COURSE JOURNAL(1)

8/7/2019 COURSE JOURNAL(1)

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COURSE JOURNAL

Erol Ali

MHF-4U-UGRADE-2011


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Table of Content

U nit 1: INTRODU CTION TO POLYNOMIALFU NCTIONS

1.1: Investigation of characteristics of polynomial functions3

1.2: Identifying a Variety of Functions,Graphically4

1.3: Graphing Polynomial Functions from aFactored Form4

1.4: Factor and Remainder Theorem 5

1.5: Finding roots of polynomial equations 6

1.6: Applications of polynomial functions 6

1.7: Introduction to Inequalities 7

1.8: Solving Inequalities Part I 8

1.9: Solving Inequalities Part II 9

U nit 2 : ADVANCED POLYNOMIAL ANDRATIONAL FU NCTIONS

2.1: Transformations of Polynomials 10

2.2: Families of Polynomial functions 11

2.3: Characteristics of Odd and Even Functions

12

2.4: Rational Functions Part I 14

2.5: Rational Functions Part II 14

2.6: Graphing rational functions, Part I 15

2.7: Graphing rational functions, Part II 15

2.8: Applications of Rational Functions 16

U nit 3: EXPONENITAL AND LOGARITHIMICFU NCTIONS

3.1: Pre-requisite Skills and LogarithmicFunctions18

3.2: Key Features of Logarithmic Functions 18

3.3: Investigating Transformations and SketchingGraphs of Logarithmic Functions19

3.4: Properties and Laws of LogarithmicFunctions20

3.5: Solving Logarithmic Equations 21

3.6: Solving Exponential Equations 21

3.7: Application of Exponential and LogarithmicFunctions22

U nit 4: TRIGONOMETRIC FU NCTIONS

4.1: Review of Sine and Cosine Functions andRadian Measure23

4.2: Trig Ratios of any Angle (Including theSpecial Angles)25

4.3: Graphing Trig and Reciprocal Trig Functions

29

4.4: Inverse Trigonometric Functions 31

4.5: Properties & Transformations of SinusoidalFunctions32

4.6: Solving Problems 33


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A polynomial function is a function whose equation is in the form of:y = anx

n + an-1xn-1 + an-2x

n-2 + ... a 2x2 + a1x + a0 where a n, an-1, ... a 1, a0 are real numbers and n is a natural number . The

coefficient which is attached to the highest degree of x is called the leading coefficient .

Degree of a polynomial function: the highest exponent.

Activity Three: Graphing Polynomial Functions from a Factored Form

Examine the DEGREE and SIGN of the given polynomial and compare it to thoseillustrated above. (This helps to determine the quadrants)

Determine the ZEROES and the y-INTERCEPT of the given polynomial.

Put zeroes and y-intercept on the graph.

For any root that is repeated an even number of times the curve will BOUNCE at that x-intercept. Eg., y = (x-2) the curve will bounce at x = 2.

For any root that is repeated an odd number of times, the curve will pass through the x-axis at that x-value.

Note: If the root is a SINGLE ROOT, the curve passes straight through the x-axis.If the root is a TRIPLE ROOT, the curve BENDSas it passes through the x-axis. Eg., y= (x+1) the curve will bend as it passes through x = -1.

Find the local maximum and minimum values for the function.

Sketch the curve.

Example 1:Sketch the following polynomial y = (x + 2) (x - 1)

Step 1: DEGREE: odd(since 3 + 2 = 5)SIGN: positive(since the numerical coefficient of the highest power of x is positive)


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To find the quadrants, we match the sign and degree without initial 4 graphs. Odd and positive matchesto y = x, which starts in quadrant 3 and finishes in quadrant 1.QUADS: 3 & 1

Step 2: To find the roots, we set each of the factors to zero:

(x - 1) = 0 therefore x = 1 double root

(x + 2) = 0 therefore x = -2 triple rootROOTS: x = -2, 1To find the y-intercept, we set x = 0

y = (0 + 2) (0 - 1)= 8

y-intercept: y = 8

Step 4:

Since (x - 1) is an even root, this means that the graph will bounce(WILL NOT CROSS THE X-AXIS AT X = 1)instead it will touch the x-axis at the specific root and form a mini-parabola with the vertex being at theroot.

