Section 4.1 Notes Page 1
4.1 Angles and Radian Measure
This section will cover how angles are drawn and also arc length and rotations.
Angles are measured a couple of different ways. The first unit of measurement is a degree in which 360 
(degrees) is equal to one revolution. Most likely the reason why we use 360 is from the Babylonians, whose
year is based on 360 days.
Another unit of measurement for angles is radians. In radians, 2 is equal to one revolution. So a conversion
between radians and degrees is 2  360  , or   180  .
When converting from degrees to radians:
Multiply your degrees by

180
When converting from radians to degrees:
Multiply your radians by
180

EXAMPLE: Convert 60  to radians.
We will take 60 and multiply it by

and you will get: 60 
180

180
. This reduces to

3
.
EXAMPLE: Convert  405  to radians.
We will take –405 and multiply it by
EXAMPLE: Convert
We will take

180
and you will get:  450 

180
. This reduces to 
5
.
2
4
into degrees.
3
180
4 180
4
and you will get:

. This reduces to 240  .
and multiply it by
3 
3

EXAMPLE: Convert 
We will take 
3
into degrees.
2
180
3 180
3
and you will get: 

. This reduces to  270 
and multiply it by
2 
2

Section 4.1 Notes Page 2
We will use  (theta) to represent an angle’s measurement. In the figure below it describes how you know if an
angle is positive or negative. The vertex of the angle is at the origin of a rectangular coordinate system. The
positive x axis is always where an angle is measure from, and this is called the initial side. An angle drawn this
way is said to be in standard form. An angle that goes counterclockwise is always positive, and clockwise
angles are negative.
EXAMPLE: Draw each angle in standard position. Indicate which quadrant the angle lies.
a.)
5
6
If you are not sure where to draw this angle, first convert it into degrees:
5 180

 150  . Our angle is measured from the positive x-axis. Since
6 
the angle is positive we need to go in the counterclockwise direction.
We see that this ends up in Quadrant II.
b.)  135 
Our angle is measured from the positive x-axis. Since we have a negative
angle, we need to go in the clockwise direction. We see that we end up in
Quadrant III.
c.)
8
3
If you are not sure where to draw this angle, first convert it into degrees:
8 180

 480  . Our angle is measured from the positive x-axis. Since
3 
the angle is positive we need to go in the counterclockwise direction.
Since this angle is more than 360 degrees, we need to subtract 360 from
480. We will get 120 degrees. So we need to got around 360 degrees and
then go an extra 120 degrees. We see that this ends up in Quadrant II.
Arc length – this is the length of the arc between the two lines shown with  .
Section 4.1 Notes Page 3
The equation is S  r , where S is the arc length, r is the radius, and  MUST be
measured in RADIANS! The  is also called the central angle.
EXAMPLE: Find the arc length of a sector whose radius is 3 inches and whose central angle is
Using our formula S  r we know that r  3 and  
S  3

3

3

3
.
. Putting this into the formula we will get:
, so S   inches.
EXAMPLE: Find the arc length of a sector whose radius is 7 inches and whose central angle is 45  .
Using our formula S  r we know that r = 7. We will not use 45 degrees as our angle because this is not in
degrees. We need to multiply it by

180
this into the formula we will get: S  7 
to get it into radians: 45 

4
, so S 
7
inches.
4

180
. Reducing this we get  

4
. Putting
Coterminal Angles – have the same initial and terminal sides but possibly different rotations. Two coterminal
angles for an angle of   can be found by adding 360  to   and subtracting 360  to   . Here’s the formula:
  360  k or   2k
EXAMPLE: Assume the following angles are in standard position. Find a positive angle less than 360  or 2
that is coterminal with each of the following:
a.) a  120  angle.
Since we want angle less than 360 degrees, we just need to add 360
to -120. You will get  120   360   240  . So a 240  angle is coterminal
with a  120  angle. The picture to the right illustrates that both of these
angles will end up at the same place.
b.) a
17
angle.
6
17
5
can be written as 2  . Since this is more than 2 then we
6
6
5
5
5
need to subtract 2 from 2  . You will get: 2   2 
. The picture to
6
6
6
the right illustrates that both of these angles will end up at the same place.
The fraction
Section 4.1 Notes Page 4
Angular Speed (  )
Angular speed is the angle  that can be swept out in time t. The formula is:

Linear Speed (v)

t
where  is in radians.
This is the speed at which a point on the circle is moving. It is measured by
what arc length is traveled in time t. The formula is:
v  r
where  is measured in radians per unit of time.
EXAMPLE: A circular gear rotates at a rate of 75 rpm (revolutions per minute). What is the angular speed (in
radians per minute)? What is the linear speed of a point on the gear 3mm from the center?
In order to find the angular speed we need to find the angle in radians. We know that 1 revolution is 2
radians, so we can multiply 75 by 2 to get 150 radians. This is our angle  . Our unit of time is 1 minute,

150
= 150 radians per minute.
so we can use the formula   . We will get  
t
1
For the linear speed we will use v  r with r = 3 and   150 . You will get v  3  150 , so v  150
millimeters per minute.
EXAMPLE: A wheel rotates at a rate of 2160  per second. What is the angular speed (in radians per minute)?
What is the linear speed of a point on the gear 30 cm from the center?
We need to first change 2160  into radians by multiplying by

180
. You will get: 2160 
180
which reduces to
12
= 12
1
t
radians per second. The question is asking us for radians per minute, so we need to multiply our answer by 60.
You will get   720 radians per minute.
Now we can find the linear speed. We will use v  r with r = 30 and   720 . You will get v  30  720
which is v  21600 centimeters per second.
12 radians. Our unit of time is 1 second, so we can use the formula  


. We will get  
Pulleys
When we have pulleys, the belt that drives the pulleys is always moving at the same speed, so actually both
pulleys will have the same linear speed. This is important when solving these kind of problems.
Section 4.1 Notes Page 5
EXAMPLE: A 2-inch radius pulley is rotating at 3 rpm. Determine the rpm of the larger pulley, which has a
radius of 8 inches (see figure)
First we need to find the linear speed of the small pulley. Then we can use
this information to find the rpm of the larger pulley. For the small pulley
we will multiply 3rpm by 2 to get 6 radians per minute. Then since
r = 2 we can put these into the formula v  r and we will have
v  2  6 so v  12 radians per minute.
We know that the larger pulley will have the same linear speed as the smaller pulley. We will still use v  r .
We know v  12 from the previous part and also r = 8. The only thing left to solve for is  . After
12
3
substituting we get: 12  8 . Solving this we get  
, or  
. The question is asking us to express
8
2
3
3
this as rpm, so now we must divide this by 2 . You will get: 2 , which is rpm.
2
4
Is there an easier way of doing this? Yes. You can just multiply by the ratio of the smaller pulley to the large
2 6 3
rpm . The same can be
one. For example, we take 3rpm and multiply it by 2in divided by 8in: 3   
8 8 4
applied to gears as well as pulleys.
EXAMPLE: To approximate the speed of the current of a river, a circular paddle wheel with a radius of 4 feet
is lowered into the water. If the current causes the wheel to rotate at a speed of 10 rpm, what is the speed of the
current in miles per hour (1 mile = 5280 feet).
We can convert 10rpm to radians per minute by multiplying it by 2 to get 20 radians per minute. This is
our angular speed,  . The water moving is a linear speed, so we need to use the formula v  r . We know
r = 4 and   20 , so v  4  20 , or v  80 feet per minute. We need to now change this into miles per
hour, so we need to use some dimensional analysis. Basically we get the units to cancel by using the
appropriate conversions:
80 ft 60 min 1 mile


 2.86 mph
1hr 5280 ft
1 min
Section 4.2 Notes Page 1
4.2 Trigonometric Functions; The Unit Circle
A unit circle is a circle centered at the origin with a radius of 1. Its equation is x 2  y 2  1 as shown in the
drawing below. Here the letter t represents an angle measure. The point P=(x, y) represents a point on the unit
circle.
The following definitions are given based on this picture.
1
sin t  y
csc t 
y
1
cos t  x
sec t 
x
y
x
tan t 
cot t 
x
y
If we start with x 2  y 2  1 and put in our definitions above we
will have cos 2   sin 2   1 .
12 
 5
EXAMPLE: Suppose a point on the unit circle is   ,   . Find all six trigonometric values.
13 
 13
We want to use our definitions to answer the questions. We know
5
12
and y  
because of the point given. This
that x  
13
13
12
5
and cos t   . We can
automatically tells us that sin t  
13
13
plug in x and y into the other equations to find the remaining trig
1
1
13
13
values. csc t 

sec t 

12
5
12
5


13
13
12
12 13 12
tan t  13   
5 13 5
5

13

5
5 13 5
cot t  13   
12 13 12 12

13

Domain and Range of Sine and Cosine
Domain is what values we can put into a trig function (t) and the range is the values it returns.
y  sin t
Domain:  ,  
Range: [-1, 1]
x  cos t
Domain:  ,  
Range: [-1, 1]
Let’s look at the angle of t  45  or

4
Section 4.2 Notes Page 2
. At this angle, we end up with the following triangle:
45 – 45 – 90 Triangle
45 
1
We can use our definitions of sine, cosine, and tangent to find exact values:
2 2
sin 45 
45 
2
2
cos 45 
2
2
tan 45 
2 2
2 2
1
2 2
From our above definitions we can also find the following:
2
2 2 2
2
2 2 2
1
1
csc 45  



 2,
sec 45  



 2,
2
2
2 2
2 2
2
2
2
2
EXAMPLE: Find the exact value without using a calculator: tan

4
 cot

4
1
cos 45    1
1
.
We just need to plug in the values of the trig functions we found above: 1 + 1 = 2.
The following Even – Odd Properties will allow you to not deal with negative angles.
Even – Odd Properties
cos(t )  cos t
sec(t )  sec t
sin(t )   sin t
csc(t )   csc t
tan(t )   tan t
cot(t )   cot t
Periodic Properties
If we start at an angle and go around one revolution ( 360  or 2 radians) we will end up at the same angle we
started with. The k value is any integer, and represents how many revolutions are going around. If you want to
use degrees, replace the 2k in the equations below with 360k.
sin(t  2k )  sin t
csc(t  2k )  csc t
cos(t  2k )  cos t
sec(t  2k )  sec t
tan(t  2k )  tan t
cot(t  2k )  cot t
Section 4.2 Notes Page 3

EXAMPLE: Find the EXACT value of sin 1485 .
We need to divide 1485 by 360 to see how many revolutions we have. If you divide 1485 by 360 you will get 4
with a remainder of 45. So we can rewrite our problem as: sin(45   360(4)) . From our definitions above we
know that sin(45   360(4))  sin 45  . From our previous values we know that sin 45  
that sin 1485 
2
. So we know
2
2
.
2
 17 
EXAMPLE: Find the EXACT value of tan 
.
 4 
First we will use our even-odd property to change the negative angle into a positive angle:
17 180
 17 
 17 

 765  . Dividing this by
tan 
 . It would be easier to change this into degrees:
   tan

4
 4 
 4 

360 we will get 2 with a remainder of 45. So our problem becomes tan(45  360(2)) . From our periodic
properties, tan(45   360(2))  tan 45  . From our previous values we know that tan 45   1 . So we know that
 17 
 tan
  1 .
 4 
Fundamental Identities
sin t 
1
csc t
csc t 
1
sin t
sin 2 t  cos 2 t  1
cos t 
1
sec t
sec t 
1
cos t
tan t 
tan t 
1
cot t
cot t 
1
tan t
EXAMPLE: Given sin t 
1  tan 2 t  sec 2 t
sin t
cos t
cot t 
1  cot 2 t  csc 2 t
cos t
sin t
2
5
and cos t 
, find the other 4 trig values using identities.
3
3
We just use the identities and plug in our values for sine and cosine where necessary:
csc t 
1
1
3


sin t 2 / 3 2
cot t 
1
1
5
5



tan t 2 5 5 2 5
2
sec t 
1
1
3
3 5



cos t
5
5 3
5
tan t 
23
sin t
2 3
2
2 5
,

 


cos t
5
5 3 3 5
5
Section 4.2 Notes Page 4
EXAMPLE: Given sin t 

7
and 0  t  , use the Pythagorean Identity sin 2 t  cos 2 t  1 to find cos t .
8
2
2
7
7
We first start with the identity sin t  cos t  1 and plug in for sin t . You will get    cos 2 t  1 .
8
8
49
49
Simplifying will give you
 cos 2 t  1 . Now isolate the cosine by subtracting
from both sides. You will
64
64
15
15

. Square root both sides to get cos t  
. Since the angle for t needs to be 0  t  , we
get cos 2 t 
64
8
2
know that this angle must be in the first quadrant. What we know from this is that the x-coordinate of the unit
15
.
circle must be positive in the first quadrant, so cos t 
8
2
2
EXAMPLE: The unit circle below has been divided into 12 equal arcs, corresponding to t-values of:
7 4 3 5 11
   2 5
0, , , ,
,
, ,
,
,
,
,
, and 2 . Use the (x, y) coordinates in the figure to
6 3 2
3
6
6
3
2
3
6
find the value of each trigonometric function at the indicated real number, t, or state that the expression is
undefined.
a.) cos

3
The cosine value is always the x, value, so we read the x value from our picture above to get cos
b.) sin
2
1
 .
3
2
2
3
The sine value is always the y, value. Since each segment is
second quadrant. Here we see that sin
2
3
.

3
2

3
, we need to go over 4 segments which is in the
Section 4.2 Notes Page 5
 
c.) cos  
 3
There are two ways to solve this one. The first method is to use the unit circle directly. A negative angle means
we need to go in the clockwise direction, so we will go down into the fourth quadrant to the corresponding point
1
3

1
 , 
 . The cosine value is always the x value, so cos    .
2

2 
 3 2

The second way to solve this is to use the Even-Odd Property to get rid of the negative. You will get

 
cos    cos . Since the negative is gone we can now go counterclockwise on the unit circle to the point
3
 3
1
 ,
2

3

1
 . The cosine value is always the x value, so cos    .

2 
 3 2
 11 
d.) tan 

 6 
There are two ways to solve this one. The first method is to use the unit circle directly. A negative angle means
we need to go in the clockwise direction, so we will go to the first quadrant to the corresponding point
 3 1
12
1 2
1
3

 . The tangent value is always y/x, so tan   11  
,
.




 2 2
6
2
3
3
2
3
3




The second way to solve this is to use the Even-Odd Property to get rid of the negative. You will get

 
cos    cos . Since the negative is gone we can now go counterclockwise on the unit circle to the point
3
 3
1
 ,
2

3

1
 . The cosine value is always the x value, so cos    .

2 
 3 2
EXAMPLE: Use an identity to find the exact value without using a calculator: 2 cos 2

6
 sin

6



This can be rewritten as: 2 cos   sin . Now substitute values by using the figure on the previous page.
6
6

2
2
 3
3
1
3 1
1
  . Now square everything inside the parenthesis: 2   . Reduce this fraction:
2
 . After

2
2 2
4 2
 2 
subtracting we get 1. Therefore, 2 cos 2

6
 sin

6
 1.
EXAMPLE: Use a calculator to find the approximate value of sin 43.5  to four decimal places.
Since we are in degrees, you want to make sure your calculator is in degrees. I can show you in class how to
make sure you are in the correct mode. For calculators where you can see what you are typing, just type in this
expression as it shows. Otherwise, type in the angle first and then the sine key. Answer: sin 43.5   0.6884 .
Section 4.3 Notes Page 1
4.3 Right Triangle Trigonometry
This is a very important section since we are giving definitions for the six trigonometric functions you be using
throughout the rest of this course and beyond. We need to first start with a drawing of a right triangle. The
following definitions only apply to RIGHT TRIANGLES.
sin  
opposite
hypotenuse
csc  
hypotenuse
opposite
cos  
adjacent
hypotenuse
sec  
hypotenuse
adjacent
tan  
opposite
adjacent
cot  
adjacent
opposite
EXAMPLE: Find the exact value of the 6 trig functions using the following figure:
5

12
First we need to find the missing side. In this case it is the hypotenuse. In order to find this we need to use the
formula a 2  b 2  c 2 . The side opposite the right angle is always side c. So we have 12 2  5 2  c 2 .
Simplifying we will get 144  25  c 2 , or 169  c 2 . We will get c  13 , however our answer is c = 13 since
we can’t have a negative side. Now we are ready to write our 6 trig functions. Here the hypotenuse is 13,
opposite is 5, and the adjacent side is 12. We will put these into the formulas and that will be our answers.
13

sin  
5
13
csc  
13
5
cos  
12
13
sec  
13
12
tan  
5
12
cot  
12
5
5
12
Section 4.3 Notes Page 2
EXAMPLE: Find the exact value of the 6 trig functions using the following figure:
7

4
First we need to find the missing side. In this case it is the hypotenuse. In order to find this we need to use the
formula a 2  b 2  c 2 . The side opposite the right angle is always side c, which is 7 in our figure. So we have
4 2  b 2  7 2 . Simplifying we will get 16  b 2  49 , or 33  c 2 . We will get c  33 . Now we are ready to
write our 6 trig functions. Here the hypotenuse is 7, opposite is 33 , and the adjacent side is 4. We will put
7
. This needs to be
these into the formulas and that will be our answers. The answer for cosecant is
33
7
33 7 33
rationalized by multiplying top and bottom by the square root of 33:


. You want to always
33
33 33
rationalize so that there is no square root in the denominator.
sin  
7
33
7
7
csc  
33
=
7 33
33
=
4 33
33
33
cos  

4
7
sec  
7
4
4
tan  
33
4
cot  
4
33
30 – 60 – 90 Triangle
60
2x

x
30
In this triangle the opposite side of 30 degrees is always half of the
hypotenuse. The adjacent side is always 3 times the opposite. From
this relationship we can get values for 30 and 60 degrees.

x 3
sin 30 
x 1

2x 2
sin 60 
x 3
3

2x
2
cos 30 
3
x 3

2x
2
cos 60 
x 1

2x 2
tan 60 
x 3
 3
x
tan 30 
x
x 3

1
3

3
3
Section 4.3 Notes Page 3
45 – 45 – 90 Triangle
45 
In this triangle the opposite and adjacent sides are the same. The
hypotenuse is always 2 times the opposite or adjacent. From
this relationship we can get values for 45 degrees.
x
x 2
45 
x
sin 45 
cos 45 
tan 45 
x
x 2

1

1
x
x 2
2

2
2

2
2
2
x
1
x
You will be referring to the following table A LOT, so please bookmark it. This summarizes what we just
covered in this section and the previous section. I am also including the 90 degree angle at this time.
Table of trigonometric values
 (degrees)  (radians) sin 
cos 
tan 
0
0
0
1
0
30

1
2
3
2
2
2
1
2
0
3
3
1
6

45
4

60
3

90
2
2
3
2
1
3
undefined
2
EXAMPLE: Find the exact value without using a calculator: 2 cos 2 30   sin 30 
This can be rewritten as: 2cos 30    sin 30  . Now substitute values by using the table.
2
2
 3
3
1
3 1
1
  . Now square everything inside the parenthesis: 2   . Reduce this fraction:  . After
2

2
2 2
4 2
 2 
subtracting we get 1.
Section 4.3 Notes Page 4
1  cos 60
.
sin 60 

EXAMPLE: Find the exact value without using a calculator:
1
2 . We can subtract on the top to get:
We can put in values right away from the table:
3
2
1
3
1 2

.
over the bottom fraction and multiply to get: 
which is
3
2 3
3
1
EXAMPLE: Find the exact value without using a calculator: cos

3
sec

3
 cot

3
1
2 . We can flip
3
2
.
We don’t have a cotangent or secant on our table, so we will need to first convert this into a sine by using the
1
1


 1

and cot 
. From the table, we know that cos  and tan  3 .
identity: sec 
3
3 cos 3
3 tan  3
3 2
We can put in these values into our expression:
3 3 3
1 1
1
1 2
3


  
 1

.
3
3
2 12
3 2 1 3
Cofunction Identities
sin   cos90   
csc   sec90   
cos   sin 90   
sec   csc90   
tan   cot 90   
cot   tan 90   
EXAMPLE: Write the following as an equivalent cosine expression: sin 33 .
We can use a Cofunction Identity for this. We will use sin   cos90    . Here   33 . Substituting we get
sin 33  cos90  33 . So our answer is: sin 33  cos 57 . If you put both of these in your calculator you should
get the same decimal.
EXAMPLE: Simplify and find the EXACT value: tan 35  sec 55  cos 35
We need to use a Cofunction Identity to get rid of the 55 angle. The only formula we can use for secant is
sec   csc90    . We will get: sec 55  csc90  55 . We will get: sec 55  csc 35 . Now our problem
becomes: tan 35  csc 35  cos 35 . Let’s now change everything into sines and cosines:
sin 35
1
cos 35


. We can cross cancel the sines. Then the cosines also cancel and we get 1.
cos 35 sin 35
1
Section 4.3 Notes Page 5
EXAMPLE: Use trigonometric functions to find x and y:
58
x
30 
y
Let’s first start with x. We want to choose a trig function that relates the side I want to find with a side that is
given. In our drawing the opposite side is x and the hypotenuse is 58. You want to choose the trig function that
relates these two sides, which is sine. Now we can use the definition of sine to set up our equation:
sin 30  
x
58
58 sin 30   x
58 
1
 29
2
Now we can solve for x by using cross multiplication
From the table we know that sin 30  
1
.
2
So we know x = 29.
Now we can solve for y. I will solve this the same way as above. Note you could also use a 2  b 2  c 2 fo find
the missing side since we have a right triangle. This time y is the adjacent side and 58 is the hypotenuse. The
trig function that relates these two sides will be cosine. The formula is:
cos 30  
y
58
58 cos 30   y
58 
3
 29 3
2
More on next page…
Cross multiply
From the table we know that cos 30  
So we know that y  29 3 .
3
2
Section 4.3 Notes Page 6
EXAMPLE: A ladder is leaning against a building and forms an angle of 72 degrees with the ground. If the
foot of the ladder is 6 feet from the base of the building how far up the building does the ladder reach? How
long is the ladder?
First let’s draw a picture that describes what is happening. A ladder leaning against a house will give us a right
triangle:
This time we want to solve for y. We can do this the same as in the
previous problem. We want to find a trig function that relates y (opposite)
and 6 (adjacent). This would be tangent. Now we will set up an equation
and solve for y, which is the first thing they are asking us to find.
x
y
72 
6
tan 72  
y
6
Now cross multiply.
6 tan 72   y
This time we need to use our calculator. Make sure your calculator is in degree mode.
y = 18.47 feet
You can just round to 2 decimal places.
For the second question we need to solve for x, which will give us the length of the ladder. We need to use
cosine this time since we have an adjacent side and a hypotenuse.
cos 72  
6
x
x cos 72   6
x
6
 19.42
cos 72 
Cross multiply.
Divide both sides by cos 72 
So the length of the ladder is 19.42 feet.
Angle of elevation and depression
When you look up at something you have an angle of elevation, and when you look down on something you
have an angle of depression.
Section 4.3 Notes Page 7
EXAMPLE: While standing 4000 feet away from the base of the CN tower in Toronto, the angle of elevation
was measured to be 24.4 degrees. Find the height of the tower.
First we need a picture. The angle of elevation is 24.4 degrees, so this is as we look up to the top of the tower.
I will call the top of the tower x.
x
24.4
Now we need an equation that relate the opposite side (x) and the adjacent
side (4000). This is tangent. So we can set up an equation.

