Precalculus Syllabus

Precalculus Syllabus

Mrs. Cuer (Rm. 192)
Phone Number: 652-3000 Ext. 7192
E-mail: kcuer@iroquoiscsd.org
Daily Materials: Text, notebook, pens/pencils, TI-83 Plus Calculator

Course Description

The Precalculus course will cover topics listed below. The course will also require the knowledge of the TI-83 Plus calculator.

1. Functions and Their Graphs
2. Polynomial and Rational Functions
3. Exponential and Logarithmic Functions
4. Trigonometric Functions
5. Analytic Trigonometry
6. Additional Topics in Trigonometry
7. Systems of Equations and Inequalities
8. Matrices and Determinants
9. Sequences, Series, and Probability
10. Topics in Analytic Geometry
11. Analytic Geometry in Three Dimensions
12. Limits and an Introduction to Calculus


Student Responsibilities

Students must be prepared, with supplies and assignments, and be on time to class daily. Everyone in Mrs. Cuer’s classroom will be respectful and courteous to one another. If a student chooses not to follow through with their responsibilities, disciplinary measures will be taken.
Tardiness: Every three tardies will result in a discipline referral. If tardiness becomes a problem, a parent/guardian will be contacted.
Absences: Students are responsible for making up any missed notes and assignments. If a student is absent for a test or a quiz, it is that student’s responsibility to request a suitable time for a make-up test within three days. If the test or quiz isn’t made up within the allotted time, then a zero will be assigned for a grade.


Grading Policy

A student’s marking period grade is determined through a mix of tests, quizzes, homework assignments, and active participation. Students will have the opportunity to earn a maximum number of points in each area listed above and the total of the points they earn will be divided by the total possible number of points to determine their average for each marking period.

Tests, Quizzes and Projects

Tests and quizzes will be given often as we move through the Precalculus curriculum. Note: if a student is absent the day before a test or quiz, they are expected to complete the test or quiz as it is scheduled, if no new material was learned on that day. Also, there are no retakes on test or quizzes.

Homework

Homework will be assigned on a daily basis. Students are expected to complete the assignment within the allotted time frame and are assessed based mainly upon effort. Late assignments will not be accepted (unless there are extenuating circumstances, in which case the student or parent/guardian should speak with me privately). If a student is absent and a homework assignment is given that day, the student should see me immediately upon return to school to get the assignment. (Even if the student doesn’t have my class that day!) They will be given an extra day to complete it, if needed. It is important for students to complete assignments, as they are a means for deeper understanding and further exploration within a topic.

Participation

Participation includes any of the following forms: answering questions, asking questions, class discussion, board work, group work and preparation for class (materials & assignments).

Final Grade

A final grade for the course will be found by averaging five grades, the four quarters and a cumulative final exam.

Extra Help

Extra help is available! I am available at various times throughout the school day and will also stay after school for additional help. Be sure to let me know if you plan on coming in for help so I can make sure I am available that particular time and day. There is also the Academic Center, room 138, where students can sign in for extra help throughout the day.

5.3 Solving Trigonometric Equations

5.3 Solving Trigonometric Equations

Goal: To isolate the trigonometric function involved in the equation.

A). Solving a Trigonometric Equation

Example 1:
4 cos²(2x) - 2 = 0
4 cos²(2x) = 2
cos²(2x) =2/4
cos²(2x) = ½
cos (2x) = ± the square root of ½= ± (the square root of 2)/2
So now Let y = 2x
Now when does cos y = ±(the square root of 2)/2
y = 45°, 135°, 225°, 315°, π/4, 3π/4, 5π/4, 7π/4

so 2x = 45°
x = 22.5° or π/8

2x = 135°
x = 67.5° or 3π/8

2x = 225°
x = 112.5° or 5π/8

2x = 315°
x = 157.5° or 7π/8

and as you can see, this only takes you from 0 radians to π radians. If the solution is between [0, π), we would be done but usually it is between [0,2π) so we have to add the following solutions:

x = {9π/8, 11π/8, 13π/8, 15π/8}

Example 2:

(the square root of 2) (sin x )+ 1 = 0
(the square root of 2) sin x = -1
sin x = -1/ (the square root of 2) = -(the square root of 2)/2
sine is negative in the III and IV quadrant so

x = 225°, 315°, 5π/4, 7π/4

Example 3:

tan²x – 1 = 0
tan²x = 1
tan x = ±1

x = 45°, 135°, 225°, 315°, π/4, 3π/4, 5π/4, 7π/4

Example 4:

2 sin²x = 2 + cos x
2(1 - cos²x) = 2 + cos x
2 - 2cos²x = 2 + cos x
0 = 2cos²x + cos x
0 = cos x(2cos x + 1)
0 = cos x and 0 = 2cos x + 1

0 = cos x
x = 90°, 270°, π/2, 3π/2

2 cos x + 1 = 0
cos x = - ½
x = 120°, 240°, 2π/3, 4π/3

Check your solutions to check for erroneous solutions
either by algebra or using the graphing utility

B) Approximate the solution

Example 5:
csc²x = 3 csc x + 4
csc²x - 3 csc x - 4 = 0
(csc x - 4)(csc x + 1) = 0
csc x = 4 and csc x = -1

csc x = 4
(1/ sin x) = 4
sin x = 1/4
x ≈ 14.48° or .2527 radians

csc x = -1
x = 270° or 3π/2

Example 6:

4 sin x = cos x - 2
Use your graphing utility and place each side of the equation into 2 different equation functions.
f(x) = 4 sin x and g(x) = cos x - 2
You will find the intersection of the two equations are
(-2.390169347, -2.730717947)
(-.2614659803,-1.033987935)
(3.89301596, -2.730717947)
(6.021719327, -1.033987935)

or

(-136.9466°, -2.730717947)
(-14.9809 °, -1.033987935)
(223.05338 °, -2.730717947)
(345.0191 °, -1.033987935)

so therefore the solutions are:
x = {-136.9466 °, -14.9809 °, 223.05338 °, 345.0191 °}

x = {-2.390169347, -.2614659803, 3.89301596, 6.021719327}

Example 7:

cos x + sin x tan x = 2

cos x + (sin x)(sin x)/ (cos x) = 2

(cos²x + sin²x)/(cos x) = 2

1/(cos x) = 2

cos x = ½

x = {60 °, 300 °, π/3, 5π/3}

Example 8:

sec x csc x = 2 csc x
sec x csc x - 2 csc x = 0
csc x (sec x - 2 ) = 0

csc x = 0
1/ sin x = 0
sin x = undefined
no solution

sec x - 2 = 0
sec x = 2
1/ cos x =2/1
cos x = ½
x = {60 °, 300 °, π/3, 5π/3}

5.3 homework #45 - page 400; 3 - 59 (by 4's), 73, 76

Precalculus 3.3 notes: Properties of Logarithms

3.3 Properties of Logarithms

Change of Base:
Let a, b, and x be positive real numbers such that a≠1 and b≠1. Then log_ax can be converted to a different base using any of the following formulas.

Base b

log_ax=(log_bx)/(log_ba)

Base 10

log_ax=(log₁₀x)/(log₁₀a)

Base e

log_ax = (ln x)/(ln a)

Examples:

log₇4 = .7124143742
log₁₅1460 = 2.690567447

log_(1/3)(0.015)=3.822736302

Properties of Logarithms:
Let a be a positive number such that a≠1, and let n be a real number. If u and v are positive real numbers, the following properties are true.

1. log_a(uv) = log_au + log_av

2. ln(uv) = ln u + ln v

3. log_a (u/v) = log_au - log_av

4. ln (u/v) = ln u - ln v

5. log_auⁿ = n log _au

6. ln uⁿ = n ln u

Examples: Expand or condense:

1. ln (x⁴ y½)/(y²z³) = 4ln x + 1/2 ln y - 2 ln y - 3 ln z = 4 ln x - 3/2 ln y - 3 ln z

2. ln (x⁴(x²+1))^¼= (1/4)( 4 ln x + ln (x² + 1))

3. 4(lnz + ln(z+5)) - 2ln(z-5) + 3ln(x-2) =

Precalculus 4.3 notes

Precalculus 5.2 notes: Verifying Trigonometric Identities

5.2 Verifying Trigonometric Identities

Example 1: sin x = 0 is a conditional equation that is true for only some values in the domain.

x = {0, π, 2 π, 3 π, etc..} so x = n π

Example 2: sin²x = 1 – cos²x is an identity and it is true for all real numbers x.

Guidelines:

1. Work with one side of the equation at a time. Usually the more complicated side first.
2. Look for opportunities to factor an expression, add fractions, square a binomial, or create a monomial denominator.
3. Look for opportunities to use the fundamental identities. Note which functions are in the final expression you want. Sines and Cosines pair up well, as do secants and tangents, and cosecants and cotangents.
4. If the preceding guidelines do not help, try converting all terms to sines and cosines.
5. Try something! Even making an attempt that leads to a dead end gives insight.

