Download Radian Measure Given any circle with radius r, if θ is a central angle

Radian Measure
Given any circle with radius r, if θ is a central angle of the circle and s
is the length of the arc sustained by θ, we define the radian measure
of θ by:
s
θ=
r
For a semi-circle with radius r, its circumfrence is πr, so the radian
measure of a semi-circle (a straight line) is
πr
=π
θ=
r
Since a (semi-circle) straight angle has measure 180◦ , π radian is equivalent to 180◦ .
π
Given an angle measurement in degree, multiply that number by
180◦
to find the radian measure.
180◦
Given an angle measurement in radian, multiply that number by
π
to find the degree measure.
E.g.
What is 55◦ in radian?
Ans:
55◦ ·
55
11
π
=
π
=
π ∼ 0.31π ∼ 0.96
180◦
180
36
Unless a decimal approximation is desired for some reason, we should
11
11π
always leave the number in exact format. That is,
π or
is the
36
36
most desirable way of writing the above angle measure in radian.
The most important point of note here is that radian measure is
a (real) number, and is more canonical than degree measure when
used in any application.
If an angle measurement is written without the ◦ symblo in the upper
right, it is a radian measure. If the ◦ is present, it is a degree measure.
Distingush the difference between 1◦ and 1 (radian).
1
E.g.
2π
radian expressed in degree measure?
3
2π 180◦
360◦
·
=
= 120◦
3
π
3
What is
With the Cartesian plane, we can define an angle in Standard Position if it has its vertex on the origin and one of its sides on the
x−axis. If we obtained the other side (Called the Terminal Side) of
the angle via a counter-clockwise rotation, we have a positive angle. If
the terminal side of the angle is obtained via a clockwise rotation, we
have a negative angle.
Using this definition, it is possible to define an angle of any (positive or negative) measurement by recognizing how its terminal side is
obtained.
π
E.g. The terminal side of is in the first quadrant.
4
π
E.g. The terminal side of − is in the fourth quadrant.
6
2π
E.g. The terminal side of
is in the second quadrant.
3
4π
E.g. The terminal side of
is in the third quadrant.
3
16π
E.g. The terminal side of
is in the first quadrant.
7
E.g. The terminal side of π is the negative x−axis.
π
E.g. The terminal side of − is the negative y−axis.
2
π
5π
E.g. The two angles and −
are co-terminal. This means they
3
3
have the same terminal side.
Notice that if θ is any angle, then the angles θ + 2π, θ + 4π, θ + 6π, θ +
8π, · · · all have the same terminal side as θ.
2
Given a Circle with radius r centered at the origin (The equation
of this circle is x2 +y 2 = r2 ), we define the terminal point of an angle
θ to be the point of intersection of the circle with the terminal side of
θ.
π
E.g. The terminal point P of θ = is in the first quadrant.
6
If the radius
of thecircle is r = 1, then the coordinate of this terminal
√
3 1
point P is  , 
2 2
If the radius
of the circle
is r = 3, then the coordinate of this terminal
 √

3 3 3
point P is 
,
2 2
π
E.g. The terminal point of θ = is on the positive side of the y−axis.
2
If the radius of the circle is r = 1, then the coordinate of the terminal
point of θ is (0, 1).
Note:
Any co-terminal angle has the same terminal point.
If θ is an angle with terminal point P , then any angle of the form
θ + 2π, θ + 4π, θ + 6π, θ + 8π, · · · all have P as the terminal point.
To find the coordinate of the terminal point, we need to know the
radius of the circle and the measurement of the angle.
3
Definition of the Trigonometric Functions:
Given an angle θ in standard position, on a circle with radius r, and
the terminal point of θ is P , and (x, y) is the coordinate of P , we define
the six trigonometric functions of θ by:
x
cos θ =
This is the cosine function.
r
sin θ =
y
r
This is the sine function.
tan θ =
y
x
This is the tangent function.
sec θ =
r
x
This is the secant function.
csc θ =
r
y
This is the cosecant function.
cot θ =
x
y
This is the cotangent function.
Sometimes, for convenience, we assume a circle of radius r = 1, called a
unit circle, when defining or evaluating the values of the trigonometric
functions.
Note that we only need the measurement of the angle (we do not need
to know the radius of the circle) to find the values of the six tri functions
(because the r cancles out when we take the ratio of x or y with each
other or with r.
4
π
, find the values of the six tri functions of θ.
3
Assuming
√  a circle of radius 1, the terminal point P of θ has coordinate

1 3
 ,
, so the values of the tri functions would be:
2 2
E.g For θ =
1
x
1
= 2 =
r
1 2
√
√
3
y
3
sin θ = = 2 =
r
1
2
cos θ =
√
y
tan θ = =
x
sec θ =
3
2
1
2
=
√
3
r
1
= 1 =2
x
2
√
r
2
2 3
1
csc θ = = √3 = √ =
y
3
3
2
x
cot θ = =
y
1
√2
3
2
√
1
3
=√ =
3
3
5
◦
◦
◦
30 − 60 − 90
45◦ − 45◦ − 90◦
π π
− − π triangles and
6 3
!
π π
− − π triangles
4 4
!