4000
tan 24.4  
x
4000
Cross multiply.
4000 tan 24.4   x
Make sure your calculator is in degree mode and solve for x.
x  1814.48 feet
EXAMPLE: An observer in a lighthouse is 66 feet above the surface of the water. The observer sees a ship and
finds the angle of depression to be 0.7  . Estimate the distance from the ship to the base of the lighthouse in
miles.
We first draw a picture. Notice that the 0.7 degrees is measured from the dotted line since it is an angle of
depression. Remember that angles of elevation and depression are always measured from the HORIZONTAL.
0.7 
66
0.7 
From geometry we know that if we have to parallel lines the
alternate interior angles are the same. So this is where the 0.7 
comes from that is inside the triangle. Now we can use tangent
again since it relates the opposite and adjacent sides.
x
tan 0.7  
66
x
x tan 0.7   66
x
66
 5401.90 feet
tan 0.7 
5401.90 ft 1 mile

 1.02 miles
1
5280 ft
Cross multiply.
Divide both sides by tan 0.7  .
This is in feet, so we need to convert it. We know 1 mile = 5280 feet.
Section 4.4 Notes Page 1
4.4 Trigonometric Functions of Any Angle
EXAMPLE: A point on the terminal side of angle  is (–5, 12). Find the exact value of each of the six
trigonometric functions of  .
First we want to plot the point and then form a triangle like below:
To get the hypotenuse, we need to use the Pythagorean theorem, which is
a 2  b 2  c 2 . Then solve for c: (12) 2  (5) 2  c 2 . The we have:
144  25  c 2 . Solving for c you will get 13. The angle  is labeled on
the graph. It is at the origin. We know that the hypotenuse is 13. The
opposite side is 12, and the adjacent side is –5. From this we can get our
six trigonometric values.
sin  
12
13
csc  
13
12
cos  
5
13
sec  
13
5
tan  
12
5
cot  
5
12
Reference Angle – an angle between 0 and 90 that is formed by the terminal side of an angle and the x-axis.
The reference angle is labeled below. It is indicated by the double curved lines. Notice that no matter where
the angle is drawn it is measured from the x-axis. Under each drawing it tells you how to find the reference
angle:
If 90     180  then
Ref. angle = 180   
If

    then
2
Ref. angle =   
If 180     270  then
Ref. angle =   180 
If 270     360  then
Ref. angle = 360   
3
then
2
Ref. angle =   
3
   2 then
2
Ref. angle = 2  
If    
If
Section 4.4 Notes Page 2

EXAMPLE: Find the reference angle for 170 .
Since 90     180  , we will use the formula 180    , so the reference angle is 180   170   10  .
9
.
5
EXAMPLE: Find the reference angle for
Since
3
9 10 9 
   2 , we will use the formula 2   , so the reference angle is 2 


 .
2
5
5
5
5
EXAMPLE: Find the reference angle for 553 .
Since this is more than 360  we need to subtract 360  from 553 and then find the reference angle for our
result. So 553  360   193 . So now we need to find the reference angle for 193 . Since 180     270  we
will use the formula   180  , so the reference angle is 193  180   13 .
EXAMPLE: Find the reference angle for
17
.
3
17 180

 1020  . This
3




is more than 2 full rotations ( 720 ) so we need to subtract this from our angle: 1020  720  300  . So now
we need to find the reference angle for 300  . Since 270     360  , we will use the formula 360    , so the
reference angle is 360   300   60  . Now since the original problem was written in degrees, we need to write
If it is difficult for you to stay in radians, we can always change this into degrees first:
our answer in degrees, so 60  

180



3
.
EXAMPLE: Draw120  in standard position and then find its reference angle.
First we will draw it in standard position. The reference angle is indicated by the double curved lines:
To find the reference angle, we use the formula above,
which says that the reference angle is 180    . So we our
reference angle is: 180   120   60  .
EXAMPLE: Draw
11
in standard position and then find its reference angle.
6
Section 4.4 Notes Page 3
11 180
11
into degrees so we know how to graph it:

 330  . Now we will draw it in
6
6

standard position. The reference angle is indicated by the double curved lines
We can change
To find the reference angle, we use the formula above,
which says that the reference angle is 360    . So we our
reference angle is: 360   330   30  . We need to change
this back into radians since the problem was originally
given in radians. Our reference angle is:

6
.
EXAMPLE: Draw  135  in standard position and then find its reference angle.
Remember when we draw it in standard position we must go clockwise this time since the angle is negative.
The reference angle is indicated by the double curved lines:
To find the reference angle, we use the formula above,
which says that the reference angle is 180    . So we our
reference angle is: 180   135   45  .
Sign values of sine, cosine, and tangent in each quadrant
sin  
sin  
cos  
cos  
tan  
tan  
Depending on which quadrant you are in the sine, cosine, and tangent
functions will be either positive or negative. You will need this for
sin  
sin  
cos  
cos  
tan  
tan  
using reference angles to find trigonometric values.
The quadrants are number from 1 to 4 counterclockwise starting with the upper right quadrant. Each quadrant
has a certain angle value: In quadrant 1: 0    90  , in quadrant 2: 90    180 , in quadrant 3:
180    270 , and in quadrant 4: 270    360 .
Section 4.4 Notes Page 4
How to find the trigonometric value for any angle:
1.) Find the reference angle.
2.) Apply the trig function to the reference angle
3.) Apply the appropriate sign.
EXAMPLE: Find the exact value of cos135  using reference angles. Draw the angle in standard position and
indicate the reference angle.
We will follow the three steps from above.
1.) First we will draw this angle in standard position. The
reference angle is indicated by the double curved lines.
We found the reference angle by taking 180   135   45 
2.) We need to apply the trig function to our reference angle, so we
2
will do cos 45  
.
2
3.) We need to apply the appropriate sign. This is where we will
use the sign chart from the last page. This angle is in the second
quadrant, so cosine needs to be negative here. So now we can
2
write our answer: cos135   
.
2
EXAMPLE: Find the exact value of sin
4
using reference angles. Draw the angle in standard position and
3
indicate the reference angle.
We can change this into degrees to see what quadrant we are in:
4 180

 240 
3 
1.) First we will draw this angle in standard position. The
reference angle is indicated by the double curved lines.
We found the reference angle by taking 240   180   60  . This is
equivalent to

3
.
2.) We need to apply the trig function to our reference angle, so we
3

will do sin 
.
3
2
3.) We need to apply the appropriate sign. This angle is in the third
quadrant, so sine needs to be negative here. So now we can write
4
3

.
our answer: sin
3
2
Section 4.4 Notes Page 5
EXAMPLE: Find the exact value of cos 330 using reference angles. Draw the angle in standard position and
indicate the reference angle.

1.) First we will draw this angle in standard position. The
reference angle is indicated by the double curved lines.
We found the reference angle by taking 360   330   30  .
2.) We need to apply the trig function to our reference angle, so we
3
.
will do cos 30  
2
3.) We need to apply the appropriate sign. This angle is in the
fourth quadrant, so cosine needs to be positive here. So now we
3
can write our answer: cos 330  
.
2
EXAMPLE: Find the exact value of tan
14
using reference angles. Draw the angle in standard position and
3
indicate the reference angle.
14 180

 840  . We can subtract two revolutions from this:

3
840   360   360   120  . From our rules, we can rewrite this problem as: tan(120   360(2))  tan 120 
2
In radian form it would look like this: tan
.
3
We can change this into degrees:
1.) First we will draw this angle in standard position. The
reference angle is indicated by the double curved lines.
2 
We found the reference angle by taking  
 . (180 – 120)
3
3
2.) We need to apply the trig function to our reference angle. We
have tan

3
 3.
3.) We need to apply the appropriate sign. This angle is in the
second quadrant, so tangent needs to be negative here. So now we
14
  3.
can write our answer: tan
3
EXAMPLE: Find the reference angle of tan(225  ) and find its EXACT value.
Using the even and odd properties, tan(225  )   tan 225  . We have changed the problem, so now we will
look at positive 225 degrees. This is in the third quadrant, so our reference angle is 225   180   45  . We will
now calculate  tan 45  . The value for tan 45  is 1. So now we have  tan(45  )  1 In the third quadrant,
tangent is positive, so we don’t need to change the sign of our answer, so tan(225  )  1 .
Section 4.4 Notes Page 6
EXAMPLE: Find the reference angle of sec(210 ) and find its EXACT value.

Using the even and odd properties, sec(210  )  sec 210  . We have changed the problem, so now we will look
at positive 210 degrees. This is in the third quadrant, so our reference angle is 210   180   30  . We will do
1
. Now we can put in the value off our table, and we will get
sec 30  . This is the same as
cos 30 
2 3
2
2 3
1
.


. Secant is negative in the 3rd quadrant, so we need to change the sign: sec(210  )  
3
3
3
3
2
 11 
EXAMPLE: Find the reference angle of sin  
 and find its EXACT value.
 3 
11 180

 660  . So now our problem becomes: sin(660  ) .

3

Using the even and odd properties, sin(660 )   sin 660  . We have changed the problem, so now we will
look at positive 660 degrees. We can rewrite the problem as  sin(300   360  ) which becomes  sin(300  ) .
We want to change our angle into degrees: 
This is in the fourth quadrant, so our reference angle is 360   300   60  , or

3
. We will now look at
3
. In the fourth quadrant, secant is
2
3
 11 
.
negative, so we need to change the sign of our answer, so sin  

2
 3 
 sin 60  . We can put in the value off our table, and we will get 
 19 
EXAMPLE: Find the reference angle of csc 
 and find its EXACT value.
 4 
19 180

 855  . Using the even and odd properties,
4



csc(855 )   csc 855 . This can be rewritten as  csc(135   360  (2)) . This reduces to  csc(135  ) . This is
in the second quadrant, so the reference angle is 180   135   45  . So now we have  csc 45  . This is the
1
1
2
2 2
same as 
. From our table we have: 


  2 . Since we are in the second

2
sin 45
2
2
Changing this into degrees we get: 
2
 19 
quadrant, sine is positive, so we don’t need to change our sign. Our answer is csc 
 2.
 4 
1
EXAMPLE: Given cos    and 180     270  , find the exact value of the six trig functions.
4
First we need to draw the triangle like we did in a previous section. This time we are told 180    270 , which
means we need to draw the triangle in the third quadrant. Our fraction is negative. That means that either 1 or
4 must be negative when we put this in our drawing. The hypotenuse is NEVER negative, so this means that 1
must be negative since this is the adjacent side. Our  is drawn at the origin, and this is always where it will be
drawn. This is like a reference angle.
Section 4.4 Notes Page 7
We can use the Pythagorean theorem to find the missing side: a  (1)  4 . Solving this you will get
2
2
2
a   15 . In our drawing, since we are in the third quadrant, we MUST use the negative answer. The
reason why is this vertical distance is really a y value, and if we think about it in terms of graphing something,
the y would be negative since we are below the x-axis. So now our drawing is complete and we can find the
six trig values:
–1

 15
4
sin   
15
,
4
csc   
4
15

EXAMPLE: Given csc   3 and
1
4 15
, cos    , sec   4 ,
15
4

2
tan   15 ,
cot  
1
15

15
15
    , find the exact value of the six trig functions.
3
. Now we know that the
1
hypotenuse is 3 and the opposite side is 1. Using the Pythagorean theorem we can find the third side:
a 2  12  3 2 . Solving this we get a   8 which can be written as a  2 2 . Since we are in the second
quadrant we want to choose  2 2 since in the second quadrant the x value is negative.
We are in the second quadrant. We can rewrite the original problem as csc  
3
1
sin  

2 2
1
,
3
csc   3 , cos   
EXAMPLE: Given tan   
2 2
,
3
sec   
3
2 2

3 2
,
4
tan   
1
2 2

2
,
4
cot   2 2
3
and sin   0 , find the exact value of the six trig functions.
4
This will be drawn in the fourth quadrant, so the opposite side must be negative. By the Pythagorean
Theorem we find: (3) 2  (4) 2  c 2 . So c = 5. Remember the hypotenuse is ALWAYS positive.
4

5
3
sin    ,
5
–3
5
4
csc    , cos   ,
3
5
sec  
5
,
4
3
tan    ,
4
cot   
4
3
Section 4.5 Notes Page 1
4.5 Graphs of Sine and Cosine Functions
This section will introduce you to the graphs of sine and cosine.
Period: How long it takes the graph to repeat itself
For sine graphs, the period is 2 .
Amplitude =
Highest value  Lowest value
2
For the regular sine graph the amplitude is 1.
The period for cosine graphs is 2
The amplitude for a regular cosine graph is 1.
General Form of a Sine or Cosine Equation:
y  A sin( Bx  C ) or y  A cos( Bx  C )
Amplitude = A , Period =
2
opp sign of C
, Phase Shift =
B
B
The phase shift is a shift of the graph to the left or to the right. The direction depends on the sign of the phase
shift:
C
 0 the graph will shift to the right.
B
C
If  0 the graph will shift to the left.
B
If
The phase shift will always be one of the five key points. In the two regular graphs of sine and cosine, the
phase shift is 0, That is why 0 is the starting key point of a cycle.
Section 4.5 Notes Page 2
EXAMPLE: Indicate the amplitude, period, and phase shift without graphing: y  3.4 sin(5 x  7)
First the amplitude is  3.4  3.4 . The period is
2
7
. The phase shift is .
5
5
1 
2 
EXAMPLE: Indicate the period, amplitude, and phase shift without graphing: y   sin x 
  1.
5 2
3 
1
1
2
2
which is . The period =
, so this is
. This simplifies to 4.

5
5
B
2
2

3   2  2   4 .
The phase shift is

3 
3
2
The amplitude is 
EXAMPLE: Graph over one period using transformations: y  4 sin x
This will have the key points as y  sin x . The only difference is that the amplitude is 4, so the highest and
lowest points on the graph will be 4 and –4.
0

2

3
2
2
EXAMPLE: Graph over one period using transformations: y  2 cos x
This one will have an amplitude of 2 and also the negative will flip over our graph. It will still have the same
key points as y  cos x .
0

2

3
2
2
Section 4.5 Notes Page 3
EXAMPLE: Graph over one period using transformations: y  3 sin x  2
This one will flip over the graph, raise the amplitude to 3, and then move the graph 2 units up. This will still
have the same key points as y  sin x . The only difference is that the axis is shifted up 2 units.

0

2
3
2
2
1 
EXAMPLE: Graph over one period using transformations: y  2 cos x 
2 
2
Here the period is
 4 . In this case the phase shift is 0, so the graph does not move vertically. In order to
1
2
Period
. In our case the
get the key points, we will use what is called the quarter point. The quarter point =
4
4
quarter point is
  . We start with the phase shift (0) and we keep adding  . The result is below. For
4
cosine graphs, we always start at the number that is in front of the cosine. In this case it is 2, so at the phase
shift the graph starts at 2.
0

2
3
4
Section 4.5 Notes Page 4
EXAMPLE: Graph over one period using transformations: y  3 sin x 
The period is
2

 2 . The phase shift once again is 0. Now we
2 1
 . You many also
4 2
use a decimal, which in this case is 0.5. We start at the phase
shift (0) and keep adding 0.5 to get the other key points:
0 + 0.5 = 0.5, 0.5 + 0.5 = 1, 1 + 0.5 = 1.5, 1.5 + 0.5 = 2
want to find the quarter point, which is


EXAMPLE: Identify the amplitude, period, phase shift and graph of y  3 cos 3x   . (Graph 1 period).
2

First the amplitude is 3  3 . The period is

2
. To find the phase shift, we take the opposite sign of C and
3


divide it by B. Then the phase shift is 2  . This tells us the graph starts at
because this is the phase shift.
3 6
6
2

We need to find our 5 key points by finding the quarter point. In this problem, the quarter point is 3  .
4
6
We will start with the left key point

6
and we will keep adding our quarter point to this to generate the other
key points:
We start with

6
. Then we have:

6


6

2
,
6
2  3
 
,
6
6
6
Now you can reduce each of your key points to the following:



6
3
2
2
3
5
6

6
,

3
,
3  4
 
,
6 6
6

2
,
4  5
 
.
6
6
6
2 5
and then graph:
,
6
3
Section 4.5 Notes Page 5
Remember the cosine graph always starts at the amplitude, which is 3 in this case.
EXAMPLE: Identify the amplitude, period, phase shift and graph of y 
1
sin x    . (Graph 1 period).
2
2

1 1
 . The period is
 2 . Then the phase shift is
 1 . This tells us the graph


2 2
2 1
starts at -1. We need to find our 5 key points by finding the quarter point. The quarter point =  . We will
4 2
start with the left key point 1 and we will keep adding our quarter point to this to generate the other key points:
First the amplitude is
We start with -1. Then we have:
1
1
1
 ,
2
2

1 1
 0,
2 2
0
1 1
 ,
2 2
1 1
  1.
2 2
1
1
The key points are  1,  , 0, , 1 . We will put these on our graph:
2
2
Notice that all sine graphs start on the x-axis.


EXAMPLE: Identify the amplitude, period, phase shift and graph of y  2 cos 2 x   . (Graph 1 period).
3


2

  . Then the phase shift is 3   . We need to find
First the amplitude is  2  2 . The period is
2
2
6

. We will start with the left key point 1
4
and we will keep adding our quarter point to this to generate the other key points:
our 5 key points by finding the quarter point. The quarter point =
We start with 

6
. Then we have:






,





,

6 4 12
12 4 3
3
7 5
. We will put these on our graph.
The key points are  ,
,
, ,
6 12 3 12
6



Notice we start this graph at -2 since there is a negative in our equation:


4

7
,
12
7  5
 
.
12 4
6
Section 4.5 Notes Page 6




6
12
3
7
12
5
6

1
EXAMPLE: Identify the amplitude, period, phase shift and graph of y   sin  x   . (Graph 1 period).
4
3

2
3
First the amplitude is  1  1 . The period is
. We need to find our 5
 6 . Then the phase shift is 4 
1
1
4
3
3
6
. There is no need to reduce this because we
key points by finding the quarter point. The quarter point =
4
already notice that this has the same denominator as the phase shift. We can keep the same denominators to
3
and we will keep adding our quarter point to this
make it easier to add. We will start with the left key point
4
3
3 6 9
9 6 15
. Then we have:


,


,
to generate the other key points: We start with
4
4
4
4
4
4
4
15 6 21
21 6 27
3 9 15 21 27


,


. The key points are
. We will put these
,
,
,
,
4
4
4
4
4
4
4
4
4
4
4
on our graph:
3
4
9
4
15
4
21
4
27
4
Section 4.6 Notes Page 1
4.6 Graphs of other Trigonometric Functions
Now we will look at other types of graphs: tan x, cot x, csc x, sec x .
Graph of y  tan x
The period is  . There are vertical
n
asymptotes at x 
where n is any
2
integer.


2
0

2

3
2
Graph of y  cot x
The period is  . There are vertical
asymptotes at x  n where n is any
integer.
0

2

3
2
2
Graph of y  csc x
In order to draw this graph we will first start
with the graph of y  sin x (dotted line).
Wherever the graph y  sin x crosses the
x-axis is where there is a vertical asymptote.
0

2

3
2
2
The period of y  csc x is 2 . The
amplitude is 1, since the graph touches
y  sin x at its amplitude.
Section 4.6 Notes Page 2
Graph of y  sec x
In order to draw this graph we will first start
with the graph of y  cos x (dotted line).
Wherever the graph y  cos x crosses the
x-axis is where there is a vertical asymptote.


0
2

3
2

2
The period of y  sec x is 2 . The
amplitude is 1, since the graph touches
y  cos x at its amplitude.


EXAMPLE: Graph y  tan x   over 2 periods.
4

If we look at the original graph of y  tan x we have one period between 
going to put everything inside the parenthesis between 

2
and

2
: 

2

2
and
 x

4

2

. In our equation we are

2
, then we solve for x:
3


to the left. So
 x  . This allows us to see where the vertical asymptotes are. They are shifted over
4
4
4
3


and . To find another vertical asymptote, just add one period to .
far we have vertical asymptotes at 
4
4
4
5

 
.
The period is  , so we will add:
4
4





4
4
3
4
5
4
3

and . You do this by finding an
4
4
4
3 

 5 3




4 4  2    . The same process is how I got 3 : 4 4  2  3 .
average:
2
2
4
4
2
2
4
Note: We got the point 

3
4
by finding the half-way point between 
Section 4.6 Notes Page 3


EXAMPLE: Graph y   cot x   over two periods.
2

If we look at the original graph of y  tan x we have one period between 0 and  . In our equation we are
going to put everything inside the parenthesis between 0 and  : 0  x 


2
  , then we solve for x:

3
. This allows us to see where the vertical asymptotes are. They are shifted over
to the left. So
2
2
2

3
3
far we have vertical asymptotes at
and
. To find another vertical asymptote, just add one period to
.
2
2
2
3
5
. Notice in the graph below that the graph is reflected over the
The period is  , so we will add:
 
2
2
horizontal axis because of the negative sign in front of the cotangent:
x

2
3
2

5
2
3
. You do this by finding an
2
2
3 5
 3


2
2
2  4  2 .
2
2

  . The same process is how I got 2 :
average:
2
2
2
2
Note: We got the point  by finding the half-way point between

2
and
1 
EXAMPLE: Graph y  2 sec x  over one period.
3 
2
2
. In this problem, the period is
 6 . This basically triples all of our
1
B
3
normal key points for the cosine graph. In order to graph this, we must first draw the cosine graph (dotted
lines). Wherever the cosine graph crosses the x-axis there will be a vertical asymptote. Where the cosine graph
hits the amplitude (2) the secant graph will touch at this point.
The period of this is: Period =
Section 4.6 Notes Page 4
3
2
0
3
9
2
6


EXAMPLE: Graph y  3 csc x   over one period.
4



We first want to graph y  3 sin  x   . The sine function has a period of 2 . Normally the graph is drawn
4

between 0 and 2 . We will put everything in parenthesis between 0 and 2 : 0  x 
we will get

4
x

4


4
,

2
4
 2 . Solving for x

9
9
. So our graph will start at
and end at
. Basically all of the normal key points
4
4
4
for the sine graph are shifted to right
0



4

3
,
4

4

. So we just add

4

5
,
4

4
to each key point of the sine graph:
3  7
 
,
2 4
4
2 

4

9
4
Then we will draw the sine graph (dotted lines). Once this is done we can draw the cosecant graph.

