Download MA 108: THINGS WE`VE LEARNED 1/9. • 1 minute = 1 = degree = 1

MA 108: THINGS WE’VE LEARNED
1/9.
◦
1
1
• 1 minute = 10 = 60
degree = 60
1 0
• 1 second = 100 = 60
180 ◦
π
•
= 1 radian ,
radians = 1 degree
π
180
• If θ is a central angle , measured in radians, of a circle of radius r, and s is the length of
the arc subtended by θ, then s = rθ
1/11.
• If θ is a central angle (in radians) of a circle of radius r, then the area A of the sector of
1
the circle swept out by θ is given by the formula A = r2 θ
2
• Angular speed ω is defined as the angle of rotation (in radians) of a circular object
θ
divided by the elapsed time t. We write ω =
t
• Linear speed v of the edge of a rotating circular object is the arclength swept out by a
s
fixed point on the edge of the object divided by the elapsed time t, written as v =
t
• Linear and angular speed of the same circular object are related by the equation v = rω ,
where r is the radius of the object.
1/12.
• The Pythagorean Theorem [MEMORIZE]: If a and b are the lengths of the legs of a
triangle with hypoteneuse c, then a2 + b2 = c2 .
• Let A be an acute angle living in a right triangle with legs a and b and hypoteneuse c
as pictured:
Then the six trigonometric functions of A are given by the following ratios [MEMORIZE]:
1
2
MA 108: THINGS WE’VE LEARNED
sin A
tan A
csc A
opposite
hypoteneuse
opposite
adjacent
hypoteneuse
opposite
a
c
a
b
c
a
cos A
cot A
sec A
adjacent
hypoteneuse
adjacent
opposite
hypoteneuse
adjacent
b
c
b
a
c
b
• Reciprocal Identities [MEMORIZE] Given a fixed angle θ:
tan θ =
cot θ =
sin θ
cos θ
, cot θ =
cos θ
sin θ
1
1
1
, csc θ =
, sec θ =
tan θ
sin θ
cos θ
• Pythagorean Identities [MEMORIZE]: Given a fixed angle θ:
(1) sin2 θ + cos2 θ = 1
(2) tan2 θ + 1 = sec2 θ
(3) cot2 θ + 1 = csc2 θ
• END OF MATERIAL FOR QUIZ 1. Formulas provided for Quiz 1 (Thursday, January
180 ◦
1
θ
s
19) will include s = rθ , A = r2 θ , ω = , v = , v = rω ,
= 1 radian ,
2
t
t
π
π
radians = 1 degree .
180
• The Complementary Angle Theorem: cofunctions of complementary angles are equal.
• Table of relations implied by the Complementary Angle Theorem
π
π
sin θ = cos
−θ
cos θ = sin
−θ
2
2
π
π
tan θ = cot
−θ
cot θ = tan
−θ
2
2
π
π
sec θ = csc
−θ
csc θ = sec
−θ
2
2
MA 108: THINGS WE’VE LEARNED
3
1/13. Today we did not introduce any new material, instead we did problems in class using
identities to simplify and evaluate expressions involving trig functions.
1/18. Today we learned some values of trig functions of some special angles (in the table
below) and worked some modeling problems using right triangles from 7.3. [PLEASE MEMORIZE WHICH TRIANGLES GO WITH THESE ANGLES. PRACTICE DRAWING THE 45-45 90 AND 30-60-90 TRIANGLES AND LABELING THE LENGTHS
OF THE SIDES]
θ in degrees θ in radians sin θ cos θ tan θ csc θ sec θ cot θ
45
30
60
π
4
π
6
π
3
1
√
2
1
2
√
3
2
1
√
2
√
3
2
1
2
1
1
√
3
√
3
√
2
√
2
1
2
2
√
3
√
2
√
3
2
3
1
√
3
1/19.
• Strategy for Right Triangle Modeling Problems:
(1)
(2)
(3)
(4)
Draw a picture that models the situation in the word problem
Find out how a right triangle appears in the model
Label all knowns and unknowns (angles, side lengths, etc.)
Use definition of trig functions or trig identities to set up an algebraic equation you
can solve for the required unknown.
• Definition of the Trig Functions of a General Angle [MEMORIZE]: Let θ be any angle
in standard position, and let (a, b) denote√
the coordinates of any point, except the origin
(0, 0), on the terminal side of θ. If r = a2 + b2 . The Six Trig Functions of θ are
given by the following ratios:
sin θ =
b
r
cos θ =
a
r
tan θ =
b
a
csc θ =
b
r
sec θ =
a
r
cot θ =
b
a
4
MA 108: THINGS WE’VE LEARNED
• Values of the Trig Functions of the Quadrantal Angles (U = undefined)
θ in degrees θ in radians sin θ cos θ tan θ csc θ sec θ cot θ
0
0
π
2
π
90
180
3π
2
270
0
1
0
U
1
U
1
0
U
1
U
0
0
-1
0
U
-1
U
-1
0
U
-1
U
0
1/20.
• The angles θ, α are coterminal if they share the same terminal side.
• If θ is a an angle then the reference angle α for θ is the angle made by the terminal
side of θ and the x-axis.
• It follows from the definition of the trig functions of θ that the signs of the trig functions
of θ (in other words, whether they are positive or negative) depend on which quadrant
contains the terminal side of θ
Quadrant of θ sin θ, csc θ cos θ, sec θ tan θ, cot θ
I
+
+
+
II
+
-
-
III
-
-
+
IV
-
+
-
• Reference angles allow us to use what we already know about acute angles and right
triangles to calculate the values of trig functions of general angles.
1/23.
