Chapter 10
Additional Topics in Trigonometry
1
Summary
• Right Triangle applications
• Law of Sines and Cosines
• Parametric equations
•
Polar coordinates
• Curves in polar coordinates
2
Chapter 10.1
Right Triangle Applications
3
Finding Sides
• We could always find sides if we could find the sin and cosine
• However, there were limited angles for which we could find
those values
• Now we are going to use a calculator to find trig functions for
arbitrary angles
4
Example
• We have a right triangle with sides 3, 4, and 5 m.
• Find the sine and cosine of the angle opposite the
shortest side
5
Solution
• We have a right triangle with sides 3, 4, and 5 m.
• Find the sine and cosine of the angle opposite the
shortest side
• The shortest side is 3, we have sin a = 3/5, need to find a
• Arcsine a = 0.64 or 36.9 deg
6
Example
• The three sides of a right triangle are 5, 12, and 13 m. find the
angle opposite the shortest side
7
Solution
• The three sides of a right triangle are 5, 12, and 13 m. find the
angle opposite the shortest side
• Sin x = 5/13 = =0.38,
8
Angle of Depression and Elevation
Horizon
Angle of
Elevation
Angle of
Depression
Horizon
Tower,
e.g.
9
Example
• Joe is standing on top of a mountain that he knows to be
1000 feet higher than the lake below him. The lake is 500 feet
from the base of the mountain. What is the angle of
depression of his sightline to the lake?
10
Solution
• Arctan(1/2) = 0.46
• Angle =  - 0.46 = 2.68 radians
Horizon
Angle of
Depression
Arctan(1/2)
Mountain,
1000 ft
500 ft
Lake
11
Chapter 10.2
Laws of sines and cosines
12
Derivation of the Law of Sines
C
b
a
h
B
c
A
Sin A = h/b, Sin B = h/a so b Sin A = a Sin B or Sin A/Sin B = a/b
Equivalently, Sin A/a = Sin B/b
Similarly, Sin C/Sin A = c/a, Sin C/c = Sin A/a, etc.
Holds if know ASA, AAS, or SSA, S = Side, A = angle
13
Example
Requires calculator:
Side b = 385 m
Angle B = 47 deg
Angle A = 108 deg
Find angle C, side a, side b
14
Solution
Requires calculator:
Side b = 385 m
Angle B = 47 deg
Angle A = 108 deg
Find angle C, side a, side b
Sin B / b = sin A / a, a = b sin A/sin B = 500
Angle C = 180 – 47 – 108 = 25
c = b sin c/sin b = 222
15
Example
Angle A = 20.4 deg
Angle B = 63.4 deg
Side c = 12.9 m
Find the rest of the sides
16
Solution
Angle A = 20.4 deg
Angle B = 63.4 deg
Side c = 12.9 m
Find the rest of the sides
Angle C = 180 – 20.4 – 63.4 = 96.2
a = c sinA/sin C = 4.5 m
b = c sin B/sin C = 11.6 m
17
SAS
B
A
C
If fix side a, Angle A, side c, then angle C and B and side b
can vary;
SAS does not give a unique solution and may have no
solution
18
Example
A
b = 100, c = 60, C = 28 deg
C
B2
B1
sin B/b = sin C/c
sin B1 = sin 28 (5/3)
B1 = 51.5 deg
B1 + B2 = 180, so B2 = 128.5 deg
Therefore A1 = 100.5 deg. A2 = 23.5 deg
A can be 125.7, or 51.0
19
Example
b = 4 2, c = 8, C = 45 deg
20
Solution
• b = 4 2, c = 8, C = 45 deg
• Sin B = sin (45) (4 2)/8 = ½
• B can be 30 deg or b can be 150 deg
• Since C = 45 deg, B cannot be 150 deg
• Now we can find a and b
• A = 180 – 45 – 30 = 105
• a = 8 sin 45 / sin 105 = 10.9
21
Can this be a triangle?
A = 34 deg
B = 73 deg
C = 52 deg
a = 14
b = 22
c = 18
22
Solution
A = 34 deg
B = 73 deg
C = 52 deg
a = 14
b = 22
c = 18
• 14 = 22 sin(34)/sin(73) = 12.9 NO
23
Can this be a triangle?
A = 42 deg
B = 57 deg
C = 81 deg
a=7
b=9
c = 22
24
Solution
A = 42 deg
B = 57 deg
C = 81 deg
a=7
b=9
c = 22
• 7 + 9 = 16 < 22, not a triangle.
