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Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9- 1
Trigonometric Identities,
Inverse Functions,
and Equations
Chapter 9
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
9.1
Identities: Pythagorean and
Sum and Difference

State the Pythagorean identities.

Simplify and manipulate expressions containing
trigonometric expressions.

Use the sum and difference identities to find function
values.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Basic Identities

An identity is an equation that is true for all possible
replacements of the variables.
1
sin x 
,
csc x
1
cos x 
,
sec x
1
tan x 
,
cot x
1
csc x 
,
sin x
1
sec x 
,
cos x
1
cot x 
,
tan x
sin( x)   sin x,
cos( x)  cos x,
tan( x)   tan x,
sin x
tan x 
,
cos x
cos x
cot x 
sin x
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9- 4
Pythagorean Identities
sin x  cos x  1,
2
2
1  cot x  csc x,
2
2
1  tan 2 x  sec 2 x
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Slide 9- 5
Example

Multiply and simplify:
a) sin x(cot x  csc x)
Solution:
sin x(cot x  csc x)
 sin x cot x  sin x csc x
cos x
1
 sin x
 sin x
sin x
sin x
 cos x  1
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Slide 9- 6
Example continued
b) Factor and simplify:
Solution:
sin 4 x  sin 2 x cos 2 x
sin x  sin x cos x
4
2
2
 sin 2 x(sin 2 x  cos 2 x)
 sin 2 x  (1)
 sin 2 x
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Slide 9- 7
Another Example

Simplify the following
trigonometric expression:
Solution:
cos x
cos x

1  sin x 1  sin x
cos x(1  sin x)
cos x(1  sin x)

(1  sin x)(1  sin x) (1  sin x)(1  sin x)
cos x  sin x cos x cos x  sin x cos x


2
1  sin 2 x
1  sin x
2cos x

1  sin 2 x
2cos x

cos 2 x
2
or 2sec x

cos x
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Slide 9- 8
Sum and Difference Identities

There are six identities here, half of them obtained by
using the signs shown in color.
sin(u  v)  sin u cos v  cos u sin v,
cos(u  v)  cos u cos v sin u sin v,
tan u  tan v
tan(u  v) 
1 tan u tan v
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Slide 9- 9
Example

Find sin 75 exactly.
sin 75  sin(30  45 )
 sin 30 cos 45  cos30 sin 45
1 2
 
2 2
2


4
2

4
3 2


2 2
6
4
6
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Slide 9- 10
9.2
Identities: Cofunction,
Double-Angle, and Half-Angle

Use cofunction identities to derive other identities.

Use the double-angle identities to find function values of
twice an angle when one function value is known for
that angle.

Use the half-angle identities to find function values of
half an angle when one function value is known for that
angle.

Simplify trigonometric expressions using the doubleangle and half-angle identities.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Cofunction Identities




sin   x   cos x,
2



tan   x   cot x,
2



cos   x   sin x,
2



cot   x   tan x,
2



sec   x   csc x,
2



csc   x   sec x
2

Cofunction Identities for the Sine and Cosine


sin  x     cos x
2



cos  x    sin x
2

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Slide 9- 12
Example



Find an identity for cot  x   .
2

Solution:


cos  x  

2


cot  x   

2


sin  x  
2

sin x

 cos x
  tan x
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Slide 9- 13
Double-Angle Identities
sin 2 x  2sin x cos x,
2 tan x
tan 2 x 
1  tan 2 x
cos 2 x  cos 2 x  sin 2 x
 1  2sin 2 x
 2cos 2 x  1
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Slide 9- 14
Example

Find an equivalent expression for cos 3x.
Solution:
cos3 x  cos(2 x  x)
 cos 2 x cos x  sin 2 x sin x
 (1  2sin 2 x)cos x  2sin x cos x sin x
 cos x  2sin 2 x cos x  2sin 2 x cos x
 cos x  4sin 2 x cos x
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Slide 9- 15
Half-Angle Identities
x
1  cos x
sin  
,
2
2
x
1  cos x
cos  
,
2
2
x
1  cos x
tan  
2
1  cos x
sin x
1  cos x