Step 5:

Since (x + 2) is an odd root, this means that the graph will bend, (WILL CROSS THE X-AXIS AT X = -2), as itpasses through the root.


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Step 6:

Calculate approximations for the local maximum point and the local minimum point for f(x):Local Maximum: find the average of the roots -2 and 1. This happens by choosing the midpoint betweenthe two roots. The midpoint between -2 and 1 will be x = -0.5 and will give a y-value of:

y = (-0.5 + 2) (-0.5 - 1)= 7.59375

We have another point: (-0.5, 7.59375)Local Minimum: In this example, our minimum just happens to land on the x-axis and we have alreadygraphed this point.

Step 7:

Draw smooth curves to connect the points you have found and create a sketch of the graph of thefunction.

Step 8:

Here is what you get when you use graphing technology to graph the given function, y = (x + 2) (x - 1)


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A factor of a polynomial divides evenly into the polynomial, with a remainder of zero.So, in the special case of the Remainder Theorem - t he case when t he remainder is zero - we know thedivisor is a factor of the dividend.

The Factor Theorem

A polynomial function, f(x), has a factor of (x - k) if and only if f(k) = 0.So, if f(k) = 0, then (x-k) is a factor of f(x).(jx-k) is a factor of f(x) if and only if f(k/j) = 0.

Steps to Factor a Polynomial Expression

1. Look for a common factor first.2. Find a factor using the factor theorem.3. Divide the factor from Step 2, to find a quotient, q(x).4. If q(x) is a quadratic (degree 2) expression, use your factoring of a trinomial methods to further

factor q(x).If q(x) is a polynomial expression of higher degree than 2, go back to step 2, and find anotherfactor using the factor theorem. Then repeat Step 3 to find a simpler quotient. Continue until youDO find a quotient of degree 2 that you can factor.

5. Express your final answer, by collecting all the factors you have discovered, into one statement.

Activity 6: Applications of Polynomial Functions

Happy Trails equestrian farm has a rectangular field that they wish to fence. There is 500 metres of fencing available and the area of the field is 10 000 m. What will be the dimensions of the fenced area?

We know the area is 10 000 m and that area = leng t h wid t h .

10 000 = w(250 - w)10 000 = 250w - w2


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w2 - 250w + 10 000 = 0

For our solution we have 2 possibilities:

w =250 + 1502

OR w =250 - 1502

w = 200 w = 50

To find the length of the rectangular field:

length = 250 - w= 250 - 200= 50

OR

The dimensions of the field will be 200 m 50 m.

For the zeroes of a polynomial equation

A polynomial equation that has a degree of n can have:

0 to n real solutions when n is even1 to n real solutions when n is odd

Activity 7: Introduction to Inequalities

The symbols < and > are inequality signs. Other signs that look similar are which means less or equal toand which means greater than or equal to.

Rule #1 : An inequality is not changed by adding the same number to both sides.If a < b, then a + c < b + cIf a > b, then a + c > b + c

Example:

We know that 3 < 7 and if we add 4 to each side we get: 3 + 4 < 7 + 4 which is still a true statement.


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Rule #2 : An inequality is not changed by subtracting the same number to both sides.If a < b, then a - c < b cIf a > b, then a - c > b - c

Example:

We know that 3 < 7 and if we subtract 4 from each side we get: 3 - 4 < 7 - 4 which is still a true statement.

Rule #3 : An inequality is not changed by multiplying by the same POSITIVE number to both sides.If a < b, then a c < b cIf a > b, then a c < b c

Example:

We know that 3 < 7 and if we multiply by 4 on each side we get: 3 4 < 7 4 which is still a truestatement.