MA 108: THINGS WE’VE LEARNED
5
• The unit circle is a circle of radius 1 centered at the origin. If (x, y) is a point on the
unit circle, then x2 + y 2 = 1. The unit circle is the machine we use to study the graph
and behavior the trig functions.
• Let (a, b) be the point on the terminal side of θ that intersects the unit circle. Then the
values of the trig functions of θ are given by [MEMORIZE]
1
b
1
cos θ = a sec θ =
a
sin θ = b
if b 6= 0
csc θ =
if a 6= 0
b
a
a
cot θ =
b
tan θ =
if a 6= 0
if b 6= 0
Note that your book will write sin t, sec t and so on to emphasize the case when we
are just thinking of t as a real number not necessarily associated to an angle.
• Recall that the domain of a function is the subset of the real numbers that we can plug
into the function, and the range is the set of possible values the function can take on its
domain. We can find the domains and ranges of our trig functions using their definitions
[MEMORIZE].
Function
Domain
Range
sin θ
(−∞, ∞)
[−1, 1]
cos θ
(−∞, ∞)
[−1, 1]
tan θ
All reals except odd integer multiples of
csc θ
All reals except integer multiples of π
sec θ
All reals except odd integer multiples of
cot θ
All reals except integer multiples of π
π
2
(−∞, ∞)
(−∞, −1] ∪ [1, ∞)
π
2
(−∞, −1] ∪ [1, ∞)
(−∞, ∞)
1/25.
• Periodic Properties of Trig Functions: in this table, t is a real number in the domain of
the given function, and k is any integer.
6
MA 108: THINGS WE’VE LEARNED
sin(t) = sin(t + k2π)
csc(t) = csc(t + k2π)
cos(t) = cos(t + k2π) sec(t) = sec(t + k2π)
tan(t) = tan(t + kπ)
cot(t) = cot(t + kπ)
• Even-Odd Properties of Trig Functions: in this table, t is a real number in the domain
of the given function.
sin(−t) = − sin(t)
csc(−t) = − csc(t)
cos(−t) = cos(t)
sec(−t) = sec(t)
tan(−t) = − tan(t) cot(−t) = − cot(t)
2/1. Today we started section 7.6 where we study the graph of the sine function in the xyplane. We draw the graphs using points on the unit circle. We studied the effect of basic graph
transformations on the sine function. A summary of graph transformations for general functions
is found on page 262 (section 3.5) of your book.
2/2.
• For a general function f , we obtain the graph of f (−x) from the graph of f (x) by
reflecting across the y-axis. Recall the four trig functions sine, cosecant, tangent, and
cotangent are odd; for any of these four functions, reflecting across the y-axis is the same
as reflecting across the x-axis. We can move the negative ”outside the function.” Refer
to the table from 1/25 ”even-odd properties”
• Because the cosine and secant functions are even, cos(−ωx) = cos(ωx) and sec(−ωx) =
sec(ωx). So of course the graphs are the same.
• For the sinusoidal functions A cos(ωx) and A sin(ωx), the amplitude is amount by which
the y-coordinate of function deviates from the x-axis, and is given by |A|. In other words
−|A| ≤ A sin(ωx) ≤ |A|, and − |A| ≤ A cos(ωx) ≤ |A|,
2π
.
ω
• How to graph one period of a sinusoidal function of the form y = A sin(ωx) or
y = A cos(ωx) using key points:
(1) Find the amplitude and period.
2π
(2) Divide the interval 0,
into 4 subintervals of equal length.
ω
(3) Use the endpoints of the subintervals to obtain the x-coordinates of five key points
on the graphs.
(4) Plot the points and connect them using the correct shape of one period either the
sine or cosine curve, whichever is relevant.
• The period of A cos(ωx) and A sin(ωx) is given by
MA 108: THINGS WE’VE LEARNED
7
• Grading graphs of sinusoidal curves f (x) on tests or quizzes: to receive full credit, you
must
(1) Find the five key points on the graph of f and clearly label them on the graph,
using x and y coordinates (complete ordered pairs (x, f (x))
(2) The graph of f must resemble the sine or cosine curve that we transform to obtain
f . For example, this means that the second key point on the graph of A sin(ωx)
must be the point on the graph with maximimum y value, e.g. the ”top of the
hump.”
2/3.
π
• The period of A tan(ωx) + B is given by .
ω
• The tangent function has no amplitude, because its range includes all real numbers. So
A is a vertical stretch factor only.
• How to graph one period of f (x) = A tan(ωx) + B using key points:
(1) Find the period, stretch
and vertical shift.
h π factor,
π i
(2) Divide the interval − ,
into 4 subintervals of equal length.
2ω 2ω
π
(3) Use the endpoints of the subintervals to identify the vertical asymptotes at x = −
2ω
π
and x =
and the x-coordinates of three key points between the asymptotes.
2ω
(4) Plot the three key points and connect them using the correct shape of one period
of the tangent curve.
• Grading graphs of f (x) = A tan(ωx) + B on tests or quizzes: to receive full credit for
the graph of one period of the function, you must
(1) Find the two asymptotes and three key points on the graph of f . Label the asympπ
π
and x = − . Clearly label the three key points using x and y
totes as x =
2ω
2ω
coordinates (complete ordered pairs (x, f (x))
(2) The graph of f must resemble the tangent curve that we transform to obtain f . Be
sure to draw arrows indicating that the graph of f is tending towards either positive
or negative infinity as the x-coordinate nears an asymptote.
2/6.
π
• The period of A cot(ωx) + B is given by .
ω
• The cotangent function has no amplitude, because its range includes all real numbers.
So A is a vertical stretch factor only. B still gives a vertical shift.
• How to graph one period of f (x) = A cot(ωx) + B using key points:
(1) Find the period, stretch
and vertical shift.
h π factor,
i
(2) Divide the interval 0,
into 4 subintervals of equal length.