To find the values of a trigonometric function of a particular angle,
we usually need to use a calculator, which gives us decimal values of
the tri function value of the angle. However, for some special angles,
π π π
namely, , , , we can find the exact values of the six tri functions
6 4 3
of these angle without the help of a calculator.
◦
For a 30◦ − 60◦ − 90◦ triangle, if the side opposite the 30
angle has
√
◦
length x, then the side opposite the 60 angle has length 3x, and the
hypotenuse has length 2x
◦
For a 45◦ − 45◦ − 90
√ triangle, the two sides have the same length and
the hypotenuse is 2 times the length of the side. I.e, if any one of
the side has length x, √
then the other side also has length x and the
hypotenuse has length 2x
Using this fact, we can find the exact value of the tri functions on some
of the special angles.
6
E.g.
5π
π
, its reference angle is , and the
6  √
6

3 1
terminal point P of θ has coordinate − , . Therefore,
2 2
!
5π
1
sin
=
6
2
√
!
3
5π
cos
=−
6
2
√
!
1
3
1
5π
= 2√3 = − √ = −
tan
6
3
3
−
Assuming a unit circle, if θ =
2
7
E.g.
π
π
Assuming a unit circle, if θ = − , its reference angle is , and the
4 √
4
√ 
2
2
terminal point P of θ has coordinate  , − . Therefore,
2
2
√
!
2
π
=−
sin −
4
2
√
!
π
2
cos −
=
4
2
√
− 22
π
tan −
= √2 = −1
4
!
2
8
Foundamental Properties of The Trigonometri Functions:
sin and csc are reciprocal functions of each other, that is:
1
csc(x) =
sin(x)
cos and sec are reciprocal functions of each other:
1
sec(x) =
cos(x)
tan and cot are reciprocal functions of each other:
1
cot(x) =
tan(x)
In addition, tan is the quotient of sin and cos:
tan(x) =
sin(x)
cos(x)
cot(x) =
cos(x)
sin(x)
The following equation can be derived from the pythagorean theorem,
and is called the Pythagorean Identity: For all real numbers x, we
have:
sin2 (x) + cos2 (x) = 1
Note: sin2 (x) means (sin x)2 . In general, to represent (sin x)n , we write
sinn (x). This notation applies to other tri functions too.
9
More properties of the tri functions:
sin (and its reciprocal, csc), is an odd function, that is,
sin(−x) = − sin(x) for all real numbers x
cos (and its reciprocal, sec), is an even function, that is,
cos(−x) = cos(x) for all real numbers x
The product (and quotient) of an odd function with an even function
is odd, so tan and cot are both odd functions.
10
Periodic Function
A function f is said to be periodic is there exists a positive number
p such that f (x + p) = f (x) for all real number x in the domain of f .
The smallest of such number is called the period of f .
All six of the tri functions are periodic. Intuitively, a periodic function
repeats itself in behavior. The period of sin, cos, csc, and sec is 2π,
and the period of tan and cot is π. Therefore, we have:
sin(x + 2π) = sin(x) for all x in the domain of sin
cos(x + 2π) = cos(x) for all x in the domain of cos
tan(x + π) = tan(x) for all x in the domain of tan
csc(x + 2π) = csc(x) for all x in the domain of csc
sec(x + 2π) = sec(x) for all x in the domain of sec
cot(x + π) = cot(x) for all x in the domain of cot
Domain and Range of Tri Functions
The domain of both the sin and cos functions is all real numbers. I.e.,
sin and cos are defined everywhere.
The range of both sin and cos is the closed interval [−1, 1]
The domain of the other four trigonometric functions are not as easily
determined, and we will discuss about that later.
11
Graphing
To draw the graph of f (x) = sin(x), we start from x = 0 and go
counter-clockwise (positive x) for a 2π period. Notice that sin(x) moves
π
from 0 up to 1 when x moves from 0 to , then sin(x) moves from 1
2
π
back to 0 as x moves from moves to π, then sin(x) becomes negative
2
3π
and moves from 0 to −1 as x moves from π to
, and sin(x) moves
2
3π
to 2π, completing the cycle.
from −1 back to 0 as x moves from
2
12
Once one cycle of the sin graph is drawn, one need only to copy and
paste this shape onto the rest of the x-axis to complete the graph of
sin, as we know that sin is periodic.
A similar technique can be used to draw the basic graph of the other
tri functions.
13
Graphing Trigonometric Functions in General Form
Graph f (x) = A sin(Bx + C) + k, where A, B, C, and k are real
constants, and B > 0
We want to form a rectangle (the envelop) which encloses one cycle of
this function. We need the following information of the rectangle:
1. The Amplitude (the distance from the middle of the envelop to
the top or bottom) of the rectangle is |A|.
C
2. The starting point of the envelop is at − (This is called the phase
B
shift)
2π
3. The period, p, of the function (length of the rectangle) is
|B|
The frequency, f , of a periodic function is the reciprocal of its period,
|B|
that is, f =
2π
4. The end point of the rectangle is at starting point + period, i.e.
C
2π
end point = − +
B |B|
5. The intersections of the function f with the middle line (x-axis if
k = 0) occurs at the points:
p
4
p
starting point +
2
3p
starting point +
4
where p is the period.
starting point +
6. The rectangle is moved up k units if k > 0 and moved down |k|
units if k < 0.