Rule #4 : An inequality is changed by multiplying by the same NEGATIVE number to both sides. Theinequality sign will need to be reversed to make the statement true.If a < b, then a c > b cIf a > b, then a c < b c

Activity 8 & 9: Solving Inequalities

Example 1: Solve and graph the solution set for 2x - 4 > 0.

Solution:

Since this is a linear inequality we can use the same techniques that we would use whensolving linear equations.

2x - 4 > 02x > 4

x > 2

Solution set: {x | x > 2}


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Activity One: Transformation of Polynomials


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Activity Two: Families of Polynomial Functions

- The degree of a polynomial function might be determined, by finding the Finite Differences in a set of dataelements belonging to the function.

- When all data elements supplied to us involve sequential domain elements , we are able to figure out thegeneral format & leading coefficient

ExamplesGiven the set of data shown below, determine a specific equation that represents the data

Fromthis Finite Difference analysis, we can determine that the function is a quartic polynomial, with a leadcoefficient of 24 / (4 3 2 1) = 1

f(x) = x4 + bx3 + cx2 + dx + e

Step 2: Determine the values of the remaining parameters.

1) Using (1,-2) we get: 1 + b + c + d + e = -2b + c + d + e = -3

2) Using (2,-4) we get: 16 + 8b + 4c + 2d + e = -48b + 4c + 2d + e = -20

3) Using (3,-6) we get: 81 + 27b + 9c + 3d + e = -627b + 9c + 3d + e = -87

4) Using (4,-8) weget:

Now, solve this system of four simultaneous equations, to find b , c , d , and e

5) [2) - 1)] produces: 7b + 3c + d = -176) [3) - 2)] produces: 19b + 5c + d = -677) [4) - 3)] produces: 37b + 7c + d = -1778) [6) - 5)] produces: 12b + 2c = -509) [7) - 6)] produces: 18b + 2c = -110

6b = -60b = -10


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Activity Three: Characteristics of Odd & Even Functions

If a function is even then

f(-x) =f(x)

The function is symmetrical about the y-axis.

If a function is odd then f(-x) =-f(x)

The function is symmetrical about the origin.

If a function is neither odd nor even then f(-x) { f(x) and f(-x) { f(x)

The function is neither odd nor even.

Without graphing the function, is f(x) =1 + x1 - x

Activity Four & Five: Rational Functions

In this activity we will look first at a number of characteristics and definitions that arecommon to the families of rational functions.


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Activity Six & Seven: Graphing Rational Functions

The following are the features we can obtain from the function expression, to aid us in sketching thecorresponding graph of the function:

y Domainy x-intercept(s)

Characteristics Function: f(x) =7x - 3

x - 16

Domain

With rational functions, the main concern is not having a zero in the denominator. The 0/0 form iscalled INDETERMINATE. You want to avoid this form. In order to avoid this form, we set thedenominator to zero:x - 16 = 0x = 16

Range

The range has a direct connection to the horizontal asymptote. For the horizontal asymptote welook at the following:

Term with the highest power of x in the numerator: 7xTerm with the highest power of x in the denominator: xNow, set-up the rational function with only these two terms:

y =7xx

y = 7

erticalasymptote

The vertical asymptote is directly related to the domain of the function. Again we would set thedenominator to zero:x - 16 = 0x = 16

Horizontalasymptote


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y y-intercept(s)y Vertical Asymptote(s)y Horizontal Asymptote(s)y Symmetry (Odd, Even, Neither)y Key Points

In the presentation, the horizontal asymptote was found by looking at the tendency of the function valuesas x and as x - .

Example:

*If a Rational Function has a numerator with a degree exactly one higher than the degree of thedenominator, there will be oblique asymptote(s) that can be found by expressing the given function as apolynomial and a simpler Rational Function, using long division.