A). Verifying a Trigonometric Identity

Example 1: cos²β – sin² β = 2 cos² β – 1

cos²β –(1 – cos² β) = 2 cos² β – 1

cos²β – 1 + cos² β = 2 cos² β – 1

2 cos² β – 1 = 2 cos² β – 1

Example 2: 2 – csc²x = 1 – cot²x

2 – (1 + cot²x) = 1 – cot²x

2 – 1 – cot²x = 1 – cot²x

1 – cot²x = 1 – cot²x

B). Combining Fractions before using Identities

Example 3: (cot α)/ (csc α – 1) = (csc α + 1)/(cot α)

(cot²α)/ ((csc α – 1)(cot α)) = (csc α + 1)/(cot α)
(csc²α – 1)/ ((csc α – 1)(cot α)) = (csc α + 1)/(cot α)

(csc α + 1) (csc α – 1)/ ((csc α – 1)(cot α)) = (csc α + 1)/(cot α)

(csc α + 1)/ (cot α) = (csc α + 1)/(cot α)

Example 4:

(cos x – cos y)/ (sin x + sin y) + (sin x – sin y)/ (cos x + cos y) = 0

((cos x – cos y)(cos x + cos y) + (sin x – sin y)(sin x + sin y))/ ((sin x + sin y)(cos x + cos y)) = 0

(cos²x – cos²y + sin²x – sin²y)/ ((sin x + sin y)(cos x + cos y)) = 0

(cos²x + sin²x – sin²y– cos²y)/ ((sin x + sin y)(cos x + cos y)) = 0

(cos²x + sin²x – (sin²y+ cos²y))/ ((sin x + sin y)(cos x + cos y)) = 0

(1 – 1)//(sin x + sin y)(cos x + cos y) = 0

0 = 0

C) Dealing with square roots and absolute value

Example 5:
(the square root of ( (1 – cos x)/(1 + cos x))) =
(1 – cos x) / (the absolute value of sin x)

(the square root of ( ((1 – cos x)(1 – cos x))/((1 + cos x)(1 – cos x))) =
(1 – cos x) / (the absolute value of sin x)

(the square root of ( (1 – cos x)²/(1 – cos²x)) =
(1 – cos x) / (the absolute value of sin x)

(the square root of ( (1 – cos x)²/(sin²x))) =
(1 – cos x) / (the absolute value of sin x)

(1 – cos x)/(the absolute value of sin x) =
(1 – cos x) / (the absolute value of sin x)

D). Using properties of logarithms and trigonometric identities to verify the identity.

Example 6: ln (absolute value of(sec x)) = - ln (absolute value of (cos x))

ln (absolute value of (sec x)) = ln (absolute value of (1/ cos x))

ln (absolute value of (sec x)) = ln (absolute value of (sec x))

E). Use the co-function identities to evaluate the expression

Example 7: cos² 14º + cos² 76º

sin²(90º - 14º) + cos² 76º

sin²(76º) + cos² 76º

1

Example 8: sin² 12º + sin² 40º + sin² 50º + sin² 78º

cos² (90º - 12º) + cos² (90º - 40º) + sin² 50º + sin² 78º

cos² 78º + cos² 50º + sin² 50º + sin² 78º

1 + 1

2

F). True or False: Is the square root of (tan²x) = tan x ?

Let x = - 45º

The square root of (tan²- 45º) = tan - 45º

The square root of ((-1)²) = -1

1 = -1 is false

Example 9:

tan⁵x = (tan³x)( sec²x) - tan³x

tan⁵x = tan³x (sec²x - 1)

tan⁵x =( tan³x)( tan²x)
tan⁵x = tan⁵x

5.2 homework: #44; pg. 389; 3 – 9 odd, 19 – 27 odd, 34, 35, 37, 43, 49, 55, 59, 71 – Quiz on 5.1 and 5.2 next class!!