ω
(3) Use the endpoints of the subintervals to identify the vertical asymptotes at x = 0
π
and x = and the x-coordinates of three key points between the asymptotes.
ω
(4) Plot the three key points and connect them using the correct shape of one period
of the cotangent curve.
• Grading graphs of f (x) = A cot(ωx) + B on tests or quizzes: to receive full credit for
the graph of one period of the function, you must
8
MA 108: THINGS WE’VE LEARNED
(1) Find the two asymptotes and three key points on the graph of f . Label the asympπ
totes as x = 0 and x = . Clearly label the three key points using x and y
ω
coordinates (complete ordered pairs (x, f (x))
(2) The graph of f must resemble the cotangent curve that we transform to obtain f .
Be sure to draw arrows indicating that the graph of f is tending towards either
positive or negative infinity as the x-coordinate nears an asymptote.
END OF MATERIAL FOR QUIZ 3, THURSDAY FEBRUARY 9
• For the quiz on Thursday you will need draw two graphs. One will either be an sine
or cosine graph, the other will be tangent or cotangent. Then you will need to do a
matching problem similar to the matching problems in Section 7.6 # 29-38
• Observations about graphing trig functions:
(1) For graphs of the form A sin(ωx) + B or A cos(ωx) + B, the five key points in our
3π
π
graphing method correspond to the first five positive quadrantal angles 0, , π, , 2π.
2
2
2π
(2) The four subintervals of 0,
of equal length that we obtain in our method are
ω
always
h π i h π π i π 3π 3π 2π 0,
,
,
,
,
,
,
2ω
2ω ω
ω 2ω
2ω ω
(3) hFor graphsiof the
A tan(ωx) + B, A cot(ωx) + B, we graph one period in either
h form
πi
π π
or 0,
respectively because we want to graph one period of a function
− ,
2ω 2ω
ω
between a pair of asymptotes.
(4) We always obtain the subintervals
h π
π i h π i h π i hπ π i
− ,−
, − , 0 , 0,
,
,
2ω 4ω
4ω
4ω
4ω 2ω
when we use our method on A tan(ωx) + B.
(5) We always obtain the subintervals
h π i h π π i π 3π 3π π ,
,
,
,
,
,
0,
4ω
4ω 2ω
2ω 4ω
4ω ω
when we use our method on A cot(ωx) + B.
(6) The three
points
on the
of the un-transformed graph tan x have coor key
π graph
π
dinates − , −1 , (0, 0),
, 1 . The three key points on the graph of the trans4
4
π
π
formed graph A tan(ωx)+B have coordinates − , −A + B , (0, B),
,A + B
4ω
4ω
(7) The three key points on
the graph
of the un-transformed graph cot x have coordi
π π 3π
nates
,1 ,
,0 ,
, −1 . The three key points on the graph of the trans4
2
4
π
π 3π
formed graph A cot(ωx)+B have coordinates
,A + B ,
,B ,
, −A + B
4ω
2
4ω
(8) In general, asymptotes for trig functions occur where the function are not defined.
Since the domains of sine and cosine include all real numbers, sine and cosine do
not have asymptotes. The tangent, cotangent, secant, and cosecant functions all
have asymptotes precisely at the places we divide by zero on the unit circle.
MA 108: THINGS WE’VE LEARNED
9
2/8.
• How to Graph A sec(ωx) + B and A csc(ωx) + B:
(1) Find the reciprocal function and graph one fundamental period using the key points
method. The reciprocal of A sec(ωx) + B is A cos(ωx) + B and the reciprocal of
A csc(ωx) + B is A sin(ωx) + B
(2) Draw a vertical asymptote wherever the reciprocal function crosses the line of its
average value (for functions without a vertical shift, this is the x-axis. For functions
with a vertical shift, this is the line y = B.
(3) Label with x and y coords the max(s) and min(s) of the reciprocal function.
(4) In each subinterval determined by the asymptotes, draw the graph of the original
function as a reflection of the reciprocal, with the reflection touching the reciprocal
at the max(s) and min(s) points. Indicate the behavior of the transformed secant
or cosecant function near the asymptotes with arrows pointing towards positive or
negative infinity.
π
. This can be verified by looked at the graph
• For all real numbers x, sin x = cos x −
2
π
of cos x −
obtained using a fundamental graph transformation from Math 107. We
2
can also show this algebraically. Recall the cofunction identity
π
sin x = cos
−x
2
Since the cosine function is even, we have
π
π
π
= cos (−1)
− x = cos
−x
cos x −
2
2
2
Because of this fundamental relationship, we can always write a transformed cosine graph
as a transformed sine graph using horizontal shifts.
• Graph transformations of sinusoidal curves help us model data for periodic phenomena.
Because the sine and cosine curves are related via a horizontal shift, this amounts to
finding A, B, ω, and φ such that the graph of
A sin(ωx − φ) + B
best fits our data points.
2/9-2/10.
• To graph y = A sin(ωx − φ) + B using key points, we use the same procedure but the
2π
horizontal interval for a fundamental period is shifted. The period is still
,
ω
φ 2π
φ
but the shifted interval is
,
+
. Divide the shifted interval into four subintervals
ω ω
ω
of equal length, and proceed as before.
• How to fit a sinusoidal curve y = A sin(ωx − φ) + B to Data:
10
MA 108: THINGS WE’VE LEARNED
(1) Find the amplitude
largest data value − smallest data value
A=
2
(2) Find the vertical shift
largest data value + smallest data value
2
(3) Find ω using the period T given in the data:
2π
.
