Slide 9- 1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 9- 1Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Trigonometric Identities, Inverse Functions,

and Equations

Chapter 9


9.1Identities: 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.


Basic Identities

An identity is an equation that is true for all possible replacements of the variables.

1 1sin , csc ,

csc sin1 1

cos , sec ,sec cos

1 1tan , cot ,

cot tan

x xx x

x xx x

x xx x

sin( ) sin ,

cos( ) cos ,

tan( ) tan ,

sintan ,

coscos

cotsin

x x

x x

x x

xx

xx

xx


Pythagorean Identities

2 2

2 2

2 2

sin cos 1,

1 cot csc ,

1 tan sec

x x

x x

x x


Example

Multiply and simplify:

a)

Solution:

sin (cot csc )x x x

sin (cot csc )

sin cot sin csc

cos 1sin sin

sin sincos 1

x x x

x x x x

xx x

x xx


Example continued

b) Factor and simplify:

Solution:

4 2 2sin sin cosx x x

4 2 2

2 2 2

2

2

sin sin cos

sin (sin cos )

sin (1)

sin

x x x

x x x

x

x


Another Example

Simplify the following

trigonometric expression:

Solution:

cos cos

1 sin 1 sin

x x

x x

2 2

2

2

cos (1 sin ) cos (1 sin )

(1 sin )(1 sin ) (1 sin )(1 sin )

cos sin cos cos sin cos

1 sin 1 sin2cos

1 sin2cos

cos2

or 2seccos

x x x x

x x x x

x x x x x x

x xx

xx

x

xx


Sum and Difference Identities

There are six identities here, half of them obtained by using the signs shown in color.

sin( ) sin cos cos sin ,

cos( ) cos cos sin sin ,

tan tantan( )

1 tan tan

u v u v u v

u v u v u v

u vu v

u v


Example

Find sin 75 exactly.

sin 75 sin(30 45 )

sin30 cos45 cos30 sin 45

1 2 3 2

2 2 2 2

2 6

4 4

2 6

4


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 double-angle and half-angle identities.


Cofunction Identities

Cofunction Identities for the Sine and Cosine

sin cos , cos sin ,2 2

tan cot , cot tan ,2 2

sec csc , csc sec2 2

x x x x

x x x x

x x x x

sin cos cos sin2 2

x x x x


Example

Find an identity for

Solution:

cot .2

x

cos2

cot2 sin

2

sin

costan

xx

x

x

xx


Double-Angle Identities

2

sin 2 2sin cos ,

2 tantan 2

1 tan

x x x

xx

x

2 2

2

2

cos2 cos sin

1 2sin

2cos 1

x x x

x

x


Example

Find an equivalent expression for cos 3x.

Solution:

2

2 2

2

cos3 cos(2 )

cos2 cos sin 2 sin

(1 2sin )cos 2sin cos sin

cos 2sin cos 2sin cos

cos 4sin cos

x x x

x x x x

x x x x x

x x x x x

x x x


Half-Angle Identities

1 cossin ,

2 2

1 coscos ,

2 2

1 costan

2 1 cossin 1 cos

1 cos sin

x x

x x

x x

xx x

x x


Example

Find sin ( /8) exactly.

Solution: 1 cos

4 4sin2 2

21

22

2 2

2


Another Example

Simplify .

Solution:

tan tan 12

xx

1 cos sintan tan 1 1

2 sin cossin (1 cos )

1sin cos

1 cos1

cos1 cos

1cos cossec 1 1

sec

x x xx

x xx x

x xx

xx

x xx

x


9.3Proving

Trigonometric Identities Prove identities using other identities.


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.


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.


Example

Prove the identity .

Solution: Start with the left side.

2 2 2(tan 1)(cos 1) tanx x x

2 2 2

22 2

2

22 2 2

2

22 2 2

2

(tan 1)(cos 1) tan

sin1 (cos 1) tan

cos

sinsin cos 1 tan

cos

sinsin cos 1 tan

cos

x x x

xx x

x

xx x x

x

xx x x

x

22

2

22

2

2 2

sin1 1 tan

cos

sintan

cos

tan tan

xx

x

xx

x

x x


Another Example

Prove the identity:

Solution: Start with the right side.