Precalculus 5.1 notes: Using Fundamental Identities

5.1 Using Fundamental Identities

The Identities can be found on the following web site:
http://www.sosmath.com/trig/Trig5/trig5/trig5.html
Reciprocal identities
Pythagorean Identities
Quotient Identities
Co-Function Identities
Even-Odd Identities

Example 1: Evaluate the trigonometric functions given
Tan x = 7/24 and sec x is negative

Tan x is positive in the I and III quadrant
Sec x is negative in the II and III quadrant
Therefore this can be evaluated by knowing we are in the III quadrant

7² + 24² = c²
49 + 576 = c²
625 = c²
25 = c

Therefore:
Sin x = -7/25
Csc x = -25/7
Cos x = -24/25
Sec x = -25/24
Tan x = 7/24
Cot x = 24/7

Example 2: Simplify
cot x sec x
((cos x)/ (sin x) ) (1/ cos x)
1/ sin x
Csc x

Example 3:

(Cos²y) / (1 – sin y)

(1 – sin²y)/ (1 – sin y)

(1 + sin y)(1 – sin y)/ (1 – sin y)

1 + sin y

Example 4:

(tan²θ)/ (sec² θ)

(sin² θ)/(cos² θ) ÷ (1 / (cos²)

(sin² θ)/(cos² θ) × (cos²)/1

sin² θ

Example 5: Verify

Cos θ sec θ – cos² θ = sin² θ

Cos θ (1/ cos θ) - cos² θ = sin² θ

1 – cos² θ = sin² θ

sin² θ = sin² θ

Example 6: verify

(sec² θ - tan² θ + tan θ)/ (sec θ) = cos θ + sin θ

(1 + tan² θ – tan² θ + tan θ)/(sec θ) = cos θ + sin θ

(1 + tan θ)/ sec θ = cos θ + sin θ

1/sec θ + tan θ / sec θ = cos θ + sin θ

cos θ + (sin θ / cos θ) / (1/ cos θ) = cos θ + sin θ

cos θ + sin θ = cos θ + sin θ

Example 7: Factor

sec²x tan²x + sec²

sec²x ( tan²x + 1)

sec²x (sec²x)

sec⁴x

5.1 Homework: #43; pg. 381; # 5, 7, 25, 27, 37 – 43 odd, 55 – 59 odd, 67 – 77 odd, 91, 93, 101, 103

Precalculus notes on 4.8 Applications and Models

4.8 Applications and Models

Solving a Right Triangle:
Recall:

Formulas: given triangle ABC with angle C = 90º

Recall: Sin A = (opposite/hypotenuse), Cos A = (adjacent/hypotenuse),
Tan A = (opposite/ adjacent) or Tan A = (Sin A)/ (Cos A)

Sin A = (CB)/ (AB)
Cos A =(AC)/(AB)
Tan A = (CB)/(AC)

Angle of Elevation and Angle of Depression:

http://www.mathsteacher.com.au/year10/ch15_trigonometry/12_elevation_depression/23elevdep.htm

The angle of elevation of an object as seen by an observer is the angle between the horizontal and the line from the object to the observer's eye (the line of sight).
If the object is below the level of the observer, then the angle between the horizontal and the observer's line of sight is called the angle of depression.

Example 1: A shadow of length L is created by an 850 foot building when the sun is θº above the horizon. Write L as a function of θ.

Drawing your triangle, you use the opposite and the adjacent sides so you would use Tangent.

Tan θ = 850 / L so writing this to have L as a function of θ, means to solve for L.

L = 850 / tan θ

What are the values of L when θ = {10º, 20º, 30º, 40º, 50º, 60º}?

Answer: {4820.6, 2335.4, 1472.2, 1013.0, 713.23, 490.75} respectively.

In surveying and navigation, directions are generally given in the terms of BEARINGS. A bearing measures the acute angle of a path or line of sight makes with a fixed north-south line.
Check out this web-site:
http://www.mathsteacher.com.au/year7/ch08_angles/07_bear/bearing.htm


Example 1: An airplane flying at 550 mph has a bearing of N 58º E. After flying 1.5 hours, how far north and how far east has the plane traveled from its point of departure?
1.5 hours times 550 mph = 825 mi.
Sin 58º = x/ 825
x = 699.6396793 miles ≈ 700 miles East
Cos 58º = y/ 825
x = 437.183393 miles ≈ 437 miles North

Harmonic Motion – check out this web site:
http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html
Amplitude – the maximum distance the ball moves vertically upward or downward from its equilibrium (or resting position) – displacement.