25
Triangles for ?
A = 38 deg
a = 432
b = 382
26
Solution
A = 38 deg
a = 432
b = 382
• 432 = 382 sin (B)/ sin (38)
• Or sin B = (382/432) sin 38 = 0.7; B = 32.9 deg or 147 deg
• Angle C = 180 – 38 – 147 = -5 deg ; not posible
• Or Angle C = 180 – 38 – 32.7 = 109.1 deg
• Is one triangle
27
One or Two Triangles
sin B/5.2 = sin 65/4.9
28
Solution
• sin B/5.2 = sin 65/4.9
• Sin B = 0.96, B = 74 deg or 105 deg
• C = 180 – 74 – 65 = 41 deg
• Or C = 180 – 74 – 105 = 1 deg
• Can be 2 triangles
29
Law of Sines
Sin A /a = Sin B /b = Sin C /c
30
Law of Cosines Derivation
C
b
a
A
y
cos A = x/b or sin A = = y/b
x = b cos A or y = b sin A
B
c
x
a2 = (x-c)2 + y2 = (b cos A – c) 2 + b2 sin2 A
= b2 cos2 A –2bc cos A - c2 + b2 sin2 A
= b2 (cos2 A + sin2 A) + c 2 - 2bc2 cos A
= b2 + c2 – 2bc cos A
31
Law of Cosines (SAS)
a2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
Use when know two sides and an angle or three sides
32
Example
B = 95 deg
a = 16
c=7
Find b:
b2 = a2 + c2 – 2ac cos B = 16x16 + 7x7 – 2x16 x 7 x cos(95)
= 256 + 49 - 272 cos (95) = 324.523
b = 18.01
33
Example Continued
B = 95 deg
a = 16
c=7
b = 18
Sin A/a = Sin B /b = 0.055
A = 62.3 deg
Sin C = 22.6 deg
34
SSS Triangles
a = 15
b = 25
c = 28
c2 = a2 + b2 - 2ab cos C
Cos C = -(c2 - a2 - b2 )/(2ab) = -(784 – 225 – 625)/750
= 0.088
C = 85 deg
Similarly A = 32.3 deg. B = 180 - 323.3 - 85 = 62.7
35
Review
Law of Sines: ASA, SAA, Sometimes SSA - ambiguous
Law of Cosines: SSS, SAS
How about AAA?
36
Review
Law of Sines: ASA, SAA, Sometimes SSA - ambiguous
If know two angles, know all; If know only one angle, possibly
two solutions
Law of Cosines: SSS, SAS
Need to know 3 of 4; 3 sides and one angle
How about AAA?
Similar triangle; need a side to determine lengths
37
Area of a Triangle
We know area = ½ base x height;
C
However, there are many variations on this.
In this triangle, h = b sin A
= ½ bc sin A, similarly
b
a
h
B
c
A
= ½ ab sin C
= ½ ac sin B
The area is one-half the product of two sides times the sine of
the angle between them.
38
Example
Find the area:
a = 16.2 cm
b = 25.6 cm
C = 28.3 deg
39
Solution
Find the area:
a = 16.2 cm
b = 25.6 cm
C = 28.3 deg
Drop perpendicular from Angle A to side b
Area is ½ ab sin C = 171.7 cm
40
Another Area Approach
Requires two angles and one side:
Since area = ½ bc sin A, b = 2 Area /(c sinA)
We have, from Area = ½ ac sin B that a = 2 Area / (csin B)
So, 2 Area = 2Area / (c sin B) x 2 Area/(c sin A) x sin C
Rearranging: c2 sin A sin B = 2 Area sin C
Or, Area = c2 sin A sin B /(2 sin C)
This holds for any combo of sides and angles.
41
Example
a= 34.5 ft
B = 87.9 deg
C = 29.3 deg
42
a= 34.5 ft
B = 87.9 deg
C = 29.3 deg
Area = a2 sin C sin B /(2 sin A)
Angle A = 180 – 87.9 – 29.3 = 62.8
Area = 34.52 sin 29.3 sin 87.9 / sin 62.8 = 654.5 sqft
1190 0.48 1 / 0.89 =
43
Yet another approach
a2 + b2 – 2ab cos C = c2
So a2 + b2 – c2 = 2ab cos C
(a2 + b2 – c2 )/2ab = cos C
Now Area = ½ ab sin C = ½ ab sqrt(1 – cos2 C)
= ½ ab sqrt{ 1- [(a2 + b2 – c2 )/2ab] 2}
This says that we can find the area by only knowing the sides!