1  cos x
sin x
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Slide 9- 16
Example

Find sin ( /8) exactly.
Solution:

sin 4  
2

1  cos

4
2
2
2
1
2
2 2

2
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Slide 9- 17
Another Example

x
Simplify tan tan x  1.
2
x
1  cos x sin x
tan
tan
x

1


1
Solution:
2
sin x cos x
sin x(1  cos x)

1
sin x cos x
1  cos x

1
cos x
1
cos x


1
cos x cos x
 sec x  1  1
 sec x
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Slide 9- 18
9.3
Proving
Trigonometric Identities

Prove identities using other identities.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
The Logic of Proving Identities

Method 1: Start with either the left or the right side of
the equation and obtain the other side.

Method 2: Work with each side separately until you
obtain the same expression.
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Slide 9- 20
Hints for Proving Identities






Use method 1 or 2.
Work with the more complex side first.
Carry out any algebraic manipulations, such as adding,
subtracting, multiplying, or factoring.
Multiplying by 1 can be helpful when rational
expressions are involved.
Converting all expressions to sines and cosines is often
helpful.
Try something! Put your pencil to work and get
involved. You will be amazed at how often this leads to
success.
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Slide 9- 21
Example

Prove the identity (tan 2 x  1)(cos 2 x  1)   tan 2 x .
Solution: Start with the left side.
(tan 2 x  1)(cos 2 x  1)   tan 2 x
 sin 2 x 
2
2
x
tan


1)

x
(cos
1



2
 cos x 
sin 2 x
2
2
2
x
tan


1

x
cos

sin x 
2
cos x
sin 2 x
2
2
2
x
tan


1

sin x  cos x 
2
cos x
sin 2 x
2
1

1


tan
x
2
cos x
sin 2 x
2



tan
x
2
cos x
 tan 2 x   tan 2 x
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Slide 9- 22
Another Example

Prove the identity:
1
 csc x  sin x
sec x tan x
Solution: Start with the
right side.
1
 csc x  sin x
sec x tan x
1

 sin x
sin x
1
sin 2 x


sin x sin x

Solution continued
1
1  sin 2 x

sec x tan x
sin x
cos 2 x

sin x
cos x cos x


sin x 1
 cot x  cos x
1
1


tan x sec x
1
1

sec x tan x sec x tan x
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Slide 9- 23
Example

tan x  cot y
 tan y  cot x.
Prove the identity
tan x cot y

Solution: Start with the left side.
tan x  cot y
 tan y  cot x
tan x cot y
tan x
cot y


tan x cot y tan x cot y
1
1


cot y tan x
tan y  cot x  tan y  cot x
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Slide 9- 24
9.4
Inverses of the Trigonometric
Functions

Find values of the inverse trigonometric functions.

Simplify expressions such as sin (sin–1 x) and
sin–1 (sin x).

Simplify expressions involving compositions such as
sin (cos–1 21 ) without using a calculator.

Simplify expressions such as sin arctan (a/b) by
making a drawing and reading off appropriate ratios.
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Inverse Trigonometric Functions
Function
Domain
Range
y  sin 1 x
 arcsin x, where x  sin y
[1, 1]
[ / 2,  / 2]
y  cos 1 x
 arccos x, where x  cos y
[1, 1]
[0, ]
y  tan 1 x
 arctan x, where x  tan y
(, )
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  / 2, / 2 
Slide 9- 26
Example

Find each of the
following:
a) sin 1
Solution: a)
 Find  such that
3
sin  
2
3
2

3
1
b) cos  

 2 


c) tan 1 (1)
 would represent a 60°
or 120° angle.
3
sin
 60 and 120
2

2
or
and
3
3
1
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Slide 9- 27
Solution continued
b) Find  such that
 3
cos 
2



 would represent a 30°
reference angle in the 2nd and
3rd quadrants.
Therefore,  = 150° or 210°

3
cos  
  150 and 210
 2 
5
7
or
and
6
6
1
c) Find  such that
tan   1.



