1 sin  50   T2 sin  32   100
Next, we solve the first of these equations for T2 as below.
95
 T1 cos  50   T2 cos  32   0
T2 cos  32   T1 cos  50 
T2 
T1 cos  50 
cos  32 
Substituting for T2 into the second equation, we obtain the following.
T1 sin  50   T2 sin  32   100
T1 sin  50  
T1 cos  50 
cos  32 
sin  32   100
T1  sin  50   cos  50  tan  32    100
T1 
100
sin  50   cos  50  tan  32 
T1  85.64 lb
Returning to our expression for T2 , we conclude as below.
T2 
T1 cos  50 
cos  32 
T2  85.64 lb
T2  64.91 lb
cos  50 
cos  32 
96
Section 7.3:
Suggested Homework in Dugopolski
#13-29 odd, #37-69 odd
Section 7.5
Section 7.6
Suggested Homework in Ratti & McWaters
#7-21 odd, #27-61 odd
#7-13 odd, #21-29 odd, #39-43 odd
Application Exercise
The Cauchy-Schwarz Inequality says that if v and w are two vectors, then v  w  v w .
Under what conditions is v  w  v w ?
Homework Problems
Find the magnitude and the direction of the vectors below.
6
 0
#1
#2  
#3  
3,1
6
 1 
#4 2i  5j
 2 
#5 

 2 3 
Let a  1,3 , b  6, 2 , and c  2, 0 and perform the indicated operations.
#6 a  c
#7 a c
#8 b  3a
#9 b a
Find the smallest positive angle between each pair of vectors.
#11 u  3, 2 , v  4,5
#12 a  9, 6 , b  4, 6
Verify the following identities.
1  cos 2  2x  1  cos x

#13
1  sin 2  2x  1  cos x
#14
1  tan 2 θ
 cos 2θ
1  tan 2 θ
#10 c
97
Complex Numbers
When we study algebra, we discover that not every quadratic equation has a solution set
of real numbers. For example, x2  1  0 has no real-number solution since there is no real
number x such that x   1 . In this lecture, we will construct a set of numbers in which
every quadratic equation has a solution starting with the definition below.
1  i
Thus, we define i to be the number whose square is negative one, that is, i 2  1 . Hence, i are
solutions for x2  1  0 .
To continue our construction to include solutions for equations of the form x2  a  0
where a  0 , we state the definition below.
If a  0 , then
a  a  i .
We call the numbers defined above imaginary numbers.
Let b be a real number. Then, we call bi an imaginary number.
If we add an imaginary number to a real number, we define the result to be a complex number.
Let a and b be real numbers. Then, we call a  bi a complex number.
Note that the definition above allows a and b to be zero. Hence, any imaginary number like 5i
is a complex number, but the converse is not true since 2  5i is complex but not imaginary.
Moreover, any real number a is complex since a  a  0i , but, again, the converse is not true
since 2  5i is complex but not real. All this leads us to the definition below.
Let a and b be real numbers. Then, we call a  bi a complex number. If
b  0 , we call a  bi a non-real complex number. If a  0 , we call a  bi an
imaginary number. If a  0  b  0 , we call a  bi a real number.
We call the form a  bi the standard form of a complex number.
We define equality among complex numbers as below.
Two complex numbers a  bi and c  di are equal if
and only if a  c and b  d .
We define addition and subtraction as follows.
 a  bi    c  di    a  c    b  d  i
 a  bi    c  di    a  c    b  d  i
98
We define multiplication and division as below.
Let c  di be a non-zero complex number. Then,
 a  bi  c  di   ac  adi  bci  bdi 2   ac  bd    ad  bc  i
and
1
c  di
1 c  di
  a  bi  

c  di c  di
c  di
  a  bi   2
c  d2
ac  adi  bci  bdi 2

c2  d 2
 ac  bd    ad  bc  i .

c2  d 2
 a  bi    c  di    a  bi  
Note that the definition for division relies on the interesting fact that the product of a complex
number and its conjugate is real.
There exists a one-to-one correspondence between the set of real numbers and the set of
imaginary numbers. Moreover, these two sets have an element in common, namely, 0  0i .
This one-to-one correspondence and the common element allow us to graph the set of complex
numbers in a coordinate plane.
For the coordinate system, the horizontal axis is the real axis while the vertical axis is the
imaginary axis. We plot real numbers as points on the horizontal axis and imaginary numbers as
points on the vertical axis. Non-real, non-imaginary complex numbers fall into one of the four
quadrants. Note the complex numbers 3, 4i , and 2  4i in Figure 1 below.
Figure 1
Figure 2
99
We sometimes use a geometric vector with the initial point at the origin and terminal point  a, b 
to represent a complex number as shown in Figure 2 above. The magnitude of the vector equals
the absolute value or modulus of the complex number.
Let a  bi be a complex number. Let r  a  bi . Then, we call r
the modulus of a  bi , and r  a 2  b2 .
Using the geometric vector representation of a complex number, we can derive a form for
complex numbers called the trigonometric form. Looking at the geometric representation below,
we see that a  r cos and b  r sin  .
Substituting into the standard form we obtain the following.
a  bi  r cos  r sin   i  r  cos  i sin  
Hence, we have the definitions below.
Let z  a  bi be a complex number with modulus r. Let  be the
angle in standard position whose terminal side contains the point
 a, b  . Then,  is the argument of z. Moreover, the trigonometric
form of z is z  r  cos   i sin   and the exponential form of z is
z  rei .
The trigonometric and exponential forms simplify the process of finding the product of
two complex numbers according to the theorem below.
Let z1  r1  cos 1  i sin 1  and z2  r2  cos 2  i sin 2  . Then,
z1 z2  r1r2 cos 1  2   i sin 1  2   r1r2e  1
i  2 
100
For proof, we multiply and employ Sum Identities as below.
z1 z2  r1  cos 1  i sin 1   r2  cos  2  i sin  2 
z1 z2  r1r2  cos 1  i sin 1  cos  2  i sin  2 
z1 z2  r1r2 cos 1 cos  2  cos 1i sin  2  cos  2i sin 1  i 2 sin 1 sin  2
z1 z2  r1r2 cos 1 cos  2  sin 1 sin  2  cos 1i sin  2  cos  2i sin 1
z1 z2  r1r2 cos 1 cos  2  sin 1 sin  2  i  cos 1 sin  2  cos  2 sin 1  
z1 z2  r1r2 cos 1   2   i sin 1   2  
The trigonometric and exponential forms also simplify the process for finding the
quotient of two complex numbers according to the theorem below whose proof so similar that it
is left as an exercise.
Let z1  r1  cos 1  i sin 1  and z2  r2  cos 2  i sin 2   0 . Then,
z1 r1
r i  
 cos 1   2   i sin 1   2    1 e  1 2 
z2 r2
r2
Finally, finding powers of complex numbers simplifies according to De Moivre’s Theorem
below.
Let z  r  cos   i sin   . Let n be any positive integer. Then,
z n  r n  cos n  i sin n   r nein
101
Section 7.4:
Suggested Homework in Dugopolski
#1-7 odd, #13-19 odd, #29-33 odd, #41-53 odd
Section 7.8:
#7-49 odd
Suggested Homework in Ratti & McWaters
Application Exercise
V
.
Z
Use Ohm’s Law to find the current given voltage V  120  cos 60  i sin 60 and
Ohm’s Law relates current I, voltage V, and impedance Z with the equation I 
impedance Z  8  cos30  i sin 30 .
Homework Problems
Find the modulus and argument of each complex number.
#1 z   3  i
#2 z  2  2i
#3 z  3  2i
Write each complex number in standard form, trigonometric form, and exponential form.
2π i
#4 2cis   π6 
#5 3  3i
#6 e 3
Let z1  3  3 3i and z2  4 3  4i . Convert the complex numbers to trigonometric form and perform
the indicated operations.
2
#7 z1 z2
#8 z2 z1
#9  z1 z2 
Verify the identity below.
#10 cos3 θ sin θ  sin 3 θ cos θ 
sin 4θ
4
102
Powers and Roots of Complex Numbers
In this lesson, we present a theorem without proof then use the theorem to find all the
roots (real or non-real) of a polynomial equation. We start with the definition below.
The complex number w  a  bi is an nth root of the complex
n
number z if  a  bi   z .
Now, we state the theorem.
For any positive integer n, then w is an nth root of the complex
number z  r  cos   i sin   if
1
    2k
w  r n cos 
n
 