ω=
T
(4) Find the phase shift φ using an extremal value in the data. If y achieves a minimum
π
π
at x0 , set − = ωx0 − φ to solve for φ. If y(x0 ) is a maximum, set = ωx0 − φ
2
2
END OF MATERIAL FOR QUIZ 4 FEBRUARY 16
B=
2/13-2/15.
• Because the trig functions are not one to one, we define inverses to the trig functions on
a restricted interval. Which interval we choose depends on the function.
• The inverse function for sin x, denoted sin−1 x or arcsin x, is the function defined by
π
π
y = sin−1 x if and only if x = sin y, for −1 ≤ x ≤ 1 and − ≤ y ≤ . The function
2h
2
h π πi
π πi
sin−1 x maps [−1, 1] to − , , and the function sin x maps − ,
to [−1, 1].
2 2
2 2
• The inverse function for cos x, denoted cos−1 x or arccosx, is the function defined by
y = cos−1 x if and only if x = cos y, for −1 ≤ x ≤ 1 and 0 ≤ y ≤ π. The function cos−1 x
maps [−1, 1] to [0, π], and the function sin x maps [0, π] to [−1, 1].
• The inverse function for tan x, denoted tan−1 x or arctan x, is the function defined by
π
π
y = tan−1 x if and only if x = tan y, for −∞ < x < ∞ and − < y < . The function
2 2
π π
π π
−1
tan x maps (−∞, ∞) to − ,
, and the function tan x maps − ,
to (−∞, ∞).
2 2
2 2
2/16. Most of the problems in 8.1 and 8.2 ask you to find the exact value of expressions involving inverse trig functions. There are different ways to approach simplifying these expressions,
depending on the type of expression.
• If f is a trig function with inverse f −1 , then the expression f −1 (f (x)) will simplify to
a real number representing an angle. If x is in the special domain over which we
invert f , then f −1 (f (x)) = x. If not, then we find f (x), a real number representing the
value of a trig function and then evaluate the function f −1 at the number y = f (x).
• For example
5π
π
−1
−1
sin (sin(1/10)) = 1/10 and sin
sin
=−
4
4
π π
Because the special interval over which we invert sin x is − 2 , 2 . The
1/10 is
number
5π
5π
1
√
in the special interval, while 4 is not. So we have to evaluate sin 4 = − 2 , and then
find the angle in − π2 , π2 whose sine function is equal to − √12 , which is − π4 .
• We briefly discussed solving simple equations like 2 cos−1 (x) = π.
MA 108: THINGS WE’VE LEARNED
11
• Simplifying expressions involving one trig function and the inverse of a different trig
function: Let f and g be trig functions. Then to simplify f (g −1 (x)), note that g −1 (x)
will be an angle called y. We will be able to determine which quadrant the angle
y = g −1 (x) lives in because of the special interval we use to invert g. Then we can draw
a reference triangle and label two of the sides using the number x written in terms of a, b,
and r. Then we can evaluate the trig function f at the angle y using the definitions
of the trig functions in terms of a, b and
r.
• For example: to evaluate, sin tan−1 21 , we draw a reference triangle for the angle
y = tan−1 12 . This angle y must live in Quadrant I because we invert the tangent
function over the right hand half of the unit circle. Then we know
√ that (2,√1) is a point√on
the terminal side of the angle y because tan y = ab . Then r = a2 + b2 = 12 + 22 = 5.
Then sin y = rb = √15 .
2/17.
• Notational note: x ∈ [a, b] means ”x is in the interval [a, b]”. The ∈ symbol just means
’in’ something, usually set of some kind.
• The inverse function for csc x, denoted sin−1 x or arccsc x, is the function
πby
i
h π defined
−1
y = csc x if and only if x = csc y, for x ∈ (−∞, −1] ∪ [1, ∞) and y ∈ − , 0 ∪ 0, .
2
2
h π πi
−1
The function csc x maps (−∞, −1] ∪ [1, ∞) to − , 0 ∪ 0, , and the function csc x
2
2
h π πi
maps − , 0 ∪ 0,
to (−∞, −1] ∪ [1, ∞).
2
2
• The inverse function for sec x, denoted sec−1 x or arcsec x, is the function
h π defined
π by
i
−1
y = sec x if and only if x = sec y, for x ∈ (−∞, −1] ∪ [1, ∞) and y ∈ 0,
∪
,π .
2
2
h π π i
−1
The function sec x maps (−∞, −1] ∪ [1, ∞) to 0,
∪
, π , and the function sec x
2
2
h π π i
∪
, π to (−∞, −1] ∪ [1, ∞).
maps 0,
2
2
• The inverse function for cot x, denoted cot−1 x or arctan x, is the function defined by
y = cot−1 x if and only if x = cot y, for −∞ < x < ∞ and 0 < y < π. The function
cot−1 x maps (−∞, ∞) to (0, π), and the function cot x maps (0, π) to (−∞, ∞).
• Since your calculator has no sec−1 x, cot−1 x, or csc−1 x buttons but it does have sin−1 x,
cos−1 x, and tan−1 x buttons , use reciprocal identities to approximate expressions like
csc−1 (20).
• Today we also learned how to rewrite expressions like sin(cos−1 (u)) as an algebraic
expression in u.
END OF MATERIAL FOR TEST 2. TEST 2 IS WEDNESDAY 2/22
2/23. Section 8.3: Establishing Trig Identities
• Definition: two functions f and g are identically equal if f (x) = g(x) for all x such
that both functions are defined. We call the expression ”f (x) = g(x)” an identity
• To establish an identity, your job is to pick one side of the identity and transform it into
the other side using a sequence of
(1) algebraic manipulations, and
(2) substitutions justified by the basic trig identities.
