chapter 6 additional topics in trigonometry. 6.1 law of sines objectives –use the law of sines to...

Chapter 6

ADDITIONAL TOPICS IN TRIGONOMETRY

6.1 Law of Sines

• Objectives– Use the Law of Sines to solve oblique

triangles– Use the Law of Sines to solve, is possible, the

triangle or triangles in the ambiguous case– Find the area of an oblique triangle using the

sine function– Solve applied problems using the Law of

Sines

Law of Sines• Previously, our relationships between sides of a

triangle and the angles were unique only to RIGHT triangles

• What about other triangles? Obtuse or acute ones? (oblique – not right!)

• The following relationship exists (A,B,C are measures of the 3 angles; a,b,c are the lengths of sides opposite those angles):

C

c

B

b

A

a

sinsinsin

Solving an oblique triangle• If given: A = 50 degrees, B = 30 degrees, b = 7 cm.

Can you solve this triangle? If so, is the solution unique?

• You know C = 100 degrees (Now that you know the measure of all 3 angles and the length of 1 side, how many triangles exist? Only one!)

• You can find “a” & “c” by law of sines:

cmccmc

cmacma

8.1330sin

7

100sin

7.10,30sin

7

50sin

What if given: a=5”,b=7”,B=45 degrees

• Law of Sines indicates sinA=.505. There are 2 angles,A, such that sin(A) = .505. A = 30 degrees, OR A = 150 degrees (WHY? Look at your unit circle!)

• Are there 2 possible triangles? NO – in this case, a (5”)is smaller than b (7”), and if angle A = 150 degrees, it must be opposite the longest side of the triangle. Clearly, it is not, therefore only 1 triangle exists. (continued)

505.sin,45sin

"7

sin

"5

A

A

Example continued

• If A = 30 degrees & B = 45 degrees, C=105 degrees

• Use law of sines again to find c.

6.9,105sin45sin

"7

c

c

What if given: a=7”,b=5”,B=45 degrees

• Law of Sines indicates sinA=.99. There are 2 angles,A, such that sin(A) = .99. A = 82 degrees, OR A = 98 degrees (WHY? Look at your unit circle!)

• Are there 2 possible triangles? YES – in this case, a (7”)is larger than b (5”), and if angle A = 82 or 98 degrees, it is a larger angle than B. Clearly, there are 2 triangles that exist. (continued)

99.sin,45sin

"5

sin

"7

A

A

2 possible triangles

• Triangle 1: use law of sines to find c:

• Triangle 2: use law of sines to find c:

53,45,82 CBA

37,45,98 CBA

"6.5,53sin45sin

"5

c

c

"3.437sin45sin

"5

c

c

Finding area of an oblique triangle

• Using Law of Sines it can be found that for any triangle, height (h) = b sin A (if c is considered to be the base), therefore Area= ½ c b sinA

6.2 Law of Cosines

• Objectives

– Use the Law of Cosines to solve oblique triangles

– Solve applied problems using the Law of Cosines

– Use Heron’s formula to find the area of a triangle.

What if know a=4”,b=6”,C=70 degrees?

• Law of Sines does NOT apply• Law of Cosines was developed:

• Use this to solve for c in the given triangle:

• b=c, so C = B = 70 degrees, thus A = 40 degrees

"6.,.35

)70cos()"6)("4(2)"6()"4(

cos2

2

222

222

cinsqc

c

Cabbac

Using Law of Cosines, you can solve a triangle with 3 given sides

• If the 3 sides are given, only 1 such triangle exists

Heron’s Formula for Area of a Triangle

• If the 3 sides of a triangle are known, the area can be found (based on Law of Cosines):

)22

1(

))()((

cbaperimeters

csbsassArea

6.3 Polar Coordinates• Objectives

– Plot points in the polar coordinate system– Find multiple sets of polar coordinate for a given

point– Convert a point from polar to rectangular

coordinates– Convert a point from rectangular to polar

coordinates– Convert an equation from rectangular to polar

coordinates– Convert an equation from polar to rectangular

coordinates

Defining points in the polar system

• Location of a point is based on radius (distance from the origin) and theta (the angle the radius moves from standard position (positive x-axis in a cartesian system))

• Any point can be described many ways. i.e. 2 units out moving pi/2 is the same as a radius of 2 units moving -3pi/2 or a radius of 2 units moving 5pi/2

What is the relationship between cartesian coordinates & polar ones?

