Download Handout 7 - UGA Math Department

Name:
MATH 1113 Precalculus
Eric Perkerson
Date: November 2, 2014
Handout No. 7
Section 5.1: Angles
An angle is given by two rays emanating from the same point. The ray the angle starts from is called the
initial side, while the ray the angle ends at is called the terminal side. Positive angles move counterclockwise
from the initial side to the terminal side, while negative angles move clockwise from the initial side to the
terminal side. Below are some examples of angles
We say that an angle is in standard position if the point the rays emanate from is the origin, (0, 0), and if
the initial side is the positive x-axis. Below are some examples of angles in standard position
35 Degrees
180 Degrees
-45 Degrees
120 Degrees
y
y
y
y
x
x
x
x
Multiple angles can have the same initial and terminal sides. Angles that share initial and terminal sides
are called coterminal angles. Note that all coterminal angles differ by an integer number of revolutions of
the circle. Below are some examples of coterminal angles in standard position
-300 Degrees
60 Degrees
420 Degrees
780 Degrees
y
y
y
y
x
x
x
x
Angles can be measured in either radians (abbreviated by rad) or degrees (abbreviated by deg or ◦ ),
where
One revolution of a circle = 2π radians = 360 degrees
◦
This relation implies that we can convert radians to degrees by multiplying by 360
2π , and degrees to radians
2π
by multiplying by 360
.
Radians
are
the
default
measure
of
angles,
and
are
unit-less,
which means that the
◦
“unit” we denote by rad can be omitted.
Why do we use radians at all? In part because of the following properties: The arclength of
Example: Convert 261.2◦ to radians.
The conversion factor above comes from the equation 2π = 360◦ , and thus we can write
1=
So to convert 261.2◦ to radians, we multiply by 1 =
261.2◦
2π
360◦
2π
360◦ ,
so we have
2π
≈ 4.5588 rad = 4.5588 (radians are unit-less, remember.)
360◦
Example: Find a negative angle coterminal with the angle of 261.2◦ in standard position.
Coterminal angles must differ by an integer number of revolutions (e.g. 1 revolution, 2 revolutions, 0
revolutions, -1 revolutions, etc.). Because one revolution is 360◦ , -1 revolutions from the terminal side of our
angle of 261.2◦ gives an angle with 261.2◦ − 360◦ = −98.8◦ . Therefore, the angle in standard position with
measure −98.8◦ is coterminal with the angle of 261.2◦ in standard position.
We generally prefer to use radians over degrees because formulas written in radians are much simpler and
easier to work with. For example, we can derive formals for arclength of pieces of a circle and for the areas
of “pie shaped” sections of disks.
A formula for the arclength of a section of a circle: The circumference Cr of a circle with radius
r is Cr = 2πr. If the angle encompassing the section of the circle we want the length of is θ, measured in
radians, then the arclength of the section s is given by s = rθ (See the picture below).
A formula for the area of a section of a circle: The area Ar of a circle with radius r is Ar = πr2 . If
the angle encompassing the section of the circle we want the area of is θ, measured in radians, then the area
of the section A is given by A = 12 θr2 (See the picture below).
r
r
θ
s
θ
A
WARNING: The angle θ in the preceding two formulas must be in radians!
Example: Supposing that two cities, A and B, are located at the same longitude, and that city A is at
latitude 10◦ N and city B is at latitude 70◦ N, find the distance between cities A and B. (Assume the radius
of the earth is 4000 mi).