4
3
4
5
4
7
4
9
4
Section 4.7 Notes Page 1
4.7 Inverse Trigonometric Functions
1
. We put in an angle and get a value as a result.
2
1
In inverse trig functions we put in the value and get an angle: sin 1  30  . So here we put in the value of one
2
half and got 30 degrees as a result. We are not allowed to put any number into our inverse trig functions. There
are restrictions on the domain that are given in the following table:
From our tables in a previous section we know that sin 30  
x
Domain
1  x  1
y  cos x
1  x  1
y  tan 1 x
  x  
y  sin
1
1
Range



y

2
2
0 y 

2
y

2
 3
.
EXAMPLE: Find the sin 1 

2


What this is really asking is: “find an angle between 

2
and

2
that has a value of
table of values, go to the sine column and go down until you see the value
3
.” If you look on your
2
3
. This corresponds to an angle of
2
60 degrees, which is the answer.
 2
.
EXAMPLE: Find the cos 1 

2


What this is really asking is: “find an angle between 0 and  that has a value of
table of values, go to the cosine column and go down until you see the value
2
.” If you look on your
2
2
. This also corresponds to an
2
angle of 45 degrees, which is the answer.
 3
.
EXAMPLE: Find the tan 1 

3


What this is really asking is: “find an angle between 

2
and

2
that has a value of
table of values, go to the tangent column and go down until you see the value
angle of 30 degrees, which is the answer.
3
.” If you look on your
3
3
. This also corresponds to an
3
Section 4.7 Notes Page 2
EXAMPLE: Use a calculator to find, cos 0.7 if possible. Round your answer to two decimal places.
Assume your angle is in radians.
1
We need to make sure our calculator is in radian mode before we proceed. The inverse cosine is above the
cosine key on your calculator. You will probably need to use your second key in order to get the inverse cosine.
Your answer should be 0.8. If you got an error, try entering 0.7 first and then get the inverse.
EXAMPLE: Use a calculator to find, sin 1 (1.2) if possible. Round your answer to two decimal places.
Assume your angle is in radians.
If you try putting this in your calculator you will get an error. This is because 1.2 is not in our domain. Recall
that the domain for the inverse sine function is  1  x  1 . This means we can only put in numbers between -1
and 1. So the answer is no solution.
Inverses and canceling
If we take cos 1 (cos x) what will we get? Well, the inverse cosine and cosine will cancel and that will leave us
with just x. However there are some restrictions on what x can be as listed below:
cos 1 (cos x)  x if 0  x  
cos(cos 1 x)  x if  1  x  1
sin 1 (sin x)  x if 
sin(sin
1

2
x

2
x)  x if  1  x  1
tan 1 (tan x)  x if 

2
x

2
1
tan(tan x)  x if    x  
EXAMPLE: Find the exact value if possible: tan(tan 1 5.3) .
According to our restrictions above, x can be any number, so tan(tan 1 5.3) = 5.3.
98 

EXAMPLE: Find the exact value if possible: sin  sin 1  .
99 

98  98

.
The fraction 98/99 is .9898, and this is less than 1, so sin  sin 1  =
99  99



EXAMPLE: Find the exact value if possible: cos cos 1 2 .
The square root changed into a decimal is 1.41, which is bigger than 1, so the answer is no solution since 1.41 is
not in our domain.
Section 4.7 Notes Page 3
2  2

EXAMPLE: Does sin 1  sin
?

3  3



2
is not in the domain   y  .
3
2
2
The answer is no, because
 

EXAMPLE: Does cos 1  cos   ?
3 3


The answer is yes, because
is in the domain 0  y   .
3
26  26

?
EXAMPLE: Does tan 1  tan

50  50

The answer is no, because
26


is not in the domain   y  .
50
2
2

 3 
EXAMPLE: Use a sketch to find the exact value: sin  cos 1    .
 5 

These problems involve drawing a triangle and labeling the sides like we did in a previous section. The inverse
trig function will tell you where to draw the triangle. In our example there is an inverse cosine. The inverse
cosine’s range will tell us where we can draw the triangle. From the last section, the range for the inverse
3
cosine is 0  y   . This corresponds to the first and second quadrant. Since the fraction is positive, the
5
only quadrant the triangle can be drawn in is the first quadrant. We know that the adjacent side is 3 and the
hypotenuse is 5. The Pythagorean Theorem will give us the opposite side, which is 4.
5

4
The sine on the outside of our problem tells us how to write our
answer. From our drawing, sine is 4 over 5, so we write our
answer as:
3

 3  4
sin  cos 1    
 5  5

Section 4.7 Notes Page 4

 1 
EXAMPLE: Use a sketch to find the exact value: cos sin 1  
  .
6



The inverse trig function will tell you where to draw the triangle, and in this case we have an inverse sine. The
inverse sine’s range will tell us where we can draw the triangle. From the last section, the range for the inverse
sine is 

 y

. This corresponds to the first and fourth quadrant. Since the fraction inside the inverse is
2
2
negative, the only quadrant the triangle can be drawn in is the fourth quadrant. We know that the opposite side
is -1 and the hypotenuse is 6 . The Pythagorean Theorem will give us the adjacent side, which is 5 .

6
The cosine on the outside of our problem tells us how to write our
answer. From our drawing, cosine is 5 over 6 , so we write
our answer as:
5
-1

 1 
cos sin 1  
  
6  


5
6

30
6

 3 
EXAMPLE: Use a sketch to find the exact value: tan cos 1     .
 4 

The inverse trig function will tell you where to draw the triangle, and in our case there is an inverse cosine.
The inverse cosine’s range will tell us where we can draw the triangle. From the last section, the range for the
inverse cosine is 0  y   . This corresponds to the first and second quadrant. Since the fraction inside the
inverse is negative, the only quadrant the triangle can be drawn in is the second quadrant. We know that the
adjacent side is -3 and the hypotenuse is 4. The Pythagorean Theorem will give us the opposite side: 7 .
4
7

-3
The tangent on the outside of our problem tells us how to write our
answer. From our drawing, tangent is 7 over -3, so:

7
 3 
tan cos 1      
3
 4 

Section 4.7 Notes Page 5

 1 
EXAMPLE: Find the exact value: csc tan 1     .
 3 

The inverse trig function will tell you where to draw the triangle, and in this case we have an inverse tangent.
The inverse tangent’s range will tell us where we can draw the triangle. From the last section, the range for the
inverse tangent is 

y

. This corresponds to the first and fourth quadrant. Since the fraction inside the
2
2
inverse is negative, the only quadrant the triangle can be drawn in is the fourth quadrant. We know that the
opposite side is -1 and the adjacent is 3. The Pythagorean Theorem will give us the adjacent side: 10 .
3
The cosecant on the outside of our problem tells us how to write
our answer. From our drawing, cosecant is 10 over -1, so we
write your answer as:

-1
10

10
 1 
csc tan 1     
  10 .
 3  1

EXAMPLE: Use right triangles to write in algebraic form: sectan 1 4 x  . Assume that x is positive and that
the given inverse trigonometric function is defined for the expression in x.
These problems involve drawing a triangle and labeling the sides with algebraic expressions. For all these
problems we will assume that x is positive and the triangle should be drawn in the first quadrant. We can
4x 

rewrite our problem as: sec tan 1  We know that the adjacent side is 1 and the opposite side is 4x. We can
1 

use the Pythagorean theorem to find the hypotenuse: c 2  (4 x) 2  (1) 2 . So we have c  16 x 2  1
16 x 2  1
4x

1
The secant on the outside of our problem tells us how to write
our answer. From our drawing, secant is 16 x 2  1 over 1 so we
write our answer as:
sectan 1 4 x   16 x 2  1
NOTE: in some textbooks, the inverse functions are written differently, for example instead of y  sin 1 x ,
some textbooks may write this as y  arcsin x . So instead of the 1 symbol, it is replaced by the word arc.
These two mean exactly the same thing. So y  arccos x would mean the same as y  cos 1 x , etc.
Section 4.8 Notes Page 1
4.8 Applications of Trigonometric Functions
We will come back to right triangles again. In these questions, if it asks you to solve the triangle then that
means you need to find all sides and angles.
EXAMPLE: Solve the triangle:
B
35 
A
200 ft
C
In this problem we need to find side AB, side AC, and mA (measurement of angle A). First we can find
mA . The sum of all the angles in a triangle is 180 degrees. We have a 35 degree angle and a 90 degree
angle, so mA  180  35  90  55  . Now we want to find side AB. Let’s call this length x. From our picture
above let’s use the 35 degree angle. The side opposite this angle is 200 ft. The hypotenuse is x, which is what
200
. We can solve for x by cross
we are trying to find. We can set up the following equation: sin 35  
x
200
multiplying. We get x 
 348.69 ft. Now we can solve for side BC, which we are calling y. Again if
sin 35 
we use the 35 degree angle, the y is the adjacent angle and 200 ft is still the opposite side. We will use the
200
following equation to solve for y: tan 35  
. We can solve for y by doing cross multiplication:
y
200
y
 285.63 ft. Now we know the measurement of all three sides and angles, so it is solved.
tan 35 
EXAMPLE: Solve the triangle:
B
I will let side BC be x. Now we can use the Pythagorean Theorem to find
 
2
find it: 3 2  x 2  3 5 . Solving this will give us: 9  x 2  45 , so
3 5
A
3 cm
C
x  6 . To find mA , we can set up the following trig equation:
3
cos A 
. So we have cos A  0.4472 . We need to take the inverse
3 5
cosine to get our answer. So A  cos 1 0.4472  63.44  . To find mB
we will subtract 90 degrees and 63.44 degrees from 180 degrees. We will
get: mB  180  63.44  90  26.56  . Now our triangle is solved.
Section 4.8 Notes Page 2
EXAMPLE: A security camera in a neighborhood bank is mounted on a wall 11 ft above the floor. What angle
of depression should be used if the camera is to be directed to a spot 6 ft above the floor and 12 ft from the
wall?
We first need to draw a picture for this one:
A
5 ft
A
12 ft
6 ft
We need to find angle A which is the angle of depression. This is the same as the angle A inside the triangle.
When looking at the angle A inside the triangle we see that the opposite side is 5ft and the adjacent is 12 feet.
5
We can set up the following equation: tan A  . This can also be written as tan A  0.4167 . In order to find
12
1
A we need the inverse tangent: A  tan 0.4167  22.62  . Therefore the angle of depression should be 22.62 
EXAMPLE: Find x, rounded to the nearest whole number:
There are two triangles in this problem. We are going to find the
opposite sides of each triangle and then subtract them to find x. Let’s first
find the opposite side of the triangle with an angle of 28  . In order to find
opposite
the opposite side we will use this equation: tan 28  
. Solving
400
this will give us: opposite  400 tan 28   213 . Now we will find the
opposite side of the triangle with an angle of 40  . We will use the
opposite
following equation: tan 40  
. Solving this will give us:
400
opposite  400 tan 40   336 . Now we subtract our two answers to get x:
336 – 213 = 123 units.
Bearing
Bearing is a way to measure direction. There are three parts to bearing. The
first part is either a N or S. The second part is an acute angle that is ALWAYS
measured from a vertical axis. The third part is an E or W. The picture to
the right shows how North, South, East, and West is orientated. The next
few examples show how to draw bearings.
Section 4.8 Notes Page 3

EXAMPLE: Draw the following: N 40 E .
North is always straight up. First we go straight north. Then is says to go 40 degrees to the east. So from the
north we will measure 40 degrees. Our picture is:
EXAMPLE: Draw the following: N 65 W .
North is always straight up. First we go straight north. Then is says to go 65 degrees to the west. So from the
north we will measure 65 degrees. Our picture is:
EXAMPLE: Draw the following: S 70  E .
South is always straight down. First we go straight south. Then is says to go 65 degrees to the east. So from
the south we will measure 70 degrees. Our picture is:
Section 4.8 Notes Page 4
EXAMPLE: An airplane takes off from a runway at a bearing of N 20 E . After flying 1 mile in this direction
the plane makes a 90 degree turn to the left. After flying 2 more miles in this new direction, what is the bearing
form the airport to the plane?
N
2 miles
First we need to draw a picture as shown. Notice the 20 degrees is
measured from the north since this is a bearing. You will end up
in the second quadrant.
20  1 mile
W
E
A
We need to find the measurement of angle A. Then we will subtract
20 degrees to find the bearing in the second quadrant. In order to find
2
angle A we can use the formula: tan A  . Again we need to use the
1
inverse tangent to find angle A: A  tan 1 2  63.43 . This is the entire
S



angle of A, so we need to subtract 20: 63.43  20  43.43 . We need
to write this as a bearing, so our answer is: N 43.43 W .

EXAMPLE: A ship leaves the port of Miami with a bearing of S 80  E and a speed of 15mph. After 1 hour the
ship turns 90 degrees towards the south. After 2.5 hours maintaining the same speed, what is the bearing from
the port to the ship?
First we draw the picture. The 80 degrees is measured from
the south. After 1 hour the ship will travel 15 miles. After
2.5 hours the boat travels (2.5)(15) = 37.5 miles as indicated
on the diagram. If we can find x, then we can subtract this
from 80 degrees to get the bearing measured from the south
axis. When looking at x, the opposite side is 37.5 and the
adjacent side is 15, so we need to use tangent again. So
37.5
. This means tan x  2.5 . To find x we need to
tan x 
15
take the inverse tangent of both sides: x  tan 1 2.5  68.2  .
This is x, but in order to find the bearing we need to subtract
this from 80 degrees since bearings are always measured from
the vertical. 80   68.2   11.8  . Finally we need to write our
bearing, which is S11.8  E .
Section 5.1 Notes Page 1
5.1 Verifying Trigonometric Identities
This section will help you practice your trigonometric identities. We are going to establish an identity. What
this means is to work out the problem and show that both sides of the identity are the same. First let’s look at a
list of identities we’ve already talked about plus a few more.
List of Identities
tan  
sin 
cos 
cot  
cos 
sin 
csc  
1
sin 
sec  
1
cos 
sin 2   cos 2   1
sin 2   1  cos 2 
cos 2   1  sin 2 
sec 2   1  tan 2 
tan 2   sec 2   1
csc 2   1  cot 2 
cot  
1
tan 
cot 2   csc 2   1
When working out these identities, you can try on or more of the following techniques. I will explain each
technique with examples:
1.) Change everything into sines and cosines.
2.) Use factoring to simplify the expression if possible.
3.) Get common denominators if there are fractions.
4.) Multiply both sides by a conjugate.
Of course, as we use the above techniques, be sure to refer back to the list of identities I gave you above. You
might need to use some of them to simplify. One you will see come up often is sin 2   cos 2   1 .
EXAMPLE: Establish the identity: csc   tan   sec  .
You want to show that one side of the equation equals the other side. In these problems you are NOT allowed
to do operations like adding or subtracting things from one side to the other. Think of each side as independent.
We are not going to do anything with the right hand side. On the left side we will use the first technique and
change the cosecant and tangent functions into sines and cosines:
1 sin 

 sec 
sin  cos 
We can now cancel the sines from the left side of the equation.
1
 sec
cos 
We can change the fraction on the left side into secant.
sec   sec 
Both sides are the same, so we are done.
Section 5.1 Notes Page 2
sin   cos 
 (cos   sin  ) .
cos   sin 
4
EXAMPLE: Establish the identity:
4
Another technique I mentioned is factoring. We can factor the top because of difference of squares.
sin
2
sin
2
  cos 2  sin 2   cos 2  
 (cos   sin  )
cos   sin 
  cos 2  
 (cos   sin  )
cos   sin 
We know sin 2   cos 2   1
We can factor the top again by difference of squares.
sin   cos sin   cos   (cos  sin  )
cos   sin 
We want to factor a negative out of the first term.
 ( sin   cos  )(sin   cos  )
 (cos   sin  )
cos   sin 
Now switch the order in the first term on top.
 (cos  sin  )(sin   cos  )
 (cos   sin  )
cos   sin 
Now we can cancel the cos   sin  terms.
 (sin   cos  )  (cos   sin  )
Both sides are equal so the proof is done.
EXAMPLE: Establish the identity:
cos x(1  2 sin x)
 cot x
cos x  sin 2 x  sin x  1
2
cos x  2 sin x cos x
 cot x .
cos 2 x  sin 2 x  sin x  1
First we can factor the numerator. Now we want to get all sines on
the bottom. We can use the identity cos 2 x  1  sin 2 x .
cos x(1  2 sin x)
 cot x
(1  sin 2 x)  sin 2 x  sin x  1
Now simplify the denominator.
cos x(1  2 sin x)
 cot x
 2 sin 2 x  sin x
Factor the denominator.
cos x(1  2 sin x)
 cot x
sin x(2 sin x  1)
The part in parenthesis on top and bottom can be cancelled.
cos x
 cot x
sin x
We will use the identity cot x 
cot x  cot x
Both sides are equal so we are done.
cos x
sin x
Section 5.1 Notes Page 3
EXAMPLE: Establish the identity: cot  
1  2 cos 
 tan  .
sin  cos 
2
Since this problem has a fraction, I will follow technique #3, which says to get common denominators if there
cos 
are fractions. At the same time I will also use the identity: cot  
.
sin 
cos   cos   1  2 cos 2 

 tan 

sin   cos   sin  cos 
Now write as a single fraction.
cos 2   1  2 cos 2 
 tan 
sin  cos 
Now simplify the numerator.
1  cos 2 
 tan 
sin  cos 
sin 2 
 tan 
sin  cos 
We will now use the identity sin 2   1  cos 2  .
We can cancel a sine from the top and bottom.
sin 
 tan 
cos 
We will use the identity tan  
tan   tan 
Both sides are equal so we are done.
EXAMPLE: Establish the identity:
sin 
.
cos 
cos x
1  sin x

 2 sec x .
1  sin x
cos x
Once again we want to first get a single fraction so we need common denominators.
cos x  cos x  1  sin x  1  sin x 



  2 sec x
1  sin x  cos x 
cos x  1  sin x 
Now multiply and write as a single fraction.
cos 2 x  (1  sin x) 2
 2 sec x
cos x(1  sin x)
We will expand the numerator.
cos 2 x  sin 2 x  2 sin x  1
 2 sec x
cos x(1  sin x)
We will use the identity cos 2 x  sin 2 x  1
1  2 sin x  1
 2 sec x
cos x(1  sin x)
Simplify the numerator.
2 sin x  2
 2 sec x
cos x(1  sin x)
Factor the numerator.
Section 5.1 Notes Page 4
2(sin x  1)
 2 sec x
cos x(1  sin x)
We can cancel the sin x  1 from the top and bottom.
2
 2 sec x
cos x
We will use the identity sec x 
2 sec x  2 sec x
Both sides are the same, so we are done.
EXAMPLE: Establish the identity:
1
.
cos x
tan x  cot x
 1.
sec x csc x
We will use technique #1 and change everything into sines and cosines. This makes it easier to reduce.
sin x cos x

cos x sin x  1
1
1

cos x sin x
We need to get common denominators in the numerator.
sin x  sin x  cos x  cos x 




cos x  sin x  sin x  cos x 
1
1
1

cos x sin x
Multiply and write as one fraction in the numerator.
sin 2 x  cos 2 x
sin x cos x  1
1
sin x cos x
Now use the identity sin 2   cos 2   1 .
1
sin x cos x  1
1
sin x cos x
Flip over the bottom fraction and multiply.
1
sin x cos x

1
sin x cos x
1
We can cancel terms.
1=1
Both sides are the same so we are done.
EXAMPLE: Establish the identity:
1  sin 
cos 

.
cos 
1  sin 
Section 5.1 Notes Page 5
We see that everything is already in terms of sine and cosine. Also notice that we can’t factor and even though
there are fractions, we don’t need common denominators. The only other technique that we can use is
technique #4, which says to multiply both sides by a conjugate. We can do this on either side. Please note that
we are NOT allowed to cross multiply because we need to treat both sides separately.
1  sin 
cos 

cos 
1  sin 
 1  sin  


 1  sin  
I chose the right side, but we can chose either side to work with.
1  sin  cos  (1  sin  )

cos 
1  sin 2 
Now use the identity cos 2 x  1  sin 2 x .
1  sin  cos  (1  sin  )

cos 
cos 2 
Now we can cancel the cosine from the top and bottom.
1  sin  1  sin 

cos 
cos 
Both sides are equal so we are done.
Let’s do this problem again, but now let’s work on the left side instead of the right side.
cos 
 1  sin   1  sin 



1  sin 
 1  sin   cos 
I chose the left side this time.
1  sin 2 
cos 

cos  (1  sin  ) 1  sin 
Now use the identity cos 2 x  1  sin 2 x .
cos 2 
cos 

cos  (1  sin  ) 1  sin 
Now we can cancel the cosine from the top and bottom.
cos
cos 

1  sin  1  sin 
Both sides are equal so we are done.
Remember you only need to show that both sides are equal. What each side is does not matter, as long as both
sides are the same. I am not looking for one exact way of doing these problems, because there may be more
than one way to show that one side equals the other. Just make sure you logically show your steps. If you start
with one statement and then jump down to the answer without showing how you got there, you will not receive
full credit. Your answers to these problems will be these logical steps you show.
Section 5.2 Notes Page 1
5.2 Sum and Difference Formulas
These formulas will allow use to find the exact value for other angles besides just 0, 30, 45, 60, and 90 degrees.
The book uses the symbols  and  which are more confusing. Instead I will use x and y.
sin( x  y )  sin x cos y  cos x sin y
sin( x  y )  sin x cos y  cos x sin y
cos( x  y )  cos x cos y  sin x sin y
cos( x  y )  cos x cos y  sin x sin y
tan( x  y ) 
tan x  tan y
1  tan x tan y
tan( x  y ) 
tan x  tan y
1  tan x tan y
 3

  .
EXAMPLE: Simplify sin 
 2

We will use the second formula above, which is sin( x  y )  sin x cos y  cos x sin y . Here the x would be
and the y is  :
3
2

 3 
 3 
 3
    sin  cos   cos  sin 
sin 
 2

 2 
 2 
 3 
 3 
We know sin    1 and cos   0 . Substitute.
 2 
 2 
 3

    1  cos   0  sin 
sin 
 2

This simplifies to  cos  .
 3

     cos  .
sin 
 2

EXAMPLE: Write as a single trig value and then find the exact value: sin 40  cos 20   cos 40  sin 20  .
We have sine cosine and then cosine sine. This is using the formula sin( x  y )  sin x cos y  cos x sin y . We
know that x  40  and y  20  . This means that the whole expression above turns into sin(40   20  ) . This
simplifies to sin(60  ) . From our table we know that sin(60  ) 
3
.
2
Section 5.2 Notes Page 2
 3    
 3    
EXAMPLE: Write as a single trig value and then find the exact value: cos  cos   sin   sin   .
 10   5 
 10   5 
We have cosine cosine and then sine sine. This is using the formula cos( x  y )  cos x cos y  sin x sin y . We

3
 3  
know that x 
  . This
and y  . This means that the whole expression above turns into cos
10
5
 10 5 
 
 
simplifies to cos  . From our table we know that cos   0 .
2
2
EXAMPLE: Write as a single trig value and then find the exact value:
tan 50   tan 20 
.
1  tan 50  tan 20 
tan x  tan y
, which is the expression for tan( x  y ) . Here, we will let
1  tan x tan y
x  50  and y  20  . So then the above fraction expression will turn into tan(50   20  ) . This simplifies into
This one is written in the form
 
tan 30  . The table of trig values tells us that this is equal to
3
.
3
EXAMPLE: Determine the exact value of sin 75  .
We don’t have 75 degrees on our table. We need to rewrite 75 degrees in terms of angle we have on our table.
One way is to rewrite this as sin 30   45  . Then we can use the formula sin( x  y )  sin x cos y  cos x sin y
where x is 30 and y is 45: sin(30   45  )  sin 30  cos 45   cos 30  sin 45  . Now everything here we can get


values from our table to get the exact answer: sin(30   45  ) 
will get: sin(30   45  ) 
1 2
3 2
. After multiplying this we



2 2
2 2
2 6
.
4
EXAMPLE: Determine the exact value of cos165 .
We need to rewrite 165 degrees in terms of angle we have on our table. One way is to rewrite this as
cos 45   120  . Then we can use the formula cos( x  y )  cos x cos y  sin x sin y where x is 45 and y is 120:
cos(45   120  )  cos 45  cos120   sin 45  sin 120  . Now everything here we can get values from our table to
get the exact answer. At 120 degrees the reference angle is 60 degrees. Since 120 degree is in the second
1
3
. So
quadrant, the cosine is negative and the sine is positive. So cos120    and sin 120  
2
2
2 1
2 3
 2 6
 

. After multiplying this we will get: sin(30   45  ) 
.
cos(45   120  ) 
2
2 2 2
4


Section 5.2 Notes Page 3

EXAMPLE: Determine the exact value: tan 105 .