for k  0,1, 2,

   2k  
  i sin 

n  


, n 1 .
Using this theorem, we can solve the equation x6  1  0 as below.
x6  1  0
x6  1
x 61
From the above work, we see that solving x6  1  0 requires finding the six sixth-roots of 1. To
apply our theorem, we first note that 1  1 cos 0  i sin 0  . Then, we substitute into the
  0  2k
expression w  1 cos 
  6
we obtain,
1
6

 0  2k
  i sin 

 6
  0  2  0 
w0  1 cos 
6
 
1
6

  for the values of k  0,1, 2,


 0  2  0 
  i sin 
6


, 5 . For k  0,
   0 
 0 
   1 cos    i sin     1 .
   6 
 6 
For k  1,
1
  0  2 1 
w1  16 cos 
6
 
3

 0  2 1       
   1
.
  i sin 
   1 cos    i sin      i
6
2


   3 
 3  2
103
For k  2,
  0  2  2  
1
3
 0  2  2      2 
 2  
.
w2  1 cos 
  i sin 
   1 cos 
  i sin 
    i
6
6
2
2


   3 
 3 
 
1
6
For k  3,
1
  0  2  3
w3  16 cos 
6
 

 0  2  3   
  i sin 
   1 cos    i sin     1 .
6



For k  4,
  0  2  4  
 0  2  4 
w4  1 cos 
  i sin 
6
6


 
1
6
   4 
 4
   1 cos 
  i sin 
   3 
 3
1
3

.
    i
2
2

   5
   1 cos 
   3
3
 1
.
   i
2
 2
Finally, for k  5 ,
  0  2  5 
w5  1 cos 
6
 
1
6

 0  2  5 
  i sin 
6


Hence, the six sixth-roots of 1 are 1, 1 ,

 5
  i sin 

 3
1
3
1
3
1
3
1
3
,  i
,  i
, and  i
.
i
2
2
2
2
2
2
2
2
104
Suggested Homework in Dugopolski
Section 7.5:
#41, #42, #43, #44, #51, #52
Suggested Homework in Ratti and McWaters
Section 7.8:
#53, #59, #65-69 odd
Application Exercise
Powers of i possess an interesting periodic property. Since i 2  1 , we obtain
the following pattern.
i 2  1
i 3  i 2  i  1  i  i
i 4  i 2  i 2  1  1  1
i 5  i 4  i  1 i  i
i 6  i 4  i 2  1  1   1
i 7  i 4  i 3  1  i   i
i 8  i 4  i 4  1 1  1
Thus, every natural number power of i can be expressed as one of the numbers i , 1 ,
i , or 1 . Use this fact and the pattern above to simplify i 228 .
Homework Problems
#1 Find the square roots of i .
#2 Find the cube roots of 4  4 3i .
#3 Find all the solutions to x4  8  8i 3  0 .
#4 Find all the solutions to x3  1  0 .
#5 Use identities to find the exact value of sin  58π  .
105
Polar Coordinates
In previous lectures, we have used the rectangular coordinate system, which employs an
ordered pair  x, y  to represent a point in a plane. The two numbers, the abscissa and the
ordinate, respectively, represent the distances between the point and two perpendicular axes (the
x and y-axes). Polar coordinates represent another way to represent points in a plane. With polar
graphing, we need only one axis called the polar axis and a point on it called the pole. These two
constructs correspond to the x-axis and the origin in the rectangular polar system. Sometimes we
include the y-axis when we represent the polar coordinate system, but we do so only to help
orient ourselves after having accustomed ourselves to the rectangular coordinate system.
With polar coordinates, we still employ an ordered pair to represent a point in a plane, but
only one of the numbers represents a linear distance. When the first number is positive, we
imagine the point as a point on the terminal side of an angle in standard position. The first
number in the ordered pair is r, the distance from the pole to the point. The second number in
the ordered pair is  , the measure of the angle on whose terminal side the point lies.
 3 
P :  2, 
 4 
r2