• The radius = r is the hypotenuse of a rt. triangle that has base = x & height=y

• Thus,

• If x = horizontal leg & y = vertical leg of a right triangle, then

222 yxr

x

ytan

Find rectangular coordinates for

)2,2)(4

)2,2)(3

)2,2)(2

)1,1)(1

)4

3,2(

Convert a rectangular equation to a polar equation

7,7

7)sin(cos

7sincos

7)sin()cos(

7

4,1/tan

2

222

2222

22

22

rr

r

rr

rr

yx

xy

xy

6.4 Graphs of Polar Equations

• Objectives

– Use point plotting to graph polar equations

– Use symmetry to graph polar equations

Graphing by Point Plotting

• Given a function, in polar coordinates, you can find corresponding values for “r” and “theta” that will make your equation true.

• Plotting several points and connecting the points with a curve provides the a graph of the function.

• Example next page 2sin2r

Example:

• Put values in for theta that range from 0 to 2pi (once around the circle..after that the values begin repeating)

2sin2r

Continued example

3,0,002sin2,

6

5,2

1,

2

1

2

12

6

5sin2,

6

5

3

2,2

3,

2

3

2

32

3

2sin2,

3

2

3,2

3,

2

3

2

32

3sin2,

3

3,2

3,

2

3

2

32

3sin2,

3

6,2

1,

2

1)2

1(2

6sin2,

6

)0,0(,00sin2,0

22

22

2

2

2

2

2

2

22

2

rr

rr

rr

rr

rr

rr

rr

“Special” curves generated by general forms

)2cos(2sin:

))cos()(sin(:

)cos(sin:lim

)sin(cos:

2222

ararlemniscate

narnarrose

barbaracon

ararcircle

r=3cos(theta)

r=2+3sin(theta)

r=3cos(5theta)

2sin92 r

6.5 Complex Numbers in Polar Form: DeMoivre’s Theorem

• Objectives– Plot complex numbers in the complex plane– Fine absolute value of a complex #– Write complex # in polar form– Convert a complex # from polar to

rectangular form– Find products & quotients of complex

numbers in polar form– Find powers of complex # in polar form– Find roots of complex # in polar form

Complex number = z = a + bi

• a is a real number

• bi is an imaginary number

• Together, the sum, a+bi is a COMPLEX #

• Complex plane has a real axis (horizontal) and an imaginary axis (vertical)

• 2 – 5i is found in the 4th quadrant of the complex plane (horiz = 2, vert = -5)

• Absolute value of 2 – 5i refers to the distance this pt. is from the origin (continued)

Find the absolute value

• Since the horizontal component = 2 and vertical = -5, we can consider the distance to that point as the same as the length of the hypotenuse of a right triangle with those respective legs

29)5(252 22 i

Expressing complex numbers in polar form

• z = a + bi

)sin(cos

sin,cos

22

irz

bar

rbra

Express z = -5 + 3i in complex form

)54.sin()54.(cos(34

54.,5

3tan

343)5( 22

iz

r

Product & Quotient of complex numbers

)sin()cos(

))sin()(cos(

)sin(cos),sin(cos

21212

1

2

1

21212121

22221111

ir

r

z

z

irrzz

irzirz

Multiplying complex numbers together leads to raising a complex

number to a given power• If r is multiplied by itself n times, it creates

• If the angle, theta, is added to itself n times, it creates the new angle, (n times theta)

• THUS, ))sin()(cos( ninrz nn

nr

Taking a root (DeMoivre’s Theorem)

• Taking the nth root can be considered as raising to the (1/n)th power

• Now finding the nth root of a complex # can be expressed easily in polar form

• HOWEVER, there are n nth roots for any complex number & they are spaced evenly around the circle.

• Once you find the 1st root, to find the others, add 2pi/n to theta until you complete the circle

If you’re working with degrees add 360/n to the angle measure to complete the circle.