We need to rewrite 105 degrees as: tan 45   60  . Using the fifth formula we get:
tan(45   60  ) 
tan(45   60  ) 
tan 45  tan 60
. Using values from our table we will get:
1  tan 45  tan 60 

1 3
1  (1) 3

. We need to rationalize the denominator.
tan(45   60  ) 
1 3 1 3

1 3 1 3
Now multiply and simplify.
tan(45   60  ) 
1 2 3  3
1 3
Simplify the top.
tan(45   60  ) 
42 3
2
Now reduce.
tan(105  )  2  3
Now we will look at solving some proofs, which was done in the previous section.
EXAMPLE: Establish the identity:
sin( x  y )
 tan x  tan y .
cos x cos y
We need to use the formula sin( x  y )  sin x cos y  cos x sin y .
sin x cos y  cos x sin y
 tan x  tan y
cos x cos y
Now divide each term on top by what is on the bottom.
sin x cos y cos x sin y

 tan x  tan y
cos x cos y cos x cos y
We can cancel.
sin x sin y

 tan x  tan y
cos x cos y
Sine over cosine is tangent.
tan x  tan y  tan x  tan y
Both sides are equal, so we are done.
EXAMPLE: Establish the identity:
Section 5.2 Notes Page 4
cos x  3 2
  tan x .
sin  x   2
For this one we will first use the sum and difference formulas on the top and bottom:
cos x cos3 2   sin x sin 3 2 
  tan x
sin x cos 2  cos x sin  2
cos x  0  sin x  1
  tan x
sin x  0  cos x  1

3
is 270 degrees, which has a reference angle of 90  .
2
Now simplify.
sin x
  tan x
cos x
 tan x   tan x
EXAMPLE: Establish the identity:
Both sides are equal, so we are done.
cos( x  y ) 1  tan x tan y

.
cos( x  y ) 1  tan x tan y
Start by using sum formulas on the right side.
sin x sin y

cos( x  y )
cos x sin y

sin x sin y
cos( x  y )
1

cos x cos y
Now get common denominators.
cos x sin y sin x sin y

cos( x  y ) cos x sin y cos x sin y

cos( x  y ) cos x sin y sin x sin y

cos x sin y cos x cos y
Combine fractions together.
cos x sin y  sin x sin y
cos( x  y )
cos x sin y

cos( x  y ) cos x sin y  sin x sin y
cos x sin y
Flip the bottom fraction and multiply.
cos( x  y ) cos x sin y  sin x sin y
cos x sin y


cos( x  y )
cos x sin y
cos x sin y  sin x sin y
Cancel the cos x sin y .
cos( x  y ) cos x sin y  sin x sin y

cos( x  y ) cos x sin y  sin x sin y
Use the sum and difference formulas.
cos( x  y ) cos( x  y )

cos( x  y ) cos( x  y )
Both sides are equal so we are done.
1
Section 5.3 Notes Page 1
5.3 Double Angle and Half Angle Formulas
If we have either a double angle 2 or a half angle

2
then these have special formulas:
sin(2 )  2 sin  cos 
cos(2 )  cos 2   sin 2 
cos(2 )  2 cos 2   1
cos(2 )  1  2 sin 2 
tan(2 ) 
sin

cos
tan
2

2

2
There are three formulas for cos(2 )
2 tan 
1  tan 2 

1  cos 
2
You will choose plus or minus depending on what quadrant

1  cos 
2
You will choose plus or minus depending on what quadrant

1  cos 
1  cos 
You will choose plus or minus depending on what quadrant
There are better formulas for tan
tan

2

sin 
,
1  cos 
tan

2


2

2

2

2
is.
is.
is.
that don’t involve a plus or minus:
1  cos 
sin 
EXAMPLE: Compute sin  , tan  , csc  , sec  , cot  , sin(2 ) , cos(2 ) , tan(2 ) , sin

2
, cos

2
, and tan

2
if you are given cos   0.8 and 90     180  . Round decimals to two decimal places.
We need to draw a triangle for this one. We are given that the triangle should be drawn in the second quadrant.
 0.8
We can rewrite our problem as cos  
. We know the adjacent side is –0.8. The hypotenuse is 1. Once
1
we label our triangle we find the third side by using the Pythagorean theorem.
1
0.6

-0.8
Section 5.3 Notes Page 2
Now we can get our first 5 trigonometric functions by reading off our triangle:
csc  
sin   0.6
tan  
0.6
 0.75
 0.8
1
 1.67
0.6
cot  
sec  
1
 1.25
 0.8
 0.8
 1.33
0.6
Next I will find sin(2 ) by using its formula: sin(2 )  2 sin  cos  . We already know sine and cosine, so we
will substitute in those decimals: sin(2 )  2(0.6)(0.8)  0.96 .
Next I will find cos(2 ) by using its formula: cos(2 )  2 cos 2   1 . Notice I had a choice of three formulas
to use. Any of them would give you the correct answer. We already know cos   0.8 , so we will substitute
in this decimal: cos(2 )  2(0.8) 2  1  0.28 .
2 tan 
. We already know tan   0.75 , so we
1  tan 2 
2(0.75)
 1.5
will substitute in this decimal: tan(2 ) 

 3.43 .
2
0.4375
1  (0.75)
Next I will find tan(2 ) by using its formula: tan(2 ) 
For the half angle formulas I need to determine which quadrant

2
is in. To do this let’s first start with our
given statement 90     180  . If I divide everything by two we get: 45  
in the first quadrant, so sine, cosine, and tangent of
2
2
 90  . This tells us that

2
is
should all be positive.

1  cos 
. We chose a positive because is in the
2
2
2
2
second quadrant. We already know cos   0.8 , so we will substitute in this decimal:
1  (0.8)
1.8

sin 

 .9  0.95 .
2
2
2
Now I will find sin






1  cos 
. We chose a positive because is in the
2
2
2
2
second quadrant. We already know cos   0.8 , so we will substitute in this decimal:
1  (0.8)
0.2

cos 

 .1  0.32 .
2
2
2
Now I will find cos

by using its formula: sin
by using its formula: cos

sin 
. We already know
2
2 1  cos 
0.6

cos   0.8 , and sin   0.6 so we will substitute in these decimals: tan 
 3.
2 1  (0.8)
Finally I will find tan


. I have three formulas to choose. I will choose: tan


Section 5.3 Notes Page 3
EXAMPLE: Compute sin  , tan  , csc  , sec  , cot  , sin(2 ) , cos(2 ) , tan(2 ) , sin
if you are given cot  

2
, cos

2
, and tan

2
1
and 180     270  .
2
We need to draw a triangle first. We are given that the triangle should be drawn in the third quadrant. We
know the adjacent side is 1 and the opposite side is 2. However because we are in the third quadrant we need to
make the 1 and 2 negative. Once we label our triangle we find the third side by using the Pythagorean theorem.
This will give us 5 .
-1

-2
5
Now we can get our first 5 trigonometric functions by reading off our triangle:
sin   
2
csc   
5
2
5

2 5
5
cos   
1
5

5
5
sec    5
tan   2
Next I will find sin(2 ) by using its formula: sin(2 )  2 sin  cos  . We already know sine and cosine, so we
 2 5 
5  20 4
 

 .
will substitute in those fractions: sin(2 )  2 


5  5  25 5

Next I will find cos(2 ) by using its formula: cos(2 )  2 cos 2   1 . Notice I had a choice of three formulas
to use. Any of them would give you the correct answer. We already know cos   
5
, so we will substitute
5
2

2
3
5
5
  1 . This simplifies to: 2   1   1   .
in this fraction: cos(2 )  2 

5
5
 25 
 5 
Next I will find tan(2 ) by using its formula: tan(2 ) 
substitute in this number: tan(2 ) 
2(2)
4
 .
2
3
1  (2)
2 tan 
. We already know tan   2 , so we will
1  tan 2 
For the half angle formulas I need to determine which quadrant

2
Section 5.3 Notes Page 4
is in. To do this let’s first start with our
given statement 180     270  . If I divide everything by two we get: 90  
is in the second quadrant, so sine of
2
2
 135  . This tells us that
should be positive, and the cosine and tangent of

2

2
should be negative.

1  cos 
. We chose a positive because is in the
2
2
2
2
second quadrant. We already know cos   0.8 , so we will substitute in this decimal:
Now I will find sin



by using its formula: sin

5

1   
5 


sin 

2
2

5 5
5 5
5
.

2
10

1  cos 
. We chose a negative because is in the
2
2
2
2
second quadrant. We already know cos   0.8 , so we will substitute in this decimal:
Now I will find cos


by using its formula: cos



5
5 5

1   


5 5
 5  
5
.
cos  

2
2
2
10
Finally I will find tan

. I have three formulas to choose. I will choose: tan


sin 
. We already know
1  cos 
2
2
cos   0.8 , and sin   0.6 so we will substitute in these decimals:
2 5
2 5



5   2 5  5   2 5  5  5   10 5  10   10 5  10   5  1 .
5

tan 
5 5 5
25  5
20
2
2

5 5 5 5
5  5 5

1   

5
 5 
EXAMPLE: Compute sin(22.5  ) and tan(22.5  ) using a half-angle formula.
 45  

1  cos 45 
 . Then we know that  is 45 degrees, so now we use: sin 
. It
We can write this as sin 
2
2
 2 
is positive since 22.5 is in the first quadrant. You get: sin
45 
sin 45 
For tan(22.5  ) we will use: tan


2
1  cos 45 

2

1
2
2
2 
2 2
2

2
2 2

4
2 2
.
2
2
2
2 2 2 2 2
2




 2 1.
2
2 2 2 2 2 2 2
1
2
2
2
2
Section 5.3 Notes Page 5
EXAMPLE: Rewrite 1  2 sin 2  12 using a double angle formula. Then find its exact value.
This one is written in the form 1  2 sin 2  , which is a form of cos 2 . In this problem, we will let    12 .
   

Therefore we will substitute this into the formula cos 2 : cos 2  . Simplifying gives us cos . Looking
6
  12 
at our table we see this has an exact value of 3 2 .
EXAMPLE: Establish the identity:
cot   tan 
 cos 2
cot   tan 
First we want to change these into sines and cosines.
cos  sin 

sin  cos   cos 2
cos  sin 

sin  cos 
Now get common denominators on the top and bottom.
 cos   cos  sin   sin  





 cos   sin  cos   sin    cos 2
 cos   cos  sin   sin  





 cos   sin  cos   sin  
Multiply and write over a single denominator
cos 2   sin 2 
cos  sin   cos 2
cos 2   sin 2 
cos  sin 
We will get rid of the double fractions and use cos 2   sin 2   1
cos 2   sin 2   cos 2
The left side is the identity for cos 2 .
cos 2  cos 2


EXAMPLE: 4 sin  cos   1  2 sin 2   sin 4
First we will use the identity 1  2 sin 2   cos 2 . Now our problem becomes:
4 sin  cos   cos 2  sin 4
We know that sin 2  2 sin  cos  . We want to rewrite our problem as the following:
2  2 sin  cos   cos 2  sin 4
Now we can use the identity sin 2  2 sin  cos  .
2 sin 2 cos 2  sin 4
We know that 2 sin 2 cos 2 is the same as sin 2  2   sin 4 .
sin 4  sin 4
Section 5.4 Notes Page 1
5.4 Product-to-Sum and Sum-to-Product Formulas
Product-to-Sum Formulas
1
cos( x  y)  cos( x  y)
2
1
cos x cos y  cos( x  y )  cos( x  y )
2
1
sin x cos y  sin( x  y )  sin( x  y )
2
sin x sin y 
EXAMPLE: Simplify: sin(6 ) sin(4 ) using a product-to-sum formula.
We will use the first formula sin x sin y 
1
cos( x  y)  cos( x  y) with x  6 and y  4
2
1
cos(6  4 )  cos(6  4 )
2
Now simplify.
1
cos(2 )  cos(10 )
2
This is as far as we can go.
EXAMPLE: Simplify: cos(3 ) cos( ) using a product-to-sum formula.
We will use the formula cos x cos y 
1
cos( x  y)  cos( x  y) with x  3 and y   .
2
1
cos(3   )  cos(3   )
2
Now simplify.
1
cos(2 )  cos(4 )
2
This is as far as we can go.
EXAMPLE: Simplify: sin(3 ) cos(5 ) using a product-to-sum formula.
We will use the formula sin x cos y 
1
sin( x  y)  sin( x  y) with x  3 and y  5 .
2
1
sin(3  5 )  sin(3  5 )
2
Now simplify.
1
sin(2 )  sin(8 )
2
We will use the identity sin(2 )  sin 2 .
1
 sin(2 )  sin(8 )
2
or
1
sin(8 )  sin(2 ) is as far as we can go.
2
EXAMPLE: Find the exact value of cos
We will use the formula cos x cos y 

5
using a product-to-sum formula.
cos
12
12
Section 5.4 Notes Page 2
1
cos( x  y)  cos( x  y) with x  5 and y   .
2
12
12
1   5  
 5  
   cos
 
cos

2   12 12 
 12 12 
Now simplify.
1   
  
cos   cos 

2 3
 2 
We can use our table to get the values of these trig functions.
1 1
 1
 0  .

2 2
 4
Sum-to-Product Formulas
x y x y
sin x  sin y  2 sin 
 cos

 2   2 
x y x y
sin x  sin y  2 sin 
 cos

 2   2 
x y x y
cos x  cos y  2 cos
 cos

 2   2 
x y x y
cos x  cos y  2 sin 
 sin 

 2   2 
EXAMPLE: Simplify: sin 5   sin 3  using a sum-to-product formula.
x y x y
We will use the formula sin x  sin y  2 sin 
 cos
 with x  5 and y  3 .
 2   2 
 5  3   5  3 
2 sin 
 cos

 2   2 
Now simplify.
 2   8 
2 sin   cos 
 2   2 
2 sin   cos4 
We can’t simplify this anymore, so we are done.
EXAMPLE: Simplify: cos3   cos2  using a sum-to-product formula.
Section 5.4 Notes Page 3
x y x y
We will use the formula cos x  cos y  2 cos
 cos
 with x  3 and y  2 .
 2   2 
 3  2   3  2 
2 cos
 cos

 2   2 
Now simplify.
 5    
2 cos  cos 
 2  2
We can’t simplify anymore, so we are done.
EXAMPLE: Simplify: cos4   cos7  using a sum-to-product formula.
x y x y
We will use the formula cos x  cos y  2 sin 
 cos
 with x  4 and y  7 .
 2   2 
 4  7   4  7 
 2 sin 
 sin

 2   2 
Now simplify.
 11    3 
 2 sin 
 sin 

 2   2 
  3 
 3 
We can use the identity sin 
   sin   .
 2 
 2 
 11 
 3 
 2 sin 
   sin  
 2 
 2 
 11   3 
2 sin 
 sin  
 2   2 
EXAMPLE: Find the exact value of sin 15   sin 75  using a sum-to-product formula.
x y x y


We will use the formula sin x  sin y  2 sin 
 cos
 with x  15 and y  75 .
 2   2 
 15   75 
2 sin 
2

  15   75 
 cos
2
 
2 sin 45  cos 30  
2
2 3
6
.


2 2
2



Simplify.
From here we can use our table to get the exact values.
Section 5.5 Notes Page 1
5.5 Trigonometric Equations
You should have received a unit circle sheet. If not, this is available on the website. This allows us to see the
exact values of certain angles between 0 and 360 degrees. Now we don’t need to use reference angles. This
section will cover how to solve trigonometric equations which is one skill you will need in calculus. The main
strategy is to isolate the trig function. The we will take the inverse trig function of both sides to get the answer.
EXAMPLE: Solve for x: cos x 
3
on [0, 360  ) .
2
The cosine is already isolated, so now we will take the inverse cosine of both sides.
cos 1 cos x   cos 1
3
.
2
3
2
x  cos 1
Now simplify.
What we need to do now is look at the unit circle sheet and find ANY angles
between 0  and 360  that give an x value of
3
. Remember x corresponds to
2
cosine and y corresponds to sine.
x  30  , 330 
Both of these will give a value of
3
.
2
EXAMPLE: Solve for x:  2 sin x  1 .
sin x  
1
2
First we isolated the sine. Now we need to take the inverse sine of both sides.
 1
sin 1 sin x   sin 1   
 2
Simplify.
 1
x  sin 1   
 2
We need to find ANY angles on the unit circle that give a y value of 
x
1
.
2
7 11
,
6
6
The above is not my complete answer. This problem did not give an interval like [0, 2 ] to find your answers.
Because of this, our answers will not only be in the first revolution of the circle. If no interval is given we need
to add a 2k to our answer. The k value represents how many times we are going around the circle until we
7
11
come to our answers. So we will write: x 
 2k ,
 2k . We could have also written our answers in
6
6
degrees as well: x  210   360  k , 330   360  k .
Section 5.5 Notes Page 2
EXAMPLE: Solve for x: 2 cos x  2  0 on [0, 2 ) .
cos x  
2
2
First we isolate the cosine. Now we need to take the inverse cosine of both sides.

2
 Simplify.
cos 1 cos x   cos 1  

2



2

x  cos 1  

2


x
We need to find ANY angles on the unit circle that give a x value of 
2
.
2
3 5
,
4
4
These are my only answer since they gave us an interval, [0, 2 ] . Since this interval is given in radians we
must write our answers in terms of radians.
EXAMPLE: Solve for x: cos 2 x 
3
on [0, 2 ).
4
To solve this one, take the square root of both sides: cos x  
3
. You can take the square root of the top and
4
3
. To solve this, look on the unit circle and find all angles that have an x value
2
 5 7 11
3
.
. The angles are: ,
,
of 
,
2
6
6
6 6
bottom separately: cos x  
EXAMPLE: Solve for x: 5 csc x  3  2 on [0, 2 ) .
First isolate the cosecant by adding 3 to both sides and then dividing both sides by 5.
1
. So now the problem becomes:
sin x
csc x  1
We will use the identity csc x 
1
1

sin x 1
Cross multiply.
sin x  1
Take the inverse sine of both sides.
sin 1 sin x   sin 1 1
Simplify.
x  sin 1 1
Look on the unit circle and find ANY angles that give a y value of 1.
x

2
This is the only single value on the unit circle.
Section 5.5 Notes Page 3
EXAMPLE: Solve for x:
3 cot   1  0 on [0, 360 )

First isolate the cotangent by subtracting one from both sides and then dividing both sides by
cot   
1
Now use the formula:
3
1
1

tan 
3
3.
1
 cot  .
tan 
Cross multiply.
tan    3
Since there is no tangent on our unit circle, look for ANY angle such that if you
divide the y by x, (y/x) you will get  3 .
  120  , 300 
EXAMPLE: Solve for x: cos 2 
1
.
2
The cosine is isolated, so now we will take the inverse cosine of both sides:
cos 1 cos 2   cos 1
2  cos 1
1
2
1
2
Simplify.
We need to find ANY angles on the unit circle that give a x value of
1
. Since
2
there is no interval given, we need to add a 360k to our answers.
2  60   360  k
2  300   360  k
For each of our answers we need to solve for  by dividing by 2.
  30   180  k
  150   180  k
These are our answers.
EXAMPLE: Solve for x: sin(2 )  
3
on [0, 2 ) .
2
We will proceed the same way we did the previous example. In this one we need to take the inverse sine of
both sides:

3

sin 1 sin 2   sin 1  

2



3
3

We
need
to
find
ANY
angles
on
the
unit
circle
that
give
a
x
value
of
2  sin 1  

. This

2
2


4
5
and
. On this one, because there is something inside the trig
will be
3
3
Section 5.5 Notes Page 4
function that is not just theta, we will also be using the k values. Because we
need to use radians, we will add 2k .
2 
4
5
 2k , 2 
 2k
3
3
For each of our answers we need to solve for  by dividing both sides by 2.
2
5
 k ,  
 k
These are our equations. To get the answers in our interval, we will be
3
6
putting values in for k. First we will start k = 0, 1, 2, …until we get a number that is outside our interval.


2
 k
3
2
2
  (0) , so  
3
3
2
5
When k = 1, we have:  
  (1) , so  
3
3
If we let k = 2, then we get something that is more than 2 , which is outside our
interval, so we will stop.

5
 k
6
5
5
  (0) , so  
6
6
5
11
When k = 1, we have:  
  (1) , so  
6
6
If we let k = 2, then we get something that is more than 2 , which is outside our
interval, so we will stop.
When k = 0, we have:  
When k = 0, we have:  
So our answers to this problem are:  
2 5 5 11
.
,
,
,
3
3
6
6
EXAMPLE: Solve for x: 2 cos(3 )  2  0 on [0, 360  ) .
2
. We will proceed the same way we did in the
2
previous example. In this one we need to take the inverse cosine of both sides:
First we need to solve the above for cosine: cos(3 ) 
 2

cos 1 cos(3 )   cos 1 

 2 
 2
2

3  cos 1 
. This
We
need
to
find
ANY
angles
on
the
unit
circle
that
give
an
x
value
of

2
 2 
will be 45 and 315 degrees. Now we set up our two equations as before.
3  45   360  k , 3  315  360  k
For each of our answers we divide both sides by 3.
  15  120  k ,   105  120  k These are our equations. To get the answers in our interval, we will be
putting values in for k. First we start with k = 0, 1, 2, … until we get an angle that is outside our interval.
Section 5.5 Notes Page 5
  15  120 k
When k = 0, we have   15  120 (0) , so   15 .
When k = 1, we have   15   120  (1) , so   135 .
When k = 2, we have   15   120  (2) , so   255  .
If we let k = 3 then we get an angle that is more than 360 degrees, which is outside our
interval, so we will stop.
  105   120  k
When k = 0, we have   105   120  (0) , so   105 .
When k = 1, we have   105   120  (1) , so   135 .
When k = 2, we have   105   120  (2) , so   345 .
If we let k = 3 then we get an angle that is more than 360 degrees, which is outside our
interval, so we will stop.





So our answers to this problem are:   15  , 105 , 135 , 225 , 255  , 345
EXAMPLE: Solve the equation: cos 2 x  cos x  0 on [0, 2 ) .
For this one the only thing we can do is factor out the common factor, which is cos x : cos x(cos x  1)  0 .
Now we need to set each factor equal to zero. We will get cos x  0 and cos x  1  0 . We need to solve each
equation separately. For the equation cos x  0 we need to look at the unit circle where the x value is zero.