pole
3
4
polar axis
As we have discovered, an angle in standard position has an infinite number of
coterminal angles. Hence, an infinite number of ordered pairs can represent a single point in the
polar coordinate system just by using coterminal measures for  . Worse, we can also imagine
points as laying on the ray opposite the terminal side of  , in which case, r  0 , as shown with
the point Q below.
y
R :  2,  6
polar axis
Q :  2,  6 
To plot a point in polar, requires two basic steps. First, look at  and orient in the correct
direction. Second, travel exactly r -units from the pole in the oriented direction. If r  0 , go in
the opposition direction. It is important to note that a point any particular quadrant does not
106
necessarily have a particular sign. For instance, all of the following coordinates describe point R:
 2,  6 ,  2,7 6 ,  2, 11 6  , and  2, 5 6  . Besides, the idea of quadrants is a
holdover from rectangular coordinates anyway since we really only have one axis in the polar
coordinate system.
In the rectangular coordinate system, y  b represents a horizontal line passing through b
on the y-axis. The polar coordinate system has its own analog to this constant function because
the equation    represents a line passing through the pole with a direction equal to  . For
instance, we can use the equations    6 or   30 or   7 6 to describe the line
containing the pole and points R and Q below.
y
R :  2,  6
polar axis
Q :  2,  6 
Similarly, x  a represents a vertical line passing through a on the x-axis in the rectangular
coordinate system, and this relation has an analog in the polar system because the equation r  a
represents a circle centered at the pole with a radius a .
r 1
polar axis
107
To aid in graphing points and relations, it is common to create polar grids using circles as
analogs to tick marks on the axes in the rectangular system as shown below.
Just as we first learn to graph relations in the rectangular coordinate system using pointby-point plotting, so can we graph relations in the polar system. Consider a function in the form
r  f   . The following items are helpful for generating the graph of the function in the polar
plane.
1. Look for symmetry and know that there are many common graphs such as
circles, cardoids, limaçons, lemniscates, spirals, and rhodonea curves (see the
function gallery in section 7.6 of the textbook).
2. Identify the values of  that maximize and minimize r .
3. Identify the zeros of the function, i.e., find   r  0 . These will be angle
values where the curve passes through the pole.
4. When in doubt about the shape of the graph, methodically plot points for
incremental increases in  .
5. Plot points over a full cycle of the un-dilated function.
6. Watch for the graph to repeat itself, tracing out the same shape again
periodically.
108
Let’s graph r  sin 2 . Looking at the graph of y  sin 2 x in the rectangular coordinate
system given on the following page, we can read values of r and  if we treat the x-values as
angles and y-values are radii.
The interval I    ,   represents a full cycle of y  sin x . The graph of y  sin 2 x passes
through the x-axis at x   , x    2 , x  0 , x   2 , and x   . Treating the x-values as
angles and y-values are radii, we see that the curve in polar form will have a radius of zero
(meaning it will pass through the pole) when  equals  ,  2 , 0 ,  2 , and  . The graph
in rectangular form takes a minimum value of 1 when x   4 and x  3 4 . Likewise,
y  sin 2 x has maximum value of 1 when x  3 4 and x   4 . Now, we generate a table
for increasing  -values making sure to include these key x-values as key  -values.


r
0

1
2
11
12

1
3
4

1
2
7
12

0

2
5
12
1

2



4
1

0
12
1

2
0


12
1
2

4
1
5
12
1
2

2
0
7
12
1

2
3
4
1
11
12
1

2
Plotting these points, the graph below takes shape as shown. Plotting further values of the
function for greater values of  will show the graph repeating itself. We call this curve the
quadrifolium. It belongs to a family of curves called rhodonea curves, sometimes referred to as a
rose curve.

0
109
We can put our familiarity with the rectangular equations to use by converting polar
equations to rectangular form or vice versa. When we convert from one system to the other, we
recall the relationships established in our definitions of the trigonometric functions, namely, the
following.
x2  y 2  r 2 , x  r cos  , and y  r sin 
For example, consider the rectangular representation y  x 2 . We start by substituting for y and
x as below.
y  x2
r sin    r cos  
2
r sin   r 2 cos 2 
Dividing both sides by r, we obtain the following.
r sin   r 2 cos 2 
sin   r cos 2 
or r  0
(If the reader is wondering where the equation r  0 came from, just remember that only Chuck
Norris can divide by zero.) Dividing by cos2  , we continue converting below using basic
identities.
sin   r cos 2 
sin 
r
cos 2 
1 sin 
r
cos  cos 
r  sec  tan 
Since r  0 , represents the pole, which is included in the graph of r  sec2  sin  when   0 ,
we may drop r  0 , and simply represent y  x 2 as r  sec tan  .
110
Let’s consider converting r  1  cos to rectangular representation. Considering the
relationship x 2  y 2  r 2 , we multiply the equation by r as follows.
r  1  cos 
r 2  r  1  cos  
r 2  r  r cos 
r 2  r cos   r
r  r 2  r cos 
Next, we square both sides and obtain the following.
r  r 2  r cos 
r 2   r 2  r cos  
2
Now, we conclude as below using substitution.
r 2   r 2  r cos  
2
x2  y 2   x2  y 2  x 
2
Thus, we have converted the relationship from polar representation to rectangular representation.
This process of conversion sometimes has the possibility of introducing extraneous roots
as above both when we squared both sides and when we multiplied the equation by the variable
r. To check, we consider both steps. For instance, we multiplied both sides by r to go from
r  1  cos to r  r 2  r cos , but in this case no extraneous roots are introduced because r  0
on the graph of r  1  cos when    . Since the graph of r  1  cos passes through the
pole, then changing the equation to r  r 2  r cos does not introduce a root not already present.
Squaring both sides of an equation can also introduce extraneous roots. When we squared both
sides of the equation, we transformed r  r 2  r cos into r    r 2  r cos   . We have already
established above that r 2  r cos  1  cos . Moreover, the graph of r  1  cos is equivalent
to the graph of r  1  cos . Hence, squaring did not introduce any extraneous roots.
Comparing the polar representation r  1  cos to the rectangular representation
x 2  y 2   x 2  y 2  x  , we begin to see how polar has certain advantages (in this particular
2
case) since graphing x 2  y 2   x 2  y 2  x  does not look like a simple task. Indeed, there was
2
a time when the polar coordinate system competed so to speak with the rectangular system for
the dominant method for representing relations in a plane. In large part, the rectangular system
won, but polar still has its particular demesne where it holds sway.
111
Suggested Homework in Dugopolski
Section 7.6:
#9-11 odd, #27, #29, #55-69 odd, #75-81 odd, #87-95 odd
Suggested Homework in Ratti and McWaters
Section 7.7:
#23-59 odd, #63-71 odd
Application Exercise
Show that a polar equation for the straight line y  mx is   tan 1 m .
Homework Problems
Convert the coordinates below from  x, y  to  r , θ  .
#1  3, 6 
#2 1, 1