14
E.g. Graph f (x) = 3 cos(2x − 1) − 2
Ans:
We have A = 3, B = 2, C = −1, and k = −2.
|A| = |3| = 3. This is the amplitude.
1
Setting 2x − 1 = 0 and solve for x, we get x = . This is our starting
2
point. (Phase shift).
2π
2π
p=
=
= π. The period of f is π.
|B|
2
1
End point = starting point + period = + π.
2
p π
= , so the intersections of the function with the middle line occurs
4
4
1 π
1 π
1 3π
at + ,
+ ,
+
2 4
2 2
2
4
k = −2, so the graph is moved 2 units down.
Don’t forget to copy and paste this rectangle to the whole x−axis.
15
Graphing Tangent and Cotangent functions:
π
The domain of tangent is all real numbers x such that x 6= + nπ,
2
where n is any integer.
Suppose θ is an angle in standard position, P = (x, y) is the terminal
point of θ, we want to find out the behavior of tangent for θ between
0 to π.
π
tan(0) = 0, and as θ moves from 0 to , the value of y becomes larger
2
and larger while the value of x becomes smaller and smaller, so the
y
y
π
ratio becomes larger and larger. In fact, → ∞ as θ →
x
x
2
π
y
At θ = , x = 0, so tan(θ) = is undefined. Tangent has a vertical
2
πx
asymptote at the point θ = .
2
π
As θ moves from toward π, y becomes smaller and smaller in value,
2
y
as x increases in magnitude. But notice that x is now negative, so
x
is now a negative number, and it becomes closer and closer to zero as
y
θ moves toward π. At θ = π, y = 0 and tan(θ) = = 0.
x
It is generally more natural to consider that one cycle of tangent starts
π
π
from − and ends at , as this gives us one connected piece of the
2
2
tangent function.
16
For a general graph of the form: f (x) = A tan(Bx + C), we set
π
Bx + C = − to solve for x, this again gives us the starting point of
2
one cycle of tangent.
π
The period of f is
. Notice that we are taking π divided by |B|,
|B|
not 2π, as the natural period for tan is π.
Tangent does not have an amplitude since it does not have a lowest or
highest point. A determines the vertical stretch of the function.
17
To draw the graph of f (x) = A csc(Bx + C) or f (x) = A cos(Bx + C),
we draw the corresponding sin and cos function and use that as a
guideline. The x intercepts of sin and cos will be where the vertical
asymptotes of the csc and cos is. Notice that the domain of csc is all
real numbers x where sin(x) 6= 0, and since sin(nπ) = 0 for all integer
n, the domain of csc is all real numbers x such that x 6= nπ, n an
integer.
π
Similarly, the domain of sec is all real numbers x such that x 6= + πn
2
18
Trigonometric Identities:
We have seen some of the identities of the trigonometri functions. We
will introduce more:
Fundamental Identities:
1
(*) sin(x) =
csc(x)
1
(*) cos(x) =
sec(x)
1
(*) tan(x) =
cot(x)
(*) tan(x) =
sin(x)
cos(x)
cos(x)
sin(x)
Even Odd properties:
(*) cot(x) =
(*) sin(−x) = − sin(x)
(*) tan(−x) = − tan(x)
(*) cos(−x) = cos(x)
19
Pythagorean Identities:
(*) sin2 (x) + cos2 (x) = 1
tan2 (x) + 1 = sec2 (x)
1 + cot2 (x) = csc2 (x)
20
Cofunction Identities:
!
π
sin
− x = cos(x)
2
!
π
cos
− x = sin(x)
2
Sum (and difference) of Angles Identities:
(*) sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
(*) cos(x + y) = cos(x) cos(y) − sin(x) sin(y)
tan(x + y) =
tan(x) + tan(y)
1 − tan(x) tan(y)
(*) sin(x − y) = sin(x) cos(y) − cos(x) sin(y)
(*) cos(x − y) = cos(x) cos(y) + sin(x) sin(y)
tan(x − y) =
tan(x) − tan(y)
1 + tan(x) tan(y)
21
Double Angle Identities:
(*) sin 2x = 2 sin x cos x
(*) cos 2x = cos2 x − sin2 x = 1 − 2 sin2 x = 2 cos2 x − 1
2 tan x
tan 2x =
1 − tan2 x
Half Angle Identities:
sin
v
u
u1
t
x
=±
2
− cos x
2
v
u
u 1 + cos x
x
cos = ±t
2
2
Product to Sum and Sum to Product Identities:
1
sin x cos y = [sin(x + y) + sin(x − y)]
2
1
sin x sin y = [cos(x − y) − cos(x + y)]
2
1
cos x cos y = [cos(x + y) + cos(x − y)]
2
x+y
x−y
sin x + sin y = 2 sin
cos
2
2
x+y
x−y
sin x − sin y = 2 cos
sin
2
2
x+y
x−y
cos x + cos y = 2 cos
cos
2
2
x+y
x−y
cos x − cos y = −2 sin
sin
2
2
22
Verifying Tri Identities:
Using the known trigonometric identities, we can transform a trigonometric expression written in one form into another. We need to use
rules of algebra and other known tri identities to verify that the two
sides of the equal sign are equal to each other.