This means that the sine and
cosine of  must be
opposites.
Therefore,  must be 135°
and 315°.
tan 1 (1)  135 and 315
3
7
or
and
4
4
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Slide 9- 28
Domains and Ranges
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Slide 9- 29
Composition of Trigonometric Functions
sin(sin 1 x)  x,
for all x in the domain of sin 1 .
cos(cos 1 x)  x,
for all x in the domain of cos 1 .
tan(tan 1 x)  x,
for all x in the domain of tan 1.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9- 30
Examples

Simplify:

 1  1  
sin  sin    
 2 


Since 1/2 is in the
domain of sin–1,
 1  1  
1
sin  sin      
2
 2 

Simplify:
 1  3 2  
cos  cos 
 

 2 


3 2
Since
is not in the
2
domain of cos–1,
 1  3 2  
cos  cos 
 

 2 

does not exist.
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Slide 9- 31
Special Cases
sin 1 (sin x)  x,
for all x in the range of sin 1 .
cos 1 (cos x)  x,
for all x in the range of cos 1 .
tan 1 (tan x)  x,
for all x in the range of tan 1.
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Slide 9- 32
Examples

Simplify:

 
sin 1  sin 
2


Since /2 is in the range
of sin–1,
  
sin 1  sin  
2 2

Simplify:


tan  tan 
3

1

Since /3 is in the range
of tan–1,
 

tan 1  tan  
3 3

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Slide 9- 33
More Examples

Simplify:

 1 
sin  cos 1    
 2 

Solution:
2
 1
cos 1     120 or
3
 2
2
3
sin

3
2

Simplify:
  2  
tan 1  sin  

3

 
Solution:
2
3
sin

3
2

3
1
tan  
  40.9
 2 
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Slide 9- 34
9.5
Solving
Trigonometric Equations

Solve trigonometric equations.
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Solving Trigonometric Equations

Trigonometric Equation—an equation that contains a
trigonometric expression with a variable.

To solve a trigonometric equation, find all values of the
variable that make the equation true.
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Slide 9- 36
Example


Solve 2 sin x  1 = 0.
Solution: First, solve for
sin x on the unit circle.
2sin x  1  0
2sin x  1

1
sin x 
2
6
1
x  sin 1  
 2
x  30 ,150 or
The values /6 and
5/6 plus any multiple
of 2 will satisfy the
equation. Thus the
solutions are
 5
 2 k and
5
 2 k
6
where k is any integer.
,
6 6
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Slide 9- 37
Graphical Solution

We can use either the Intersect method or the Zero
method to solve trigonometric equations. We graph the
equations y1 = 2 sin x  1 and y2 = 0.
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Slide 9- 38
Another Example

Solve 2 cos2 x  1 = 0.
Solution: First, solve for cos x on the unit circle.
x  45 ,135 , 225
 3 5
or ,
,
,
4 4
4
 3
2cos 2 x  1  0
2cos 2 x  1
1
2
cos x 
2
cos x  
The values
1
2
2
cos x  
2
,
,
,315
7
4
5 7
,
4 4
4
4
any multiple of 2 will satisfy
the equation.
plus
The solution can be written
as

4


2
k where k is any integer.
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Slide 9- 39
Graphical Solution


Solve 2 cos2 x  1 = 0.
One graphical solution shown.
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Slide 9- 40
One More Example

Solve 2 cos x + sec x = 0
Solution:
2cos x  sec x  0
1
2cos x 
0
cos x
1
2cos 2 x  1  0

cos x
1
0
cos x
1
0
cos x
or
Since neither factor of the equation
can equal zero, the equation has no
solution.

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
2cos 2 x  1  0
2cos 2 x  1
1
2
cos x 
2
1
cos x   
2
1
cos x  
2
Slide 9- 41
Graphical Solution

2 cos x + sec x
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Slide 9- 42
Last Example


Solve 2 sin2 x + 3sin x + 1 = 0.
Solution: First solve for sin x on the unit circle.
2sin x  3sin x  1  0
(2sin x  1)(sin x  1)  0
2
2sin x  1  0
2sin x  1
1
sin x  
2
1
1 
x  sin   
 2
7 11
x
,
6 6
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Slide 9- 43
Last Example continued

sin x  1  0
sin x  1

One Graphical Solution
1
x  sin ( 1)
3
x
2

x
7
11
3
 2 k ,
 2 k ,
 2 k
6
6
2
where k is any integer.
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Slide 9- 44
9.5
Rotation of Axes