#3  3,1
Convert the coordinates below from polar to rectangular.
#5  6, 76π 
#6  3, π2 
#7  2,  56π 
#4
2
#8
 1, 210
3, 2
Convert each polar equation to rectangular form.
#9 r  2cos θ
#10 r cos θ  1
#11 r 
1
cos θ  sin θ
Convert each equation to polar form.
#13 x2  y 2  16
#14 x 2  y 2  6 y
#15 x2  y 2  3x  4
Graph the curve using polar coordinates.
#17 r  4cos3θ
#18 r  9
#19 θ  π3
#20 Verify the identity:
cos  2θ   1
1

.
tan θ
sin  2θ 
#12 r  3
#16 y  x

112
Parametric Equations
We are familiar with functions of the form y  f  x  or z  f  x, y  . In either case, we
have a direct relationship between x and y. Sometimes, however, expressing x and y in terms of
a third variable called a parameter is convenient. In other words, using t as the parameter, we
let x  f  t  and y  g  t  , and we call these equations parametric equations.
For instance, consider the parametric equations x  sin t and y  cos t . If t  0 , then
x  0 and y  1. Thus, the point  0,1 is a point of the graph described parametrically by
x  sin t and y  cos t . Note that we still use the x-axis and y-axis in the coordinate system. The
parameter t does not appear on the graph. Finding more points, we generate the table and graph
below.
t
0
 6
 3
 2
2 3
x  sin t
0
12
3 2
1
5 6
3 2
12

7 6
0
1 2
4 3
 3 2
1
3 2
5 3
11 6
 3 2
1 2
2
0
y  cos t
1
3 2
12
0
1 2
 3 2
1
 3 2
1 2
y
=t
t=
=t
t=
=t
=t
t=
x
=t
t=
0
12
t=
3 2
1
=t
=t
It looks as if we are generating the unit circle. To prove this, we can recall our unit circle
definitions for sine and cosine, or we can eliminate the parameter. There are many tricks to
eliminating a parameter, one is to square both equations as below.
x  sin t
y  cos t
x  sin t
2
2
y 2  cos 2 t
Adding the two equations and applying the Pythagorean Identity, we conclude that the function
is indeed a circle with radius equal to one.
113
x 2  y 2  sin 2 t  cos 2 t
x2  y 2  1
Consider the parametric equations x  t  1 and y  t 2  3t  2 . We could sketch the
graph point-by-point using a table of incremental t-values and the corresponding x and y-values,
or we can eliminate the parameter to return to the more familiar rectangular representation.
Solving the first equation for t gives us t  x  1 . Then, we conclude by substituting x  1 for t
into y  t 2  3t  2 as below.
y  t 2  3t  2
y   x  1  3  x  1  2
2
y  x 2  2 x  1  3x  3  2
y  x 2  2 x  3x  1  3  2
y  x2  x
y  x  x  1
Hence, we see that we have a familiar second-degree polynomial with x-intercepts at x  1 and
x 0.
Parametric equations provide a means for representing polar graphs in the rectangular
coordinate system. Consider the cardoid r  1  cos and recall the equations x  r cos and
y  r sin  . Substituting 1  cos  for r into these two equations, we obtain x  1  cos   cos 
and x  1  cos   sin  , which gives a pair of parametric equations using  as the parameter.
These equations describe a cardoid in the rectangular coordinate system.
y
2
x
A natural question at this point might ask what advantage we gain by defining a relation
with parametric equations. Well, one advantage is the direction of motion suggested by the
parameter. For instance, consider the parametric equations below, which describe the motion of
a particle in a plane.
x  t 2  t and y  2t  1
114
If we assume t represents seconds elapsed with t  0 corresponding to the particle’s original
position, then when we plot points on the curve using incremental t-values, the resulting curve
traces out the motion of the particle as time elapses over a one-second interval as shown below.
t
x
0
0.5
1
0
0.75
2
y
1
0
1
115
Suggested Homework in Dugopolski
Section 7.7:
#7, #13-19 odd
Suggested Homework in Ratti and McWaters
The Ratti and McWaters text does not discuss parametric equations, so students with this
textbook should work the homework problems below.
Application Exercise
Suppose that the parametric equations below describe the position of one
particle at time t .
x1  3sin t , y1  2cos t , 0  t  2
Suppose further that the next pair of parametric equations describe the position of a
second particle at time t .
x2  3  cos t , y  1  sin t , 0  t  2
Graphs the paths of both particles and determine the number of literal intersection
points in the paths of the particles. Will the particles collide?
Homework Problems
#1
Suppose x  4t  5 and y  3  4t . Eliminate the parameters and identify the graph.
#2
Let t 
#3
Suppose x  4sin  3t  and y  4cos  3t  . Identify the graph by eliminating parameters.
#4
Suppose x 
#5
Suppose x  tan t and y  3  2 tan t . Identify the graph by eliminating parameters.
#6
Verify the identity: sin x sin y  12 cos  x  y   cos  x  y  .
. Graph the curve in the Cartesian plane described by x  3t  2 and y  t 2 .
t
and y  et . Identify the graph by eliminating parameters.
4
116
Sequences
In general, a sequence is a list of numbers indexed by the natural numbers. The list, 1, 3,
5, 7, 9, . . . is an example of an arithmetic sequence. The ellipsis (the three dots) indicates that
the pattern will continue according to the establish pattern signified in the first few elements of
the sequence. In the example 1, 3, 5, 7, 9, . . . the sequence grows by a common difference of
two, so the next number after nine will be eleven. Each number in the sequence is called a term.
In the sequence 1, 3, 5, 7, 9, . . . the first element, 1, is the first term. The second element, 3, is
the second term while 5 is the third term, and so on. The notation a1, a2, a3, . . ., denotes the
different terms in a sequence. Thus, a15 denotes the fifteenth term in a sequence. The expression
an is referred to as the general or nth term of the sequence.
Considering the notation just discussed, the formal definition of a sequence given below
makes sense.
A sequence, denoted an nm , is a function that has a specified value

