THE UNIT CIRCLE

Hey everyone hope you had a fun Halloween and enjoyed your weekend! But remember to keep up with your homework and don't forget about Pre-Calculus.

THE UNIT CIRCLE

The "Unit Circle" is a circle with a radius of 1.

Using the unit circle makes it easy to find the values of trigonometric functions at quadrantal angles.                                                                                                                    

 For example, a 90º rotation from the positive x-axis puts you on the positive y-axis, which intersects the unit circle at the point (0, 1). From this, you know that (cos 90º, sin 90º) = (0, 1).

This is a graph of the values of all three trigonometric functions at each quadrantal angle:

Important Angles 30º , 45º , 60º

Most angles on the unit circle are based on the reference angles of either 30º , 45º , 60º

You should remember the special triangles from grade 11 , If not here is a labeled diagrams to help you.

Finding the missing side of a right triangles

Finding the missing side of a right triangle is a pretty simple matter if two sides are known. One of the more famous mathematical formulas is a2+b2=c2, which is known as the Pythagorean Theorem. The theorem states that the hypotenuse of a right triangle can be easily calculated from the lengths of the sides. The hypotenuse is the longest side of a right triangle.

If you're given the lengths of the two sides it is easy to find the hypotenuse. Just square the sides, add them, and then take the square root. Here's an example:

Since we are given that the two legs of the triangle are 3 and 4, plug those into the Pythagorean equation and solve for the hypotenuse:

a2+b2=c2

32+42=c2

9+16=c2

c=5

If you are given the hypotenuse and one of the legs it's going to be slightly more complicated, but only because you have to do some algebra first. Suppose you know that one leg is 5 and the hypotenuse (longest side) is 13. Plug those into the appropriate places in the Pythagorean equation:

a2+b2=c2

52+b2=132

25+b2=169

b2=144

b=12

Thank you for reading my blog post, I hope this post helped you review/ learn about the unit circle and the special triangles involved. 

-Ravinder Sehira 

sources: 

Pre-calculus 12 McGraw-Hill Ryerson- Text Book

Pre-Calculus 40s Unit 4 Circular Functions- Notes

sites:

http://www.freemathhelp.com/q6-right-triangle.html

http://www.mathsisfun.com/geometry/unit-circle.html

http://www.wyzant.com/resources/lessons/math/trigonometry/unit-circle

Radians, Degrees and You...

I'd like to invite the lovers of circles to read this blog. Today, our class was taught about radians and how we can use them along with degrees to confuse us about circles even further than what was already possible.

A radian is another way to measure all the angles in a circle. When the arc of a circle has the same length as the radius, the central angle that intercepts the arc is measured as 1 radian. 

You can find the arc length of a circle with this equation: a=ϴr

     a is arc length

     ϴ is the angle in radians

     r is the radius

THIS IS THE ONE GOLDEN RULE OF LIFE:

     To convert radians to degrees, multiply the radian measure by 180˚/π

     To convert degrees to radians, multiply the degree measure by π/180˚

Example 1.

Degree measure: 360˚

-To find the radian measure of 360˚, you need to convert it using the equation that was literally just stated as the one golden rule of life. (Remember to simplify, children)

360˚(π/180˚)= 360π/180= 2π                   2π is our answer in EXACT radians

                                                                 6.28 is our APPROXIMATE radian measure

Example 2.

Radian measure: π/3

-To find the degree measure, you must (again) use our golden equation.

π/3(180˚/π)=180/3=60˚                           60˚is our answer in degrees

Now that you have a general idea of how to convert radians to degrees, the clock in Mr.Piatek's room should be easier to read than a childrens book. (I know this isn't his clock, but I am not spending more than 10 minutes on google looking for an exact replica)

FRACTIONS ARE PRETTIER THAN DECIMALS

Once you figure out the angles of what each radians represents, I'm sure you have now realized that the order of the radians represents the angles on a circle instead of the time. (Unless you're lazy and read ahead...shame on you) As I'm sure you've noticed, π alone is equal to 180˚ and 2π is 360˚. because the number we've all come to love (3.14159) is actually an approximation, it can only be so accurate, this is why we like to leave our radians as fractions.

