4.1 Angles and Radian Measure

114 Pages·2013·2.34 MB·English

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:

180
Multiply your radians by


EXAMPLE: Convert 60 to radians.

  
We will take 60 and multiply it by and you will get: 60 . This reduces to .
180 180 3

EXAMPLE: Convert 405 to radians.

  5
We will take –405 and multiply it by and you will get: 450 . This reduces to  .
180 180 2

4
EXAMPLE: Convert into degrees.
3

4 180 4 180
We will take and multiply it by and you will get:  . This reduces to 240.
3  3 

3
EXAMPLE: Convert  into degrees.
2

3 180 3 180
We will take  and multiply it by and you will get:   . This reduces to 270
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.

5
a.)
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.

8
c.)
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.
Section 4.1 Notes Page 3
Arc length – this is the length of the arc between the two lines shown with .

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 .
3


Using our formula S  r we know that r 3 and  . Putting this into the formula we will get:
3

S 3 , so S  inches.
3

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 to get it into radians: 45 . Reducing this we get  . Putting
180 180 4
 7
this into the formula we will get:S 7 , so S  inches.
4 4

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.

17
b.) a angle.
6

17 5
The fraction 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.
Section 4.1 Notes Page 4
Angular Speed ()

Angular speed is the angle  that can be swept out in time t. The formula is:


 where is in radians.
t

Linear Speed (v) 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
so we can use the formula  . We will get  = 150 radians per minute.
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 . You will get: 2160 which reduces to
180 180
 12
12 radians. Our unit of time is 1 second, so we can use the formula  . We will get  = 12
t 1
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.

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
2 3
this as rpm, so now we must divide this by 2. You will get: , 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
one. For example, we take 3rpm and multiply it by 2in divided by 8in: 3   rpm. The same can be
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 60min 1 mile
   2.86 mph
1 min 1hr 5280 ft
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 x2  y2 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
sint  y csct 
y
1
cost  x sect 
x
y x
tant  cott 
x y

If we start with x2  y2 1 and put in our definitions above we
will have cos2sin21.

 5 12
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
that x   and y   because of the point given. This
13 13
12 5
automatically tells us that sint   and cost   . We can
13 13
plug in x and y into the other equations to find the remaining trig
1 13 1 13
values. csct    sect   
12 12 5 5
 
13 13

12 5
 
13 12 13 12 13 5 13 5
tant     cott    
5 13 5 5 12 13 12 12
 
13 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 sint x cost
   
Domain: ,  Domain: , 
Range: [-1, 1] Range: [-1, 1]
Section 4.2 Notes Page 2

Let’s look at the angle of t  45 or . At this angle, we end up with the following triangle:
4

45 – 45 – 90 Triangle

45 We can use our definitions of sine, cosine, and tangent to find exact values:
1 2 2
2 2 2 2
45 sin45 cos45 tan45 1
2 2 2 2
2 2

From our above definitions we can also find the following:
1 2 2 2 2 1 2 2 2 2 1
csc45      2 , sec45      2 , cos45  1
2 2 2 2 2 2 2 2 1
2 2

 
EXAMPLE: Find the exact value without using a calculator: tan cot .
4 4

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) cost sec(t) sect

sin(t)  sint csc(t)  csct

tan(t)  tant cot(t)  cott

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) sint csc(t 2k) csct

cos(t 2k) cost sec(t 2k) sect

tan(t 2k)  tant cot(t 2k) cott
Section 4.2 Notes Page 3
EXAMPLE: Find the EXACT value of sin1485.

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
2
know that sin(45 360(4)) sin45. From our previous values we know that sin45  . So we know
2
2
that sin1485  .
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 17 17 180
tan   tan . It would be easier to change this into degrees:  765. Dividing this by
 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))  tan45. From our previous values we know that tan45 1. So we know that
17
tan   1.
 4 

Fundamental Identities

1 1
sint  csct  sin2t cos2t 1 1tan2t sec2t 1cot2t csc2t
csct sint

1 1 sint cost
cost  sect  tant  cott 
sect cost cost sint

1 1
tant  cott 
cott tant

2 5
EXAMPLE: Given sint  and cost  , 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:

1 1 3 1 1 3 3 5 sint 2 3 2 3 2 2 5
csct    sect     tant       ,
sint 2/3 2 cost 5 3 5 5 cost 5 3 3 5 5 5
1 1 5 5
cott    
tant 2 5 5 2 5 2
Section 4.2 Notes Page 4
7 
EXAMPLE: Given sint  and 0t  , use the Pythagorean Identity sin2t cos2t 1 to find cost.
8 2

2
7 7
We first start with the identity sin2t cos2t 1 and plug in for sint. You will get   cos2t 1.
8 8
49 49
Simplifying will give you cos2t 1. Now isolate the cosine by subtracting from both sides. You will
64 64
15 15 
get cos2t  . Square root both sides to get cost   . Since the angle for t needs to be 0t  , we
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 cost  .
8

EXAMPLE: The unit circle below has been divided into 12 equal arcs, corresponding to t-values of:
   2 5 7 4 3 5 11
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

2 1
The cosine value is always the x, value, so we read the x value from our picture above to get cos   .
3 2

2
b.) sin
3


The sine value is always the y, value. Since each segment is , we need to go over 4 segments which is in the
3
2 3
second quadrant. Here we see that sin  .
3 2
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 3   1
 , . The cosine value is always the x value, so cos   .
 
2 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  11 1 2 1 2 1 3
 , . The tangent value is always y/x, so tan       .
 
 2 2  6  3 2 2 3 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 3   1
 , . The cosine value is always the x value, so cos   .
 
2 2   3 2

 
EXAMPLE: Use an identity to find the exact value without using a calculator: 2cos2 sin
6 6

2
  
This can be rewritten as: 2cos  sin . Now substitute values by using the figure on the previous page.
 6 6
2
 3 1 3 1 3 1
2   . Now square everything inside the parenthesis: 2  . Reduce this fraction:  . After
 
 2  2 4 2 2 2
 
subtracting we get 1. Therefore, 2cos2 sin 1.
6 6

EXAMPLE: Use a calculator to find the approximate value of sin43.5 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: sin43.5  0.6884.