12
MA 108: THINGS WE’VE LEARNED
• Useful factoring formulas include
(1) Difference of Squares
a2 − b2 = (a − b)(a + b)
(2) Sum of Cubes
a3 + b3 = (a + b)(a2 − ab + b2 )
(3) Difference of Cubes
a3 − b3 = (a − b)(a2 + ab + b2 )
• While all of these factoring formulas will make your life easier, the Difference of Squares
will be the most important for this class.
• Advice for Establishing Identites:
(1) Usually it’s easier to start with the most complicated side of the identity.
(2) If both sides are equally complex, start with the one you like better.
(3) Look for patterns in useful algebra tricks (rewriting sums/differences of quotients
as one, multiplying quotients top and bottom by other expressions, factoring)
(4) If you’re stuck, try writing everything in the your expression in terms of sines and
cosines using the Quotient and Reciprocal identities.
(5) Always keep the desired result in mind. Try to make substitutions that make your
expression look more like the goal expression.
2/24. How to get full credit for establishing a trigonometric identity on a test or quiz:
(1) Write down one side of the identity. (Don’t assume what you want to prove!!!)
(2) Write a sequence of equalities obtained from your starting expression where you make a
single substitution or algebraic manipulation at each step.
(3) For every substitution made using a trigonometric identity (Quotient, Pythagorean etc)
refer to the relevant identity.
(4) Stop when your expression matches the other side of the identity.
2/27.
• Today we covered the sum and difference identities for the sine, cosine and tangent
functions. By assuming the difference identity for cosine, we can derive the rest. These
identities can be found in Section 8.4. You will not need to memorize them, but you
will need to know how to use them both in establishing identities and in application
problems.
• Sum and Difference formulas:
cos(α + β) = cos α cos β − sin α sin β
cos(α − β) = cos α cos β + sin α sin β
sin(α + β) = sin α cos β + cos α sin β
sin(α − β) = sin α cos β − cos α sin β
tan(α + β) =
tan α + tan β
1 − tan α tan β
tan(α − β) =
tan α − tan β
1 + tan α tan β
2/29. Today we did an example of using sum and difference identities to calculate exact value.
Then students worked in groups proving identities and presented proofs to the class. Also, I
handed out Quiz 5 which is take-home (a copy is posted on the course website)
MA 108: THINGS WE’VE LEARNED
13
3/01. Today we covered the double angle and half angle formulas. We will use these to find
exact values and to establish trig identities.
• The double angle formulas follow easily from the sum and difference formulas:
sin(2θ) =
2 sin θ cos θ
1 − 2 sin2 θ = 2 cos2 θ − 1
2 tan θ
tan(2θ) =
1 − tan2 θ
• Using the double angle formulas, we can derive these half-angle formulas, which are
useful for calculating exact values:
r
α
1 − cos α
sin
=±
2
2
r
α
1 + cos α
cos
=±
2
2
r
α
1 − cos α
=±
tan
2
1 + cos α
We choose the sign of the square root based on which quadrant contains the terminal
α
.
side of the angle
2
• These formulas involvlng half-angles are more useful for establishing trigonometric identities:
cos(2θ) =
sin2 θ − cos2 θ =
1 − cos α
2
2
α 1 + cos α
cos2
=
2
2
α 1 − cos α
tan2
=
2
1 + cos α
• In particular, there are two nice formulas for the tangent function of a half angle.
α
1 − cos α
sin α
tan
=
=
2
sin α
1 + cos α
We can use these in establishing identities where we don’t want to choose a sign for the
tangent of the half-angle.
sin2
α
=
3/2. Today we practiced proving identities again, and learned the Product to Sum and Sum to
Product identities
• Product to Sum Identities
1
sin α sin β = [cos(α − β) − cos(α + β)]
2
1
cos α cos β = [cos(α − β) + cos(α + β)]
2
1
sin α cos β = [sin(α + β) − sin(α − β)]
2
14
MA 108: THINGS WE’VE LEARNED
• Sum to Product Identities
sin α + sin β = 2 sin
α+β
2
cos
α−β
2
α+β
α−β
cos
2
2
α+β
α−β
cos α + cos β = 2 cos
cos
2
2
α+β
α−β
cos α − cos β = −2 sin
sin
2
2
END OF MATERIAL FOR QUIZ 6. QUIZ 6 IS FRIDAY MARCH 16 IN CLASS
sin α − sin β = 2 sin
3/12. Trigonometric Equations : Sections 8.7 and 8.8
• The fundamental problem for sections 8.7 and 8.8 is solving equations like sin(f (θ)) = a
where a is some real number. We solve equations with a trig function on one side, and
a real number on the other. The argument of the trig function might be just θ or it
might be a function of θ. For example, the argument of cos(2πθ − 1) is 2πθ − 1. The
argument of cot θ is θ.
5
• When we solve an equation like x + 1 = 4, for x, we obtain one unique solution by
2
getting x on one side of the equation and a number on the other. A key difference
when we solve trigonometric equations is that once we isolate the trig function on one
side, there is an infinite set of angles θ that make the equation true. So when we solve
sin(3θ) = 1, we want to find a concise description of the infinite set of angles θ such that
sin(3θ) = 1.
3/14. Solving Trigonometric Equations
θ π
−
− 1 = 0 for θ.
3 4
(1) First we want the trig function ”by itself” so we add 1 to both sides and divide by
2:
θ π
θ π
θ π
1
2 cos
−
− 1 = 0 ⇐⇒ 2 cos
−
= 1 ⇐⇒ cos
−
=
3 4
3 4
3 4
2
• A Step by Step approach to solving 2 cos
θ π
1
− and a = .