44
Chapter 10.5
Parametric Equations
45
Example of Parametric Equations
• Let x = 2t and y =
𝑡2
,
2
t ≥0
• The points x and y give us a location, a point, at any time t
• Let t = 1, x = 2 and y = ½, P is at (2, ½)
• The definitions of x and y are parametric equations, and t is
the parameter
– Often t is time, and the equations constrain an object’s
motion
46
Working with Parametric Equations
• Let x = 2t and y =
𝑡2
,
2
t ≥0
• We often want to eliminate t so have equations in x and y
• Let t = x/2, then y = t2/2 = (x/2)2/2 = x2/4, a parabola
47
Example
• x = 6 cos t, y = 3 sin t, 0 ≤ t ≤ 2π
• Let us see if we can eliminate the parameter t
x/6 = cos t, y/3 = sin t
sin2 t + cos2 t = 1,
so (x/6) 2 + (y/3) 2 = 1
• This is the equation of an ellipse
• The parametric equations for an ellipse are
x = a cos t and y = b sin t
48
In each of these plots, the arrow is the direction of increasing t
49
A Circle and an Ellipse
• A circle is x2 + y2 = r2
• You can also translate a circle:
(x – x1) 2 + (y – y1) 2 = r2
• An ellipse just changes the x/y ratio:
(x/a) 2 + (y/b) 2 = r2
50
Example
• Let x = 1 + 2t and y = 2 + 4t, 0 ≤ t
• Find an equation for x and y for the path traced out
51
Solution
• t = (x-1)/2, so y = 2 + 4 (
𝑥 −1
2
) = 2 + 2x – 2 = 2x
• y = 2x is just a line
52
Example
• Let x = 1+ t, y = t2
53
Solution
• Let x = 1+ t, y = t2
• t = x-1, so y = (x-1)2
54
Example
• x = 2 sin t, y = 3 cos t
55
Example
• x = 2 sin t, y = 3 cos t
• Sin t = x/2, cos t = y/3; cos2t + sin2t = 1 = x/2 + y/3
56
Example
• x = cos3t, y = sin3t, 0 ≤ t ≤ 2 π
• Write in terms of x and y
57
Solution
• x= cos3t, so cos t =
3
𝑥, cos2t =
• Now, cos2t + sin2t = 1, so
3
3
𝑥2 +
𝑥 2 , and similarly for y
3
𝑦2 = 1
58
Example
• x = -4 + 3tan t
• y = 7 – 2 sec t
59
Solution
• x = -4 + 3tan t
• y = 7 – 2 sec t
• Tan t = (x+4)/3
• Sec t = -(y-7)/2
sec2t - tan2t = 1 = (y-7)2/4- (x+4) 2 = 1
60
Solution
• x = 2cos t + 2t, y = 2sint – sin 2t
• Sin 2t = 2 cos t sin t = (y-2 sin t)(x – 2t)/2
3
2
1
0
-2
-1
0
1
2
3
4
-1
-2
-3
61
More Examples
• x = 4 cos 2t, y = 6 sin 2t
• X = 2cos t, y = 2 sin t
• x = 3t + 2, y = 3t -2
• X = 2 tan t, y = 2 cos2t
62
Section 10.6
Polar Coordinates
63
Introduction
• Points are represented in terms of an angle and a radius:
P(r,θ)
• The angle is measured from the positive x axis
• A negative value is a long the same line, but in the opposite
direction
P(r, )
r

P(-r, )
64
Examples
Plot the following given in polar notation:
• (3, 2/3)
• (-5, - /4)
• (0, )
65
Equivalencies
• cos  = x/r so x = r cos 
• sin  = x/r so x = r sin 
• x2 + y2 = r2
• tan  = y/x
66
To Convert from Rectangular to Polar
• Given x, y,
• r=
𝑥2 + 𝑦2
•  = arctan(y/x)
67
Rectangular to Polar
• x = r cos , y = r sin ; usually we want r  0, 0   < 2
• Convert to Polar:
(-1, -1)
(1, 3)
(-2, 0)
68
Solution
• Convert to Polar:
(-1, -1) becomes (1, 5/4)
(1, 3) becomes (2, /3 )
(-2, 0) becomes (2, ) (why  and not zero?)