Example 1: lets say it took 4 seconds for the ball to move from resting position to resting position. This means the period is 4. The maximum displacement is 10 cm.
Using the formula: d = a sin ωt
The absolute value of a = 10
Period = (2π/ω) = 4
4ω = 2 π
ω = π / 2
Therefore: d = 10 sin (πt)/2

Definition: A point that moves on a coordinate line is said to be in Simple Harmonic Motion if its distance “d” from the origin at time “t” is given by either
d = a sin ωt or d = a cos ωt where “a” and “ω” are real numbers such that ω is greater than 0. The motion has the amplitude equal to the absolute value of a, period is equal to (2π)/ω and the frequency is equal to ω/ (2π)

Example 1: d = .5 cos 20πt
a. find the displacement: a = .5
b. find the frequency: ω/ (2π) = 20π/ (2π) = 10
c. least positive value of t when d = 0
0 = .5 cos 20πt
0 = cos 20πt now a thought: when does 0 = cos x x = π/2 so
20π t = π/2
t = 1/40

Example 2: d = (1/64) sin 792πt
a. find the displacement: a = 1/64
b. find the frequency: ω/ (2π) = 792π/ (2π) = 396
c. least positive value of t when d = 0
0 = (1/64) sin 792πt
0 = sin 792πt now a thought: when does 0 = cos x x = π/2 so
792πt = π/2
t = 1/792

Example 3: find a model for simple harmonic motion given:
Displacement (t = 0) of 2 feet
Amplitude of 2 feet
Period of 10 seconds
d = a cos ωt
a = 2
period = (2π/ω) = 10
(2π/10) = ω
π/5 = ω
Therefore: d = 2 cos πt/5

Section 4.8 Homework #41; pg. 361; #1 – 21 odd, 25, 29, 33, 37, 45, 49 – 59 odd, 65
pg. 351; #8, 18, 34, 59, 91, 93

Review Homework due the day after 4.8, Homework #42; pg. 370; # 111 – 123 odd, 127, 135, 143, 155, 159 – 171 odd.

Precalculus - notes for 4.7 Inverse Trigonometric Functions

4.7 Inverse Trigonometric Functions

I. The inverse sine function:
y = arc sin x if and only if sin y = x

We will use a capital A for Arcsin to show when the range is restricted so therefore:

y = Arc sin x
Domain: [-1, 1]
Range: restricted to [-π/2, π/2 ]

y = sin x some points are {(- π/2, -1), (0,0), (π/2, 1)}

so for the inverse, switch the x-values and y-values to get

y = Arc sin x some points are {(-1, - π/2), (0, 0), (1, π/2)}

Sketch the graph

Evaluating the inverse sine function:

Example 1: Arc sin (-1/2) = x
x = {-30º, 330º, 210º, - π/6, 11π/6, 7 π/6}

Example 2: Arc sin √(2)/2 = x
x = {45º, 135º, π/4, 3 π/4}

Example 3: Arc sin -√(3)/2 = x
x = {-60º, -120º, 300º, 240º, -π/3, -2π/3, 5π/3, 4π/3}
---------------------------------------------------------------------------------
II. The inverse cosine function:

y = arc cos x if and only if cos y = x

Again, to show a restricted range:

y = Arc cos x
Domain: [-1, 1]
Range: [0, π]
---------------------------------------------------------------------------------
II. The inverse tangent function:

y = Arc tan x if and only if tan y = x

Domain: all the real numbers
Range: (-π/2, π/2)
---------------------------------------------------------------------------------

Recall f ( f^-1 (x)) = x and f^-1 (f (x)) = x

Therefore: sin (Arc sin x) = x
Arc sin (sin x) = x
cos (Arc cos x) = x
Arc cos (cos x) = x
tan (Arc tan x) = x
Arc tan (tan x) = x

This only applies with sine in the interval [-π/2, π/2]
This only applies with cosine in the interval [0, π]
---------------------------------------------------------------------------------
Example 1: sin (Arc sin 0.7) = 0.7

Example 2: tan (Arc tan 35) = 35

Example 3: cos [ Arc cos (-0.3)] = -0.3

Example 4: Arc sin (sin 3 π)
Recall 3 π is not in the interval [0, π] so sin (3π) = sin (π) = 0 so
Arc sin (0) in the interval [-π/2, π/2] is equal to 0.

Example 5: sin (Arc tan 4/3) =
Using Pythagorean Theorem and since tangent is opposite divided by adjacent. Therefore 4 is the opposite and 3 is the adjacent to the angle. Using Pythagorean Theorem, that means that the hypotenuse is 5. So the sine would equal 4/5.