It is important to note that, while verifying a tri-identity, one may
not assume the identity as equal. Instead, you must start with one
side of the identity (either the left or right hand side is fine) and, by
using correct algebra and other tri identites, change the expression to
look the same as the other side. You may not work the expression as
if you are solving an equation.
23
E.g.
Verify
sin t cos t
+
=1
csc t sec t
Ans: We may start with either the left or right hand side, but for this
example, the right hand side is too simple and does not allow for any
clue as to where we should go, it is easier that we start with the left
hand side:
1
1
sin t cos t
+
= sin t ·
+ cos t ·
L.H.S =
csc t sec t
csc t
sec t
= sin t sin t + cos t cos t = sin2 t + cos2 t = 1 = R.H.S.
24
Verify
1 − cos x sec x − 1
=
1 + cos x
sec x + 1
Ans: Both sides are equally complicated. Let’s start with the left hand
side:
1 − cos x (1 − cos x)(sec x) sec x − cos x sec x
L.H.S. =
=
=
1 + cos x
(1 + cos x)(sec x) sec x + cos x sec x
sec x − 1
=
= R.H.S.
sec x + 1
25
An example of an incorrect way of trying to “verify” the above identity would be something like this:
1 − cos x sec x − 1
=
1 + cos x
sec x + 1
(1 − cos x)(sec x + 1) = (sec x − 1)(1 + cos x)
sec x + 1 − cos x sec x − cos x = sec x + sec x cos x − 1 − cos x
sec x + 1 − 1 − cos x = sec x + 1 − 1 − cos x
sec x − cos x = sec x − cos x
In doing the cross multiply and whatever steps follow, you have already
assumed that the two sides are equal to each other. This is logically
invalid and does not constitute a valid argument for varifying anything.
(In logic, this is called a circular argument)
26
E.g. Verify the identity:
(sin x + cos x)2 = 1 + sin 2x
Ans: We start with the left hand side:
L.H.S = (sin x + cos x)2 = sin2 x + 2 sin x cos x + cos2 x
= sin2 x + cos2 x + 2 sin x cos x = 1 + 2 sin x cos x = 1 + sin 2x = R.H.S.
27
E.g. Verify:
cos(x + y)
sin x cos y
This time, it is easier to start with the right hand side:
cos(x + y) cos x cos y − sin x sin y
=
R.H.S. =
sin x cos y
sin x cos y
cos x cos y sin x sin y
cos x sin y
=
−
=
−
= cot x − tan y = L.H.S.
sin x cos y sin x cos y
sin x cos y
cot x − tan y =
28
Combining the pythagorean theorem and using the tri-identities, one
can find the exact values of the other trigonometric functions on a
given angel knowing minimal information.
E.g.
Find the exact value of the other trigonometric functions of θ if
2
sin θ = − and the terminal point of θ is in the third quadrant.
5
Ans: Since the terminal point of θ is in the third quadrant, using a
circle of radius√5, we see that the coordinate of the terminal point P
of θ is P = (− 21, −2). We get:
√
21
cos θ = −
5
−2
2
tan θ = √ = √
− 21
21
5
1
=−
csc θ =
sin θ
2
1
5
sec θ =
= −√
cos θ
21
√
√
− 21
21
cot θ =
=
−2
2
29
Inverse Trigonometric Functions:
The inverse sine function, denoted by f (x) = arcsin(x) or f (x) =
sin−1 (x) is defined by:
π
π
y = sin−1 (x) if and only if sin(y) = x and − ≤ y ≤
2
2
I.e., the range of f (x) = arcsin(x) is all real numbers y such that
π
π
− ≤y≤
2
2
The inverse cosine function, denoted by f (x) = arccos(x) or f (x) =
cos−1 (x) is defined by:
y = cos−1 (x) if and only if cos(y) = x and 0 ≤ y ≤ π
I.e., the range of f (x) = arccos(x) is all real numbers y such that
0≤y≤π
The inverse tangent function, denoted by f (x) = arctan(x) or f (x) =
tan−1 (x) is defined by:
π
π
y = tan−1 (x) if and only if tan(y) = x and − < y <
2
2
I.e., the range of f (x) = arctan(x) is all real numbers y such that
π
π
− <y<
2
2
The inverse trigonometric functions are the inverse functions of the
trigonometric functions. They undo the effect of the original tri functions. Keep in mind that since a trigonometric function takes an angle
measurement (in radian) as input and gives a real number as output,
an inverse tri function (such as arccos or arcsin takes a real number as
input and produces an angle measurement (in radian) as output.
30
The domain of the sine function is all real numbers. Ideally, to undo
the effect of sine, we would like the range of the f (x) = arcsin(x)
function to be all real numbers too. Unfortunately, since sine is NOT
a one-to-one function, this is not possible.
More precisely, since:
sin(0) = 0
sin(π) = 0
sin(2π) = 0
We would like
arcsin(0) = 0
arcsin(0) = π
arcsin(0) = 2π
But this would make f (x) = arcsin(x) fails to be a function. As a
result, we need to restrict the range for arcsin such that, within this
range, arcsin returns only one output for each input. We make the
choice "based on
# convenience. As a result, we chose the range of arcsin
π π
to be − , . Using similar guidelines, we defined the range of the
2 2
other inverse trigonometric functions the way we did.