I would like to conclude this post on radians, degrees and you by wishing all my fellow pre-calculus students a successful provincial exam and a great year. I hope my post was helpful as a review/teaching tool to those of you reading this. Good luck and math on. 

Division of Polynomials

Hey guys! Jeoffrey here. I am here to teach you how to divide a polynomial by a binomial. There are 2 ways to divide a polynomial by a binomial: long division and synthetic division. The first way we are going to look at it long division.


Long Division

Do you guys remember the terms divisor, dividend, quotient, and remainder? Well, if you don't here is a quick reminder.

 

 

 

Dividend is the number we want to be divided, divisor is the number we are dividing by, quotient is the result of the division, and remainder is the left over after all of the calculations.

 

 

Now, in the example above, we see that we are doing the question 9÷4. The first step we need to do is divde 4 by 9, and see what is the closest whole number that we get as a result. We see that the answer is 2. So we put that as our quotient. Next, we multiply 4 and 2 together, get 8, and then do 9-8, to see if there will be a remainder at the end. We see that there is a remainder of 1, so we state that in our final answer.

 

With polynomials, it is the same thing. lets say we have (5n²+33n-22)÷(n+7). First, we would format it like any other long divison question.

 

 

 

 

First, we would first do (5n²÷n) in order to get the first part of our quotient. once we get that, we take that quotient, in this case 5n, and multiply it with (n+7). Then we take our answer, 5n²+35 and subract it from 5n²+33. In the end, we will be left with -2.

 

 

 

**note: even though (5n)(7)=35n, it turns to negative because we always subtract when doing long division. same thing with the 5n². It turns into -5n²

 

 

After this step, you take your answer from the previous step and divide it by n to get the next part of your quotient. In this case, it would be (-2n)÷n= -2

 

From there, you would bring down the (-22) and put it beside the -2n. Then, you would repeat the previous step where you would take -2 and multiply it by the divisor, and then take that answer and subtract it from (33n-22) in order to see if there is a remainder. In this case, (-2)x(n+7)= -2n-14. Following this, (33n-22)-(-2n-14)= -8

Finally, in order to state our answer, we re-write the question, then state our quotient as our answer. if there is a remainder, in this case, (-8), then we write is as the quotient+ R/(x-a) or the quotient multiplied by (x-a)+R.


 

 

 

 

**note. Since the remainder is (-8), and the (x-a) is (n+7), then the equation changes to quotient-R/n=7 or quotient*(n+7)-8

 

 

 

 

Synthetic Division

The other way to divde is called synthetic division. It is kind of similar to long division, but it is a lot easier and quicker to do. Let's say we are given the question (2x³+3x²-4x+15)÷(x+3)

The first thing you need to do is determine your a value. To do this, you put your divisor equal to zero, and solve for x. That will be your "a" value.

 

 

 

 

 

 

Next, when you format your synthetic division, it is kind of like long division, except you put your a value on the outside (where the quotient usually goes), and then you put the coefficient in front of each term in the dividend inside (instead of the whole dividend) **note: make sure you don't forget your positives and negative!!

 

 

 

 

 

 

 

Once you format this, you bring down the first coefficient. ( because you don't have to multiply by anything yet). In this case, we bring down the 2. Once we do that, we multiply the 2 with our "a" value, which is -3, and then put our answer under the next coefficient. From here, we just add the coefficient, which in this case is a 3, and the -6 together.

 

 

 

 

 

**Note: make sure you are adding the 3 and -6 together. ( In synthetic division, we add, where as long division, we are subtracting.

 

 

 

 

Once we get -3 as an answer, we repeat the previous step, but we would multiply our "a"value, (-3) with the answer we just got, (-3). From this, we would go (-3*-3) to get 9, then add that to -4 to get 5. Then we do the same thing again; multiply 5 with (-3, which is our "a" value) to get -15. And then we add 15 and (-15) together to get 0.