3 4
2
(2) Since the period of the cosine function is 2π, I will find the infinite solution set by
1
first determining which u ∈ [0, 2π) satisfy cos u = . Such u must have reference
2
π
1
π
angle . Also, since is positive, u must be in either Quadrants I or IV. So, u =
3
2
3
5π
or
. The equation will be true whenever the argument of our trigonometric
3
function is equal to u, in other words when f (θ) = u.
(3) Now letting k be any integer, we set
Now, my equation is in the form cos(f (θ)) = a where f (θ) =
θ π
− = u + 2πk
3 4
MA 108: THINGS WE’VE LEARNED
15
for each choice of u. Solving for θ will give a description of the infinite solution set
in each case. Note that if the argument was simply θ and not a more complicated
expression, we would be done at this step, obtaining a family of solutions for each
choice of u.
(4) The first equation we solve is
π
θ
π π
θ π
− = + 2πk ⇐⇒
= + + 2πk
3 4
3
3
4
3
θ
7π
3 π
4 π
θ
=3
=
+
+ 2πk ⇐⇒ 3
+ 2πk ⇐⇒
3
3 4
4 3
3
12
7π
θ=
+ 6πk
4
5π
Then we solve the second equation obtained by letting u =
:
3
θ π
5π
θ
π 5π
− =
+ 2πk ⇐⇒
= +
+ 2πk
3 4
3
3
4
3
θ
3 π
4 5π
θ
23π
=
+
+ 2πk ⇐⇒ 3
=3
+ 2πk ⇐⇒
3
3 4
4 3
3
12
23π
θ=
+ 6πk
4
So we can describe the solution set in terms of the two families
7π
23π
θ=
+ 6πk , θ =
+ 6πk
4
4
where k is any integer.
• Note that in Step 2 and Step 3 we used the fact the period of cosine is 2π. If we were
solving an equation of the form tan(f (θ)) = a or cot(f (θ)) = a, then we would look for
u in the interval [0, π) in Step 2, and in Step 3 we would solve equations of the form
f (θ) = u + πk.
3/15-3/16. Strategies for Solving Trigonometric Equations
• Recall that we solve x2 − 3x + 2 = 0 by factoring and using the zero product principle:
e.g. (x − 1)(x − 2) = 0 is the factored form which gives us two solutions x = 1 and x = 2.
We can solving equations that are quadratic in a trig function by factoring and setting
each factor to zero:
2 cos2 θ + cos θ − 1 = 0 ⇐⇒ (2 cos θ − 1)(cos θ + 1) = 0
This gives us two equations
2 cos θ − 1 = 0,
cos θ + 1 − 0
That we can solve for θ. Each equation will give us a family or families of solutions.
• We can use Trigonometric Identities to solve equations that contain a mixture of trig
functions. We use the identities to convert the equation into an equation that is either
linear or quadratic in one function. For example, using one version of the double angle
formula for cosine, we can solve cos(2θ) = sin θ:
cos(2θ) = sin θ ⇐⇒ 1 − 2 sin2 θ = sin θ ⇐⇒ 2 sin2 θ + sin θ − 1 = 0
16
MA 108: THINGS WE’VE LEARNED
The last form of the equation can be factored to obtain (2 sin θ − 1)(sin θ + 1) = 0. At
this point we use the same method as in the previous example.
• Sometimes if I have a product of terms with mixed trig functions on one side and zero
on the other, I can factor to solve without using identities. For example, the equation
sin θ cos θ = 0 can be solved by setting either factor equal to zero. So we solve sin θ = 0
and cos θ = 0 to obtain the families of solutions.
• Sometimes algebraic manipulations help us put equations into forms we can tackle using
identities, factoring, and the like. The goal is always to get a trig function on one side
of the equation and a number on the other. For example: Solve cos θ = sec θ for θ:
1
⇐⇒ cos2 θ = 1 ⇐⇒ cos θ = ±1
cos θ
The we solve each of the equations cos θ = 1 and cos θ = −1 for θ.
• Finally, remember there are often multiple correct descriptions of a solution set. Often
the simplest way to check if two descriptions are equivalent is to draw the unit circle
and examine what happens for different values of k in the solution set.
END OF MATERIAL FOR TEST 3. TEST 3 IS FRIDAY MARCH 23 AND
COVERS 8.3-8.8
cos θ = sec θ
ReciprocalIdentity
⇐⇒
cos θ =
3/21. Applications of Right Triangles: Section 9.1
• Solving Right Triangles: Given a right triangle with hypoteneuse c, leg a across from
angle A and leg b across from angle B, we use the facts that
(1) a2 + b2 = c2 (Pythagorean Theorem)
(2) A + B = 90
to determine all values of A, B, a, b, c given partial information.
• Bearing/Direction for Navigation and Surveying. The bearing or direction from a
point O to a point P is given by stating the direction along the North-South axis through
O, the angle (in degrees) formed by the North-South Axis through O and the line OP ,
and the direction of rotation away from the North-South axis given by this angle. For
example, some bearings include N 30◦ E and S80◦ W .
• We can use right triangles to model many physical situations. We saw some at the
beginning of the semester. Additional topics include Bearing for Navigation, Average
Incline of a Section of Road, and Stellar Parallax.
3/23. Section 9.2: The Law of Sines
• An oblique triangle is any triangle that is not a right triangle. We need some new tools
to solve oblique triangles.
• We always label the angles of our oblique triangles with capital letters A, B and C,
and then the side across from each letter with the corresponding lowercase letter. For
example, a is the side across from A.
• The Law of Sines: For a triangle with sides a, b, c and opposite angles labeled A, B, C
respectively,
sin A
sin B
sin C
=
=
a
b
c
• We also need the fact that A + B + C = 180◦ to solve these triangles.