3
This will happen at
and at
. For the second equation cos x  1  0 this is the same as cos x  1 . Again
2
2
we look at the unit circle and find all places where the x value is 1. We will get 0. The angle of zero radians is
also the same as 2 radians, but we don’t write the 2 because that is not included on our interval.
EXAMPLE: Solve the equation: 2 cos 2 x  cos x  1  0 on [0, 2 ) .
We can factor this one: (2 cos x  1)(cos x  1)  0 . Now set each part equal to zero. We get 2 cos x  1  0 and
1
cos x  1  0 . Solving the first equation we will get cos x  . We look on the unit circle and look for any x
2

1
5
values that are . This will happen at
and
. Solving the second equation you will get cos x  1 . The
2
3
3

5
angle that gives an x value of negative one is  . Therefore our answers are ,  , and
.
3
3
EXAMPLE: Solve the equation: sin x cos 2 x  2 sin x on [0, 360  ) .
We need to set this equal to zero: sin x cos 2 x  2 sin x  0 . Now factor out a common factor of sin x :
sin x(cos 2 x  2)  0 . Now set each factor equal to zero. We have sin x  0 . Looking at our unit circle we see
that x is 0  and 180  . The other equation gives us cos 2 x  2 . Taking the square root we get cos x   2 .
This means that cos x  1.41 . Since this number is larger than one, this will not give us any solutions because
of the domain of the cosine. So our answers for x are 0  and 180  .
Section 5.5 Notes Page 6
EXAMPLE: Solve the equation: tan x  tan x  3 tan x  3  0 on [0, 2 ) .
3
2
This one can be factored by grouping. I will factor out a tan 2 x from the first two terms and then a negative 3
from the second two terms. This will give us: tan 2 x(tan x  1)  3(tan x  1)  0 . There is a common factor of
tan x  1 that I will factor out: (tan x  1)(tan 2 x  3)  0 . Now we set both factors individually equal to zero.

5
from the unit circle. For
For the first equation we have tan x  1  0 , in which tan x  1 , so x  and x 
4
4
the second equation we have tan 2 x  3  0 , in which tan x   3 . Since we get a positive or a negative this
means that we will get an answer in all four quadrants of the unit circle. We get the following:
 2 4 5
 2 4 5  5
,
,
,
,
,
, ,
. Our final answer if all 6 of the following angles: ,
.
3 3
3
3
3 3
3
3 4 4
EXAMPLE: Solve the equation: 2 sin 2   cos 2  0 on [0, 360  ) .
Since there are both sines and cosines we need to use an identity to get all the terms to have the same trig value.
Since I notice there is already a sine in the problem I want to use cos 2  1  2 sin 2  . So now the problem is:
1
2 sin 2   (1  2 sin 2  )  0 . Simplifying we get: 4 sin 2  1  0 . When we solve this we get sin 2    so
4
1
sin    . Then values off the unit circle are: 30  , 150  , 210  , 330  . Notice that since our interval was
2
given in degrees we can write our answers in degrees.
EXAMPLE: Solve the equation: sin 2  cos  on [0, 2 ) .
We need to use another identity on this one. This time we will use sin 2  2 sin  cos  . So now our problem
becomes: 2 sin  cos   cos  . Setting it equal to zero will give us: 2 sin  cos   cos   0 . We can factor
out a cosine to get: cos  (2 sin   1)  0 . Setting the first term equal to zero will give us cos   0 , so we
 3
. Setting the second term equal to zero we will get 2 cos   1  0 , so
know from the unit circle   ,
2 2
 5
 3  5
1
sin   . Then we know from the unit circle   ,
, ,
. So our final answer is:   ,
.
2
6 6
2 2 6 6
EXAMPLE: Solve the equation: 2 cot 2 x  csc 2 x  2  0 on [0, 360  ) .
Once again we need to use a trig identity so that all of these are in terms of the same trig value. We can either
use all cotangents or all cosecants. I will use the identity cot 2 x  csc 2 x  1 : 2(csc 2 x  1)  csc 2 x  2  0 .
4
Simplifying we get: 3 csc 2 x  4  0 . Solving for cosecant we get csc 2 x  . I will now use the identity
3
1
4
1
csc 2 x 
. This will change the problem into:
 . Cross multiplying gives us 4 sin 2 x  3 , so
2
2
sin x
sin x 3
3
3
sin 2 x  . Solving for sine we get sin x  
. Then we know from the unit circle:
4
2
x  60  , 120  , 240  , 300 
Section 6.1 Notes Page 1
6.1 The Law of Sines
The law of sines is used to solve for missing sides or angles of triangles when we have the following three
cases:
ASA – Angle Side Angle
Law of Sines
SAA – Side Angle Angle
SSA – Side Side Angle
A
sin A sin B sin C


a
b
c
b
c
B
a
C
Usually you will only use two parts of the above formula, but all three ratios are equal.
EXAMPLE: Solve the triangle:
B
60 
4
A
40 
C
The first thing we can find is the measurement of angle C by subtracting the given angles from 180 degrees:
mC  180  40  60  80  .
Next we can find side AC, which I will label as b. We will use the law of sines for this. You always need to
start with a known side and a side opposite the known side. I will use the 40 degrees and the 4. This will be
one side of the equation. The other side is for the side you want to find. Since we want to find side AC we
sin 40  sin 60 
need to use the angle opposite of this side, so we will use the 60 degrees. The equation is:

.
4
b
4 sin 60 


Cross multiplying will give us b sin 40  4 sin 60 . Solving for b we get: b 
 5.39 .
sin 40 
Now we want to find side AB, which I will call c. Once again we will start with a known angle and side
sin 40  sin 80 
opposite this angle. The equation is:

. Cross multiplying will give us c sin 40   4 sin 80  .
c
4

4 sin 80
Solving for c we get c 
 6.13 . Now the triangle is solved.
sin 40 
Section 6.1 Notes Page 2
EXAMPLE: Solve the triangle:
B
15 
5
35 
A
C
The first thing we can find is the measurement of angle C by subtracting the given angles from 180 degrees:
mC  180  35  15  130  .
Next we can find side AC, which I will label as b. We will use the law of sines for this. I will use the 130
sin 130  sin 15 

. Cross multiplying will
degrees and the 5 as my known angle and side. The equation is:
5
b
5 sin 15 
give us b sin 130   5 sin 15  . Solving for b we get: b 
 1.69 .
sin 130 
Now we want to find side BC, which I will call a. Once again we will start with a known angle and side
sin 130  sin 35 

. Cross multiplying will give us a sin 130   5 sin 35  .
opposite this angle. The equation is:
5
a

5 sin 35
Solving for c we get a 
 3.74 . Now the triangle is solved.
sin 130 
EXAMPLE: Solve the triangle:
C
2
A 40 
3
B
We first want to find one of the missing angles. The only one I can solve for is angle B since I have a side
sin 40  sin B
opposite that is given:

. Now cross multiply: 2 sin 40   3 sin B . Solving for sin B we get:
3
2

2 sin 40
sin B 
. So sin B  0.4285 . Now we will take the inverse sine of both sides to get mB  25.37  .
3
Now we can find angle C: mC  180  40  25.37  114.63  .
sin 40  sin 114.63
Now that we know angle C we can find side AB. I will call this c:

. Cross multiplying
c
3
3 sin 114.63
 4.24 .
gives us c sin 40   3 sin 114.63  . Solving for c we get c 
sin 40 
Section 6.1 Notes Page 3
EXAMPLE: Solve the triangle:
B
2
C
1
50 
A
I want to solve for angle A first since there is a side opposite this angle given. We can use the equation:
sin 50  sin A

. Cross multiplying will give us: 2 sin 50   sin A . So sin A  1.53 . If we try and take the
1
2
inverse of this in our calculator we will get an error. This is because the domain of the inverse sine function
must be between -1 and 1. There is no solution for this problem. That means it is impossible to draw this
triangle. The drawing above is not to scale. If we did draw it to scale it would look something like this:
1
2
Notice that the side with a 1 is not long enough, so we can’t
complete the triangle. Again the answer is no solution.
50 
EXAMPLE: Solve the triangle:
B
2
C
1
25 
A
I showed on the board that the triangle above could be drawn two different ways. The first way is the above
drawing. If we take side AB and swing it to the left, we will get a second triangle:
B
2
C
25 
1
A
sin 25  sin A

.
1
2
Cross multiplying gives us: 2 sin 25   sin A . This will give us sin A  0.8452 . The inverse sine will give us:
A  57.7  . Now the calculator just gives us this one angle, which would correspond to our first drawing.
Remember that sine is positive in the first and second quadrant, so if 57.7  is a reference angle, then we find
another angle in the second quadrant by subtracting this from 180 degrees: mA  180   57.7   122.3 . So
we are going to get two solutions. We will find two different angle B’s and two different side AC’s .
You will actually get two solutions algebraically. Let’s try and solve for angle A again:
Go to next page to see both solutions worked out.
Section 6.1 Notes Page 4
Triangle 1:
If mA  57.7  then we can find angle B: mB  180  25  57.7  97.3 
sin 25  sin 97.3

. Cross multiplying will give us:
1
b
sin 97.3


 2.35 .
b sin 25  sin 97.3 . Solving for b we get: b 
sin 25 
We can use this to find side AC, which I will call b:
Triangle 2:
If mA  122.3 then we can find angle B: mB  180  25  122.3  32.7 
sin 25  sin 32.7 

. Cross multiplying will give us:
b
1
sin 32.7 
 1.28 .
b sin 25   sin 32.7  . Solving for b we get: b 
sin 25 
We can use this to find side AC, which I will call b:
EXAMPLE: Solve the triangle:
B
40 
3
A
4
C
The only angle we can solve for here is angle C since we have a side opposite angle C that is given:
sin 40  sin C
3 sin 40 

. Cross multiplying will give us 3 sin 40   4 sin C . Then we have: sin C 
. This
4
3
4
will give us: sin C  0.4821 . When we find the inverse sine our calculator gives us mC  28.82  . Now we
need to see if there will be another solution. Once again sine is positive in the second quadrant, so we can find
a second solution by subtracting our answer from 180 degrees: mC  180  28.82  151 .18  . Notice that we
already have an angle in the triangle that is 40 degrees, so it is impossible to also have an angle of 151.18
degrees because then the sum of the angles would be more than 180 degrees. Therefore we can ignore this
second solution. We know for sure that mC  28.82  . Now we can find angle A by subtracting from 180
degrees: mA  180  28.82  40  111.18  . Finally we can find side BC, which I will label as a:
sin 40  sin 111.18 

. Cross multiplying gives us: a sin 40   4 sin 111.18  . Solving for a will give us:
a
4
4 sin 111.18 
a
 5.8 .
sin 40 
More on next page…
Section 6.1 Notes Page 5
EXAMPLE: Solve the triangle:
A
5
B
4
40 
C
We want to find angle C since we have a side opposite angle C that is given to us:
sin 40  sin C

. Cross
4
5
5 sin 40 
multiplying will give us: 5 sin 40  4 sin C . Then we have: sin C 
. So sin C  0.8035 . The
4
inverse sine gives us mC  53.46  . Our other solution for angle C is: mC  180   53.46   126.54  . If
we add this to the 40 degree angle already in the triangle we will get an angle less than 180 degrees, so this tells
us there will definitely by two solutions to this triangle.

The first solution is the drawing shown above. If we take side AC and swing it to the left, we will get a second
triangle:
A
5
B
40 
4
C
Now we will solve both triangle separately:
Triangle 1:
If mC  53.46  then we can find angle A: mA  180  40  53.46  86.54 
sin 40  sin 86.54 

. Cross multiplying will give us:
a
4
4 sin 86.54 
 6.21 .
a sin 40   4 sin 86.54  . Solving for b we get: a 
sin 40 
We can use this to find side BC, which I will label a:
Triangle 2:
If mC  126.54  then we can find angle A: mB  180  40  126.54  13.46 
sin 40  sin 13.46 

. Cross multiplying will give us:
a
4
4 sin 13.46 


 1.45 .
a sin 40  4 sin 13.46 . Solving for a we get: a 
sin 40 
We can use this to find side BC, which I will label a:
Section 6.2 Notes Page 1
6.2 The Law of Cosines
The law of cosines is used to solve for missing sides or angles of triangles when we have the following three
cases:
SAS – Side Angle Side
SSS – Side Side Side
The law of cosines can be written three different ways depending on what you are trying to solve for.
Law of Cosines
B
a  b  c  2bc cos A
b 2  a 2  c 2  2ac cos B
c 2  a 2  b 2  2ab cos C
2
2
2
c
a
A
C
b
EXAMPLE: Solve the triangle:
B
2
60 
C
3
A
The only side we can find right now is side c since this is opposite angle C which is given. We need to use this
version of the law of sines: c 2  a 2  b 2  2ab cos C . Here a = 2, b = 3, and C = 60  . So we have:
c 2  2 2  3 2  2(2)(3) cos 60  . Simplifying we get: c 2  4  9  12(0.5) . So c 2  7 and so c  7 .
In order to solve for angle A we can either use the law of cosines again or we can use the law of sines. I will
sin 60  sin A

. Cross multiplying will give us: 2 sin 60   7 sin A . So
use the law of sines:
2
7
sin A 
2 sin 60 
. We get sin A  0.6547 . After taking the inverse sine we get mA  40.89 . We also get a
7
second answer for A: mA  180  40.89  139.11 . This is not possible since there is already an angle of 60
degrees inside the triangle. Also if we already know three sides of the triangle there can only be one possible
solution.
Lastly we will find angle B: mB  180  60  40.89  79.1 .
Section 6.2 Notes Page 2
EXAMPLE: Solve the triangle:
B
91
2
3
A
C
The only side we can find right now is side b since this is opposite angle B which is given. We need to use this
version of the law of sines: b 2  a 2  c 2  2ac cos B . Here a = 3, b = 2, and C = 91 . So we have:
b 2  3 2  2 2  2(3)(2) cos 91 . Simplifying we get: b 2  9  4  12(0.0175) . So c 2  13.21 and so c  3.63 .
In order to solve for angle A we can either use the law of cosines again or we can use the law of sines. I will
sin 91 sin A

. Cross multiplying will give us: 3 sin 91  3.63 sin A . So
use the law of sines:
3.63
3

3 sin 91
. We get sin A  0.8263 . After taking the inverse sine we get mA  55.72 .
sin A 
3.63
Lastly we will find angle B: mB  180  91  55.72  33.28  .
EXAMPLE: Solve the triangle:
B
4
6
C
A
3
We can solve for whatever angle we want since all three sides are given. So I will just solve for angle A first.
We need to use a 2  b 2  c 2  2bc cos A . Here a = 4, b = 3, and c = 6. So we have 4 2  3 2  6 2  2(3)(6) cos A
which after simplification is: 16  9  36  36 cos A . Then we have: 16  45  36 cos A , and  29  36 cos A ,
so cos A  0.8055 . The inverse cosine will give us mA  36.34  . Will there be a second answer? The
answer is no because cosine is only positive in the first and fourth quadrant. The fourth quadrant has angles
between 270 and 360 degrees, which are impossible in a triangle. Once again since all three sides are given we
will only have one solution.
Let’s now solve for angle C. We can either use the law of cosines again or we can use the law of sines. I will
sin 36.34  sin C
. Cross multiplying will give us: 6 sin 36.34   4 sin C . So

use the law of sines:
4
6

6 sin 36.34
. We get sin C  0.8889 . After taking the inverse sine we get mC  62.73 . What is
sin C 
4
wrong with this? If we look at our picture it shows that angle C should be more than 90 degrees. We know
there should only be one solution to this triangle so we know that the angle we found is not our solution, so
mC  180  62.73  117.27  .
Section 6.2 Notes Page 3
In order to avoid this and get the direct answer it is best to use the law of cosines a second time whenever you
have a triangle where three sides are given. You will get the answer right away. Let’s use the law of cosines to
get angle C. We will use: 6 2  4 2  3 2  2(4)(3) cos C . This simplifies to: 36  16  9  24 cos C . Then we
have cos C  0.4583 . After taking the inverse cosine we get mC  117.28  .
Finally we can find angle B: mB  180  36.34  117.28  26.38 .
EXAMPLE: Solve the triangle:
B
2
3
A
C
3
Since side BC and side AC are the same this means that angle A and angle B are equal. So we will find A first
and then we know this will automatically be angle B. We will use a 2  b 2  c 2  2bc cos A . In this problem we
know a = 3, b = 3, and c = 2. Putting this into the formula we have: 3 2  3 2  2 2  2(3)(2) cos A . Then we
have: 9  9  4  12 cos A . Solving for cosine will give us cos A  0.3333 . The inverse cosine will give us
mA  70.53 . So because we have an isosceles triangle we know that mB  70.53 . Now we can find angle
C: mC  180  70.53  70.53  38.94  .
EXAMPLE: A plane flies due north from Ft. Myers to Sarasota, a distance of 150 miles. Then the plane flies
at a bearing N 50  E and flies to Orlando, a distance of 100 miles.
a.) How far is it from Ft. Myers to Orlando?
C
100 mi
First we need to draw a picture. I will let point A = Ft. Myers
B = Sarasota, and C = Orlando. Notice that the 50 degrees is
measured from the north. Then we know that the angle B inside
the triangle is 180   50   130  .
50

B
130 
150 mi
A
We need to find the distance from A to C, and we will call this b. We need to use this version of the Law of
Cosines: b 2  a 2  c 2  2ac cos B . We know a = 100, c = 150 and B = 50 degrees. Putting this into the
formula we will have: b 2  100 2  150 2  2(100)(150) cos130 . Then we have
b 2  10000  22500  30000(0.6427) . This give us b 2  51783.63 . After taking the square root of both sides
we get b  227.56 . So the distance from Ft. Myers to Orlando is approximately 227.56 miles.
Section 6.2 Notes Page 4
b.) What bearing should the pilot use to fly directly from Ft. Myers to Orlando?
We need to find angle A since this is measured from the north. We can use the law of cosines again but this
time we will use a 2  b 2  c 2  2bc cos A . Now we know that a = 100, b = 227.56, and c = 150. Putting this
into the formula will give us 100 2  227.56 2  150 2  2(227.56)(150) cos A . Simplifying will give us:
10000  51783.5536  22500  68268 cos A . Combining like terms gives us:
10000  74283.5536  68268 cos A . Then we have  64283.5536  68268 cos A . Solving for cosine we will
get: cos A  0.9416 . Then mA  19.67  .
We need to write our answer as a bearing, so N19.67  E is the bearing the pilot must use to fly from Ft. Myers
to Orlando.
Section 6.3 Notes Page 1
6.3 Polar Coordinates


In this section we will learn a new coordinate system. In this system we plot a point in the form r ,  . As
shown in the picture below you first draw angle  in standard form. Then you label how long r is:
 3 
EXAMPLE: Plot  4,
 in the polar coordinate system.
4 

180
3
.
into degrees by multiplying by
4

You will get 135 degrees. This is in the second quadrant.
First draw the angle and then mark off 4 units to represent
the radius.
We can convert
 5 
EXAMPLE: Plot  5,
 in the polar coordinate system.
3 

180
5
into degrees by multiplying by
.
3

You will get 300 degrees. This is in the fourth quadrant.
First draw the angle and then mark off 5 units to represent
the radius.
We can convert


EXAMPLE: Plot  3, 120  in the polar coordinate system.
We already have it in degrees. This is in the second quadrant.
First draw the angle in standard position. Since we have a negative
radius, we must plot this differently. Instead of marking along our
original angle, we will draw another angle exactly 180 degrees
away from our original angle. Then we mark off 3 units.
Section 6.3 Notes Page 2


EXAMPLE: Plot   3,   in the polar coordinate system.
2

We need to convert this into degrees by multiplying by
180

.
You will get -90 degrees. First draw the angle in standard position.
Remember that negative angles are drawn clockwise in standard
position. Since we have a negative radius instead of marking along
our original angle, we will draw another angle exactly 180 degrees
away from our original angle. Then we mark off 3 units.
3 

EXAMPLE: Plot   3,   in the polar coordinate system.
4 

We need to convert this into degrees by multiplying by
180

.
You will get -135 degrees. First draw the angle in standard position.
Remember that negative angles are drawn clockwise in standard
position. Since we have a negative radius instead of marking along
our original angle, we will draw another angle exactly 180 degrees
away from our original angle. Then we mark off 3 units.
Equivalent Angles
There are more than one way to arrive at the same angle. For example in the previous problem, -135 degrees is
the same as 360    135   225  . If we have 120 degrees then this is the same as 120    360   240  .
So for negative angles, just add 360 degrees. For positive angles add negative 360 degrees to find the
equivalent angle. So basically we can either move clockwise or counterclockwise to arrive at the same angle.




r ,    r ,   2  or r ,    r ,   360 
r ,     r ,     or r ,     r ,   180 




EXAMPLE: Given the polar coordinate 5, 300  , find an equivalent polar coordinate that has the following
characteristics: a.)  360     0 , r > 0 b.) 0    360  , r < 0, c.) 360     720  , r > 0
a.) In this problem we are told to work in degrees. We want an angle that is negative that will lead us to the
same point. We are allowed to add or subtract a 360  and that won’t change our problem. So we can do
300   360   60  . This is an equivalent angle. So our equivalent point is 5, 60  . Our r is positive, so we
are done.


b.) Now we want r to be negative. The formula above r ,     r ,   180   tells us that we can add a 180
degrees to our angle and this will change the r to a –r. Now if we add 180 degrees then we will get an angle
more than 360 degrees, so we must subtract 180 degrees: 300   180   120  . Our equivalent point is:
 5, 120   .
Section 6.3 Notes Page 3
c.) We want r to be positive and we need an angle that is more than one revolution, so we just need to add 360 
to our angle: 300   360   660  . Our equivalent point is: 5, 660   .
 3 
EXAMPLE: Given the polar coordinate  4,
 , find an equivalent polar coordinate that has the following
4 

characteristics: a.)  2    0 , r > 0 b.) 0    2 , r < 0, c.) 2    4 , r > 0
a.) In this problem we are told to work in radians. We want an angle that is negative that will lead us to the
same point. We are allowed to add or subtract a 2 and that won’t change our problem. So we can do
3
5 
5

 2  
. This is an equivalent angle. So our equivalent point is  4,   . Our r is positive, so we
4
4
4 

are done.
b.) Now we want r to be negative. The formula above r ,     r ,     tells us that we can add a pi to our
angle and this will change the r to a –r. Now if we subtract a pi then we will get a negative angle, and our
3
7
 
. Adding the pi
question tells us we must have a positive angle, so we will add pi to our angle:
4
4
7 

will change our r to a negative r, so our equivalent point is:   4,
.
4 

c.) We want r to be positive and we need an angle that is more than one revolution, so we just need to add a 2
3
11
 11 
to our angle:
 2 
. Our equivalent point is:  4,
.
4
4
4 

EXAMPLE: Given the polar coordinate  2, 120   , find an equivalent polar coordinate that has the
following characteristics: a.)  360     0 , r > 0 b.) 0    360  , r < 0, c.) 360     720  , r > 0
a.) For this problem we are working all in degrees. We want an angle that is negative but an r that is positive.
Our formula says that we can either add or subtract 180  from our original angle. If we add 180  then we
won’t have a negative angle anymore so we need to subtract:  120   180   300  . Now we have the
equivalent point: 2, 300   .
b.) We already have a negative r. Now we need a positive angle. We don’t want to change r, so we need to add
360  to our original angle:  120  360   240  . So our equivalent point is:  2, 240   .
c.) We need r to be positive and our angle needs to be 360     720  . We need to add 180  to change the r
into a negative r.  120   180   60  . We are not done yet because this is not between 360 degrees and 720
degrees. We can add 360 degrees and this won’t change our r: 60   360   420  .
Section 6.3 Notes Page 4
Conversion formulas from polar to rectangular coordinates.
x  r cos 
y  r sin 
x2  y2  r 2
 
EXAMPLE: Convert  5,  into a rectangular point.
3

We can use the above formulas and plug in a 5 for r and a

3
for  . We will have: x  5 cos

3
. This equals:
 3 5 3

1 3

x  5   . Now we will find y: y  3 sin . This equals: y  5
. So our rectangular point is

3
2
2
2 2


5 5 3
 ,

2

2




EXAMPLE: Convert   3,   into a rectangular point.
4

 
for  . We will have: x  3 cos   . This
4
 4
 2

3 2
2 3 3
 



equals: x  3
.
Now
we
will
find
y:


y
.
This
equals:
y
3


3
sin



 2  2 .