Activity Eight: Applications of Rational Functions

The Student Activity Council (SAC) of the local secondary school is raising money by selling school T-shirts. The originalcost of the graphics machine and other set-up costs amount to $400. The SAC estimates the cost to to print each T-shirt will be $5.

a. Write a cost function C(x) for producing x T-shirts. You must include the set-up cost.b. Write an average cost function A(x) for producing x T-shirts.c. What is the domain of A(x)? Explain the elements of the domain in the context of the problem.d. Does A(x) have a vertical asymptote? If so, what is the equation?e. Find the horizontal asymptote for A(x). In the context of the problem, what is the meaning of the value of

the asymptote?


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U nit 3E xponen t ial and Logari t hmic Func t ions

Activity 1: Pre-requisite Skills and Logarithmic Functions

In this activity we learn the exponent laws as well as functions and their inverses. We also gain anunderstanding of logarithmic functions as they relate to the exponential function, (i.e. the inverse).

Key concepts

- understand the relationship between exponential expressions and logarithmic expressions,evaluate logarithms, and apply the laws of logarithms to simplify numeric expressions- Identify and describe key features of the graphs of logarithmic functions, make connectionsamong the numeric, graphical, and algebraic representations of logarithmic functions, and solverelated problems graphically;- Solve exponential and simple logarithmic equations in one variable algebraically, including thosein problems arising from real-world applications.

f(x) = logbx

Change of base formula:

Examples:

(x^3)(x^5) = x^3+5 = x^8(x^9)/(x^3) = x^9-3 = x^6(x^5)^2 = x^10


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Activity 2: Key Features of Logarithmic Functions

This activity focuses on the domain, range, intercepts, intervals of increase/decrease, and asymptotes of logarithmic functions

Key Concepts:

Identify and describe key features of the graphs of logarithmic functions, making connections among thenumeric, graphical, and algebraic representations of logarithmic functions, and solving relatedproblems graphically;

Solving exponential and simple logarithmic equations in one variable algebraically, including those fromreal-world applications

Example :For the graph of g(x) = Log2xFind the:

Domain {x|x>0}Range {y|y R}X-int = 1Y-int = noneIntervals of increase/decrease = increasing when x>0Vertical/horizontal asymptote = VerticalEquation of asymptote = x = 0


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Activity 3: Investigating Transformations and Sketching Graphs of Logarithmic Functions

This activity investigates the transformations of the functions and how the parameters a, k, d, and c affectthe curve, also we learn to sketch transformed functions.

Key Concepts:

Identify and describe key features of the graphs of logarithmic functions, make connections among thenumeric, graphical, and algebraic representations of logarithmic functions, and solve relatedproblems graphically.

Period =

Amplitude = Max-Min/2

d is a horizontal translation left or right "d" units.- right when d > 0- left when d < 0

c is a vertical translation up or down "c" units.- up when c > 0- down when c < 0

a can be a vertical flip, stretch, or compression.

- a is a vertical flip when a is negative (a < 0)- a is a vertical stretch when a > 1 or a < 1- a is a vertical compression when -1 < a < 0 or 1 > a > 0 [a is between -1 and 0 or a is between 0

and 1]

k can be a horizontal flip, stretch, or compression.

- k is a horizontal flip when k is negative (k < 0)- k is a horizontal compression when k > 1 or k < -1- k is a horizontal stretch when -1 < k < 0 or 1 > k > 0 [k is between -1 and 0 or k is between 0 and

1]


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Activity 5: Solving Logarithmic Equations

In this activity we converted between exponential and logarithmic forms to solve simple logarithmicequations.

Key Concepts:

- Understand the relationship between exponential expressions and logarithmic expressions,evaluate logarithms, and know how to apply the laws of logarithms to simplify numericexpressions

- Solve exponential and simple logarithmic equations in one variable algebraically.

y = logb x (log form) is equivalent to x = by (exponential form)

Examples :

Solve for x. log9(x - 6) = 2

x - 6 = 9x - 6 = 81

x = 87

Solve for x.

log(x + 1) + log(x - 2) = 1

log(x + 1)(x - 2) = 1log(x - x - 2) = 1

(x - x - 2) = 10x - x - 2 - 10 = 0

x - x - 12 = 0(x + 3)(x - 4) = 0

x = -3 or x = 4x = 4 only since x > 2


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Activity 6: Solving Exponential Equations

In this activity we learned that there are many different types of exponential equations and more thanone method to solve them.

