Post on 13-Jan-2016
transcript
Slide 10- 1Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Applications of Trigonometry
Chapter 7
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
Slide 10- 4Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
Slide 10- 5Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
5 Possible Triangles
Slide 10- 6Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 7Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 8Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 9Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 10Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 11Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 12Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
Slide 10- 14Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 15Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 16Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 17Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
Slide 10- 19Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 20Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 21Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 22Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 23Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 24Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 25Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 26Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 27Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 28Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
Slide 10- 30Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
Slide 10- 31Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 32Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Conversion Formulas
Slide 10- 33Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 34Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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) =
Slide 10- 35Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 36Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 37Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 38Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Graphing with graphing calculator
Graph the equation
r + 2 = 2 sin 2.
Solution solve for r.
2sin 2 2r
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
Slide 10- 40Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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��������������
Slide 10- 41Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 42Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Vector Addition
The sum of two vectors is called the resultant.
If and = , then + = .AB BC AB BC AC u v u v����������������������������������������������������������������������
Slide 10- 43Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Parallelogram Law
Vector addition is commutative.
Slide 10- 44Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 45Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
Slide 10- 46Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
Slide 10- 48Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 49Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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 ��������������
Slide 10- 50Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
Slide 10- 51Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Operations on Vectors
Slide 10- 52Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Operations on Vectors continued
Slide 10- 53Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 54Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Properties of Vector Addition and Scalar Multiplication
Slide 10- 55Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 56Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 57Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 58Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 59Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 60Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 10- 61Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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