Solution continued1

csc sinsec tan

x xx x

2

1csc sin

sec tan1

sinsin

1 sin

sin sin

x xx x

xx

x

x x

2

2

1 1 sin

sec tan sin

cos

sincos cos

sin 1cot cos

1 1

tan sec1 1

sec tan sec tan

x

x x x

x

xx x

xx x

x x

x x x x


Example

Prove the identity .

Solution: Start with the left side.

tan cottan cot

tan cot

x yy x

x y

tan cottan cot

tan cot

tan cot

tan cot tan cot

1 1

cot tan

tan cot tan cot

x yy x

x y

x y

x y x y

y x

y x y x


9.4Inverses 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 ) without using a calculator.

Simplify expressions such as sin arctan (a/b) by making a drawing and reading off appropriate ratios.

1

2


Inverse Trigonometric Functions

[0, ][1, 1]

[1, 1]

RangeDomainFunction

1sin

arcsin , where sin

y x

x x y

1cos

arccos , where cos

y x

x x y

1tan

arctan , where tan

y x

x x y

( , )

[ / 2, / 2]

/ 2, / 2


Example

Find each of the following:

a)

b)

c)

Solution: a) Find such that

would represent a 60° or 120° angle.

1 3sin

2

1 3cos

2

1tan ( 1)

3sin

2

1 3sin 60 and 120

22

or and 3 3


Solution continued

b) Find such that

would represent a 30° reference angle in the 2nd and 3rd quadrants.

Therefore, = 150° or 210°

c) Find such that

This means that the sine and cosine of must be opposites.

Therefore, must be 135° and 315°.

3cos

2

1 3cos 150 and 210

2

5 7or and

6 6

tan 1.

1tan ( 1) 135 and 315

3 7or and

4 4


Domains and Ranges


Composition of Trigonometric Functions

1 1

1 1

1 1

sin(sin ) , for all in the domain of sin .

cos(cos ) , for all in the domain of cos .

tan(tan ) , for all in the domain of tan .

x x x

x x x

x x x


Examples

Simplify:

Since 1/2 is in the domain of sin–1,

Simplify:

Since is not in the

domain of cos–1,

1 1sin sin

2

1 1 1sin sin

2 2

1 3 2cos cos

2

3 2

2

1 3 2cos cos

2

does not exist.


Special Cases

1 1

1 1

1 1

sin (sin ) , for all in the range of sin .

cos (cos ) , for all in the range of cos .

tan (tan ) , for all in the range of tan .

x x x

x x x

x x x


Examples

Simplify:

Since /2 is in the range of sin–1,

Simplify:

Since /3 is in the range of tan–1,

1sin sin2

1tan tan3

1sin sin2 2

1tan tan3 3


More Examples

Simplify:

Solution:

Simplify:

Solution:

1 1sin cos

2

1 1 2cos 120 or

2 3

2 3sin

3 2

1 2tan sin

3

1

2 3sin

3 2

3tan 40.9

2


9.5 Solving

Trigonometric Equations Solve trigonometric equations.


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.


Example

Solve 2 sin x 1 = 0. Solution: First, solve for

sin x on the unit circle.

The values /6 and 5/6 plus any multiple of 2 will satisfy the equation. Thus the solutions are

where k is any integer.1

2sin 1 0

2sin 1

1sin

21

sin2

530 ,150 or ,

6 6

x

x

x

x

x

52 and 2

6 6k k


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.


Another Example

Solve 2 cos2 x 1 = 0.

Solution: First, solve for cos x on the unit circle.2

2

2

2cos 1 0

2cos 1

1cos

2

1cos

2

2cos

2

x

x

x

x

x

45 ,135 ,225 ,315

3 5 7 , , ,

4 4 4 4

x

or

3 5 7The values , , , plus

4 4 4 4any multiple of 2 will satisfy

the equation.

The solution can be written

as where is any integer.4 2

k k


Graphical Solution

Solve 2 cos2 x 1 = 0. One graphical solution shown.


One More Example

Solve 2 cos x + sec x = 0

Solution:

2

2cos sec 0

12cos 0

cos1

2cos 1 0cos

x x

xx

xx

2

2

2

2cos 1 0

2cos 1

1cos

2

1cos

2

x

x

x

x

1cos

2x

Since neither factor of the equation can equal zero, the equation has no solution.