Use rotation of axes to graph conic sections.
Use the discriminant to determine the type of conic
represented by a given equation.
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Rotation of Axes
When B is nonzero, the graph of
Ax 2  Bxy  Cy2  Dx  Ey  F  0
is a conic section with an axis that is not parallel to the
x– or y–axis. Rotating the axes through a positive angle
 yields an x y  coordinate system.
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Slide 9- 46
Rotation of Axes Formulas
If the x– and y–axes are rotated about the origin
through a positive acute angle then the coordinates
of (x, y)and ( x , y )of a point P in the xy  and x y 
coordinates system are related by the following
formulas.
x   x cos  ysin  : y  x sin   y cos ,
x  x  cos  y sin  : y  x  sin   y cos .
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Slide 9- 47
Example
Suppose that the xy  axes are rotated through an angle of 45º.
Write the equation xy  1 in the x y  coordinate system.
Solution:
Substitute 45º for  in the formulas
x  x  cos  y sin  : y  x  sin   y cos .
x  x  cos 45º  y sin 45º : y  x  sin 45º  y cos 45.
2
2
2
x  x
 y

x  y 
2
2
2
2
2
2
y  x
 y

x  y 
2
2
2
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Slide 9- 48
Example continued
Now, substitute these into the equation
xy  1.
2
2


x

y


 x  y    1
2
2
1
2
2
 1


x

y





2
2
2


x
y
     1
2
2
2
2


x    y  1
2
2
2
2
   
This is the equation of a hyperbola in the x y  coordinate system.
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Slide 9- 49
Eliminating the xy-Term
To eliminate the xy-term from the equation
Ax  Bxy  Cy  Dx  Ey  F  0, B  0,
2
2
select an angle  such that
AC
cot 2 
, 0º  2  180º ,
B
and use the rotation of axes formulas.
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Slide 9- 50
Example
Graph the equation 3x 2  2 3xy  y 2  2x  2 3y  0.
Solution: We have
A  3, B  2 3, C  1, D  2, E  2 3, F  0.
31
2
1
cot 2 


, 2  120º ,   60º
2 3 2 3
3
x  x  cos 60º  y sin 60º , y  x  sin 60º  y cos 60º
1
3 x  y 3
x  x   y
 
2
2
2
2
3
1 x  3 y
y  x
 y 

2
2
2
2
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Slide 9- 51
Example continued
Substitute these into 3x 2  2 3xy  y2  2x  2 3y  0
and simplify.
4 y   4 x   0
2
y    x
2
Parabola with vertex at (0, 0) in the x y  coordinate
system and axis of symmetry y  0.
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Slide 9- 52
The Discriminant
2
The expression B  4AC is the discriminant of
the equation Ax 2  Bxy  Cy2  Dx  Ey  F  0.
The graph of the the equation
Ax 2  Bxy  Cy 2  Dx  Ey  F  0
is, except in degenerate cases,
1. an ellipse or circle if B2  4AC  0,
2
2. a hyperbola if B  4AC  0, and
2
3. a parabola if B  4AC  0.
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Slide 9- 53
Example
Graph the equation 3x 2  2xy  3y2  16.
Solution:
We have A  3, B  2, C  3,
so
B2  4 AC  22  4 3 3  4  36  32.
The discriminant is negative so it’s a circle or ellipse.
A  C 3 3
cot 2 

 0 2  90º   45º
B
2
Determine

Substitute into the rotation formulas
x  x  cos 45º  y sin 45º : y  x  sin 45º  y cos 45º .
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9- 54
Example continued
x  x  cos 45º  y sin 45º , y  x  sin 45º  y cos 45º
2
2
2
x  x
 y

x  y 
2
2
2
2
2
2
y  x
 y

x  y 
2
2
2
After substituting into the equation and simplifying we
have:
2
2
4 x   2 y   16
x 
2
4
y 


2
8
 1.
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Slide 9- 55
Example continued
x 
2
y 


2
1
is an ellipse,
4
8
with vertices 0,2 2 and 0,2 2 on the y-axis
and x   intercepts 2, 0  and 2, 0 .