Example 6: tan (arc cos x/5) = θ
This means that x is the adjacent and 5 is the hypotenuse.

a² + b² = c²
x² + b² = 5²
b² = 25 - x²
b = the square root of (25 - x²)

so the tan θ = (the square root of (25 - x²))/5

Example 7: cot (arc tan 1/x)
This means that the tangent is equal to 1/x ,
tangent/1 = 1/x
taking the reciprocal of both sides gives us
1/tangent = x/1
1/tangent is equal to cotangent so

so the cotangent is equal to x.

Example 8: Arc cos 3/ (the square root of (x² - 2x + 10)) = Arc sin ___

Using Pythagorean Theorem:
3² + b² = (the square root of (x² - 2x + 10))²

9 + b² = x² - 2x + 10
b² = x² - 2x + 1
b² = (x - 1)(x - 1)
b² = (x - 1)²
b = x - 1

Therefore sin θ = (x - 1)/(the square root of (x² - 2x + 10))

so the Arc cos 3/(the square root of (x² - 2x + 10)) = Arc sin ((x - 1)/(the square root of (x² - 2x + 10))

Prove the identity:

Arc sin x + Arc cos x = π/2

Let θ = Arc sin x and β = Arc cos x

By substitution: θ + β = π/2

Now since: Arc sin x = θ and β = Arc cos x
This means sin θ = x and cos β = x.

Since x = x, then sin θ = cos β
For this to happen, the triangle has to be a right triangle.

Therefore θ + β = π/2

Since this is the same answer as our substitution above, we now know

Arc sin x + Arc cos x = π/2

4.7 homework #34; pg. 351; #1 - 15 odd, 19 - 23 odd, 27 - 53 odd, 57, 59, 71, 87, 88
pg. 341; #19, 45, 52, 64, 91
Quiz on Sections 4.5 - 4.6 next class

Precalculus 4.6 notes - Graphs of Other Trigonometric Functions

4.6 Graphs of Other Trigonometric Functions

I. Graph of the Tangent Function: y = a tan (bx - c) + d

A. The tangent function is odd. tan (-x) = - tan x
B. Therefore the graph of the y = tan x is symmetric with respect to the origin.
C. Period = π
D. Domain: all the reals except x ≠ π /2 + n π
E. Range: all the reals
F. Vertical Asymptotes: x = π /2 + n π

II. To sketch the graph y = a tan (bx - c) + d
A. Locate key points that identify the intercepts and asymptotes.
1. to find the asymptotes:

bx - c = -π /2 and bx - c = π /2

2. The midpoint between two consecutive asymptotes is an x-intercept of the graph.

3. The period of the tangent function is the distance between two consecutive asymptotes. This can also be found by: period = π /b

4. The amplitude of a tangent function is not defined. But if “a” is a negative number, the graph is a reflection in the x-axis

Example 1: y = tan 2x

bx - c = -π /2 OR bx - c = π / 2
2x = -π /2 OR 2x = π /2
x = -π /4 OR x = π /4

(-π /2 + π /2)/2 = 0 so the midpoint is (0,0)

period = π /b = π /2

So therefore some of the key points are: (-π /2 , 0 ), (0,0) , (π /2 , 0)

and some of the asymptotes are x = -3π /4 , x = -π /4, x = π /4, x = 3π /4

Example 2: y = -3 tan (x/4)

bx - c = -π /2 OR bx - c = π /2
x/4 = -π /2 OR x/4 = π /2
x = -2π OR x = 4π

(-2π + 2π )/2 = 0 so the midpoint is (0,0)

period = π /b = π /(1/4) = 4π Or (2π --2π ) = 4π

So therefore some of the key points are: (-π , 3 ), (0,0) , (π , -3) and some of the asymptotes are
x = -2π , x = 2π

III. Graph of the Cotangent function y = cot x = (cos x)/(sin x)

has vertical asymptotes wherever sin x = 0 so... x = n π

A. The cotangent function is odd. cot (-x) = -cot x
B. Therefore the graph of the y = cot x is symmetric with respect to the origin.
C. Period = π
D. Domain: all the reals except x ≠ n π
E. Range: all the reals
F. Vertical Asymptotes: x = n π

Some key points are: (-7π/4 , 1), (-3π/2, 0), (-5π/4, -1), (-3π/4, 1), (-π/2, 0), (-π/4, -1)

IV. Cosecant function is y = csc x = 1/(sin x)
so to sketch the graph of the cosecant function, first make a sketch of the sine curve, then take the reciprocals of the y-coordinates to obtain the points needed.