for each integer n  m where m usually equals 1 or 0.
Above we define a sequence to be a function that maps the natural numbers (or sometimes the
nonnegative integers) to real numbers.
An arithmetic sequence is a progression of numbers that grow with regular spacing, that
is, they grow with a common difference. For example, 6, 9, 12, 15, . . . is an arithmetic sequence
with a common difference of three. Each number is three higher than the number that precedes
it.
Let an  be a sequence of real numbers. If
an   a1  d   n  1 , then
an  is an arithmetic sequence with common difference equal to d.
A geometric sequence is a progression of numbers that grow with regular relative
spacing, that is, they grow (or diminish) according to multiplication by a common ratio. For
example, 7, 14, 28, 56, . . . is a geometric sequence with a common ratio of two. Each number is
double the previous number. Geometric sequences grow/diminish much more rapidly than
arithmetic sequences.
Let an  be a sequence of real numbers. Let r  0  r  1 . If
an   a1  r n1 ,
then an  is a geometric sequence with common ratio equal to r .
A sequence progresses (or regresses) according to a rule, but the rule operates on a
domain restricted to the natural numbers (sometimes the whole numbers). For example, a


sequence could progress according to the rule an n1  5n  2n1 where n represents the natural
numbers and an represents the nth term of the sequence. We An astute student might note the
similarity with a linear function. This similarity exists because the sequence exemplifies an
arithmetic sequence, which progresses according to a common difference. The slope of a line
117
serves as a common difference creating the "regular spacing" (the common difference) between
one term and the next. Examine the first three terms of the sequence an  5n  2 .
a1  5(1)  2  7
a 2  5( 2)  2  12
a 3  5(3)  2  17
The first three terms of this sequence are 7, 12, 17. Adding five to a term predicts the subsequent
term. Thus, the fourth term of the sequence would be 22 because 17 + 5 = 22. Sequences, like
this one, that progress according to a common difference are arithmetic.
The rule of a sequence often involves a common difference as shown in the above
example or a common ratio. Sequences that progress according to a common ratio are geometric
sequences. For example a geometric sequence might progress according to the rule an = 4(3n)
where n represents the natural numbers and an represents the nth term of the sequence. An astute
student might note the similarity with an exponential function. This similarity exists because the
sequence progresses according to a common ratio, that is, a constant factor that is multiplied by
the previous term to produce the next term, creating a "regular relative spacing." Examine the
n
first three terms of the sequence an  4  3 .
a1  43  12
1
a 2  43  36
2
a 3  43  108
3
The first three terms of this sequence are 12, 36, 108. Multiplying a term by three predicts the
subsequent term. Thus, the fourth term of the sequence would be 324 because 3 × 108 = 324.
The factor, in this case 3, is a "common ratio" because it equals the ratio of a term and the
previous term. Sequences, like this one, that progress according to multiplication are geometric.
The General Term of Sequences
An arithmetic sequence is a sequence in which adding the same value to the previous
term finds each successive term (after the second term). Its general term is described by the
formula an = a1 + (n – 1)d. The number d is the common difference. Taking any term in the
sequence and subtracting its preceding term finds the common difference, d. Consider the
sequence, 6, 13, 20, 27, . . . Find the common difference by taking any term after the first and
subtracting the preceding term. Thus, the common difference is seven because 20 – 13 = 7. The
equation an = a1 + (n – 1)d gives any particular term of an arithmetic sequence, so the general
term of the sequence 6, 13, 20, 27, . . . is as follows.
118
a n  a1  n  1d
a n  6  n  17
a n  6  7n  7
a n  7n  1
We find the twentieth term in particular below.
a n  7n  1
a 20  720  1
a 20  140  1
a 20  139
A geometric sequence is a sequence in which multiplying a term by the same factor finds
the next term. Its general term is an = a1rn–1. The value r is the common ratio. Taking any term
in the sequence and dividing by its preceding term gives r. Consider the sequence 40, 80, 160,
320, . . . Find the common ratio by finding the ratio of a term to its preceding term. Thus, the
a
80
common ratio is two because r  2 
 2 . The equation a n  a1r n 1 gives any particular
a1 40
term of a geometric sequence, so the general term of the sequence 40, 80, 160, 320, . . . is below.
a n  a1r n 1
a n  402 
n 1
In particular, we find the seventh term as follows.
a7  402 
7 1
a7  40  2 6
a7  40  64
a7  2,560
Recursive sequences are sequences in which later terms depend upon earlier terms.
Defining the first term or the first few terms then writing an equation that depends on these
definitions creates a recursive formula for the nth term of a recursive sequence. Consider the
sequence 8, 18, 33, 53, . . . Here, the first term is defined as a1 = 8. Then, the equation
an = an-1 + 5n defines the sequence. The fifth term is below.
an  an 1  5n
a5  a4  5(5)
a5  53  25
a5  78
119
Suggested Homework in Dugopolski
Section 11.1: #13, #15, #63-73 odd
Section 11.3: #1-13 odd, #21-29 odd
Suggested Homework in Ratti & McWaters
Section 11.1: #7, #9, #23, #27, #31, #39, & #40
Section 11.2: #15, #23
Section 11.3: #7-11, #27, #33, #35
Application Exercise
A retirement plan gives the recipient a fixed raise of d dollars each year. If the
fifth year income was $24,500 and the ninth year income was $25,700 , then what
was the first year income?
Homework Problems
Find the first four terms of each sequence below.
#1
 2
1
4
k 
k 1
Find the nth term of the sequences below.
#3 621, 207, 69,
#2 2k  1k 0
100
#4 3, 3, 9,15,
Write the first five terms of the recursive sequences below.
#5 an  2an1  an2 where a1  1 and a2  3
#6 an  an1  an2 where a1  1 and a2  4
Solve the conditional identities below.
#7 sin  2 x   sin x
#8 2sin 2 x  3sin x  1  0
120
Series
We use a special notation called summation notation as a shorthand way to indicate the
m
sum of several terms in a sequence. The notation
a
i n
n
indicates an  an1  an2    am . The
variable i, called the index of summation, assumes consecutive integer values beginning with n
4
and ending with m. For example, consider  3k . In this example, the index of summation is k.
k 0
The beginning value of the index is zero, and the terminating value is four. Each successive term
4
is given by 3k. Our example,  3k reads, "The summation of 3k as k progresses by consecutive
k 0
integers from 0 to 4." To compute the sum, add each of the terms as shown here.
4
3
k
 30  31  32  33  34
k
 1  3  9  27  81
k
 121
k 0
4
3
k 0
4
3
k 0
Using summation notation, we define a very special type of sequence called a series
defined in the box below.
Let an n k be a sequence in