 

 

 

 

**Note: the last digit when you are using synthetic division is your remainder, and the number before the remainder is your constant!

 

 

 

Now, after you have done all of these calculations, you would look at your original equation, take the x's with their exponents, and reduce all of their powers by 1. In this case, since it was x³+x²+x, it would turn into just x²+x. Once you do that, you take the answers that you got from the synthetic calculations, and put them in as your new coefficients for the x-values with the reduced power.

 

 

 

 

In this case, it would look like 2x²-3x+5 as your final answer!!

 

 

 

 

 

 

 

 

 

 

Thanks for reading my blog! I hoped you learned how to divide polynomials with a binomial! :D

BYE!!!!!!!!!!!!!!!!!!





Factoring

Hey, what's up guys, it's Chris Tabios and I am going to explain the different methods on how to factor polynomials which we covered a awhile ago. 

The first method that we learned was finding the GCF (Greatest Common Factor). To summarize, the Greatest Common Factor is the greatest term that can be divided into each of the terms of the the given polynomial. 

                     Ex. 5x²+35x

 The GCF would be 5x because 5x goes into 5 and 35 without having to deal with any difficult numbers. Once you come up with the GCF all you have to do is divide each term of the polynomial by it. 

For this example, the final answer would be 5x(x+7).

The second factoring method we covered is Factoring By Grouping. Sometimes with polynomials with four or more terms, the easiest way to completely factor the polynomial is by grouping up the terms.In order to do this you would have to split-up the polynomial into groups.

Ex. 5x+5y+zx+zy 

(5x+5y)      (zx+zy) 

*Remember to keep the (+) and (-) signs as they are *

After you group up the terms, you will have to once again have to find the GCF of the terms. Following the final answer format of the GCF example, you would get:

5(x+y) and  z(x+y)

The final step is to write it out as 

(x+y)(5+z) 

The next factoring method we learned is Factoring the Difference of Two Squares. To do this, all you have to do is remember (ax)²-b²= (ax+ b)(ax-b)

Ex.  x²-9 =
 (x+3)(x-3) 

Another method of factoring polynomials is Factoring the Difference of Cubes. to put it simply, it is used for binomials that are a difference of two perfect cubes.  The form to remember is  x³-y³=(x-y)(x²+xy+y²)


Ex. x³-27
 x³-3³=(x-3)(x²+3x+9)
*Turn terms into x³*

There is also a method called Factoring the Sum of Cubes, which is the same concept, except the form is x³+y³=(x+y)(x²-xy+y²)

The final two methods of factoring polynomials are Factoring Perfect Square Trinomials and Factoring Trinomials with a leading coefficient of other than 1

Factoring Perfect Square Trinomials:

 x² + 2xy + y² and x² - 2xy + y² are called perfect square trinomials
 This is the form that will be used:

x² + 2xy + y² = (x + y)² and x² - 2xy + y² = (x - y)²

 16x² + 24 + 9 = (4x)² + (2)(4x )( 3 )+ (3)²=
(4x + 3)²


Factoring Trinomials With a Leading Coefficient Other Than 1:

 To factor these trinomials you must first multiply the the first and last numbers, then you would have to find two numbers that would add up to the middle term and multiply together to turn into the new number.



Ex. 5x² +13x-6 

(5) x (-6) = -30 

5x² +15x-2x-6

5x(x+3) -2(x+3)

(5x-2)(x+3)

AAAAAHHHHHH!!!!

IT'S MR. PIATEK!!!

Okay, just kidding, now that I have your attention, keep reading.




Hello there, and my name is John... La! Obviously.

Today I will be explaining on how to write an equation of a transformed function graph. What? That's possible? Well yes. After I explain it, this should be as easy as 3.14. ;)

Now let's start, shall we?

Firstly, you would determine the equation of g(x) in the form of
  y = af [ b ( x - h ) ] + k


I probably need pictures..

There we go, okay. First we should plot the points of both f(x) and the transformed function, which we will call g(x).

