Slide 10- 1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

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

Applications of Trigonometry

Chapter 7


7.1 The Law of Sines

Use the law of sines to solve triangles.

Find the area of any triangle given the lengths of two sides and the measure of the included angle.


Solving Oblique Triangles

To solve a triangle means to find the lengths of all its sides and the measures of all its angles.

Any triangle that is not a right triangle is called oblique.

Any triangle, right or oblique, can be solved if at least one side and any other two measures are known.


5 Possible Triangles


The Law of Sines

In any triangle ABC,

Thus in any triangle, the sides are proportional to the sines of the opposite angles.

.sin sin sin

a b c

A B C

B

A C

ac

b


Solving a Triangle (AAS)

In triangle ABC, A = 57, B = 43, and b = 11.2. Solve the triangle.

C = (180° (57° + 43°))

C = 180° 100° = 80°

Find a and c by using the law of sines:

B

A C

ac

11.2

43

57

11.2

sin57 sin 4311.2sin57

sin 4313.8

a

a

a

11.2

sin80 sin 4311.2sin80

sin 4316.2

c

c

c

Therefore,

57 13.8

43 11.2

80 16.2

A a

B b

C c


Solving a Triangle (SSA)

If the given angle is acute, then there may be: a) no solution, b) one solution, or c) two solutions

If the given angle is obtuse, then there may be: a) no solution, or b) one solution


Example

In triangle ABC, b = 8.6, c = 6.2, and C = 35. Solve the triangle.

Solution:

sin sin35

8.6 6.28.6sin35

sin6.2

sin .7956

52.7 , 127.3

B

B

B

B

180

180 52.7 35 92.3

or

180 127.3 35 17.7

A B C

A

A

C

B

A?

a

8.6

6.2

35


Example continued

There are two solutions:

6.2

sin92.3 sin356.2sin92.3

sin3510.8

a

a

a

6.2or

sin17.7 sin356.2sin17.7

sin353.3

a

a

a

92.3 10.8

52.7 8.6

35 6.2

A a

B b

C c

17.7 3.3

127.3 8.6

35 6.2

A a

B b

C c

or


The Area of a Triangle

The area K of any ABC is one half the product of the lengths of two sides and the sine of the included angle:

1 1 1sin sin sin .

2 2 2K bc A ab C ac B


Example

Find the area of the triangle, ABC with A = 72, b = 16, and c = 10.

Solution:

1sin

21

(16)(10)sin 72276.1

K bc A

K

K

10

A

B

C1672


7.2 The Law of Cosines

Use the law of cosines to solve triangles.

Determine whether the law of sines or the law of cosines should be applied to solve a triangle.


In any triangle ABC,

Thus, in any triangle, the square of a side is the sum of the squares of the other two sides, minus twice the product of those sides and the cosine of the included angle. When the included angle is 90, the law of cosines reduces to the Pythagorean theorem.

The Law of Cosines

C

BA

ba

c

2 2 2

2 2 2

2 2 2

2 cos ,

2 cos ,

2 cos .

a b c bc A

b a c ac B

c a b ab C


Solving Triangles (SAS)

Solve ABC if a = 4, c = 6, and B = 105.2.

2 2 2

2

4 6 2(4)(6)cos105.2

64.585

8.0

b

b

b

sin sin105.2

4 84sin105.2

sin8

sin .4825

28.8 or 151.2

151.2, so 28.8

A

A

A

A

A A

Therefore,

28.8 105.2 46

4 8.0 6

A B C

a b c

A

C

B

b

6105.2

4


Solving Triangles (SSS)

Solve ABC if a = 15, b = 11, and c = 8.

Solution: Solve for A first.

15

11

8

A

B

C

2 2 2

2 2 2

15 11 8 2(11)(8)cos

11 8 15cos

2(11)(8)

cos .227

103.1

A

A

A

A


Solving Triangles (SSS) continued

sin sin103.1

11 1511sin103.1

sin15

sin .7142

45.6

B

B

B

B

180

180 103.1 45.6

31.3

C A B

C

C

103.1 15

45.6 11

31.3 8

A a

B b

C c

15

11

8

A

B

C


7.3 Complex Numbers: Trigonometric Form

Graph complex numbers.

Given a complex number in standard form, find trigonometric, or polar, notation; and given a complex number in trigonometric form, find standard notation.

Use trigonometric notation to multiply and divide complex numbers.

Use DeMoivre’s theorem to raise complex numbers to powers.

Find the nth roots of a complex number.


Absolute Value of a Complex Number

The absolute value of a complex number a + bi is

Example: Find the absolute value of:

a) 4 + 5i

b) 1 i

2 2 .a bi a b

2 24 5 4 5 41i

2 21 ( 1) ( 1) 2i


Trigonometric Notation for Complex Numbers

a + bi = r(cos + i sin )

Example: Find trigonometric notation for 1 i.