2
2
4






 3 2 3 2
.
So our rectangular point is  
,

2
2


We can use the above formulas and plug in a -3 for r and a 

2 

EXAMPLE: Convert   2,
 into a rectangular point.
3 

2
 2 
for  . We will have: x  2 cos
 . This
3
 3 
 3
 2 
 1
   3 . So our
equals: x  2    1 . Now we will find y: y  2 sin 
 . This equals: y  2

 3 
 2
 2 
rectangular point is 1,  3 .
We can use the above formulas and plug in a -2 for r and a


EXAMPLE: Convert the equation r  1  sin  into a rectangular equation.
First let’s multiply the both sides of the equation by r. This will allow us to put in substitutions for the sine:
r 2  r  r sin  . Now we will replace the r 2 with x 2  y 2 and we can replace the r sin  with y. Then we
have: x 2  y 2  x 2  y 2  y . There is nothing more we can do with this.
Section 6.3 Notes Page 5
EXAMPLE: Convert the equation r  4 into a rectangular equation.
We can square both sides by r to get: r 2  16 . So x 2  y 2  16 , which is a circle.
EXAMPLE: Convert the equation r 
3
into a rectangular equation.
3  cos 
First we can cross multiply to get: 3r  r cos   3 . Now replace the r with
x 2  y 2 and replace the r cos 
with an x: 3 x 2  y 2  x  3 . Not much more to do here.
EXAMPLE: Convert the equation r  2 sin   4 cos  into a polar equation.
First let’s multiply the both sides of the equation by r. This will allow us to put in substitutions for the sine and
cosine: r 2  2r sin   4r cos  . Now we will replace the r 2 with x 2  y 2 and we can replace the r sin  with
y and the r cos  with x. Then we have: x 2  y 2  2 y  4 x . On the quiz and test this type of question will be
given as multiple choice. You would notice that our answer above would not appear as one of the choices.
There is more we can do with this one. First set it equal to zero and group all the x and y terms together:
x 2  4 x  y 2  2 y  0 . Now we can complete the square on both sides: x 2  4 x  4  y 2  2 y  1  0  4  1 .
Now we factor and our answer is: ( x  2) 2   y  1  5 . This is a circle.
2
Conversion formulas from rectangular to polar coordinates
x2  y2  r 2
If (x, y) is in the first or fourth quadrant, then   tan 1
y
.
x
If (x, y) is in the second or third quadrant, then   tan 1

y
 .
x

EXAMPLE: Convert  3, 3 into a polar coordinate. Express your angle in radians.
We can use the above formulas and plug in a -3 for x and a 3 for y. This will give us r: (3) 2  (3) 2  r 2 . This


will give us r  3 2 . If we plot  3, 3 we will end up in the second quadrant. So we will use
3
3

y
  tan 1   , so   tan 1
  . This equals:      
. So our polar coordinates are
x
4
4
3
3 

3 2,
.
4 

Section 6.3 Notes Page 6
EXAMPLE: Convert  2, 2 3 into a polar coordinate. Express your angle in radians.


We can use the above formulas and plug in a -2 for x and a  2 3 for y. This will give us r:

(2) 2   2 3

2


 r 2 . This will give us r 2  4  12 , so r = 4. If we plot  2, 2 3 we will end up in the
third quadrant. So we will use   tan 1
4 

our polar coordinates are  4,
.
3 

2 3
4

y
  , so   tan 1
  . This equals:     
. So
2
x
3
3
1
3
 into a polar coordinate. Express your angle in radians.
EXAMPLE: Convert  , 

2
2


We can use the above formulas and plug in a
1
3
for y. This will give us r:
for x and a 
2
2
2
2
1
3
3
1 3
 1  
 we will end up in the
  r 2 . This will give us r 2   , so r = 1. If we plot  , 


  

2
4 4
2
 2   2 


3

y
fourth quadrant. So we will use   tan 1 , so   tan 1 2 . This equals:   tan 1  3 . Therefore,
1
x
2




   . So our polar coordinates are 1,   .
3
3

EXAMPLE: Convert the equation x 2  y 2  x into a polar equation.
In this equation we will replace x 2  y 2 with r 2 and we will replace x with r cos  . We get: r 2  r cos  .
We need to set this equal to zero and solve for r. We will get: r 2  r cos   0 . Now factor out an r:
r (r  cos  )  0 . Solving for r we get: r  0 and r  cos  .
EXAMPLE: Convert the equation 4 x 2 y  1 into a polar equation.
In this equation we will replace x 2 with r 2 cos 2  and we will replace y with r sin  . We get:
4r 2 cos 2   r sin   1 . This equals 4r 3 cos 2   sin   1 . Can’t do much more with this.
EXAMPLE: Convert the equation y 2  2 x into a polar equation.
In this equation we will replace y 2 with r 2 sin 2  and we will replace x with r cos  . You will get:
r 2 sin 2   2r cos  . On the quiz and test this type of question will be given as multiple choice. You would
notice that our answer above would not appear as one of our choices. That means we need to simplify this
Section 6.3 Notes Page 7
further. First set this equal to zero: r sin   2r cos   0 . Now factor our an r: r (r sin 2   2 cos  )  0 .
Now set both factors equal to zero. You will get r = 0 and r sin 2   2 cos   0 . We need to solve the second
1
2 cos 
cos 
equation for r. You will get: r 
. Then our final answer
. This can be written as: r  2 

2
sin  sin 
sin 
is r  2 cot  csc  .
2
2
Section 6.6 Notes Page 1
6.6 Vectors
Vectors are needed in physics and engineering courses. A vector is a quantity that has magnitude (size) in a
certain direction. You indicate a vector by a ray. The length of the arrow represents the magnitude and the
arrowhead indicates the direction of the vector as shown below:
Algebraic Vector
This is expressed as v  a, b . The a is the horizontal component and b is the vertical component.
Position Vector
This is an algebraic vector with the starting point at the origin. Suppose that v is a vector with a starting point
of P1  ( x1 , y1 ) and ending point of P2  ( x 2 , y 2 ) . Then v  x 2  x1 , y 2  y1 .
EXAMPLE: Find the position vector of the vector v  P1 P2 if P1  (1, 2) and P2  (4, 6) .
We will use the formula v  x 2  x1 , y 2  y1 and plug in the given information: v  4  (1), 6  2 . Then
we will simplify: v  5, 4 .
Now I will show you what we just did. As in the picture we started with
a vector not at the origin and we found an equivalent vector that starts at
the origin. These vectors are the same length and are going in the same
direction except they start at a different place.
EXAMPLE: Find the position vector of the vector v  P1 P2 if P1  (1, 4) and P2  (6, 2) .
We will use the formula v  x 2  x1 , y 2  y1 and plug in the given information: v  6  (1), 2  4 . Then
we will simplify: v  7, 6 .
Section 6.6 Notes Page 2
Alternate form of writing v  a, b
We can also write this as v = ai + bj
How to find a vector’s magnitude v
To find the magnitude, use the formula v  a 2  b 2
EXAMPLE: Given v = –3i + 4j, find v .
Here a is -3 and b is 4. We will use the formula v  a 2  b 2 and put in our given information:
v  (3) 2  (4) 2 . Now square the terms: v  9  16 . We will get v  25 , so v  5 .
EXAMPLE: Given v = 6i + 3j, find v .
Here a is 6 and b is 3. We will use the formula v  a 2  b 2 and put in our given information:
v  (6) 2  (3) 2 . Now square the terms: v  36  9 . We will get v  45 , so v  3 5 .
EXAMPLE: Given v = i + 2j and w = –3i + 4j, find the following:
a.) -2v + 3w
We can multiply both components in v by -2 and also multiply both components in w by 3:
-2(i + 2j ) + 3(–3i + 4j). Now multiply and we get: -2i + 4j – 9i + 12j. Now add the like terms to get:
–11i + 8j.
b.) v – 4w
We will take v and add this top -4 times w: i + 2j – 4(–3i + 4j) . Now multiply: i + 2j + 12i – 16j). Now add
the like terms to get: 13i – 14j.
c.) v  w
Inside we need to add v and w: || i + 2j + –3i + 4j ||. Now add the like terms: ||–2i + 6j ||. Now we need to find
the magnitude using the formula v  a 2  b 2 and put in our given information: v  (2) 2  (6) 2 . Now
square the terms: v  4  36 . We will get v  40 , so v  2 10 .
Section 6.6 Notes Page 3
Unit Vector – a vector with a magnitude of 1.
v
v
This finds a vector u that is going in the same direction as a given vector v but has a magnitude of 1.
Finding a unit vector u in the same direction as v:
u
EXAMPLE: Find a unit vector u that has the same direction as the vector v = 12i + 5j.
We need to first find v . Here a is 12 and b is 5. We will use the formula v  a 2  b 2 and put in our given
information: v  (12) 2  (5) 2 . Now square the terms: v  144  25 . We will get v  169 , so
12iˆ  5 ˆj
v
and put in or information: u 
. Then we can split this
v  13 . Now we will use the formula u 
13
v
up to get: u 
12 ˆ 5 ˆ
i
j . (Some books use this notation instead of the bold letters).
13 13
EXAMPLE: Find a unit vector u that has the same direction as the vector v = –2i + 3j.
We need to first find v , which is v  (2) 2  (3) 2 . We will get v  13 . Now we will use the formula
u
 2iˆ  3 ˆj
v
and put in or information: u 
. After rationalizing we get:
v
13
Vector components
A vector is made up of a horizontal and vertical part. These parts are called
components. The picture to the right shows how to break up a vector into
components if an angle is given:
EXAMPLE: Given v  8 and   30  , write the vector v in terms of i and j.
Following the formula we have v  8 cos 30  i + 8 sin 30  j. Now we simplify:
1
3
i + 8  j. So our final answer is v  4 3 i + 4j.
v  8
2
2
Resultant vector: the result when two vectors are added together.
Direction of Resultant: If the resultant vector is in the form F  aiˆ  bˆj ,
then you can find the angle between this vector and the x axis by using the
b
formula:   tan 1   if the resultant vector is quadrant 1 or 4. Also
a
b
  tan 1    180  if the resultant vector is in quadrant 2 or 3.
a
u
2 13 ˆ 3 13 ˆ
i
j.
13
13
Section 6.6 Notes Page 4
EXAMPLE: Two forces of magnitude 30 newtons (N) and 70 newtons act on an object at angle of 45 degrees
and 120 degrees with the positive x-axis as shown in the figure. Find the direction and magnitude of the
resultant force, that is, find F1  F2 .
We need to break this up into components.
F1  30 cos 45  iˆ  30 sin 45  ˆj = 21.21iˆ  21.21 ˆj
F2  70 cos120  iˆ  70 sin 120  ˆj =  35iˆ  60.62 ˆj
Now let’s add these together. First add the iˆ components and then the ĵ components. We will get:
F1  F2  13.79iˆ  81.83 ˆj . To find the magnitude we use the formula: F1  F2  (13.79) 2  (81.83) 2 .
 81.83 


You will get: F1  F2  82.98 . To find the direction we will use:   tan 1 
  180  99.57 . We
  13.79 
need to add 180 degrees since our resultant F1  F2  13.79iˆ  81.81 ĵ ends up in the second quadrant.
EXAMPLE: The magnitude and direction exerted by two tugboats towing a ship are 4200 pounds at N 65  E
and 3000 pounds at S 58 E respectively. Find the magnitude and the bearing of the resultant force.
F1 = 4200 lbs
65
We need to break this up into components.

58 
F2 = 3000 lbs
When you do components the angle must be measured from the x-axis. For the first force we will subtract 65
from 90 to get 25 degrees. For the second force we can subtract 58 from 90 to get 32 degrees. However, since
this is below the x-axis, this angle is actually negative, so we want to use -32 degrees (or 328  ).
F1  4200 cos 25  iˆ  4200 sin 25  ˆj = 3806.49iˆ  1774.99 ˆj
F2  3000 cos  32  iˆ  3000 sin  32  ˆj = 2544.14iˆ  1589.76 ˆj




Now let’s add these together. First add the iˆ components and then the ĵ components. We will get:
F1  F2  6350.63iˆ  185.23 ˆj . To find the magnitude we use the formula:
F1  F2  (6350.63) 2  (185.23) 2 . You will get: F1  F2  6353.33 pounds. To find the direction we will
 185.23 

use:   tan 1 
  1.67 . This is measured from the x-axis, so we need to subtract this from 90
 6360.63 
degrees to get the bearing since the bearing is always measured from the vertical axis: 90  1.67  88.3 . Our
bearing is N 88.3 E .
Section 6.6 Notes Page 5
EXAMPLE: A 9.73 pound picture is hung from 2 wires as shown below. The tension on A is 3.4 pounds at
161 degrees. The tension on B is 9.2 pounds at 69.55 degrees. The tension on C is 9.73 pounds at -90 degrees.
Find the magnitude and direction of the resultant force.
A
B
C
We will write the vectors A, B, and C in component form using the given information:
A: 3.4 cos161 iˆ  3.4 sin 161 ˆj =  3.21iˆ  1.11 ˆj
B: 9.2 cos 69.55  iˆ  9.2 sin 69.55  ˆj = 3.21iˆ  8.62 ˆj
C: 9.73 cos 90  iˆ  9.73 sin  90   ˆj = 0iˆ  9.73 ˆj
If we add A, B, and C we get 0iˆ  0 ˆj . This means the magnitude and direction are all 0. Whenever this
happens we have what is called static equilibrium. This is something you will cover more in physics and also
engineering statics courses.
Section 6.7 Notes Page 1
6.7 The Dot Product
The dot product is one way to multiply vectors. You will get a number for an answer. In physics you will also
learn the cross product in which a vector is the answer.
The Dot Product
If u  a1 i + b1 j and v  a 2 i + b2 j then the dot product u  v  a1  a 2  b1  b2 .
EXAMPLE: Given u = –3i + 4j and v = 6i + 5j, find the following:
a.) u  v
We will use the dot product formula. Here, a1  3 , a 2  6 , b1  4 , b2  5 . Plug these into the formula:
u  v  (3)(6)  (4)(5) . Simplifying we get u  v  2 .
b.) v  u
We will use the dot product formula. Here, a1  4 , a 2  5 , b1  3 , b2  6 . Plug these into the formula:
v  u  (4)(5)  (3)(6) . Simplifying we get v  u  2 . Notice that it doesn’t matter what order we multiply
these vectors when using the dot product. We get the same answer.
c.) u  u
We will use the dot product formula. Here, a1  3 , a 2  3 , b1  4 , b2  4 . Plug these into the formula:
u  u  (3)(3)  (4)(4) . Simplifying we get u  u  25 .
d.) v  v
We will use the dot product formula. Here, a1  6 , a 2  6 , b1  5 , b2  5 . Plug these into the formula:
v  v  (6)(6)  (5)(5) . Simplifying we get v  v  61 .
e.) u
We will use the formula u  a 2  b 2 : u  (3) 2  (4) 2 = 5.
f.) v
We will use the formula v  a 2  b 2 : v  (6) 2  (5) 2 =
61 .
Notice that if you multiply and a vector by itself using the dot product this is equal to magnitude squared. This
2
2
is a property: u  u  u . Also, v  v  v . Another property is u  v  v  u .
Section 6.7 Notes Page 2
Angle between two vectors
Given two vectors u and v, the angle between the two vectors is given as:
cos  
u v
u  v
EXAMPLE: u = –3i + 4j and v = 6i + 5j, find the angle between u and v.
We already know from the previous problem that u  v  2 . We also know u  5 and v  61 . We can put
these into our formula to find the angle: cos  
2
.
5  61
Putting this into a calculator will give: cos   0.0512 , so
after taking the inverse cosine we will get our answer:   87.06  .
In the picture to the right we drew the vectors and you can see the
angle between them looks like it’s about 87 degrees.

If the angle between the vectors is 0 or 180 degrees, then the vectors are parallel.
If the angle between the vectors is 90 degrees then the vectors are orthogonal.
EXAMPLE: Given u = –3i + 2j and v = 4i + 6j, find the following:
a.) u  v
We will use the dot product formula. Here, a1  3 , a 2  4 , b1  2 , b2  6 . Plug these into the formula:
u  v  (3)(4)  (2)(6) . Simplifying we get u  v  0 .
b.) u
We will use the formula u  a 2  b 2 : u  (3) 2  (2) 2 = 13 .
c.) v
We will use the formula v  a 2  b 2 : v  (4) 2  (6) 2 =
52 .
d.) The angle between u and v.
We will use the formula cos  
0
u v
and plug in our known values: cos  
. So cos   0 . After
u  v
13 52
taking the inverse cosine we get   90  , so this means that  is orthogonal.
Section 6.7 Notes Page 3
EXAMPLE: Is the angle between u = 2i + 5j and v = 3i – 7j orthogonal, parallel, or neither?
We need to use the formula cos  
u v
. First we will find u  v  2(3)  5(7)  29 . Then we can find u
u  v
by using the formula: u  (2) 2  (5) 2  29 . Then we can find v by using the formula:
v  (3) 2  (7) 2  58 . Now we can put these into the formula to find our angle: cos  
 29
. We
29 58
can put this into our calculator to get: cos   0.7071 . So   135 . This is not orthogonal or parallel, so we
will answer neither.
EXAMPLE: Let u = 2i – j, v = 3i + j, and w = i + 4j. Find the following scalar: v  u  w .
We will start by substituting into the formula: (3i + j)(2i – j + i + 4j). First we simplify inside the parenthesis:
(3i + j) (3i + 3j). Now we use the dot product to multiply these: 3(3) + (1)(3) = 9 + 3 = 12.
EXAMPLE: Let u = 2i – j, v = 3i + j, and w = i + 4j. Find the following scalar: 5v   w .
We will start by substituting into the formula: 5(3i + j)( i + 4j). First we distribute the 5:
(15i +5j)( i + 4j). Now we use the dot product to multiply these: 15(1) + (5)(4) = 15 + 20 = 35.
Section 9.1 Notes Page 1
9.1 The Ellipse
In this section we will be looking at the ellipse which is basically an elongated circle. The pictures below show
the two different ways an ellipse can be drawn that are both centered at the origin. The formulas below contain
a and b. The a is the length of the major axis, or the longest axis. The b is the length of the shortest axis. It is
very important to note that a is ALWAYS larger than b. If the larger number is under the x then the ellipse is
drawn horizontally. If the larger number is under the y then the graph is drawn vertically.
Ellipses centered at (0, 0).
(a)
x2 y2

1
a2 b2
(b)
x2 y2

1
b2 a2
In both of these cases the length of the major axis is 2a. The length of the minor axis is 2b. To find the c value
in any of these graphs, use the equation c  a 2  b 2 .
Eccentricity: a measure of how much the ellipse resembles a circle. The formula is e 
c
. If e = 0 then the
a
ellipse is a circle. The larger the ellipse the skinnier it becomes.
x2 y2

 1 and identify the foci, eccentricity, center, length of the major axis, length of
25 9
the minor axis, and the two vertices on the major axis.
EXAMPLE: Graph
The larger number is under the x, so we know this ellipse will
be drawn horizontally. From our equation we know that a is 5
and b is 3. We can findc by the formula c  a 2  b 2 :
c  5 2  3 2 . So c = 4. We start at the center, which is (0, 0).
Since this is drawn horizontally I will add 5 and subtract 5 from
the x value of the center since a is 5. This will give us the vertices
(5, 0) . Then I will go up and down 3 since this is our b. The foci
run along the same line as the major axis, so from my center I will add
4 and subtract 4 from the x value of the center to find the foci. We
find that the foci are at (4, 0) . We know the length of the major axis
is 2(5) = 10. The length of the minor axis is 2(3) = 6. Finally we can
4
find the eccentricity, which is e   0.8 .
5
Section 9.1 Notes Page 2
EXAMPLE: Graph 36 x  4 y  144 and identify the foci, eccentricity, center, length of the major axis,
length of the minor axis, and the two vertices on the major axis.
2
2
We need to first put this in the proper form. We need a 1 on the
right hand side so we can divide the whole equation by 144 to get:
36 x 2 4 y 2
x2 y2

 1 . After reducing we get:

 1.
144 144
4 36
The larger number is under the y, so we know this ellipse will
be drawn vertically. From our equation we know that a is 6
and b is 2. We can find c by the formula c  a 2  b 2 :
c  6 2  2 2 . So c  32  4 2  5.66 . We start at the center,
which is (0, 0). Since this is drawn vertically I will add 6 and subtract
6 from the y value of the center since a is 6. This will give us the
vertices (0, 6) . Then I will go up and down 2 since this is our b.
The foci run along the same line as the major axis, so from my center
I will add 5.66 and subtract 5.66 from the y value of the center to find
the foci. We find that the foci are at (0, 5.66) .We know the length of
the major axis is 2(6) = 12. The length of the minor axis is 2(2) = 4.
5.66
Finally we can find the eccentricity, which is e 
 0.94 .
6
Now we will look at ellipses that are centered at (h, k). Once again if the larger number is under the x then it
will be drawn horizontally. If the larger number is under the y then it will go vertically. The pictures below
include the formulas to find the foci and the vertices.
Ellipses centered at (h, k).
(a)
( x  h) 2 ( y  k ) 2

1
a2
b2
(b)
( x  h) 2 ( y  k ) 2

1
b2
a2
Once again in both of these cases the length of the major axis is 2a. The length of the minor axis is 2b. To find
the c value in any of these graphs, use the equation c  a 2  b 2 .
The eccentricity is still e 
c
.
a
Section 9.1 Notes Page 3
( x  2)
( y  1)

