Tuesday, December 9, 2014

Trigonometric Identities

Trigonometric Identities
 

Trigonometric identities is a form of proof that all permissible values is true of the variable for which both sides of the equation are defined.

Reciprocal Identities


  In calculation we may replace either member of the identity with the other.
 
 
 
  Tangent and Cotangent Identities
Pythagorean Identities
 
Verifying the identity or proving that the given equation is an identity, it is show that the left hand side of the equation is equal to the right hand side. 
Example:

Sum and Difference Formulas
 
 Evaluate sin 15°.
Solution.sin 15°trigonometric identities 
 
 trigonometric identities 
 
 trigonometric identities 
 
 trigonometric identities
 
 
 

Monday, December 1, 2014

Determining Equations for Sinusoidal Functions

In order to determine the equation of a sinusoidal function, you need to find:

a - amplitude
b - affects the period
Period -    2π
                 /b/
c - horizontal shift
- vertical shift

f(x)=asinb(x-c) + d           OR             f(x)=acosb(x+c) + d

The equation could either be for a sine or cosine function

Example 
State a sine equation for the function graphed below.


Steps

  1. First, we need to identify the Middle Axis. This determines if there is any up or down shifting - the d value. 


                  Middle Axis = Maximum + Minimum
                                                           2
                  Middle Axis = 3 + (-1)  = 1
                                                2
                  Middle Axis is at  y  =  1

We can see that there is a vertical translation 1 unit up therefore,  d = 1

     2. Next, find the amplitude or the a value. The amplitude is the distance between the middle axis and the maximum value. In this graph, the amplitude is 2.






     3.  Next find b. But in order to find b, we must determine the period of the function. To find the period, measure the distance along the x-axis between two maximum points or two minimum points.


Once you got the period, you can calculate for b.

           period = 2π                         b =  2π          b = 2π        b = 1

                                                           period             2π



4. Lastly, let's find the phase shift (horizontal shift) or c. This is the trickiest value to find since there are many choices possible for c. But the question is asking for a sine equation. so we know that at
 x = 0, the y-intercept is on the middle axis.


Let's recap the information we've determined:
a = 2
b = 1
cπ/2
d = 1

 Since the question is asking for an equation for a sine function, we will use the form    
 f(x)=asinb(x-c) + d


Just plug the a, b, c and d values into the equation.

The final answer will be f(x) = 2sin(x-π/2) + 1


Thank you for reading my post. I hope this helped you a bit and I wish you all the best in our final exam. Good luck and have a nice day!




Tuesday, November 18, 2014

Graphing Sine function

Sketch the graph y= 3sinΘ for 0 < x < 2π

In order to graph sine or cosine function, you need to determine:
·        -  Amplitude (a)
·         - Period (b)

General Form:
     y= asinb(
Θ-c)+d

Determine the Amplitude of a sine Function
- Amplitude is the distance from the center of the axis to the highest or lowest point for a sin
Θ or cosΘ function.

     Amplitude = Max- min
                                 2
     Amplitude= IaI

So




 Determine the period of sine function


- To find the period, you first need to determine b

Then plug it in the formula of a period which is    2π
                                                                                IbI



- the period of the sin function is 2π, which means the function repeats itself every 2π











You are now ready to graph a sine function!    
1.      Plot the period (make sure to have 5 coordinates)
2.      The amplitude will be the maximum and minimum y-coordinates




















3.      If the equation is positive, after plotting zero the next point will be plot on the positive side. If the equation is negative, it will be on the negative side.

4.      then, the next point will be on the x-axis since the amplitude on the positive and negative side should equidistant, the graph should look like a wave



 


 








y= 3sinΘ


 

Thursday, November 13, 2014

Solving Trigonometric Equations Part 1

        Hi Pre Calc. class! Last time you guys learned about the unit circle, and all of it's properties, special angles, etc. Well now you are ready to continue to the next step, which is solving Trigonometric Equations with a specified interval! How exciting right? I'll walk you through the the steps though, so don't be worried.

THINGS TO REMEMBER:
- Special triangles
-Quadrantal Angles
SOH CAH TOA
-CAST rule
- ΘR angles

Example 1
To solve for  Θ:
1) See if the interval is set in either radians or degrees. That will determine how you will write your final answer. 



2)Simplify the equation by isolating the unknown, which is the theta, and factor.
3)Find  ΘR 
and find the quadrants that the answer is in using the CAST rule.

4)Find values for  Θ in degrees or radians. 
Since the interval is in degrees, I will keep it my answer in degrees. 
You see how there are two answers? That's because there are two quadrants that my answer for  Θ could be in! 
* If there is a particular restriction, such as " Where sin  Θ> 0" in  a question such as " tan Θ= -1/ 3 " tan would be possible in TWO quadrants, Q. 2 and 4. BUT because of restriction, sin cannot be in Q. 4, only Q. 2 therefore there will only be ONE answer.*
Here is the full question for you! 

So I hope that this gave you a little more understanding about solving trigonometric equations with a specific interval. Try to review your note as well, as they are very useful and important! 
Enjoy my fellow Pre calc mates!



Thursday, November 6, 2014

Trigonometric Ratios

Hello class. I am going to be teaching you all about trigonometric ratios.. Just kidding Mr.P already taught us. I'm just going to be recapping what we may or may not already know. Bad joke? Yeah I know, anyways lets get started.


The first thing you'll need to remember are these: 
The 6 Trigonometric Ratios


Also SOH CAH TOA will come very handy in this unit. 



So for example, lets say the given points are A(-3/5, -4/5), we need to find 6 trigonometric ratios for theta(θ)

So first off we're going to need a diagram, 



Then you'll need to look at the given points again which is (-3/5,-4/5) and figure out which quadrant point A would go when using the CAST rule. You also need to remember that cos is x and sin is y. 
So since cos and sin are both negative, the point would go in quadrant 3. 

Cos is adjacent/hypotenuse which makes -3 the adjacent and sin is opposite/hypotenuse which makes -4 is opposite and the hypotenuse is 5



If you understand the CAST rule then you are able to plug the points in sin, cos and tan plus their reciprocals. 
So here are the following answers:




This is just one of many examples that are almost exactly the same as this. So now that you know this, the world is in your hands... haha just kidding but in all seriousness, you will be able to figure out any trigonometric ratio question that comes your way.













Tuesday, November 4, 2014

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

Tuesday, October 28, 2014

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 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. 






Friday, October 17, 2014

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!!!!!!!!!!!!!!!!!!



Tuesday, October 14, 2014

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)



Sunday, October 12, 2014

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 - ) ] + 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, 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 - ) ] + k

y½¼ ( 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!