The distance between cities A and B is the length of the shortest path between them along the surface of
the earth (a sphere). The solution to this problem is to use the formula for arclength of a section of a circle
above, s = rθ. To use this formula, we have to find a circle that both cities are located on, i.e. the longitude
they are both on (longitude “lines” are really circles going through the north and south poles). Looking at
the picture below, r is the radius of the earth, θ is the angle between the two cities, φ is the angle from the
equator to city A, and s is the distance between the two cities. To solve for θ in radians, we find θ in degrees
2π
by taking 70◦ − 10◦ = 60◦ and then multiplying by 360
to convert to radians. Once we have θ is radians,
s = rθ, so s = 4000(60 ∗ π/180)
City B
s
θ
r
ϕ
City A
Sections 5.2 and 5.3: Trigonometric Functions
The trigonometric functions arise as the ratios of the sides of a right triangle. Given a right triangle as shown
below with an angle given as x, the hypotenuse labeled “hyp,” the side of the triangle opposite x labeled
“opp,” and the side of the triangle adjacent to x not the hypotenuse labeled “adj,” we have:
sin(x) =
opp
hyp
cos(x) =
adj
hyp
tan(x) =
opp
adj
csc(x) =
hyp
opp
sec(x) =
hyp
adj
cot(x) =
adj
opp
hyp
x
hyp
opp
adj
x
adj
opp
A useful mnemonic for the first row is SOHCAHTOA, where Sin is Opposite over Hypotenuse, etc. The sec1
1
1
ond row consists of reciprocals of the first row, as csc(x) = sin(x)
, sec(x) = cos(x)
, and tan(x) = tan(x)
.
WARNING:
1
sin(x)
is not the same as sin(1/x),
1
cos(x)
is not the same as cos(1/x), and so forth.
The graphs of the six trig functions we use are shown below:
sin(x)
cos(x)
1
-
3π
2
-π
-
1
π
π
2
2
π
3π
2
-
3π
2
-π
-
3π
2
-π
-
1
π
π
2
2
π
3π
2
-
3π
2
-π
-
π
π
2
2
-1
-1
-1
csc(x)
sec(x)
cot(x)
1
-
tan(x)
1
π
π
2
2
-1
π
3π
2
-
3π
2
-π
-
3π
2
1
π
π
2
2
-1
π
π
3π
2
-
3π
2
-π
-
π
π
2
2
-1
π
3π
2
The domains and ranges of the six trig functions we use are shown below:
Function
sin(x)
cos(x)
tan(x)
csc(x)
sec(x)
cot(x)
Domain
(−∞, ∞)
(−∞, ∞)
(−∞, ∞)
(−∞, ∞)
(−∞, ∞)
(−∞, ∞)
Range
[−1, 1]
[−1, 1]
(−∞, ∞)
(−∞, −1] ∪ [1, ∞)
(−∞, −1] ∪ [1, ∞)
(−∞, ∞)
except π2 + kπ for integers k
except kπ for integers k
except π2 + kπ for integers k
except kπ for integers k
Perhaps the most important interpretation of the functions sin(x) and cos(x) is that the point (cos(x), sin(x))
is the point on the unit circle (circle centered at the origin with radius 1) that is also on the terminal side
of the angle in standard position with measure x. See the picture below:
1.0
(cos(x), sin(x))
0.5
x
-1.0
0.5
-0.5
1.0
1.5
2.0
-0.5
-1.0
Example: Find all values of sin(t) given that cos(t) = 57 .
Because the point (cos(t), sin(t)) is a point on the unit circle, we intersect the unit circle and the vertical line
given by x = 57 . This gives us two intersection points which we can use the Pythagorean theorem to solve for.
p
Using the fact that the hypotenuse has length 1 and the x-coordinate is 5/7, we have that y = 1 − (5/7)2 ,
and the other solution is −y.
1
y
x
-x
5
7
1
The Pythagorean identities also arise from the fact that the point (cos(x), sin(x)) is on the unit circle. Using
the distance formula, we have that
cos2 (x) + sin2 (x) = 1
This is the first Pythagorean identity. To get the other two we do the following:
cos2 (x) + sin2 (x) = 1
1
cos2 (x) + sin2 (x)
=
cos2 (x)
cos2 (x)
cos2 (x)
sin2 (x)
1
+
=
2
2
cos (x) cos (x)
cos2 (x)
1 + tan2 (x) = sec2 (x)
and
cos2 (x) + sin2 (x) = 1
cos2 (x) + sin2 (x)
1
=
2
2
sin (x)
sin (x)
cos2 (x) sin2 (x)
1
+
=
sin2 (x)
sin2 (x)
sin2 (x)
cot2 (x) + 1 = csc2 (x)
So the three Pythagorean identities are
cos2 (x) + sin2 (x) = 1
1 + tan2 (x) = sec2 (x)
cot2 (x) + 1 = csc2 (x)