The vertical asymptotes are where sin x = 0 so therefore at x = nπ

A. The cosecant function is odd. csc (-x) = -csc x
B. Therefore the graph of the y = csc x is symmetric with respect to the origin.
C. Period = 2π
D. Domain: all the reals except x ≠ n π
E. Range: all the reals except y ≠ [-1,1]
F. Vertical Asymptotes: x = n π

V. Secant function is y = sec x = 1/(cos x)
so to sketch the graph of the secant function, first make a sketch of the cosine curve, then take the reciprocals of the y-coordinates to obtain the points needed.

The vertical asymptotes are where cos x = 0 so therefore at x = π/2 + nπ

A. The secant function is even. sec (-x) = sec x
B. Therefore the graph of the y = sec x is symmetric with respect to the y - axis.
C. Period = 2π
D. Domain: all the reals except x ≠ π/2 + n π
E. Range: all the reals except y ≠ [-1,1]
F. Vertical Asymptotes: x = π/2 + n π

VI. Damped Trigonometric Graphs:

A product of 2 functions can be graphed using properties of the individual functions.

Example 1: f(x) = e^-x cos (x)

key points are (-3 π /2 , 0), (-3.93, -35.89), (-π/2 , 0), (0,1), (π/2 , 0)

The damping factor is e^-x because:

-e^-x ≤ e^-x cos x ≤ e^-x

As the y-values approach zero, the x-values approach infinity.

Example: h(x) = 2^(-x/4) sin x

graph this and see what happens as x approaches zero and as y approaches zero.

Homework: #33; pg. 341; #1 - 17 odd, 41, 43, 47, 49, 51-55, 57, 63 - 67 odd, 75,
pg. 330; #12, 15, 21, 29, 45, 99, 101

Precalculus 4.5b - Translations of Sine and Cosine Curves

4.5b Translations of Sine and Cosine Curves

Given the functions:

y = a sin (bx – c) + d and y = a cos (bx – c) + d creates horizontal and vertical translations of the basic sine and cosine curves.

The following explains what happens:

1. a is the amplitude
2. (2π)/b is the period
3. bc is the horizontal shift
4. d is the vertical shift

One complete cycle interval – the left and right endpoints can be determined by:

Left endpoint is bx – c = 0
Right endpoint is bx – c = 2π

Example 1: y = 2 sin (3x – π/6)

Amplitude = 2

bx – c = 0
3x – π/6 = 0
3x = π/6
x = π/18

bx – c = 2π
3x – π/6 = 2π
3x = 13π/6
x = 13π/18

So therefore [π/18, 13π/18 ] corresponds to one cycle.

13π/18 - π/18 = (12π/18)/ 4 = 3π/18

π/18 + 3π/18 = 4π/18 + 3π/18 = 7π/18 + 3π/18 = 10π/18 + 3π/18 = 13π/18

Therefore 5 key points are:

(π/18 , 0), (4π/18 , 2), (7π/18 , 0), (10π/18 , -2), (13π/18 , )

Example 2: y = 4 cos (2x + π/2)

Amplitude = 4

bx – c = 0
2x + π/2= 0
2x = - π/2
x =- π/4

bx – c = 2π
2x + π/2= 2π
2x = 3π/2
x = 3π/4

So therefore [-π/4, 3π/4 ] corresponds to one cycle.

3π/4 - -π/4 = (4π/4)/ 4 = π/4

-π/4 + π/4 = 0 + π/4 = π/4 + π/4 = 2π/4 + π/4 = 3π/4

Therefore 5 key points are:

(-π/4 , 4), (0 , 0), (π/4 , -4), (π/2 , 0), (3π/4 , 4 )

Example 3: y = 2 sin (3x - π/4 ) + 1

Amplitude = 2

bx – c = 0
3x - π/4 = 0
3x = π/4
x = π/12

bx – c = 2π
3x - π/4 = 2π
3x = 9π/4
x = 9π/12 = 3π/4

So therefore [π/12, 3π/4 ] corresponds to one cycle.

3π/4 - π/12 = (8π/12)/ 4 = π/6

π/12 + π/6 = π/4 + π/6 = 5π/12 + π/6 = 7π/12 + π/6 = 3π/4

Do the horizontal shift first, then the vertical. Therefore 5 key points are:
(π/12 , 0), (π/4 , 2), (5π/12 , 0), (7π/12 , -2), (3π/4 , 0 )

Now do the vertical shift up one so the points become
(π/12 , 1), (π/4 , 3), (5π/12 , 1), (7π/12 , -1), (3π/4 , 1 )
Mathematical Modeling:

Sine and Cosine curves can be used to model many real-life situations, including electric currents, musical tones, radio waves, tides, and weather patterns.