. Construct the associated sequence
sn nk as below.

sk  ak
sk 1  ak  ak 1
sk  2  ak  ak 1  ak  2
s j  ak  ak 1 
 aj
sn  ak  ak 1 
 a j 1  a j  a j 1 
 an
n
In other words, sn   ai .
i k

We call sn n k the series defined by the sequence an  . Moreover, the

number sn is called the nth partial sum of the infinite series  ai .
i k
If an is arithmetic, we call sn an arithmetic series. If an is geometric, we
call sn a geometric series.
121
k
 1
Let’s consider the series, sn     . This is a geometric series of the terms of the
k 0  2 

geometric sequence an n 0

n

1 1 1 1
 1  

     1, , , , ,
2 4 8 16

 2  
n  0
. Hence, the terms of sn are below.
3 7 15 31
1, , , , ,
2 4 8 16
We obtain each term of the sequence by adding the appropriate number of terms of an n 0 . For

a geometric series like this, we can use a formula to obtain any particular term of sn .
Let s n represent the sum of the first n terms of a geometric sequence a1r n 1 as below.
sn  a1r11  a1r 21  a1r 31 
 a1r n1
If we multiply by the common ratio, we obtain the following.
rsn  a1r  r11  a1r  r 21  a1r  r 31 
rsn  a1r1  a1r 2  a1r 3 
 a1r  r n1
 a1r n
Subtracting rs n from s n eliminates most of the terms as below.
sn  rsn  a1r11  a1r 21  a1r 31 
sn  rsn  a1r 0  a1r1  a1r 2 
 a1r n1   a1r1  a1r 2  a1r 3 
 a1r n 1  a1r1  a1r 2  a1r 3 
sn  rsn  a1r 0  a1r n
Now, we factor and divide to obtain the sum of a finite geometric series.
sn  rsn  a1r 0  a1r n
sn 1  r   a1 1  r n 
n
sn 1  r  a1 1  r 

1  r 
1  r 
sn 
a1 1  r n 
1  r 
Thus, we have the theorem stated on the next page.
 a1r n 
 a1r n
122
n
Sum of a Finite Geometric Series Theorem: If sn   ai and ai is a
i k
geometric sequence, then
sn 
a1 1  r n 
1  r 
.
k
1
For example, let’s find the fifth partial sum of the series sn     . Since the index of
k 0  2 
summation starts at zero, to find the fifth partial sum, we let the index of summation progress to

k
1
four. In other words, we need    . Plugging into the formula, we find s5 as follows.
k 0  2 
4
sn 
s5 
a1 1  r n 
1  r 
1
 
2
  1 5 
1     11  1  32  1
  2     32   32 32  31  1  31  2  31
2 1
1
32 2 32
16
 1

1  
2 2
2
 2
0
Similarly, there exists a formula for a finite arithmetic series sn as shown below.
 a1  d    a1  2d  
sn   an 
 an  d    an  2d  
2sn   a1  an    a1  an    a1  an  
2sn  n  a1  an 
n  a1  an 
s 
sn  a1 
n
 an
 a1
  a1  an 
2
Therefore, we have the sum of the first n terms of an arithmetic sequence.
n
Sum of a Finite Arithmetic Series Theorem: If sn   ai and ai is
i k
an arithmetic sequence, then
sn 
n
n
 a1  an   2a1  d  n  1
2
2
123
For example, consider the sequence 2, 7, 12, 17, . . . To compute the sum of the first twenty
n
n
terms, we can use S n  a1  a n  or Sn   2a1  d  n  1 . We will use the latter formula
2
2
since we do not have a20 (although we could find it easily).
n
 2a1  d  n  1 
2
20
S20 
 2  2   5  20  1   10  4  5 19    990
2 
Sn 
Both of the previous two theorems provide formulas for the value of a finite series. It
seems counterintuitive to think we could find the value of an infinite series, but sometimes this is
the case. For our purposes, we will simply accept the following theorem without proof.
Sum of an Infinite Geometric Series Theorem: If s is an infinite
geometric series, a  ar  ar 2  ar 3 
s 
, with r  1 , then
a
.
1 r
For example, consider the series 300 + 225 + 675/4 + · · ·. We calculate the sum below.
S
a1
300
300


 1,200
1  r 1  .75 .25
124
Suggested Homework in Dugopolski
Section 11.2: #1-25 odd, #53-65 odd
Section 11.3: #61, #63, #77, #81
Suggested Homework in Ratti & McWaters
Section 11.1: #69, #77, & #79
Section 11.2: #37, #41, #47
Section 11.3: #47, #51, #71, #73
Application Exercise
If a patient starts taking 100 mg of a particular drug every 8 hours, then  100  0.69 
n
k 1
k 1
gives the amount of the drug in the patient’s bloodstream after taking the nth pill. Calculate
the amount of the drug in the patient’s bloodstream after 25 pills.
Homework Problems
Use summation notation to express the sums below.
#1 8  4  2   18
#2 2  7  12  17 
Evaluate the sums below.

6
#3
1
4
 3

900
k 1
#4

 9k  12 
#5
k 1
k 1
Verify the identities below.
cos  x  y  cot y  tan x

#6
cos  x  y  cot y  tan x
k 1
#7
cos  2 x 
 csc2  x   2
2
sin  x 
Which expression is greater?