31
E.g.

√

2
Find the value of sin
− 
2
√
 √ 
2
2
−1 
, then by definition, sin(θ) = −
and
Ans: Let θ = sin
−
2
2
π
π
− ≤ θ ≤ . The only angle that satisfies both of these requirements
2
π 2
is θ = − .
4
 √ 
2
π
Therefore, sin−1 −  = −
2
4
−1 
32
E.g.

Find cos
−1 
√

3
− 
2

−1 
Ans: Let θ = cos
√
√

3
− , by definition, we want
2
3
and 0 ≤ θ ≤ π. The only angle that satisfies both of
2
5π
.
these requirements is θ =
6
 √ 
5π
3
Therefore, cos−1 −  =
2
6
cos(θ) = −
33
E.g.
31π
Find arcsin sin
4
3π
31π
= 7π + .
Ans:
4
4
!!
3π
3π
Since sine is periodic, sin 7π +
= sin π +
.
4
4
!!
31π
Let θ = arcsin sin
. We want:
4
!
π
π
3π
and − ≤ θ ≤
sin(θ) = sin π +
4
2
2
!
3π
The terminal side of π +
is in the fourth quadrant, therefore,
4
!
π
3π
is negative. We need sin(θ) to be negative and − ≤
sin π +
4
2
π
θ ≤ . This tells us that θ must be an angle in the fourth quadrant
2
π
and the only angle θ that satisfies this requirement is θ = − .
4
!
34
!
E.g.
Find arccos (cos (23π + 3))
Ans: (23π + 3 = 22π + π + 3)
Since cosine is periodic, cos (23π + 3) = cos (π + 3)
Let θ = arccos (cos (23π + 3)). We want:
cos(θ) = cos (π + 3) and 0 ≤ θ ≤ π
The terminal side of (π + 3) is in the fourth quadrant (why?), which
makes cos(π + 3) positive. Therefore, cos(θ) must also be positive and
0 ≤ θ ≤ π. The terminal side of θ must be in the first quadrant and
the only angle θ that satisfies this requirement is θ = π − 3 (why)?
35
E.g.
5
Find cos arctan −
7
!!
5
Ans: Let θ = arctan − , we have
7
π
π
5
tan(θ) = − and − < θ < . If we draw θ in standard position, the
7
2
2
terminal side of θ is in quadrant IV. Using the Pythegorean Theorem,
we see that the coordinate of the terminal point P√of θ is P = (7, −5)
where the radius of the circle we are using is r = 74.
!!
7
5
= cos θ = √
Using this information, cos arctan −
7
74
!
36
2
Find sin arccos −
3
!!
2
Ans: Let θ = arccos − , we have
3
2
cos(θ) = − and 0 ≤ θ ≤ π. If we draw θ in standard position, the
3
terminal side of θ is in quadrant II. Using the Pythegorean Theorem,
√ we see that the coordinate of the terminal point P of θ is P = −2, 5
where the radius of the circle we are using is r = 3.
√
!!
2
5
Using this information, sin arccos −
= sin θ =
3
3
!
37
Express tan(arcsin(x)) in terms of x so that the expression is free of
any trigonometric function. Assume that x is a positive number.
Ans: Let θ = arcsin(x), we have
sin(θ) = x and since x is positive, the terminal side of θ must be in
the first quadrant. Using the Pythegorean Theorem,
we see
√
that the
coordinate of the terminal point P of θ is P =
1 − x2 , x where the
radius of the circle we are using is r = x.
x
Using this information, tan(arcsin x) = tan(θ) = √
1 − x2
38
Solving Trigonometric Equations To solve equations that involve
trigonometric functions, one wants to isolate the (hopefully one) tri
function then use special triangle or the inverse tri functions to isolate the variable. Note that since a trigonometric function is not an
algebraic function (i.e. to find sin(x) we cannot find the value by performing algebraic operations on x), we cannot isolate x by performing
algebraic operations. We can only use our knowledge of special triangles, or using the inverse functions.
E.g.
Solve the equation:
√
3
sin(x) =
2
Ans: There are two angles in [0, 2π) that solve the equation, namely
π
2π
and
3
3
In addition, since sine is periodic with a period of 2π, if x is a solution
to the equation, then x + 2π, x + 4π, x + 6π, x − 2π, x − 4π, ... are
all solutions to the equation. Therefore, our complete solution set will
be:
π
2π
x = + 2πn or x =
+ 2πn
3
3
where n is an integer.
39
E.g.
Solve the equation:
tan(x) = −1
Ans: Tangent is negative in the second and fourth quadrant. In the
3π
second quadrant the angle that would solve this equation is x =
4
Since tan has a period of π instead of 2π, we don’t need to add 2π to
the angle to find all the solutions. Instead, we have:
3π
+ πn, where n is an integer.
x=
4
40
Solve the equation:
√
3
cos(2x + 1) = −
2
Ans: We treat 2x − 1 as a single quantity. I.e.,√substitue y = 2x − 1
3
and we are looking at the equation cos(y) = − .