Key Concepts:

- Solve exponential and simple logarithmic equations in one variable algebraically.

To solve exponential equations without logarithms, you need to have equations with comparableexponential expressions on either side of the "equals" sign, so you can compare the powers and solve. Inother words, you have to have "(some base) to (some power) equals (the same base) to (some otherpower)", where you set the two powers equal to each other, and solve the resulting equation.

Examples :

Solve for x using a common base. 3x = 27(x+1)

3x = (33)(x+1)

3x = (3)3(x+1)

x = 3(x + 1)x = 3x + 3

-3 = 2x-3/2 = x

Solve for x using a common base. 252x = 125 (x - 5)

(52)2x = (53)(x - 5)

(5)4x = (5)3(x - 5)

4x = 3(x - 5)4x = 3x - 15

x = -15

Solve for x using a common base. 500(1.02) x = 1750

(1.02) x = 3.5


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log(1.02) x = log3.5xlog(1.02) = log3.5

x = log3.5/log1.02x 63.26

Activity 7: Application of Exponential and Logarithmic Functions

In this activity we learn about the Richter scale for earthquakes

Key Concepts:

- identify and describe key features of the graphs of logarithmic functions, make connectionsamong the numeric, graphical, and algebraic representations of logarithmic functions, and solverelated problems graphically

- Solve exponential and simple logarithmic equations in one variable algebraically

The magnitude of an earthquake is defined as:

Review of Sine and Cosine Functions and Radian Measure

Introductory Problem

An angle is a measure of how far something has turned. It makes sense to devise a system of measurement based on one full turn. There are actually three systems for measuring angles. A scientificcalculator is able to work in all three. You will have a button commonly labelled DRGand the symbolsDEG, RAD, GRA, appear in the display.

You will be most familiar with degrees. There are 360 degrees in a full circle. We write this as 360 .

You will be less familiar with the gradian. This is a decimal system in which there are 400 grad in a fullcircle. (i.e. 100 grad in a right angle.)

In this lesson, we will look at the third measure, radians , in detail. Radians are useful in many differentformulae and applications.

Example:


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One example is the radian s relationship between angles and linear measurement. Radian measure allowsus to determine both angular velocity and linear velocity in several applications.

A telecommunication satellite in circular orbit 1250 kilometers above the earth s surface makes onecomplete revolution every 90 minutes. Use 6400 kilometers for the radius of the earth.

A. How fast is this satellite traveling?(Linear speed)

= a/r2pi = a/6400a = 40121.39 km (arc length for one completerevolution)

The satellite completes this distance in 90minutes or 1.5 hours

Speed = distance/time= 40121.39km/1.5hrs= 26808.26km/hour

The satellite travels 26,808.26 km/hr

B. How fast is angle of the satellitechanging with respect to the earth?(Angular velocity)

One complete revolution, the satellite rotates 2radians in 1.5 hours

Since speed = distance/TimeAngularSpeed = Angle/Time

AngularSpeed = 2 rad/1.5 hours

= 2 rad/3/2 hours= 2 rad 2/3 hours= 4/3 rad/hour

or in seconds

= 4/3 rad/hr 1hr/60min 1min/60s= 0.01 rad/s

The angle is changing at 0.01rad/s with respectto the earth.


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Trig Ratios of any Angle (Including the Special Angles)

Review:

Before we can do this we must first define some key features of angles

In grade 11 you learned how to find the trigonometric ratios of any angle

Angles can either be positive or negativeIf the terminal arm rotates Counter clockwise=POSTIVE, Clockwise=NEGATIVE

To understand angles we also need to know the terms: Principal Angle and Acute Angle Finally, let us review the trigonometric ratios:


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Trig Ratios of any Angle (Including the Special Angles) cont d

Example:Question 1: Point P(-5, -3) is on the terminal arm of an angle, ?, in standard position.Sketch the principle angle, ?

x

y

AD

OP r x

HY AD

r y

HY OP

!!