 1 and identify the foci, eccentricity, center, length of the major axis,
16
4
length of the minor axis, and the two vertices on the major axis.
2
2
EXAMPLE: Graph
The larger number is under the x, so we know this ellipse will
be drawn horizontally. From our equation we know that a is 4
and b is 2. We can find c by the formula c  a 2  b 2 :
c  4 2  2 2 . So c  12  2 3  3.46 . We start at the center,
which is (2, 1). Remember to take the opposite sign of what is given.
Since this is drawn horizontally I will add 4 and subtract 4 from
the x value of the center since a is 4. This will give us the vertices
(-3, 1) and (6, 1). Then I will go up and down 3 from the y value of
our vertex since this is our b. The foci run along the same line as the
major axis, so from my center I will add 3.46 and subtract 3.46 from
the x value of the center to find the foci. We can write the foci as
(2  2 3 , 0) . We know the length of the major axis is 2(4) = 8. The
length of the minor axis is 2(2) = 4. Finally we can find the
3.46
eccentricity, which is e 
 0.87 .
4
EXAMPLE: Graph 16 x 2  9 y 2  64 x  54 y  1  0 and identify the foci, eccentricity, center, length of the
major axis, length of the minor axis, and the two vertices on the major axis.
First we need to get this into the correct form. The first thing to do is subtract the 1 from both sides and group
the x and y terms together: 16 x 2  64 x  9 y 2  54 y  1 . Now I will factor the two x terms and the two y
terms separately: 16( x 2  4 x)  9( y 2  6 y )  1 . Now it is time to complete the square. In the first term we
will divide the 4 by 2 and square it. You will get 4. We will add 4 to the left hand side however we do NOT
add 4 to the right side. The 4 is really being multiplied by the 16, so we will add 4(16) = 64 to both sides. In
the second term we will divide the 6 by 2 and square it. You will get 9. We will add 9 to the left hand side
however we do NOT add 9 to the right side. The 9 is really being multiplied by the 9, so we will add 9(9) = 81
to both sides: 16( x 2  4 x  4)  9( y 2  6 y  9)  1  64  81 . We can simplify the right side and factor the left
side: 16( x  2) 2  9( y  3) 2  144 . Now divide both sides by 144 and reduce to get:
The larger number is under the y, so we know this ellipse will
be drawn vertically. From our equation we know that a is 4
and b is 3. We can find c by the formula c  a 2  b 2 :
c  4 2  3 2 . So c  7  2.65 . We start at the center, which is
(-2, 3). Since this is drawn vertically I will add 4 and subtract 4
from the y value of the center since a is 4. This will gives us the
vertices (-2, 7) and (-2, -1). Then I will go left and right 3 since this
is our b. The foci run along the same line as the major axis, so from
my center I will add 2.65 and subtract 2.65 from the y value of the
center to find the foci. We find that the foci are at (2, 3  7 ) .We
know the length of the major axis is 2(4) = 8. The length of the
2.65
minor axis is 2(3) = 6. The eccentricity is e 
 0.66 .
4
( x  2) 2 ( y  3) 2

 1.
9
16
Section 9.1 Notes Page 4
EXAMPLE: Graph x  16 y  160 y  384  0 and identify the foci, eccentricity, center, length of the major
axis, length of the minor axis, and the two vertices on the major axis.
2
2
First we need to get this into the correct form. The first thing to do is subtract the 384 from both sides and
group the x and y terms together: x 2  16 y 2  160 y  384 . Now I will factor the two x terms and the two y
terms separately: x 2  16( y 2  10 y )  384 . Now it is time to complete the square. We can’t do anything with
the first term. In the second term we will divide the 10 by 2 and square it. You will get 25. We will add 25 to
the left hand side however we do NOT add 25 to the right side. The 25 is really being multiplied by the 16, so
we will add 16(25) = 400 to both sides: x 2  16( y 2  10 y  25)  384  400 . We can simplify the right side
and factor the left side. I can rewrite the x as the quantity x – 0: ( x  0) 2  16( y  5) 2  16 . Now divide both
( x  0) 2 ( y  5) 2

 1.
16
1
The larger number is under the x, so we know this ellipse will
be drawn horizontally. From our equation we know that a is 4
sides by 16 and reduce to get:
and b is 1. We can find c by the formula c  a 2  b 2 :
c  4 2  12 . So c  15  3.87 . We start at the center, which is
(0, 5). Since this is drawn horizontally I will add 4 and subtract 4
from the x value of the center since a is 4. This will gives us the
vertices (-4, 5) and (4, 5). Then I will add and subtract 1 from the
y coordinate of our vertex since this is our b. The foci run along the
same line as the major axis, so I will add 3.87 and subtract 3.87 from
the x value of the center to find the foci. We find that the foci are at
( 15 , 5) .We know the length of the major axis is 2(4) = 8. The
length of the minor axis is 2(1) = 2. The eccentricity is
3.87
e
 0.97 .
4
EXAMPLE: Find the equation of an ellipse with foci (0, 1) and vertices (0, 4) .
First let’s plot the above points. From this we can conclude that the
ellipse must be centered at the origin. We also know that c = 1
because this is how far the foci are from the center. We also know
a = 4 since that is how far the vertices are from the center. We can use
c  a 2  b 2 to find b 2 : 1  4 2  b 2 . After squaring both sides we
get: 1  16  b 2 . Solving for b 2 we get b 2  15 . Now we have
everything we need to find our equation. From our graph we know
x2 y2
this ellipse is drawn vertically, so it must use the formula 2  2  1 .
b
a
2
2
x
y

 1.
Plugging in our known variables will give us:
15 16
Section 9.1 Notes Page 5
3
EXAMPLE: Find the equation of an ellipse centered at the origin with an eccentricity of and with one
5
vertice at (-5, 0).
If the eccentricity is
3
then we automatically know that c is 3 and a is 5. We can use c  a 2  b 2 to find b:
5
3  5 2  b 2 . Again we square both sides: 9  25  b 2 . Solving for b 2 we get b 2  16 . If the vertice is
(-5, 0) then this is 5 units to the left of the origin, which means the ellipse is drawn horizontally, so we can use
x2 y2
x2 y2


1
.
Plugging
in
our
unknowns
will
give
us:

 1 . We don’t need a graph for this one.
25 16
a2 b2
EXAMPLE: Find the equation of an ellipse with foci at (1, 2) and (-3, 2) and with one vertex at (-4, 2).
First let’s plot the above points. From this we can conclude that the
ellipse must be centered at (-1, 2) since this point is halfway between
our foci. We also know that c = 2 because this is how far the foci are
from the center. We also know a = 3 since that is how far the vertices
are from the center. We can use c  a 2  b 2 to find b 2 : 2  3 2  b 2 .
After squaring both sides we get: 4  9  b 2 . Solving for b 2 we get
b 2  5 . Now we have everything we need to find our equation. From
our graph we know this ellipse is drawn horizontally, so it must use the
( x  h) 2 y  k ) 2
formula

 1 . Plugging in our known variables will
b2
a2
( x  1) 2 ( y  2) 2

 1.
give us:
9
5
Section 9.2 Notes Page 1
9.2 The Hyperbola
In this section we will be looking at the hyperbola whose shape is shown below. The pictures below show the
two different ways a hyperbole can be drawn that are both centered at the origin. The equations contain a and b.
You will use this to create a box as shown. The boxes are used as an aid in graphing the asymptotes. This time
it does not matter if a or b is bigger. If the equation starts with x then it opens to the right and left then it is
drawn horizontally. If the equation starts with y then it opens up and down and up then it is drawn vertically.
The first number in the denominator of the equation is always a regardless if x or y comes first.
Hyperbolas centered at (0, 0).
(a)
x2 y2

1
a2 b2
Asymptotes: y  
b
x
a
(b)
y2 x2

1
a2 b2
Asymptotes: y  
a
x
b
In both of these cases the length of the transverse axis is 2a. The length of the conjugate axis is 2b. To find
the c value in any of these graphs, use the equation c  a 2  b 2 . This is used to find the foci. To find the
c
eccentricity use the formula is e  . The larger the e the wider the hyperbola is.
a
x2 y2

 1 and identify the foci, eccentricity, center, length of the transverse and
9
4
conjugate axis, vertices, and the equations of the asymptotes.
EXAMPLE: Graph
Since x comes first in the equation, we know this hyperbola will
be drawn horizontally. From our equation we know that a is 3
and b is 2. We can find c by the formula c  a 2  b 2 :
c  3 2  2 2 . So c  13  3.61 . We start at the center, at (0, 0).
Since this is drawn horizontally I will add 3 and subtract 3 from
the x value of the center since a is 3. This will give us the vertices
(3, 0) . Then I will go up and down 2 since this is our b. This
gives us the box. We can connect the corners of the box to create
the asymptotes. The foci run along the same line as the transverse
axis, so from my center I will add 3.61 and subtract 3.61 from the x
value of the center to find the foci. The foci are at ( 13 , 0) . We
know the length of the transverse axis is 2(5) = 10. The length of the conjugate axis is 2(3) = 6. Finally we can
3.61
2
find the eccentricity, which is e 
 1.2 . The equation of the asymptotes are: y   x .
3
3
Section 9.2 Notes Page 2
EXAMPLE: Graph 4 y  16 x  64 and identify the foci, eccentricity, center, length of the transverse and
conjugate axis, vertices, and the equations of the asymptotes.
2
2
We need to first put this in the proper form. We need a 1 on the
right hand side so we can divide the whole equation by 64 to get:
y2 x2
4 y 2 16 x 2

 1 . After reducing we get:

 1.
64
64
16 4
The y comes first, so we know this hyperbola will be drawn vertically.
The a is always first, so a = 4 and b = 2. We can find c by the
by the formula c  a 2  b 2 : c  4 2  2 2 .
So c  20  2 5  4.47 . We start at the center, which is (0, 0)
Since this is drawn vertically I will add 4 and subtract 4 from the y
value of the center since a is 4. This will give us the vertices (0, 4)
Then I will go to the left and right 2 since this is our b. This creates
the box in which I can connect the corners to graph the asymptotes.
The foci run along the same line as the major axis, so from my center
I will add 4.47 and subtract 4.47 from the y value of the center to find the foci. We find that the foci are at
(0, 2 5 ) .We know the length of the transverse axis is 2(4) = 8. The length of the conjugate axis is 2(2) = 4.
4.47
4
 1.12 . The equation of the asymptotes are: y   x .
Finally we can find the eccentricity, which is e 
4
2
This reduces to y  2 x .
Now we will look at hyperbolas that are centered at (h, k). Once again if the x comes first then it will be drawn
horizontally. If y comes first it will go vertically. The pictures below include the formulas to find the foci and
the vertices.
Hyperbolas centered at (h, k).
( x  h) 2 ( y  k ) 2

1
a2
b2
b
Asymptotes: y  k   ( x  h)
a
Vertices: (h  a, k ) , Foci: (h  c, k )
(a)
( y  k ) 2 ( x  h) 2

1
a2
b2
a
Asymptotes: y  k   ( x  h)
b
Vertices: (h, k  a) , Foci: (h, k  c)
(b)
Once again in both of these cases the length of the major axis is 2a. The length of the minor axis is 2b. To find
c
the c value in any of these graphs, use the equation c  a 2  b 2 . The eccentricity is still e  .
a
Section 9.2 Notes Page 3
( x  3)
( y  1)

 1 and identify the foci, eccentricity, center, length of the transverse
1
4
and conjugate axis, vertices, and the equations of the asymptotes.
2
2
EXAMPLE: Graph
The x comes first, so we know this hyperbole will be drawn
horizontally. Since a always comes first we know that a = 1 and b = 2.
We can find c by the formula c  a 2  b 2 : c  12  2 2
So c  5  2.24 . We start at the center, which is (3, -1). Remember
to take the opposite sign of what is given. Since this is drawn
horizontally I will add 1 and subtract 1 from the x value of the center
since a is 1. The a always goes in the direction the hyperbola is
opening up. This will give us the vertices (2, -1) and (4, -1). Then I
will go up and down 2 from the y value of our vertex since this is our
b. The foci run along the same line as the transverse axis, so from the
x value of the center I will add 2.24 and subtract 2.24 to find the foci.
We can write the foci as (3  5 , 1) . We know the length of the transverse axis is 2(1) = 2. The length of the
2.24
conjugate axis is 2(2) = 4. Then we can find the eccentricity, which is e 
 2.24 . The equation of the
1
asymptotes is y  1  2( x  3) .
EXAMPLE: Graph 16 x 2  9 y 2  32 x  90 y  353  0 and identify the foci, eccentricity, center, length of the
transverse and conjugate axis, vertices, and the equations of the asymptotes.
First we need to get this into the correct form. The first thing to do is subtract the 1 from both sides and group
the x and y terms together: 16 x 2  32 x  9 y 2  90 y  353 . Now I will factor the two x terms and the two y
terms separately: 16( x 2  2 x)  9( y 2  10 y )  353 . Now it is time to complete the square. In the first term we
will divide the -2 by 2 and square it. You will get 1. We will add 1 to the left hand side however we do NOT
add 1 to the right side. The 1 is really being multiplied by the 16, so we will add 1(16) = 16 to both sides. In
the second term we will divide the -10 by 2 and square it. You will get 25. We will add 25 to the left hand side
however we do NOT add 25 to the right side. The 25 is really being multiplied by the -9, so we will add -9(25)
= -225 to both sides: 16( x 2  2 x  1)  9( y 2  10 y  25)  353  16  225 . We can simplify the right side and
factor the left side: 16( x  1) 2  9( y  5) 2  144 . Now divide both sides by 144 and reduce to get:
( x  1) 2 ( y  5) 2

 1 . The x comes first, so we know this hyperbola will be drawn horizontally. From our
9
16
From our equation we know that a is 3 and b is 4. We can find c
by the formula c  a 2  b 2 : c  3 2  4 2 . So c  25  5 .
We start at the center, which is (1, 5). Since this is drawn horizontally
I will add 3 and subtract 3 from the x value of the center since a is 3.
This will gives us the vertices (-2, 5) and (4, 5). Then I will go up and
down 4 since this is our b. To find the foci I will add 5 and subtract 5
from the x value of the center. The foci can be written as: (6, 5) & (-4, 5).
The length of the transverse axis is 2(3) = 6. The length of the
5
conjugate axis is 2(4) = 8. The eccentricity is e   1.67 . The equation
3
4
of the asymptotes is: y  5   ( x  1) .
3
Section 9.2 Notes Page 4
EXAMPLE: Graph 9 y  18 y  4 x  16 x  43  0 and identify the foci, eccentricity, center, length of the
transverse and conjugate axis, vertices, and the equations of the asymptotes.
2
2
First we need to get this into the correct form. The first thing to do is subtract the 1 from both sides and group
the x and y terms together: 9 y 2  18 y  4 x 2  16 x  43 . Now I will factor the two x terms and the two y terms
separately: 9( y 2  2 x)  4( x 2  4 x)  43 . Now it is time to complete the square. In the first term we will
divide the -2 by 2 and square it. You will get 1. We will add 1 to the left hand side however we do NOT add 1
to the right side. The 1 is really being multiplied by the 9, so we will add 1(9) = 16 to both sides. In the second
term we will divide the 4 by 2 and square it. You will get 4. We will add 4 to the left hand side however we do
NOT add 4 to the right side. The 4 is really being multiplied by the -4, so we will add -4(4) = -16 to both sides:
9( y 2  2 x  1)  4( x 2  4 x  4)  43  9  16 . We can simplify the right side and factor the left side:
9( y  1) 2  4( x  2) 2  36 . Now divide both sides by 36 and reduce to get:
( y  1) 2 ( x  2) 2

 1 . The y
4
9
comes first, so we know this hyperbola will be drawn vertically. From our
From our equation we know that a is 2 and b is 3. We can find c
by the formula c  a 2  b 2 : c  2 2  3 2 . So c  13  3.61 .
We start at the center, which is (1, -2). Since this is drawn vertically
I will add 2 and subtract 2 from the y value of the center since a is 2.
This will gives us the vertices (-2, 3) and (-2, -1). Then I will go left and
right 3 since this is our b. To find the foci I will add 3.61 and subtract 3.61
from the y value of the center. The foci can be written as: (2, 1  13 ) .
The length of the transverse axis is 2(2) = 4. The length of the
13
conjugate axis is 2(3) = 6. The eccentricity is e 
 1.8 . The equation
2
2
of the asymptotes is: y  1   ( x  2) .
3
EXAMPLE: Find an equation for the hyperbola centered at the origin with focus at (-3, 0) and vertex at (2, 0).
We know that c = 3 because this is how far the focus is from (0, 0).
We also know a = 2 since that is how far the vertex is from the center.
We can use c  a 2  b 2 to find b 2 : 3  2 2  b 2 . After squaring
both sides we get: 9  4  b 2 . Solving for b 2 we get b 2  5 . Now we
have everything we need to find our equation. From our graph we know
this hyperbola is drawn horizontally, so it must use the
x2 y2
formula 2  2  1 . Plugging in our known variables will
a
b
2
x
y2
give us:

 1.
5
22
Section 9.2 Notes Page 5
EXAMPLE: Find an equation for the hyperbola with a focus at (0, 6) and vertex at (0, 2) .
First let’s plot the above points. From this we can conclude that the
hyperbola must be centered at (0, 0) since this point is halfway between
our vertices. We also know that c = 6 because this is how far the focus is
from the center. We also know a = 2 since that is how far the vertices
are from the center. We can use c  a 2  b 2 to find b 2 : 6  2 2  b 2 .
After squaring both sides we get: 36  4  b 2 . Solving for b 2 we get
b 2  32 . Now we have everything we need to find our equation. From
our graph we know this hyperbola is drawn vertically, so it must use the
y2 x2
formula 2  2  1 . Plugging in our known variables will
a
b
2
y
x2
give us:

 1.
2 2 32
EXAMPLE: Find an equation for the hyperbola with foci at (0, 2) and with an asymptote of y  x .
First let’s plot the above points. From this we can conclude that the
hyperbola must be centered at (0, 0) since this point is halfway between
our foci. We also know that c = 2 because this is how far the focus is
from the center. The slope of the asymptote is 1. From plotting our
points we know that this hyperbola is drawn vertically. This means
a
that the equation of the asymptote is y   x . So we know that
b
a
 1 . This tells us a = b. Since the foci is at (0, 2) , we know that
b
c = 2. We will use the formula c  a 2  b 2 . Since we know that
a = b, we can substitute an a for b. We can also put in a 2 for c:
2  a 2  a 2 . This simplifies to: 2  2a 2 . Now square both sides
to get: 4  2a 2 . So we know that a 2  2 . Since the foci are drawn
vertically, we know the hyperbola opens up and down. Therefore we
y2 x2
y2 x2

 1.
use the equation 2  2  1 . Our equation is
2
2
a
b
Section 9.3 Notes Page 1
9.3 The Parabola
In this section we will be looking at the parabola in more detail. The pictures below show the four different
ways a parabola can be drawn that are all centered at the origin. The V represents the vertex of the parabola.
The F is called the focus. All rays that hit the parabola will be directed through the focus. That is why satellite
dishes are made this way. The focus the incoming signals. The D is called the directrix, and this is always
behind the parabola. Notice below that the parabola can have four different equations depending on which way
it is orientated. Each formula has a letter ‘a’ in it. This a value is important since it will tell you how far the
vertex is from the focus and how far the vertex is from the directrix.
The quantity 4a is called the focal width. A focal width is the length of a vertical or horizontal line that
passes through the focus and touches the parabola on each end.
EXAMPLE: Graph y 2  8 x and identify the directrix, focus, and focal width.
First we need to find out the ‘a’ value. From the picture above
we are going to use the general equation y 2  4ax . Therefore
we know that 4a  8 . Then we know that a = 2. We can find
the focal width by using 4(2)  8 . According to our general
equation it says that the directrix is x  2 . We also know that
the focus is at (2, 0). The vertex is at (0, 0). To draw the graph
we first plot the vertex, focus, and draw the directrix. Then we
will put in the focal width. Since the focal width is 8 this means
from our vertex we will go up 4 and down 4 for a total distance
of 8. This will give us two points on the graph so it will give us
an aid in sketching. We know the graph opens to the right
because we are using model (a) from the above pictures since it
matches our equation.
Section 9.3 Notes Page 2
EXAMPLE: Graph x  4 y and identify the directrix, focus, and focal width.
2
First we need to find out the ‘a’ value. From the picture above
we are going to use the general equation x 2  4ay . Therefore
we know that  4a  4 . Then we know that a = 1. We can find
the focal width by using  4(1)  4 . According to our general
equation it says that the directrix is y  1 . We also know that
the focus is at (0, -1). The vertex is at (0, 0). To draw the graph
we first plot the vertex, focus, and draw the directrix. Since we
are looking at graph model (d) above we know that our graph
must open down. For the focal width we will go to the focus and
go to the left and to the right a distance of 2 so that our total focal
width is 4 as we found earlier.
Equations for the Parabola centered at (h, k).
Section 9.3 Notes Page 3
EXAMPLE: Graph ( y  3)  16( x  2) and identify the directrix, focus, and focal width.
2
First we need to find out the ‘a’ value. From the picture above
we are going to use the general equation ( y  k ) 2  4a( x  h) .
We know that  4a  16 . Then we know that a = 4. We can find
the focal width by using  4(4)  16 . We can also find the vertex.
It is the opposite sign of what appears in the equation. So our vertex
is (-2, 3). We want to first plot the vertex. From the model we are
using we know that the parabola opens to the left. That means the
focus is to the left. From the vertex we will go 4 places to the left
since a is 4. So we know our focus is at (-6, 3). From the vertex we
can go 4 places to the right and this will be our directrix. The equation
of the directix will be x = 2. Through our focus we will go up 8 and
down 8 since the focal width is 16. This will give us two more points
on the graph so this will help us draw the sketch.
EXAMPLE: Graph ( x  5) 2  20( y  3) and identify the directrix, focus, and focal width.
First we need to find out the ‘a’ value. From the picture above
we are going to use the general equation ( x  h) 2  4a( y  k ) .
We know that 4a  20 . Then we know that a = 5. We can find
the focal width by using  4(5)  20 . We can also find the vertex.
It is the opposite sign of what appears in the equation. So our vertex
is (5, 3). We want to first plot the vertex. From the model we are
using we know that the parabola opens up. That means the focus is
above the vertex. From the vertex we will go up 5 places since a
is 5. So we know our focus is at (5, 8). From the vertex we can go
down 5 and this will be our directrix. The equation of the directrix is
y = -2. Through our focus we will go to the left 10 and to the right 10
since the focal width is 20. This will give us two more points on the
graph so this will help us draw the sketch.
EXAMPLE: Graph y 2  4 y  12 x  40  0 and identify the directrix, focus, and focal width.
This is not in the proper form for us to graph, so we need to do something to this equation first. I will leave the
y terms on the right hand side. Then I will move the other terms to the right side. Then I will have:
y 2  4 y  12 x  40 . You need to complete the square on the left hand side. We take the number 4 and divide it
by 2 and then square it. You will get 4. We want to add this to both sides of the equation:
y 2  4 y  4  12 x  40  4 . Now we can simplify the right hand side and factor the left side:
( y  2) 2  12 x  36 . Finally we can factor the right side: ( y  2) 2  12( x  3)
Now we have it in the proper form, so we can now graph it on the next page…
Section 9.3 Notes Page 4
First we need to find out the ‘a’ value. From the picture above
we are going to use the general equation ( y  k ) 2  4a( x  h) .
We know that 4a  12 . Then we know that a = 3. We can find
the focal width by using  4(3)  12 . We can also find the vertex.
It is the opposite sign of what appears in the equation. So our vertex
is (3, -2). We want to first plot the vertex. From the model we are
using we know that the parabola opens to the right. That means the
focus is to the right of the vertex. From the vertex we will go 3 places
to the right. So we know our focus is at (6, -2). From the vertex
we can go 3 places to the left and this will be our directrix. The
equation of the directrix is x = 0. Through our focus we will go up
6 and down 6 since the focal width is 12. This will give us two
more points on the graph so this will help us draw the sketch.
EXAMPLE: Graph 2 x 2  4 x  20 y  38  0 and identify the directrix, focus, and focal width.
This is not in the proper form for us to graph, so we need to do something to this equation first. First we need to
divide the entire equation by 2: x 2  2 x  10 y  19  0 . I will leave the x terms on the right hand side. Then I
will move the other terms to the right side. Then I will have: x 2  2 x  10 y  19 . You need to complete the
square on the left hand side. We take the number 2 and divide it by 2 and then square it. You will get 1. We
want to add this to both sides of the equation: x 2  2 x  1  10 y  19  1 . Now we can simplify the right hand
side and factor the left side: ( x  1) 2  10 y  20 . Finally we can factor the right side: ( x  1) 2  10( y  2)
First we need to find out the ‘a’ value. From the picture above
we are going to use the general equation ( x  h) 2  4a ( y  k ) .
We know that  4a  10 . Then we know that a = 2.5. We can find
the focal width by using  4(2.5)  10 . We can also find the vertex.
It is the opposite sign of what appears in the equation. So our vertex
is (-1, 2). We want to first plot the vertex. From the model we are
using we know that the parabola opens down. That means the
focus is below the vertex. From the vertex we will go down 2.5 places
So we know our focus is at (-1, -0.5). From the vertex we can go
up 2.5 places and this will be our directrix. The equation of the
directrix is y = 4.5. Through our focus we will go to the left 5 and to
the right 5 since the focal width is 10. This will give us two more
points on the graph so this will help us draw the sketch.
Section 9.3 Notes Page 5
EXAMPLE: Find the equation of the parabola if it is known the focus is (-4, 0) and the vertex is (0, 0).
For this one we know that the focus is 4 units away from
the vertex. This means that a is 4. Since it is to the left of
the vertex (see graph) we know the graph is opening up to
the left. A parabola opening up to the left has the equation
y 2  4ax . If we put in a 4 for a we will get our solution:
y 2  16 x .
EXAMPLE: Find the equation of the parabola if it is known the focus is (2, 4) and the directrix is at x = -4.
We can first plot (2, 4) and then draw in the directrix.
The vertex has to be exactly half way between the
focus and directrix. Then we know the vertex is at
(-1, 4). The distance from the vertex to either the
focus or the directrix is 3, so we know that a is 3.
The graph will open towards the right since the focus
is to the right of the vertex. Then we know the general
equation is: ( y  k ) 2  4a( x  h) . We know that a is
3, h is -1 and k is 4. We plug these into the formula to
get: ( y  4) 2  12( x  1) .
EXAMPLE: Find the equation of the parabola if it is known the vertex is (3, 0) and the directrix is at y = 2.
We first plot (3, 0) and draw y = 2. The distance from
the vertex to the directrix is 2. This tells us that a = 2.
Since the directrix is horizontal and the vertex is below
this line we know the graph opens down, so the general
equation is: ( x  h) 2  4a ( y  k ) . We can now plug in
2 for a, 3 for h, and 0 for y. Our equation will be:
( x  3) 2  8 y .
Section 9.5 Notes Page 1
9.5 Parametric Equations
If we take a point (x, y) and move it on the x–y plane after a time t, we have a pair of equations:
x  f (t ) and y  g (t )
These equations are called parametric equations with parameter t.
EXAMPLE: Graph the following equations x  3t and y  t 2 where  2  t  2 . Eliminate the parameter.
In order to do this we will make a table and then plot the point.
We will start with t = -2 and go to t = 2. You can use any values
between -2 and 2. Each value for t will give you an x and a y.
When put together this gives us a point to plot.
x  3t
y  t2
(x, y)
-2 3(-2) = -6
(2) 2  4
(-6, 4)
-1 3(-1) = -3
2
(1)  1
(-3, 1)
0
3(0) = 0
(0) 2  0
(0, 0)
1
3(1) = 3
(1) 2  1
(3, 1)
3(2) = 6
(2)  4
(6, 4)
t
2
2
The graph also needs directional arrows. When t is -2 we
started at (6, 4). Each time t increases the point moves from
left to right, so we indicate that on our graph be arrows.
Now we need to eliminate the parameter. This means you want to write an equation the same as what we
graphed that does not contain a t. Usually you want to take one of the equations and solve for t. Then substitute
x
this into the other equation. Our first equation is x  3t . If we solve this for t we get: t  . Now we want to
3
2
x2
 x
. This is our answer. If we made
substitute this into our other equation for y: y    . We will get y 
9
3
a table with this equation we would get the same graph as above.
EXAMPLE: Graph the following equations x  t and y 
1
t  1 where 0  t  4 . Eliminate the parameter.
2
In order to do this we will make a table and then plot the points. We will start with t = 0 and go to t = 4. You
can use any values between 0 and 4. Each value for t will give you an x and a y. When put together this gives
us a point to plot. Make sure you remember to put the correct directional arrows. When t is 0 you are at the
point (0, 1). The point is traveling up. To eliminate the parameter on this one, first solve for t in x  t . After
1
squaring both sides we get t  x 2 . Then we plug this into the y equation to get: y  x 2  1 .
2
Section 9.5 Notes Page 2
t
x t
0
x 0 0
1
x  1 1
2
x  2  1.41
3
x  3  1.73
4
x 42
y
1
t 1
2
1
(0)  1  1
2
1
3
(1)  1 
2
2
1
(2)  1  2
2
1
5
(3)  1 
2
2
1
(4)  1  3
2
(x, y)
(0, 1)
 3
1, 
 2
(1.41, 2)
5

