Download Chapter 3 Trigonometric Functions

3 Trigonometric Functions
Chapter 3 Trigonometric Functions
In chapter 2 we have discussed about algebraic functions and their domain and range. In
this chapter we will study six different trigonometric functions namely sine, cosine,
tangent, cosecant, secant and cotangent, and also their graph, domain and range.
3.1 Measuring Angles and Unit Circles
The trigonometric functions act as an operator on the variable (angle) x, resulting in an
output value y. In this section we will study trigonometry looking at angles and methods
of measuring them.
Different Positive Angles: Acute angle, Right angle, Straight angle, Obtuse angle and
Reflex angle
Types of Angle
Acute Angle
Right angle
Description
Example
An angle, which is less than 90 degree or less than
An angle equal to 90 degree or

2

2
is right angle
Straight angle
Obtuse angle
An angle equal to 180 degree or 
An angle greater than 90 degree but less than 180
degree
Reflex angle
An angle, which is greater than 180 degree but
less than 360 degree
60 , 10.55 ,
 
,
6 3
90 degree or

2
180 degree or 
2
100 , 93.55 ,
3
3
200 , 193.55 ,
2
Trigonometric angles
Two rays drawn from a common vertex form an angle. One ray of the angle is called the
initial side while the other is referred to as the terminal side. The angle formed is
identified by showing the direction and amount of rotation between initial and terminal
sides. If the rotation of terminal side from initial side is in the counter clockwise direction
we call it positive angle. If the rotation is in clockwise direction the angle is negative.
Terminal side
Initial side
Terminal side
Vertex
Vertex
Initial side
Fig 1a
Positive angle
Initial side
Fig 1b
Negative angle
Vertex
Initial side
Fig 1c
Positive angle
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3 Trigonometric Functions
Angle measured in degrees
The angle from between initial and terminal sides by one complete revolution is equal to
360 degrees, we write as 360 , which is equal to 4 right angles. Thus one right angle is
equal to 90 . On the other hand one straight angle is equal to 2 right angles or 1/2 of a
complete revolution. Thus one straight angle is equal to 180 .
Terminal side
Vertex
Initial side
Fig 2a
Right angle
Terminal side
Initial side
Terminal side
Vertex
Vertex
Initial side
Fig 2b
Fig 2c
Straight angle
Complete revolution
Angle measured in radians
In the radian system of measuring an angle, the measure of one revolution is 2  360 .
Thus   180 = right angle, and  / 2  90 = right angle.
A central angle is an angle whose vertex is at the center of a circle. If the central angle
formed between the rays is equal to the radius of the circle, we call it one radian.
 /2

r
2
2
1
r
3 / 2
1
2
Central angle = 1 radian
Central angle = 1 radian
Relation between degree and radian measures
We have found that 2 radian = 360 degrees. Now we have 1 radian =
where   3.14159265359 . Also 1 degree =
2

.

360 180
Conversion between degrees and radians

To convert degrees to radians, multiply given degrees by

180
360 180
,

2

and your result is
in radians

To convert radians to degrees, multiply given radians by
180

and your result is in
degrees
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3 Trigonometric Functions
Coterminal angles
Two angles with the same initial and terminal sides are coterminal angles. Every angle
has infinitely many coterminal angles. The coterminal angle of any angle   is
   360 k , where k is an integer showing number of rotations of the terminal side. For
radian measure the angle  radian has coterminal angle   2 k radians.
Two coterminal angles for an angle of   can be found by adding 360 to   and also by
subtracting 360 from   . In this section we will find only positive coterminal angles.
If you graph angles x = 30o and y = - 330o in standard position, these angles will have the
same terminal side.
The meaning and value of 
Suppose the measure of circumference is C and the diameter is D then
C
   3.142
D
The meaning of   3.142 is that the circumference of the circle is little more than 3
times the diameter.
Calculator: In this section we advise students not to use calculator to solve a problem
completely. A calculator is a tool that can be used in the process of learning and
understanding in mathematics. It should be used to enhance the understanding of
concepts, not to replace the learning skills.
Examples:
1. Convert 720 to radian measure.
2. Find the positive coterminal angles of 730
3. Convert 6 to degree measures
4. Find the positive coterminal angles of 7
Solutions:


 4 radians
180
180
2. The positive coterminal angles of 730 are 10 , and 370
180
180
3. Multiply by
to get 6 
 1080 degrees
1. Multiply by

to get 720 

4. The positive coterminal angles of 7 are  , 3 and 5
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3 Trigonometric Functions
Arc length: Let r be the radius of a circle
and  be the central angle measured in
radians. Then the length of the arc
intercepted by the central angle is given by

s
r
s  r
5. Find the arc length of a circle of radius 9 cm intercepted by the central angle of
120 degrees.
Solution: We need to convert the central angle of 120 degrees to radian measures and
then multiply by the radius to get the arc length.

2
Now 120 degrees =
.
120 
180
3
2
The arc length is s  r  9 
 6  18.85 cm.
3
6. A circle has radius 2 meters. Find the arc length intercepted by a central angle of
2 radians.
Solution: The arc length is s  r  2  2  4 meters.
7. The radius of the Earth is given as 3960 miles. If the central angle between two
cities A and B is 20 degrees, find the distance between the city A and city B.
Solution: We use distance
s  r  3960  0.3491  1382 miles.
Where   20  20 

180
 0.3491
8. If two points on the arc of a circle are 600 miles apart, express the central angle x
both in degree and radian measures., when radius of the circle is 400 miles.
Solution: We use the formula
s  r
600  400

400

600
600
180
 1.5 radians = 1.5 
 85.94
400

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3 Trigonometric Functions
Area of a sector of a circle
A
Consider a circle of radius r. If  is
measured in radian is the central angle of
the circle, then the area of the section
O 
OAB is given by
B
1
Area of section OAB  r 2 square unit.
2
9. The radius of a circle is given 10 cm. Find the area of the section formed by the
initial and terminal sides with central angle 30 degrees.
1
1
30
Solution: The area of the section is OAB  r 2  102 
 26.18 cm2
2
2
180
Linear and angular speed
Suppose a circular wheel of radius r makes a complete rotation. We know that for this
complete rotation the wheel travels a horizontal distance 2 r and one complete
revolution makes central angle equals 2 radian. For a central angle  radians the wheel
travels a horizontal distance equals s  r and its angular speed at time t is  

t
. The
s r

 r .
t
t
10. A wheel of radius 10 cm rotating at 4 revolutions per minute. Find the linear
speed, in centimeter per minute.
linear speed is given by v 
Solution: The four complete revolutions per minute = 4  2  8 radian central
angles per minute.
The angular speed  

t
 8 . The linear speed v  r  80  25.13 cm per
minute.
Reference number or reference angle
Let  be a nonacute angle, the reference angle   is the acute angle formed by the
terminal side of  and the x-axis.
11. Find the reference angle of 135
Solution: The reference angle of 135 is 135  90  45
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3 Trigonometric Functions
3.2 Trigonometric Functions and Their Graphs
In this section we will discuss three very important functions namely sine, cosine and
tangent functions. If x is any angle then sine of x is written sin x . Similarly cosine of
angle x is cos x and tangent of angle x is tan x . Following are the definitions for sine,
cosine and tangent functions. We have the following common ideas.
The right triangle has hypotenuse h, adjacent a and opposite o.
Then, by Pythagorean we have h2  a2  o2 .
If the endpoint of the terminal side in a unit circle that makes
an angle x with initial side soh, cah and toa stand for the
following important results:
o
opposite
soh: sin x  
h hypotenuse
a
adjacent
cah: cos x  
h hypotenuse
o sin x opposite
toa: tan x  

a cos x adjacent
h
o
x
a
Some standard values of sine, cosine and tangent functions: Let us produce a table, which
is useful in this case.
How do we make the first row:






Write successively 0
1
2
3
4
Divide each number by 4
Take square root
The values are for sine function
Write sine values from backward for cosine function
Divide sine values by corresponding cosine values to get values for tangent
function.
0
0
4
0
sin
0
cos
1
tan
0
1 1

4 2
2
2

4
2
3
3

4
2
4
1
4




6
1
2
4
3
2
2
2
0
Undefined
3
2
1
2
2
3
2
1
2
3
1
3
1
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3 Trigonometric Functions
Unit Circle: A circle of radius 1 is a unit circle center at origin. For rectangular
coordinate system, if P(x, y) is any point on the unit circle then x 2  y 2  1 . If t is a real
number and P( x, y)  P(t )  P(cos t,sin t ) is a point on the unit circle that corresponds to
the number t, then we have the following results:
x  cos t,
y  sin t, y  0
2
2
sin t  cos t  1 for all values of t
Important results to remember:
sin(n  x) sin x
sin positive
all positive
cos(n  x) cos x
tan positive cos positive


sin   x  cos x
2

tan(n  x) tan x
sin( x)   sin x


cos   x  sin x
2



tan   x  cot
2

cos( x)  cos x
You need to decide a + or
- sign to put inside the 
tan( x)   tan x
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3 Trigonometric Functions

find the location of the point P(t )  P( x, y)
3
Solution: We know that
 
  1
x  cos t  cos     cos   
 3
3 2
Example 1. If t  
3
 
 
y  sin t  sin      sin    
2
 3
3
Example 2. If t  
Solution:

3
find the location of the point P(  t )
P(  t )  P(cos(   / 3),sin(   / 3))
 P( cos( / 3),  sin( / 3))  (1/ 2,  3 / 2)
Example 3. If P(t )  (0.5,0.866) find P(  t )
Solution:
P(  t )  P(cos(  t ),sin(  t ))
 P( cos(t ), sin(t ))  (0.5, 0.866)
Example 4. If P(t )  (0.5,0.866) find P (360  t )
Solution:
P(360  t )  (cos(360  t ),sin(360  t ))
 (cos t ,  sin t )  (0.5, 0.866)
Example 5. If the point P(t )  ( x,0.866) lies on the unit circle find x rounded to three
decimal places.
Solution: To find x we remember that
cos 2 t  sin 2 t  1
x 2  0.866 2  1
x  1  0.8662  0.5
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3 Trigonometric Functions
Example 6. Find all vales of t in the interval [0, 2 ] satisfying the equation
2 cos 2 t  sin t  2 .
Solution:
2 cos 2 t  sin t  2
sin t  2  2 sin 2 t  2 , using cos 2 t  1  sin 2 t
sin t (1  2 sin t )  0
sin t  0  t  0, 2
sin t 
1
2
 3

t  ,
2
4 4
2
The Graphs of Sine, Cosine and Tangent Functions
Sine function: Remember that the domain of sine function is the entire real line or
(, ) and the range is [-1, 1]. Sine is an odd function. It has origin symmetry.
The sine graph and the period:
The sine graph on the interval [4 , 4 ]
The period of sine function is 2 . sin x  sin(2  x)  sin(4  x)    sin(2n  x)
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3 Trigonometric Functions
Cosine function: The domain of cosine function is the entire real line or (, ) and the
range is [-1, 1]. Cosine is an even function. It has y - axis symmetry.
The cosine graph and the period:
The period of cosine function is also 2 .
cos x  cos(2  x)  cos(4  x)    cos(2n  x)
Tangent Function: The domain of tangent function is the set of all real numbers that are
not the odd multiples of