10 or

cos1

0cos

x

x


Graphical Solution

2 cos x + sec x


Last Example

Solve 2 sin2 x + 3sin x + 1 = 0. Solution: First solve for sin x on the unit circle.

22sin 3sin 1 0

(2sin 1)(sin 1) 0

x x

x x

1

2sin 1 0

2sin 1

1sin

21

sin2

7 11,

6 6

x

x

x

x

x


Last Example continued

where k is any integer.

One Graphical Solution

1

sin 1 0

sin 1

sin ( 1)

3

2

x

x

x

x

7 11 32 , 2 , 2

6 6 2x k k k


9.5Rotation of Axes

Use rotation of axes to graph conic sections. Use the discriminant to determine the type of conic

represented by a given equation.


Rotation of Axes

When B is nonzero, the graph of

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 coordinate system.

Ax2 Bxy Cy2 Dx Ey F 0

x y


Rotation of Axes Formulas

If the x– and y–axes are rotated about the origin through a positive acute angle then the coordinates of and of a point P in the and coordinates system are related by the following formulas.

x x cos ysin : y x sin ycos,

x x cos y sin : y x sin y cos.

(x, y) ( x , y ) xy x y

jlb

The text includes the statement about the origin."are rotated about the origin through a positive acute angle"

jlb

The scripte places a colon after formulas, a comma at the end of the first line of formulas,and a period at the end.

jlb

There should be no "s" after second coordinate.It should be "coordinate systems"


Example

Suppose that the axes are rotated through an angle of 45º.

Write the equation in the 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.

x x2

2 y

2

2

2

2x y

y x2

2 y

2

2

2

2x y

xy x y xy 1


Example continued

Now, substitute these into the equation

2 2

2 2

2 2

2 2

2 21

2 21

12

12 2

12 2

x y x y

x y

x y

x y

This is the equation of a hyperbola in the coordinate system. x y

xy 1.


Eliminating the xy-Term

To eliminate the xy-term from the equation

select an angle such that

Ax2 Bxy Cy2 Dx Ey F 0, B 0,

cot 2 A C

B, 0º 2 180º ,

and use the rotation of axes formulas.


Example

Graph the equation

Solution: We have

3x2 2 3xy y2 2x 2 3y 0.

cot 2 3 1

2 3

2

2 3

1

3, 2 120º , 60º

A 3, B 2 3, C 1, D 2, E 2 3, F 0.

x x cos60º y sin 60º , y x sin60º y cos60º

x x1

2 y

3

2

x

2

y 3

2

y x3

2 y

1

2

x 3

2

y

2


Example continued

Substitute these into

and simplify.

3x2 2 3xy y2 2x 2 3y 0

4 y 2 4 x 0

y 2 x

Parabola with vertex at (0, 0) in the coordinate system and axis of symmetry

x y y 0.


The Discriminant

The expression

Ax2 Bxy Cy2 Dx Ey F 0.

B2 4AC

is, except in degenerate cases,

is the discriminant of

the equation

The graph of the the equationAx2 Bxy Cy2 Dx Ey F 0

1. an ellipse or circle if

2. a hyperbola if and

3. a parabola if

B2 4AC 0,

B2 4AC 0,

B2 4AC 0.

jlb


Example

Graph the equation 3x2 2xy 3y2 16.

so

Solution:

We have A 3, B 2, C 3,

The discriminant is negative so it’s a circle or ellipse.

Determine

Substitute into the rotation formulas

2 24 2 4 3 3 4 36 32.B AC

cot 2 A C

B

3 3

20 2 90º 45º

x x cos 45º y sin 45º : y x sin 45º y cos 45º .


Example continued

After substituting into the equation and simplifying we have:

4 x 2 2 y 2 16

x 2

4

y 2

81.

x x cos 45º y sin 45º , y x sin 45º y cos 45º

x x2

2 y

2

2

2

2x y

y x2

2 y

2

2

2

2x y


Example continued

with vertices

and

0, 2 2 and 0,2 2 on the y -axis

x intercepts 2,0 and 2,0 .

x 2

4

y 2

81 is an ellipse,


9.6Polar Equations of Conics

Graph polar equations of conics. Convert from polar to rectangular equations of conics. Find polar equations of conics.