2
The two angles in [0, 2π) that solves the equation is:
5π
7π
y=
and y =
6
6
Since we are solving for x, not for y, we must back substitue x into the
equation:
5π
2x + 1 =
6
7π
2x + 1 =
6
Since cos is a periodic function with period of 2π, any integral multiple
of 2π added to the angles will solve the equation too, so the complete
solutions will be:
5π
2x + 1 =
+ 2πn
6
7π
2x + 1 =
+ 2πn
6
where n is an integer.
We still need to isolate x in the above equations:
5π 1
− + πn
x=
12 2
or
7π 1
x=
− + πn
12 2
41
Note from the example that you add the 2πn to the angles immediately after you isolated the argument (2x + 1) of cos, but before
you isolated x. Some people, not understanding the reason for adding
the 2πn, would just always add the 2πn after they isolated x. This
is √
incorrect. The angles that cosine must take to produce a value of
!
!
2
5π
7π
−
is
+ 2πn or
+ 2πn .
2
6
6
Since (2x + 1) is the !argument to cosine
in the equation, (2x + 1) must
!
5π
7π
equal to
+ 2πn or
+ 2πn , not x itself. We will, instead,
6
6
isolate x using algebraic means after we solved for 2x + 1.
42
Solve: cos(2x) − 3 cos(x) − 1 = 0
Ans: We use the double angle formula to change cos(2x) to
cos(2x) = 2 cos2 (x) − 1.
We have:
2 cos2 (x) − 1 − 3 cos(x) − 1 = 0
2 cos2 (x) − 3 cos(x) − 2 = 0
This is a quadratic equation in cos(x), factoring we get:
(2 cos(x) + 1)(cos(x) − 2) = 0
This gives:
2 cos(x) + 1 = 0
or
cos(x) − 2 = 0
We can ignore the second equation as it will not give us any solution.
(why?) The first equation gives us:
1
cos(x) = −
2
The two angles in [0, 2π) that solves the equation is:
2π
4π
x=
and x =
3
3
Adding any integral multiple of 2π will give us the same solutions too,
so the complete solution set will be:
2π
x=
+ 2πn
3
or
4π
x=
+ 2πn
3
where n is an integer
43
Solve for exact value:
1
sin(x) = − , 0 ≤ x < 2π
3
Ans: Since 1/3 is the not the ratio of a special triangle, we have to use
sin−1 . Notice that
!
1
π
−1
x = sin
−
is an angle between − and 0 (why?). This value is
3
2
outside of the interval where we want our solution for x to be. Instead,
note that sin will be negative in the third and fourth quadrant. In the
third quadrant, the angle that is within the desired interval is:
!
1
−1
x = π − sin
− .
3
The other angle that
! would work is the angle in the fourth quadrant,
1
x = 2π + sin−1 − .
3
The solution set is thus:
!
!
1
1
−1
−1
x = π − sin
−
or x = 2π + sin
− .
3
3
44
Solve for exact value:
1
cos(x) = −
5
Ans: Another ratio that does not come from a special triangle. We
use:
!
1
−1
x = cos
−
5
This will produce the angle in the second quadrant. Since cos is an
even function,
!
1
−1
will also work. Adding our 2πn gives the complete
x = − cos
−
5
solution:
!
1
−1
x = ± cos
− + 2πn
5
45
Law of Sine:
If ∆ABC is any triangle whose angles are A, B, C and the sides
opposite these angles are, correspondingly, a, b, c, then:
sin(A) sin(B) sin(C)
=
=
a
b
c
The formula tells us that, as long as we know the value of two angles in
a triangle and a side of the triangle, we can find the value of the other
missing angles and sides. In other words, if two angles and a side of a
triangle is fixed, the triangle is fixed. This corresponds to the (ASA)
(angle-side-angle) and (AAS) (angle-angle-side) triangle congruency
in geometry.
46
Law of Cosine:
If ∆ABC is any triangle whose angles are A, B, C and the sides
opposite these angles are, correspondingly, a, b, c, then:
c2 = a2 + b2 − 2ab cos(C)
b2 = a2 + c2 − 2ac cos(B)
a2 = b2 + c2 − 2bc cos(A)
π
, then cos(C) = cos π2 = 0 and the first formula
2
2
2
becomes: c = a + b2 , which is just the Pythegorean Theorem.
Note that if C =
You may think of the law of cosine as an adjustment to the Pythegorean
Theorem for an arbitrary triangle.
The law of cosine tells us that, if two sides of a triangle and the angle
between the two sides is known (SAS) or if all three sides of a triangle
is known (SSS), the other parts of the triangle can be found.
47
Vectors
Intuitively, a vector is a mathematical object that has both a magnitude and a direction. An arrow is a vector. The tip of the arror
points to the direction of the vector, and the length of the arror is the
magnitude of the vector.
Many quantities in physics have both a direction and a magnitude.
Velocity is an example. It matters not just how fast you are going, but
also where you are going.
In R2 , we use v = hx, yi to represent a vector. The coordinates x, y
gives the location of the tip of the vector, and we always assume that
the tail of the vector is at the origin.
Example: The vector v = h2, 1i is the vector with its tail at the origin
and its tip at the point (2, 1).