2
, and the range is (, ) .
The Tangent graph and the period:
The period of tangent function is  . tan x  tan(  x)  tan(2  x)    tan(n  x)
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3 Trigonometric Functions
3.3 Other Trigonometric Functions and Right Triangles
The three main trigonometric functions are sine, cosine, and tangent. In the section we
will discuss three more trigonometric functions cosecant, secant and cotangent, which are
the reciprocal of sine, cosine and tangents functions respectively. For an angle x the
functions cosecant, secant and cotangent are written as csc x,sec x and cot x respectively.
1
h

sin x o
1
h
sec x 

cos x a
1
a
cot x 

sin x o
csc x 
h
o
a
Following are the graphs of csc x,sec x and tan x :
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3 Trigonometric Functions
cot x
Example 1. For the given angle x in the right triangle given in the graph
find the following trigonometric functions:
sin x,cos x, tan x,csc x,sec x,cot x
Solution: From the adjacent figure we have
5
h  5, a  4, then o  3
o 3
1
h 5
3
sin x  
csc x 
 
h 5
sin x o 3
a 4
1
5
cos x  
sec x 

h 5
cos x 4
o 3
cos x a 4
tan x  
cot x 
 
a 4
sin x o 3
x


 19 
Example 2. Find exact value of csc  2   , sec  7  , cot 

3

 6 


 19 
Solution: We may consider the known function like sin  2   , cos  7  , tan 

3

 6 


3
 2 2 3


sin  2    sin 
 csc  2   

3
3
2
3
3
3


cos  7   cos 0  sec(7 ) does not exist
 19
tan 
 6


1


 19
 cot 
  tan  3    tan 
6
6
3


 6

 3

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3 Trigonometric Functions
3.4 Transformations of Trigonometric Functions
In this section we will study the graph and properties of following transformation of
functions. Remember the graphs of sin x,cos x, and tan x we have discussed in section
3.2. We have the following functions to discuss.
y  Asin(Bx  C)  D, y  A cos(Bx  C)  D, y  A tan(Bx  C)  D ,
For the given functions y  Asin( Bx  C)  D, y  A cos(Bx  C)  D
Amplitude  A
2
, B0
B
C
Phase shift 
B
Horizontal shift  D
The period is 
Note: if A is negative the graph has horizontal reflection. Look at example 3.
Example 1. Study the transformation of y  3sin(2x  3)  4
Solution: Find that A = 3, B = 2, C = 3 and D = 4. The amplitude is  A  3
The period of the function is
2 2



B
2
C 3
Phase shift  
B 2
And horizontal shift is = 4
Period
Example 2. Study the transformation of y  3sin(2x  3)  4
Solution: Find that A = 3, B = 2, C = - 3 and D = - 4. The amplitude is  A  3
The period of the function is
2 2



B
2
C
3
Phase shift   
B
2
And horizontal shift is = - 4
Period
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3 Trigonometric Functions
Example 3. Study the transformation of y  3sin(2x  3)  4
Solution: Rewrite the function y  3sin(2x  3)  4  3sin(2x  3)  4
Find that A = -3, B = 2, C = - 3 and D = - 4. The amplitude is  A  3
The period of the function is
2 2



B
2
C
3
Phase shift   
B
2
And horizontal shift is = - 4
We have produced the graph of
y  3sin(2x  3)  4 and y  3sin(2x  3)  4
Period
Graphing y  A tan(Bx  C)  D
We now discuss about y  A tan(Bx  C)  D .

C
Period  , phase shift  , horizontal shift is D.
B
B




 Bx  C  .
2
2
2
2
The x intercept is the mid point between two consecutive asymptotes. Two points having
1
3
y coordinate –A have x coordinates
and of the way between the consecutive
4
4
asymptotes.
Two asymptotes are Bx  C  
, Bx  C 
, within
Example 4. Discuss the graph y  2 tan(0.75x  3)  1
Solution: Find that A = 2, B = 0.75, C = 3 and D = 1.
The period of the function is


B


2
C
3

4
B 0.75
And vertical shift is = 1
Phase shift 
We have produced the graph of y  tan( x) and y  2 tan(0.75x  3)  1
64