Example: Chicago’s high temperatures are the following. Find a trigonometric model.

t= { 1 ,2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
C ={29, 33.5, 45.8, 58.6, 70.1, 79.6, 83.7, 81.8, 74.8, 63.3, 48.4, 34.0 }

Using your graphing calculator, place the t-values in List One and the C-values in List Two.

Next using sin-reg you will find the function:

y = a sin (bx + c) + d

y = 28.58 sin (.47x – 1.81) + 55.71 rounding to the second decimal place. By graphing the original equation and plotting the points, how well did the model fit?

Homework #32: pg. 330; #15, 21, 23, 33, 45, 47, 53 – 61 odd, 63, 79, 80
pg. 320 #52, 80, 91, 93

Precalculus 4.5a Graphs of Sine and Cosine Functions

Graphs of Sine and Cosine FunctionsI.

Sine Curve: The graph of the sine function is a sine curve.

Recall: the period of the sine function is 2π and the sine function is an odd function.

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On the unit circle, the sine function was the y-value. If we cut the unit circle at the 0, 2π place, and used the angle measure as the x-value and the sine function (or y-values from the unit circle) stays as the new y-values, you will get the sine curve.

Some of the points are (0, 0), (π / 2, 1), (π, 0), (3 π / 2, -1), (2 π, 0).

The sine curve is symmetric to origin.
Domain: all the real numbersRange: [-1, 1]

II. Cosine curve:The graph of the cosine function is the cosine curve.

Recall: The period of the cosine function is 2 π and the cosine function is an even function.

Some of the points are (0, 1), (π / 2, 0), (π, -1), (3 π / 2, 0), (2 π, 1).

The cosine curve is symmetric with respect to the y-axis.
Domain: all real numbersRange: [-1, 1]

To sketch the graph: Note 5 key points; maximums, minimums, and intercepts.

Divide the period into four equal parts to get the key points.

Amplitude:The amplitude of y = a sin x and y = a cos x represents half the distance between the maximum and minimum values of the functions and is given by:

Amplitude a or (1/2 (maximum value - minimum value)

Example 1: y = 32 sin x
a = 32

Period = (2 π)/ 1 = 2 π

Dividing 2 π into 4 equal parts, you get the points(0,0), (π / 2, 1), (π, 0), (3 π / 2, -1), (2 π, 0)

Recall: y = - f(x) is a reflection in the x-axis of the graph of y = f(x)

Example 2: y = -32 sin x

It is the same as example 1 only reflected in the x-axis
so the points are(0,0), (π / 2, -1), (π, 0), (3 π / 2, 1), (2 π, 0)

Period of Sine and Cosine Functions

Let "b" be a positive real number. The period of y= a sin bx and y = a cos bx is given by:

Period = (2 π) / b

If b is between 0 and 1 then horizontal stretch
If b is less than 1 then horizontal shrink
If b is negative then use:sin(-x) = - sin x and cos (-x) = cos (x)

Example 1: y = 2 cos (3x)

a = 2

period =(2π / 3)

(2 π / 3) / 4 = π / 6

so the points are:(0, 2), (π / 6, 0), (π / 3, -2), (π / 2, 0), (2π / 3, 2)

Example 2: y = (3/2) sin (π / 2 x)

a = 3/2

period = (2 π)/ (π / 2) = 4

So the points are:(0, 0), (1, 3/2), (2, 0), (3, -3/2), (4, 0)

Example 3: y = -2 cos (12π x)

a = 2

period = (2π)/ (12π) = 1/6

so dividing 1/6 by 4 you get 1/24

so the points are:(0, -2), (1/24, 0), (1/12, 2), (1/8, 0), (1/6, -2)

Example 4: y = 3 sin (-3x)

Recall y = sin (-x) = -sin x

so we would havey = -3 sin(3x)

a = 3

Period = (2π)/ 3

so dividing that into 4 equal parts

you will get the points:(0, 0), (π / 6, -3), (π / 3, 0), (π / 2, 3), (2π / 3, 0)

Homework: #31 pg 330; # 1- 13 odd, 17, 19, 27, 39, 41, 43, 49, 71, 73, 75
pg. 320; #9, 11, 25, 32