• There are five ways that we are given partial information about the triangle. Let A
mean ”angle” and S mean ”side.” Then ASA, SAA, SSA SAS and SSS denote the
MA 108: THINGS WE’VE LEARNED
17
cases ”angle-side-angle”, ”side-angle-angle” and so on. ”angle-side-angle” means that
the known side is between the two known angles.
• The cases ASA and SAA will produce one triangle. We can solve these by setting up
equations for the unknown quantities using the Law of Sines and the fact A + B + C =
180◦ .
• We can solve SSA using the Law of Sines, but depending on the values of a, b and sin A,
there will be either 1, 2 or 0 triangles satisfying the given conditions.
3/26. Today we went over an example in detail of an SSA case that produced two triangles.
Then we worked on solving triangles in groups.
3/28.
• The Law of Cosines : For any triangle with sides a, b, c opposite angles A, B, C, the
following formulas hold:
– a2 = b2 + c2 − 2bc cos A
– b2 = a2 + c2 − 2ac cos B
– c2 = a2 + b2 − 2ab cos C
• We worked Problem 38 from Section 9.2 and Problem 44 from Section 9.3.
• The Law of Cosines is used to solve the SSS and SAS cases of triangles.
• Sometimes once we obtain part of the triangle using the Law of Cosines, we can switch
to the Law of Sines. However, since all angles in Quadrants I and II (which are the
angles that can live in a triangle, being ≤ 180), have postive sine function, we will need
to check two angles to see which fits our problem when we use the Law of Sines. This
ambiguity does not arise when we use the Law of Cosines because obtuse angles live in
Quadrant II and have negative cosine function.
• Quiz 7 covers 9.1 and 9.2 only. See Practice Problems.
3/29. Today we did practice problems in groups. We found out that when students start SSS
and SAS cases with the Law of Cosines and finish with the Law of Sines, they tend to forget to
check both possible angles that the sin−1 (x) indicates. My advice: if you are solving a SSS or
SAS triangle, stick with the Law of Cosines through the whole problem.
3/30. Today we finished Section 9.2 by doing some more word problems from that section.
4/2. Today we started Section 9.4 which covers applications of the area of a triangle.
• The area K of a triangle with base b and height h is given by the formula
1
K = bh
2
This is the formula you already know. We can use this formula, along with the Law of
Cosines and trigonometric identities to find other formulas for the area of a triangle that
might be useful when we don’t know h.
• The area K of the SAS triangle with sides a, b and included angle C is given by the
formula
1
K = ab sin C
2
• The above formula is still true if we change the names of the two sides and their included
angle:
1
1
K = ac sin B
K = bc sin A
2
2
18
MA 108: THINGS WE’VE LEARNED
• Heron’s formula lets us calculate the area of a SSS triangle. The area K of a triangle
with sides a, b, and c is given by the formula
p
K = s(s − a)(s − b)(s − c)
1
where s = (a + b + c).
2
4/4. Today we derived Heron’s formula in class. Then we did problems 34 and 42 from Section
9.4.
END OF MATERIAL FOR QUIZ 8
4/9. Today we began Section 9.5.
• An object moves with simple harmonic motion if it moves on a coordinate axis
where the distance d from the rest position at time t is given by either d = a cos ωt or
d = a sin ωt, where a, ω are constants with ω > 0. The motion has amplitude |a|, period
ω
2π
and moves with frequency
.
ω
2π
• When making an equation to model simple harmonic motion, the y-axis represents displacement from equilibrium. We let y = 0 correspond to equilibrium, and choose an
orientation for the motion, e.g. increasing y coordinate could indicate rightward or
upward motion for our object.
• We use d = a sin ωt when our model starts with the object at equilibrium when t = 0.
We use d = a cos ωt when the object is at maximum displacement from equilibrium when
t = 0. Whether or not a is positive or negative depends on how we choose the orientation
in our coordinate system.
• Recall that we fit curves of the form A sin(ωx − φ) + B to data in Section 7.8. Then φ
represented a phase shift, and B represented a vertical shift. In 9.5 we use the choice
of cosine or sine functions in simple harmonic motion models instead of a phase shift.
Also, the vertical shift B is not needed for simple harmonic motion models because we
choose our coordinate system so that the equilibrium position corresponds to y = 0.
4/11. Today we covered Damped Harmonic Motion:
• Let a be the displacement at t = 0 of an object of mass m with damping factor b. Assume
2π
is the period of the simple harmonic motion model for the object. Then the
that
ω
displacement d of the oscillating object from its equilibrium position at time t is given
by
" r
! #
2
b
d(t) = ae−bt/2m cos
ω2 −
t
4m2
• In simple harmonic motion, the amplitude of the oscillation is given by a constant. In
damped motion models, the amplitude is given by the decreasing function e−bt/2m . The
amplitude decreases over time as the friction fights the oscillation.
• We can predict how changing the parameters in the model will affect its behavior. The
amplitude will decrease faster as the coefficient of friction increases. As the weight
increases, the amplitude will decrease more slowly, because the momentum of the object
will be greater.
MA 108: THINGS WE’VE LEARNED
19
4/12. Graphing the Sum of Two Functions
• Let f (x) be a function that can be written as a sum of two functions f1 (x) and f2 (x),
in other words f (x) = f1 (x) + f2 (x). Then we can draw the graph of the two functions
by plotting ordered pairs (x, f (x)) and inferring the shape between the plotted points
based on our knowledge of the behavior of the functions.
• For example, we can graph the function f (x) = x + sin x by calculating the values of x
and sin x at points x, then adding them together to get the y-value for f (x).
• Good choices of x for plotting the functions will depend on the components f1 (x) and
π
3π
f2 (x). To graph f (x) = x + sin x, we chose the points 0, , π, , 2π because sin x has
2
2
nice values at the quadrantal angles.