It is important to note that, even though by assumption we said that
the vector v = hx, yi always has its tail at the origin, the location of
a vector is irrelevant. The reason we need to specify that its tail is at
the origin is because we want to know to which direction that vector is
pointing. Therefore, the vector v = h3, −1i (with its tail at the origin)
and the vector w with its tail at (1, 2) and its tip at (4, 1) are exactly
the same vector, because they have the same length and points to the
same direction.
In other words, a vector is distingushed only by its magnitude and
direction, not by its location in space.
We distingush between a point and a vector by using bold type when
printing vector. So p = (1, 2) represents the point located at that
particular coordinate, while v = h1, 2i represents the vector that has
its tail at the origin and its tip at the point (1, 2). Sometimes we also
write ~v to mean a vector.
For a point (x, y), we say that the vector v = hx, yi is the position
vector of the given point. In short, the position vector of a given point
in space is a vector whose tip points to that point (and its tail is at
the origin).
48
Vector Addition
Let v = hx1 , y1 i and w = hx2 , y2 i, we define
v + w = hx1 + x2 , y1 + y2 i
In other words, to add two vectors, add their components correspondingly.
Geometrically, when adding two vectors, the resulting vector is the
vector obtained by placing the tail of the second vector (w) on the tip
of the first vector, and the result is the vector with its tail at the first
vector (v) and its tip at the second vector (w).
49
Scalar Multiplication
If v = hx, yi is a vector, and c is a real number (a scalar), then we
define
cv = hcx, cyi
To multiply a number to a vector, we multiply the number to each of
the components of the vector.
Geometrically, multiplying a scalar c to a vector stretches or shrinks
the vector by a factor of |c| units, and reverses the direction of the
vector if c is negative.
Example:
Let v = h2, 3i , w = h−1, 1i
then
v + w = h2 + −1, 3 + 1i = h1, 4i
3v = h3 · 2, 3 · 3i = h6, 9i
Notation: The vector 0 = h0, 0i is the zero vector. It has both it tip
and tail at the origin. The magnitude (length) of the zero vector is 0,
and it is the only vector whose direction is arbitrary.
Notice that for any scalar c and any vector v, v + 0 = v, and c0 = 0.
In this sense, the zero vector serves the same function as the number
0 when considering vectors.
We also write −v for the vector (−1)v. When we write v − w, we
mean v + −w.
We say that two vectors v and w are parallel to each other if there is
a scalar c > 0 such that v = cw.
Geometrically, two vectors are parallel to each other if they point to
the same direction.
Two vectors are anti-parallel if they point to opposite direction. In
other words, v and w are anti-parallel if there exists a number c < 0
such that v = cw.
Notation: Sometimes we want to specify a particular vector that has
its tail not at the origin. Suppose A = (x1 , y1 ) and B = (x2 , y2 ) are
50
two points in R2 , then the vector reprsented by
→
AB= hx2 − x1 , y2 − y1 i
is the vector with its tip at B and its tail at A.
51
The magnitude, or length of a vector v = hx, yi is represented by
|v| and is given by the formula:
q
|v| = x2 + y 2
This should be expected since it is just the distance formula restated.
A vector whose magnitude is 1 is called a unit vector.
Given a non-zero vector v, we can form a unit vector that points to
the same direction as v. We do that by dividing v by its magnitude.
√
√
Example:
v
=
h1,
3i,
so
|v|
=
1
+
9
=
10, and the vector u =
√v = √1 , √3
is a unit vector that points to the same direction as
10
10
10
v.
Important Note: It is very important to distingush the difference between a number (scalar) and a vector. We can add two numbers, or
add two vectors, and we can multiply a number to a vector. However,
we cannot add a number with a vector, and we cannot multiply or div
, because the magnitude of
vide two vectors. It makes sense to write |v|
a vector is a number, and multiplying a number to a vector is defined.
However, it never makes sense to write wv , where both v and w are
vectors.
52
Dot Product
If v = hx1 , y1 i and w = hx2 , y2 i are vectors, then the dot product of
v and w, denoted by v · w, is given by
v · w = x 1 x 2 + y1 y 2
Example:
v = h2, −2i, w = h1, −3i, then
v · w = 2 · 1 + −2 · −3 = 8
Notice that the dot product of two vectors is a scalar.
The following can be easily proved using the definition of the dot products:
If v, w, s are vectors and c is a scalar, then:
v · v = |v|2
v·w =w·v
v · (w + s) = v · w + v · s
(cv) · w = c(v · w) = v · (cw)
0·v =0
v · v ≥ 0,
and v · v = 0
53
if and only if
v=0
One of the most useful feature of the dot product between two vectors
is that it gives us geometry of the vectors. If θ is the angle formed
between two vectors v and v, then we have:
v · w = |v||w| cos θ
This theorem tells us that the dot product of two vectors is just the
product of the length of the vectors times the cosine of the angle between the vectors.
We know from geometry that two lines are perpendicular if the angles
they formed is π/2, we use the same concept for vectors. We say that
two vectors are perpendicular, or orthogonal, if the angle between
them is π/2. Since cos(π/2) = 0, we have the following:
Two vectors v and w are orthogonal if and only if v · w = 0
54
Polar Coordinates:
Instead of the rectangular coordinate system (x, y), which uses intersecting perpendicular lines to represent locations in a plane, sometimes
it is useful to represent the location of a point in the coordinate plane
using concentric circles and angles the point makes with the x− axis.