#8

k 1
1
2
 
1 k 1
2
or csc 
5π
6

9
 3k  1 or sec  
100
#9
k 1
#10 Explain the Gauss trick for evaluating an arithmetic series.
30, 000
π
3


2 k 1
3
125
Combinations
Factorial provides a means for expressing the answer to any natural number
multiplied by all the preceding natural numbers. The symbol for factorial is an
exclamation point. Thus, 5! = 5∙4∙3∙2∙1 = 120. Since zero is not a natural number, 0! has
no meaning according to the definition above, but we define 0! as 1.
For any natural number n, we read the notation n ! as “n factorial.”
We define n ! as below.
n!  n   n  1   n  1 
 3  2 1
By convention, we say 0!  1.
Counting methods called permutations and combinations use factorials.
Combinations are selections of distinct objects in which the order of the objects is
not important. If a club with twenty members, for example, is to select which three
members must attend a national conference, the arrangements are combinations because
choosing Alice, Bobby, and Charlene is not different from selecting Charlene, Bobby,
Alice. In either case, the same three members will attend the conference.
The number of combinations of n objects taken r at a time is given by the
Combinations Theorem below.
Combinations Theorem: If set S has n elements, then we denote the number of
n
combinations of r distinct objects taken from S as C  n, r  or n Cr or   . The
r 
formula below gives the number of combinations of r distinct objects taken from S.
n
n!
.
  
 r  r! n  r !
Consider a marketing director planning a promotion for a product sold in twelve
different cities. If the director's budget only allows a promotional blitz in three cities,
how many different ways can the director select three cities from the twelve target cities?
The selection of Dallas, San Diego, and Nashville is no different from the selection of
Nashville, San Diego, and Dallas because the order of the selection of those three cities
does not change the fact that those three cities will be the focus of the marketing
promotion. Since order does not make a difference, the Combination Formula gives the
answer as follows.
2
12!
12 1110  9! 12 1110
C 12,3 


 2  11 10  220 .
3!12  3! 3  2 1  9!
3 2
126
Binomial Expansion
When whole number exponents are applied to a binomial, the coefficients of the
terms form a pattern.
(a + b)0 =
(a + b)1 =
(a + b)2 =
(a + b)3 =
(a + b)4 =
(a + b)5 =
1
1a + 1b
1a2 + 2ab + 1b2
1a3 + 3a2b + 3ab2 + 1b3
1a4 + 4a3b + 6a2b2 + 4ab3 + 1b4
5
1a + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + 1b5
The binomial expansion exhibits several patterns. First, the number of terms of each
expansion equals n + 1 where n equals the degree of the expansion (i.e., the exponent on
the binomial). Second, the sum of the exponents in each term in the expansion equals the
degree of the expansion. Third, the exponents on a in the expansion decrease by one,
while the exponents on b increase by one. Fourth, the coefficients form a symmetrical
pattern. Fifth, each entry below the second row is the sum of the closest pair of numbers
in the line directly above it (other than the 1s at the end of each row).
These patterns give us the Binomial Theorem below.
If n is a natural number, then for any real numbers a and b, we have the
following binomial expansion.
a  bn  
n
n  n 0  n  n 1 1  n  n 2 2
n
n
a b   a b   a b     a 0 b n    a n k b k
k 0  k 
0
1 
2
n
n
Recall that   , the number of combinations of n objects taken k at a time, equals
k 
n!
.
k! n  k !
 n  n k 1 k 1
In general, the expression 
b equals the kth term of any binomial
a
 k  1
expansion.
.
127
Suggested Homework in Dugopolski
Section 11.1: #21-25 odd
Section 11.5: #5-13 odd, #47, #49, #57
Suggested Homework in Ratti & McWaters
Section 11.5: #11, #13, #15, #17, #21, #39
Application Exercise 1
Given eight astronauts qualified as pilots and eighteen astronauts qualified as
specialists, how many crews with one pilot and two specialists can be chosen for a
mission?
Application Exercise 2
Let success be defined as any observable outcome of an experiment. Let
represent exactly X number of occurrences of success in n trials. If the probability of
success remains constant throughout n trials, then the formula below gives the
probability of
.
P
  C  n, X    p  1  p
X
n X
where p equals the probability of success.
Only three out of four patients who undergo a particular surgical procedure are
known to survive five years without having to undergo the surgery a second time. Of
six patients who receive this treatment, what is the probability that all will survive
five years without needing another surgery?
Homework Problems
 n 
#1 Let n be a positive integer. Prove 
  n.
 n  1
5
#2 Evaluate   .
3
#3 A book club has 11 members, but can only afford to send 9 members to a seminar
hosted by their favorite author. How many different sets of members can be chosen to
attend the seminar?
Expand the following binomials.
5
#4  x  2 
#6 Verify the identity
#5  2 x  3
1
1

 2 tan x .
tan x  sec x tan x  sec x
4
128
Appendix: Answers to Odd Homework Problems
Lecture 1: Angle Measurement
#1 3.75 radians
#7 20
#3 1 radian
#9 28π inches  88 inches
Lecture 2: Trigonometric Functions Defined
#1 30
#3 π3 radians
#7 0
#9
Lecture 3: Winding Function
#1  0,1

#7 
1
2
,
1
2

2
3

#9  
#3 
or
#5
3π
4
#5 
radians
1
2
or 
2
2
2 3
3
3
2
,  12
1
2
,
3
2


#5  1, 0 
Lecture 4: Right Triangle Trigonometry
#1 ≈528.8 feet
#3 ≈39.5 additional feet
Lecture 5: Graphing Trigonometric Functions
1
10
1
#1
#5
#3
10
1
#7
Lecture 6: Dilations and Reflections
#1
#3
1
#5
1
π
#7
π
π
π
#9
1
1