!!

!!

U

U

U

tan

cos

sin


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a. What is the measure of angle ? in radians?

ACUTE ANGLEtan = y/xtan = 3/5

= tan-1 (3/5) = tan-1 (3/5)

= 0.54 radPRINCIPAL ANGLE = pi + Acute Angle (third quadrant)= 3.68 rad

Graphing Trig and Reciprocal Trig Functions

y = cscx is equal to y = 1/sin x


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y = secx is equal to y = 1/cosx

y = cotx is equal to y = 1/cosx

Inverse Trigonometric Functions

Recall that on your calculator, there exist buttons sin -1, cos -1, and tan -1.

These are the buttons that help you to find an angle when you have the ratio of the sides. This is theinverse of finding the ratio of sides when you have the angle.

Find the first quadrant angle when sinx = 0.707

The buttons you need to press include the second function of the sin button, or sin -1, to produce an angleof approximately 45 or pi/4 radians. Notice that this is the inverse operation of finding the value of sin45.

This, once again, is where we find the inverse notation sin -1x.

It is very interesting to note, that the inverse function is given another notation-sin -1x is alsocalled arcsinx. This is because there is a possibility of confuing sin -1x with its reciprocal (sinx) -1. Why??


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Properties & Transformations of Sinusoidal Functions

Before we can proceed with graphing trig functions we must review some important terms.


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Terms: Cycle: The smallest part of a graph that contains one complete repeating pattern.Period: The length of one cycle.Amplitude: The vertical distance between the axis of the curve and either the maximum or minimumvalue.

a =maximum value - minimum value

2

Roles of a, k d and c in f (x) = a sin(k(x - d)) + c or f(x) = a cos(k(x - d)) + c

Vertical Stretch

y a is the vertical stretch factor and is also called the amplitude.


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y For a > 1, the graph stretches vertically by a factor of a.y For 0 < a < 1 , the graph compresses vertically by a factor of a.y For a < 0, the graph is reflected in the x-axis.

Horizontal Stretch

y

k is the horizontal stretch factor which changes the period.y For k > 1, the graph compress horizontally by a factor of 1/k.y For 0 < k < 1, the graph stretched horizontally by a factor of 1/k.y For k < 0, the graph is reflected in the y-axis.y To determine the period of a graph:

Vertical Translations

y For d > 1, the graph is translated d units up .y For d < 1, the graph is translated d units down .

Horizontal Translations

y For c > 1, the graph is translated c units left .y For c < 1, the graph is translated d units right .

Graphing f(x) = a sin(k(x - d)) + c or f(x) = a cos(k(x - d)) + c

y When graphing sinusoidal functions it is important to understand the behaviour of the basefunctions: f(x) = sinx and g(x) = cosx.

y When transforming these functions it is important to identify five special points for each of thebase functions:

Function Point 1 Point 2 Point 3 Point 4 Point 5

y = sinx ZERO(0,0)MAX(/2,1)

ZERO(,0)

MIN(3/2,-1)

ZERO(2,0)

y = cosxMAX(0,1)

ZERO(/2,0)

MIN(,-1)

ZERO(3/2,0)

MAX(2,1)

y When transforming trig functions, these are the special points we will translate. This is called the5-point method of graphing trigonometric

Solving Problems

Example: The SkyWheel is a ferris wheel in Niagara Falls. It has a diameter of 53 metres and the ride lastsfor 12 minutes for a total of 6 revolutions. It has a total of 42 gondolas that can each hold 6 passengers.Assuming that the height of the gondola follows a sinusoidal model, if you enter the gondola at a height of 2 m above the ground, what is your altitude at 11 minutes and 21 seconds? At what time(s) is your height21 m high?

Start by finding the period, k value, amplitude and half the diameter of the wheel.


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The next item we must determine is the phase shift. Since our base function is a sine graph, the functionstarts at zero. We are going to graph this relation based on the following data.