E.g. Think of the radar screen, or when people describe the location
by saying, “2 o’clock”.
Like the rectagular coordinate system, a point in polar coordinate consists of an ordered pair of numbers, (r, θ). The first coordinate is the
distance of the point from the origin (0, 0), and the second coordinate
is the angle, in standard position, and the point we want to locate is
the terminal point of this angle.
55
E.g.
√
The point P = (1, 1) (in rectangular coordinate) has a distance
of
2
!
π
, therefore,
from the origin, and is the terminal point of the angle
4
the equivalent polar coordinate of point P is:
√ π!
P =
2,
4
Note that, unlike the rectangular coordinate system, the polar representation of a point is not unique. For!example, the above! point P
√ 9π
√
7π
can also be represented by P =
2,
or P =
2, −
4
4
If r is a negative number, it is understood that the point referenced has
the same distance |r| from the origin, but is in the opposite direction
of the terminal point of θ.
E.g.
!
√
π
The polar point −2,
is the retangular point (− 3, −1).
6
56
Given the definition of cosine and sine:
x
cos θ =
r
y
sin θ =
r
We readily have the following conversion from polar to rectangular
coordinate system:
x = r cos θ
y = r sin θ
And given the formula for the equation of a circle with radius r centered
at the origin:
x2 + y 2 = r2
we can interchange between polar and rectangular coordinate:
57
E.g. Express the rectangular point (1, 7) in polar coordinate:
This is a point
√ in the first
√quadrant.
√ The distance the point from the
2
2
origin is r = 1 + 7 = 50 = 5 2
We can use either the arcsine or arccosine function to find the exact
value of the angle, it is customary to use arcsine, so:
√
7 = 5 2 sin θ
7
sin θ = √
5 2
!
7
−1
√
θ = sin
5 2
This angle θ returned by arcsine is in the first quadrant (why?), which
is where the point (1, 7) is located, this is what we want. Therefore, the
!!
√
7
−1
√
polar equivalent of the rectangular point (1, 7) is 5 2, sin
5 2
58
E.g. Express the rectangular point (−2, −5) in polar coordinate:
q
√
√
r = (−2)2 + (−5)2 = 4 + 25 = 29
−5
sin θ = √
29
−5
√
has a terminal point in the fourth quadrant
the angle sin
29
(why?), but the point we want is in the third quadrant, the angle we
want is:
!
−5
−1 √
θ = π − sin
29
The polar equivalent of the rectangular point (−2, −5) is
!!
√
−5
−1 √
29, π − sin
29
−1
!
59
In general, to change a rectangular coordinate (x, y) to a polar coordinate (r, θ), we have:
√
r = x2 + y 2 and
!
−1 y
θ = sin
if x ≥ 0
r
and
!
−1 y
if x < 0
θ = π − sin
r
60
π
in rectangular form:
E.g. Express the polar coordinate 3, −
3
Ans:
!
!
π
1
3
x = 3 cos −
=3
=
3
2
2
√
 √ 
!
3
3
3
π
y = 3 sin −
= 3 −  = −
3
2
2
√ 

3
3
3
The coresponding rectangular point is  , −
2
2
!
61
In general, to change a polar coordinate point (r, θ) into rectangular
coordinate (x, y), use
x = r cos θ
y = r sin θ
The origin, in either rectangular or polar form, will be (0, 0)
62
Equations in Polar Form:
In rectangular coordinate, we know that the equations x = 2 or y = 3
are equations of vertical and horizontal lines, respectively, what do the
equations
r=2
and
θ=3
in polar coordinate represent?
Note that in the rectangular equation, x = 2, y is a free variable,
meaning that y can be any value. In the equation r = 2, θ is also a
free variable and can assume any value, therefore, any point that has
a distance of 2 units from the origin will be a solution to the equation
r = 2, this is the graph of a circle with radius 2 centered at the origin.
In θ = 3 (3 is 3 radian, not 3 degrees), r is the free variable, so any
points, as long as they are the terminal point of the angle (in standard
position) that makes 3 radian angle with the x axis, will be on the
curve, regardless of its distance from the origin. This is a straight line
throught the origin with slope equal to m = tan(3).
In general, the polar equation r = c, (c > 0 a constant), is the equation
of a circle with radius c centered at the origin.
The polar equation θ = c, (c a constant), is the equation of a straight
line through the origin with slope m = tan(c)
63
E.g. Describe the graph of the polar equation:
r = θ, 0 ≤ θ < ∞
As θ increases, so does r, so we have a curve that keeps on expanding
from the origin in a circular manner. This is a spiral.
64
E.g. Describe the graph of the polar equation:
r = 6 cos θ
It may be a little easier if we change this back to rectangular equation:
!
x
r=6
r
r2 = 6x
x2 + y 2 = 6x
x2 − 6x + y 2 = 0
x2 − 6x + 9 + y 2 = 9
(x − 3)2 + y 2 = 9
This is a circle with radius 3 centered at (rectangular point) (3, 0)
65