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Slide 9- 56
9.6
Polar Equations of Conics



Graph polar equations of conics.
Convert from polar to rectangular equations of conics.
Find polar equations of conics.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
An Alternative Definition of Conics
Let L be a fixed line (the directrix); let F be a fixed point (the
focus), not on L; and let e be a positive constant (the
eccentricity). A conic is the set of all points P in the plane such
that
PF
 e,
PL
where PF is the distance from P to F and PL is the
distance from P to L. The conic is a parabola if e = 1,
an ellipse if e < 1, and a hyperbola if e > 1.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9- 58
Polar Equations of Conics
To derive equations position focus F at the pole, and
the directrix L either perpendicular or parallel to the
polar axis. In this figure L is perpendicular to the polar
axis and p units to the right of the pole.
Note: PL  p  r cos
PF  r
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Slide 9- 59
Polar Equations of Conics
Simplified:
PL  p  r cos
PF  r
PF
r

e
PL p  r cos
ep
r
1  ecos
For an ellipse and a hyperbola, the eccentricity e is
given by e = c/a, where c is the distance from the
center to a focus and a is the distance from the center
to a vertex.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9- 60
Example
Describe and graph the conic r 
18
.
6  3cos
Solution:
3
r
, e  0.5,
1 0.5 cos
p6
Since e < 1, it’s an ellipse with vertical directrix 6 units to the
right of the pole. The major axis lies along the polar axis.
Let  = 0, and π to find vertices: (2, 0) and (6, π).
The center is (2, π).
The major axis = 8, so a = 4.
Since e = c/a, then 0.5 = c/4, so c = 2
The length of the minor axis is given by b:
b=
a 2  c 2  4 2  2 2  16  4  12  2 3
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9- 61
Example continued
Sketch the graph r 
18
.
6  3cos
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Slide 9- 62
Polar Equations of Conics
A polar equation of any of the four forms
ep
r
,
1  ecos
ep
r
1  esin 
is a conic section. The conic is a parabola if e = 1, an
ellipse if 0 < e < 1, and a hyperbola if e >1.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9- 63
Polar Equations of Conics
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Slide 9- 64
Converting from Polar to
Rectangular Equations
Use the relationships between polar and rectangular
coordinates.
Remember:
x  r cos , y  r sin  , r  x  y
2
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2
2
Slide 9- 65
Example
2
.
Convert to a rectangular equation: r 
1 sin 
Solution:
2
We have
r
1  sin 
r  r sin   2
r  r sin   2
Substitute.
x 2  y2  y  2
x 2  y 2  y 2  4y  2
x 2  4y  4  0
This is the equation of a parabola.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9- 66
Finding Polar Equations of Conics
We can find the polar equation of a conic with a focus
at the pole if we know its eccentricity and the
equation of the directrix.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9- 67
Example
Find a polar equation of a conic with focus at the
pole, eccentricity 1/3 and directrix r  2 csc .
Solution:
2
r  2 csc 
or r sin   2, which is y  2.
sin 
Choose an equation for a directrix that is a
horizontal line above the polar axis and substitute
e = 1/3 and p = 2
1 2
ep
2
3
r


1
1 esin  1 sin  3  sin 
3
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Slide 9- 68
9.7
Parametric Equations