Solution: First, find r.

Thus,

2 2

2 2( 1) ( 1)

2

r a b

r

r

1 2 1 2sin cos

2 22 25

4

5 51 2 cos sin or 2 cos225 sin 225

4 4i i i


Complex Numbers: Multiplication

For any complex numbers 1 1 1cos sin and r i

2 2 2cos sin ,r i

1 1 1 2 2 2

1 2 1 2 1 2

cos sin cos sin

cos sin .

r i r i

r r i


Example

Multiply and express in standard notation.

4(cos50 sin50 ) and 2(cos10 sin10 )i i

4(cos50 sin50 ) 2(cos10 sin10 )

8 cos(50 10 ) sin(50 10 )

8(cos60 sin 60 )

1 38

2 2

4 4 3

i i

i

i

i

i


Complex Numbers: Division

For any complex numbers 1 1 1cos sin and r i

2 2 2 2cos sin , 0,r i r

1 1 1 1

1 2 1 22 2 2 2

cos sincos sin .

cos sin

r i ri

r i r


Example

Divide and express in standard notation.

16(cos70 sin 70 ) and 4(cos40 sin 40 )i i

16(cos70 sin 70 ) 16 = cos(70 40 ) sin(70 40 )

4(cos40 sin 40 ) 4

4(cos30 sin30 )

3 14

2 2

2 3 2

ii

i

i


DeMoivre’s Theorem

For any complex number

any natural number n,

cos sin and r i

cos sin cos sin .n nr i r n i n


Example

Find (1 i)5.

Solution: First, find trigonometric notation for 1 i.

Then

1 2 cos225 sin 225i i

55

5

1 2 cos225 sin 225

2 cos5(225 ) sin5(225 )

4 2 cos1125 sin1125

2 24 2

2 2

4 4

i i

i

i

i

i


Roots of Complex Numbers

The nth roots of a complex number

, are given by

where k = 0, 1, 2, …, n 1.

cos sin , 0r i r

1/ 360 360cos sin ,nr k i k

n n n n


Example

Find the square roots of .

Trigonometric notation:

For k = 0, the root is For k = 1, the root is

1 3 i

1 3 2 cos60 sin 60i i

1122

60 360 60 3602 cos60 sin 60 2 cos sin

2 2 2 2

2 cos 30 180 sin 30 180

i k i k

k i k

2 cos30 sin30i

2 cos210 sin 210i


7.4 Polar Coordinates and Graphs

Graph points given their polar coordinates.

Convert from rectangular to polar coordinates and from polar to rectangular coordinates.

Convert from rectangular to polar equations and from polar to rectangular equations.

Graph polar equations.


Polar Coordinates

Any point has rectangular coordinates (x, y) and polar coordinates (r, ).

To plot points on a polar graph: Locate the directed angle . Move a directed distance r

from the pole. If r > 0, move along ray OP. If r < 0, move in the opposite direction of ray OP.


Example

Graph each of the following points.

a) A(3, 60) b) B(0, 10) c) C(5, 120) d) D(1, 60)

e) E

f) F

32,

2

4,3


Conversion Formulas


Example

Convert (4, 2) to polar coordinates.

Thus (r, ) =

2 2

2 24 2

16 4

20 2 5

r x y

r

r

r

2 1tan

4 2

26.6

2 5,26.6


Another Example

Convert to rectangular coordinates.

5,4

cos

5cos4

25

2

5 2

2

x r

x

x

x

sin

5sin4

25

2

5 2

2

y r

y

y

y

5 2 5 2

,2 2

(x, y) =


Polar and Rectangular Equations

Some curves have simpler equations in polar coordinates than in rectangular coordinates. For others, the reverse is true.

Convert x + 2y = 10 into a polar equation.

x + 2y = 10cos 2 sin 10

(cos 2sin ) 10

r r

r


Example

Convert r = 3 cos sin into a rectangular equation.

2

2 2

3cos sin

3 cos sin

3

r

r r r

x y x y


Graphing Polar Equations

Graph r = 2 sin

1.414

1

.5176

0

-.5176

-1

-1.414

-1.732

r

225

210

195

180

165

150

135

120

.5176

1

1.414

1.732

1.932

2

1.932

1.732

r

345

330

315

300

285

270

255

240

-1.932105

-290

-1.93275

-1.73260

-1.41445

-1

-.5176

0

r

30

15

0


Graphing with graphing calculator

Graph the equation

r + 2 = 2 sin 2.

Solution solve for r.

2sin 2 2r


7.5 Vectors and Applications

Determine whether two vectors are equivalent.

Find the sum, or resultant, of two vectors.

Resolve a vector into its horizontal and vertical components.

Solve applied problems involving vectors.