• Example: Find the 6th roots of z= -2 + 2i• Express in polar form, find the 1st root, then add

60 degrees successively to find the other 5 roots.

)5.322sin5.322(cos8

)5.262sin5.262(cos8

)5.202sin5.202(cos8

)5.142sin5.142(cos8

)5.82sin5.82(cos8

)5.22sin5.22(cos8)6

135sin

6

135(cos8

)135sin135(cos22,135,2

2tan

606

360,222)2(

126

126

126

126

126

1266

22

iz

iz

iz

iz

iz

iiz

iz

r

6.6 Vectors• Objectives

– Use magnitude & direction to show vectors are equal

– Visualize scalar multiplication, vector addition, & vector subtraction as geometric vectors

– Represent vectors in the rectangular coordinate system

– Perform operations with vectors in terms of i & j– Find the unit vector in the direction of v.– Write a vector in terms of its magnitude & direction– Solve applied problems involving vectors

Vectors have length and direction• Vectors can be represented on the rectangular

coordinate system• Vectors have a horizontal & a vertical component• A vector starting at the origin and extending left 2

units and down 3 units is given below, along with the magnitude of the vector (distance from the origin):

22)2()2(

223,2

22

v

jiv

Adding & subtracting vectorsScalar Multiplication

• Add (subtract) horizontal components together & add (subtract) vertical components together

• v = 3i + 7j, w = -i + 2j, v – w = 4i + 5j• Scalar Multiplication: multiply the i & j

components by the constant• v = 3i + 7j, 4v = 12i + 28j (the new vector is 4

times as long as the original vector)

Unit Vector has the same direction as a given vector, but is 1 unit long

• Unit vector = (original vector)/length of vector

• Simply involves scalar multiplication once the length of the vector is determined (recall the length = length of hypotenuse if legs have lengths = a & b)

• Given vector, v = -2i + 7j, find the unit vector:

jijiunit

length

53

537

53

532

53

7

53

2:

537)2( 22

Writing a Vector in terms of its Magnitude & Direction

• v is a nonzero vector. The vector makes an angle measured from the positive x-axis to v, and we can talk about the magnitude & direction angle of this vector:

jvivvthus

v

b

v

a

sincos:

sin,cos

Velocity Vector: vector representing speed & direction of object in

motion• Example: The wind is blowing 30 miles per

hour in the direction N20 degrees E. Express its velocity as a vector v.

• If the wind is N20 degrees E, it’s 70 degrees from the positive x-axis, so the angle=70 degrees and the magnitude is 30 mph.

jijiv

jiv

2.283.10)94(.30)34(.30

)70sin(30)70cos(30

Resultant Force

• Adding 2 force vector together

• Add horizontal components together, add vertical components together

6.7 Dot Product• Objectives

– Find dot product of 2 vectors

– Find angle between 2 vectors

– Use dot product to determine if 2 vectors are orthogonal

– Find projection of a vector onto another vector

– Express a vector as the sum of 2 orthogonal vectors

– Compute work.

Definition of Dot Product

• The dot product of 2 vectors is the sum of the products of their horizontal components and their vertical components

2121

2211 ,

bbaawv

jbiawjbiav

Find the dot product of v&w if v=3i+j and w= -2i - j

1. 7

2. -5

3. -7

4. -4

Properties of Dot Product

• If u,v, & w are vectors and c is scalar, then

)()())(5

)4

00)3

)()2

)1

2

cvuvucvcu

vvv

v

wuvuwvu

uvvu

Angle between vectors, v and w

wv

wvcos

Parallel Vectors

• Parallel: the angle between the vectors is either 0 (the vectors on top of each other) or 180 (vectors are in opposite directions), in either case, cos(0)=1, cos(180) = -1, this will be true if the dot product of v & w = (plus/minus)product of their magnitudes

Orthogonal Vectors

• Vectors that are perpendicular to each other. The angle between vectors is 90 degrees or 270 degrees. cos(90)=cos(270)=0

• Since they are orthogonal if the numerator = 0, thus the dot products of the 2 vectors = 0 if they are orthogonal

wv

wvcos

Vector projection of v onto w

ww

wvvprojw 2

Work done by a force F moving an object from A to B

• Force and distance are both vectors

cosBAFBAFW