Graph parametric equations.
Determine an equivalent rectangular equation for
parametric equations.
Determine parametric equations for a rectangular
equation.
Determine the location of a moving object at a specific
time.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Graphing Parametric Equations
We have graphed plane curves that are composed of
sets of ordered pairs (x, y) in the rectangular
coordinate plane. Now we discuss a way to represent
plane curves in which x and y are functions of a third
variable t.
One method will be to construct a table in which we
choose values of t and then determine the values of x
and y.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9- 70
Example
Graph the curve represented by the equations
1
x  t,
y  t 2  3;
 3  t  3.
2
The rectangular equation is y  4x  3,
2
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
3
3
 x .
2
2
Slide 9- 71
Parametric Equations
If f and g are continuous functions of t on an interval I,
then the set of ordered pairs (x, y) such that x = f(t)
and y = g(t) is a plane curve.
The equations x = f(t) and y
equations for the curve.
= g(t) are parametric
The variable t is the parameter.
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Slide 9- 72
Determining a Rectangular Equation
for Given Parametric Equations
Solve either equation for t.
Then substitute that value of t into the other equation.
Calculate the restrictions on the variables x and y
based on the restrictions on t.
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Slide 9- 73
Example
Find a rectangular equation equivalent to
x  t 2 , y  t 1; 1  t  4
Solution
y  t 1
t  y 1
Calculate the restrictions:
1  t  4
Substitute t  y  1 into x  t 2 .
x  t 2 ; 0  x  16
y  t  1;  2  y  3
x  y  1
2
The rectangular equation is: x  y  1 ; 0  x  16.
2
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Slide 9- 74
Determining Parametric Equations
for a Given Rectangular Equation
Many sets of parametric equations can represent the
same plane curve. In fact, there are infinitely many
such equations.
The most simple case is to let either x (or y) equal t
and then determine y (or x).
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9- 75
Example
Find three sets of parametric equations for the
2
parabola y  4  x  3 .
Solution
If x  t, then y  4  t  3  t  6t  5.
2
2
If x  t  3, then y  4  t  3  3  t 2  4.
2
2
2
t
t

t
If x  , then y  4    3    2t  5.
3 
9
3
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9- 76
Applications
The motion of an object that is propelled upward can be
described with parametric equations. Such motion is
called projectile motion. It can be shown that,
neglecting air resistance, the following equations
describe the path of a projectile propelled upward at an
angle  with the horizontal from a height h, in feet, at
an initial speed v0, in feet per second:
x  v0 cos t,
y  h  v0 sin t  16t 2 .
We can use these equations to determine the location
of the object at time t, in seconds.
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Slide 9- 77
Example
A baseball is thrown from a height of 6 ft with an initial
speed of 100 ft/sec at an angle of 45º with the
horizontal.
a)
b)
c)
d)
e)
Find parametric equations that give the position of
the ball at time t, in seconds.
Find the height of the ball after 1 sec, 2 sec and 3
sec.
Determine how long the ball is in the air.
Determine the horizontal distance that the ball
travels.
Find the maximum height of the ball.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9- 78
Example continued
Solution
a)
x  v0 cos t,
y  h  v0 sin  t  16t 2
x  100 cos 45º t y  6  100 sin 45º t  16t 2
x  50 2t
y  16t 2  50 2t  6
b) The height of the ball at time t is represented by y.
2
If t  1, y  16 1  50 2 1  6  60.7 ft.
If t  2, y  16 2   50 2 2   6  83.4 ft.
2
If t  3, y  16 3  50 2 3  6  74.1 ft.
2
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Slide 9- 79
Example continued
Solution
c) The ball hits the ground when y = 0.
y  16t 2  50 2t  6
0  16t 2  50 2t  6
t
50 2 
50 2   4 16 6 
2
2 16 
t   0.1 or t  4.5
The ball is in the air for about 4.5 sec.
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Slide 9- 80
Example continued
Solution
d) Substitute t = 4.5
x  50 2t  50 2 4.5   318.2 ft.
So the horizontal distance the ball travels is 318.2 ft.
e) Find the maximum value of y (vertex).
b
50 2
t

 2.2
2a
2 16 
y  16 2.2   50 2 2.2   6  84.1 ft.
2
So the maximum height is about 84.1 ft.
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Slide 9- 81
Applications
The path of a fixed point on the circumference of a
circle as it rolls along a line is called a cycloid. For
example, a point on the rim of a bicycle wheel traces a
cycloid curve.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9- 82
Applications
The parametric equations of a cycloid are
x  a t  sint ,
y  a 1 cost ,
where a is the radius of the circle that traces the
curve and t is in radian measure.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9- 83
Example
The graph of the cycloid described by the parametric
equations
x  3t  sint , y  31 cost ; 0  t  6
is shown below.
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Slide 9- 84