                            f(x)                                     g(x)

                            (8,10)                                                      (4,4)

                            (7,2)                                                        (0,0)

                            (6,10)                                                      (-4,4)

Still with me? Let's continue.

Now remember that formula I told you about? Yes? No?  Here : y = af [ b ( x - h ) ] + k

We can find the Translations first, which are h and k. 

    

h = -7  

To find out what h is, we look at the difference from the original vertex and the transformed vertex of each function, horizontally.

k = -2

To find k we look at the difference from the original vertex and the transformed vertex of each function, vertically.

Now we can look at the stretches, a and b. 

a = 1/2 (one half) 

We get a by taking g(x)'s vertical distance and divide it by f(x)'s vertical distance. 

4 ÷ 8 = 0.5 (or 1/2)

b = 1/4 (one fourth) 

We get b by taking the horizontal distance of g(x) and dividing it by f(x)'s horizontal distance. 

8 ÷ 2 = 4 , BUT we read b as opposite/ inverse. That is why it is 1/4. 

Before we plug the formula in, h is also read as opposite/inverse.

Now we just simply plug it into the formula. y = af [ b ( x - h ) ] + k

y = ½f [ ¼ ( x + 7 ) ] - 2

Now you're all done!

That wasn't so hard, right?

Hope everybody did well on their tests that was on Friday!

PEACE OUT Y'ALL! 

Just kidding,

Bye! 

Going Backwards with transformations

Hi class, This is Greg! A few classes ago, we learned about going backwards with transformations. Today I will be doing a re-cap of everything we covered that day.

In order to go backwards, you must do all the steps in reverse order. So in this case, you first figure out the transformations. Then move along with the stretches and reflections.

Example 1: The function  y=f(x-2)+3 is graphed below. How would you sketch a graph of the function f(x)?

Keep in mind.

Outside the function à affects the y-values à the effect is the same.

Inside the function à affects the x-values à the effect is the opposite.

But since you’re working backwards, the effect would be opposite.

First off, instead of adding 3 to the y-value of the coordinates (+3) you must subtract 3 instead. For the x-value (-2) you must subtract 2 instead of adding 2.

Sketch the new graph to represent f(x) using the new coordinates.

Example 2: (No Graph)

The function y=3f(-2[x-2])-2 = (-2,4), Then what is f(x)?

For this question, you must figure out the new coordinates for the function f(x).

First you start of with the transformations, and then move on with the stretches and reflections. For stretches you would originally multiply, but in this case you would have to do the opposite and divide.

For the x-value (-2) you must subtract 2 (=-4) Then divide it by -½ (=8).

For the y-value (4) you must add 2 (=6) Then divide it by 3 (=2)

f(x)= (8,2)

HAVE A GREAT DAY!!!

Reflections and Inverse Functions

Hay guyz I'm going to teach you about reflections and inverse functions

• In the equation, y= Af(B (x+C))+D , when ever the "A" value is a negative number, the function will reflect in the x-axis changing all y values to the opposite sign. 

                                 
          Once all the y values have switched their sign, it would like the orange function below, where the purple function was f(x) = x


                                                               

• In the same equation y= Af(B (x+C))+D , when the "B" value is a negative number, the function will reflect in the y axis changing all x values to the opposite sign.                                                         

      In this graph, the original function was the green function, but since the function has been reflected in the y axis the x values became opposite.

• If the function is to be inverted, the function will reflect along the y=x line. To perform that sort of function you'd need to replace f(x) with a y

                                                 f(x) = (3x+3)+1 -------> y= (3x+3)+1

        The next step would be to replace the "y" with the "x" and vice-versa 

                                                y=  (3x+3)+1 ------->  x= (3y+3)+1

         Then from there you'd solve for y

                     

                     x= (3y+3)+1 ------->  x -1 = 3y+3  ------->   x - 4 = 3y   ------->  1x - 4  = y

                                                                                                       3                              3     3 

          And there is your inverse equation!

                                                                           1x - 4  = y

                                                                            3     3 

         Hopefully with all of this you can solve your equations.