π
2
1
π
2
Lecture 7: Translations
#1
2
#3
#5
1
1

π
2
π
2
π
2
π
1
1
π
π
1
1

#7
#9
Lecture 8: Inverse Functions
#1  π2
#3
π
6
π
6
π
2
π
6
#5
π
3
Lecture 9: Basic Identities
sin x
#1 LHS  sin1 x  cos1 x  cos
x  tan x  RHS
#3 RHS  tan 2 λ  sec2 λ  1   sec λ  1 sec λ 1  LHS
#5
tan x sin x
sec x 1
sec x 1
 sec
x 1 
tan x sin x sec x 1
sec2 x 1

tan x sin x sec x 1
tan 2 x
#7 LHS  csc2 x  1  cot 2 x  1  tan12 x 
sin x sec x 1
tan x

tan 2 x
tan 2 x
 tan12 x 
x
 sin x  sec x  1  cos
sin x   sec x  1 cos x  1  cos x
tan 2 x 1
tan 2 x
 RHS
sin θ
cos θ
sin θ
cos θ sin θ
#9 LHS  1  tan θ  1  cos
 RHS
θ  cos θ  cos θ 
cos θ
2
2
sin x
cos x
sin x
cos x
1
1
#11 RHS  sin x1cos x  tan x1 cot x  sin x1cos x   cos
x  sin x   sin x cos2 x  sin 2 x cos x  cos2 x  sin 2 x  sec x  csc x  LHS
Lecture 10: Sum and Difference Identities
#1 2 4 6 or 12 23
#3 6 4
2
or
3 1
2 2
#5 LHS  cos  α  180  cos α cos180  sin α sin180  cos α  1  sin α  0   cos α  RHS
π  tan x
0 tan x
#7 RHS  tan  π  x   1tan
 tan π tan x  10tan x 
tan x
10

tan x
1
 tan x  LHS
#9 RHS  12 cos x cos y  sin x sin y  cos x cos y  sin x sin y   12 2cos x cos y   cos x cos y  LHS
#11 cos 2 x  cos  x  x   cos x cos x  sin x sin x  cos 2 x  sin 2 x  1  sin 2 x  sin 2 x  1  2sin 2 x
Lecture 11-12: Double-Angle & Half-Angle Identities
#1
1
2
#3
2 2
2
x
1 2cos x 1
2cos x
cos x
#5 RHS  1sincos2
2 x  2sin x cos x  2sin x cos x  sin x  cot x  LHS
2
2
#7 RHS   sin x  cos x    sin x  cos x  sin x  cos x   sin 2 x  2sin x cos x  cos 2 x  1  sin 2 x  LHS
2
x
#9 LHS  1sincos2
2x 

1 1 2sin 2 x
2sin x cos x

1 2sin x
2sin x
sin x
 12sin
x cos x  2sin x cos x  cos x  tan x  RHS
Lecture 13: Conditional Identities
#1 x  π6 , 56π
#7 θ  arctan  12   kπ
2
2
#3 x  π3 , 56π
#5 x  π6 , 32π ( 56π is extraneous)
#9 x   π6  2kπ , x  76π  2kπ
130
Lecture 14: Law of Sines
#1 ≈2.5 km
#3 h  9.5  c no triangle
Lecture 15: Law of Cosines
#1 BC  21.7 m
#3 B  114.5 , C  45.9 , A  19.6
Lecture 16: Parabolas
2
#1 y   x  4   13 , vertex:  4, 13 , focus:  4, 12.75 , directrix: y  13.25
#3 y  2  x  3  13 , vertex:  3, 13 , focus:  3, 12 78  , directrix: y  13 18
2
#5 x   12  y  1  72 , vertex:   72 , 1 , focus:  4, 1 , directrix: x  3
2
#7 LHS 
1
1
1  sin x 1  sin x 1  sin x




 RHS
1  sin x 1  sin x 1  sin x 1  sin 2 x cos 2 x
Lecture 17: Circles and Ellipses
2
2
#1  x  2    y  3  4 ,  2,3 , r  2
#3 center: 1, 2  , foci: 1, 2  5
Lecture 18: Hyperbolas
#1 y   32 x
#3 center:  0, 0  , foci:  2, 0 , asymptotes: y   x

#5 center:  1, 2  , vertices:
#7 LHS 


#5 e 

 3, 2   5, 2  , foci:  4, 2   6, 2 , asymptotes:
y   34  x  1  2
csc x  sin x csc x sin x


 csc x  csc x  1  csc2 x  1  cot 2 x  RHS
sin x
sin x sin x
Lecture 19: Vectors
#1 r  2, θ  π6 or 30
#7 2
#13 LHS 


 1 
1  cos 2  2x 
1  sin
2

x
2
1
Lecture 20: Complex Numbers
#1 z  2, θ  56π or 150
1 cos x
2
1 cos x
2


#3 r  1, θ   π2 or  90
#9 0
#5 r  5, θ  43π or 240
#11  95
2
2

x
1 cos x
2
1  1cos
2
2 
2


x
1 cos x
2
1  1cos

2
2
2
1 cos x
2
1 cos x
2

1  cos x
 RHS
1  cos x
#3 z  13, θ  arctan  23   180  213.7
#5 trigonometric form: 3 2cis  π4  ; exponential form: 3 2e 4 ; standard form: 3  3i
iπ
#7 trigonometric form: 48cis  56π  ; standard form: 24 3  24i
#9 trigonometric form:
9
16
cis  π  ; standard form:  169
Lecture 21: Roots
#1 trigonometric form: cis  π4  , cis  54π  ; standard form:
#3 standard form:
#5
2
3
2 2
2
3  i,  1  3i,  3  i,1  3i
2
2

2
2
i, 
2
2

2
2
i
131
Lecture 22: Polar Coordinates
#1 3 5,arctan  2   3 5, 63.4

#5  3
3, 3
 


#3  2, 56π  or  2,150  or  2, 210
#7


3,1
#9  x  1  y 2  1
#11 y  1  x
#13 r  4
#17
#15 r 2  3r cos θ  4  0
#19
θ  π3
2
r  4cos3θ
Lecture 23: Parametric Equations
#1
y  x  2
x2  y 2  16
#3
#5
1
1
1
1
1
1
Lecture 24: Sequences
n1
#1 12 ,1, 2, 4,
#3 621 13 
#5 1, 3, 7,17, 41,
#7 x  kπ, x  π3  2πk , x  53π  2πk
Lecture 25: Series
 8 
7
#1
k 1

1 k 1
2


#3 91
#5 27
cos  2 x  1  2sin 2 x
1
2sin 2 x



 csc2 x  2  RHS
sin 2 x
sin 2 x
sin 2 x sin 2 x
30, 000
.
 3k  1  15, 250 is greater than 15, 000 
sec   π3 
#7 LHS 

100
#9
k 1
y  2x  3
Lecture 26: Combinations and Binomial Theorem
n  n  1!
n  n  1!
 n 
n!
n
n



 n
#1 

 n  1  n  1! n   n  1 !  n  1! n  n  1!  n  1!  0  1! 1! 1
#3 110
4
#5  2 x  3  16 x 4  96 x3  216 x 2  216 x  81