1.73, 
2

(2, 3)
EXAMPLE: Graph the following equations x  e t and y  e t where 0  t  2 . Eliminate the parameter.
First we make our table and plot the points. Since t is between 0 and 2 I wanted to do more than just plot three
points. That is why I chose 0.5 and 1.5. I wanted to have enough points to plot.
t
x  et
y  e t
(x, y)
0
.5
1
1.5
2
x  e0  1
x  e 0.5  1.65
x  e1  2.72
x  e1.5  4.48
x  e 2  7.4
x  e 0  1
x  e 0.5  0.61
x  e 1  0.37
x  e 1.5  0.22
x  e 2  0.1353
(1, 1)
(1.65, 0.61)
(2.72, 0.37)
(4.48, 0.22)
(7.4, 0.1353)
To eliminate the parameter, solve for t in the first equation.
In order to do this we must take the natural log of both
sides: ln x  ln e t . We get t  ln x . Now plug this into
the second equation for t. You will get: y  e  ln x . This
1
simplifies to: y  x 1  .
x
Section 9.5 Notes Page 3
EXAMPLE: Graph the following equations x  4 cos t and y  2 sin t where 0  t  2 . Eliminate the
parameter.
We make our table. I picked values off the unit circle that would give either a -1, 0, or 1 and an answer.
t
x  4 cos t
y  2 sin t
(x, y)
0
x  4 cos 0  4(1)  4
y  2 sin 0  2(0)  0
(4, 0)
(0, 2)

2

3
2
2
x  4 cos

 4(0)  0
2
x  4 cos   4(1)  4
3
x  4 cos
 4(0)  0
2
x  4 cos 2  4(1)  4
y  2 sin

2
 2(1)  2
y  2 sin   2(0)  0
3
y  2 sin
 2(1)  2
2
y  2 sin 2  2(0)  0
(-4, 0)
(0, -2)
(4, 0)
What kind of graph do we get? That’s right, it’s an ellipse. In order
to eliminate the parameter we just need to come up with the equation
of this ellipse. It is going horizontally, so we know that a is 4 and b is 2
x2 y2
and this is centered at the origin. The equation is: 2  2  1 .
4
2
EXAMPLE: Graph the following equations x  3 cos t and y  3 sin t where 0  t  2 . Eliminate the
parameter.
We make our table. I picked values off the unit circle that would give either a -1, 0, or 1 and an answer.
t
x  3 cos t
y  3 sin t
(x, y)
0
x  3 cos 0  3(1)  3
y  3 sin 0  3(0)  0
(3, 0)
(0, 3)

2

3
2
2
x  3 cos

 3(0)  0
2
x  3 cos   3(1)  3
3
x  3 cos
 3(0)  0
2
x  3 cos 2  3(1)  3
y  3 sin

2
 3(1)  3
y  3 sin   3(0)  0
3
y  3 sin
 3(1)  3
2
y  3 sin 2  3(0)  0
(-3, 0)
(0, -3)
(3, 0)
What kind of graph do we get? That’s right, it’s a circle. In order
to eliminate the parameter we just need to come up with the equation
of this circle. It has a radius of 3, centered at the origin, so the equation
is: x 2  y 2  3 2 .
Section 7.3 Notes Page 1
7.3 Partial Fractions
4
3

. We would need to get
x 1 x  2
4  x 2
3  x 1
4( x  1)  3( x  1)


. Distributing on top
common denominators:

 . You will get:
x 1  x  2  x  2  x 1
( x  1)( x  2)
x  11
4 x  8  3x  3
will give you:
. This simplifies to:
. Therefore we come to the following:
( x  1)( x  2)
( x  1)( x  2)
4
3
x  11

. In this section we will be doing the reverse of what we just did. We will start
=
( x  1)( x  2)
x 1 x  2
with the single fraction and break it up (decompose it) into separate fractions. This is something you will do in
a calculus 2 course. Sometimes it is easier to do calculus operations on two smaller fractions instead of one big
fraction.
Suppose we were asked to write the following as a single fraction:
Three Rules of How a Fraction Decomposes
Let P(x) be a polynomial.
Rule 1:
An
A1
A2
P ( x)


 ... 
( x  a1 )( x  a 2 )...( x  a n ) x  a1 x  a 2
x  an
Rule 2:
A3
An
A
A2
P( x)
 1 

 ... 
n
2
3
x  a ( x  a)
( x  a)
( x  a)
( x  a) n
Rule 3:
An x  Bn
A1 x  B1
A2 x  B2
 ... 

2
2
2
ax  bx  c
ax  bx  c
ax  bx  c
ax 2  bx  c
Where ax 2  bx  c is irreducible, or nonfactorable.
P ( x)


n
2


 


EXAMPLE: Set up the following for decomposition but DO NOT SOLVE:
We first need to factor:

n
3x  5
.
x  6x  5
2
A
A
3x  5
3x  5
 1  2 .
. Now we will use rule 1:
( x  5)( x  1)
( x  5)( x  1) x  5 x  1
EXAMPLE: Set up the following for decomposition but DO NOT SOLVE:
x
This time we can’t factor the denominator. This means we need to use rule 3:
x
2x  4
2

A3 x  B3
A1 x  B1
A2 x  B2


3
2
2
x  x 1
x2  x 1
x2  x 1
 
 x 1
3
 
 

2x  4
2
 x  1
3
.
Section 7.3 Notes Page 2
2x  1
2
EXAMPLE: Set up the following for decomposition but DO NOT SOLVE:
This one requires rule 2:
2x 2  1
3x  5
4

3x  54
.
A3
A1
A2
A4



2
3
3 x  5 (3 x  5)
(3 x  5)
(3x  5) 4
EXAMPLE: Set up the following for decomposition but DO NOT SOLVE:
4 x 2  3x  1
.
x( x  3) 2
Sometimes you may need to combine more than one rule. This one will use rule 1 and rule 2:
A3
A
4 x 2  3x  1 A1

 2 
2
x x  3 ( x  3) 2
x( x  3)
EXAMPLE: Determine the partial fraction decomposition:
5 x  27
.
x2  9
The first thing we should do with all these kind of problems is to factor if possible:
5 x  27
. Now we
( x  3)( x  3)
can use rule 1 to set it up:
A
A
5 x  27
 1  2 . Now we want to get common denominators with the
( x  3)( x  3) x  3 x  3
right side of the equation:
A  x  3
A  x  3
5 x  27
 1 
  2 
 . This will give us:
( x  3)( x  3) x  3  x  3  x  3  x  3 
A ( x  3)  A2 ( x  3)
5 x  27
. Since the denominator of each fraction is the same, we can just set the
 1
( x  3)( x  3)
( x  3)( x  3)
numerators equal to each other. You will get the following equation: 5 x  27  A1 ( x  3)  A2 ( x  3) . From
here the book shows two methods of solving this. I will only focus on one method. This one involves choosing
a value for x and plugging it into both sides of the equation. You want to choose a number that will cancel part
of the equation so you only have one variable left to solve for. In the problem above I want to let x be 3 and -3.
Let x = 3.
We will put 3 into both sides of the equation: 5(3)  27  A1 (3  3)  A2 (3  3) . You want to
simplify inside the parenthesis first: 42  A1 (0)  A2 (6) . This simplifies further to: 42  6 A2 .
Solving for A2  7 .
Let x = -3.
We will put 3 into both sides of the equation: 5(3)  27  A1 (3  3)  A2 (3  3) . You want
to simplify inside the parenthesis first: 12  A1 (6)  A2 (0) . This simplifies further to:
12  6 A1 . Solving for A1 we get: A1  2 .
Our answer is written as:
5 x  27
2
7
.


( x  3)( x  3) x  3 x  3
EXAMPLE: Determine the partial fraction decomposition:
6  4x
.
x  x 2  4x  4
Section 7.3 Notes Page 3
3
The first thing we should do is to factor the denominator. This involves the grouping method. Factor the first
two terms and the second two terms separately: x 2 ( x  1)  4( x  1) . Now factor out the common factor of
x – 1: ( x  1)( x 2  4) . We can factor this one more time to get: ( x  1)( x  2)( x  2) . So now our problem
6  4x
. Now we can use rule 1 to set it up:
becomes:
( x  1)( x  2)( x  2)
A
A
A
6  4x
 1  2  3 . Now we want to get common denominators
( x  1)( x  2)( x  2) x  1 x  2 x  2
A
A ( x  2)( x  2)
A
6  4x
( x  1)( x  2)
( x  1)( x  2)
. This will give us:
 1 
 2 
 3 
( x  1)( x  2)( x  2) x  1 ( x  2)( x  2) x  2 ( x  1)( x  2) x  2 ( x  1)( x  2)
A ( x  2)( x  2)  A2 ( x  1)( x  2)  A3 ( x  1)( x  2)
6  4x
. Since the denominator of each
 1
( x  1)( x  2)( x  2)
( x  1)( x  2)( x  2)
fraction is the same, we can just set the numerators equal to each other. You will get the following equation:
6  4 x  A1 ( x  2)( x  2)  A2 ( x  1)( x  2)  A3 ( x  1)( x  2) . You want to choose a number that will cancel
part of the equation so you only have one variable left to solve for. In the problem above I want to let x be 1, 2,
and -2.
Let x = 1.
We will put 3 in for x: 6  4(1)  A1 (1  2)(1  2)  A2 (1  1)(1  2)  A3 (1  1)(1  2) . You want to
simplify inside the parenthesis first: 2  A1 (3)(1)  0  0 . This simplifies further to:
2  3 A2 . Solving for A1 we get: A1  2 / 3 .
Let x = 2.
We will put 2 in for x: 6  4(2)  A1 (2  2)(2  2)  A2 (2  1)(2  2)  A3 (2  1)(2  2) . You want
to simplify inside the parenthesis first:  2  0  0  A3 (1)(4) . This simplifies further to:
 2  4 A2 . Solving for A2 we get: A3  1 / 2 .
Let x = -2.
We will put 3 in for x: 6  4(2)  A1 (2  2)(2  2)  A2 (2  1)(2  2)  A3 (2  1)(2  2) .
You want to simplify inside the parenthesis first: 14  0  A2 (3)(4)  0 . This simplifies
further to: 14  12 A2 . Solving for A2 we get: A2  7 / 6 .
Our answer is written as:
6  4x
7
1
 2 / 3 7 / 6  1/ 2
2
or





x 1 x  2 x  2
( x  1)( x  2)( x  2)
3( x  1) 6( x  2) 2( x  2)
EXAMPLE: Determine the partial fraction decomposition:
We can use rule 2 to set this up:
7x
.
( x  5) 2
A
A2
7x
 1 
. Now we want to get common denominators with
2
x  5 ( x  5) 2
( x  5)
Section 7.3 Notes Page 4
the right side of the equation:
A  x 5
A2
7x
 1 
. This will give us:

2
x  5  x  5  ( x  5) 2
( x  5)
A ( x  5)  A2
7x
. Since the denominator of each fraction is the same, we can just set the numerators
 1
2
( x  5)
( x  5) 2
equal to each other. You will get the following equation: 7 x  A1 ( x  5)  A2 . You want to choose a number
that will cancel part of the equation so you only have one variable left to solve for. In the problem above I want
to let x be 5.
Let x = 5.
We will put 5 into both sides of the equation: 7(5)  A1 (5  5)  A2 . You want to simplify inside
inside the parenthesis first: 35  A1 (0)  A2 . This simplifies further to: 35  A2 .
So now our problem becomes: 7 x  A1 ( x  5)  35 . We don’t have another value to plug in for x to cancel
something out. So now we can let x be ANY number. It doesn’t matter which one you choose because you will
get the same answer for A1 . Let’s let x = 0 since this is an easy one to plug in:
Let x = 0.
We will put 0 into both sides of the equation: 7(0)  A1 (0  5)  35 . You want to simplify inside
the parenthesis first: 0  5 A1  35 . Solving for A1 we get: A1  7 .
7x
7
35
.


2
x  5 ( x  5) 2
( x  5)
Our answer is written as:
EXAMPLE: Determine the partial fraction decomposition:
x2
.
x3  x2
We want to factor a common factor out of the denominator. We will get:
and rule 2 to set it up:
x2
. Now we can use rule 1
x ( x  1)
2
A
A A
x2
 1  22  3 . Now we want to get common denominators
x x
x 1
x ( x  1)
2
A1 x( x  1)  A2 ( x  1)  A3 x 2
A1 x( x  1) A2 x  1 A3 x 2
x2
x2


.




 . This will give us: 2
x x( x  1) x 2 x  1 x  1 x 2
x ( x  1)
x 2 ( x  1)
x 2 ( x  1)
Since the denominator of each fraction is the same, we can just set the numerators equal to each other. You will
get the following equation: x  2  A1  x( x  1)  A2 ( x  1)  A3 x 2 . You want to choose a number that will
cancel part of the equation so you only have one variable left to solve for. In the problem above I want to let x
be 0, and 1.
Let x = 0.
We will put 0 in for x: 0  2  A1  0(0  1)  A2 (0  1)  A3 0 2 . You want to
simplify inside the parenthesis first: 2  0  A2 (1)  0 . This simplifies further to:
2   A2 . Solving for A2 we get: A2  2 .
Section 7.3 Notes Page 5
Let x = 1.
We will put 1 in for x: 1  2  A1  1(1  1)  A2 (1  1)  A3 1 . You want
to simplify inside the parenthesis first: 3  0  0  A3  1 . This gives us: A3  3 .
2
So now our problem becomes: x  2  A1  x( x  1)  2( x  1)  3x 2 Since we have run out of numbers to plug in
for x to cancel terms, we can now choose ANY number to plug in. I will choose x = 2.
Let x = 2.
We will put 3 in for x: 2  2  A1  2(2  1)  2(2  1)  3  2 2 . You want
to simplify inside the parenthesis first: 4  2 A1  2  12 . This simplifies further to:
4  2 A1  10 . Solving for A1 we get: A1  3 .
Our answer is written as:
x2
3 2
3
.

 2 
x
x 1
x ( x  1)
x
2
x2  3
EXAMPLE: Determine the partial fraction decomposition: 3
.
x  6x 2  9x
We want to factor a common factor out of the denominator. We will get:
x2  3
. We can factor this
x( x 2  6 x  9)
A3
A
A
x2  3
x2  3
Now
we
can
use
rule
1
and
rule
2
to
set
it
up:
.
 1 2 
2
2
x x  3 ( x  3) 2
x( x  3)
x( x  3)
Now we want to get common denominators:
A3
A1 ( x  3) 2
A
x2  3
x
x( x  3)


 2 

 . This will give us:
2
2
2
x ( x  3)
x  3 x( x  3) ( x  3) x
x( x  3)
one more time:
A1 ( x  3) 2  A2 x( x  3)  A3 x
x2  3
.

x( x  3) 2
x( x  3) 2
Setting the numerators equal we get: x 2  3  A1  ( x  3) 2  A2 x( x  3)  A3 x . Now choose values for x.
Let x = 0.
We will put 0 in for x: 0 2  3  A1  (0  3) 2  A2  0(0  3)  A3  0 . This gives us 3  9 A1 , so
A1  1 / 3 .
Let x = -3.
We will put -3 in for x: (3) 2  3  A1  (3  3) 2  A2 (3)(3  3)  A3 (3) . This gives us:
12  3 A3 , so A3  4
So now our problem becomes: x 2  3  (1 / 3)  ( x  3) 2  A2 x( x  3)  4 x Since we have run out of numbers to
plug in for x to cancel terms, we can now choose ANY number to plug in. I will choose x = 3.
Let x = 3.
We will put 3 in for x: 3 2  3  (1 / 3)  (3  3) 2  A2  3(3  3)  4(3) . This gives us
12  12  18 A2  12 . Solving for A 2 we get A2  2 / 3
x2  3
1/ 3 2 / 3
4
x2  3
1
2
4



, or you can write:



.
Our answer is:
2
2
2
x
x  3 ( x  3)
3 x 3( x  3) ( x  3) 2
x( x  3)
x( x  3)
EXAMPLE: Determine the partial fraction decomposition:
x7
.
x3  2x
We want to factor a common factor out of the denominator. We will get:
and rule 3 to set it up:
Section 7.3 Notes Page 6
x7
. Now we can use rule 1
x( x 2  2)
A A x  B2
x7
. Now we want to get common denominators
 1  22
2
x
x( x  2)
x 2
A1 x 2  2 A2 x  B2 x
x7


 2
 . This will give us:
x x2  2
x( x 2  2)
x 2 x
A1 ( x 2  2)  ( A2 x  B2 ) x
x7

.
x( x 2  2)
x( x 2  2)
Setting the numerators equal we get: x  7  A1  ( x 2  2)  ( A2 x  B2 ) x . Now choose values for x.
Let x = 0.
We will put 0 in for x: 0  7  A1  (0 2  2)  ( A2 (0)  B2 )(0) . This gives us  7  2 A1 , so
7
A1   .
2
 7( x 2  2)
 ( A2 x  B2 ) x Since we have run out of numbers to plug in
So now our problem becomes: x  7 
2
for x to cancel terms, we can now choose ANY number to plug in. I will choose x = 2.
Let x = 2.
 7(2 2  2)
 ( A2 (2)  B2 )(2) . This gives us
2
 5  21  4 A2  2 B2 . We can write this as: 4 A2  2 B2  16 . This equation can be reduced to:
2 A2  B2  8 .
We will put 3 in for x: 2  7 
But we don’t have enough information to solve for the two unknowns, so we need one more equation. We need
to pick another value for x. I will let x = 4.
Let x = 4.
 7(4 2  2)
 ( A2 (4)  B2 )(4) . This gives us
2
 3  63  16 A2  4 B2 . We can write this as: 16 A2  4 B2  60 . This equation can be reduced
to 4 A2  B2  15 .
We will put 3 in for x: 4  7 
2 A2  B2  8
. If we subtract the two equations we get  2 A2  7 , so
4 A2  B2  15
A2  7 / 2 . Going to the top equation we have 2(7 / 2)  B2  8 , or 7  B2  8 . Then B2  1 .
We now need to solve the system:
Our answer is:
7
x7
7x  2
x7
 7 / 2 (7 / 2) x  1

.


, or you can write:

2
2
2
x
x( x  2) 2 x 2( x 2  2)
x( x  2)
x 2