• To graph the sum of two sinusoidal curves, we need to determine the fundamental period
and key points for each component functions and make strategic choices for plotting
points from there.
4/18. Today we introduced Conic Sections (Section 11.1). Conic sections are the twodimensional geometric figures that arise as the intersection of a right circular cone and a plane.
• Let a and g be two lines that intersect in a point V . Fix a and rotate g about a
maintaining a constant angle between them. The collection of points swept out by g is
a right circular cone. The fixed line a is the axis of the cone, V is the vertex, the
lines swept out by g are the generators, the two pieces of the cone that intersect at the
vertex are the nappes.
• Circles arise when the plane is perpendicular to the axis of the cone and does not
intersect the vertex.
• We get an ellipse when the plane intersects one nappe of the cone in all of the generators
but is not perpendicular to the axis.
• Parabolas arise when the plane is parallel to exactly one generator (and so intersects
one nappe but not all of the generators in the nappe)
• Hyperbolas occur when the plane intersects both nappes.
• When the plane intersects the vertex we get degenerate conics, which are points, lines,
or pairs of intersecting lines.
4/23.
• A parabola is the collection of all points P = (x, y) in the plane that are the same
distance from a fixed point F as they are from a fixed line D. The point F is called the
focus and the line D is the directrix. So a parabola is the set of all points P satisfying
d(F, P ) = d(P, D). The axis of symmetry is the line through which is perpendicular
to the directrix D. The vertex V is the intersection of the parabola with the axis of
symmetry.
• Consider the case where V = (0, 0), F = (a, 0) (a is positive) and D has equation x = −a.
Then the equation of the parabola is y 2 = 4ax. The parabola opens to the right.
• Still assume a > 0. Then if F = (−a, 0), the directrix is x = a, and we have y 2 = −4ax.
Example: find the equation of parabola with vertex at (0, 0), focus at (−2, 0). Draw
the picture. Formula: a = 2, y 2 = −4(2)x ⇐⇒ y 2 = −8x. To sketch, draw focus,
vertex, and directrix. Then find two points on the parabola by plugging x = −2:
y 2 = −8(−2) = 16 ⇐⇒ y = ±4. Plot the two points on the parabola (−2, −4) and
(2, 4).
20
MA 108: THINGS WE’VE LEARNED
• Remark: the line segment across the parabola through F perpendicular through the
axis of symmetry is called the latus rectum. Always has length 4a. Verify line from
(−2, −4) to (2, 4) has length 4a = 8.
• Parabolas that open up or down: we expect the equations for these parabolas to look
the same except with the roles of x and y switched. Let a > 0. For the parabola with
equation x2 = 4ay, we have V = (0, 0), F = (0, a), D given by y = −a, axis of symmetry
x = 0.
• : This summary captures the four cases. There are two examples written out after the
table.
Vertex Focus Directrix Equation
Direction
Axis of Symmetry
2
(0, 0)
(a, 0)
x = −a
y = 4ax
Opens Right
x-axis
(0, 0) (−a, 0)
x=a
y 2 = −4ax
Opens Left
x-axis
(0, 0)
(0, a)
y = −a
x2 = 4ay
Opens Upward
y-axis
2
(0, 0) (0, −a)
y=a
x = −4ay Opens Downward
y-axis
2
2
• The equation of a parabola with vertex (0, 0) focus at 0,
: x2 = 4
y ⇐⇒ x2 =
3
3
8
2
yThe axis of symmetry is now the y-axis and directrix will be y = − . Sketch the
3
3
parabola with labeled V , F and D.
• Example: Find the equation of the parabola with focus at (0, 4) and vertex (0, 0). What
is the directrix, points that define the latus rectum? Sketch the graph. Solution: We
know that the focus is on the y-axis and above the y-axis so that the parabola opens
upward with axis of symmetry x = 0. Therefore the equation must be of the form
x2 = 4ay. Since the focus is (0, a) = (0, 4), the equation is x2 = 4(4)y = 16y. The
MA 108: THINGS WE’VE LEARNED
21
directrix is y = −4. The points of the latus rectum are the points on the parabola with
y-coords equal to 4. To find the x-coords x2 = 16(4) ⇐⇒ x2 = 64 ⇐⇒ x = ±8 so the
points are (−8, 4) and (8, 4). Sketch the picture.
• Consider the parabola with equation y 2 = −7x and vertex (0, 0). Find the focus, directrix, axis of symmetry, latus rectum. Sketch the graph and label F, D, and the latus
rectum.
Solution: We know that this parabola opens left with focus (0, −a)
directrix
and 7
7
x = a. We have y 2 = −4ax ⇐⇒ −7 = −4a ⇐⇒ a = . Then F = 0, −
and D
4
4 7
7
has equation x = . The y-coords of latus rectum points satisfy y 2 = −7 −
⇐⇒
4
4
7
y = ± . Here is the sketch:
2
22
MA 108: THINGS WE’VE LEARNED
4/25. This table summarizes the important facts about parabolas with vertex at (h, k) and axis
of symmetry parallel to a coordinate axis. (We will not cover parabolas that are ”tilted.”)
Vertex
(h, k)
(h, k)
(h, k)
(h, k)
Focus
(h + a, k)
(h − a, k)
(h, k + a)
(h, k − a)
Directrix
Equation
Direction
Axis of Symmetry
x = h − a (y − k)2 = 4a(x − h)
Opens Right
y=k
x = h + a (y − k)2 = −4a(x − h)
Opens Left
y=k
2
y = k − a (x − h) = 4a(y − k)
Opens Upward
x=h
y = k + a (x − h)2 = −4a(y − k) Opens Downward
x=h