Let us graph this relation for the duration of the ride.

Based on this graph, how many units is the phase shift to ghe right since the Axis of Curve representsthe base axis of y = sinx?Based on all of the above information, the function that represents this model?Based on this graph, the phase shift is 0.5 units to the right since the Axis of Curve represents the baseaxis of y = sinx.Next, let us find the height at 11 minutes and 21 seconds. What is the time converted into minutes? Basedon all of the above information, the function that represents this model is h(x) = 26.5sin((x - 0.5)) + 28.5

Substitute the values into the function.

Next, let us find the height at 11 minutes and 21 seconds. This time converted into minutes is clickhere minutes. Substitute the values into the function:h(11.35) = 26.5sin((11.35 - 0.5)) + 28.5

= 40.53 m

What is the time at 21 m high?

21 = 26.5sin((11.35 - 0.5)) + 28.5-0.283 = sin((x - 0.5))

sin-1(-0.283) = (x - 0.5)-0.287 = (x - 0.5)

x = 0.409 min

Since the cycle is 2 minutes, the gondola is 21 m at 0.409 min, 2.409min, 4.409min, 6.409min, 8.409min,10.409min.


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Double Angle Formula for Sine

Let A = B = xsin(x + x) = sinxcosx + sinxcosxsin(2x) = 2sinxcosx

Double Angle Formula for Cosine

Let A = B = xcos(x + x) = cosxcosx - sinxsinxcos(2x) = cosx - sinxORcos(2x) = 2cosx - 1ORcos(2x) = 1 - 2sinx

Trigonometric Identities

Illustration of this interference is easily demonstrated by graphing the sound waves. Below are two soundwaves: y = sinx in blue and y = sin(x -pi) in red . The x-axis represent time and the y-axis represent the

amplitude or sound intensity. If we graph these on the same axis weget:


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f the sound in a room is represented by the red graph, your noise cancelling headphones would producethe blue graph to cancel this sound. The resulting graph is shown below in green , in which all values

(sound intensity) of this new resulting function is zero:

Algebraic Approach

The above graph can be written as an equation:How can we illustrate that the left side of the equation is equal to the ride side of the equation?Highlight the data in between the two brackets, in order to reveal the information.lThe example is called a trigonometric identity where we showed that the left side of the equation isequal to the right side of the equation. Notice, we DO NOT SOLVE these identities, we just work on eachside until LEFT SIDE (LS) = RIGHT SIDE (RS) We are next going to view the different trigonometric identitiesand their proofs.

Tips for Proving TrigonometricIdentities

1. Separate the two sides of theexpression as L.S. and R.S.

2. Simplify the more complicated sideuntil it is identical to the other side ORsimplify both sides at the same time.

3. Express all tangent functions in termsof sine or cosine.

4. Express all reciprocal as sine, cosineand tangent.

5. Use Pythagorean identity wherepossible.

6. Where possible find a common

denominator.

Summary of Trigonometric Identities,Reciprocal Relationships andCompound Angles

Quotient Identity

tan =sin cos

Pythagorean Identity

sin + cos = 1 or sin = 1 - cos or cos =1 - sin

Reciprocal Trigonometric Functions

csc =1sin

sec =1cos

cot =1tan

Compound Angle Formulae

csc =1sin


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y Write all in terms of cosy + = 0y Add the two terms using a common

denominatory = 0y Multiply out the denominator

y

cos x + cos 2x = 0y Use the double angle formulae for

cos2xy cos x + (2cos x - 1) = 0y Expand and factory 2cos x + cos x - 1 = 0y Let u = cosxy 2u + u - 1 = 0y (2u - 1)(u + 1) = 0y u = , u = -1y let u = cosxy cosx = 1/2 and cosx = -1

y

Solve for xy cosx = 1/2 (Acute angle of /3 in

quadrant 1 and 4) .: x = /3 and 5/3y cosx = -1, x =

y Place values in order

y x = /3, and 5/3

COURSE JOURNAL(1)

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