Vector

A vector in the plane is a directed line segment. Two vectors are equivalent if they have the same magnitude and direction.

Consider a vector drawn from point A to point B A is called the initial point B is called the terminal point

Equivalent vectors: vectors with the same length and direction.

Magnitude: length of a vector, expressed as

:AB��

.AB��


Example

Find the magnitude of the vector, v, with initial point (1, 2) and terminal point (3, 2).

2 2

2 2

(3 ( 1)) ( 2 2)

4 ( 4)

16 16

32 4 2

v

v

v

v


Vector Addition

The sum of two vectors is called the resultant.

If and = , then + = .AB BC AB BC AC u v u v��


Parallelogram Law

Vector addition is commutative.


Example

Forces of 10 newtons and 50 newtons act on an object at right angles to each other. Find the magnitude of the resultant and the angle of the resultant that it makes with the larger force.

The resultant vector, v, has magnitude 51 and makes an angle of 11.3 with the larger force.

10

50

10

v

2 210 50

100 2500

2600

10 26 51

v

v

v

v

1

10tan

5010

tan50

11.3


Components

Resolving a vector into its vector components—write a vector, w, as a sum of two vectors, u and v, which are the components.

Most often, the two components will be the horizontal component and the vertical component of the vector.


Example

A vector w has a magnitude of 45 and rests on an incline of 20. Resolve the vector into its horizontal and vertical components.

The horizontal component of w is 42.3 right and the vertical component of w is 15.4 up.

cos2045

45cos20

42.3

u

u

u

sin 2045

45sin 20

15.4

v

v

v

v

u

45

20


7.6 Vector Operations

Perform calculations with vectors in component form.

Express a vector as a linear combination of unit vectors.

Express a vector in terms of its magnitude and its direction.

Solve applied problems involving forces in equilibrium.


Position Vectors

Standard position—the initial point is the origin and the terminal point is (a, b).

Notation: v = a is the scalar horizontal component b is the scalar vertical component

Scalar—numerical quantity rather than a vector quantity

,a b


Component Form of a Vector

The component form of with A = (x1, y1) and C = (x2, y2) is

Example: Find the component form of if A = (3, 2) and B = (5, 6).

Solution:

AC��

2 1 2 1, .AC x x y y ��

AB��

5 3, 6 ( 2) 8, 4AB ��


Length of a Vector

The length, or magnitude, of a vector v =

is given by

Equivalent Vectors

Let u =

1 2,v v2 2

1 2 .v v v

1 2 1 2, and = , . Thenu u v vv

1 2 1 2, ,u u v v if and only if u1 = v1 and u2 = v2.


Operations on Vectors


Operations on Vectors continued


Example

a) 4v

b) 2u + v

c) 2u 3v

d) |u + 4v| 4 5, 3 20, 12

2 4,10 5, 3

13,17

2 4,10 3 5, 3

7,29

2 2

4,10 4 5, 3

24, 2

(24) ( 2)

576 4

580 24.1

Let u = and v = . Find each of the following.

5, 34,10


Properties of Vector Addition and Scalar Multiplication


Unit Vectors

A vector of magnitude 1.

If v is a vector and v O, then

is a unit vector in the direction of v.

1, or ,

vv

v v


Example

Find a unit vector that has the same direction as

5,12 . w

2 2( 5) (12)

25 144

169 13

w

w

w

1

131

5,1213

5 12,

13 13

u w


Linear Combinations

i = is a vector parallel to the x-axis. j = is a vector parallel to the y-axis.

The linear combination, v, of unit vectors i and j, where

Example: The linear combination of i and j for

is 2i + 3j.

1,00,1

1 2 1 2, isv v v v v , i j.

2,3


Direction Angles

A vector u in standard position on the unit circle can be written as

where is called the direction angle. It is measured counterclockwise from the positive x-axis.

, cos ,sin (cos ) (sin )x y u i j


Example

Determine the direction angle of the vector w = 3i + 4j.

Solution:

is a 2nd quadrant angle, so = 53.1 + 180 = 126.9.

3 4 3,4

4tan

353.1

w i j


Definitions

The dot product of two vectors u =

and v =

(Note that u1v1 + u2v2 is a scalar, not a vector.)

Angle Between Two Vectors

If is the angle between two nonzero vectors u and v, then cos .

u v

u v

1 2,u u

1 2, isv v 1 1 2 2.u v u v u v


Example

Solution: Find u • v, |u|, |v|.

u • v = 4(3) + 2(1) =14

Find the angle between u = and v = .

2 2( 4) (2)

16 4 20

u

2 2(3) ( 1)

9 1 10

v1

14cos

20 10

14cos

20 10

171.9

u v

u v

4,2 3, 1

Slide 10- 1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

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Transcript of Slide 10- 1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.