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

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.

PF

PLe,


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



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.

PL p r cos PF r

PF

PL

r

p r cose

r ep

1 ecosSimplified:


Example

Describe and graph the conic r 18

6 3cos.

Solution:

r 3

1 0.5 cos, e 0.5, 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:

a2 c2 42 22 16 4 12 2 3b =


Example continued

Sketch the graph r 18

6 3cos.



A polar equation of any of the four forms

is a conic section. The conic is a parabola if e = 1, an ellipse if 0 < e < 1, and a hyperbola if e >1.

r ep

1ecos, r

ep

1esin




Converting from Polar to Rectangular Equations

Use the relationships between polar and rectangular coordinates.

Remember:

x r cos, y r sin, r2 x2 y2


Example

Convert to a rectangular equation: r 2

1 sin.

Solution:

We have r 2

1 sinr r sin 2

r r sin 2 Substitute.

x2 y2 y 2

x2 y2 y2 4y 2

x2 4y 4 0

This is the equation of a parabola.


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.


Example

Find a polar equation of a conic with focus at the pole, eccentricity 1/3 and directrix

r ep

1 esin

13 2

1 13sin

2

3 sin

r 2csc.

Solution:

r 2 csc 2

sin or r sin 2, which is y 2.

Choose an equation for a directrix that is a horizontal line above the polar axis and substitutee = 1/3 and p = 2


9.7Parametric 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.


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.


Example

Graph the curve represented by the equations

x 1

2t, y t 2 3; 3 t 3.

The rectangular equation is y 4x2 3, 3

2x

3

2.


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 = g(t) are parametric equations for the curve.

The variable t is the parameter.


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.


Example

Find a rectangular equation equivalent to

Solution

x t 2 , y t 1; 1 t 4

The rectangular equation is:

y t 1

t y 1

Substitute t y 1 into x t 2 .

x y 1 2

Calculate the restrictions:

1 t 4

x t 2; 0 x 16

y t 1; 2 y 3

x y 1 2; 0 x 16.


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).


Example

Find three sets of parametric equations for the parabola y 4 x 3 2

.

Solution

If x t, then y 4 t 3 2 t 2 6t 5.

If x t 3, then y 4 t 3 3 2 t 2 4.

If x t

3, then y 4

t

3 3

2

t

9

2

2t 5.


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.


ExampleA 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) Find parametric equations that give the position of the ball at time t, in seconds.

b) Find the height of the ball after 1 sec, 2 sec and 3 sec.

c) Determine how long the ball is in the air.d) Determine the horizontal distance that the ball

travels.e) Find the maximum height of the ball.


Example continued

Solution

a)

b) The height of the ball at time t is represented by y.

x v0 cos t, y h v0 sin t 16t 2

x 100 cos 45º t y 6 100sin 45º t 16t 2

x 50 2t y 16t 2 50 2t 6

If t 1, y 16 1 2 50 2 1 6 60.7 ft.

If t 2, y 16 2 2 50 2 2 6 83.4 ft.

If t 3, y 16 3 2 50 2 3 6 74.1 ft.


Example continued

Solution

c) The ball hits the ground when y = 0.

The ball is in the air for about 4.5 sec.

y 16t 2 50 2t 6

0 16t 2 50 2t 6

t 50 2 50 2 2

4 16 6 2 16

t 0.1 or t 4.5


Example continued

Solution

d) Substitute t = 4.5

So the horizontal distance the ball travels is 318.2 ft.

e) Find the maximum value of y (vertex).

So the maximum height is about 84.1 ft.

x 50 2t 50 2 4.5 318.2 ft.

t b

2a

50 2

2 16 2.2

y 16 2.2 2 50 2 2.2 6 84.1 ft.


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.


Applications

The parametric equations of a cycloid are

x a t sin t , y a 1 cos t ,where a is the radius of the circle that traces the curve and t is in radian measure.


Example

The graph of the cycloid described by the parametric equations

x 3 t sin t , y 3 1 cos t ; 0 t 6

is shown below.

Date post:	03-Jan-2016
Category:	Documents
Upload:	claribel-carter
View:	215 times
Download:	0 times

Slide 9- 1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Documents