69
Polar to Rectangular
Convert to rectangular coordinates
• P = (5, 5/6)
• r = 2 cos 
• r=4
70
Solution
• P = (5, 5/6); (5 3/2, 5/2)
• r = 2 cos ; x = r cos , so cos  = x/r. Multiply both sides by r
𝑥 2 + 𝑦 2 = 2x
• r=4;
𝑥2 + 𝑦2 = 4
71
Convert to Rectangular
• r2 = sin 2
72
Solution
• r2 = sin 2
• sin 2 = 2sin cos
• = (2/ r2) x y
• So (x2 + y2 ) 2 = 2xy
73
Converting an Equation
• r = cos + 2sin 
74
Solution
• r = cos + 2sin 
• Multiply both sides by r
• r2 = rcos + 2rsin 
• x2 + y2 = x + 2y
75
Distances
• Distance between two points in polar coordinates,
(r1, 1) and (r2, 2) :
d=
r12 + r22 −2r1r2cos (2− 1)
• See next page for derivation
76
Distance Derivation
• The distance between two points (x1, y1) and (x2 – y2) in
rectangular coordinates is
• Let r1 =
(𝑥1 − 𝑥2)2 +(𝑦1 − 𝑦2)2
𝑥12 + 𝑦12 and r2 =
𝑥22 + 𝑦22 and
x1 = r1 cos 1, etc.
• Then x1 – x2 = r1 cos 1 – r2 cos 2 and
y1 – y 2 = r1 sin 1 – r2 sin 2
• Squaring each, we have:
(x1 – x2)2 = r12 cos2 1 - 2r1r2 cos 1 cos 2 + r22 cos2 2
(y1 – y2)2 = r12 sin2 1 - 2r1r2 sin 1 sin 2 + r22 sin2 2
• Adding gives r12 + r22 - 2r1r2cos 1 cos 2 sin 1 sin 2 =
r12 + r22 -2r1r2cos (2- 1)
77
An example
• Find the distance between (2, 5/6) and (4, /6)
78
Solution
• Find the distance between (2, 5/6) and (4, /6)
• r12 + r22 -2r1r2cos (1- 2) =
• 4 + 16 – 16 cos(5/6 - /6) = 20 – 16 cos (-2/3)
= 20 – 16(-1/2) = 28
• Note: the order of the angles is immaterial since
cos x = cos (-x)
79
Circles
• A circle centered at the origin has the equation r = a, where a
is the radius
• A circle of radius a, centered at (r1, 1) has the equation:
r2 + r12 – 2r r1 cos( - 1) = a2
80
Polar Equation of a Line?
• If a line goes through the origin, then we just have
 = c, where c is a constant
• For an arbitrary line, cos( - ) = d/r or d = r cos( - ), where
d is the distance from the origin to the line and (d, ) is the
foot of the perpendicular to the line
P(r, )
d


x axis
81
Chapter 10.7
Curves in Polar Coordinates
82
Graphing
• Just as with rectangular coordinates, create a table of r, .
A “compass rose” of angles (e.g., the unit circle) might
help
83
Example; Spiral of Archimedes
• Graph r = /

0
/4
/2
3/4

5/4
3/2
7/4
2
9/4
5/2
11/4
3
13/4
7/2
15/4
4
r
0
1/4
1/2
3/4
1
5/4
3/2
7/4
2
9/4
5/2
11/4
12
13/4
7/2
15/4
4
3
2
1
0
-4
-2
0
2
4
6
-1
-2
-3
-4
From r and  you might just find rcos and
rsin and plot x and y
84
Symmetry
1. If in the equation substituting - for  changes nothing, then
the graph is symmetric about the x axis
2. If in the equation substituting - for  and –r for r changes
nothing, then the graph is symmetric about the y axis
3. If in the equation substituting - for  changes nothing, then
the graph is symmetric about the y axis
4. If in the equation substituting –r for r changes nothing, then
the graph is symmetric about the origin
HOWEVER: A graph may be symmetric without these holding
85
Examples
• r2 = 4cos 2 
1. 4cos 2  = 4cos (-2 ), symmetric about x
2. 4cos 2  = 4cos (-2 ), and (-r) 2 = r2, symmetric about y
3. cos (  - ) = cos , symmetric about y
4. Interchanging –r and r no change, symmetric abut origin
86
Which Goes with Which?
1) r = 3 + 3 sin θ
3) r = 3 + 3 cos θ
2) r = 3 -3 cos θ
4) r = 3 – 3 sin θ
C
A
B
D
D
87
Solution
• 1: D
• 2: B
• 3: A
• 4: C
88